Notes on Combinatorics. Peter J. Cameron

Transcription

1 Notes o Combiatorics Peter J Camero

2 ii Preface: What is Combiatorics? Combiatorics, the mathematics of patters,, helps us desig computer etwors, crac security codes, or solve sudous Ursula Marti, Vice-Pricipal (Sciece ad Egieerig), Quee Mary, Uiversity of Lodo These otes accompaied the course MAS219, Combiatorics, at Quee Mary, Uiversity of Lodo, i the Autum semester 2007 It is impossible to defie combiatorics, but a approximate descriptio would go lie this We are give the job of arragig certai objects or items accordig to a specified patter Some of the questios that arise iclude: Is the arragemet possible? I how may ways ca the arragemet be made? How do we go about fidig such a arragemet? This is best illustrated by examples Example 1: Sudou You are give a 9 9 grid, divided ito ie 3 3 squares Your job is to put the umbers 1,2,,9 ito the cells of the grid i such a way that each umber occurs just oce i each row, oce i each colum, ad oce i each 3 3 subsquare It is ot hard to see that the arragemet is ideed possible A heroic calculatio by Bertram Felgehauer ad Frazer Jarvis i 2005 showed that there are 6,670,903,752,021,072,936,960 differet ways of fillig the grid Now suppose that someoe has complicated the problem by writig some umbers ito the grid already I geeral it may or may ot be possible to complete the grid; ad eve if it is, it may be very difficult to fid a solutio Nevertheless, may people aroud the world ejoy egagig with this combiatorial problem every day Example 2: Euler s officers 1782: The great mathematicia Leohard Euler ased i

3 Six differet regimets have six officers, each oe holdig a differet ra (of six differet ras altogether) Ca these 36 officers be arraged i a square formatio so that each row ad colum cotais oe officer of each ra ad oe from each regimet? Euler cojectured that the aswer is o, ad this guess was evetually proved correct i 1900 However Euler also cojectured that the aswer is o if six is replaced by 10, 14, or ay umber cogruet to 2 mod 4 He was completely wrog about this, but this was ot discovered util the 1960s Example 3: Kirma s schoolgirls I 1843, Thomas Kirma ased: Fiftee schoolgirls go for a wal every day for a wee i five rows of three Is it possible to arrage the wals so that every two girls wal together exactly oce durig the wee? This is certaily plausible Each girl has to wal with fourtee others; every day there are two other girls i her row, so seve days would be the right umber for the schedule However, this does ot prove that the arragemet is possible I fact, it ca be doe; Kirma himself foud a schedule satisfyig the coditios Examples ad reality The examples may give you the impressio that combiatorics is a collectio of charmig puzzles of little relevace to our moder techological world I fact this is completely wrog The course is ot really about applicatios, but i the digital world this subject is of eormous sigificace People (ad computers!) sped a lot of time sortig data, sedig messages through etwors, correctig faulty data or ecodig data to eep it safe from uauthorised access, desigig better etwors, looig for ew combiatios of atoms to form molecules which will provide us with better drugs, ad so o We eed to decide whe such a problem has a solutio, ad to fid the solutio efficietly These otes These otes reflect the cotets of the course i 2007 I have added a couple of proofs of major theorems ot covered i the course The otes have bee provided with exercises (some of them with wored solutios) ad a idex The recommeded textboo for the course was my ow boo Combiatorics: Topics, Techiques, Algorithms, first published i 1994; but rather tha followig the boo I have writte everythig aew The course covers roughly the first half of the boo; if you ejoyed this, you may wat to read more, or to loo at my Notes o coutig o the Web I am grateful to Vola Yildiz who spotted a umber of misprits i a prelimiary versio of the otes iii

4 iv Further readig Either of the two level 4 courses at Quee Mary ca be tae by studets who have doe the Combiatorics course: MAS408: Graphs, Colourigs ad Desig MAS439: Eumerative ad Asymptotic Combiatorics I metioed above my Notes o coutig which are o the web i the same place as these otes Some other boos which cotai further material (icludig the recommeded course text) are: Marti Aiger, Combiatorial Theory, Spriger, 1979 Norma Biggs, Discrete Mathematics (2d editio), Oxford Uiversity Press, 2002 Peter J Camero, Combiatorics: Topics, Techiques, Algorithms (2d editio), Cambridge Uiversity Press, 1996 J H va Lit ad R M Wilso, A Course i Combiatorics, Cambridge Uiversity Press, 1992 Jiri Matouse ad Jaroslav Nešetřil, Ivitatio to Discrete Mathematics, Oxford Uiversity Press, 1998

5 Cotets Preface ii 1 Subsets ad biomial coefficiets 1 11 Subsets 1 12 Subsets of fixed size 2 13 Properties of biomial coefficiets 3 14 The Biomial Theorem 6 15 Further properties of biomial coefficiets 7 16 Appedix: Proof of Lucas Theorem 10 2 Selectios ad arragemets The formulae Proofs Balls i urs Maig words from letters 18 3 Power series Power series Operatios o power series The Biomial Theorem Other power series 28 4 Recurrece relatios Fiboacci umbers Liear recurreces with costat coefficiets Liear recurreces with o-costat coefficiets No-liear recurreces 41 5 Partitios ad permutatios Partitios: Bell umbers Partitios: Stirlig umbers 49 v

6 vi CONTENTS 53 Permutatios: cycle decompositio Permutatios: Stirlig umbers 52 6 The Priciple of Iclusio ad Exclusio PIE Surjectios ad Stirlig umbers Deragemets 61 7 Families of sets Sperer s theorem Itersectig families The de Bruij Erdős theorem Fiite fields ad projective plaes Appedix: Proof of the Erdős Ko Rado Theorem 72 8 Systems of distict represetatives Hall s Theorem How may SDRs? Sudou 81 9 Lati squares Row by row Youde squares Orthogoal Lati squares Sets of mutually orthogoal Lati squares Appedix: Proof of Bose s Theorem Steier triple systems Existece of STS() Kirma s schoolgirls Appedix: Proof of Kirma s Theorem 101 Solutios to odd-umbered exercises 107 Miscellaeous problems 119 Idex 123

7 Chapter 1 Subsets ad biomial coefficiets Oe of the features of combiatorics is that there are usually several differet ways to prove somethig: typically, by a coutig argumet, or by aalytic methods There are lots of examples below If two proofs are give, study them both Combiatorics is about techiques as much as, or eve more tha, theorems 11 Subsets Let be a o-egative iteger, ad let X be a set with elemets How may subsets does X have? Propositio 11 The umber of subsets of a -elemet set is 2 First proof We ecode subsets by sequeces (e 1,e 2,,e ), where each e i is either 0 or 1 There are 2 choices for e 1, 2 choices for e 2,, 2 choices for e ; so altogether 2 sequeces So we are doe if we ca establish a bijectio betwee subsets ad sequeces To each subset Y of X, we associate the sequece (e 1,e 2,,e ) where e i = { 1 if i Y, 0 if i / Y It is easy to see that each sequece arises from a subset, ad distict sequeces arise from distict subsets; so the correspodece is a bijectio Secod proof This is a proof by iductio Let f () be the umber of subsets of {1,2,,} We see that f (0) = 1 (the empty set has just oe subset, amely itself) Also, f ( + 1) = 2 f (); for each subset Y of {1,2,,} ca be exteded i two ways to a subset of {1,2,, + 1}: we ca choose whether or ot to 1

8 2 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS iclude + 1 i the subset Now we ca easily prove by iductio that f () = 2 The iductio starts because f (0) = 1 = 2 0 For the iductive step, assume that f () = 2 ; the f ( + 1) = 2 f () = 2 2 = 2 +1 So the iductio goes through, ad the proof is complete 12 Subsets of fixed size If ad are itegers satisfyig 0, how may -elemet subsets does a -elemet set X have? ( ) Defie the biomial coefficiet by ( ) = ( 1) ( + 1) ( 1) 1 (There are factors i both the umerator ad the deomiator, the i-th factors beig i + 1 ad i + 1) For 0, the umber of -elemet subsets of a -elemet set is ( ) Proof We choose distict elemets of the -elemet set X There are choices for the first elemet; 1 choices for the secod; i+1 choices for the i-th; ad + 1 choices for the -th Multiply these umbers ( together ) to get that the total umber of choices is the umerator of the fractio This is ot the aswer, sice choosig the same elemets i a differet order would give the same subset for example, 1, the 4, the 3 would be the same as 3, the 1, the 4 So we have to divide by the umber of differet orders i which we could choose the elemets There are choices for the first; 1 for the secod; i + 1 for the i-th; ad + 1 = 1 choice (really o choice at all!) for the last Multiplyig these umbers gives the deomiator of the fractio So the result is proved ( ) It will sometimes be coveiet to give a meaig to the symbol eve if is greater tha We specify: ( ) If >, the = 0

9 13 PROPERTIES OF BINOMIAL COEFFICIENTS 3 This is a reasoable choice sice, if >, there are( o) -elemet subsets of a -elemet set You should chec that our formula for remais correct i this case: if >, the oe of the factors i the umerator is equal to 0 13 Properties of biomial coefficiets 131 Sum of biomial coefficiets The total umber of subsets of a -elemet set is 2 We ow the umber of subsets of size, for each value of : addig these up must give the total I other words, =0 ( ) = Biomial coefficiets ad factorials Here is a alterative formula for the biomial coefficiets This uses the factorial fuctio, defied by! = ( 1)( 2) 1, the product of all the itegers from 1 to iclusive Now we have ( )! =!( )! For if we tae the defiitio of the biomial coefficiet, ad multiply top ad bottom by ( )!, the i the umerator we have the product of all the itegers from 1 to, that is,!; the deomiator is!( )! I order to mae this formula valid i the limitig cases = 0 ad =, we have to adopt the covetio that 0! = 1 This may seem strage, but if we wat the recurrece! = ( ( ) 1)! ( ) to hold for = 1, the it ( is) forced upo us! This 0 the correctly gives = = 1, ad i particular = However, the formula does ot wor if >, sice the < 0 ad we caot defie factorials of egative umbers 133 A recurrece relatio There is a simple recurrece relatio for the biomial coefficiets, which eables big oes to be calculated from smaller oes by additio: ( ) ( ) ( ) = 1

10 4 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS First proof Cosider the problem of coutig the -elemet subsets of a - elemet set X, which cotais oe special elemet called x First we cout the sets which cotai x ( Each ) of these must have 1 out of 1 the remaiig 1 elemets So there are such sets 1 Next we cout the sets which do ot cotai x Each of these must ecessarily have ( ) elemets chose from the 1 elemets differet from x; so there are 1 such sets Addig these umbers together gives all the Secod proof ( ) sets We ca prove the result by calculatio, usig our formula: ( ) ( ) = = = = ( 1)! ( 1)!( )! ( 1)!( )! ( 1)!!( )! ( ), + ( 1)!!( 1)! + ( 1)! ( )!( )! where we have used the facts that! = ( 1)!,! = ( 1)!, ad ( )! = ( ) ( 1)! I mae o secret of the fact that I lie the first proof better! 134 Symmetry We have ( ) ( ) = For the first proof, we fid a bijective correspodece betwee the -elemet sets ad the ( )-elemet sets i a set of size ; this is easily doe by simply matchig each set with its complemet The secod proof, usig the formula i 2 above, is a simple exercise for the reader

11 13 PROPERTIES OF BINOMIAL COEFFICIENTS Pascal s Triagle It is possible to arrage the biomial coefficiets i a symmetrical ( ) triagular ( ) patter, i which the ( + 1)-st row cotais the + 1 umbers,, 0 The triagle begis as follows: Although we call this Pascal s Triagle, Pascal was ot the first perso to write it dow Below is a versio due to Chu-Shi-Chieh (Zhu Shijie), tae from wor of Yag Hui, i his boo Ssu Yua Yü Chie, dated 1303 Jia Xia ew it about 250 years earlier Other people who ew about it at roughly the same time iclude Halayudha i Idia, ad Al-Karaji ad Omar Khayyam i Ira We do t ow who iveted it! The property i 134 above shows that the triagle has left-right symmetry The recurrece relatio 133 shows that each etry of the triagle is the sum of the two etries immediately above it This gives a very quic method to geerate as much of the triagle as required

12 6 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS 14 The Biomial Theorem We ow come to the Biomial Theorem, a geeralisatio of the property 1 of the precedig paragraph (put x = y = 1 to see this) Theorem 12 (Biomial Theorem) (x + y) = =0 ( ) x y First proof We have (x + y) = (x + y)(x + y) (x + y), where there are factors o the right-had side of the equatio If all the bracets are expaded, we get a sum of very may terms; but each term is obtaied by choosig x from some of the bracets ad y from the remaiig oes If we choose x from bracets ad y from the remaiig, we obtai a term x y So the coefficiet of this term is the umber of ways we ca do this, i other words, the umber ( of) choices of out of the bracets from which x is selected This umber is So the theorem is proved Secod proof We prove the theorem by iductio o For = 0, the left-had side ( is )(x+y) 0 = 1, while the right-had side has just the sigle term = 0, which 0 is x 0 y 0 = 1 So the iductio starts 0 Suppose that the Biomial Theorem holds for a value The (x + y) +1 = (x + y)(x + y) = x ( =0x y ) + y ( =0x y ) ( ) For = m, secod term gives us a cotributio x m y +1 m What is the cotributio to of the first term to the coefficiet ( of x) m y +1 m? To get this term, we m must put = m 1, ad the coefficiet is m 1 So the coefficiet of x m y +1 m i (x + y) +1 is ( ) ( ) ( ) =, m 1 m m

13 15 FURTHER PROPERTIES OF BINOMIAL COEFFICIENTS 7 which is just what we require to mae the iductio wor So the proof is complete Sometimes it is coveiet to have a oe-variable form of the Biomial Theorem Puttig y = 1, we obtai ( ) (1 + x) = x =0 15 Further properties of biomial coefficiets 151 Eve ad odd ( ) We ow that, for fixed, the sum of the biomial coefficiets over all values of from 0 to is 2 What if we add them up just for eve, or just for odd? For > 0, /2 i=0 ( ) = 2i ( 1)/2 ( ) = 2 1 i=0 2i + 1 Proof Let S e ad S o be the sums of the eve ad odd biomial coefficiets respectively The S e +S o is the sum of all the biomial coefficiets; i other words, S e + S o = 2 If we put x = 1 i the oe-variable Biomial Theorem, we obtai =0( 1) ( ) = ( 1 + 1) = 0 Now i this sum, the eve biomial coefficiets have coefficiet +1 ad the odd oes have coefficiet 1; so the equatio says that S e S o = 0 The two displayed equatios show that S e = S o = 2 /2 = Biomial idetities There are a huge umber of other equatios coectig biomial coefficiets Here is oe Let m,, be positive itegers The i=0 ( i )( m i ) ( m + = (This result is sometimes called the Vadermode covolutio) )

14 8 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS First proof Suppose a school class cosists of m girls ad boys, ad we eed to choose a team of childre I how may ways ( ca ) this be doe? We cout m the umber of teams cotaiig i girls: there are ways to choose the girls, ( ) i ad ways of choosig the remaiig i team members from the boys i Multiplyig these umbers gives us the umber of possible teams cotaiig i girls, ad ( summig ) over i gives the total umber of teams But we ow that the m + total is Secod proof Cosider the equatio (1 + x) m (1 + x) = (1 + x) m+ ( ) m + What is the term i x? O the right, it is, by the Biomial Theorem O the left, we could choose the term x i from the first factor ad( x i ) from( the secod ) m ad multiply them The coefficiets of these two terms are ad ; so i i we multiply these umbers, ad the sum over i ( ) Puttig m = =, ad otig that = i i=0 153 Sum of sizes of sets ( ) 2 = i ( i ( ) 2 Here are a some further results ad proof techiques ( ) ( ) 1 First result: = 1 ), the equatio reduces to First proof From a class of childre, ( we ) have to choose a team of members, ad a captai for the team There are teams, ad choices of a captai for ( ) ay team; altogether choices But we could proceed differetly: we could choose the captai first (i ways), ad ( the) the remaiig ( 1 team) members 1 1 from the remaiig 1 childre (i ways), givig i all 1 1

15 15 FURTHER PROPERTIES OF BINOMIAL COEFFICIENTS 9 Secod proof ( )! =!( )! ( 1)! = ( 1)!( )! ( ) 1 = 1 ( ) Secod result: = 2 1 =1 First proof ( ) =1 = = =1 1 l=0 = 2 1 ( ) 1 1 ( ) 1 l Secod proof We have Differetiatig gives (1 + x) = =0 ( ) x (1 + x) 1 ( ) = x 1 =1 (We omit the = 0 term sice it is zero) Now put x = 1 to get the result ( ) Third proof There are subsets of size of a -elemet set X; so the sum o the left simply adds up the sizes of all these subsets But we ca calculate this sum aother way Pair up each subset A with its complemet X \ A; these two sets cotai elemets betwee them There are 2 subsets, ad so they fall ito 2 /2 = 2 1 pairs Thus the value of the sum is Cogrueces Here is a picture of part of Pascal s triagle I have put to mea that the etry is odd, ad left a bla if the etry is eve Notice the fractal structure of the

16 10 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS diagram: If we ow the triagle formed from the first 2 rows, we obtai the first 2 +1 rows by puttig two copies of the triagle side by side below the first oe, ad leavig the positios i the middle triagle bla This is explaied by a result called Lucas Theorem: Theorem 13 (Lucas Theorem) Let p be a prime umber Write ad to the base p: = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 The ( ) I particular, d i=0 ( ai b i ) (mod p) ( ) is divisible by p if ad oly if a i < b i for some value of i The proof of the theorem is i the ext sectio You should try to explai how this justifies the fractal shape of the diagram showig the parities i Pascal s triagle 16 Appedix: Proof of Lucas Theorem Recall the statemet of Lucas Theorem: Theorem (Lucas Theorem) base p: Let p be a prime umber Write ad to the = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 The ( ) d i=0 ( ai The proof comes from the followig lemma: b i ) (mod p)

17 16 APPENDIX: PROOF OF LUCAS THEOREM 11 Lemma Let p be prime, ad let = cp + a, = d p + b, with 0 a,b p 1 The ( ) ( )( ) c a (mod p) d b Proof Here is a short proof usig the Biomial Theorem The ey is the fact that, if p is prime, the (1 + x) p 1 + x p (mod p) ( ) p For each biomial coefficiet, for 1 i p 1, is a multiple of p, so all i ( ) p itermediate terms i the Biomial Theorem vaish mod p (We have = i p!/i!(p i)!, ad p divides the umerator but ot the deomiator) Thus (cogruece mod p): (1 + x) = (1 + x) cp (1 + x) a = (1 + x p c (1 + x) a ( ) c x pi i c i=0 a j=0 ( a j ) x j Sice 0 a,b < p, the oly way to obtai a term i t = t d p+b i this expressio is to tae the term i = d i the first sum ad the term j = b i the secod; this gives as required ( ) ( c d )( ) a b (mod p), Proof of the theorem The proof is by iductio o d The iductio starts with d = 1 sice, the = a 0, = b 0, ad there is othig to prove Suppose that the theorem holds with d 1 replacig d As i the statemet of the theorem, let = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 Put a = a 0, c = a 1 +a 2 p+ +a d p d 1, b = b 0, d = b 1 + b 2 p + + b d p d 1 The = cp + a, = d p + b, ad we have (with cogrueces mod p): ( ) ( )( ) c a (by the Lemma) d b

18 12 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS Corollary = ( d ( ai i=1 b i d ( ) ai i=0 b i ) ) ( a0 b 0 With the hypotheses of the theorem, if a i < b i for some i with 0 i d Proof If a i < b i, the ( ai b i ) (by the iductio hypothesis) ( ) is divisible by p if ad oly ) = 0 So oe of the factors o the right-had side of the theorem is zero, whece the product is zero ( ) ai If a i b i, the the biomial coefficiet is ot divisible by p (it is ozero ad there are o factors p i the umerator sice a i p 1 Now a product of d umbers ot divisible by p is itself ot divisible by p Example Let = 2 m 1 The all the digits a i of i base 2 are equal to 1, so we have b i a i for ay This meas that every etry i row of Pascal s triagle is odd Exercises 1 Write 1001 as a biomial coefficiet b i ( ) with 20 2 If X is a set of 8 elemets, the the umber of 3-elemet subsets of X is twice the umber of 2-elemet subsets Is there ay other size of the set X for which this holds? 3 Calculate =0 2( ) 4 Let X be a -elemet set Fid a bijectio F betwee the set of -elemet subsets of X ad the set of all ( )-elemet subsets of X Deduce that ( ) ( ) = 5 This exercise exteds the result about summatio of eve ad odd biomial coefficiets i 151 Similar methods ca deal with sums of biomial coefficiets where lies i ay fixed cogruece class of positive itegers

19 16 APPENDIX: PROOF OF LUCAS THEOREM 13 Let i deote the square root of 1, ad ote that 1+i = 2e π/4 Hece fid the real ad imagiary parts of (1 + i) for ay atural umber (You will probably fid it coveiet to cosider the differet cogruece classes mod 8 separately) Expadig (1 + i) by the Biomial Theorem, fid expressios for ( t)/4 ( ), j=0 4 j +t for t = 0, 1, 2, 3, agai separatig the cogruece classes mod 8 (This ivolves a lot of repetitious wor You should at least do all the calculatios for 0 (mod 8)) 6 By calculatig the coefficiet of x o the two sides of the idetity or otherwise, prove that 7 (1 + x) (1 x) = (1 x 2 ), =0( 1) ( ) { 2 0 if is odd, = ( ) 2m ( 1) m if = 2m m (a) Prove that ( ) = ( ) ( ) ( ) (b) Prove that, if > 2+1, the > ; if = 2+1, the ( ) ( + 1 ) ( ) ; ad if < 2 + 1, the < + 1 ( ) = + 1 (c) Hece show that, for fixed ad = 0,1,,, the biomial coefficiets icrease, the remai costat for oe step (if is odd), the decrease (Such a sequece is said to be uimodal) ( ) 2m (d) Show further that the largest biomial coefficiet is ( ) 2m + 1 while if = 2m+1 is odd, the ad m (e) Deduce that, if = 2m, the 2 2m 2m + 1 ( ) 2m 2 2m m ( 2m + 1 m + 1 if = 2m is eve, m ) are equal largest

20 14 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS 8 (a) Show that the biomial coefficiet ( ) 2m is divisible by every prime p sat- m isfyig m + 1 p 2m (b) Usig the estimate o Problem Sheet 1, Questio 2, show that the umber 2m of primes betwee m + 1 ad 2m is at most log 2 m Remar: This is a wea versio of the famous Prime Number Theorem, which says that the umber of prime umbers p satisfyig 1 p is asymptotically log

