Universal Cycles. Yu She. Wirral Grammar School for Girls. Department of Mathematical Sciences. University of Liverpool

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Universal Cycles. Yu She. Wirral Grammar School for Girls. Department of Mathematical Sciences. University of Liverpool"

Transcription

1 Univesal Cycles 2011 Yu She Wial Gamma School fo Gils Depatment of Mathematical Sciences Univesity of Livepool Supeviso: Pofesso P. J. Giblin

2 Contents 1 Intoduction 2 2 De Buijn sequences and Euleian Gaphs Constucting a De Buijn Sequence Euleian Gaphs Fleuy s Algoithm The numbe of De Buijn sequences Constuction using Modula Aithmetic Applications of De Buijn sequences Pemutations Univesal Cycles fo Pemutations Multiplying Univesal Cycles 13 5 Patitions of a Set Univesal Cycles of Patitions of a Set Bell Numbes The Exponential Geneating Function of Bell Numbes Stiling Numbes of the Second Kind Patitions of a numbe 26 7 Acknowledgements 31 1

3 1 Intoduction De Buijn sequences ae sequences whee each possible binay / tenay / quatenay... sequence of length n appeas exactly once. Univesal cycles ae genealisations of De Buijn sequences to othe combinatoial stuctues such as pemutations and patitions of a set. Although univesal cycles do not exist fo patitions of a numbe, this was an inteesting extension to the poject. 2 De Buijn sequences and Euleian Gaphs Fo a given alphabet whee each digit can take k diffeent values, a De Buijn sequence B(k,n) is a sequence of numbes in which evey possible set of n digits appeas exactly once. Example In the case of a binay De Buijn sequence of ode 3 (meaning each subsequence has length 3) whee each digit is eithe 0 o 1: The 8 possible sets of 3 digits ae: 000, 001, 011, 111, 110, 101, 010, 100 A De Buijn sequence B(2,3) would be: By unning a window of length 3 along the above sequence, we can see that each of the possible tiplets appeas exactly once (this includes going aound the cone at the end of the sequence to obtain the tiplets 010 and 100). A De Buijn sequence has length =k n as each digit can take k diffeent values and thee ae n digits in each set (called an n-tuple). Fo the example above, the sequence has length 2 3 = Constucting a De Buijn Sequence De Buijn sequences can be constucted fom diected Euleian Gaphs. Consideing the case k = 2 and n = 3: To find a De Buijn sequence of ode 3, we wite down all the possible binay sets of length 2 and use them as the vetices of the gaph. 00, 01, 11, 10 We daw aows fom a vetex to a second vetex when the second digit of the fist vetex is the same as the fist digit of the second vetex. These aows can also stat and finish at the same vetex, as in the case of the vetices 00 and 11 as the second digit of both these vetices ae the same as the fist digit. By labelling an aow AB (tavelling fom vetex A to vetex B) with the digits in vetex A as well as the last digit of vetex B o the fist digit of vetex A as well as the digits in vetex B (the same sequence of digits) and finding an Euleian cycle aound the gaph, a De Buijn sequence can be found. Fo any vetex A, thee ae 2 aows leaving the vetex because the sequence is binay, meaning thee ae 2 choices fo the next digit in the sequence (i.e. 0 and 1), which will be satisfied by joining vetex A to 2 othe vetices. Similaly, thee will be 2 aows enteing vetex A because thee ae 2 choices fo the pevious digit in the sequence (i.e. 0 and 1), which will be satisfied by joining 2 othe vetices to vetex A. Theefoe, the numbe of aows enteing the vetex equals the numbe of aows leaving the vetex the vetices ae all even. We can wite out all the binay 3-tuples by witing out each binay 2-tuple and adding the digit 0 to the end, then witing out each binay 2-tuple again and adding the digit 1 to the end. This is effectively the same pocess as the one used when dawing and labelling the aows of the gaph, so each of the aows is distinct and epesents a unique binay 3-tuple. As a esult of the way we 2

4 Figue 1: Digaph fo the constuction of B(2,3) have constucted the gaph, the last 2 digits of any aow enteing a given vetex ae the same as the fist 2 digits of any aow leaving the same vetex. To wite out a De Buijn sequence fom the gaph, we find a oute which tavels along each aow exactly once and etuns to the stating vetex (this is called an Euleian cicuit). We wite out the labels on each aow as we tavel along it, deleting the 2 digits fom the end of the fist aow that ae epeated at the stat of the second aow. Fo the gaph above, a suitable oute would be: 000, 001, 011, 111, 110, 101, 010, 100. This gives the De Buijn sequence B(2,3): To find a De Buijn sequence of ode 4, we epeat the above pocess using the binay sets of length 3 as the vetices of the gaph and labelling the aows between vetices with sequences of 4 digits. Thee ae still 2 paths enteing and 2 paths leaving each vetex, but the vetices of the gaph fo B(2,4) now have the same labels as the aows of the gaph fo B(2,3). This new gaph N is fomed by doubling the oiginal gaph N. In this case, an Euleian cicuit fo the gaph would be: 0000, 0001, 0011, 0111, 1111, 1110, 1100, 1001, 0010, 0101, 1011, 0110, 1101, 1010, 0100, This gives the De Buijn sequence B(2,4): Similaly, the simplest gaph fo De Buijn sequences fo k = 2 (when n = 2) has only 2 vetices, with each vetex labelled with just 1 digit i.e. one is labelled 0 and the othe 1. In this case thee is only one Euleian cicuit (if the vetex fom which we stat does not matte): 00, 01, 11, 10. This leads to the (only) De Buijn sequence B(2,2): 0011 This method can also be used when k 2. Fo k = 3 and n = 3, the vetices of the gaph ae all the possible combinations of 2 digits whee each digit can take one of the 3 values 0, 1 o 2. Thee ae 3 2 = 9 such combinations so the gaph has 9 vetices. The aows ae dawn and labelled in the same way as befoe and an Euleian cicuit aound the gaph is found. The De Buijn sequence in this case would have length 3 3 = 27. An Euleian cicuit would be: 011, 111, 112, 121, 210, 101, 012, 120, 201, 010, 102, 021, 211, 110, 100, 002, 022, 221, 212, 122, 222, 220, 202, 020, 200, 000, 001. This gives the De Buijn sequence B(3,3):

5 Figue 2: Digaph fo the constuction of B(2,4) Figue 3: Digaph fo the constuction of B(2,2) A gaph to find B(k,n) can be constucted in the same way fo any k and n values and will always have the following popeties: k n 1 vetices k n aows k aows enteing each vetex (in degee = k) k aows leaving each vetex (out degee = k) 4

6 Figue 4: Digaph fo the constuction of B(3,3) each vetex is even 2.2 Euleian Gaphs An Euleian gaph is a gaph in which an Euleian cicuit can be found. To ensue that this method woks fo De Buijn sequences of any k and n values, we must pove that an Euleian cicuit can always be found fo the gaphs we use to constuct De Buijn sequences. Theoem 1 Let D be a connected digaph. Then D is Euleian if and only if the out degee of each vetex equals the in degee. Poof 1. If D is Euleian, the out degee of each vetex equals the in degee. If D has an Euleian cicuit, we can tavel along the cicuit using each aow exactly once and etun to ou stating point. Wheneve we pass though a vetex of G thee is a contibution of 1 to both the in degee and the out degee of the vetex (meaning they ae the same) - this includes the initial vetex, which we etun to at the end of the cicuit. Since each in and out aow of D is used just once, the in and out degee of each vetex is the same. 2. If the out degee of each vetex equals the in degee, the gaph is Euleian. Conside a connected digaph whee each vetex is even: A 1 A 2 whee the in degee = out degee fo each vetex Stat at any vetex and follow diected paths to othe vetices. When a vetex A i is eached, thee ae 2 possible situations: 1. Thee is an aow out of A i, in which case follow this aow. 2. Thee is no edge out of A i as the in degee of each vetex equals the out degee, this vetex A i must have been eached befoe (i.e. the path stated at A i ) as we have followed all edges out of A i. This means a diected cycle C has been fomed in the gaph. 5

7 As thee ae a finite numbe of vetices, situation 2 must at some point aise, so it is always possible to find a diected cycle C in the gaph. Let D be a gaph whee the out degee and in degee of each vetex ae equal and the total numbe of aows = m. D contains a diected cycle C. When m = 0, D consists of one vetex only and it is theefoe Euleian. Assume tue fo any connected digaph whee m < n. Now conside a digaph with n aows. Delete the edges of the diected cycle C fom D. The esulting digaph H has m < n and evey vetex H has an in degee = out degee, theefoe each component of H is Euleian. By following the diected cycle C, taking Euleian tails fo the components of H when we meet them and etuning to cycle C, we can find an Euleian tail fo the gaph. This explains why De Buijn sequences always exist fo any k,n as they ae fomed fom Euleian cicuits of gaphs whee the vetices ae even and of ode n Fleuy s Algoithm Fleuy s algoithm is a way of finding an Euleian cicuit in an Euleian gaph. In the case of De Buijn sequences, this will be a diected gaph with aows instead of edges connecting the vetices. 1. Choose a stating vetex u. 2. At each stage, tavese any available edge, choosing a bidge (an edge whose emoval disconnects a vetex fom the est of the gaph) only if thee is no altenative. 3. Afte tavesing each edge, ease it (also ease any vetices of degee 0 which esult) and then tavese anothe available edge. 4. Stop when thee ae no moe edges - an Euleian cicuit has been found. 2.4 The numbe of De Buijn sequences Thee is moe than one De Buijn sequence when n > 2 o k > 2 and these can be found by taking diffeent Euleian cicuits in the appopiate digaph. The numbe of distinct binay sequences, whee the evese of a sequence is counted as a diffeent sequence but cyclic pemutations ae egaded as the same, is P n = 2 2n 1 n. Let the numbe of diffeent Euleian cicuits in an Euleian gaph N be denoted by N. Similaly, the numbe of Euleian cicuits in N (the doubled gaph of N) is denoted by N. It was poved by N.G. de Buijn in [2] that N = 2 m 1 N whee m is the ode of N (N has m vetices and 2m aows). This esult can be used to pove the numbe of De Buijn sequences fo any value of n when k = 2. Theoem 2 The numbe of binay De Buijn sequences P n = 2 2n 1 n Poof When n = 1, P 1 = 1(this is evident fom the gaph) Using the fomula: P 1 = = 2 0 = 1 Theefoe the fomula is tue fo n = 1. Assume tue fo n y. 6

