Solutions to Exercises Chapter 3: Subsets, partitions, permutations

Transcription

1 Solutios to Exercises Chapter 3: Subsets, partitios, permutatios 1 A restaurat ear Vacouver offered Dutch pacaes with a thousad ad oe combiatios of toppigs. What do you coclude? Sice 1001 ( 14 4, it is plausible that there were fourtee differet toppigs, of which ay four could be chose. (I fact, 1001 simply meat a large umber; there were 25 differet toppigs, ay subset beig permitted. 2 Usig the umberig of subsets of {0,1,..., 1} defied i Sectio 3.1, prove that, if X X l, the l (but ot coversely. Let (a 0,...,a 1 ad (b 0,...,b 1 be the characteristic fuctios of X ad X l (the biary digits of ad l respectively. Each of a i ad b i is either zero or oe. If X X l, the it is impossible that a i 1 ad b i 0; hece a i b i for all i, ad so a i 2 i b i 2 i l. The coverse is false: 1 < 2, but X 1 {0} is ot a subset of X 2 {1}. 3 Prove that, for fixed, the greatest biomial coefficiet ( occurs whe /2 or /2. The ratio of successive biomial coefficiets is give by ( ( +1 / ( /( +1. Now, if < ( 1/2, the + > +1, ad this ratio is greater tha 1, so the biomial coefficiets icrease. If is odd, the the ratio is equal to 1 for ( 1/2, so that ( ( ( 1/2 (+1/2 ; after that, the ratio is less tha 1 ad the coefficiets decrease agai. So the two middle coefficiets are the largest. If is eve, the ratio is ever equal to 1, ad so the largest biomial coefficiet is the first for which the ratio is less tha 1, amely ( /2. 1

2 4 Prove the followig idetities: ( ( ( ( l (a. l l l (b (c (d (e ( ( ( m m + i i (recall the covetio that ( 0 if < 0 or >. ( ( + i i i0 i0 1 ( ( 1 ( 2 { 0 if is odd; ( 1 m( 2m m if 2m. (a Let X be a -set. Cout pairs (Y,Z with Z Y X ad Y, Z l. There are ( ( choices for Y, ad the l choices of a l-subset Z of Y. Alterately, there are ( l choices of Z; the we obtai Y by choosig l poits from X \ Z to adjoi to Z, ad sice X \ Z l, this ca be doe i ( l l ways. If < l, the o choices are possible, ad both sides are zero, as a result of our covetios about biomial coefficiets. (b Choose a team of players from a class with m girls ad boys. (c The result ca be proved by iductio, beig clear whe 0. Assumig it for 1, we have i0 ( + i i ( ( ( ad the iductio step is proved. (d Differetiatig the Biomial Theorem gives 1 ( t 1 (1 + t 1 (the term with 0 is costat ad its derivative is zero. Now put t 1. (e Cosider the idetity (1 + t (1 t (1 t 2, ad calculate the coefficiet of t. O the left, we multiply the coefficiet of t i (1 +t (which is ( ( by the coefficiet of t i (1 t (which is ( 1 (, ad sum over 2,

3 . O the right, it is zero if is odd (sice oly eve powers of t appear, ad ( 1 /2( /2 if is eve, as required. 5 Followig the method i the text, calculate the umber of subsets of a -set of size cogruet to m (mod 3 (m 0,1,2 for each value of (mod 6. Let α e 2πi/3 ad β e 4πi/3 be the two cube roots of uity. The 1+α+β 0. We have ( ( , 0 ( α (1 + α ( β, 0 ( β (1 + β ( α. 0 The values of ( β are 1, β,α, 1,β, α accordig as 0,1,2,3,4,5 (mod 6. Also, 1 + α + β 3 if 0 (mod 3, ad 0 otherwise. Addig the three equatios shows that the umber of subsets of size divisible by 3 is equal to (2 +( β +( α /3, which is (2 +2/3, (2 +1/3, (2 1/3, (2 2/3, (2 1/3, or (2 + 1/3, depedig o the cogruece of modulo 6. For sets of size cogruet to 1 mod 3, multiply the secod equatio by α 2 ad the third by β 2 before addig; for 2 mod 3, use the multipliers α ad β istead. 6 Let be a give positive iteger. Show that ay o-egative iteger N ca be writte uiquely i the form ( ( ( x x 1 x1 N , 1 1 where 0 x 1 <... < x 1 < x. [HINT: Let x be such that ( x ( N < x+1. The ay possible represetatio has x x. Now use iductio ad the fact that N ( x ( < x 1 (Fact to show the existece ad uiqueess of the represetatio.] Show that the order of -subsets correspodig i this way to the usual order of the atural umbers is the same as the reverse lexicographic order geerated by the algorithm i Sectio [HINT: i ( j ( j0 i j +1 i.] 3