21 Chapter 2 Selectios ad arragemets 21 The formulae We have a hat cotaiig ames, ad we are goig to draw out ames I how may ways ca we do this? To aswer the questio, we have to clarify the strategy a bit First, do we care about the order i which the ames are draw, or ot? Secod, whe we have draw a ame, do we put it bac i the hat ad shae it up before the ext draw, or do we discard it? The aswers to this questio correspod to samplyig with order sigificat or ot, ad with repetitio allowed or ot allowed If the order is sigificat, we have a -tuple of ames; if ot, we have a set (if repetitio is ot allowed), or what might be called a multiset if repetitio is allowed We will write multisets i square bracets to distiguish them from sets For example, if the ames are a,b, ad we draw two of them, the order importat, repetitio allowed: there are four possibilities, (a, a), (a, b), (b,a) ad (b,b) order importat, repetitio ot allowed: there are two possibilities, (a, b) ad (b,a) order uimportat, repetitio allowed: there are three possibilities, [a, a], [a,b], ad [b,b] (Choosig a the b is the same as choosig b the a) order uimportat, repetitio ot allowed: just oe possibility, amely {a, b} I geeral, the umbers of selectios are give by the etries i the followig table We use the otatio () ( for ) the umber ( 1) ( + 1) This is the umerator i our defiitio of, ad is ofte called the fallig factorial 15

22 16 CHAPTER 2 SELECTIONS AND ARRANGEMENTS Repetitio allowed Repetitio ot allowed Order Order ot sigificat sigificat ( ) + 1 ( ) () Note that the umerator i the top right etry is ( + 1) ( + 1), the so-called risig factorial 22 Proofs Order sigificat, repetitio allowed: We get to mae choices, ad there are ames to choose at each step So there are possibilities Order sigificat, repetitio ot allowed: This time, there are ames to choose at the first step; 1 at the secod step (sice we discarded the first ame after we chose it); 2 at the third step; ad + 1 at the -th step Multiplyig these umbers gives the aswer Order ot sigificat, repetitio ot allowed: We simply choose a set with ( ) elemets from the elemets i the hat The umber of ways of doig this is, by defiitio Alteratively choose with order sigificat, ad repetitio allowed, ad ote that each uordered sample has! differet orderigs; so the aswer is () /! Order ot sigificat, repetitio allowed: This case is the most difficult But ote, before we begi, that we caot just use the argumet i the precedig paragraph to get /! [WHY NOT?] Step 1: The umber of choices of objects from, with order ot sigificat ad repetitio allowed, is equal to the umber of ways of choosig o-egative itegers x 1,,x satisfyig x x = For give the selectio, we ca let x i be the umber of times that the ith ame was selected; clearly x 1,,x satisfy the stated coditios Coversely, give x 1,,x satisfyig the coditios, form a selectio i which the i-th ame is chose x i times Thus, for example, suppose that = 3 ad = 6 If the ames are a,b,c, the the selectio [a,a,b,b,b,c] correspods to x 1 = 2, x 2 = 3, x 3 = 1

23 23 BALLS IN URNS 17 Step 2: So we have to cout the umber of choices of o-egative itegers x 1,,x with sum To do this, tae a row( of + ) 1 cells; ( choose ) 1 of them ad put marers i them There are = ways of 1 maig this choice Havig made the choice, defie x 1,,x as follows: Let x 1 be the umber of cells before the first mared cell Let x 2 be the umber of cells betwee the first ad secod mared cell Let x 1 be the umber of cells betwee the 1-st ad -th mared cell Let x be the umber of cells after the -th mared cell The clearly the umbers x 1,,x are o-egative itegers; they add up to the umber of umared cells, which is ( + 1) ( 1) = Moreover, every way of choosig o-egative itegers addig up to is represeted uiquely by such a marig of 1 out of + 1 boxes So the result is proved For example, our choice x 1 = 2, x 2 = 3, x 3 = 1 would come from a marig of the followig 1 = 2 out of + 1 = 8 boxes: To mae this clearer, here is a table which gives both steps i the case = 3, = 2 Let the ames i the hat be a,b,c The first colum gives a selectio of two ames (with repetitio allowed ad order uimportat) The secod gives three umbers addig up to 2 The third gives four boxes with a choice of two of them mared aa (2, 0, 0) ab (1, 1, 0) ac (1, 0, 1) bb (0, 2, 0) bc (0, 1, 1) cc (0, 0, 2) 23 Balls i urs There is aother way to loo at the mai result of the last sectio Suppose that we have urs, or vases, U 1,,U We have idistiguishable balls How may ways ca we put the balls i the urs? [Of course this problem ca be put ito

24 18 CHAPTER 2 SELECTIONS AND ARRANGEMENTS may disguises I have idetical sweets I how may ways ca I distribute them to a class of childre?] If x i is the umber of balls I put ito the ith ur [or the umber of sweets I give to the ith child], the x 1,,x are o-egative itegers ( which add ) up to So + 1 the umber of ways of puttig the balls ito the urs is The coditios ca be varied i may ways Suppose, for example, that I have to distribute balls amog urs as above, but with the requiremet that o ur should be empty This ass that x i 1 for all i If we defie ew variables y 1,,y by y i = x i 1, the the sum of the ys is ; so the umber of choices of the y s is ( ) ( ) + ( ) 1 1 = 1 The simple way to thi about this is: Suppose each ur is to be o-empty The I first tae balls ad put oe i each ur The I distribute the remaiig balls ito the urs i ay way This gives the same result as above Example How may ways ca I distribute 100 sweets to a class of 30 boys ad 20 girls, if it is required that each boy has at least oe sweet ad each girl has at least two sweets? To solve this, I first give oe sweet to each boy ad two to each girl, usig up = 70 sweets The I( distribute the) remaiig ( ) 30 sweets amog the childre, which ca be doe i = ways Maig words from letters How may ways ca we arrage distict objects i order? By the formula i the bottom left of the box, the aswer is simply () =! Aother way of seeig this is as follows Let F() be this umber The F(0) = 1, F() = F( 1) for > 0 (We tae F(0) = 1 because there is just oe list with o etries, the empty or bla list To get the secod equatio, we choose oe of the objects to be first o the list (there are ways of doig this), ad the we have to put the remaiig 1 i order after the first oe) Now a easy iductio argumet shows that F() =! for all Now we mae the questio a bit harder How may ways of arragig some (possibly all) of the objects i a list?

25 24 MAKING WORDS FROM LETTERS 19 Example How may words ca I mae from the letters of the word FACE- TIOUS? (A word is simply a strig of letters chose from those i the word; we do ot require that it maes sese i Eglish or ay other laguage The order of the origial letters i the word is irrelevat; a better aalogy is that you are playig Scrabble ad you have these letters By covetio we iclude the empty word, which is the strig cotaiig o letters) I the case give, the letters are all distict; this maes life easier, so we start with this case Suppose that we are give letters, all differet How may - letter words ca we costruct? These words are just selectios of letters from the give, with order importat ad repetitio ot allowed; so the umber is () = ( 1) ( + 1) So the total umber W() of words is W() = =0 () We ca express this aother way Note that () =!/( )! So W() =! ( =0 Now recall from calculus that ) ( 1 =! ( )! m=0 m=0 1 m! = e ) 1 m! Iside the bracets of the formula for W(), we see the sum of the reciprocals of the factorials from 0 to, i other words, the sum of the first + 1 terms of the ifiite series So we see that W() is approximately e! We ca be more precise: e! W() = = <! m=+1 m! ( + 1)( + 2) + 1 ( + 1)( + 2)( + 3) ( + 1) ( + 1) 3 + = 1 (I the last term we summed a geometric series)

26 20 CHAPTER 2 SELECTIONS AND ARRANGEMENTS I other words, e! is bigger tha the iteger W() but smaller tha W() + 1/; so we get W() by calculatig e! ad roudig dow to the iteger below So fially we coclude, usig the floor or iteger part fuctio, that W() = e! For the word FACETIOUS, we have = 9, ad W() = e 9! = Remar Although W() = e! is a beautiful simple formula, ad gives us a very good estimate for the size of W(), it is ot so good for the purpose of calculatio For example, 70! is a umber with about 100 digits, so i order to decide whether W(70) is odd or eve we would eed to ow e to 100 places of decimals (at least) For exact calculatio it is better to use the formula W() = () = ( 1) + ( 1)( 2) + =0 We ca also fid W() by a recurrece method We have W(0) = 1, W() = 1 + W( 1) for > 0 (The coditio W(0) = 1 is because of the empty word I geeral, to form a word of from letters, we choose oe letter to go first (i ways), ad mae a word from the remaiig 1 letters (i W( 1) ways) to follow it; but we have missed out oe word, amely the empty word, so we eed to add 1) A easy iductio ow gives the formula for W() If the letters we are give cotai repetitios, it is more difficult to write dow a formula Here, we will simply do a example Example How may words ca be made from the letters of SYZYGY? For the case whe we use all the letters, the aswer is ot too hard There are 6! ways of arragig the six letters, but ay rearragemet of the three Ys will give the same word So the umber of arragemets is 6!/3! = 120 If we allow words of arbitrary legth, it is a bit more difficult To solve it, we subdivide the words accordig to the umber of occurreces of the letter Y At most oe Y We have to mae words out of the four letters S, Z, G ad Y This ca be doe i W(4) = = 65 ways

27 24 MAKING WORDS FROM LETTERS 21 Two Ys Temporarily label the Ys as Y 1 ad Y 2 so we ca distiguish them Now we have five letters S, Z, G, Y 1 ad Y 2, but we must use the two Ys Choose some of the other three letters, order all letters icludig the two Ys i ay way, ad add up all possibilities; fially divide by 2 sice the Ys are really idistiguishable We get (( ) ( ) ( ) ( ) / 2! + 3! + 4! + 5!) 2 = All three Ys Similarly the total for this case is (( ) ( ) ( ) ( ) / 3! + 4! + 5! + 6!) 3! = So the total is = 364 Exercises 1 (a) How may ordered sequeces of legth 5 ca be made usig the elemets {1,2,3,4,5,6,7} if repetitios are allowed? How may of these cotai exactly two of the umbers 1,2,3? I how may of them do eve ad odd umbers alterate? (b) What are the aswers to these questios if repetitios are ot allowed? 2 How may words ca be made usig the letters of the word STARTS? How may of these are palidromes (that is, read the same bacward as forward)? 3 Let X ad Y be sets with X = ad Y = m (a) Determie the umber of fuctios f mappig X ito Y (b) How may of these fuctios are ijectios, ie oe-to-oe? (c) How may of these fuctios are bijectios, ie oe-to-oe ad oto? (d) (much harder) How may of these fuctios are surjectios, ie oto? 4 How may permutatios of the set {1,2,,} are there? How may of these are cyclic permutatios, that is, their cycle decompositio cosists of a sigle cycle of legth? 5 For which values of is W() odd?

28 22 CHAPTER 2 SELECTIONS AND ARRANGEMENTS

29 Chapter 3 Power series A lot of combiatorics is about sequeces of umbers: We ll see such sequeces as (Fiboacci umbers), or (a 0,a 1,a 2,) (1,1,2,3,5,8,13,21,34,) (1,1,2,6,24,120,720,5040,) (factorials) A very useful device to represet such a sequece of umbers is to tae the umbers to be the coefficiet i a power series a x = a 0 + a 1 x + a 2 x 2 + a 3 x We call this power series the geeratig series or geeratig fuctio for the sequece of umbers I this chapter we loo at power series ad some of their uses i combiatorics Example We saw that the umber of subsets of a -elemet set is 2 This gives us a sequece of umbers, amely whose geeratig fuctio is (2 0 = 1,2 1 = 2,2 2 = 4,2 3 = 8,) 2 x = x usig the formula for the sum of a geometric series 23

30 24 CHAPTER 3 POWER SERIES 31 Power series You ve met power series i calculus, ad maybe i aalysis also So how do they compare with combiatorics: First, the good ews We are ot doig calculus here, so we do t have to worry whether the sequeces coverge or ot For us, a power series is just a booeepig device, to wrap up ifiitely may terms ito a sigle mathematical object For example, if our sequece is the factorials above, the the power series is!x 0 ad if you remember the ratio test from calculus, you should be able to show that this series ever coverges uless x = 0 (The ratio of successive terms is ( + 1)!x +1 /!x = ( + 1)x, which teds to ifiity as ) But this power series might still be useful! Secod, the good ews If a power series does coverge, ad if you ow somethig about the properties of the fuctio A(x) it defies, the you ca use those properties i combiatorics also! We ll see some examples later I the example above, the sum of the series is 1/(1 2x); the series coverges if x < 1/2 We deote the set of all power series with iteger coefficiets by Z[[x]] This should remid you of the otatio Z[x] for the set of polyomials with iteger coefficiets; power series are very similar to polyomials, but ca have ifiitely may coefficiets Similarly, if we wat the coefficiets to be real umbers, we write R[[x]], with similar modificatios for the other umber systems 32 Operatios o power series There are various operatios that ca be doe to power series If you studied Algebra, you have met the idea of a rig; the first two operatios below (additio ad multiplicatio) mae R[[x]] ito a rig, for ay rig R (though we wo t stop to prove this) Additio We add two power series term by term: ( ) ( a x ) b x = (a + b )x 0

31 33 THE BINOMIAL THEOREM 25 Multiplicatio We multiply power series i the same way as we multiply polyomials To get a term i x i the product, we multiply the term i x i the first factor by the term i x i the secod, ad sum over all values of from 1 to Thus Substitutio ( ) ( ) a x b x 0 0 where c = = c x, 0 a b =0 Let A(x) = a x ad B(x) = b x Suppose that a 0 = The we ca substitute A(x) for x i the secod series: B(A(x)) = b (A(x)), 0 where A(x) is calculated usig the multiplicatio rule Why do we eed the costat term of A(x) to be zero? Cosider the costat term of the series B(A(x)) It would be b 0 +b 1 a 0 +b 2 a 2 0 +, ad we would have a ifiite series of umbers, ad would have to worry about covergece But if a 0 = 0, the the smallest power of x occurrig i A(x) is at least x ; so whe we come to calculate the coefficiet of x i B(A(x)), we oly have to cosider fiitely may terms b A(x) for 0 I other words, we oly eed fiitely may additios ad multiplicatios to wor out ay term Differetiatio the We ca also differetiate power series If A(x) = a x, 0 d dx A(x) = a x 1 = (m + 1)a m+1 x 1 m 0 Notice what has happeed here The term = 0 is zero, so we leave it out i the first step; the we use a ew summatio variable m = 1, so that as rus from 1 to ifiity, m rus from 0 to ifiity 33 The Biomial Theorem We saw the Biomial Theorem, a formula for (1+x) for positive itegers Here is a geeralisatio of it, first proved by Isaac Newto

32 26 CHAPTER 3 POWER SERIES We eed to geeralise the defiitio of biomial coefficiets first Let a be ay umber, positive or egative, ratioal or irratioal, real or complex Let be a atural umber (a positive iteger or zero) Defie ( a ) = a(a 1) (a + 1) ( 1) 1 This has the properties ( ) a if a is a atural umber, the = 0 for ; ( ) a otherwise, 0 for all a ( ) a For the oly way we ca have = 0 is for oe of the factors i the umerator to be zero, that is, a i = 0 (that is, a = i) for some i 1 Now we have: Theorem 31 (The Biomial Theorem) For ay complex umber a, ( ) a (1 + x) a = x 0 There are two ways to iterpret this theorem I terms of calculus: the series o the right coverges for x < 1, ad its sum is (1 + x) a Secod, i terms of combiatorics: The usual rules of expoets hold A calculus proof of the Biomial Theorem (without all the tricy details about covergece) is give i a appedix Example 1 The first law of expoets says that By the Biomial Theorem, ( ( ) a )x 0 (1 + x) a (1 + x) b = (1 + x) a+b ( 0 ( ) b )x = 0 Now by the rule for multiplicatio of power series, ( )( ) ( a b a + b = i i i=0 ( a + b ) ) x This is the Vadermode covolutio We saw it for atural umbers a ad b i Sectio 152; but ow we ow that it holds for ay a ad b at all

33 33 THE BINOMIAL THEOREM 27 Example 2 We get some iterestig examples by choosig expoets which are ot atural umbers so Case a = 1 We have ( ) 1 = ( 1)( 2) ( ) ( 1) 1 ( 1 (1 x) 1 = 0 ) = ( 1), ( x) = x, 0 so we have the formula for the sum of a geometric series We already used this i calculatig the geeratig fuctio for the powers of 2 Case a = 1/2 We have ( ) 1/2 = = = = ( 1/2)( 3/2) ( (2 1)/2) ( 1) 1 ( ) 1 (2 1)(2 3) 1 2 ( 1) 1 ( ) 1 1 2(2 1) 1 4!! ( ) 1 ( ) 2, 4 where we have used the fact that 2(2 2) 2 = 2! Thus ( 1/2 (1 4x) 1/2 = 0 ) ( 4x) = 0 ( 1 4 ) ( ) 2 ( 4x) = 0 So the geeratig fuctio for the cetral biomial coefficiets =0 ( ) 2 x ( ) 2 is 1/ 1 4x Exampe 2, cotiued We ca use what we just leared to prove the followig idetity for the cetral biomial coefficiets: ( )( ) 2 2( ) = 4

34 28 CHAPTER 3 POWER SERIES Proof We start from the idetity (1 4x) 1/2 (1 4x) 1/2 = (1 4x) 1 Now the coefficiet of x o the left is obtaied by taig the coefficiet of x i the first factor (1 4x) 1/2, multiplyig by the coefficiet of x i the secod factor, ad summig over from 0 to This gives precisely the left-had side of the result we are provig O the right, (1 4x) 1 = 4 x, 0 so the coefficiet of x is 4, ad we are doe Example 3 Here is a simple example of the use of power series to solve a recurrece We will have more complicated examples later Suppose that a sequece of umbers a 0,a 1,a 2 satisfy a 0 = 1 ad a = 2a 1 for 1 Of course it is clear that these umbers are the powers of 2 But let us see this aother way The geeratig fuctio is A(x) = a x 0 = 1 + 2a 1 x 1 = 1 + 2a m x m+1 m 0 = 1 + 2xA(x) (Chec that you ca follow all these steps I the third step we have used a ew summatio variable m = 1) This equatio ca be rearraged to give A(x) = 1 1 2x = (2x) = 2 x 0 Now if two power series are equal the their coefficiets must be the same; so we have a = 2 for all 0 34 Other power series Apart from the Biomial Theorem, there are a couple of other famous power series which crop up from time to time:

35 34 OTHER POWER SERIES 29 The expoetial fuctio I calculus this is usually writte as e x I will usually write it as exp(x); this meas the same thig The power series is The most importat properties are x exp(x) = 0! d exp(x) = exp(x) This is easy to prove from the power series sice dx d dx x! = x 1 ( 1)! exp(x + y) = exp(x) exp(y) (We prove this below) The logarithm fuctio The fuctio log(x) is ot defied at x = 0 so we caot write it as a power series Istead, we have log(1 + x) = 1 If we differetiate term by term we get ( 1) 1 x d dx log(x) = ( x) 1 = (1 + x) 1 1 The logarithm is the iverse of the expoetial: exp(log(1 + x)) = 1 + x, log(exp(x)) = x (Remember that we ca substitute oe power series i aother if the first oe has costat term zero This is OK for the first result above I the secod case, it is really log(1 + y), where y = exp(x) 1, which does ideed have costat term zero) Example is Cosider the equatio exp(x + y) = exp(x) exp(y) The left had side (x + y) 0! = 0 1! =0 ( )( x = 0! l 0 = exp(x) exp(y)!!( )! x y ) y l l!