8 We must now pove it is tue fo n = y + 1, i.e. P y+1 = 2 2y y 1 A De Buijn sequence of ode n can be found fom an Euleian gaph of ode 2 n 1, so fo a De Buijn sequence whee n = y + 1 we must conside an Euleian gaph of ode 2 y. N y 1 has ode 2 y 1 and can can be used to constuct a De Buijn sequence of ode y N y has ode 2 y and is theefoe the doubled vesion of N y 1 P = N y = N y 1 = 2 m 1 N y 1 = 2 2y y 1 y = 2 2 2y 1 y 1 = 2 2y y 1 We have shown that the fomula fo P is coect fo n = 1 and that if it is tue fo n = y, then it is also tue fo n = y + 1. Theefoe it is tue, by induction, fo all n 1. This esult was extended by T. van Aadenne-Ehenfest and N. G. de Buijn in [1] and thee is now a fomula fo the numbe of distinct De Buijn sequences fo any values of k and n: P = k! kn 1 n 2.5 Constuction using Modula Aithmetic An altenative way of finding a De Buijn sequence when k = 2 was given in [6]. Fo a given n and stating with a 1 = 2 n 1, a sequence of numbes can be geneated by epeatedly substituting the pevious numbe in the sequence into the fomula: If fo some i j, a i = 2a j, then in this case: a i+1 2a i (mod 2 n ) a i+1 2a i + 1 (mod 2 n ) This means that if the fist fomula gives a numbe aleady geneated, then add 1 to this numbe to obtain the next numbe in the sequence, substituting this new numbe back into the fist equation to obtain futhe numbes in the sequence. All these numbes of the sequence should be witten in binay fom (to base 2 with 3 digits - put 0s befoe the numbe if thee ae fewe than 3 digits) and consecutive numbes in the sequence will join togethe to fom a De Buijn sequence B(2,n). Fo n = 3, we have mod 2 3 = 8. This means that if the numbe 8, we take the emainde when it is divided by 8. Anothe way to think of this would be a clock with 8 hous on the face - when it is 9 o clock the clock face looks the same as it does when it is 1 o clock i.e. 9=1 (mod 8). a 1 = = 8 1 = 7 = 111(in binay fom) a 2 = 2 7 = 14 = 6 = 110 a 3 = 2 6 = 12 = 4 = 100 (hee we can do 2 6 instead 2 14 as it is simple and both calculations give the same answe mod 8) a 4 = 2 4 = 8 = 0 = 000 a 5 = 2 0 = 0 which has aleady been geneated a 5 = = 1 = 001 a 6 = 2 1 = 2 = 010 a 7 = 2 2 = 4 which has aleady been geneated a 7 = 4+1 = 5 = 101 a 8 = 2 5 = 10 = 2 which has aleady been geneated a 8 = = 3 = 011 We can now stop as we have obtained the fist 2 n numbes in the sequence needed to fom a De Buijn sequence. Putting these numbes togethe and omitting the ovelapping digits between consecutive numbes (including the last 2 digits which ovelap with the fist 2 digits) gives the sequence B(2, 3):

9 Why does this method wok? Effectively, fo a given n, we ae geneating the numbes 0 to 2 n 1 in a special ode which means that when they ae witten in binay fom, the fist n 1 digits of a numbe ae the same as the last n 1 digits of the numbe geneated befoe it. This popety must hold in ode fo the constuction of a De Buijn sequence to be possible. When using this method only the fist 2 n numbes need to be geneated as the De Buijn sequence will have length 2 n. Because we ae using modula aithmetic mod 2 n and do not allow epeated numbes to be counted in the sequence, this means the fist 2 n numbes geneated must be the numbes 0 to 2 n 1. Multiplying a i by 2 to find a i+1 shifts the last n 1 digits of a i one place to the left (like multiplying by 10 in base 10), so consecutive numbes geneated by the fist fomula always ovelap by n 1 digits. The fist fomula alway esults in a numbe ending in 0 so to obtain altenative endings i.e. a last digit of 1 we add 1 to 2a i (fomula 2). This does not affect the fist n 1 digits of the numbe, so it still ovelaps with the pevious numbe geneated. We must be caeful with the choice of a 1 because othewise we may need to add 1 twice consecutively to a i to get a numbe which has not peviously been geneated. This would mean the numbe geneated would not ovelap with the pevious numbe in the sequence. Let us conside the lagest numbe to be geneated: 2 n 1, which is a numbe consisting of n 1 s when witten in binay fom. If the intege 2 n 1 is not at the beginning, then it must be peceded by 2 n 1 1 (binay fom: a 0 followed by n 1 1 s) and 2 n 2 (binay fom: n 1 1 s followed by a 0 ) must have aleady occued in the sequence so that we add 1 using fomula 2. Howeve, the numbe 2 n 2 must be peceded by eithe 2 n 1 o 2 n 1 1. This means that fo this method of constuction to wok, we must have a 1 = 2 n 1 o a 1 = 2 n 2. If we wee to stat with the second option fo a 1 athe than the fist, we would obtain the same sequence of numbes, except that the numbe 2 n 1 would be the last numbe geneated as opposed to the fist. Both values fo a 1 theefoe give the same De Buijn cycle. Fo n = 4, we have mod 2 4 = 16. a 1 = = 16 1 = 15 = 1111 a 2 = 2 15 = 30 = 14 = 1110 a 3 = 2 14 = 28 = 12 = 1100 a 4 = 2 12 = 24 = 8 = 1000 a 5 = 2 8 = 16 = 0 = 0000 a 6 = 2 0 = 0 a 6 = = 1 = 000 a 7 = 2 1 = 2 = 0010 a 8 = 2 2 = 4 = 0100 a 9 = 2 4 = 8 a 9 = = 9 = 1001 a 1 0 = 2 9 = 18 = 2 a 1 0 = = 3 = 001 a 11 = 2 3 = 6 = 0110 a 12 = 2 6 = 12 a 12 = = 13 = 110 a 13 = 2 13 = 26 = 10 = 1010 a 14 = 2 10 = 20 = 4 a 14 = = 5 = 010 a 15 = 2 5 = 10 a 15 = 10+1 = 11 = 1011 a 16 = 2 11 = 22 = 6 a 16 = 6+1 = 7 = 011 We now have the 2 n numbes needed to fom a De Buijn sequence B(2,4): Does this method wok fo k 3? To see if modula aithmetic could be used to constuct De Buijn sequences with lage alphabets i.e. k 3, I tied using this method to constuct B(3,2) by stating with a 1 = 3 n 1, witing numbes geneated in the sequence in tenay fom and adapting the pevious fomulae: If fo some i j, a i = 3a j, then in this case: a i+1 3a i (mod 3 n ) a i+1 3a i + 1 (mod 3 n ) 8

10 Fo k = 3,n = 2, we have mod 3 2 = 9. a 1 = = 9 1 = 8 = 22 a 2 = 3 8 = 24 = 6 = 20 a 3 = 3 6 = 18 = 0 = 00 a 4 = 3 0 = 0 a 4 = = 1 = 01 a 5 = 3 1 = 3 = 10 a 6 = 3 3 = 9 = 0 a 6 = = 1 a 6 = = 2 = 02 a 7 = 3 2 = 6 a 7 = = 7 = 21 a 8 = 3 7 = 21 = 3 a 8 = = 4 = 11 a 9 = 3 4 = 12 = 3 a 9 = = 4 a 9 = = 5 = 12 We now have the 3 n numbes needed to fom a De Buijn sequence B(3,2): Theefoe the method has been successful in poducing a De Buijn sequence when k 2. Note how in this example, we wee able to add 1 twice consecutively, wheeas in the pevious example we only added 1 once to obtain a paticula numbe in the sequence. This is because in this example we wote the numbes in base 3, so could add 1 o 2 afte multiplying by 3 without changing the fist 2 digits of the final numbe geneated. This meant that the fist 2 digits of this numbe would still ovelap with the final 2 digits of the pevious numbe in the sequence. Adding 3 in this situation would have changed the second digit in the sequence, meaning it would no longe fully ovelap with the pevious numbe (a popety equied fo the fomation of a De Buijn sequence), but thee was no need to do this. Remak 3 I believe that this method could be extended fo all values of k, with a 1 = k n 1 and: a i+1 ka i (mod k n ) Unless fo some i j, a i = ka j, then in this case: a i+1 ka i + 1 (mod k n ) Howeve, a disadvantage of this method is that only one De Buijn sequence fo a given n and k can be fomed, wheeas the method of constuction using Euleian gaphs can be used to find all possible distinct De Buijn cycles by finding all the diffeent Euleian cicuits aound the gaph. 2.6 Applications of De Buijn sequences De Buijn sequences have many uses, including in the positioning of obots. If a obot is moving along a tack maked with a De Buijn sequence e.g.b(2,3), then by looking at the neaest 3 numbes in the sequence, the obot can detemine its location on the tack as each tiplet is unique. Euleian gaphs wee used to solve the famous poblem the Seven Bidges of Konigsbeg. The Russian city of Konigsbeg is located on the Rive Pegel. The city has 2 lage islands, which wee connected to the mainland aea by 7 bidges. The people of the city often wondeed whethe it was possible to coss each of the 7 bidges exactly once on a single oute, but Leonhad Eule poved that this was impossible. By dawing each land mass as a vetex of a gaph and the bidges as edges of the gaph, he showed that the esulting gaph was not Euleian and so cossing each bidge exactly once was not possible. De Buijn sequences can be used to minimise the effot needed to guess a code in locks that do not have an ente key but instead accept the last n digits enteed. Fo example, to ty all the possible combinations of a 4 digit PIN like code, a De Buijn sequence B(10,4) could be used. This univesal cycle would have length 10 4 and enteing digits in the ode given by the cycle would equie only = pesses, wheeas tying all the possible combinations fo the code sepaately would take = pesses. 9