4 Give N ad, choose x such that ( x ( N < x+1. We must have x x. Moreover, ( ( ( xi x + i x + 1 1, i i i1 i1 (the first iequality holdig sice x i x + i ad the secod by Exercise 3(c; so x < x would be impossible. Thus we must have x x. By iductio o, the iteger N ( x has a uique expressio i the same form with 1 replacig ; ad, as i the hit, Fact shows that x 1 < x. So the result is proved. The last part is proved by iductio; the iductive step ivolves cosiderig the algorithm ( Suppose that, i the expressio for N, we have x i + i for i 1,...,r 1, ad x r > + r. The r 1 ( xi ( i1 i +r r 1, by Exercise 3(c (ote that the term i 0 is missig from the sum. So, if we icrease x r 1 by 1 ad set x i i 1 for i < r 1, the correspodig sum is icreased by 1, ad represets N + 1. So the -set represetig N + 1 is the same as the ext oe produced by (3.12.3, that is, the ext i reverse lexicographic order. 7 Use the fact that (1 +t p 1 +t p (mod p to prove by iductio that p (mod p for all positive itegers. Clearly 1 p 1 (mod p. Assumig that the result is true for, we have (1 + p 1 + p 1 + (mod p, the first cogruece by the give fact ad the secod by the iductive hypothesis. So the iductive step holds. The give fact is proved o page 28. 4

5 8 A computer is to be used to calculate values of biomial coefficiets. The largest iteger which ca be hadled by the computer is Four possible methods are proposed: ( (1!/!(!; ( (2 ( 1...( + 1/!; ( ( ( (3 1, + 1 for > 0; ( 1 + for 0 < < (i.e., Pas- 0 ( ( (4 1, 0 cal s Triagle. 1 ( ( 1 1 For which values of ad ca ( be calculated by each method? What ca you say about the relative speed of the differet methods? Method 1 will fail if! > 32767, that is, if > 7; so it computes ( for 7. Method 2 requires the umerator to be at most 32767; so, for give, it wors for up to some value (, where ( 32767, 181, 33, 15, 10, 8, 7 for 1,...,7. Method 3 requires ( ( , which bouds by a similar fuctio m(; we fid that m( 32767, 181, 41, 22,.... Method 4 reaches each biomial coefficiet from below, so it calculates all those ( which do ot exceed For example, for 2, 3 it wors up to 256, 59 respectively. (I additio, it is the oly method which wors for close to. Of the best two, methods 3 ad 4, we see that 3 is iferior i terms of the rage of values for which it wors. It is faster for computig a sigle biomial coefficiet, requirig oly 1 multiplicatios ad 1 divisios, whereas method 4 eeds roughly additios. But method 4 is preferable if the etire Pascal triagle is required, sice each additio yields a ew piece of data. 9 Show that there are ( 1! cyclic permutatios of a set of poits. A cyclic permutatio, writte i cycle form, cosists of the poits writte i some order iside a bracet. There are! orders. However, the cycle ca start at 5