36 30 CHAPTER 3 POWER SERIES (I the secod lie we used a dummy variable l = We have to chec the rages of summatio: taig all values ad ruig from 0 to is the same as ad l idepedetly taig all o-egative values) We could have reversed the procedure ad derived the Biomial Theorem from the property of the expoetial fuctio Actually there is a lot of very iterestig combiatorics hidde i the power series for the expoetial ad logarithm fuctios If you are iterested i this, see my Notes o Coutig o the Web 341 Appedix: Proof of the Biomial Theorem This proof is a bit of a cheat, sice all the hard wor is i the calculus Suppose we have a power series a x whose sum is a ow fuctio f (x) 0 How do we wor out the coefficiets a? If we differetiate the series times, we get d dx f (x) = a ( 1) ( + 1)x (We start the sum at = because the -th derivative of ay smaller power of x is zero) The if we put x = 0, we fid [ d ] dx f (x) =!a, x=0 so that a = [(d /dx ) f (x)] x=0 /! Taig f (x) = (1 + x) a, whe we differetiate times we get Puttig x = 0, we get [ d d dx (1 + x)a = a(a 1) (a + 1)(1 + x) a (1 + x)a dx ] x=0 = a(a 1) (a + 1) So the coefficiet of x i the power series for (1 + x) a is ( ) a(a 1) (a + 1) a =,! ( ) a so that (1 + x) a = x 0

37 34 OTHER POWER SERIES 31 Exercises 1 The purpose of this exercise is to show you that, eve whe a power series fails to coverge, algebraic maipulatios o it ca still give us somethig iterestig (a) Let π be a permutatio of the set {1,,} We say that π is decomposable if there is a umber, with 1 1, such that π maps the umbers 1,, to themselves If o such exists the π is idecomposable There are! permutatios of the set {1,,} Suppose that g() of them are idecomposable (By covetio we tae 0! = 1 but we do ot defie g(0)) For ay permutatio π, let be the smallest umber such that π maps 1,, to themselves (so that = if π is idecomposable) Show that there are g()( )! permutatios with ay give value of Hece show that g()( )! =! =1 Now let F(x) =!x ad G(x) = g()x be the geeratig fuctios 0 1 for the factorial umbers ad the umbers g() respectively Note that G(x) has costat term zero sice we start at 1 Prove that F(x)(1 G(x)) = 1 Note that this equatio maes sese eve though the power series do ot coverge for ay o-zero value of x

38 32 CHAPTER 3 POWER SERIES

39 Chapter 4 Recurrece relatios Recurrece relatios are a very powerful method of calculatig combiatorial umbers But there are ot may geeral methods for dealig with them, so mostly we will just loo at a few importat examples The mai idea is that we ca tur a recurrece relatio for a sequece of umbers ito a equatio (algebraic or differetial) for the geeratig fuctio 41 Fiboacci umbers Leoardo Fiboacci was a Italia mathematicia of the 13th cetury His most importat wor was the itroductio of the Arabic umerals 0,1,2,3,4,5,6,7,8,9 to Europe I order to show how much easier it is to calculate with these tha with the Roma umerals previously used, he posed the followig problem as a exercise i his boo Liber Abaci (The Boo of Calculatio): A pair of rabbits do ot breed i their first moth of life, but at the ed of the secod ad every subsequet moth they produce oe pair of offsprig If I acquire a ew-bor pair of rabbits at the begiig of the year, how may pairs of rabbits will I have at the ed of the year? Uder these coditios, the umber of pairs of rabbits after moths is called the th Fiboacci umber F How do we calculate these umbers? First, we have F 0 = 1, F 1 = 1 For we are give that we have oe pair of rabbits at the start of moth 0, ad they do ot produce offsprig i moth 1 Next, F = F 1 + F 2 for 2 33

40 34 CHAPTER 4 RECURRENCE RELATIONS To show this, let G be the umber of pairs of rabbits which are old eough to breed at the ed of moth Now by the coditios of the problem, we have G = F 2 (sice the rabbits breedig i moth are all those bor i moth 2 or earlier) Also, F F 1 = G, sice G pairs are bor i moth ad are those cotributig to F but ot to F 1 Elimiatig G from these two equatios gives the result So the aswer to Fiboacci s exercise ca be foud by a doze additios, a simple job usig Arabic umerals Fiboacci did ot ivet these umbers, which had bee ow to Idia mathematicias icludig Pigala, Virahaa ad Hemachadra for early 1500 years whe he wrote his boo The coditio F = F 1 + F 2 is a example of a recurrece relatio This is a relatio which eables ay term of the sequece to be calculated if the earlier terms are ow I this case we oly eed to ow the two precedig terms Usually, a recurrece relatio eeds to be supplemeted with iitial coditios, tellig us how the sequece starts I this case the recurrece relatio oly applies for 2, so we eed to be give the values of F 0 ad F 1 separately I the ext sectio, we will solve this recurrece relatio to fid a explicit formula for the th Fiboacci umber First, though, we give a couple more coutig problems for which the Fiboacci umbers are the solutio Example I have a staircase with steps At a sigle stride, I ca go up either oe or two of the steps I how may differet ways ca I wal up the staircase? Let a be this umber The a 0 = 1 (sice if there are o steps, the there is oly oe way to do othig!) ad a 1 = 1 (obviously) We claim that a = a 1 +a 2 for 2 For let S be the set of all ways of walig up the steps The last step we use before we reach the top is either umber 1 or umber 2 (sice we asced either oe or two steps i the last stride); so let S 1 be the set of ways i which the peultimate step is umber 1, ad S 2 those i which it is umber 2 The S 1 ad S 2 are disjoit ad have uio S Moreover, clearly we have S 1 = a 1 while S 2 = a 2, ad S = a So the recurrece relatio holds Now a straightforward iductio shows that a = F for all atural umbers This represetatio of the Fiboacci umbers was discussed by Virahaa i the 6th cetury, i coectio with Sasrit poetry A vowel i Sasrit ca be log or short If we assume that a log vowel is twice as log as a short vowel, i how may ways ca we mae a lie of poetry of legth out of log ad short vowels? Clearly this is the same problem, ad the aswer is the th Fiboacci umber F

41 41 FIBONACCI NUMBERS 35 From this theorem we get a curious formula for F : F = /2 =0 ( For aother way of statig our result is that the umber of ways of writig as a ordered sum of oes ad twos is F Now we ca cout these expressios aother way Suppose that we have twos i the sum The we must have 2 oes, so there are terms altogether (ad we see that /2 ) So the umber of expressios with twos is the umber of selectios of elemets from ( ) (the positios i the sequece where the 2s occur), of which there are Summig over gives the result For example, whe = 4, we have ( ) 4 = 1 (correspodig to ); 0 ( ) 3 = 3 (correspodig to , ad ); 1 ( ) 2 = 1 (correspodig to 2 + 2) 2 Summig, we have F 4 = 5 Example How may sequeces of legth are there cosistsig of zeros ad oes with o two cosecutive oes? (Call such a sequece admissible) Let b be this umber Clearly b 0 = 1 (oly the empty sequece), ad b 1 = 2 (the sequeces 0 ad 1 are both admissible) Partitio the set T of all admissible sequeces ito two subsets T 0 ad T 1, where T 0 is the set of sequeces edig i 0, ad T 1 is the set of sequeces edig i 1 Now give ay admissible sequece of legth 1, we ca add a zero to it to get a admissible sequece of legth ; so T 0 = b 1 But we may oly add a 1 to a admissible sequece if it eds i zero; so T 1 is the umber of admissible sequeces of legth 1 edig i zero, which by the precedig argumet is b 2 Thus, b = b 1 + b 2 We have the same recurrece relatio as for the Fiboacci umbers, but differet iitial coditios However, we have b = F +1 for all The proof is by iductio We have b 0 = 1 = F 1, b 1 = 2 = F 2, ad for 2, ) b = b 1 + b 2 = F + F 1 = F +1

42 36 CHAPTER 4 RECURRENCE RELATIONS 42 Liear recurreces with costat coefficiets I this sectio we will fid a formula for the th Fiboacci umber The two methods we use ca be exteded to a wider class of recurrece relatios Method 1 We are tryig to solve the recurrece relatio with iitial coditios F 0 = 1, F 1 = 1, F = F 1 + F 2 for 2 We begi by observig that there is a uique solutio For F 0 ad F 1 are give, ad the the recurrece determies F 2,F 3, (This is really a argumet by iductio!) So, if we ca fid by ay method at all a solutio, the we ow it is the uique solutio We will cosider just the recurrece relatio a = a 1 + a 2, ad worry about the iitial coditios later The ext observatio that we mae is that the recurrece relatio is liear That meas that, if two sequeces (a ) ad (b ) satisfy it, the so does ay liear combiatio (c ) with c = pa + qb for ay umbers p ad q So we cocetrate o fidig specific solutios We try a solutio of the form a = α for some umber α [Why? Oe aswer is that it wors, as we will see A better aswer is that, if you cosider a oe-term recurrece relatio lie a = αa 1, it is obvious that there will be a solutio a = α ] Now a = α will satisfy the recurrece relatio if ad oly if α = α 1 + α 2 for 2 This will be the case if ad oly if α 2 = α + 1 The quadratic equatio x 2 = x + 1 has two solutios α = , β = So a = α ad a = β both satisfy our recurrece relatio, ad by the liearity priciple, so does a = pα + qβ for ay p ad q Fially, we try to choose p ad q such that this solutio also satisfies the iitial coditios a 0 = a 1 = 1 This gives us two equatios p + q = 1, pα + qβ = 1

43 42 LINEAR RECURRENCES WITH CONSTANT COEFFICIENTS 37 Solvig these equatios we fid that p = , q = So we coclude that F = 1 5 ( 1 + ) +1 ( ) Now this is ot a very good formula for calculatio, sice we eed to ow 5 to a high degree of accuracy to use it But it has oe advatage We have α = 1618 ad β = 0618 So α > 1 while β < 1 This meas that the th power of α grows expoetially, while the th power of β teds expoetially to zero So we get a very good approximatio F 1 5 ( 1 + ) The umber (1 + 5)/2 is called the golde ratio It has a log history i Wester art, music ad botay Method 2 This method wors with the geeratig fuctio f (x) = F x Recall our coditios: We claim that F 0 = 1, F 1 = 1, F = F 1 + F 2 for 2 (1 x x 2 ) f (x) = 1 For the costat term i (1 x x 2 ) f (x) is F 0 = 1, ad the coefficiet of x is F 1 F 0 = 0; while, for 2, the coefficiet of x is F F 1 F 2 = 0 So 1 f (x) = 1 x x 2 To proceed we use the method of partial fractios First, we factorise the deomiator: 1 x x 2 = (1 αx)(1 βx), where α ad β are as i the last sectio The we write 1 1 x x 2 = p 1 αx + q 1 βx 0

44 38 CHAPTER 4 RECURRENCE RELATIONS Multiplyig up by the deomiator, 1 = p(1 βx) + q(1 αx), givig two equatios p + q = 1, pβ + qα = 0 Solvig these two equatios gives the same values of p ad q as we foud i the last sectio Fially, we use the fact that 1 1 αx = α x, 0 ad similarly for β, usig the formula for a geometric series So we have f (x) = (pα + qβ )x, 0 so that F = pα + qβ, exactly as we foud by the other method The methods used here wor more geerally A th order liear recurrece with costat coefficiets is a relatio a = c 1 a 1 + c 2 a c a, for fixed costats c 1,,c, coectig the terms of a sequece (a ) I order to specify the terms completely, we eed to specify the values of a 0,a 1,,a 1 ; the the recurrece expresses all later terms uiquely We ca use either of the above methods The umbers α ad β earlier are replaced by the solutios α 1,,α of the equatio x = c 1 x 1 + c 2 x c There is oe complicatio If this polyomial has repeated roots, we do t fid eough solutios of the form a = α to use the first method Istead, if the umber α is a r-fold root, the the r fuctios all tur out to be solutios a = α,α,, r 1 α

45 43 LINEAR RECURRENCES WITH NON-CONSTANT COEFFICIENTS 39 Example The umber f () steps required to solve the Chiese rigs puzzle with rigs satisfies the recurrece f (1) = 1, f (2) = 2, f () = f ( 1) + 2 f ( 2) + 1 for 3 There is a awward 1 o the right, which ca be removed by puttig g() = f () + 1/2; we fid that g(1) = 3/2, g(2) = 5/2, g() = g( 1) + 2g( 2) for 3 Now the equatio for α ad β is x 2 = x + 2, with solutios α = 2, β = 1 So The iitial values give g() = p 2 + q( 1) 2p q = 3/2, 4p + q = 5/2, with solutio p = 2/3, q = 1/6 So the solutio to the origial problem is f () = (2/3)2 (1/6)( 1) (1/2) 43 Liear recurreces with o-costat coefficiets The ext complicatio is that a recurrece relatio ca have coefficiets which are ot costats but fuctios of The simplest example is the recurrece for the factorial umbers!: 0! = 1,! = ( 1)! for 1 A closely related example cocers W(), the umber of words that ca be formed of distict letters We saw that W(0) = 1, W() = 1 + W( 1) for 1 There are geeral methods for solvig recurreces of this type (if the coefficiets are polyomials i ) i terms of so-called hypergeometric fuctios Here I will simply discuss oe example, which illustrates aother techique

46 40 CHAPTER 4 RECURRENCE RELATIONS Deragemets A permutatio of {1,,} is a bijective fuctio from the set {1,,} to itself The umber of permutatios is! We will say more about permutatios later Here we loo at a special type of permutatio A deragemet is a permutatio which leaves o poit fixed That is, the permutatio π is a deragemet if π(i) i for i = 1,, How may deragemets are there? Let this umber be d() Trivially d(0) = 1, sice there are o poits to fix Also, d(1) = 0 (there is oly oe permutatio of {1}, ad it obviously fixes the poit 1), ad d(2) = 1 (the uique deragemet beig the permutatio which swaps 1 ad 2) We show that the followig recurrece relatio holds: d() = ( 1)(d( 1) + d( 2)) for 2 To see this, cosider deragemets π of {1,,} Sice the poit is ot fixed, we must have π() = i for some i, with 1 i 1 Now by symmetry, the umber of deragemets satisfyig π() = i is idepedet of i; so we oly have to cout the deragemets with a fixed value of i, ad multiply the umber of these by 1 We divide the deragemets satisfyig π() = i ito two types: Type 1: Those with π(i) =, that is, swappig i with Such a permutatio π is a deragemet of the 2 poits differet from i ad There are d( 2) such deragemets; each of them ca be exteded to the whole set so that it swaps i ad Type 2: Those with π(i) The π( j) = for some j i Now π maps j i We ca tae a short-cut by goig straight from j to i, givig a permutatio of {1,, 1}; this permutatio is a deragemet, so there are d( 1) choices Give ay deragemet of {1,, 1}, we ca exted it to {1,,} by iterpolatig just before i So the umber of deragemets mappig to i is d( 1) + d( 2), ad the total umber of deragemets is ( 1)(d( 1) + d( 2)) It is possible to use this recurrece to fid a formula for the umbers d(), or to fid a geeratig fuctio for them; ad there is a completely differet approach usig the Iclusio Exclusio Priciple that I will discuss later i the otes Here I will merely quote the formula: d() =! =0 ( 1)!

47 44 NON-LINEAR RECURRENCES 41 Let us loo at this formula It is very similar to the formula for W(), ad ca be aalysed similarly Note that, from the expoetial series, we see that so e 1 = 0 ( 1),!!e 1 d() =! +1 ( 1)! Just as for W(), the right-had side of this equatio has modulus smaller tha 1, ideed, smaller tha 1/2 if 1 We coclude that d() is the iteger earest to!/e the differece beig alterately positive ad egative This has a iterestig iterpretatio Suppose that people go to the theatre, ad leave their hats at the cloaroom After the performace, whe they go to collect their hats, the cloaroom attedat gives them out at radom The the probability that obody gets his or her correct hat is very close to 1/e = For we ca regard the allocatio of the hats as a radom permutatio of the correct allocatio; ad the evet that obody gets the correct hat is just that the radom permutatio is a deragemet 44 No-liear recurreces A recurrece relatio is really ay expressio, however complicated, which expresses the th term of a sequece i terms of smaller terms There is o geeral method for solvig a arbitrary recurrece relatio Here I will just cosider oe importat example Catala umbers The Catala umbers appear as the solutio of may differet coutig problems For example, suppose that we have to calculate a product x 1 x 2 x If we ca oly multiply two factors at a time, we have to put i bracets to mae the expressio well-defied how may ways ca we bracet such a product? Let C be this umber If = 1, o bracets are eeded, ad C 1 = 1 If 2, the we ca bracet together the first terms i C ways, ad the last terms i C ways, ad fially multiply together these two expressios ad sum over ; so C = 1 C C =1

48 42 CHAPTER 4 RECURRENCE RELATIONS For example, there are five bracetigs for = 4, amely ((ab)c)d, (a(bc))d, (ab)(cd), a((bc)d), a(b(cd)) Thus, the Catala umbers are the umbers C 1,C 2, satisfyig C 1 = 1, C = 1 C C for 2 =1 We have C 2 = 1 1 = 1; C 3 = = 2; C 4 = = 5 (as illustrated above); ad so o Each value C for 1 is uiquely determied by these coditios Let c(x) be the geeratig fuctio: We claim that c(x) = C x 1 c(x) 2 = c(x) x For cosider the coefficiet of x o the left-had side If 2, we obtai a cotributio to this term by taig the term i x i the first factor c(x), ad the term i x i the secod; multiplyig the coefficiets (givig C C ); ad summig over The sum rus from 1 to 1 sice the lowest degree of a term is 1, ot 0 For = 1, this argumet is wrog; there is o term i x o the left, where as c(x) starts with the term x So we have to subtract x to mae the coefficiets equal This gives the stated relatio We ca write this relatio as c(x) 2 c(x) + x = 0 Thi of this as a quadratic i the uow c(x) The solutio is c(x) = 1 ± 1 4x 2 Now we seem to have two solutios, whereas there should oly be oe But we ow that c(0) = 0, sice the series has o costat term If we too the plus sig, we would get c(0) = 1 So we have to tae the mius sig: c(x) = 1 1 4x 2 We ca use this to fid a formula for the Catala umbers, usig the Biomial Theorem: ( ) 1/2 (1 4x) 1/2 = ( 4) x 0

49 44 NON-LINEAR RECURRENCES 43 We have ( 1/2 ) ( 4) 1 2 = 1 (2 3) 2 2 ( 4)! 1 3 (2 3) 4 = 2! (2 2) 22 = = !( 1)! ) ( (We used the fact that 2 4 (2 2) = 2 1 ( 1)!) Now C is the coefficiet i x i this series multiplied by 1/2; so we have C = 1 ( ) Exercises 1 Let s() be the umber of expressios for as a sum of positive itegers For example, 4 = = = = = = = , so s(4) = 8 (a) Show that s(1) = 1 ad for 2 1 s() = 1 + s() =1 (b) Deduce that s(1) = 1 ad s() = 2s( 1) for 2 (c) Hece show that s() = 2 1 for 1 Notice how we have coverted a rather complicated recurrece relatio ito a much simpler oe! 2 Solve the recurrece relatio ad iitial coditios a 0 = 2, a 1 = 4, a 2 = 7, a = 4a 1 5a 2 + 2a 3 for 3 3 I purchase a item costig pece I have a large umber of 1 ad 2 pece cois at my disposal I how may ways ca I pay for the item

50 44 CHAPTER 4 RECURRENCE RELATIONS (a) if I am buyig it from a machie ad have to isert the cois oe at a time; (b) if I am buyig it i a shop ad ca had the moey over all at oce? 4 Solve the recurrece relatio ad iitial coditios a 0 = 1, a 1 = 1, a = 3a 1 2a 2 for 2 5 Solve the recurrece relatio ad iitial coditios a 0 = 2, a = a 2 1 for 1 6 Let Prove by iductio that A = ( ) ( ) A +1 F 1 F = F F +1 for 1, where F is the th Fiboacci umber 7 (a) Use the recurrece relatio i the text to prove that the deragemet umbers d() satisfy the simpler recurrece d(0) = 1, d() = d( 1) + ( 1) for 1 (b) Now put f () = d()/! Show that f (0) = 1, Hece show that f () = =0 (c) Use this formula to show that f () = f ( 1) + ( 1)! for 1 ( 1) /!, ad deduce the formula i the text for d() d x = e x 0! 1 x The series o the left is the expoetial geeratig fuctio of the deragemet umbers 8 Tae a circle ad put 2 poits P 1,Q 1,P 2,Q 2,,P,Q equally spaced aroud the circumferece A matchig is a set of chords to the circle such that

51 44 NON-LINEAR RECURRENCES 45 each chord jois a poit P i to a poit Q j, for some i, j; each of the 2 poits lies o exactly oe of the chords; o two chords cross Let A be the umber of differet matchigs (a) Show that A 0 = 1, A 1 = 2, A 2 = 3, A 3 = 5 (b) Show that, for 1, we hvae A = i=1 A i 1 A i [Hit: Cosider the matchigs i which P 1 is joied to the poit Q i, ad show that there are A i 1 A i of these] (c) Hece show by iductio that A is the ( + 1)st Catala umber C +1

52 46 CHAPTER 4 RECURRENCE RELATIONS

53 Chapter 5 Partitios ad permutatios It ca be argued that combiatorics is about three thigs: subsets, partitios, ad permutatios I the first sectio of the otes we couted the subsets of a set I this sectio we cout partitios ad permutatios 51 Partitios: Bell umbers A partitio of a set X is a set P of subsets of X with the properties: ay set i P is o-empty; ay two sets i P are disjoit; the uio of all the sets i P is X I other words, the sets of the partitio cover X without ay overlap By the Equivalece Relatio Theorem, if R is a equivalece relatio o X (a reflexive, symmetric ad trasitive relatio), the the equivalece classes of R form a partitio of X Coversely, ay partitio is the set of equivalece classes of a (uique) equivalece relatio So the umber of partitios of X is equal to the umber of equivalece relatios o X Let B() be the umber of partitios of a -elemet set, say {1,2,,} The umber B() is the th Bell umber It is easy to see that B(0) = B(1) = 1, B(2) = 2, ad B(3) = 5 The five partitios of {1,2,3} are {123},{12,3},{13,2},{23,1},{1,2,3}, where we have writte 12 istead of {1,2} to avoid a proliferatio of curly bracets 47

54 48 CHAPTER 5 PARTITIONS AND PERMUTATIONS Propositio 51 The Bell umbers satisfy the recurrece ( ) 1 B(0) = 1, B() = B( ) 1 =1 Proof We have see that the iitial coditio holds For the recurrece, we as: how may partitios are there such that the part cotaiig has exactly elemets? We must have 1 We have to choose this part, which ivolves choosig( 1) of the remaiig 1 elemets to go i a part with ; this ca be 1 doe i ways The we must partitio the remaiig poits, which 1 ca be doe i B ways Multiplyig, ad summig over, gives the result This recurrece ca be used to fid a geeratig fuctio for the Bell umbers The type of geeratig fuctio we use is called a expoetial geeratig fuctio, or egf for short This has the form F(x) = 0 B()x! The ame is because of the relatio to the expoetial fuctio We claim that For o the left we have x exp(x) = 0! d F(x) = exp(x)f(x) dx d dx F(x) = B()x 1 1 ( 1)! o cacellig the from the derivative ito! So the coefficiet of x 1 is B()/( 1)! O the right, to obtai the coefficiet of x 1, we tae the coefficiet of x 1 i exp(x) (which is 1/( 1)!, multiply by the coefficiet of x i F(x) (which is B( )), ad sum, obtaiig =1 1 B( ) ( 1)! ( )! = 1 ( 1)! So the two sides are equal =1 ( ) B( ) = B() ( 1)!