11 De Buijn sequences also have applications in geneating sequences in DNA, with an alphabet consisting of the 4 types of nucleotide which make up DNA: adenine(a), thymine(t), cytosine(c), guanine(g). The uses of De Buijn sequences in cad ticks ae given in the section on multiplying univesal cycles. 3 Pemutations When consideing pemutations, it is the elative size of each digit which we ae concened with athe than the actual value of a digit. Fo example, fo pemutations of 3 digits, we can think of each of the 3 digits as being low (L), medium (M) o high (H). In this case the numbe of pemutations would be 3! = 6 and can be expessed by the lettes above as LMH, LHM, MLH, MHL, HLM, HML, which coespond to the numbe pemutations 123, 132, 213, 231, 312, 321. Fo pemutations of n objects, thee ae n! distinct pemutations as we have a choice of n values fo the fist digit, n 1 values fo the second digit and so on until thee is only one possible numbe left fo the nth digit. 3.1 Univesal Cycles fo Pemutations We say that the n-tuples a and b ae ode-isomophic, witten ā b if a i < a j b i < b j. Univesal cycles fo pemutations ae cycles of length n! whee each of the n! pemutations of n distinct integes is ode-isomophic to exactly one n-tuple in the cycle. That is to say, any n consecutive digits in the cycle will have a distinct elative size ode. A univesal cycle fo pemutations of n digits whee n 3 will consist of at least n + 1 digits. This is because if, fo example, we ty constucting a univesal cycle fo n = 3 using just 3 digits: 123 the next digit must be 1 as each symbol in a window of length 3 must be diffeent 1231 the next digit must be the next digit must be 3, but this means we have epeated the fist window of 3 symbols, so this is not a univesal cycle. The same poblem aises if we ty constucting a univesal cycle fo any n 3 with just n symbols the sequence always ends up epeating itself, so this cannot be done. It was poved in [5] that fo n 3, it is always possible to find a univesal cycle fo pemutations using exactly n + 1 symbols. The method of constuction of univesal cycles fo pemutations is simila to that used to fom De Buijn sequences as gaphs ae used. This time we use the n! pemutations of n integes as the vetices of the gaph. This gaph is called a tansition gaph. Next we wok out which vetices ae connected by dawing aows fom each vetex to the vetices whee the last n 1 digits of the oiginal vetex and the fist n 1 digits of the second vetex ae ode-isomophic. When n = 3, thee ae 3! = 6 pemutations and so the gaph has 6 vetices. We can stat with any 3 numbes (even non-integes) as we ae only concened with the elative size of the integes. e.g. 679, We conside the possible size of the next digit by looking at the last n 1 = 3 1 = 2 digits: 79x. x could lie in n = n = 3 diffeent anges: 1. x < 7 < 9 79x < x < 9 79x < 9 < x 79x 123 This means thee ae n aows leaving and n aows enteing each vetex and a total of n n! aows 10

12 in the tansition gaph Figue 5: Tansition gaph fo n = 3 To fom a univesal cycle, we would need to take a Hamiltonian cicuit in the gaph (a path which visits each vetex exactly once, etuning to the stating vetex). Howeve, Hamiltonian cicuits ae difficult to find, especially fo moe complicated gaphs and thee is no way of knowing if such a path exists without tying out all the possible paths. To avoid this poblem, we convet the tansition gaph to an Euleian gaph by gouping togethe pemutations whee the fist n 1 digits ae ode-isomophic into new lage vetices. Thee should be (n 1)! vetices in the new gaph Figue 6: Euleian gaph fo n = 3 In the Euleian gaph, an aow fom a pemutation to a vetex means that it is possible to tavel fom that pemutation to all the pemutations in the vetex. Next we find an Euleian cicuit in the gaph, e.g. 231, 312, 123, 132, 321, 213, 231. Clealy, the maximum numbe of symbols we would need fo a sequence of length n! is n! and fo this example we will use the 3! = 6 lettes: abcdef. By unning windows of length n along the sequence of lettes and compaing this to the pemutations of length 3 in the Euleian cicuit, inequalities can be witten down to compae the elative sizes of the lettes. abc : 231 c < a < b bcd : 312 c < d < b cde : 123 c < d < e def : 132 d < f < e efa : 321 a < f < e fab : 213 a < f < b Combining these inequalities togethe gives: a and d can be combined and b and e can be combined whilst still satisfying all the inequalities, 11

13 c a d f b e so only 4 symbols wee needed fo the pemutation cycle (which is the least possible numbe as we need at least n + 1 = = 4 symbols). Finally we wite down the numbes coesponding to the sequence abcdef to obtain the univesal cycle fo pemutations of length 3: This combining of lettes is not always possible and sometimes we ae foced to use moe than n+1 symbols. Fo the Euleian cicuit: 312, 231, 213, 123, 132, 321, 312 we have the inequalities: abc : 312 b < c < a bcd : 231 d < b < c cde : 213 d < c < e def : 123 d < e < f efa : 132 e < a < f fab : 321 b < a < f Combining these inequalities togethe gives: d < b < c < e < a < f so in this case none of the lettes can be combined and we get the univesal cycle: Remak 4 I found seveal Euleian cicuits in the gaph and conveted these into univesal cycles fo pemutations when n = 3 and an inteesting patten began to emege. As has aleady been mentioned, the numbe of symbols used in a univesal cycles fo n = 3 is a minimum of 4 and a maximum of 6. Howeve, I noticed that thee did not seem to be any cycles consisting of exactly 5 diffeent symbols. Looking at this poblem in moe detail, I found that the Euleian cicuits which led to a cycle using 4 symbols had 4 of what I thought of as good connections and cicuits with 4 good connections always led to a cycle with 4 symbols. These good connections occued when the final 2 digits of the fist pemutation wee exactly the same as the fist 2 digits of the second pemutation as opposed to being just ode-isomophic. These good connections seemed to lead to fewe symbols being needed fo the cycle as it meant some of the 6 lettes initially used could be combined. I also found that the Euleian cicuits which led to a cycle using 6 symbols had only 1 good connection and cicuits with 1 good connection always led to a cycle with 6 symbols. I thought that pehaps Euleian cicuits with 2 o 3 good connections would lead to a univesal cycle with 5 symbols. Howeve, I was unable to find such a cicuit in the gaph and theefoe think that pehaps it is not possible to obtain a 5 symbol cycle fo pemutations (obviously this excludes the case whee one of the numbe 4s in a cycle with 4 symbols is simply changed to a bigge numbe). In the tansition gaph below, the good connections ae those with dashed aows. This method can be used to constuct univesal cycles fo pemutations of 4 o moe objects. Fo n = 4, the tansition gaph has 3! = 6 vetices and the pemutations whee the fist n 1 = 4 1 = 3 digits ae ode-isomophic ae gouped in the same vetex. Aows ae dawn in the same way as befoe, with a single aow to a vetex epesenting paths to all the pemutations in that vetex. In this case, an Euleian cicuit would be: 1234, 2341, 2314, 3142, 1423, 4231, 3421, 3214, 2143, 1432, 12

14 , 4213, 2134, 1243, 2431, 4312, 3124, 1342, 2413, 4132, 1324, 3241, 3412, 4123, Using this to wite inequalities gives: a b c d e g h I j l m o n q s u v w a b c f k p t x 5 This in tun gives the univesal cycle: This cycle using the smallest possible numbe of diffeent symbols fo n = 4: 5. [5] gave an altenative method of constucting univesal cycles fo pemutations which guaanteed that the esulting cycle had the smallest possible numbe (n + 1) of diffeent symbols. This involves constucting sub cycles whee the pemutations had what I peviously called good connections. These sub cycles ae then connected togethe to fom a cycle of n! pemutations which can then be conveted into a univesal cycle of length n!. 4 Multiplying Univesal Cycles Multiplying togethe 2 univesal cycles x and y which have lengths R and S espectively is possible when one of the cycles with window length k finishes with a block of k epeated symbols. The 13