6 ay poit, ad so orders yield the same cyclic permutatio. Thus, the umber of cyclic permutatios is!/ ( 1!, as required. 10 The order of a permutatio π is the least positive iteger m such that π m is the idetity permutatio. Prove that the order of a cycle o poits is. Prove that the order of a arbitrary permutatio is the least commo multiple of the legths of the cycles i its cycle decompositio. If the poit x lies i a -cycle of π, the o successive applicatios of π, x visits i tur all the poits of the cycle, returig to its startig poit after applicatios. So xπ x ad xπ x for 0 < <. Thus, if π is a sigle -cycle, the π is ot the idetity for 0 < < ; but π maps every poit to itself, that is, π is the idetity. So the order of π is. More geerally, we see that xπ x if ad oly if is a multiple of the legth of the cycle i which x lies. So, for a arbitrary permutatio π, we see that π is the idetity if ad oly if is a multiple of the legth of every cycle of π. So the order of π, beig the least positive for which this holds, is the least commo multiple of the cycle legths. 11 How may words ca be made from the letters of the word ESTATE? There are words with o repeated letters. If E but ot T is repeated, there are words. (If there are 2 further letters, the there are ( 2 positios for the Es, ad (3 2 choices for the positios of the other letters, where ( ( 1 ( + 1. The same umber occur if T but ot E is repeated. If both pairs are repeated, there are words. (Choose the positios for the Es, the those for the Ts, the those for the remaiig letters. I total, 523 words. 12 Give letters, of which m are idetical ad the rest are all distict, fid a formula for the umber of words which ca be made. If we use of the m idetical letters ad l of the others, there are ( +l ( ml possible words. Now sum over ad l. 13 Show that, for 2,3,4,5,6, the umber of ulabelled trees o vertices is 1, 1, 2, 3, 6 respectively. This is doe by drawig the trees: for 4, the two trees are a path with three edges, or three edges radiatig from a cetral vertex. As a chec that othig has bee forgotte, the umber of labelligs of a tree is equal to! divided by the umber of automorphisms (symmetries of the tree. (So, for example, the 4-vertex 6

7 path has two symmetries, the idetity ad reversal, ad the other tree has six, sice the three edges ca be permuted arbitrarily; ad we have 4!/2 + 4!/ , i accordace with Cayley s Theorem. You ca fid drawigs of the trees i N. J. A. Sloae ad S. Plouffe, The Ecyclopedia of Iteger Sequeces, Academic Press, 1995, Figure M0791. The sequece is olie here. 14 The lie segmets from (i,logi to (i + 1,log(i + 1 lie below the curve y log x. (This is because the curve is covex, i.e., its secod derivative 1/x 2 is egative. The area uder these lie segmets from i 1 to i is log! log( + 1, sice it cosists of the rectagles of Fig. 3.1(b together with triagles with width 1 ad heights summig to log( + 1. Deduce that! e ( + 1. e This is mostly spelt out i the questio. The area of the rectagles is i2 logi log!, while the triagles add up to half a rectagle of height log( + 1. So log! log( logxdx ( + 1log( + 1 ( , givig! + 1(( + 1/e. The fial result is obtaied usig the fact that ( + 1 e (see Exercise 3(b of Chapter Use Stirlig s Formula to prove that ( / π. By Stirlig s formula, ( 2 (2! (! 2 4π(2/e 2 2π(/e 2 22 π. (Chec the defiitio of the relatio to esure that this is valid. 7

8 16 (a Let 2 be eve, ad X a set of elemets. Defie a factor to be a partitio of X ito sets of size 2. Show that the umber of factors is equal to (2 1. This umber is sometimes called a double factorial, writte (2 1!! (with!! regarded as a sigle symbol, the two exclamatio mars suggestig the gap of two, ot the factorial fuctio iterated! (b Show that a permutatio of X iterchages some -subset with its complemet if ad oly if all its cycles have eve legth. Prove that the umber of such permutatios is ((2 1!! 2. (c Deduce that the probability that a radom elemet of S iterchages some 1 2 -set with its complemet is O(1/. We get a 1-factor by writig /2 boxes each with room for two etries, ad fillig them with the elemets 1,...,. These ca be writte i the boxes i! ways. However, permutig the boxes (i! ways, or the elemets withi the boxes (i 2 ways does t chage the 1-factor. The product of these umbers is (2; dividig, we obtai (2 1 (2 1!! for the umber of 1-factors. (b Suppose that the -set A is exchaged with its complemet B by a permutatio. The elemets of A ad B alterate aroud each cycle, which thus has eve legth. Coversely, if all cycles have eve legth, we may colour the elemets i each cycle alterately red ad blue; the red ad blue sets are the exchaged. From a permutatio with all cycles eve we obtai a pair of 1-factors as follows. A 2-cycle is assiged to both 1-factors; i a loger cycle, the cosecutive pairs are assiged alterately to the two 1-factors. The process is t uique, sice the startig poit is t specified; ideed, a permutatio gives rise to 2 d ordered pairs of 1-factors, where d is the umber of cycles of legth greater tha 2. Coversely, let a pair of 1-factors be give. Their uio is a graph with all vertices of valecy 2, thus a disjoit uio of circuits, all of eve legth (sice the 1-factors alterate aroud a circuit. We tae these circuits to be the cycles of a permutatio. I fact, for a circuit of legth greater tha 2, there are two choices for the directio of traversal. So the umber of permutatios obtaied is 2 d, where d is as before. (c The proportio is ((2 1!! 2 /(2! i1 e 1/2i i1 8 (1 1/2i