55 52 PARTITIONS: STIRLING NUMBERS 49 Now the differetial equatio ca be separated as Itegratig, we obtai 1 d F(x) = exp(x) F(x) dx logf(x) = exp(x) + c, so F(x) = e c exp(exp(x)) But F(0) = 1 (sice B(0) = 1), so c = 1, ad we coclude that F(x) = exp(exp(x) 1) Ufortuately this simple formula for the egf does t help us fid a formula for B() Eve its asymptotic behaviour for large is very complicated 52 Partitios: Stirlig umbers We saw that the subsets of a -elemet set (which are 2 i umber) ( ) ca be split up accordig to the umber of elemets they cotai There are -elemet subsets, ad so we have ( ) = 2 =0 I the same way, the partitios of a set ca be split up For 1, let S(,) be the umber of partitios of a -elemet set havig parts The umbers S(,) are called Stirlig umbers of the secod id (We meet Stirlig umbers of the first id later) Thus, we have S(,) = B() =1 I the last sectio we listed the partitios of a 3-elemet set; from the list we see that S(3,1) = 1, S(3,2) = 3, S(3,3) = 1 There is a recurrece relatio for Stirlig umbers, similar to that for biomial coefficiets: Propositio 52 S(,1) = S(,) = 1 ad for 1 < < S(,) = S( 1, 1) + S( 1,)

56 50 CHAPTER 5 PARTITIONS AND PERMUTATIONS Proof The iitial values are clear: there is a uique partitio with a sigle part, ad a uique partitio with parts (each part has oe elemet) Now cosider the partitios of {1,,} with parts, ad divide them ito two classes: Those i which {} is a part These are obtaied by addig the set {} to a partitio of {1,, 1} with 1 parts So there are S( 1, 1) of them Those i which belogs to a part of size bigger tha 1 If we delete from this part, we get a partitio of {1,, 1} with parts But ow, to go bac, we have to choose a partitio, ad also choose oe of its parts to add the elemet to So there are S( 1,) of these Addig gives the result We ca arrage the Stirlig umbers i a triagle lie Pascal s, except that we will lie it up o the left S(,) is the th umber i the th row, startig both couts at The rule is a little differet from Pascal s To fid the ext elemet i colum, we multiply the umber immediately above it by ad add the umber above ad to the left The Stirlig umbers have a remarable property Recall the fallig factorial: Propositio 53 For 1, (x) = x(x 1)(x 2) (x + 1) x = S(,)(x) =1 First proof We use the fact that (x) +1 = (x) (x ) Now our proof is by iductio, the result beig clear for = 1 Assumig the result for 1, we have x = x 1 x

57 52 PARTITIONS: STIRLING NUMBERS 51 = = 1 =1 1 =1 S( 1,)(x) (x + ) 1 S( 1,)(x) +1 + =1 S( 1,)(x) For 1, the coefficiet of (x) is S( 1, 1) + S( 1,) = S(,) (We have to shift the argumet dow by oe i the first term) For =, the coefficiet of (x) is S( 1, 1); but this is equal to S(,), sice both are 1 The iductio is complete Secod proof Here is a completely differet proof Let m ad be positive itegers We ow that m is the umber of ordered selectios of objects from a set of m objects, with repetitios allowed; if repetitios are ot allowed, the umber is (m) Let us cout the selectios with repetitios allowed i aother way Tae ay such selectio, say (x 1,x 2,,x ) Defie a relatio o the set {1,,} by the rule that i j if x i = x j This is a equivalece relatio, so it correspods to a uique partitio of the set {1,,} If this partitio has classes, say, the there are distict elemets amog x 1,,x, which we ca regard as a selectio of thigs from a set of m with repetitios ot allowed Now give a partitio of {1,,} with parts, ad a selectio (y 1,,y ) of from m with repetitios ot allowed, we ca recover the origial selectio: put x i = y j if i belogs to the jth part of the partitio So the umber of selectios with distict elemets is S(,)m, ad we coclude that the total umber (which we ow to be m ) is the sum of all these values: m = =1 (m) Now cosider the two polyomials x ad =1 (x) We ow that they tae the same value if ay positive iteger m is substituted for x So they are equal as polyomials For their differece is a polyomial of degree at most ; if it is ot idetically zero, it could have at most roots So we have as required x = =1 (x), I Exercise 1 at the ed of this chapter, we will see that some values of S(,) ca be calculated A geeral formula will be give i a later chapter of the otes

58 52 CHAPTER 5 PARTITIONS AND PERMUTATIONS 53 Permutatios: cycle decompositio A permutatio is a oe-to-oe ad oto fuctio from a set to itself; i other words, a arragemet of the elemets of the set I this sectio ad the ext we will be coutig permutatios The total umber of permutatios of a set of size is! But we will subdivide the permutatios, much as we did for partitios, ad cout the parts Recall the cycle decompositio of a permutatio, which we do by example Cosider the permutatio π = ( ) of the set {1,,10} (This two-lie otatio idicates the permutatio which maps 1 to 4, 2 to 3, 3 to 5, ad so o) To compute the cycle decompositio, we start aywhere (say 1), ad follow the iterates of the fuctio applied to our startig poit util we retur there If we have used every poit, we are fiished; otherwise, we close the cycle, ad start aother oe at a uused poit Cotiue util every poit has bee used We write a cycle as a list of poits i bracets, separated by commas For our example above, the cycle decompositio is π = (1,4,7,8,10)(2,3,5)(6)(9) (Note that poits fixed by the permutatio show up as cycles of legth i) The cycle decompositio of a permutatio is ot uique We could start each cycle at ay poit, ad write the cycles i ay order For example, the permutatio above could also be writte π = (3,5,2)(9)(4,7,8,10,1)(6) 54 Permutatios: Stirlig umbers The parity of a permutatio π is the parity (odd or eve) of the umber c(π), where c(π) is the umber of cycles of π (icludig fixed poits) A permutatio is called eve or odd accordig to its parity Sometimes we tal about the sig of π: this is ( 1) c(π), that is, +1 if π is eve ad 1 if π is odd Now the usiged Stirlig umber of the first id, u(,), is defied to be the umber of permutatios of {1,,} which have cycles; the Stirlig umber of the first id is s(,) = ( 1) u(,) (The sig we put i frot is the sig of the permutatios we are coutig)

59 54 PERMUTATIONS: STIRLING NUMBERS 53 We have s(, ) = u(, ) ad (sice the sum couts all permutatios) s(,) =! =1 Propositio 54 s(,1) = ( 1) 1 ( 1)!, s(,) = 1, ad s(,) = s( 1, 1) ( 1)s( 1,) Proof How may permutatios of {1,,} have a sigle cycle? Sice the cycle ca start aywhere, we may as well begi with 1 The we ca visit the other 1 umbers i ay order So the umber of cyclic permutatios is ( 1)! Each has sig ( 1) 1, so s(,1) = ( 1) 1 ( 1)! There is oly a sigle permutatio with cycles, amely the idetity permutatio which fixes every poit; it has sig +1 So s(,) = 1 For the recursio, tae the set of permutatios with cycles, ad divide ito two classes: Those which fix the poit (that is, which have a cycle ()) Deletig this cycle gives a permutatio of {1,, 1} with 1 cycles Clearly the procedure reverses Sice ( 1) ( 1) ( 1) = ( 1), the cotributio to s(, ) from these permutatios is s( 1, 1) Those which move the poit, (that is, i which is i a cycle of legth greater tha 1) Deletig from the cycle cotaiig it gives a permutatio of {1,, 1}, also with cycles Whe we reverse the costructio, for each permutatio of {1,, 1}, there are 1 places i which we could isert Sice ( 1) ( 1) = ( 1), the cotributio of these permutatios to s(, ) is ( 1)s( 1, ) Addig these two terms gives the result We ca write these Stirlig umbers, lie the others, i a triagular array This time, igorig sigs, a give etry is obtaied by multiplyig the etry above by its row umber (rather tha its colum umber, as before) ad addig the etry

60 54 CHAPTER 5 PARTITIONS AND PERMUTATIONS above ad to the left The put i sigs i a chessboard fashio We obtai: From Propositio 54 we ca prove a result which is the iverse of Propositio 53: Propositio 55 For 1, (x) = s(,)x =1 For example, sice (x) 3 = x(x 1)(x 2) = x 3 3x 2 + 2x, we have s(3,3) = 1, s(3,2) = 3, s(3,1) = 2 Proof Agai the proof is by iductio For = 1, both sides of the equatio are equal to x So suppose that the result is true for 1 The we have (x) = (x) 1 (x + 1) = = 1 =1 1 =1 s( 1,)x (x + 1) s( 1,)x +1 1 =1 ( 1)s( 1,)x The coefficiet of x for < is s( 1, 1) ( 1)s( 1,) (movig the idex dow by oe i the first term) The coefficiet of x is s( 1, 1) = 1 = s(,) Oe cosequece of Propositios 53 ad 55 is the followig result Propositio 56 The lower triagular matrices of Stirlig umbers of the first ad secod types are iverses of each other This applies whether we regard them as ifiite matrices or chop them off after a fixed umber of rows Because the matrices are lower triagular, eve multiplyig the ifiite matrices oly ivolves fiite sums

61 54 PERMUTATIONS: STIRLING NUMBERS 55 Proof The polyomials with costat term zero form a vector space V The followig two sequeces are bases for V : x, x 2, x 3, (x) 1, (x) 2, (x) 3, Propositios 53 ad 55 show that the two matrices of Stirlig umbers are the trasitio matrices betwee these two bases Aother cosequece of Propositio 55 is: Propositio 57 For 2, the umbers of eve ad odd permutatios of {1,,} are equal Proof Because 2, we see that (x) has a factor (x 1), ad so is zero whe we put x = 1 Substitutig x = 1 ito Propositio 55, we have s(,) = 0 =1 for 2 Now a eve permutatio cotributes +1 to this sum, ad a odd permutatio cotributes 1; the cotributios must match Remar For those who have doe some abstract algebra, here is a completely differet proof of this result The set of all permutatios of {1,,} forms a group (with the operatio of compositio), called the symmetric group ad writte S The mappig that taes a permutatio to its sig is a homomorphism from S to the multiplicative group {±1} of order 2; the eve permutatios form the erel of this homomorphism, ad therefore comprise a ormal subgroup of S of idex 2, called the alteratig group ad writte A The odd permutatios form a coset of A Now Lagrage s Theorem tells us that the subgroup ad its coset have equally may elemets Exercises 1 (a) Prove each of the followig statemets (i) by directly coutig the partitios, (ii) by usig the recurrece relatio: S(,2) = 2 1 1; ( ) S(, 1) = 2

62 56 CHAPTER 5 PARTITIONS AND PERMUTATIONS (b) Fid a formula for S(, 2) 2 (a) Prove (i) by directly coutig ( ) the permutatios, (ii) by usig the recurrece relatio, that s(, 1) = 2 (b) Fid a formula for s(, 2) 3 Calculate the umber of permutatios of {1,2,3,4,5,6} with three cycles (a) by usig the recursio formula for the appropriate Stirlig umbers; (b) by listig the possible cycle legths of such a permutatio ad coutig the umber of permutatios with each possible cycle structure 4 Let be give, ad let p () be the probability that a radomly-chose ( ) permutatio of {1,,} has exactly fixed poits Show that p () = d( )/!, where d( ) is the ( )th deragemet umber, ad hece show that ( ) (a) d( ) =!; =0 (b) lim p () = e 1! [If you have studied some probability theory, the last statemet says that the umber of fixed poits of a radom permutatio of the set {1,,} approaches a Poisso distributio with parameter 1 as ]

63 Chapter 6 The Priciple of Iclusio ad Exclusio Suppose that, i a class of 100 pupils, we are give the followig iformatio: 50 play football, 48 play music ad 42 play chess; 23 play football ad music, 22 play football ad chess, ad 21 play music ad chess; 10 play all three How may pupils do oe of the three activities? We ca illustrate the eight possible combiatios of activities with a Ve diagram Startig from the iside ad worig out, it is possible to see that the umbers are as show i the diagram So the aswer is 16 Football Music Chess 9 16 I this sectio we are goig to develop a formula for this umber, so that the aswer ca be calculated directly from the give data We will the use this formula to cout deragemets ad to fid a formula for the Stirlig umbers of the secod id 57

64 58 CHAPTER 6 THE PRINCIPLE OF INCLUSION AND EXCLUSION 61 PIE The set-up is as follows We have a uiversal set X, ad a collectio A 1,A 2,,A of subsets of X (I the example, X is the set of 100 pupils, = 3, ad A 1,A 2,A 3 are the sets of pupils who tae part i each of the three activities We are give X, A 1, A 2 ad A 3, ad also the sizes of the itersectios A 1 A 2, A 1 A 3, A 2 A 3 ad A 1 A 2 A 3 For example, A 1 A 2 is the set of pupils who play both football ad music I order to simplify the otatio, we will deote A 1 A 2 by A {1,2} More geerally, for every subset I of the idex set {1,2,,}, we let A I = i I A i Thus, A I is the set of elemets belogig to all the sets A i for which the idex i belogs to I, ad possibly some others By covetio, A {i} = A i, ad A /0 = X Theorem 61 (Priciple of Iclusio ad Exclusio) The umber of elemets of X which lie i oe of the sets A 1,,A is equal to ( 1) I A I I {1,2,,} I our example, the umber of childre taig part i o activity is = 16, agreeig with what we foud directly Proof We loo at the expressio i the theorem It is the sum of cardialities of various subsets of X with plus ad mius sigs We evaluate this by looig at each elemet x X ad seeig how much it cotributes to the sum A elemet x which lies i oe of the sets A i gives a cotributio +1 from A /0 = X, ad o cotributio from ay of the other sets Now cosider a elemet x which lies i some of the A i, ad let J = {i {1,,} : x A j } be the set of idices of sets cotaiig x The x A I if ad oly if I J So the cotributio of X is ( 1) I I J

65 62 SURJECTIONS AND STIRLING NUMBERS 59 ( ) j Let J = j The there are subsets of J of size i, ad each of them gets the i sig ( 1) i So the cotributio is i j=0 ( ) j ( 1) i = (1 1) j = 0, i by the Biomial Theorem So the elemets i oe of the sets A i cotribute +1 to the sum, while the elemets lyig i some of these sets cotribute 0 Hece the sum is just equal to the umber of elemets lyig i oe of the A i, as was to be proved As a corollary we have the followig result: Propositio 62 Suppose that A 1,,A are subsets of a set X Suppose that X = m 0 ad that A i = m 1 for i = 1,, Suppose further that the itersectio of ay j of the sets A i has cardiality m j The the umber of elemets i oe of the sets is j=0( 1) j( ) m j j ( ) For i the sum i Theorem 61, there are sets A I with I = j; each of j them has cardiality m j ad cotributes ( 1) j m j to the sum 62 Surjectios ad Stirlig umbers Let A = ad B = How may fuctios are there from A to B? To specify a fuctio f, we simply have to defie the values f (a) for a A, which ca be arbitrary; so the umber is How may of these fuctios are ijective (oe-to-oe)? To cout these, we proceed as above, maig sure that the values are all distict; that is, we sample without replacemet The aswer is () = ( 1) ( + 1) Note that this umber is zero if > ; there ca be o ijective fuctio from a set to a smaller set How may of the fuctios are surjective (oto)? This is more difficult to cout by elemetary meas; but PIE allows us to fid the aswer Let X be the set of all fuctios from {a 1,,a } to {b 1,,b m } The X =

66 60 CHAPTER 6 THE PRINCIPLE OF INCLUSION AND EXCLUSION For i = 1,,, let A i be the set of all those fuctios which ever tae the value b i (We are tryig to cout fuctios that tae all values; to apply PIE we eed to remove all the fuctios that miss at least oe value) The the fuctios i A i tae 1 possible values, so there are ( 1) of them Now suppose that I {1,,} with I = j The A I is the itersectio of the sets A i for i I, so it cosists of all the fuctios which tae o value b i for i I These fuctios have i possible values, so there are ( i) of them Sice the surjectios are the fuctios lyig i oe of the sets A i, Propositio 62 gives: Theorem 63 The umber of surjectios from a -elemet set to a -elemet set is j=0( 1) j( ) ( j) j Remar This formula is useful but has its drawbacs For example, it should give zero whe >, sice there caot be a surjectio from a set to a larger set But this is quite hard to show directly have a try! Also, it is ot obvious that j=0( 1) j( ) ( j) =!, j though of course this must hold sice if = the the surjectios are bijectios ad there are! of them This theorem allows us to fid a formula for the Stirlig umber S(,) of the secod id Remember that S(,) is the umber of partitios of a -elemet set ito parts Give such a partitio of {1,,}, say, we ca defie a surjectio from {1,,} to {1,,} as follows: if P 1,,P are the parts, map the elemets of part P i to the value i Sice P i /0, some elemet is mapped to i for all i {1,,}, so we do have a surjectio I fact, a give partitio gives! surjectios, sice we ca order the parts i ay way we lie Coversely, ay surjectio f gives a partitio ito parts, where the ith part is the iverse image of i uder f Thus the umber of surjectios is! times the umber of partitios, ad so we have: Propositio 64 S(,) = 1! j=0( 1) j( ) ( j) j

67 63 DERANGEMENTS Deragemets We ca use a similar method to fid a formula for the umber of deragemets of {1,,} (permutatios with o fixed poits) Let X be the set of all permutatios of {1,,} The X =! Now for i = 1,,, let A i be the set of permutatios which fix the poit i There are ( 1)! such permutatios, sice they ca permute the other 1 elemets arbitrarily For ay I {1,,}, A I cosists of permutatios fixig every poit i I If I = j, these permutatios move the other j poits arbitrarily, so A I = ( j)! Now a permutatio is a deragemet if ad oly if it does ot fix ay poit; so, if d is the umber of deragemets, the Propositio 62 gives: Propositio 65 d = j=0( 1) j( ) ( j)! =! j j=0 ( 1) j j! This is the formula we saw i Chapter 5 Exercises 1 A opiio poll reports that the percetage of voters who would be satisfied with each of three cadidates A, B, C for Presidet is 65%, 57%, 58% respectively Further, 28% would accept A or B, 30% A or C, 27% B or C, ad 12% would be cotet with ay of the three What do you coclude? 2 I have 25 sweets to distribute to a class of 10 childre (a) I how may ways ca I distribute the sweets? (b) I how may ways ca I distribute the sweets if I give Alice at least four sweets? (c) I how may ways ca I distribute the sweets if I give both Alice ad Bob at least four sweets? (d) Use the Priciple of Iclusio ad Exclusio to cout the umber of ways I ca distribute the sweet if o child is to have more tha three sweets

68 62 CHAPTER 6 THE PRINCIPLE OF INCLUSION AND EXCLUSION

69 Chapter 7 Families of sets We ow ow how may subsets of the set {1,2,,} there ( ) are altogether (amely 2 ), ad the umber of subsets of fixed size (amely ) We are ow goig to tur to collectios of sets satisfyig various other coditios Of course we are really talig about sets of sets here, but to avoid cofusio we will refer to them as families of sets Typically we will deote a family of sets by script capital F, thus: F The set of all subsets of a set X is called the power set of X, ad deoted by P(X) Thus, a family of sets is a subset of P({1,,}) The mai questios will be: Suppose that we place some restrictio o the relatioships betwee sets i a family What is the largest umber of sets that we ca have? Which families reach this upper boud? We will examie oe case i detail, ad state ad prove Sperer s Theorem The we will loo more briefly at itersectig families ad at families where ay two sets itersect i a sigle elemet 71 Sperer s theorem A family F of subsets of {1,,} is called a Sperer family if the followig is true: For ay two distict sets A,B F, either of them cotais the other; that is, A B ad B A Our first questio is: What is the largest Sperer family? The cheapest way to build ( a) Sperer family is to tae all the subsets of some fixed size This gives us such sets, ad clearly o such set ca cotai 63

70 64 CHAPTER 7 FAMILIES OF SETS aother ( We) saw i Assigmet 1, Questio 2, that for fixed the biomial coefficiet is greatest whe = /2 (if is eve) or whe = ( 1)/2 or = ( + 1)/2 (whe is odd these two biomial coefficiets are equal) Other Sperer families ca be costructed: {{1,2},{1,3,4},{2,3,4}} is a example Perhaps it is possible to fid a larger family cotaiig sets of differet sizes? The first part of Sperer s Theorem tells us that it is ot: Theorem 71 Let F be a Sperer family of subsets of {1,,} The ( ) F /2 Proof The followig igeious proof is ow as the LYM method, sice it was iveted by Lubell, Yamamoto ad Meshali (ad also by Bollobás; we have see may istaces of mis-amed theorems i mathematics!) A chai is a sequece of subsets of {1,2,,}, i which each set is cotaied i the ext set i the sequece The maximal umber of sets i a chai is + 1; i such a maximal chai (A 0,A 1,,A ), the set A has elemets Such a chai starts with the empty set ad adds oe ew elemet each time Thus, ay maximal chai is described by a permutatio π of {1,,}; we have A i = {π(1),,π(i)} It follows that the umber of maximal chais is equal to the umber of permutatios of {1,,}, amely! Next we as: how may maximal chais cotai a give set A? If A =, the the first umbers π(1),,π() i the permutatio must be the elemets of A, ad the last must be the remaiig elemets; so the umber of maximal chais cotaiig A is!( )! Now, let F be a Sperer family The, for A,B F, o maximal chai ca cotai both A ad B For if so, such a maximal chai would be (,A,,B,), say, ad the A B, cotradictig the defiitio of a Sperer family So if we add up the umbers of maximal chais cotaiig all the sets i F, the total caot be more tha the total umber of maximal chais: A!( A )!!, A F from which we deduce (dividig both sides by!) that A F 1 ( ) 1 A

71 71 SPERNER S THEOREM 65 This result (which is valid for ay Sperer family) is ow ( as) the LYM iequality Now we saw, the largest biomial coefficiet is So if we replace /2 the deomiators o the left by larger quatities, we mae the sum smaller, ad we fid F 1 ( ) = ( ) 1, A F /2 /2 ad so we coclude that F ( ) /2 Our secod questio is: What are the families which meet this boud? We have see that, if is eve, the set of all -elemet subsets meets the boud; if is odd, the set of all ( 1)/2-elemet subsets meets the boud, ad so does the set of all ( + 1)/2-elemet subsets The secod part of Sperer s Theorem tells us that there are o others: Theorem ( 72 ) Suppose that F is a Sperer family of subsets of {1,,} with F = The /2 (a) if is eve, the F cosists of all the /2-elemet subsets; (b) if is odd, the either F cosists of all the ( 1)/2-elemet subsets, or it cosists of all the ( + 1)/2-elemet subsets Proof We have to loo bac at the precedig proof; i particular, the step from the ( LYM ) iequality to the ext lie This ( ivolved) replacig the deomiators by the possibly larger deomiators If it ever occurred that the A /2 ew deomiator was strictly larger, the we would ( have ) strict iequality i the ext step, ad we would coclude that F < So all the sets i the /2 family F must have size /2 (if is eve), or size ( 1)/2 or ( + 1)/2 (if is odd) So the theorem is proved i the case where is eve If is odd, however, we have oe more job to do: we must rule out the possibility that there are sets of both possible sizes i F So let = 2m + 1 Looig at the proof agai we see that, if the boud is met, the every maximal chai must cotai a set of F If A ad B are sets of sizes m ad m+1 respectively,

72 66 CHAPTER 7 FAMILIES OF SETS the there is a maximal chai cotaiig both A ad B; so F must cotai exactly oe of these two sets So if A F the B / F ad coversely Now if C ad D are two sets of size m, the it is possible to fid a sequece of sets of sizes alterately m ad m + 1 from C to D If C F, the we fid that the sets of size m + 1 do t belog to F, while the sets of size m all do So every set of size m belogs to F This implies that o set of size m + 1 ca belog to F, so F is the set of all m-elemet sets I a similar way, if F cotais a (m + 1)-elemet set, the it cosists of all (m + 1)-elemet sets Here is a example to illustrate this proof Suppose that = 7, m = 3, ad F cotais the set {1,2,3} We wat to show that it cotais {4,5,6} We produce the sequece ({1,2,3},{1,2,3,4},{2,3,4},{2,3,4,5},{3,4,5},{3,4,5,6},{4,5,6}) Now {1,2,3} F ; so {1,2,3,4} / F ; so {2,3,4} F ; ad so o Fially, {4,5,6} F, as required 72 Itersectig families We say that two sets A ad B itersect if their itersectio is ot empty: A B /0 A family of subsets of {1,,} is a itersectig family if ay two of its members itersect Theorem 73 The maximum size of a itersectig family of subsets of {1,, } is 2 1 Proof First we ote that there do exist itersectig families cotaiig 2 1 sets For cosider all the subsets of {1,,} which cotai the elemet Such a subset has the form {} A, where A is a arbitrary subset of {1,, 1}; there are 2 1 choices for A Clearly the resultig family is itersectig sice ay two members have at least the elemet i commo We have to show that it is ot possible to have more tha 2 1 sets i a itersectig family To see this, we list the sets i complemetary pairs {A,A }, where A = {1,,} \ A There are 2 /2 = 2 1 such pairs Now a itersectig family F ca cotai at most oe set from each pair, sice A ad A are disjoit So F 2 1, as required Followig the patter of Sperer s Theorem, we should ow go o to fid all the itersectig families which meet this boud But this is ot possible; there are may differet itersectig families meetig the boud Here are a few examples