15 method fo doing this was given in [4]. The 2 univesal cycles being multiplied togethe need not have the same window length but in most cases they do. Multiplying these cycles togethe gives a cycle of pais of length RS, with values fom x on the top ow and values of y on the bottom. If both univesal cycles have window length k, each k consecutive pais in the poduct cycle will be unique (i.e. a unique combination of the top and bottom ows). The cycles multiplied togethe can be of diffeent types, e.g. a De Buijn sequence can be multiplied with a univesal cycle fo pemutations. When a pemutation cycle fo 3 objects: is multiplied by a De Buijn sequence fo binay stings of length 3: , we get: Running a window of length 3 along the poduct cycle gives a unique coupling of a low, medium, high pemutation with a binay sting of length 3. This poduct cycle can be used in cad ticks. It has a length of 48 so the 4 kings must be emoved fom a deck of 52 cads. Next, the pemutation cycle fo window length k on the top ow of the poduct is lifted so that it contains 6 distinct symbols athe than just 4 (lifting is the pocess of inceasing the alphabet of a univesal cycle whilst etaining the cycle s popeties). This is done by finding the highest digit (it may appea seveal times) and changing one appeaance of this digit to the numbe k!. Then we take the next highest digit (not including the k! digit) and change it to the digit k! 1. This pocess continues until thee ae as many distinct digits in the cycle as equied. Fo example, a univesal cycle fo pemutations of 3 objects is: Woking fom left to ight in the case of the same digit appeaing moe than once, we get: (which now contains 6 diffeent digits). We now look at 2 adjacent copies of this cycle: Each digit 1 can be assigned eithe 1 o 2, each digit n can be assigned eithe the numbe 2n-1 o the numbe 2n up until the digit 6 can be assigned eithe 11 o 12. Choosing the lowe numbe fist in each case gives: (We can change whethe we choose the lowe o highe of the 2 numbes to make the sequence look moe andom.) In the poduct cycle above, this patten is epeated 4 times so each of the 12 cads in each suit is included exactly once, with ed cads (Heats and Diamonds) fo 1 s in the De Buijn sequence on the bottom ow and black cads (Clubs and Spades) fo 0 s in the bottom ow. An example of a sequence which could be used fo a cad tick would be: D H H S D C C C D D D S H S S C H D H C D S S D D H C H C C S D H H S H C S C D H H S D S C C S By asking people to choose 3 adjacent cads in the sequence and finding out which cads ae ed, as well as which cads ae the highest and lowest, we can deduce the value and suit of each of the 3 cads chosen by efeing to the oiginal poduct sequence and then the lifted sequence. 14

16 Conside 2 cycles x and y of lengths R and S. If R and S ae elatively pime (have a highest common facto of 1), then to multiply the 2 cycles, simply wite down x S times and below wite down y R times. Fo example, to multiply a univesal cycle fo patitions of set of 4 numbes (x): daabbbbcbccbadb (R = 15) with a De Buijn sequence fo binay stings of length 4 (y): (S = 16), we simply wite out x 16 times and y 15 times as 15 and 16 ae co-pime. This gives the poduct consisting of = 240 pais: d a a b b b b c b c c b a d b d a a b b b b c b c c b a d b Running a window of length 4 along this poduct gives a unique patition of a set of 4 elements plus binay sting of length 4 combination. If R and S ae not elatively pime, and thei highest common facto is d (d 1), then we have: R = d and S = sd whee and s ae co-pime. To obtain a cycle of RS pais: 1. wite down x S times 2. wite down y times 3. emove the last epeated digit in the epeated y sequence, foming the sequence y 4. wite down y a total of d times below the epeated sequence of x 5. add d of the digit emoved to the end of the epeated y sequence This gives a sequence of length: (S 1) d + d = Sd d + d = Sd = RS as equied. Fo example, to multiply a De Buijn sequence fo binay stings of length 2: 0011 by itself, R = S = 4, d = 4, R = S = 1 d. Fist, we wite down x 4 times: Next, wite down y once and emove the last 1 to fom y : Finally, wite down y 4 times below the epeated sequence of x and add 4 1 s at the end to give the poduct sequence: Running a window of length 2 along the poduct gives 16 unique 2 2 squaes. This pocess can be continued and the poduct cycle can be multiplied by anothe univesal cycle, fo example multiplying the De Buijn sequence fo n = 1: 01 by itself gives:

17 Multiplying the poduct by 01 again gives: Multiplying the poduct by 01 again gives: and so on. 5 Patitions of a Set A patition of any set A consisting of n elements {... n} is fomed when A is divided into subsets which ae mutually exclusive (have no elements in common) and collectively contain all the elements in the set. Each subset must be non-empty (must contain at least 1 element fom the set). A patition of a set can be expessed by using vetical bas to show the sepaation of elements of the set into subsets. The subsets ae independent of ode, meaning the ode in which the sepaate subsets of a patition ae witten and the ode of the elements within a subset do not matte. Fo example, the set A{} has 15 patitions, which can be witten as: The numbe of patitions of a set with n elements is given by the Bell numbes B n. 5.1 Univesal Cycles of Patitions of a Set Univesal cycles of patitions of a set with n elements can be fomed. These univesal cycles ae vey diffeent fom those fomed fo pemutations and De Buijn sequences. They usually consist of a sequence of B n lettes. When a window of length n is moved along the sequence, we numbe each lette in the window 1 to n fom left to ight. If the lettes of 2 o moe numbes ae the same, this means these numbes ae in the same subset in the patition. This means that if 2 numbes coespond to diffeent lettes in the sequence, these 2 numbes ae in sepaate subsets. The method used to constuct these univesal cycles is simila to that used to poduce univesal cycles of pemutations and De Buijn sequences as we must again use Euleian gaphs. Fist, we constuct a tansition gaph by witing down the patitions of the set as the vetices of the gaph. Aows ae dawn fom one vetex to anothe vetex when the elationship between the numbes 2 to n of the fist vetex is the same as the elationship between the numbes 1 to n 1 in the second vetex (in tems of whethe o not they ae in the same subset of the patition). 16

18 Fo the case whee n = 3, we ae constucting a cycle fo patitions of the set {1 2 3}. Thee ae 5 patitions of this set: These patitions ae the 5 vetices of the tansition gaph. The vetex 123 would be epesented by the lettes aaa as all 3 numbes ae in the same patition and when dawing aows fom this vetex we look at the the numbes 2 to n = 3 which ae epesented by aa. This can be followed by eithe aaa o aab which coespond to the vetices 123 and 12 3 so aows can be dawn fom the oiginal vetex to these vetices. It is impotant to note that the actual lettes that epesent a patition ae not of geat impotance - it is whethe these lettes ae the same o diffeent that mattes. Fo example, an aow could be dawn fom (abc) to 1 23 (abb) Figue 7: Tansition gaph fo patitions of a set with 3 elements This tansition gaph is then conveted to anothe gaph by gouping togethe patitions whee the fist n 1 numbes have the same elationship to fom a lage single vetex. This means the new gaph will have B n 1 vetices. Once again an aow fom a patition to a vetex means that it is possible to tavel fom that pemutation to all the pemutations in the vetex Figue 8: Euleian gaph fo patitions of a set with 3 elements Next, we must find an Euleian cicuit in this gaph. Fo n = 3 the only 2 possible cicuits ae: 123, 3 12, 2 13, 1 2 3, 1 23, 123 and 123, 3 12, 1 2 3, 2 13, 1 23,

19 Finally we need to lift the Euleian cicuit to a univesal cycle. This is done by assigning x Bn symbols to the numbes in the patitions and witing equalities to show whethe o not these symbols ae equal (i.e. whethe they ae in the same subset of a patition). Fo each patition, the numbes ae eaanged in inceasing ode, then we look at whethe o not thei coesponding symbols should be equal. x 1 x 2 x 3 x 4 x 5 x 1 x x 1 x 2 x 3 x 4 x 5 x 1 x Figue 9: Lifting Euleian cicuits fo n = 3 Fo the fist Euleian cicuit, we obtain the following set of equalities: x 1 = x 2 = x 3 x 2 = x 3 x 4 x 3 = x 5 x 4 x 4 x 5 x 5 x 1 x 1 x 4 x 1 = x 2 x 5 Howeve, this leads to a contadiction, as we can deduce that x 5 x 1 = x 3 = x 5. Theefoe we conclude that this paticula Euleian cicuit cannot be lifted successfully to fom a univesal cycle. Fo the second Euleian cicuit, we obtain the following set of equalities: x 1 = x 2 = x 3 x 2 = x 3 x 4 x 3 x 4 x 4 x 5 x 5 x 3 x 4 = x 1 x 5 x 1 = x 2 x 5 Howeve, this also leads to a contadiction, as we can deduce that x 4 = x 1 = x 3 x 4. We conclude that this Euleian cicuit cannot fom a univesal cycle. As thee ae only 2 Euleian cicuits fo the gaph of patitions of a set of 3 elements and we have shown that neithe can be used to fom a univesal cycle, we have poved by exhaustion that no univesal cycles exist fo patitions of a set of 3 elements. Univesal cycles fo patitions of a set of 4 elements do exist and can be constucted by skipping the tansition gaph and dawing the Euleian gaph staight away and combining the patitions in which the numbes 1 to n 1 = 4 1 = 3 have the same elationship into a single lage vetex. Dawing aows between the vetices in the same way as in the pevious example gives an Euleian gaph with 5 vetices and 15 aows. In [3] it was stated that in ode to pevent equalities leading to a contadiction and to guaantee an Euleian cicuit can be lifted to fom a univesal cycle, a sequence of patitions called a beake must occu in the Euleian cicuit. Using the beake they gave: , 12 34, 1 234, 1234, 4 123, I found an Euleian cicuit fo the gaph and used it to poduce a univesal cycle fo n = 4. The Euleian cicuit was: 18

20 Figue 10: Euleian gaph fo patitions of a set with 4 elements , 12 34, 1 234, 1234, 4 123, 3 124, 13 24, 2 134, 14 23, , , , , , , x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 x 1 x 2 x Figue 11: Lifting an Euleian cicuit fo n = 4 This gave the set of equalities: x 1 x 2 = x 3 x 4 x 1 x 4 x 2 = x 3 x 4 = x 5 x 3 x 4 = x 5 = x 6 x 4 = x 5 = x 6 = x 7 x 5 = x 6 = x 7 x 8 x 6 = x 7 = x 9 x 8 x 7 = x 9 x 8 = x 10 x 8 = x 10 = x 11 x 9 19