9 e i1 (1/2i e log/2+o(1 O( 1/2. 17 How may relatios o a -set are there? How may are (a reflexive, (b symmetric, (c reflexive ad symmetric, (d reflexive ad atisymmetric? A relatio is a set of ordered pairs, so the first part of the questio ass for the umber of subsets of a set of size 2 (the umber of ordered pairs. (a For a reflexive relatio, we must iclude all pairs (x,x, together with a arbitrary subset of the pairs (x,y with x y; there are ( 1 uequal pairs. (b There are ( + 1/2 uordered selectios of two elemets from, with repetitios allowed. A symmetric relatio is determied by a subset of these, sice it must iclude both or either of (x,y ad (y,x for all x,y. (c We must put i all pairs (x,x, ad decide about pairs (x,y with x y. So the relatio is determied by a subset of the set of uordered selectios without repetitios. (d Agai we iclude all pairs (x,x. For each x,y with x y, there are three possibilities: iclude (x,y, or (y,x, or either. 18 Give a relatio R o X, defie R + {(x,y : (x,y R or x y}. Prove that the map R R + is a bijectio betwee the irreflexive, atisymmetric ad trasitive relatios o X, ad the reflexive, atisymmetric ad trasitive relatios o X. Show further that this bijectio preserves the property of trichotomy. Let R be irreflexive, atisymmetric ad trasitive. We claim first that R + is reflexive, atisymmetric ad trasitive. The first two assertios are clear; we must verify the trasitive law. So suppose that (x,y,(y,z R +. If (x,y,(y,z R the (x,z R by the trasitivity of R; if (x,y R ad y z the (x,z R; ad the other two cases are similar. Coversely, let S be reflexive, atisymmetric ad trasitive, ad let S {(x,y : (x,y S ad x y}. Clearly S is irreflexive ad atisymmetric. We show that it is trasitive. Suppose that (x,y,(y,z S. The (x,y,(y,z S, so (x,z S, by trasitivity of S. Could we have x z? No, sice the (x,y,(y,x S with x y, cotradictig atisymmetry. So (x,z S. Moreover, (R + R ad (S + S, so we have a bijectio as claimed. 9

10 For the fial part, we must show that R + satisfies trichotomy if ad oly if R does. This is clear: for trichotomy ca be expressed as if x y, the oe of (x,y ad (y,x satisfies the relatio, ad the pairs (x,y with x y are the same i R ad R Recall that a partial preorder is a relatio R o X which is reflexive ad trasitive. Let R be a partial preorder. Defie a relatio S by the rule that (x,y S if ad oly if both (x,y ad (y,x belog to R. Prove that S is a equivalece relatio. Show further that R iduces a partial order R o the set of equivalece classes of S i a atural way: if (x,y R, the (x,y R, where x is the S- equivalece class cotaiig x, etc. (You should verify that this defiitio is idepedet of the choice of represetatives x ad y. Coversely, let X be a set carryig a partitio, ad R a partial order o the parts of the partitio. Prove that there is a uique partial preorder o X givig rise to this partitio ad partial order as i the first part of the questio. Show further that the results of this questio remai valid if we replace partial preorder ad partial order by preorder ad order respectively, where a preorder is a partial preorder satisfyig trichotomy. Now Let R be a partial preorder (a reflexive ad trasitive relatio, ad let S {(x,y : (x,y R ad (y,x R}. S is reflexive: for (x,x R for all x. S is symmetric, by defiitio. S is trasitive: for, give (x,y,(y,z S, we have (x,y,(y,z R so (x,z R, ad also (y,x,(z,y R so (z,x R, whece (x,z S. Thus S is a equivalece relatio. Let x be the equivalece class cotaiig x. We put R {(x,y : (x,y R}. Note that the defiitio is idepedet of the choice of represetatives of the equivalece classes. For, if x x ad y y, ad (x,y R, the (x,x,(x,y,(y,y R, so (x,y R. Now the verificatio that R is reflexive ad trasitive is straightforward. To show that R is atisymmetric, suppose that (x,y,(y,x R. The (x,y,(y,x R, ad so (x,y S, whece x y. The verificatio that trichotomy carries over from R to R is more of the same. 10