73 72 INTERSECTING FAMILIES 67 As i the theorem, if we tae all the subsets which cotai some fixed elemet of {1,,}, we obtai a itersectig family of size 2 1 If is odd, choose all the subsets which have size strictly greater tha /2 Ay two such subsets must itersect; ad there are 2 1 of them, sice out of each complemetary pair we tae the larger oe If is eve we ca modify this argumet Tae all subsets with more tha /2 elemets, ad out of the sets of size /2 pic oe of each complemetary pair For example, for = 4, we could have either of the followig (where I write {1,2,3} as 123 for brevity): {1234,123,124,134,234,12,13,14} {1234,123,124,134,234,12,13,23} May other examples are possible For = 7, it ca be showed that there are 2 6 = 64 sets which cotai at least oe of 123,145,167,246,257,347,356 Sice these seve sets itersect, ay sets cotaiig them will also itersect Let s as a differet questio What is the maximum size of a itersectig family of -elemet subsets of {1,,}? If < 2, the ay -elemet set cotais ( ) more tha half of {1,,}, so ay two of them itersect The aswer the is, which is ot very iterestig So we assume that 2 I this case, let F (i) cosist of all the -elemet subsets ( of ){1,,} which 1 cotai the elemet i, for i {1,,} The F (i) =, sice we have 1 to pic 1 more elemets from {1,,} \ {i} A famous theorem called the Erdős Ko Rado theorem shows that this is the best we ca do: Theorem 74 (a) If 2, the ( a itersectig ) family of -elemet subsets of 1 {1,,} has size at most 1 ( ) 1 (b) If > 2, the the oly itersectig families which have size are 1 the sets F (i), for i {1,,} The proof of this theorem is beyod the scope of the course, but I have writte out the mai part i the fial sectio of this chapter If you are iterested i combiatorics, you are ecouraged to study this proof, which has a lot of importat ideas i it

74 68 CHAPTER 7 FAMILIES OF SETS I will give here just the proof of part (a) of the theorem i the case whe = 2 This is similar to what we did before Arrage the -elemet subsets of {1,,2} ito complemetary pairs {A,A }, where A = {1,,2}\A The a itersectig family cotais at most oe of each complemetary pair, so the size of such a family is at most 1 2 ( ) 2 = ( ) The de Bruij Erdős theorem Now we will be eve more specific What is the maximum size of a family of subsets of {1,,} havig the property that ay two of them have exactly oe poit i commo? First we give some examples of such families For i {1,,}, let A (i) cosist of the set {i} together with all sets of the form {i, j} for j i The A (i) = ad ay two members of A (i) meet i the poit {i} For i {1,,}, let B(i) cosist of the set {1,,} \ {i} together with all sets of the form {i, j} for j i The B(i) =, ad ay two members of B(i) meet i oe poit For = 7, the seve sets 123, 145, 167, 246, 257, 347, 356 have the required property This cofiguratio of seve sets is ow as the Fao plae The diagram shows the secod ad third examples The de Bruij Erdős theorem shows that is the maximum size, ad describes the families which meet the boud: Theorem 75 Let F be a family of subsets of {1,,} with the property that A B = 1 for all A,B F The F Equality holds if ad oly if oe of the followig occurs: (a) F = A (i) or F = B(i) for some i {1,,};

75 74 FINITE FIELDS AND PROJECTIVE PLANES 69 (b) there is a positive iteger q such that = q 2 +q+1, every set i F cotais q + 1 elemets, ad every elemet of {1,,} is cotaied i q + 1 of the sets of F Note that the third example i our list above satisfies coclusio (c), with q = 2: we have = = 7, each set has 2+1 = 3 elemets ad each poit lies i = 3 sets of F The proof will ot be give here: it ca be foud i the recommeded textboo for the course We will ed this chapter by a brief loo at some families satisfyig coclusio (c) of the theorem 74 Fiite fields ad projective plaes We first do a little bit of liear algebra Recall that a field is a algebraic structure i which additio, subtractio, multiplicatio ad divisio (except by zero) are all possible, ad the commutative, associative ad distributive laws apply If F is a field, the we ca tal about vector spaces over F; the stadard m-dimesioal vector space is the set F m of all m-tuples of elemets of F, with coordiatewise operatios We are iterested i fiite fields Do ay exist? Yes, the itegers mod p form a field wheever p is a prime umber; this field is deoted by Z p There are others too, as we will see later Theorem 76 Let F be a fiite field with q elemets, ad V a m-dimesioal vector space over F The (a) V = q m ; (b) the umber of 1-dimesioal subspaces of V is (q m 1)/(q 1); (c) the umber of 2-dimesioal subspaces of V is (q m 1)(q m 1 1) (q 1)(q 2 1) Proof (a) Ay m-dimesioal vector space ca be coordiatised by choosig a basis; so we ca tae it to be F m The result is clear (b) Ay 1-dimesioal subspace is spaed by a o-zero vector, of which there are q m 1 i V But, if c is ay o-zero elemet of F, the v ad cv spa the same 1-dimesioal subspace So each such subspace has q 1 spaig vectors, ad the umber of such subspaces is (q m 1)/(q 1)

76 70 CHAPTER 7 FAMILIES OF SETS (c) Ay 2-dimesioal subspace is spaed by two liearly idepedet vectors v ad w There are q m 1 choices for v (sice it must be o-zero) ad q m q choices for w (sice it caot be a multiple of v) Thus there are (q m 1)(q m q) pairs (v,w) But by the same argumet (puttig m = 2) we see that ay 2- dimesioal subspace cotais (q 2 1)(q 2 q) spaig pairs of vectors So we get the umber of subspaces by dividig, ad cacellig a factor q We see that a 3-dimesioal vector space has (q 3 1)/(q 1) = q 2 + q dimesioal subspaces, ad (q 3 1)(q 2 1) (q 1)(q 2 1) = q2 + q dimesioal subspaces Furthermore, if V is 3-dimesioal, the ay 2-dimesioal subspace cotais q dimesioal subspaces; ay 1-dimesioal subspace is cotaied i q dimesioal subspaces; ay two 2-dimesioal subspaces itersect i a 1-dimesioal subspace The first part follows from the case m = 2 i the theorem; the secod is proved by a similar argumet For the third, let dim(v ) = 3 ad let W 1,W 2 be subspaces with dim(w 1 ) = dim(w 2 ) = 2 By the dimesio formula, dim(w 1 +W 2 ) + dim(w 1 W 2 ) = dim(w 1 ) + dim(w 2 ) Now the right-had side is equal to 4 We must have dim(w 1 +W 2 ) = 3 (it caot be larger sice dim(v ) = 3, ad it caot be smaller sice W 1 +W 2 properly cotais W 1 ) So dim(w 1 W 2 ) = 1, as required Now let = q 2 + q + 1, ad umber the 1-dimesioal subspaces of V = F 3 as U 1,,U, ad the 2-dimesioal subspaces as W 1,,W Now let A i be the set { j : U j W i }, the set of idices of the 1-dimesioal subspaces i W i The A 1,,A are subsets of {1,,}, each havig size q+1, ad ay elemet lyig i q + 1 of them; ay two of A 1,,A itersect i just oe elemet So we have a cofiguratio satisfyig case (c) of the de Bruij Erdős theorem A family of sets satisfyig case (c) of the de Bruij Erdős theorem is called a projective plae The umber q is called the order of the projective plae So the example i the last sectio (the Fao plae) is a projective plae of order 2; ad the costructio of this sectio shows that there exists a projective plae of

77 74 FINITE FIELDS AND PROJECTIVE PLANES 71 ay order q for which there is a fiite field with q elemets, i particular, if q is a prime umber A theorem of Galois (which we will ot prove here) says that there exists a fiite field with q elemets if ad oly if q is a power of a prime umber; ad moreover, such a fiite field is uique Here, for example, are the additio ad multiplicatio tables for a field with 4 elemets, called 0,1,α,β:idexfield!of order α β α β β α α α β 0 1 β β α α β α β α 0 α β 1 β 0 β 1 α Here is a outlie of the costructio We build this field from the field Z 2 (itegers mod 2) by addig the root of a irreducible polyomial, i the same way that we costruct the complex umbers from the real umbers by addig i, a root of the polyomial x = 0 Some trial ad error shows that the polyomial x 2 + x + 1 is irreducible over Z 2 ; let α be a root of this polyomial, so that α 2 + α + 1 = 0, or α 2 = α + 1 (Remember that = 0 i Z 2, ad so x + x = 0 for ay x) The the elemets of the field are all liear combiatios of 1 ad α, ie 0,1,α,α + 1 We have put β = α + 1 i the tables For example, α + β = α + (α + 1) = (α + α) + 1 = 1, αβ = α(α + 1) = α 2 + α = α α = 1 For which umbers q does there exist a projective plae of order q? This seems to be oe of the hardest questios i combiatorics We have see that they exist wheever q is a prime power No example is ow if q is ot a prime power A famous theorem called the Bruc Ryser Theorem says that, if q is cogruet to 1 or 2 mod 4 ad q is ot the sum of two squares, the there is o projective plae of order q Thus, there is o projective plae of order 6, sice 6 is cogruet to 2 mod 4 but is ot the sum of two squares The umber 10 is also cogruet to 2 mod 4, but is the sum of two squares (10 = ), so the theorem does t tell us whether there is a projective plae or ot But a huge computatio, icludig two years o a Cray supercomputer, showed i 1989 that there is o projective plae of order 10 The ext umber i doubt is = 12 (this is cogruet to 0 mod 4, so the Bruc Ryser theorem does ot apply to it) We have o idea whether a projective plae of order 12 exists or ot!

78 72 CHAPTER 7 FAMILIES OF SETS 75 Appedix: Proof of the Erdős Ko Rado Theorem We will prove this theorem usig a result about graphs First we develop some termiology A graph cosists of a set V of vertices, ad a set E of edges; each edge is a set of two vertices, ad is regarded as coectig those two vertices Here is a picture of a graph A clique i a graph is a set of vertices, ay two of which are joied by a edge A coclique is a set of vertices, o two of which are joied by a edge I the example draw above, the largest clique has three vertices, ad the largest coclique has two vertices A automorphism of a graph is a permutatio of the vertices which maps every edge to a edge The set of all automorphisms is a group (a subgroup of the symmtric group), called the automorphism group of the graph I our example, the left-to-right reflectio is a automorphism; the automorphism group cotais four elemets (icludig the idetity) A graph is said to be vertex-trasitive if, for ay two vertices, there is a automorphism of the graph which carries the first to the secod More geerally, for ay group G actig o ay set X, we say that G acts trasitively o X if we ca carry ay elemet of X to ay other by some elemet of the group Our example graph is ot vertex-trasitive sice o automorphism ca tae the left-had vertex (which lies o two edges) to the top vertex (which lies o three edges) We use the fact that, if a group G acts trasitively o a set X, with X =, the for ay x,y X, the umber of elemets of G mappig x to y is equal to G / For it is clear that the average umber (if x is fixed ad y varies over all poits i X) is G /; ad the elemets of the group which map x to y form a coset of a subgroup H, where H is the stabiliser of x, the set of all elemets of G fixig x By Lagrage s Theorem, all cosets cotai the same umber of elemets We first prove a theorem which is too wea for our purposes, but illustrates the proof techique The we stregthe it to get the result we wat Theorem 77 Let Γ be a vertex-trasitive graph o vertices Let C be a clique ad D a coclique i G The C D

79 75 APPENDIX: PROOF OF THE ERDŐS KO RADO THEOREM 73 Remar Our earlier example shows that the assumptio of vertex-trasitivity is ecessary for this theorem The graph there has four vertices ad has a 3-vertex clique ad a 2-vertex coclique Proof Let G be the automoorphism group of the graph Γ Cout triples (x,y,g), where x C, y D, g G, ad g maps x to y There are C choices of x, D choices of y, ad (by the remar before the theorem) G / choices for G, so C D G / such triples Now cout aother way For each elemet g G, there is at most oe elemet of C which ca be mapped ito D by g For ay two elemets of C are joied; ay two elemets of D are ot joied; so if g mapped two elemets of C ito D it would chage a edge ito a o-edge, which is impossible for a automorphism So for every elemet of G there is at most oe triple of the type we are coutig So C D G / G, from which the result follows Theorem 78 Let Γ be a vertex-trasitive graph o vertices Let Y be a subset of the vertex set such that ay clique cotaied i Y has size at most m Y The ay clique i G has size at most m To see that the precedig theorem follows from this oe, tae X to be a coclique Ay clique cotaied i D has at most oe vertex, ie size at most m D, where m = 1/ D ; so ay clique C i G satisfies C m = / D, from which the required result follows Proof Agai let G be the automorphism group of the graph Γ Let C be a clique As i the precedig proof, we cout triples (x,y,g), where x C, y Y, g G, ad g maps x to y As before there are C Y G / such triples Now choose first the elemet g Suppose that it maps poits of C ito Y Their images form a clique, so m Y Thus there are at most m Y G such triples So C Y G / m Y G, whece C m Theorem 79 (Erdős Ko Rado) Suppose that ad are give, with( 2 ) 1 The the size of a itersectig family of -subsets of {1,,} is at most 1 Proof We mae a graph Γ as follows: the vertices are the -elemet subsets of {1,,}; two vertices are joied if the correspodig subsets have o-empty itersectio The a itersectig family is just a clique i this graph Moreover, it is clear that the symmetric group o {1,,} acts vertex-trasitively o this

80 74 CHAPTER 7 FAMILIES OF SETS graph (This says, give ay two -elemet subsets A ad B of {1,,}, there is a permutatio of {1,,} which carries A to B Note that ay permutatio maps a itersectig pair of -elemet sets to a itersectig pair, so is a automorphism of the graph Γ) Now we fid a collectio of subsets such that a itersectig family cotais at most of them To do this, we tae poits p 0, p 1,, p 1 equally spaced aroud a circle Number the itervals betwee the poits as I 0,,I 1, where I j = (p j, p j+1 ) (We read the subscripts modulo where ecessary) The set Y cosists of all itervals of legth, that is, sets of the form { j, j + 1,, j + 1} There are such sets; we have to show that a itersectig family cotais at most of them (Note that we have replaced {1,,} by {0,, 1}; this does ot affect the argumet) So suppose that Z is a subset of Y, ay two of whose sets itersect Each poit p j is the edpoit of two itervals i Y, amely { j, j + 1,, j + 1} ad { j,, j 2, j 1} These two sets are disjoit, because of our assumptio 2; so Z cotais at most oe of them Now tae a set i Z, say A = { j, j + 1,, j + 1} Ay other set i Z itersects this oe, so must have a ed poit i the set {p j+1,, p j+ 1 } (the set of p s which are iterior poits of the iterval correspodig to A) Sice each of these poits ca be the ed poit of at most oe iterval correspodig to sets i Z, there are at most 1 more such sets, that is, at most altogether Now it follows from Theorem 78 that the size of ay itersectig family of -sets (that is, ay clique i the graph G) is at most ( ) = ( ) 1 1 Remar It is possible to show that, if > 2, the ay itersectig family of - sets attaiig the boud of the theorem must cosist of all -sets cotaiig some give poit of {1,,} Exercises ( 2 1 ) 1 Let = 2 Show that there are 2 1 itersectig ( families ) of -elemet 2 1 subsets of {1,,} havig the maximum umber of members Show 1 that oly 2 of them have the form F (i) for i {1,,} Hece show that the secod part of the Erdős Ko Rado Theorem goes badly wrog whe = 2

81 75 APPENDIX: PROOF OF THE ERDŐS KO RADO THEOREM 75 2 Idetify the right-had picture before Theorem 75 with the example costructed usig the fiite field F = Z 2 [Hit: the subspace spaed by the vector v = (a,b,c) correspods to the poit with label 4a + 2b + c i the figure] 3 Costruct a fiite field with 8 elemets 4 Let X = {1,,} Show that, for every o-empty subset A of X, there is a itersectig family F of subsets of X of size 2 1 with A F Show further that ay two subsets A,B with A B /0 are cotaied i a family with these properties What about three pairwise itersectig sets? 5 Show that the largest Sperer family of subsets of {1,2,3,4,5} cotaiig the sets {1,2} ad {3,4,5} cotais eight sets How does this compare with Sperer s Theorem? 6 Let S be the family {{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} of subsets of {1,2,3,4,5,6,7} Let F be the family of all subsets of {1,,7} which cotai a member of S Show that (a) F is a itersectig family; (b) F cotais 7 sets of size 3, 28 of size 4, 21 of size 5, 7 of size 6, ad oe of size 7: i all, 64 sets 7 Let X = {1,2,,} ad S = {A 1,A 2,,A b } be a family of distict subsets of X such that A j A = 1 for all j For each i X, let r i be the umber of subsets of S which cotai i Prove that i=1 r i (r i 1) = b(b 1) [Hit: Cout the umber of ordered triples (i,a j,a ), where A j,a are distict sets i S ad A j A = {i}, i two differet ways]

82 76 CHAPTER 7 FAMILIES OF SETS

83 Chapter 8 Systems of distict represetatives The Studets Uio has affiliated clubs Each club elects a delegate to the Executive Committee The delegate must be a member of the club (s)he represets, ad o perso ca represet more tha oe club The two atural questios for a combiatorialist are: is it possible to choose the represetatives accordig to these rules? ad I how may ways ca this be doe? We will aswer the first questio here; the secod is more difficult Of course the aswer depeds o the membership of the clubs If, for example, Sid ad Doris are the oly members of the Football Club, the Music Club, ad the Chess Club, the the electio is clearly ot possible More geerally, we see that if ay m clubs cotai altogether less tha m members, the electio is ot possible So a ecessary coditio for the electio is that ay m clubs have betwee them at least m members Surprisigly, this obvious coditio also turs out to be sufficiet; this is the cotet of Hall s Theorem 81 Hall s Theorem Let us express these ideas more mathematically Let A 1,A 2,,A m be sets (Suppose that they are all subsets of a uiversal set X) We allow here the possibility that some of the sets are equal A system of distict represetatives for the sets (A 1,A 2,,A ) is a -tuple (x 1,x 2,,x ) of elemets of X with the properties (a) x i A i for i = 1,, (this says that the elemets are represetatives of the sets); (b) x i x j for i j (this says that the represetatives are distict ) We abbreviate system of distict represetatives to SDR 77

84 78 CHAPTER 8 SYSTEMS OF DISTINCT REPRESENTATIVES For J {1,2,,}, defie A(J) = i J A i (Do ot cofuse this with A J, which we met i the chapter o PIE; A J is the itersectio of the sets with idex i J, A(J) is their uio) We say that the family of sets satifsies Hall s Coditio if the followig holds: for ay subset J of {1,2,,}, A(J) J Theorem 81 Let (A 1,,A ) be a family of subsets of X The there exists a system of distict represetatives for A 1,,A ) if ad oly if Hall s coditio holds Proof First suppose that there is a SDR, say (x 1,,x ) for the sets Tae ay subset J of {1,,} The A(J) cotais all the sets A i for i J, ad hece cotais all the represetatives x i for i J; sice the represetatives are distict, we have A(J) J So Hall s coditio holds (This is the argumet we saw earlier If the choice of represetatives is possible, the ay m sets must cotai at least eough members to act as their represetatives) Now we prove the coverse, which is more difficult Let (A 1,,A ) be a family of sets satisfyig Hall s coditio We have to show that a SDR ca be foud Our proof will be by iductio o ; we assume that a family of fewer tha sets which satisfies Hall s coditio has a SDR The iductio begis with = 1 sice Hall s coditio guaratees that A 1 is ot empty [WHY?] We say that a set J {1,,} of idices is critical if A(J) = J (The all of the members of the sets A i for i J must be used as their represetatives) We divide the proof ito two cases: Case 1: No set is critical except for /0 ad possibly {1,,} This meas that A(J) > J for every o-empty proper set of idices Choose ay elemet x of A to be its represetative The x caot be the represetative of ay other set; so we remove it Let A i = A i \ {x } for i = 1,, 1 Now for ay o-empty subset J of {1,, 1}, we have A (J) A(J) 1 > J 1, where the strict iequality holds by the case assumptio This meas that A (J) J, so that the family (A 1,,A 1 ) satisfies Hall s coditio By the iductive hypothesis, this family has a SDR, say (x 1,,x 1 )

85 81 HALL S THEOREM 79 But the (x 1,,x 1,x ) is a SDR for the origial family; for x A, ad x x i for i <, sice x i belogs to A i but x does t Case 2: There is a critical set, say J By the iductio hypothesis, we ca choose a SDR (x j : j J) for the sets idexed by J For / J, let A = A \ A J (We must remove the elemets of A J : they have all bee used as represetatives) We chec Hall s coditio for this family For K {1,,} \ J, we have A (K) = A(J K) A(J) = K, J K J where i the secod lie, A(J K) J K by Hall s coditio, ad A(J) = J by the case assumptio So the sets A for / J satisfy Hall s Coditio By iductio we ca fid a SDR for them, say (x : / J) Puttig this together with the previously chose SDR for the sets A j for j J gives a SDR for all of the sets This cocludes the iductive step ad so the proof There is a lot of checig to do to verify Hall s coditio, though it is possible to do this i a systematic way quite efficietly But there is oe ice situatio i which we ca guaratee that it holds Propositio 82 Let (A 1,,A ) be a famly of subsets of the set {1,,} Suppose that there is a positive umber such that (a) A i = for = 1,,; (b) each elemet of {1,,} is cotaied i exactly of these sets The the family has a SDR Proof We show that Hall s coditio holds Tae J {1,,} ad cout pairs (i,x) with i J ad x A i Clearly there are J such pairs O the other had, x ca be ay elemet of A(J), ad for each x there are at most sets A i cotaiig x with i J (sice there are just suchj sets altogether) So the umber of pairs is at most A(J) Thus J A(J), ad so A(J) J For example, the Fao plae discussed i the last chapter of the otes cosists of seve 3-elemet subsets of {1,,7}, so that each elemet of {1,,7} lies i exactly three of them So it has a SDR