21 x 9 = x 12 x 10 = x 11 x 10 = x 11 x 12 x 13 x 10 = x 11 x 13 x 11 x 12 x 13 x 14 x 11 x 13 x 11 x 14 x 12 x 14 x 12 = x 15 x 13 x 14 x 12 = x 15 x 14 x 13 x 14 = x 1 x 15 x 13 x 15 x 14 = x 1 x 15 x 2 x 14 = x 1 x 2 x 2 = x 3 x 15 x 1 x 2 = x 3 x 1 I assigned a lette to each goup of x symbols which wee equal: a : x 2 = x 3 b : x 4 = x 5 = x 6 = x 7 = x 9 = x 12 = x 15 c : x 8 = x 10 = x 11 d : x 14 = x 1 e : x 13 I found that e could be combined with a whilst still satisfying all the equalities. This meant that the univesal cycle fo n = 4 would only contain 4 lettes as opposed to 5. Finally, I wote out the lettes coesponding to each x symbol when the x symbols wee aanged in ode. This gave the univesal cycle: daabbbbcbccbadb. Remak 5 Based on the popeties of the beake given in [3], I tied to constuct a diffeent beake fo n = 4. The oiginal beake has the popeties: x 1 x 2 x 3 x 4 x 5 x 6 x 7 x Figue 12: Popeties of the oiginal beake x 1 x 2 = x 3 x 4 x 1 x 4 x 2 = x 3 x 4 = x 5 x 3 x 4 = x 5 = x 6 x 4 = x 5 = x 6 = x 7 x 5 = x 6 = x 7 x 8 In [3], it is claimed that the inclusion of such a beake in an Euleian cicuit will always lead to a cicuit which can be lifted to fom a univesal cycle. The eason given fo this is that a x symbol occuing befoe x 1 e.g. x 15 cannot be foced to be eithe equal o unequal to an x symbol afte x 7. This is tue when looking at the beake alone. Howeve, the patitions fom a cicuit, so when the full cicuit of patitions ae witten out, it will be possible to deduce whethe o not most, if not all the x symbols ae equal o unequal. Fo example, in the constuction of a univesal cycle fo n = 4 above, it was possible to deduce fom the full set of equalities that x 15 x 8. Theefoe it would seem that this claim is not tue. 20

22 I constucted the following beake which has the same popeties as the oiginal beake, except the equalities ae evesed: 1 234, 1234, 4 123, 12 34, x 1 x 2 x 3 x 4 x 5 x 6 x 7 x Figue 13: Popeties of diffeent beake x 1 x 2 = x 3 = x 4 x 2 = x 3 = x 4 = x 5 x 3 = x 4 = x 5 x 6 x 4 = x 5 x 6 = x 7 x 5 x 6 = x 7 x 8 x 5 x 8 This beake was successful when used in the Euleian cicuit: 1 234, 1234, 4 123, 12 34, , 3 124, 13 24, , , 2 134, 14 23, , , , , giving the univesal cycle: daaaabbcbcaccab. Howeve, when put in the Euleian cicuit: 1 234, 1234, 4 123, 12 34, , , , , , , , 14 23, 3 124, 13 24, 2 134, this cicuit could not be lifted successfully. x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 x 1 x 2 x Figue 14: Failue using the new beake Which gave the equalities: x 1 x 2 = x 3 = x 4 21

23 x 2 = x 3 = x 4 = x 5 x 3 = x 4 = x 5 x 6 x 4 = x 5 x 6 = x 7 x 5 x 6 = x 7 x 8 x 5 x 8 x 6 = x 7 x 8 x 9 x 6 = x 7 x 9 x 7 = x 10 x 8 x 9 x 7 = x 10 x 9 x 8 x 9 x 10 x 11 x 8 x 10 x 8 x 11 x 9 x 11 x 9 x 10 = x 12 x 11 x 9 x 11 x 10 = x 12 x 11 x 13 x 10 = x 12 x 13 x 13 = x 14 x 11 x 12 x 13 = x 14 x 12 x 12 = x 15 x 13 = x 14 x 13 = x 14 = x 1 x 15 x 14 = x 1 x 15 = x 2 x 1 x 15 = x 2 = x 3 Howeve, this leads to the contadiction that x 5 x 6 = x 7 = x 10 = x 12 = x 15 = x 2 = x 3 = x 4 = x 5. We conclude that this Euleian cicuit cannot fom a univesal cycle. The evesal of equalities compaed to the oiginal beake should not affect the beake s effectiveness, yet in this case it failed to wok. This casts futhe doubt on the effectiveness of beakes and it would seem that beakes need to be defined in moe detail to guaantee they always wok. 5.2 Bell Numbes We aleady know that the numbe of patitions of a set with n elements is given by the Bell numbes B n. Conside a set of n + 1 numbes: { n+1}. If n + 1 is on its own in a subset, thee ae B n ways to patition the emaining n elements in the set. This can be witten as: ( n) 0 Bn as by definition ( n 0) is equal to 1. If n + 1 is with 1 othe element in a subset, thee ae ( n 1) = n ways of choosing this othe element fom the n available elements in the set and thee ae B n 1 ways of patitioning the emaining n 1 elements in the set. This means the numbe of patitions which contain such a subset can be witten as: ( n 1) Bn 1. If n + 1 is with 2 othe elements in a subset, thee ae ( n 2) ways of choosing the othe 2 elements and thee ae B n 2 ways of patitioning the emaining n 2 elements in the set. This means the numbe of patitions which contain such a subset can be witten as: ( n 2) Bn 2.. Thee ae ( n n 1) B1 patitions which contain a subset whee n + 1 is with n 1 othe elements in a subset. Thee ae ( n n) B0 patitions which contain a subset whee n+1 is with n othe elements in a subset. (By definition, B 0 = 1, so thee is only one patition containing all the elements in a single subset.) Theefoe the Bell numbes satisfy the ecuence elation: ( B n+1 = n =0 ( ) n B = n due to the symmety of combinations: n =0 ( ) n n ( ) n B Using this fomula, we can calculate the fist few Bell numbes: B 0 = 1 22 = n!!(n )! = ( )) n

24 B 1 = B 0 = 1 B 2 = B 0 + B 1 = = 2 B 3 = B 0 + 2B 1 + B 2 = = = 5 B 4 = B 0 + 3B 1 + 3B 2 + B 3 = = = 15 B 5 = B 0 + 4B 1 + 6B 2 + 4B 3 + B 4 = = = 52 and so on The Bell tiangle can be used to geneate Bell numbes (like Pascal s tiangle can be used to geneate binomial coefficients). B 0 B 1 B 2 B 3 B 4 B The fist tem in each ow is the same as the last tem in the pevious ow Each tem (apat fom the fist in evey ow) is obtained by adding the pevious tem in the ow and the tem above the pevious tem Figue 15: Finding Bell numbes using the Bell tiangle 5.3 The Exponential Geneating Function of Bell Numbes A geneating function is a fomal powe seies whee the coefficients of x n give infomation about the numbes in a sequence a n an infinite seies whee the vaiable x is geneally egaded as a place holde athe than assigned an actual value. Geneating functions ae vey useful as they can epesent sequences as functions and can be used to solve counting poblems. Thee ae 2 main types of geneating functions: odinay geneating functions and exponential geneating functions. The odinay geneating function G(x) fo an infinite sequence (a 0,a 1,a 2,a 3...) would be: G(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3... The exponential geneating function E(x) fo the same sequence would be: E(x) = a 0 + a 1 1! x + a 2 2! x2 + a 3 3! x3... In this case, thee is an exponential geneating function fo the Bell numbes B n. Theoem 6 The exponential geneating function of the Bell numbes is e ex 1, i.e. the coefficient of xn n! in the powe seies expansion of e ex 1 is the numbe of patitions of a set of n elements. 23

25 Poof We aleady know the ecuence elation fo Bell numbes: B n+1 = n =0 ( ) n B B n = n 1 =0 ( ) n 1 To find B 1 (which is the coefficient of x), we diffeentiate the exponential geneating function y = E(x) and take the value of the function when x = 0. The kth Bell numbe B k should equal the kth deivative y (k) of e ex 1 when x = 0. When n = 1, we know that B 1 = 1 y = e ex 1 lny = ln(e ex 1 ) lny = (e x 1) 1 1 y dy dx = ex dy dx = yex When x = 0, dy dx = 1 eex e x = e e0 1 e 0 = e 0 e 0 = 1 1 = 1 tue fo n = 1 Assume tue fo n = k, i.e. ( ) k 1 y (k) = e x k 1 y () =0 hee e x = 1 as we put x = 0 to obtain the Bell numbes We need to pove it is tue fo n = k + 1, i.e. ( ) k k y (k+1) = e x y () =0 B y (k+1) = d ( dx k 1 e x =0 ( ) k 1 )y () ( ) ( ) k 1 = e x k 1 k 1 y () + e x k 1 y (+1) =0 =0 [( ) ( ) ( ) ( ] k 1 k 1 = e x y (0) k 1 k 1 k 2 + y () + y (k) k 1 + )y (+1) 0 k 1 =1 =0 ( ) ( ] k 1 = e [y x (0) k 1 k 1 + y () k 1 + )y () + y (k) 1 =1 =1 [( ) ( } k 1 = e {y x (0) k 1 k )]y () + y (k) 1 =1 24