11 20 List the (a partial preorders, (b preorders, (c partial orders, (d orders o the set {1,2,3}. The umbers are (a 29, (b 13, (c 19, (d 6. (The umbers of ulabelled structures are 9, 4, 5, 1 respectively. It is ot too much labour to draw all the ulabelled orders, etc., ad cout the umbers of labelligs of each. For partial orders o up to four poits, see N. J. A. Sloae ad S. Plouffe, The Ecyclopedia of Iteger Sequeces, Academic Press, 1995, Figure M1495, olie here. 21 Prove that B <! for all > 2. With every permutatio of a set, there is associated the partitio of the set ito disjoit cycles of the permutatio (see page 30. Every partitio arises from a permutatio i this way: tae a cyclic permutatio of each part of the partitio, ad compose them. So the umber! of permutatios is at least as great as the umber B of partitios. The partitio with a sigle part is associated with the cyclic permutatios, of which there are ( 1! (Exercise 8; this umber is greater tha 1 if 3, so B <! for > Verify, theoretically or practically, the followig algorithm for geeratig all partial permutatios of {1,...,}: ( Algorithm: Partial permutatios of {1,..., } FIRST PARTIAL PERMUTATION is the empty sequece. NEXT OBJECT after (x 1,...,x m : If the legth m of the curret sequece is less tha, exted it by adjoiig the least elemet it does t cotai. Otherwise, proceed as i the algorithm for permutatios, up to the poit where x j ad x are iterchaged; the, istead of reversig the terms after x j, remove them from the sequece. Use iductio o : chec by had that the algorithm is correct for small. (For example, whe 3, it geerates i tur /0, 1, 12, 123, 13, 132, 2, 21, 213, 23, 231, 3, 31, 312, 32, 321, ad the termiates. Whe it reaches i, it the geerates i order all partial permutatios which begi with i (sice the same algorithm is beig applied to the last 1 positios usig the symbols differet from i, ad the proceeds to i

12 23 Verify the followig recursive procedure for geeratig the set of partitios of a set X. ( Recursive algorithm: Partitios of X If X /0, the /0 is the oly partitio. If X /0, the select a elemet x X; geerate all subsets of X \ {x}; for each subset Y, geerate all partitios of X \ ({x} Y, ad adjoi to each the additioal part {x} Y. This algorithm is just the costructive form of the proof of the recurrece ( Let A (a i j ad B (b i j be ( + 1 ( + 1 matrices (with rows ad colums idexed from 0 to defied by a i j ( i j, bi j ( 1 i+ j( i j (where 0 if i < j. Prove that B A 1. ( i j The vector (x 0,...,x represets (with respect to the basis cosistig of powers of t the polyomial f (t x( i t i. Now let f (t + 1 x i (t + 1 i y i t i. By the Biomial Theorem, y j i i j x j ; so the trasformatio f (t f (t + 1 is represeted by the matrix A. Similarly, the trasformatio f (t f (t 1 is represeted by B. But these trasformatios are obviously iverses of each other. A direct proof would ru as follows. We are required to show that j ( i j ( j ( 1 j { 1 if i 0 if i. Now ( i j 0 uless j i, ad ( j 0 uless j; so the sum is zero if i >, ad is oe if i (sice the oly the term j i is o-zero. Try to prove that the sum is zero for i <. (Your coclusio may well be that this is harder tha the coceptual proof outlied above. No solutios will be give for the projects. A crib for 25 is I. P. Goulde ad D. M. Jacso, Combiatorial Eumeratio, Joh Wiley & Sos, New Yor, I 27, the limitig ratio was show by A. Réyi to be e: see Publ. Math. Ist. Hugar. Acad. Sci. 4 (1959,