86 80 CHAPTER 8 SYSTEMS OF DISTINCT REPRESENTATIVES 82 How may SDRs? If the sets A 1,,A do ot satisfy Hall s coditio, they have o SDR But if they do, ad they are ot too small, the they have may differet SDRs This is the cotet of the ext result Propositio 83 Let A 1,,A be subsets of a set X which satisfy Hall s coditio Suppose that A i for i = 1,,, where is a positive iteger The the umber of SDRs of the family is at least {! if, () if >, where () is the fallig factorial ( 1) ( + 1) Proof The proof follows closely the proof of Hall s Theorem We prove the result by iductio o If = 1, there is oly oe set, havig elemets; there are at least () 1 = SDRs (Note that i this case, so we are i the secod case i the statemet of the result) So let A 1,,A be sets satisfyig the hypothesis We divide ito two cases as i the proof of Hall s Theorem Case 1: There are o o-empty critical sets except possibly {1,,} Choose ay elemet x of A as its represetative (there are at least choices for x) The the family (A 1,,A 1 ) cosists of 1 sets, each of cardiality at least 1 So the umber of SDRs of this family is at least { ( 1)! if 1 1, ( 1) 1 if 1 > 1, by the iductio hypothesis Multiplyig by gives the correct lower boud for the umber of SDRs of the origial family (Note that ( 1)! =! ad ( 1) 1 = () ) Case 2: There is a o-empty proper critical set, say J We have A J = J, so by the iductio hypothesis the family (A j : j J) has at least! SDRs As i the proof of Hall s Theorem, ay such SDR ca be exteded to a SDR for the whole family So there are at least! SDRs, ad the iductio is complete As a cosequece, we ca improve Propositio 82: Propositio 84 Suppose that the hypotheses of the precedig Propositio are satisfied The there are at least! distict SDRs of the family of sets

87 83 SUDOKU 81 This follows immediately from Propositios 83 ad 82 For example, the family ({1,2},{1,3},{2,3}) of sets has two SDRs, amely (1,3,2) ad (2,1,3); the seve lies of the Fao plae have at least 3! = 6 SDRs 83 Sudou Hall s Marriage Theorem, ad i particular the idea of a critical set which we met i the proof, is relevat to solvig Sudou puzzles This is somethig which every Sudou player ows to some degree Loo at the empty cells i ay row, colum or subsquare of a Sudou puzzle Let A i be the set of etries which could appear i the ith empty cell (ie those which do ot already appear i the same row, colum or subsquare) The the etries which we put there must form a SDR for the sets A i Moreover, if we ca fid a critical set, the as i the proof of Hall s Theorem, we ca remove its elemets from the other sets, which simplifies the search for a SDR Here is a example The Times, 14 September 2005 Ratig: Fiedish Loo at the 3 3 square i the bottom left of the puzzle It has five empty cells, whose row ad colum umbers are (1,1), (1,2), (1,3), (2,1) ad (3,3) Cell (1,1) has 8 ad 7 i the same row, 1 ad 2 i the same colum, ad 1,2,3,5 i the same subsquare So the umber we put there must be oe of 4,6,9 Similarly we fid the possibilities for (1,2) are 4,6,9, for (1,3) also 4,6,9, for (2,1) are 4,6,7,8,9, ad for (3,3) are 4,6,7,9

88 82 CHAPTER 8 SYSTEMS OF DISTINCT REPRESENTATIVES Cells (1,1), (1,2) ad (1,3) form a critical set, sice betwee them they oly cotai three elemets 4,6,9 So we ca delete 4,6,9 from the other sets We fid that 7 must go i (3,3) ad the 7 or 8 i (2,1) So it must be 8 i cell (2,1) I fact, this etire puzzle ca be solved by this method of fidig critical sets ad removig their elemets from other sets Exercises 1 (a) Write dow a SDR for the Fao plae (b) How may differet SDRs ca you fid? 2 Let A 1,,A be subsets of {1,,} Let M be the matrix whose (i, j) etry is 1 if j A i, ad 0 otherwise Prove that the umber of SDRs of (A 1,,A ) is at least det(m) [Hit: Use the formula for the determiat as a sum over permutatios Each SDR cotributes a term ±1 to the sum] Deduce that the Fao plae has at least 24 SDRs 3 Costruct five families, F 1, F 2, F 3, F 4, F 6, each cosistig of three subsets of the set {1,2,3}, such that F i has exactly i differet SDRs, for each i {1,2,3,4,6} Does there exist a family of three subsets of {1,,6} with five SDRs? 4 This exercise gives the deficit form of Hall s Theorem It is a geeralisatio of Hall s Theorem, but ca be deduced from Hall s Theorem Theorem Let A 1,,A be subsets of a set X Suppose that, for some positive iteger m, we have A(J) J m for all J {1,,}, where A(J) = j J A j The it is possible to fid m of the sets A 1,,A which have a SDR Prove this [Hit: Tae m dummy elemets z 1,,z m, ad add them to all the sets A i ]

89 Chapter 9 Lati squares A Lati square of order is a array, each cell cotaiig oe etry from the set {1,,}, with the property that each elemet of {1,,} occurs oce i each row ad oce i each colum of the array Here is a example Notice that each row ad each colum is a permutatio of {1,,} Sometimes we use a differet set of symbols as the etries of a Lati square The defiitio is just the same For example, the Cayley table of a group of order is a Lati square whose symbols are the group elemets We will sometimes loo at more geeral structures Two of these are: A Lati rectagle is a array (where ) with etries from {1,,} such that each symbol occurs oce i each row ad at most oce i each colum A partial Lati square is a array (where ) with each cell either empty or cotaiig a symbol from {1,,} such that each symbol occurs at most oce i each row ad at most oce i each colum Thus, the first rows of a Lati square form a Lati rectagle; ad if we tae a Lati square ad bla out some of the etries we obtai a partial Lati square There is o shortage of partial Lati squares; the ewspapers publish examples every day! Lati squares occur i algebra as Cayley tables of groups If G = {g 1,,g } is a fiite group of order, the its Cayley table is the matrix whose (i, j) 83

90 84 CHAPTER 9 LATIN SQUARES etry is g i g j (the product of g i ad g j i the group) This matrix is a Lati square this follows from the group axioms, ad is ot discussed here 91 Row by row It is ot difficult to show that there exists a Lati square of order for each We are goig to show somethig stroger, amely, a Lati square ca be costructed by addig a row at a time; it is ot possible to get stuc The proof uses Hall s Theorem Propositio 91 Let L be a Lati rectagle, where < The L ca be exteded to a ( + 1) Lati rectagle Proof For i = 1,,, let A i be the set of symbols which do ot occur i the ith colum of L The A i =, sice there are distict symbols i ay colum of L Pic a symbol j How may sets A i cotai j? We ow that j occurs times i L, oce i each row; these occurreces are i differet colums, so there are colums i which j does ot occur, that is, sets A i cotaiig j We have ow verified the hypotheses of Propositio 82 From that propositio we coclude that the family (A 1,,A ) has a system of distict represetatives, say (x 1,,x ) We claim that we ca add the row (x 1,,x ) to L to obtai a larger Lati rectagle This is true because the x i are all distict, so o symbol is repeated i the ew row; ad x i A i, so x i does t occur i colum i of L, so o repeated symbol is itriduced i ay colum So the result is proved We ca do more; we ca give a lower boud for the umber of Lati squares of order Theorem 92 The umber of Lati squares of order is at least! ( 1)! 2! 1! Proof I the precedig proof, we replace Propositio 82 by the stroger Propositio 84, to coclude that the umber of SDRs of the sets (A 1,,A ) is at least ( )! So the umber of ways of extedig a Lati rectagle to a (+1) Lati rectagle is at least ( )! (Each SDR gives a extesio) Now there are! 1 Lati rectagles (these are just permutatios); each ca be exteded to a 2 Lati rectagle i at least ( 1)! ways; each of these ca be exteded to a 3 Lati rectagle i at least ( 2)! ways; ad so o The result follows

91 92 YOUDEN SQUARES 85 Ca we put a upper boud o the sumber of Lati squares? There is a trivial upper boud, amely 2 ; this is just the umber of ways we ca put symbols ito the 2 cells without worryig about the rules This ca be improved slightly Each row is a permutatio, so the umber of Lati squares is at most (!) Ad, havig chose the first row, each subsequet row is a deragemet of it, so the umber of Lati squares is at most! (d()) 1, where d() is the deragemet umber (remember that d() is approximately!/e) The exact aswer has bee calculated for 11 by exhaustive search The literature o this cotais a lot of mistaes; the most reliable paper is by McKay, Meyert ad Waless i the Joural of Combiatorial Desigs i 2007, which gives these values: Number of Lati squares Beyod this obody ows the exact value Eve the best ow upper ad lower bouds are quite a log way apart! 92 Youde squares Youde squares form a class of desigs used i statistics As we will preset them (ad as they were first described by Youde) they are Lati rectagles; the ame comes from a differet represetatio used by Fisher, i which they are partial Lati squares (but we wo t go ito that) Essetially, a Youde square is a way of represetig a family of sets satisfyig the coditios of Propositio 82 as a Lati rectagle Strictly speaig, statisticias oly use the term whe a extra coditio is satisfied by the family of sets, but that does ot affect the result below Propositio 93 Let A 1,,A be subsets of {1,,} Suppose that, for some > 0, every set A i has elemets, ad every elemet x {1,,} lies i of

92 86 CHAPTER 9 LATIN SQUARES the sets A i The there is a Lati rectagle M such that the etries i the ith colum of M form the set A i Example The Fao plae 123,145,167,246,257,347,365 ca be represeted as Proof By Propositio 82, the family of sets has a SDR (x 1,,x ), which we ca tae to be the first row of the rectagle Now let A i = A i \ {x i } for i = 1,, Clearly A i = 1 Aos, give ay x {1,,}, we have used x as the represetative for oe of the sets A i, so it lies i just 1 of the sets A j So the ew family satisfies the coditios of the propositio with 1 replacig Cotiue the process, with each SDR formig a ew row, util = 0 93 Orthogoal Lati squares Two Lati squares A = (a i j ) ad B = (b i j ) are said to be orthogoal if they have the followig property: give a pair (,l) of symbols from the set {1,,}, there is exactly oe cell (i, j) such that a i j = ad b i j = l Here is a example: Sometimes a pair of orthogoal Lati squares is called a Graeco-Lati square If we replace the symbols i the first square by Lati letters ad those i the secod by Gree letters, the the orthogoality coditio says that every pair cosistig of a Lati ad a Gree letter occurs exactly oce i the array For our above example, we would get the followig: Aα Bβ Cγ Dδ Bγ Aδ Dα Cβ Cδ Dγ Aβ Bα Dβ Cα Bδ Aγ Euler posed the followig questio i 1782

93 93 ORTHOGONAL LATIN SQUARES 87 Of 36 officers, oe holds each combiatio of six ras ad six regimets Ca they be arraged i a 6 6 square o a parade groud, so that each ra ad each regimet is represeted oce i each row ad oce i each colum? This questio ass whether there exists a Graeco-Lati square of order 6 If so, the taig the ras as Lati letters ad regimets as Gree letters we could produce the required parade Euler iveted Graeco-Lati squares i his researches o magic squares A magic square is a square cotaiig the umbers 1,, 2 each oce, so that the sum of the umbers i ay row, colum or diagoal is the same (ecessarily ( 2 + 1)/2 Euler oticed that it is ofte possible to costruct a magic square from a Graeco-Lati square as follows: Replace each of the two sets of symbols by the umbers 0,1,, 1 Regard a pair of umbers i j as beig the base represetatio of a sigle umber i + j Now the etries ru from 0 to 1 Add oe to each so that the rage is 1 to The resultig square has all row ad colum sums costat [WHY?] With some extra care it is possible to mae the diagoal sums costat as well Here is a example Cβ Aα Bγ Aγ Bβ Cα Bα Cγ Aβ Euler ew how to costruct two orthogoal Lati squares of ay order ot cogruet to 2 mod 4 We ow outlie a algebraic costructio which is similar to the oe Euler used First we give the costructio usig modular arithmetic Remember that Z deotes the itegers modulo We ca tae the elemets to be 0,1,, 1 To add or multiply two elemets, we add or multiply i the usual way as itegers, ad the divide by ad tae the remaider So, i Z 7, we have = 2, 4 5 = 6 A elemet a Z is a uit if there exists b Z such that a b = 1 The followig fact is proved i elemetary algebra: The elemet a is a uit i Z if ad oly if gcd(a,) = 1

94 88 CHAPTER 9 LATIN SQUARES Now, for ay a Z, let L(a) be the matrix whose (x,y) etry is a x+y Here are the matrices L(1),L(2),L(3) over Z 4 : We see that L(1) ad L(3) are Lati squares but L(2) is ot Moreover L(1) ad L(3) are ot orthogoal: the pair (0,0) occurs twice, the pair (0,1) ot at all Propositio 94 (a) L(a) is a Lati square if a is a uit i Z (b) L(a 1 ) ad L(a 2 ) are orthogoal if a 1 a 2 is a uit Proof (a) I L(a), if we loo for symbol c i row x, we fid it i colum y where ax + y = c This has a uique solutio y = c ax Similarly, if we loo for c i colum y, it will be i row x is ax+y = c This implies ax = c y, so x = b(c y), where b is the iverse of a (b) Suppose we are looig for a cell i which the first square has the etry c ad the secod square has the etry d The we have to solve the simultaeous equatios a 1 x + y = c, a 2 x + y = d From these equatios we deduce that (a 1 a 2 )x = (c d) So x = b(c d), where b is the iverse of a 1 a 2 The either of the equatios ca be used to fid y This theorem is o use for costructig orthogoal Lati squares of eve order For suppose that is eve If a is a uit i Z, the gcd(a,) = 1, so a must be odd But if a 1 ad a 2 are both odd, the a 1 a 2 is eve, ad gcd(a 1 a 2, = 1 However, for odd, gcd(1,) = gcd(2,) = 1, so we get a pair of orthogoal Lati squares for every odd Euler ew that it is possible to costruct orthogoal Lati squares of order also if is a multiple of 4 Here s why This argumet shows oe of the beefits of abstract algebra If we loo at the costructio we just gave, we see that there is othig special about the itegers mod The costructio wors for ay commutative rig with idetity, that is, ay structure i which we ca add, subtract, ad multiply, so that the associative, commutative, distributive ad idetity laws hold Propositio 94

95 93 ORTHOGONAL LATIN SQUARES 89 holds i this geerality Thus, if R is a commutative rig with idetity, the we ca defie the matrix L(a) for ay a R; ad L(a) is a Lati square if a is a uit i R, while L(a 1 ) ad L(a 2 ) are orthogoal if a 1 a 2 is a uit Now we use the followig facts: (a) The direct product of two commutative rigs with idetity is a commutative rig with idetity [The direct product of R 1 ad R 2 is the set of all ordered pairs (r 1,r 2 ), with r 1 R 1 ad r 2 R 2, with coordiatewise additio ad multiplicatio] If a 1 is a uit i R 1 with iverse b 1, ad a 2 is a uit i R 2 with iverse b 2, the (a 1,a 2 )(b 1,b 2 ) = (1,1), so (a 1,a 2 ) is a uit i R 1 R 2 (b) There is a fiite field (a commutative rig with idetity i which every ozero elemet is a uit) of every prime power order Usig (b), we ca costruct orthogoal Lati squares of order 4, 8, ad ay larger power of 2 The usig (a), we ca costruct Lati squares of order 4m, 8m,, for ay odd umber m Here is a example for (b) We give first the additio ad multiplicatio tables for a field with four elemets 0,1,α,β (We saw this already i chapter 6 of the otes; there is a coectio which we will see i the ext sectio of this chapter) The we give the three Lati squares L(1), L(α) ad L(β) α β α β β α α α β 0 1 β β α α β α β α 0 α β 1 β 0 β 1 α 0 1 α β 1 0 β α α β 0 1 β α β α α β 0 1 β α β α 0 1 α β β α β α α β 0 1 Euler ew this, ad he ased about the 36 officers because his costructios could ot deal with the cases 2, 6, 10, or ay umber cogruet to 2 mod 4 He cojectured that orthogoal Lati squares of these orders could ot exist He was right about 2 (this is easy to show directly) ad 6 (this was proved by a log caseby-case argumet by Tarry i 1900), but wrog about the rest Bose, Shrihade ad Parer (the Euler spoilers ) showed i 1960 that orthogoal Lati squares of order exist for every except = 2 ad = 6

96 90 CHAPTER 9 LATIN SQUARES 94 Sets of mutually orthogoal Lati squares A set of mutually orthogoal Lati squares or set of MOLS of order is a set {L 1,L 2,,L r } such that (a) each L i is a Lati square of order ; (b) L i ad L j are orthogoal if i j We let N() deote the maximum size of a set of MOLS of order, for 2 [WHY NOT = 1?] This is a very importat fuctio Our coclusio from the last sectio ca be expressed as follows: N(2) = 1, N(6) = 1, N() 2 for 2,6 Propositio 95 N() 1 Proof Tae a set {L 1,,L r } of MOLS of order We ca chage the symbols i each Lati square to be aythig we lie So let us agree that each square has etry 1 i the top right-had cell (i row 1 ad colum 1) Now each square cotais 1 further etries 1 Noe of them ca be i the first row or first colum, sice this would violate the Lati square coditio Also, o two of the etries 1 i differet squares ca be i the same cell For ay two cells L i ad L j have etries (1,1) i the top right-had cell, so this combiatio caot occur aywhere else So the r( 1) further etries 1 i the r squares fit i to the ( 1) 2 positios outside the first row ad colum without overlap So we must have r( 1) ( 1) 2, whece r 1 (sice > 1) Here is a importat theorem of Bose about whe equality ca hold Recall the defiitio of a projective plae of order from Chapter 7: a set of m subsets of {1,,m}, where m = , such that each subset cotais + 1 elemets, each elemet of {1,,m} lies i + 1 subsets, ad ay two subsets itersect i exactly oe poit Theorem 96 We have N() = 1 if ad oly if there is a projective plae of order The proof of this theorem is give i the ext sectio If is a prime power, we have see that there exists a field F with elemets Now every o-zero elemet of F is a uit (by defiitio of a field) So all the matrices L(a) for a F, a 0, are Lati squares; ad ay two of them are orthogoal, sice if a 1 a 2 the a 1 a 2 is a uit Thus:

97 95 APPENDIX: PROOF OF BOSE S THEOREM 91 Propositio 97 If there exists a field of order (that is, if is a prime power), the N() = 1 It is thought that the coverse of this propositio is also true But this is ot proved, ad seems to be oe of the hardest ope problems i combiatorics We ow that N(6) = 2, ad that 3 N(10) 6 But apart from = 6, there is ot a sigle o-prime-power value of for which N() is ow! 95 Appedix: Proof of Bose s Theorem This proof is closely coected with classical topics i projective geometry, goig bac to the ivetio of perspective by Reaissace artists ad mathematicias Recall the statemet of the theorem: Theorem 98 For a iteger 2, the followig are equivalet: (a) there exists a projective plae of order ; (b) there exists a set of 1 mutually orthogoal Lati squares of order Proof The two costructios are just the reverse of each other (a) implies (b): We are give a projective plae of order, cosistig of poits ad the same umber of ( + 1)-elemet subsets called lies, so that ay two lies itersect i a uique poit Pic a lie L This will play a special role i our costructio, correspodig to the lie at ifiity where parallel lies meet Also, select two special poits X ad Y i L, ad umber the remaiig poits as Z 1,,Z 1 The poit X lies o + 1 lies, oe of which is L Number the others as x 1,,x Similarly umber the remaiig lies through Y as y 1,,y, ad the remaiig lies through Z i as z i1,,z i for i = 1,, 1 Ay poit P ot o L lies o a uique lie x j through X ad a uique lie y through Y The lies x j ad y itersect just i {P} We idetify P with the cell ( j,) i row j ad colum of a grid Now we ca defie arrays M(1),,M( 1) as follows Let P be the poit correspodig to cell ( j,) as above There is a uique ie joiig P to Z i If this lie is z is, the we put symbol s i this cell ii the array M(i) Claim: M(i) is a Lati square For if the symbol s occurred twice i the same row, say i positios ( j,) ad ( j,l), the the lies z is ad x j would have the correspodig two poits i commo, which is ot the case

98 92 CHAPTER 9 LATIN SQUARES Claim: M(i) ad M( j) are orthogoal for i j For suppose we are looig for a cell cotaiig etry s i M(i) ad etry t i M( j) The correspodig poit lies o the lies z is ad z jt, ad so is uiquely defied as their itersectio So there is just oe such cell So, from a projective plae of order, we have costructed a set of 1 MOLS of order (b) implies (a): Reverse the above costructio Here is a setch, with the details left out Suppose that M(1),M(2),,M( 1) be MOLS We build a projective plae The poits are of two types First, the 2 ordered pairs ( j,), for 1 j, The + 1 special poits X,Y,Z 1,,Z 1 This maes altogether, the right umber The lies are of several types: lies x j for j = 1,,: x j cotais the poits ( j,) for = 1,, ad X lies y for = 1,,: y cotais the poits ( j,) for j = 1,, ad Y ( 1) lies z is for i = 1,, 1 ad s = 1,,: z i,s cotais all poits ( j,) for which M(i) j = s, ad Z i Fially, a lie {X,Y,Z 1,Z 1 } Oe ca chec that this really is a projective plae of order Here is a example We start with a pair of orthogoal Lati squares of order 3 ad costruct a projective plae The lies are writte i the same order as i the above proof that (b) implies (a) The Lati squares (which we have met before) are: The 13 poits of the projective plae are the pairs i j for 1 i, j 3 together with

99 95 APPENDIX: PROOF OF BOSE S THEOREM 93 X,Y,Z 1,Z 2 The thirtee lies are: X X X Y Y Y Z Z Z Z Z Z 2 X Y Z 1 Z 2 Here it is i diagrammatic form The lies through X ad Y are the vertical ad horizotal lies of the grid; the lies through Z 1 ad Z 2 pass through the positios of the three symbols i the squares L 1 ad L 2 Exercises 1 Let A ad B be orthogoal Lati squares of order, which uses the symbols 0,1,, 1 Costruct a matrix S i which the ith etry cosists of the pair (a i j,b i j ), regarded as a two-digit umber writte i base Show that S has the properties (a) its etries are all the itegers from 0 to 2 1 iclusive;

100 94 CHAPTER 9 LATIN SQUARES (b) the sum of the etries i ay row or colum is ( 2 1) (This costructio is due to Euler) 2 Show that, up to permutatios of rows ad colums ad chages i the ames of the symbols, there are just two differet Lati squares of order 4 Show that oe, but ot the other, has a orthogoal mate (a Lati square orthogoal to it) 3 Prove that the Lati square give by the additio table of the itegers mod has a orthogoal mate if ad oly if is odd 4 Let A be a Lati square of order Suppose that, for some positive iteger r <, oly the umbers 1,,r occur i the first r rows ad colums of A (a) Show that the submatrix of A formed by the first r rows ad colums is a Lati square of order r (b) Show that 2r (c) Give a example with r = 3 ad = 7 5 Let m be a iteger greater tha 1 Let X(m) be the multiplicatio table of the o-zero itegers mod m, that is, the (m 1) (m 1) matrix defied as follows: rows ad colums are idexed by 1,2,,m 1, ad the (i, j) etry is i j mod m Prove that X(m) is a Lati square if ad oly if m is prime ( ) Does it have a orthogoal mate?