26 ( ] k 1 = e [y x (0) k + )y () + y (k) =1 ( ) k k = e x y () =0 We have shown that the esult is tue fo n = 1 and that if it is tue fo n = k then it is also tue fo n = k + 1. Theefoe it is tue, by induction, fo all n 1. Poof *Hee the popety used was: = ( ) k = ( ) k ( ) k ( ) k 1 ( ) k 1 (k 1)! ( 1)!(k )! + (k 1)!!(k 1)! = (k 1)! (k 1)!(k ) +!(k )!!(k )! = (k + )(k 1)!!(k )! = k(k 1)!!(k )! k! =!(k )! ( ) k = 5.4 Stiling Numbes of the Second Kind A second way of finding the numbe patitions of a set ae using the Stiling numbes of the second kind. { n } k = the numbe of ways of patitioning a set with n elements into k non-empty subsets. Think of an element n in a set with n elements: If n is on its own, the numbe of patitions = { n 1 k 1}. If n is not on its own, it is pat of one of the k subsets fomed fom n 1 elements, the numbe of patitions = k { n 1} k. Theefoe the Stiling numbes of a second kind have the popety: { } { } { } n n 1 n 1 = k + k k k 1 { } n n B n = k k=0 25

27 6 Patitions of a numbe A patition of a numbe is a way of epesenting a non negative intege n as the sum of positive integes called pats, with these pats witten in non inceasing ode. e.g. n = a + b + c whee a b c The patition function p(n) is the numbe of distinct ways of witing n as the sum of positive integes, whee the ode of the pats does not matte. To find p(n), we can use the geneating function of patitions of a numbe: k=1 1 1 x k Using the binomial expansion, this can be witten as: (1 + x + x 2 + x 3...)(1 + x 2 + x 4 + x 6...)(1 + x 3 + x 6 + x 9...)... This expansion is only valid when 1 < x < 1, but because we do not usually assign values to x in geneating functions, we do not need to woy about the issue of convegence. The coefficient of x k is the numbe of patitions of the numbe k. Patitions can be epesented by Fees diagams which show the pats of each patition. Fo example, the Fees diagams below show the 22 patitions fo when k= The above patitions can be categoised: Figue 16: Patitions of 8 (i) The pats ae all odd i.e. each pat of the patition is an odd numbe: 7+1, 5+3, , , , Thee ae 6 such patitions. 26

Symmetric polynomials and partitions Eugene Mukhin

Symmetric polynomials and partitions Eugene Mukhin Symmetic polynomials and patitions Eugene Mukhin. Symmetic polynomials.. Definition. We will conside polynomials in n vaiables x,..., x n and use the shotcut p(x) instead of p(x,..., x n ). A pemutation

More information

Week 3-4: Permutations and Combinations

Week 3-4: Permutations and Combinations Week 3-4: Pemutations and Combinations Febuay 24, 2016 1 Two Counting Pinciples Addition Pinciple Let S 1, S 2,, S m be disjoint subsets of a finite set S If S S 1 S 2 S m, then S S 1 + S 2 + + S m Multiplication

More information

The Binomial Distribution

The Binomial Distribution The Binomial Distibution A. It would be vey tedious if, evey time we had a slightly diffeent poblem, we had to detemine the pobability distibutions fom scatch. Luckily, thee ae enough similaities between

More information

UNIT CIRCLE TRIGONOMETRY

UNIT CIRCLE TRIGONOMETRY UNIT CIRCLE TRIGONOMETRY The Unit Cicle is the cicle centeed at the oigin with adius unit (hence, the unit cicle. The equation of this cicle is + =. A diagam of the unit cicle is shown below: + = - - -

More information

Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360!

Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360! 1. What ae angles? Last time, we looked at how the Geeks intepeted measument of lengths. Howeve, as fascinated as they wee with geomety, thee was a shape that was much moe enticing than any othe : the

More information

2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses,

2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses, 3.4. KEPLER S LAWS 145 3.4 Keple s laws You ae familia with the idea that one can solve some mechanics poblems using only consevation of enegy and (linea) momentum. Thus, some of what we see as objects

More information

Episode 401: Newton s law of universal gravitation

Episode 401: Newton s law of universal gravitation Episode 401: Newton s law of univesal gavitation This episode intoduces Newton s law of univesal gavitation fo point masses, and fo spheical masses, and gets students pactising calculations of the foce

More information

Algebra and Trig. I. A point is a location or position that has no size or dimension.

Algebra and Trig. I. A point is a location or position that has no size or dimension. Algeba and Tig. I 4.1 Angles and Radian Measues A Point A A B Line AB AB A point is a location o position that has no size o dimension. A line extends indefinitely in both diections and contains an infinite

More information

The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges

The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges The foce between electic chages Coulomb s Law Two chaged objects, of chage q and Q, sepaated by a distance, exet a foce on one anothe. The magnitude of this foce is given by: kqq Coulomb s Law: F whee

More information

Chapter 3 Savings, Present Value and Ricardian Equivalence

Chapter 3 Savings, Present Value and Ricardian Equivalence Chapte 3 Savings, Pesent Value and Ricadian Equivalence Chapte Oveview In the pevious chapte we studied the decision of households to supply hous to the labo maket. This decision was a static decision,

More information

Saturated and weakly saturated hypergraphs

Saturated and weakly saturated hypergraphs Satuated and weakly satuated hypegaphs Algebaic Methods in Combinatoics, Lectues 6-7 Satuated hypegaphs Recall the following Definition. A family A P([n]) is said to be an antichain if we neve have A B

More information

An Introduction to Omega

An Introduction to Omega An Intoduction to Omega Con Keating and William F. Shadwick These distibutions have the same mean and vaiance. Ae you indiffeent to thei isk-ewad chaacteistics? The Finance Development Cente 2002 1 Fom

More information

Spirotechnics! September 7, 2011. Amanda Zeringue, Michael Spannuth and Amanda Zeringue Dierential Geometry Project

Spirotechnics! September 7, 2011. Amanda Zeringue, Michael Spannuth and Amanda Zeringue Dierential Geometry Project Spiotechnics! Septembe 7, 2011 Amanda Zeingue, Michael Spannuth and Amanda Zeingue Dieential Geomety Poject 1 The Beginning The geneal consensus of ou goup began with one thought: Spiogaphs ae awesome.

More information

Semipartial (Part) and Partial Correlation

Semipartial (Part) and Partial Correlation Semipatial (Pat) and Patial Coelation his discussion boows heavily fom Applied Multiple egession/coelation Analysis fo the Behavioal Sciences, by Jacob and Paticia Cohen (975 edition; thee is also an updated

More information

Nontrivial lower bounds for the least common multiple of some finite sequences of integers

Nontrivial lower bounds for the least common multiple of some finite sequences of integers J. Numbe Theoy, 15 (007), p. 393-411. Nontivial lowe bounds fo the least common multiple of some finite sequences of integes Bai FARHI bai.fahi@gmail.com Abstact We pesent hee a method which allows to

More information

Continuous Compounding and Annualization

Continuous Compounding and Annualization Continuous Compounding and Annualization Philip A. Viton Januay 11, 2006 Contents 1 Intoduction 1 2 Continuous Compounding 2 3 Pesent Value with Continuous Compounding 4 4 Annualization 5 5 A Special Poblem

More information

CHAPTER 10 Aggregate Demand I

CHAPTER 10 Aggregate Demand I CHAPTR 10 Aggegate Demand I Questions fo Review 1. The Keynesian coss tells us that fiscal policy has a multiplied effect on income. The eason is that accoding to the consumption function, highe income

More information

Power and Sample Size Calculations for the 2-Sample Z-Statistic

Power and Sample Size Calculations for the 2-Sample Z-Statistic Powe and Sample Size Calculations fo the -Sample Z-Statistic James H. Steige ovembe 4, 004 Topics fo this Module. Reviewing Results fo the -Sample Z (a) Powe and Sample Size in Tems of a oncentality Paamete.

More information

Displacement, Velocity And Acceleration

Displacement, Velocity And Acceleration Displacement, Velocity And Acceleation Vectos and Scalas Position Vectos Displacement Speed and Velocity Acceleation Complete Motion Diagams Outline Scala vs. Vecto Scalas vs. vectos Scala : a eal numbe,

More information

8-1 Newton s Law of Universal Gravitation

8-1 Newton s Law of Universal Gravitation 8-1 Newton s Law of Univesal Gavitation One of the most famous stoies of all time is the stoy of Isaac Newton sitting unde an apple tee and being hit on the head by a falling apple. It was this event,

More information

1.4 Phase Line and Bifurcation Diag

1.4 Phase Line and Bifurcation Diag Dynamical Systems: Pat 2 2 Bifucation Theoy In pactical applications that involve diffeential equations it vey often happens that the diffeential equation contains paametes and the value of these paametes

More information

Financing Terms in the EOQ Model

Financing Terms in the EOQ Model Financing Tems in the EOQ Model Habone W. Stuat, J. Columbia Business School New Yok, NY 1007 hws7@columbia.edu August 6, 004 1 Intoduction This note discusses two tems that ae often omitted fom the standad

More information

1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2

1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2 Chapte 5 Example The helium atom has 2 electonic enegy levels: E 3p = 23.1 ev and E 2s = 20.6 ev whee the gound state is E = 0. If an electon makes a tansition fom 3p to 2s, what is the wavelength of the

More information

Questions & Answers Chapter 10 Software Reliability Prediction, Allocation and Demonstration Testing

Questions & Answers Chapter 10 Software Reliability Prediction, Allocation and Demonstration Testing M13914 Questions & Answes Chapte 10 Softwae Reliability Pediction, Allocation and Demonstation Testing 1. Homewok: How to deive the fomula of failue ate estimate. λ = χ α,+ t When the failue times follow

More information

2. SCALARS, VECTORS, TENSORS, AND DYADS

2. SCALARS, VECTORS, TENSORS, AND DYADS 2. SCALARS, VECTORS, TENSORS, AND DYADS This section is a eview of the popeties of scalas, vectos, and tensos. We also intoduce the concept of a dyad, which is useful in MHD. A scala is a quantity that

More information

On Some Functions Involving the lcm and gcd of Integer Tuples

On Some Functions Involving the lcm and gcd of Integer Tuples SCIENTIFIC PUBLICATIONS OF THE STATE UNIVERSITY OF NOVI PAZAR SER. A: APPL. MATH. INFORM. AND MECH. vol. 6, 2 (2014), 91-100. On Some Functions Involving the lcm and gcd of Intege Tuples O. Bagdasa Abstact:

More information

Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review

Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review Vecto Calculus: Ae you eady? Vectos in D and 3D Space: Review Pupose: Make cetain that you can define, and use in context, vecto tems, concepts and fomulas listed below: Section 7.-7. find the vecto defined

More information

Hour Exam No.1. p 1 v. p = e 0 + v^b. Note that the probe is moving in the direction of the unit vector ^b so the velocity vector is just ~v = v^b and

Hour Exam No.1. p 1 v. p = e 0 + v^b. Note that the probe is moving in the direction of the unit vector ^b so the velocity vector is just ~v = v^b and Hou Exam No. Please attempt all of the following poblems befoe the due date. All poblems count the same even though some ae moe complex than othes. Assume that c units ae used thoughout. Poblem A photon

More information

FXA 2008. Candidates should be able to : Describe how a mass creates a gravitational field in the space around it.

FXA 2008. Candidates should be able to : Describe how a mass creates a gravitational field in the space around it. Candidates should be able to : Descibe how a mass ceates a gavitational field in the space aound it. Define gavitational field stength as foce pe unit mass. Define and use the peiod of an object descibing

More information

STUDENT RESPONSE TO ANNUITY FORMULA DERIVATION

STUDENT RESPONSE TO ANNUITY FORMULA DERIVATION Page 1 STUDENT RESPONSE TO ANNUITY FORMULA DERIVATION C. Alan Blaylock, Hendeson State Univesity ABSTRACT This pape pesents an intuitive appoach to deiving annuity fomulas fo classoom use and attempts

More information

Revision Guide for Chapter 11

Revision Guide for Chapter 11 Revision Guide fo Chapte 11 Contents Student s Checklist Revision Notes Momentum... 4 Newton's laws of motion... 4 Gavitational field... 5 Gavitational potential... 6 Motion in a cicle... 7 Summay Diagams

More information

Graphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities.

Graphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities. Gaphs of Equations CHAT Pe-Calculus A coodinate sstem is a wa to gaphicall show the elationship between quantities. Definition: A solution of an equation in two vaiables and is an odeed pai (a, b) such

More information

Experiment MF Magnetic Force

Experiment MF Magnetic Force Expeiment MF Magnetic Foce Intoduction The magnetic foce on a cuent-caying conducto is basic to evey electic moto -- tuning the hands of electic watches and clocks, tanspoting tape in Walkmans, stating

More information

est using the formula I = Prt, where I is the interest earned, P is the principal, r is the interest rate, and t is the time in years.

est using the formula I = Prt, where I is the interest earned, P is the principal, r is the interest rate, and t is the time in years. 9.2 Inteest Objectives 1. Undestand the simple inteest fomula. 2. Use the compound inteest fomula to find futue value. 3. Solve the compound inteest fomula fo diffeent unknowns, such as the pesent value,

More information

Converting knowledge Into Practice

Converting knowledge Into Practice Conveting knowledge Into Pactice Boke Nightmae srs Tend Ride By Vladimi Ribakov Ceato of Pips Caie 20 of June 2010 2 0 1 0 C o p y i g h t s V l a d i m i R i b a k o v 1 Disclaime and Risk Wanings Tading

More information

4a 4ab b 4 2 4 2 5 5 16 40 25. 5.6 10 6 (count number of places from first non-zero digit to

4a 4ab b 4 2 4 2 5 5 16 40 25. 5.6 10 6 (count number of places from first non-zero digit to . Simplify: 0 4 ( 8) 0 64 ( 8) 0 ( 8) = (Ode of opeations fom left to ight: Paenthesis, Exponents, Multiplication, Division, Addition Subtaction). Simplify: (a 4) + (a ) (a+) = a 4 + a 0 a = a 7. Evaluate

More information

Coordinate Systems L. M. Kalnins, March 2009

Coordinate Systems L. M. Kalnins, March 2009 Coodinate Sstems L. M. Kalnins, Mach 2009 Pupose of a Coodinate Sstem The pupose of a coodinate sstem is to uniquel detemine the position of an object o data point in space. B space we ma liteall mean

More information

On Correlation Coefficient. The correlation coefficient indicates the degree of linear dependence of two random variables.

On Correlation Coefficient. The correlation coefficient indicates the degree of linear dependence of two random variables. C.Candan EE3/53-METU On Coelation Coefficient The coelation coefficient indicates the degee of linea dependence of two andom vaiables. It is defined as ( )( )} σ σ Popeties: 1. 1. (See appendi fo the poof

More information

Concept and Experiences on using a Wiki-based System for Software-related Seminar Papers

Concept and Experiences on using a Wiki-based System for Software-related Seminar Papers Concept and Expeiences on using a Wiki-based System fo Softwae-elated Semina Papes Dominik Fanke and Stefan Kowalewski RWTH Aachen Univesity, 52074 Aachen, Gemany, {fanke, kowalewski}@embedded.wth-aachen.de,

More information

Economics 326: Input Demands. Ethan Kaplan

Economics 326: Input Demands. Ethan Kaplan Economics 326: Input Demands Ethan Kaplan Octobe 24, 202 Outline. Tems 2. Input Demands Tems Labo Poductivity: Output pe unit of labo. Y (K; L) L What is the labo poductivity of the US? Output is ouhgly

More information

The LCOE is defined as the energy price ($ per unit of energy output) for which the Net Present Value of the investment is zero.

The LCOE is defined as the energy price ($ per unit of energy output) for which the Net Present Value of the investment is zero. Poject Decision Metics: Levelized Cost of Enegy (LCOE) Let s etun to ou wind powe and natual gas powe plant example fom ealie in this lesson. Suppose that both powe plants wee selling electicity into the

More information

Carter-Penrose diagrams and black holes

Carter-Penrose diagrams and black holes Cate-Penose diagams and black holes Ewa Felinska The basic intoduction to the method of building Penose diagams has been pesented, stating with obtaining a Penose diagam fom Minkowski space. An example

More information

Over-encryption: Management of Access Control Evolution on Outsourced Data

Over-encryption: Management of Access Control Evolution on Outsourced Data Ove-encyption: Management of Access Contol Evolution on Outsouced Data Sabina De Capitani di Vimecati DTI - Univesità di Milano 26013 Cema - Italy decapita@dti.unimi.it Stefano Paaboschi DIIMM - Univesità

More information

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing

More information

Skills Needed for Success in Calculus 1

Skills Needed for Success in Calculus 1 Skills Needed fo Success in Calculus Thee is much appehension fom students taking Calculus. It seems that fo man people, "Calculus" is snonmous with "difficult." Howeve, an teache of Calculus will tell

More information

Things to Remember. r Complete all of the sections on the Retirement Benefit Options form that apply to your request.

Things to Remember. r Complete all of the sections on the Retirement Benefit Options form that apply to your request. Retiement Benefit 1 Things to Remembe Complete all of the sections on the Retiement Benefit fom that apply to you equest. If this is an initial equest, and not a change in a cuent distibution, emembe to

More information

INITIAL MARGIN CALCULATION ON DERIVATIVE MARKETS OPTION VALUATION FORMULAS

INITIAL MARGIN CALCULATION ON DERIVATIVE MARKETS OPTION VALUATION FORMULAS INITIAL MARGIN CALCULATION ON DERIVATIVE MARKETS OPTION VALUATION FORMULAS Vesion:.0 Date: June 0 Disclaime This document is solely intended as infomation fo cleaing membes and othes who ae inteested in

More information

Voltage ( = Electric Potential )

Voltage ( = Electric Potential ) V-1 Voltage ( = Electic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage is

More information

An Immunological Approach to Change Detection: Algorithms, Analysis and Implications

An Immunological Approach to Change Detection: Algorithms, Analysis and Implications An Immunological Appoach to Change Detection: Algoithms, Analysis and Implications Patik D haeselee Dept. of Compute Science Univesity of New Mexico Albuqueque, NM, 87131 patik@cs.unm.edu Stephanie Foest

More information

Reduced Pattern Training Based on Task Decomposition Using Pattern Distributor

Reduced Pattern Training Based on Task Decomposition Using Pattern Distributor > PNN05-P762 < Reduced Patten Taining Based on Task Decomposition Using Patten Distibuto Sheng-Uei Guan, Chunyu Bao, and TseNgee Neo Abstact Task Decomposition with Patten Distibuto (PD) is a new task

More information

Physics 505 Homework No. 5 Solutions S5-1. 1. Angular momentum uncertainty relations. A system is in the lm eigenstate of L 2, L z.