101 Chapter 10 Steier triple systems A Steier triple system is a very special id of family of sets Here is the defiitio A Steier triple system is a family B of subsets of the -elemet set X = {1,,} with the properties (a) every set i B has three elemets; (b) every two poits of X are cotaied i exactly oe member of B We ofte call the elemets of X poits ad the elemets of B blocs or triples Examples Here are some examples The first three are trivial, the last two are more iterestig Tae X = /0, B = /0 Tae X = {1}, B = /0 Tae X = {1,2,3}, B = {{1,2,3}} Tae X = {1,2,3,4,5,6,7} ad B to be the Fao plae: Tae X = {1,,9} ad arrage the poits of X i a 3 3 grid Now tae B to cosist of the horizotal ad vertical lies ad the positios of the six 95

102 96 CHAPTER 10 STEINER TRIPLE SYSTEMS terms i the formula for the determiat of a 3 3 matrix Aother way of sayig the same thig is that these six sets are the positios of the symbols i a pair of orthogoal Lati squares of order 3 Steier triple systems have various uses For example, a Lati square L = (l i j ) of order is called idempotet if l ii = i, ad totally symmetric if l i j = implies l ji = ad l j = i, for ay i, j, Now, give a Steier triple system (X,B), where X = {1,,}, we costruct a matrix L = (l i j ), where l ii = i ad, if i j, the l i j = if {i, j,} B Coversely, ay idempotet ad totally symmetric Lati square arises i this way from a Steier triple system 101 Existece of STS() For which umbers does there exist a STS()? Theorem 101 Let (X,B) be a Steier triple system of order, where > 0 The: (a) Ay elemet of X is cotaied i ( 1)/2 members of B (b) B = ( 1)/6 (c) 1 or 3 mod 6 Proof (a) Tae x X, ad let r be the umber of members of B cotaiig x Cout pairs (y,b) where B B, y X, ad y x There are 1 poits y x, ad for each such y, there is a uique set B B cotaiig x ad y; so there are 1 such pairs O the other had, there are r choices of B cotaiig x, ad

103 101 EXISTENCE OF STS(N) 97 the two choices of y B with y x (sice B = 3) So the umber of pairs is 2r Thus 2r = 1, whece r = ( 1)/2 (b) Now cout pairs (x,b) where x X, B B, ad x B There are choices for x ad ( 1)/2 choices for a set B cotaiig it (by (a)) O the other had, there are B choices for B, ad 3 choices for a poit x B So whece B = ( 1)/6 ( 1)/2 = 3 B, (c) Sice ( 1)/2 must be a iteger, we see that must be odd So 1, 3, or 5 mod 6 We have to exclude the last case So suppose that = 6m + 5 The, by (b), B = (6m + 5)(6m + 4) 6 = (6m + 5)(3m + 2), 3 which is impossible sice 3 does ot divide 6m + 5 or 3m + 2 We saw that there exist Steier triple systems of orders 1, 3, 7 ad 9 The above theorem shows that they do ot exist for ay other positive order less tha 10 I order to show that there is a STS(), we eed to give a costructio of oe To prove the ext result, we have to give ifiitely may costructios So the proof is quite complicated! This theorem was first proved by Kirma Theorem 102 There exists a Steier triple system of order if ad oly if either = 0 or 1 or 3 mod 6 Proof We have see the oly if part already So we have to tae a umber cogruet to 1 or 3 mod 6, ad costruct a STS() For the cases where is cogruet to 3 mod 6, we ca give a direct costructio For the other cases, we have to use a rather complicated recursive costructio, buildig up large systems from smaller oes I wish there were a simpler proof! Case 3 mod 6: Let = 3m, where m is odd We tae the poits of the STS to be symbols a i,b i,c i, where i = 0,,m 1: there are 3m = such poits The blocs are of two types: (a) sets {a i,b i,c i }, for i = 0,1,,m 1; (b) sets {a i,a j,b }, {b i,b j,c } ad {c i,c j,a }, where i j ad i+ j 2 mod m

104 98 CHAPTER 10 STEINER TRIPLE SYSTEMS All additio i this proof will be mod m, so the last coditio will simply be writte as i + j = 2 We ote that, i this equatio, ay two of i, j, determie the third (Clearly i ad determie j = 2 i, ad similarly j ad determie i Suppose i ad j are give Sice m is odd, we have gcd(2,m) = 1, so 2 is a uit i Z m ; so there is a uique satisfyig 2 = i + j, amely = h(i + j), where h is the iverse of 2 mod m) Moreover, if the two give values are uequal, the the third value is differet from both (Clearly i = implies j = If i = j the the equatio reads 2i = 2 which has the solutio i = ) Clearly every bloc is a set of size 3, so coditio (a) of the defiitio holds We have to verify (b) So choose two poits p ad q; we have to show that there is a uique bloc cotaiig them There are several cases: (a) p = a i, q = a j ( j i) Now i ad j determie a uique such that i+ j = 2; ad {a i,a j,b } is the uique bloc cotaiig p ad q (b) p = b i, q = b j, or p = c i, q = c j : the argumet is similar (c) p = a i, q = b i : clearly there is a uique bloc {a i,b i,c i } of type (a) cotaiig p ad q (d) p = b i, q = c i, or p = c i, q = a i : the argumet is similar (e) p = a i, q = b where i There is a uique j satisfyig i + j = 2; ad j i So the uique bloc is {a i,a j,b } (f) p = b i, q = c, or p = c i, q = a, where i: the argumet is similar For = 9, we obtai the twelve blocs {a 0,b 0,c 0 },{a 1,b 1,c 1 },{a 2,b 2,c 2 },{a 0,a 1,b 2 },{a 0,a 2,b 1 },{a 1,a 2,b 0 }, {b 0,b 1,c 2 },{b 0,b 2,c 1 },{b 1,b 2,c 0 },{c 0,c 1,a 2 },{c 0,c 2,a 1 },{c 1,c 2,a 0 } Case 1 mod 6: This case is much harder I will give oe example of a recursive costructio here, ad put the complete proof of the theorem i a appedix to this chapter Propositio 103 Suppose there exists a STS() The there exists a STS(2+1)

105 101 EXISTENCE OF STS(N) 99 Proof Let (X,B) be a STS(), where X = {1,2,,} We tae a ew set Y = {a 1,,a,b 1,,b,c}, with Y = 2+1, ad costruct a set C of blocs; we show that these blocs form a STS(2 + 1) The blocs are of two types: (a) {a i,b i,c}, for i = 1,, (b) For every bloc {i, j,} B, the four blocs {a i,a j,b }, {a j,a,b i }, {a,a i,b j }, ad {b i,b j,b } Clearly every bloc cotais three poits, so (a) of the defiitio holds Tae two poits p ad q; we have to show that just oe bloc cotais them There are several cases: (a) p = c, q = a i : a uique bloc {a i,b i,c} C cotais p ad q (b) p = c, q = b i, or p = a i, q = b i : the argumet is similar (c) p = a i, b = a j with i j: there is a uique bloc {i, j,} B cotaiig i ad j, ad the a uique bloc {a i,a j,b } C cotaiig p ad q (d) p = a i, q = b j, with i j: the argumet is similar (e) p = b i, q = b j, with i j: there is a uique bloc {i, j,} B cotaiig i ad j, ad the a uique bloc {b i,b j,b } C cotaiig p ad q For = 3, startig with the sigle bloc {1,2,3}, we obtai seve blocs {a 1,b 1,c},{a 2,b 2,c},{a 3,b 3,c},{a 1,a 2,b 3 },{a 2,a 3,b 1 },{a 3,a 1,b 2 },{b 1,b 2,b 3 } of STS(7) This method costructs Steier triple systems of orders 7, 15, 19, 31,, but leaves several values udecided, such as 13, 25, 29, 37, These will be settled i the Appedix The geeral priciple is always the same (we build larger systems out of smaller oes) except i oe case, where we have to give a direct costructio: this is = 13, where we ca tae the poit set to be {0,,12} (the itegers mod 13), ad the blocs to be {0,1,4},{1,2,5},{2,3,6},{3,4,7},{4,5,8},{5,6,9},{6,7,10}, {7,8,11},{8,9,12},{0,9,10},{1,10,11},{2,11,12},{0,3,12}, {0,2,8},{1,3,9},{2,4,10},{3,5,11},{4,6,12},{0,5,7},{1,6,8}, {2,7,9},{3,8,10},{4,9,11},{5,10,12},{0,6,11},{1,7,12}

106 100 CHAPTER 10 STEINER TRIPLE SYSTEMS Note that it is ot ecessary to remember all these blocs We start with the two blocs {0,1,4} ad {0,2,8}, ad produce the rest of the system by addig x to each of their elemets mod 13, for x = 1,,12 (This process is called the developmet of the two blocs mod 13) There is a simple test for a startig set of blocs i this costructio: Propositio 104 Let B 1,,B be 3-elemet subsets of Z The the developmet of {B 1,,B } is a Steier triple system if ad oly if every o-zero elemet of Z has a uique represetatio i the form x y, where x,y B i for some i with 1 i If this holds, the = Proof We will prove it oe way roud Suppose that the coditio of the Propositio holds, ad let u,v be distict elemets of Z We wat to show that there are uique elemets i,z with 1 i ad z Z such that u,v B i + z If this is to hold, we must have u z,v z B i But (u z) (v z) = u v, ad by assumptio there are uique i,x,y such that u v = x y with x,y B i So we must have u z = x ad v z = y, whece z ad i are uiquely determied Moreover, the 1 o-zero elemets must be give by the 6 differeces for each of the blocs, so 1 = 6, as required I the above example, we have the followig expressios for elemets of Z 13 as differeces from {0,1,4} ad {0,2,8}: 1 = = = = = = = = = = = = 0 1 For a simpler example, we get STS(7) as the developmet of a sigle bloc {0,1,3} i Z 7 : 1 = = = = = = Kirma s schoolgirls Despite their ame, Steier triple systems were iveted, ot by Steier, but by Kirma; he gave the defiitio ad proved Theorem 102 several years before Steier published a paper asig whether or ot they exist The reaso we do ot call them Kirma triple systems is that this ame has bee used for somethig a bit differet Kirma posed the followig problem:

107 103 APPENDIX: PROOF OF KIRKMAN S THEOREM 101 Fiftee schoolgirls go for a wal every day for a wee i five rows of three Is it possible to arrage the wals so that every two girls wal together exactly oce durig the wee? Let X = {1,,15}, ad let B be the set of all groups of three girls who wal together durig the course of the wee The the terms of the problem require that (X,B) is a Steier triple system But there is more structure The 15 14/6 = 35 blocs of the Steier triple system must have a partitio ito seve sets of five (correspodig to the days of the wee) such that the five blocs i each part of the partitio themselves form a partitio of X We say that a Steier triple system (X,B) is resolvable, or is a Kirma triple system, if the set B ca be partitioed ito subsets B 1,,B r such that B i is a partitio of X for i = 1,,r Here is a example of a Kirma triple system of order 9 The twelve blocs are arraged ito four rows formig the required partitio B 1 : {{1,2,3},{4,5,6},{7,8,9}} B 2 : {{1,4,7},{2,5,8},{3,6,9}} B 3 : {{1,5,9},{2,6,7},{3,4,8}} B 4 : {{1,6,8},{2,4,9},{3,5,7}} Sice the poits must be partitioed by the blocs, each of which has size 3, we see that must be divisible by 3 for such a system to exist Sice we ow that is odd, we coclude that must be cogruet to 3 mod 6 Now each class B i cotais /3 blocs Sice there are ( 1)/6 blocs altogether, we see that the umber of classes is equal to ( 1)/2 This ca be verified aother way We ow that each poit lies i ( 1)/2 blocs; but exactly oe of these blocs belogs to each class B i, so there must be ( 1)/2 classes Kirma himself costructed a solutio to the problem with = 15 (his origial schoolgirls problem ) It too 120 years before the geeral case was fially solved by Ray-Chaudhuri ad Wilso i the 1970s They proved the followig theorem (which is too complicated for this course!) Theorem 105 For > 0, there exists a Kirma triple system of order if ad oly if 3 mod Appedix: Proof of Kirma s Theorem Kirma s Theorem states that a Steier triple system of order exists if ad oly if = 0 or is cogruet to 1 or 3 mod 6 We have see that this coditio is ecessary, ad we have to show that it is sufficiet: i other words, if satisfies the cogruece coditio, the we ca costruct a STS of order

108 102 CHAPTER 10 STEINER TRIPLE SYSTEMS This is a good example of a combiatorial costructio It cotais both a direct part (for umbers cogruet to 3 mod 6, which we saw already) ad a recursive part; the recursive part shows that oe small case ( = 13) remais to be dealt with (we already gave a costructio for this value); ad as we will see below, it requires the costructio of a auxiliary structure for use i the mai costructio (this is a STS cotaiig a subsystem of order 7, which we ow defie) Let (X,B) be a Steier triple system, ad let Y be a subset of X We say that Y is a subsystem if, for ay two poits y 1,y 2 Y, the uique bloc of (X,B) cotaiig y 1 ad y 2 is cotaied i Y If Y is a subsystem, ad C is the set of blocs which are cotaied i Y, the (Y,C ) is a Steier triple system i its ow right Give ay Steier triple system (X,B) with X 3, there are some obvious subsystems which always exist: the empty set; ay 1-elemet set {x}; ad ay bloc of B Our mai recursive costructio is give by the followig result: Theorem 106 Let (X,B) be a Steier triple system of order v cotaiig a subsystem Y of order u, ad let (Z,D) be a Steier triple system of order w The there exists a Steier triple system of order u + w(v u) Moreover, if 0 < u < v ad w > 1, the we may assume that this system cotais a subsystem of order 7 Remar If we tae v = 3 ad u = 1 (that is, (X,B) cosists of a sigle bloc ad the subsystem is a sigle poit), the we obtai a Steier triple system of order 1 + 2w This is precisely the costructio of Propositio 103 So the above theorem geeralises that propositio Proof We tae the poit set of the ew system to be Y ((X \Y ) Z), which does ideed have u + (v u)w poits, sice Y = u, X \Y = v u, ad Z = w We set m = v u = X \Y, ad umber the poits of X \Y with the elemets of Z m The blocs of the ew system are of the followig types: (a) All blocs cotaied i Y (b) All blocs of the form {y,(x,z),(x,z)}, where y Y, x,x X \Y, z Z, ad {y,x,x } B (c) All blocs of the form {(x,z),(x,z),(x,z)}, where x,x,x X \Y, z Z, ad {x,x,x } B

109 103 APPENDIX: PROOF OF KIRKMAN S THEOREM 103 (d) All blocs of the form {(x i,z),(x j,z ),(x,z )}, where x i,x j,x X \ Y, z,z,z Z {z,z,z } is a bloc of D, ad i + j + = 0 i Z m [Recall that poits of X \Y are idexed by elemets of Z m ] Now we have to show that ay two poits lie i a uique bloc There are several cases: Two poits of Y lie i a uique bloc of type (a) A poit of Y ad a poit of (X \Y ) Z lie i a uique bloc of type (b) Two poits of (X \Y ) Z with the same Z-coordiate lie i a uique bloc of type either (b) or (c) Two poits of (X \Y ) Z with differet Z-coordiates lie i a uique bloc of type (d) [Note that ay two of i, j, uiquely determie the third] So we do have a STS For the last part, assume that u 0 The u ad v are odd, so m = v u is eve Choose a bloc of B of the form {y,x,x }, where y Y ad x,x X \Y The choose the idexig of X \Y by Z m such that x = x 0 ad x = x m/2 Let {z,z,z } be a bloc of D The the seve poits ad the seve blocs y,(x 0,z),(x 0,z ),(x 0,z ),(x m/2,z),(x m/2,z ),(x m/2,z ) {y,(x 0,z),(x m/2,z)},{y,(x 0,z ),(x m/2,z )},{y,(x 0,z ),(x m/2,z )} {(x 0,z),(x m/2,z ),(x m/2,z )},{(x m/2,z),(x 0,z ),(x m/2,z )}, {(x m/2,z),(x m/2,z ),(x 0,z )},{(x 0,z),(x 0,z ),(x 0,z )} form a subsystem [Note that = 0 + (m/2) + (m/2) = 0 i Z m ] Example Earlier, we were uable to costruct STS of orders 25 or 37 With Theorem 106, we ca ow costruct these, usig 25 = 1 + 3(9 1), 37 = 1 + 3(13 1) (This is shorthad for sayig: there exists a STS(9) cotaiig a STS(1) subsystem, ad also a STS(3), so by Theorem 106 there is a STS(1 + 3(9 1)), ad similarly for the other oe

110 104 CHAPTER 10 STEINER TRIPLE SYSTEMS We ow tur to the proof of Kirma s Theorem I will write the proof as a argumet by cotradictio That is, let A() be the statemet that a STS() exists We say that is admissible if is cogruet to 1 or 3 mod 6 I will assume that is the smallest admissible umber for which a STS() does ot exist, ad deduce a cotradictio It is ecessary to have aother iductio goig o at the same time Let B() be the statemet that there exists a STS() cotaiig a subsystem of order 7 I will prove that if is cogruet to 9 mod 12 ad 15, the B() holds First let us cosider A() Suppose that A(m) is true for all admissible umbers m < If is cogruet to 3 mod 6, the the direct costructio i Theorem 102 gives a STS() So we oly have to deal with umbers cogruet to 1 mod 6 If is cogruet to 7 mod 12, the = = 1 + 2(6 + 3) By assumptio, A(6 + 3) holds; the A(12 + 7) follows from Propositio 103 So we oly have to deal with umbers cogruet to 1 mod 12 We separate these accordig to their cogruece mod 36 If is cogruet to 1 mod 36, the = = 1 + 3( ), ad A(12 + 1) is true, so A(36 + 1) is true If is cogruet to 25 mod 36, the = = 1+3(12+9 1), ad A(12 + 9) is true, so A( ) is true If is cogruet to 13 mod 36, the = = 7+3(12+9 7) so we eed a STS(12 + 9) with a subsystem of order 7, that is, we eed to ow that B(12 + 9) is true, ad A( ) will follow We are goig to prove this for 1 For = 0, we gave a direct costructio of STS(13) o pp (Note that we already ow that a STS of order exists by Theorem 102, but that oe does t have the required subsystem) So we have to prove B(12 + 9) for 1 Agai we split ito cogrueces mod 36 If is cogruet to 9 mod 36, the = = 3 + (9 3)(6 + 1) So A(6 + 1) together with Theorem 106 give the result If is cogruet to 21 mod 36, the = = 3+(9 3)(6 +3) So A(6 + 3) together with Theorem 106 gives the result

111 103 APPENDIX: PROOF OF KIRKMAN S THEOREM 105 If is cogruet to 33 mod 36, the = = 3+( )3, So A( ) together with Theorem 106 give the result Although phrased as a miimal couterexample argumet, this proof is essetially costructive For example, how to costruct a STS(625)? We have 625 is cogruet to 13 mod 36, so we write 625 = 7 + 3(213 7); we eed a STS(213) with a subsystem of order is cogruet to 33 mod 36, so we write 213 = 3 + 3(73 3); we eed a STS(73) 73 is cogruet to 1 mod 36, so we write 73 = 1 + 3(25 1); we eed a STS(25) We already saw how to costruct this: write 25 = 1 + 3(9 1), so we eed a STS(9), which of course we ow (it is give by the direct costructio at the start of Theorem 102) I fact, we could costruct STS(625) more easily by usig the fact that 625 = 25 2 = (25 0) ad the existece of STS(25) But a geeral proof caot rely o lucy accidets lie this! Exercises 1 Suppose that a STS(v) of order v o a set X has a subsystem Y of order u, with u < v Show that v 2u + 1 [Hit: Let x be a poit ot i Y Show that the triples cotaiig x ad a poit of Y are all distict] 2 Prove directly that if a STS(v) ad a STS(w) exist, the a STS(vw) exists Show further that, if v,w 3, the we ca fid a STS(vw) cotaiig a subsystem of order 9 3 Let Z 2 deote the itegers mod 2 Let X be the set of all o-zero vectors i the -dimesioal vector space (Z 2 ) Let B be the set of all triples {x,y,z} of vectors of X satisfyig x + y + z = 0 (a) Prove that (X,B) is a Steier triple system of order 2 1 (b) Idetify the Fao plae as a Steier triple system of this form 4 Let X = {1,,} ad suppose that B is a collectio of 3-elemet subsets of X with the property that ay two members of B itersect i at most oe elemet

112 106 CHAPTER 10 STEINER TRIPLE SYSTEMS (a) Let r x deote the umber of members of B cotaiig the elemet x {1,,} Show that r x ( 1)/2 (b) By coutig pairs (x,b), with x X ad B B, show that B = x {1,,} r x 3 (c) Deduce that B (d) Hece or otherwise show that, if = 6, the B 4 (e) Fid a example of four 3-elemet subsets of {1,,6} which satisfy the hypothesis B i B j 1 for B i,b j B, i j

113 Appedix A Solutios to odd-umbered exercises Chapter = So the umerator has to cotai umbers divisible by all three primes So 13 A little trial ad error shows that 1001 = = ( ) 14 4 ( ) 3 We show first that ( 1) = ( 1)2 2 This ca be show i two =0 ways: (a) Tae the Biomial Theorem =0 ( ) x = (1 + x), differetiate twice, ad put x = 1 ( ) ( ) 2 (b) Use the fact that ( 1) = ( 1) for 2 (The terms for 2 = 0,1 are zero) Now sum over Thus =0 2( ) = =0 ( ) ( 1) + =0 ( ) = ( 1) = (+1)2 2 5 Here it is for cogruet to 0 mod 8 I this case, we have (1 + i) = 2 /2 107

114 108 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES Equatig real ad imagiary parts gives ( 1)/2 j=0 /2 j=0 ) ( ( 1) j 2 j ( ) ( 1) j 2 j + 1 = 2 /2, = 0 So, if S 0,S 1,S 2,S 3 deote the sums of biomial coefficiets for cogruet to 0,1,2,3 mod 4 respectively, we have Sice we already ow that we coclude that S 0 S 2 = 2 /2, S 1 S 3 = 0 S 0 + S 2 = S 1 + S 3 = 2 1, S 0 = ( 2)/2, S 1 = 2 2, S 2 = ( 2)/2, S 3 = 2 2 As a chec, whe = 8, we have S 0 = = 72, S 1 = 8+56 = 64, S 2 = = 56, S 3 = 56+8 = 64 7 (a) ( ) ( ) ( ) ( 1) ( + 1) = = + 1 ( + 1) ( 1) ( ) ( ) (b) The ratio of to is ( )/( + 1) This ratio is >,=,<1 + 1 accordig as is >,=,< + 1, that is, as is >,=,<2 + 1 (c) By part (b), the biomial coefficiets icrease util the poit where = (if this occurs), at which poit they remai costat for oe step ad the decrease This happes if is odd If is eve, the there is o value of for which = 2 + 1, so the biomial coefficiets icrease util = /2 ad the decrease (d) This follows immediately from (c) ( ) 2m (e) Suppose that = 2m The is the largest of the 2m + 1 biomial ( ) ( ) m 2m 2m coefficiets,, Now the sum of these biomial coefficiets is 0 2m 2 2m (see Property 1 i sectio 13 of the otes) The largest biomial coefficiet is smaller tha the sum, ad larger tha the average This gives the stated result ( ) 20 Here is a chart of the biomial coefficiets, for = 0,,20

115 109 We have = , ( ) 20 = , 2 20 = Chapter 2 1 (a) 7 5 = 16807; 7650 (see below) ; = 1008 The argumet for the secod umber goes lie this There are three choices for which of 1,2,3 ca occur; so we do the calculatio for the case that 1 ad 2 but ot 3 occur ad multiply by 3 ( ) 5 Suppose that 1 ad 2 occur i positios There are ways to choose these positios; 2 2 ways to fill them with 1s ad 2s (we are ot allowed to use all 1s or all 2s) ad 4 5 ways to fill the remaiig positios with 4,5,6,7 So the total is (2 5 2) + 5(2 4 2)4 + 10(2 3 2) (2 2 2)4 3 = 2550 I (c) there are two terms i the sum because the sequece might begi with a eve umber or a odd umber I the first case, there are three eve umbers which ca be chose from {2,4,6} i 3 3 ways, ad two odd umbers which ca be chose from {1,3,5,7} i 4 2 ways The other term is similar (b) (7) 5 = 2520; 1440 (see below); (3) 3 (4) 2 + (4) 3 (3) 2 = 216 For the first ad third parts, simply replace the formula by () everywhere For the secod part, there are 3 choices for which of 1,2,3 to use, ad 5 4 = 20 choices for their positios; the remaiig etries are filled from the other four umbers, i (4) 3 ways So there are 1440 such sequeces

116 110 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES 3 (a) m ; (b) (m) ; (c)! (we must have m = i this case); (d) see Chapter 6 5 We have W() = ( 1) + ( 1)( 2) + +! Every term except the first two cotais the product of two cosecutive itegers ad so is eve So the parity of W() is the same as that of 1 +, that is, eve if is odd ad vice versa Chapter 3 1 Divide up the permutatios π accordig to the value of the smallest umber for which π maps {1,,} ito itself Note that taes values 1,2,,, ad that = if ad oly if π is idecomposable (O the other had, = 1 if ad oly if π fixes 1) The π iduces a idecomposable permutatio o the set {1,,}, ad a arbitrary permutatio o the set { + 1,,} So there are g()( )! permutatios with a give value of Summig over, we get all permutatios, so the total is! Cosider the product F(x)(1 G(x)) The costat term is 1 1 = 1 For > 1, the coefficiet of x is obtaied by taig the term i x from F(x) ad the term i x from 1 G(x) for = 1,,, together with oe more term which is the term i x from F(x) ad the costat term from 1 G(x) So the coefficiet of x i the product is g()( )! +! = 0 So F(x)(1 G(x)) = 1 Chapter 4 =1 1 Clearly s(1) = 1 A expressio with sum could simply be Otherwise, if the last term is (where = 1,, 1), the the terms before the last sum to, ad form a arbitrary expressio summig to So we have Now for > 1, we see that 1 s() = 1 + s() =1 2 s( 1) = 1 + s(), =1

117 ad so 2 s() = 1 + s() + s( 1) = 2s( 1) =1 111 This recurrece with the iitial coditio s(1) = 1 has solutio s() = 2 1, as is easily show by iductio 3 (a) We have see that the aswer is the th Fiboacci umber F() (b) Suppose that I had over cois of value 2 pece The we must have 0 /2, ad I must also iclude 2 cois of value 1 pey So the total umber of ways of payig is the umber of choices of, which is 1 + /2 5 It is clear that a is a power of 2 for all So put a = 2 b The we fid that b 0 = 1 ad b = 2b 1 for all So b = 2, ad we coclude that a = Begi by recallig the recurrece relatio for the deragemet umbers: d 0 = 1, d 1 = 0, d = ( 1)(d 1 + d 2 ) for 2 (a) Iductio o For = 1, d 1 = 0 = 1 + ( 1) 1 so the result holds Now assume that d 1 = ( 1)d 2 + ( 1) 1 The d = ( 1)d 1 + ( 1)d 2 = ( 1)d 1 + d 1 ( 1) 1 = d 1 + ( 1), so it holds for Thus the formula is proved for all (b) Iductio o The formula is clear for = 0 Suppose that it holds for 1, that is, d 1 = ( 1)! 1 =0 ( 1) /! The d = d 1 + ( 1) 1 =! =! =0 =0 ( 1)! ( 1)! + ( 1)!! as required (c) We have (e x )(1 x) 1 = ( 0 ( 1) x! )( x ) 0

118 112 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES The coefficiet of x i this product is ( 1) =0! = d!, by (b); but this is the same as the coefficiet of x i D(x) Chapter 5 1 (i) We have S(2,2) = 1 ad S(,2) = S( 1,1) + 2S( 1,2) = 1 + 2S( 1,2) for 3 By iductio we ow get S(,2) = for all 2 (The iductio starts at = 2 Assumig S( 1,2) = 2 2 1, we have S(,2) = 1 + 2(2 2 1) = 2 1 1) We have S(2,1) = 1 ad S(, 1) = S( 1, 2) + ( 1)S( 1, 1) = S( 1, 2) + ( 1) Agai the required result follows by iductio, the details of which are left to you (ii) The set {1,,} has 2 subsets Two of these (the empty set ad the whole set) caot occur as parts i a partitio with two parts Each of the other 2 2 sets occurs with its complemet i a uique partitio So the umber of partitios is (2 2)/2 = A partitio with ( ) 1 parts must have oe part of size 2 ad all the rest of size 1 There are ways to choose the part of size 2; all the other poits lie i 2 parts of size 1 A partitio with 2 parts either has two parts of size 2 or oe of size 3, with all remaiig parts of size 1 So we have S(, 2) = ( )( )/ ( ) = ( 1)( 2)(3 1) 24 (The divisio by 2 i the first term is because the two parts of size 2 ca be chose i either order yieldig the same partitio) 3 (a) See the calculatio of the table of values of Stirlig umbers of the first id o p 54 of the text We see that s(6,3) = 225, so the umber of permutatios is 225 (the mius sig idicates that they are all odd permutatios)

119 113 The possible cycle legths are three positive umbers summig to 6; these are [1,1,4], or [1,2,3], ( ) or [2,2,2] 6 There are = 15 ways to choose the four poits to form the 4-cycle, ad 4 3! = 6 possible ( 4-cycles ) o this set So 90 permutatios with cycle legths [1,1,4] 6 There are = 20 choices of three poits for a 3-cycle, ad two 3-cycles o ( 3) 3 this set The choices of two poits for the 2-cycle, ad the rest is determied 2 uiquely So 120 ( )( permutatios )( ) with cycle legths [1,2,3] There are = 90 choices of three 2-cycles; but they could be chose i ay order, so we have to divide by 3! = 6, givig 15 permutatios with cycle legths [2,2,2] Total = 225 Chapter 6 1 The percetage of people satisfied with oe of the cadidates is = 7, which is impossible So the data is icorrect Chapter 7 1 Let = 2 Out of each complemetary pair of sets of size, F cotais exactly oe The resultig family satisfies F = 1 2 ( ) 2 = ( ) 2 1, 1 sice it cotais oe out of each complemetary pair of -sets by costructio Moreover, F is a itersectig family For tae A,B F ; the A ad B caot be disjoit, sice disjoit -sets are complemetary ad we too just oe out of each complemetary( pair ) ( ) There are 2 1 = complemetary pairs of -sets, ad so 2 raised 1 to this power umber of choices of oe set from each complemetary pair So this is the umber of maximum itersectig families of this form

120 114 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES Clearly there are oly 2 sets of the form F (i), sice i = 1,, = 2 This is much much smaller tha the umber of families just costructed Eve for = 6, we costructed 2 10 = 1024 families of which oly six are of the form F (i) The Erdős Ko Rado Theorem ( says ) that every itersectig family of -subsets 1 of a -set of maximum size has the form F (i) if > 2 The above 1 costructio shows that this is ot true if = 2 3 By trial ad error, or otherwise, the polyomial x 3 + x + 1 is irreducible over Z 2, so we adjoi a root α of this polyomial Elemets of the resultig field have the form a + bα + cα 2, where a,b,c Z 2 ; so there are eight elemets, amely 0,1,α,α + 1,α 2,α 2 + 1,α 2 + α,α Additio is doe by addig the coefficiets a,b,c mod 2, so that, for example, (α 2 + 1) + (α 2 + α) = α + 1 To multiply, we use the fact that α 3 = α + 1 (sice α is a root of x 3 + x + 1 = 0), ad so α 4 = α 2 + α So for example (α 2 + 1) (α 2 + α) = α 4 + α 3 + α 2 + α = α 2 + α + α α 2 + α = α Ay further set must cotai at least oe of 1 ad 2 (if ot, it would be cotaied i {3,4,5}) ad ca t cotai both (or it would cotai {1,2}) Similarly, such a set must cotai at least oe of 3,4,5 but caot cotai them all So to get such a set we must iclude oe of {1} ad {2}, ad oe of the six subsets {3}, {4}, {5}, {3,4}, {3,5}, {4,5}, ad tae their uio This gives 2 6 = 12 possible sets However, we caot tae all of these sets If we tae {1,3}, the {1,3,4} ad {1,3,5} are ot permitted I fact, we ca tae at most six of these sets: for if we arrage the six sets cotaiig 1 i pairs ({1,3},{1,3,4}),({1,4},{1,4,5}),({1,5},{1,3,5}), ad similarly for the sets cotaiig 2, we ca choose at most oe of each pair So we ca t have more tha = 8 sets altogether However, we ca obtai a Sperer family with 8 sets, by combiig 1 with the 1-elemet subsets of {3,4,5}, ad 2 with the 2-elemet subsets: {1,2},{3,4,5},{1,3},{1,4},{1,5},{2,3,4},{2,3,5},{2,4,5}

121 115 7 Follow the hit O oe had, if we choose the poit i, there are r i choices of A j cotaiig it, ad (r i 1) choices of A (which must be differet from A j ), so r i (r i 1) such triples begiig with i To cout all the triples, we have to sum this expressio over all values of i from 1 to O the other had, if we choose A j ad A first, there are b choices for A j ad (b 1) choices for A ; the A j A = 1, so there is just oe choice for i belogig to both these sets So the umber of triples is b(b 1) Equatig these two expressios gives the result I the family S of Questio 6, each poit lies i three sets of S So each term i the sum is 3 2 = 6, ad the sum is 7 6 = 42 O the other had, b = 7, so the right-had side is also 7 6 = 42 Chapter 8 1 Tae the Fao plae to have the sets 123,145,167,246,257,347,356 Clearly (1,4,6,2,7,3,5) is a SDR There are i all 24 SDRs for the Fao plae Our argumets for this will be based to some extet o symmetry Thus, for example, we ca choose ay elemet of 123 to be its represetative, so by symmetry we may choose 1 Agai by symmetry, we ca choose either of 4 ad 5 as the represetative of the secod set, ad either of 6 ad 7 as the represetative of the third Suppose that we choose 4 ad 6 Removig these represetatives from the last four sets gives 2,257,37,35 We must use 2 as represetative of the fourth set The it is easy to see that there are just two choices for the other three, amely (5,7,3) or (7,3,5) So altogether there are = 24 SDRs 3 (a) Clearly {{1},{2},{3}} has just oe SDR (b) {{1,2},{1,2},{3}} has two SDRs, sice both 1 ad 2 ad be represetatives for the first two sets (c) {{1,2},{2,3},{1,2,3}} has three SDRs Chec that there are three choices of represetatives for the first two sets; the the uused elemet ca be the represetative of the last set

122 116 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES (d) {{1,2},{1,2,3},{1,2,3}} has four SDRs For there are two choices for the represetative of the first set; if 1 is chose, the the remaiig elemets of the other two sets are {{2,3},{2,3}, with two SDRs, ad similarly if 2 is chose (e) For {{1,2,3},{1,2,3},{1,2,3}}, ay permutatio of (1,2,3) will be a valid SDR Suppose that we have a family of three subsets of {1,2,3} with five SDRs, say all except (1,2,3) The sice (1,3,2), (2,1,3) ad (3,2,1) are all SDRs, we see that each of 1,2,3 ca appear as represetative of each set, so each of the three sets must be {1,2,3} But the (1,2,3) would be a SDR, after all Chapter 9 1 (a) Sice A ad B are orthogoal, each pair (i, j) for i, j = 0,, 1 occurs just oce These pairs represet i base all the umbers from 00 = 0 to ( 1)( 1) = ( 1) + ( 1) = 2 1, each oce (b) I each row or colum, each of the s digits 0,, 1 occurs oce, ad each of the uits digits 0,, 1 occurs oce So the row or colum sum is as required (0 + + ( 1)) + (0 + + ( 1)) = ( + 1)( 1)/2, 3 The additio table of Z is the Lati square L(1) defied i this chapter If is odd, the L(2) is a Lati square orthogoal to it Suppose that is eve, ad suppose (for a cotradictio) that M is a Lati square orthogoal to L(1) We begi by observig that the sum of all the elemets i Z is /2 if is eve (For the sum i Z is ( 1)/2, ad is eve, so divides 2 /2) Loo at the positios (x i,y i ) i which a give symbol occurs i M The Each row occurs oce as x i, so x i = /2 Each colum occurs as y i, so y i = /2 The (x i,y i ) etry of L(1) is x i + y i Sice L(1) is orthogoal to M, each symbol occurs oce as x i + y i, so (x i + y i ) = /2 But this is a cotradictio, sice /2 + /2 = 0 /2 5 Suppose that m is ot a prime; say m = ab, where 1 < a,b < m The (a + 1)b = ab + b b = 1b mod m,

123 117 so the elemet b occurs twice i colum b, i rows 1 ad a + 1 Thus M(m) is ot a Lati square Coversely, suppose that m is prime I this case we show that M(m) is a Lati square with etries 1,,m 1 (a) First ote that all etries belog to this set For if the (i, j) etry were ot i the set {1,,m 1}, the it would ecessarily be zero; so m would divide i j, cotrary to the fact that m is a prime which does ot divide either i or j (b) Suppose that the same etry x occurs twice i a row, say i positios (i, j) ad (i,), with j < The i j i x mod m; so i( j) 0 mod m This is impossible for the same reaso as i (a) (c) A very similar argumet to (b) shows that a etry caot occur twice i a colum The last part of the questio is more difficult There is a theorem of algebra sayig that, if p is prime, there exists a primitive root mod p, that is, a elemet g whose powers give all the o-zero elemets of Z p Thus the multiplicative group of Z p is isomorphic to the additive group of Z p 1 Now by the solutio to Questio 3 above, we see that the Cayley table of this group does ot have a orthogoal mate if p is odd Chapter 10 1 Follow the hit We ow that the umber of triples of the STS cotaiig the poit x is (v 1)/2 (Theorem 101(a)) For each y Y, there is a uique triple cotaiig x ad y No two of these triples are equal, sice if the same triple cotaied {x,y 1 } ad {x,y 2 } for y 1,y 2 Y, the this triple would cotai two poits of Y ad hece would be cotaied i Y, cotradictig the fact that it cotais x So Y = u (v 1)/2, whece v 2u (a) We have to show that, give ay two distict o-zero vectors x ad y, the uique solutio of x+y+z = 0 (amely z = (x+y)) is o-zero ad is ot equal to either x or y It will the follow that x ad y lie i a uique triple i B Note that x = x for ay vector i (Z 2 ), so we ca write z = x + y Now: If z = 0 the x + y = 0, so y = x = x, cotrary to assumptio If z = x, the y = 0, cotrary to assumptio If z = y, the x = 0, cotrary to assumptio

124 118 APPENDIX A SOLUTIONS TO ODD-NUMBERED EXERCISES Fially, the order of the STS is the umber of o-zero vectors, which is 2 1 (b) All seve equatios x + y + z = 0 ca be checed from the picture:

125 Appedix B Miscellaeous problems ( ) 1 Show that the coefficiet of x i (x + x 2 ) is equal to Hece show that the coefficiet of x i (1 x x 2 ) 1 is equal to Deduce that /2 =0 ( ) = F /2 =0 ( ( 2 a ) + b 2 Prove that is odd if ad oly if b = 2 a 1 [This exercise is due to 2b + 1 Thomas Müller] 3 How may words ca be made usig some or all (possibly oe) of the letters of the word MAMMAL? 4 5 (a) How may permutatios of {1,,9} are there? (b) How may of them cosist of a sigle cycle? (c) How may of them have exactly three cycles, oe of which is of legth 1? (a) I how may ways ca 25 sweets be distributed to a class of 12 childre? (b) How may ways are there if each child is to have at least oe sweet? (c) How may ways are there if each child is to have at least two sweets? 119 )

126 120 APPENDIX B MISCELLANEOUS PROBLEMS 6 Solve the recurrece relatio ad iitial coditios a 0 = 2, a 1 = 4, a 2 = 7, a = 4a 1 5a 2 + 2a 3 for 3 7 Let B() be the th Bell umber, the umber of partitios of {1,,} Prove directly that B()! for all 8 Let F be the th Fiboacci umber Prove by iductio that for 1 F 2 F 1 F +1 = ( 1) 9 Let S(,) be the Stirlig umber of the secod id (the umber of partitios of {1,,} ito parts, ad let (a) Prove that F 1 (x) = 1/(1 x) F (x) = S(,)x (b) Write dow a recurrece relatio for S(,) (c) Use it to show that for 1 F (x) = x 1 x F 1(x) (d) Hece show by iductio o that for 1 F (x) = x (1 x)(1 2x) (1 x) 10 Show that a permutatio is eve if ad oly if it has a eve umber of cycles of eve legth (with o restrictio o cycles of odd legth) 11 Let A 1,,A be subsets of a set X For J {1,,}, let A J = i J A i be the set of elemets which lie i the sets A i for i J (ad possibly i some other sets as well) Let B J be the set of elemets which lie i the sets A i for i J, ad

127 121 do ot lie i the sets A for / J Use the Priciple of Iclusio ad Exclusio to show that B J = ( 1) K\J A K K J 12 Let B be the matrix obtaied from Pascal s Triagle by movig the rows( right ) so that the left-had side is vertical So the etry i row ad colum is ( ) Let B be the matrix i which the etry i row ad colum is ( 1) Prove that the matrices B ad B are iverses of each other, by fidig two bases for the space of all polyomials such that B ad B are the trasitio matrices [Hit: The Biomial Theorem] 13 (a) Show that there does ot exist a Lati square orthogoal to the square [Hit: Suppose that B is orthogoal to the give square ad has etry 1 i positio (1,1) Where ca the other 1s i B occur?] (b) Write dow two orthogoal Lati squares of order 5 14 Let F = Z 3, the itegers mod 3 Let V = F be the vector space of all -tuples of elemets of F Let B = {{x,y,z} : x,y,z B,x,y,z distict,x + y + z = 0} Show that (V,B) is a Steier triple system of order 3 15 Let F be a itersectig family of subsets of X = {1, 2,, } (a) Show that F 2 1 (b) Show that, if F = 2 1, the for ay subset A of X, either A F, or X \ A F (c) Show that, if F = 2 1, A F, ad B is a subset of X cotaiig A, the B F

128 122 APPENDIX B MISCELLANEOUS PROBLEMS 16 Let F be a itersectig family of 2-elemet subsets of {1,,} Show that either (a) there is a elemet x {1,,} cotaied i every set i F ; or (b) F = {{x,y},{y,z},{x,z}} for some x,y,z {1,,} 17 Let F = {123,456,789,147,258,369,159,267,348}, where, for example, 123 meas {1,2,3} (a) Prove directly that F satisfies Hall s coditio (b) Fid a 3 9 Lati rectagle whose colums are the sets of F

129 Idex additio, 24 alteratig group, 55 arragemet, 15 automorphism, 72 automorphism group, 72 Bell umbers, 47 biomial coefficiet, 2 Biomial Theorem, 6, 25, 30 Bose s Theorem, 90 Bruc Ryser Theorem, 71 Catala umbers, 41 Cayley table, 83, 117 cetral biomial coefficiets, 14, 27 Chiese rigs puzzle, 39 clique, 72 coclique, 72 commutative rig with idetity, 88 coset, 55, 72 critical set, 78 cycle decompositio, 52 de Bruij Erdős Theorem, 68 deragemet, 40 deragemet umbers, 40, 61 developmet, 100 differetiatio, 25 edge, 72 equivalece relatio, 47 Equivalece Relatio Theorem, 47 Erdős Ko Rado Theorem, 67, 73, 114 Euler s officers, ii, 87 expoetial fuctio, 29 factorial, 3 families of sets, 63 Fao plae, 68, 79, 95, 115 Fiboacci umbers, 33 field, 69, 89, 91, 114 geeratig fuctio, 23 expoetial, 44, 48 geometric series, 23 graph, 72 vertex-trasitive, 72 group, 72 Hall s Coditio, 78 Hall s Theorem, 78, 84 deficit form, 82 itersectig family, 66 isomorphic, 117 Kirma triple system, 101 Kirma s schoolgirls, iii, 101 Kirma s Theorem, 97 Lagrage s Theorem, 55, 72 Lati square, 83 Lati squares orthogoal, 86 logarithm fuctio, 29 Lucas Theorem, 10 LYM iequality, 65 modular arithmetic, 87 multiplicatio, 25 order 123

130 124 INDEX of projective plae, 71 orthogoal Lati squares, 86 orthogoal mate, 94 parity of biomial coefficiets, 10 of permutatio, 52 partial fractios, 37 partitio, 47 Pascal s triagle, 5 permutatio, 52 perspective, 91 power series, 23 power set, 63 primitive root, 117 Priciple of Iclusio ad Exclusio, 58 projective plae, 71, 90 substitutio, 25 subsystem, 102 Sudou, ii, 81 surjectios, 60 symmetric group, 55 system of distict represetatives, 77 uimodality of biomial coefficiets, 13 uit, 87 Vadermode covolutio, 7, 26 vector space, 69 vertex, 72 Youde square, 85 recurrece relatio, 3, 33 for Bell umbers, 48 for biomial coefficiets, 4 for Catala umbers, 42 for deragemet umbers, 40 for factorials, 3 for Fiboacci umbers, 34 for Stirlig umbers, 49, 53 liear, 36, 39 recursive costructio, 102 SDR, 77 selectio, 15 sequece, 23 sig of permutatio, 52 Sperer family, 63 Sperer s Theorem, 64, 65 Steier triple system, 95 Stirlig umbers of first id, 52, 112 of secod id, 49, 60, 112 subset, 1