Physics 505 Homework No. 5 Solutions S5-1. 1. Angular momentum uncertainty relations. A system is in the lm eigenstate of L 2, L z. Physics 55 Homewok No. 5 s S5-. Angula momentum uncetainty elations. A system is in the lm eigenstate of L 2, L z. a Show that the expectation values of L ± = L x ± il y, L x, and L y all vanish. ψ lm

More information

Ilona V. Tregub, ScD., Professor

Ilona V. Tregub, ScD., Professor Investment Potfolio Fomation fo the Pension Fund of Russia Ilona V. egub, ScD., Pofesso Mathematical Modeling of Economic Pocesses Depatment he Financial Univesity unde the Govenment of the Russian Fedeation

More information

MULTIPLE SOLUTIONS OF THE PRESCRIBED MEAN CURVATURE EQUATION

MULTIPLE SOLUTIONS OF THE PRESCRIBED MEAN CURVATURE EQUATION MULTIPLE SOLUTIONS OF THE PRESCRIBED MEAN CURVATURE EQUATION K.C. CHANG AND TAN ZHANG In memoy of Pofesso S.S. Chen Abstact. We combine heat flow method with Mose theoy, supe- and subsolution method with

More information

2. TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES

2. TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES . TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES In ode to etend the definitions of the si tigonometic functions to geneal angles, we shall make use of the following ideas: In a Catesian coodinate sstem, an

More information

The Role of Gravity in Orbital Motion

The Role of Gravity in Orbital Motion ! The Role of Gavity in Obital Motion Pat of: Inquiy Science with Datmouth Developed by: Chistophe Caoll, Depatment of Physics & Astonomy, Datmouth College Adapted fom: How Gavity Affects Obits (Ohio State

More information

Life Insurance Purchasing to Reach a Bequest. Erhan Bayraktar Department of Mathematics, University of Michigan Ann Arbor, Michigan, USA, 48109

Life Insurance Purchasing to Reach a Bequest. Erhan Bayraktar Department of Mathematics, University of Michigan Ann Arbor, Michigan, USA, 48109 Life Insuance Puchasing to Reach a Bequest Ehan Bayakta Depatment of Mathematics, Univesity of Michigan Ann Abo, Michigan, USA, 48109 S. David Pomislow Depatment of Mathematics, Yok Univesity Toonto, Ontaio,

More information

Mechanics 1: Work, Power and Kinetic Energy

Mechanics 1: Work, Power and Kinetic Energy Mechanics 1: Wok, Powe and Kinetic Eneg We fist intoduce the ideas of wok and powe. The notion of wok can be viewed as the bidge between Newton s second law, and eneg (which we have et to define and discuss).

More information

Software Engineering and Development

Software Engineering and Development I T H E A 67 Softwae Engineeing and Development SOFTWARE DEVELOPMENT PROCESS DYNAMICS MODELING AS STATE MACHINE Leonid Lyubchyk, Vasyl Soloshchuk Abstact: Softwae development pocess modeling is gaining

More information

How to create RAID 1 mirroring with a hard disk that already has data or an operating system on it

How to create RAID 1 mirroring with a hard disk that already has data or an operating system on it AnswesThatWok TM How to set up a RAID1 mio with a dive which aleady has Windows installed How to ceate RAID 1 mioing with a had disk that aleady has data o an opeating system on it Date Company PC / Seve

More information

Quantity Formula Meaning of variables. 5 C 1 32 F 5 degrees Fahrenheit, 1 bh A 5 area, b 5 base, h 5 height. P 5 2l 1 2w

Quantity Formula Meaning of variables. 5 C 1 32 F 5 degrees Fahrenheit, 1 bh A 5 area, b 5 base, h 5 height. P 5 2l 1 2w 1.4 Rewite Fomulas and Equations Befoe You solved equations. Now You will ewite and evaluate fomulas and equations. Why? So you can apply geometic fomulas, as in Ex. 36. Key Vocabulay fomula solve fo a

More information

Physics HSC Course Stage 6. Space. Part 1: Earth s gravitational field

Physics HSC Course Stage 6. Space. Part 1: Earth s gravitational field Physics HSC Couse Stage 6 Space Pat 1: Eath s gavitational field Contents Intoduction... Weight... 4 The value of g... 7 Measuing g...8 Vaiations in g...11 Calculating g and W...13 You weight on othe

More information

92.131 Calculus 1 Optimization Problems

92.131 Calculus 1 Optimization Problems 9 Calculus Optimization Poblems ) A Noman window has the outline of a semicicle on top of a ectangle as shown in the figue Suppose thee is 8 + π feet of wood tim available fo all 4 sides of the ectangle

More information

Fast FPT-algorithms for cleaning grids

Fast FPT-algorithms for cleaning grids Fast FPT-algoithms fo cleaning gids Josep Diaz Dimitios M. Thilikos Abstact We conside the poblem that given a gaph G and a paamete k asks whethe the edit distance of G and a ectangula gid is at most k.

More information

Model Question Paper Mathematics Class XII

Model Question Paper Mathematics Class XII Model Question Pape Mathematics Class XII Time Allowed : 3 hous Maks: 100 Ma: Geneal Instuctions (i) The question pape consists of thee pats A, B and C. Each question of each pat is compulsoy. (ii) Pat

More information

867 Product Transfer and Resale Report

867 Product Transfer and Resale Report 867 Poduct Tansfe and Resale Repot Functional Goup ID=PT Intoduction: This X12 Tansaction Set contains the fomat and establishes the data contents of the Poduct Tansfe and Resale Repot Tansaction Set (867)

More information

LINES AND TANGENTS IN POLAR COORDINATES

LINES AND TANGENTS IN POLAR COORDINATES LINES AND TANGENTS IN POLAR COORDINATES ROGER ALEXANDER DEPARTMENT OF MATHEMATICS 1. Pola-coodinate equations fo lines A pola coodinate system in the plane is detemined by a point P, called the pole, and

More information

Risk Sensitive Portfolio Management With Cox-Ingersoll-Ross Interest Rates: the HJB Equation

Risk Sensitive Portfolio Management With Cox-Ingersoll-Ross Interest Rates: the HJB Equation Risk Sensitive Potfolio Management With Cox-Ingesoll-Ross Inteest Rates: the HJB Equation Tomasz R. Bielecki Depatment of Mathematics, The Notheasten Illinois Univesity 55 Noth St. Louis Avenue, Chicago,

More information

VISCOSITY OF BIO-DIESEL FUELS

VISCOSITY OF BIO-DIESEL FUELS VISCOSITY OF BIO-DIESEL FUELS One of the key assumptions fo ideal gases is that the motion of a given paticle is independent of any othe paticles in the system. With this assumption in place, one can use

More information

Problem Set 6: Solutions

Problem Set 6: Solutions UNIVESITY OF ALABAMA Depatment of Physics and Astonomy PH 16-4 / LeClai Fall 28 Poblem Set 6: Solutions 1. Seway 29.55 Potons having a kinetic enegy of 5. MeV ae moving in the positive x diection and ente

More information

Lecture 16: Color and Intensity. and he made him a coat of many colours. Genesis 37:3

Lecture 16: Color and Intensity. and he made him a coat of many colours. Genesis 37:3 Lectue 16: Colo and Intensity and he made him a coat of many colous. Genesis 37:3 1. Intoduction To display a pictue using Compute Gaphics, we need to compute the colo and intensity of the light at each

More information

Performance Analysis of an Inverse Notch Filter and Its Application to F 0 Estimation

Performance Analysis of an Inverse Notch Filter and Its Application to F 0 Estimation Cicuits and Systems, 013, 4, 117-1 http://dx.doi.og/10.436/cs.013.41017 Published Online Januay 013 (http://www.scip.og/jounal/cs) Pefomance Analysis of an Invese Notch Filte and Its Application to F 0

More information

Define What Type of Trader Are you?

Define What Type of Trader Are you? Define What Type of Tade Ae you? Boke Nightmae srs Tend Ride By Vladimi Ribakov Ceato of Pips Caie 20 of June 2010 1 Disclaime and Risk Wanings Tading any financial maket involves isk. The content of this

More information

The Supply of Loanable Funds: A Comment on the Misconception and Its Implications

The Supply of Loanable Funds: A Comment on the Misconception and Its Implications JOURNL OF ECONOMICS ND FINNCE EDUCTION Volume 7 Numbe 2 Winte 2008 39 The Supply of Loanable Funds: Comment on the Misconception and Its Implications. Wahhab Khandke and mena Khandke* STRCT Recently Fields-Hat

More information

Introduction to Electric Potential

Introduction to Electric Potential Univesiti Teknologi MARA Fakulti Sains Gunaan Intoduction to Electic Potential : A Physical Science Activity Name: HP: Lab # 3: The goal of today s activity is fo you to exploe and descibe the electic

More information

PHYSICS 111 HOMEWORK SOLUTION #5. March 3, 2013

PHYSICS 111 HOMEWORK SOLUTION #5. March 3, 2013 PHYSICS 111 HOMEWORK SOLUTION #5 Mach 3, 2013 0.1 You 3.80-kg physics book is placed next to you on the hoizontal seat of you ca. The coefficient of static fiction between the book and the seat is 0.650,

More information

Uncertain Version Control in Open Collaborative Editing of Tree-Structured Documents

Uncertain Version Control in Open Collaborative Editing of Tree-Structured Documents Uncetain Vesion Contol in Open Collaboative Editing of Tee-Stuctued Documents M. Lamine Ba Institut Mines Télécom; Télécom PaisTech; LTCI Pais, Fance mouhamadou.ba@ telecom-paistech.f Talel Abdessalem

More information

Chris J. Skinner The probability of identification: applying ideas from forensic statistics to disclosure risk assessment

Chris J. Skinner The probability of identification: applying ideas from forensic statistics to disclosure risk assessment Chis J. Skinne The pobability of identification: applying ideas fom foensic statistics to disclosue isk assessment Aticle (Accepted vesion) (Refeeed) Oiginal citation: Skinne, Chis J. (2007) The pobability

More information

Japan s trading losses reach JPY20 trillion

Japan s trading losses reach JPY20 trillion IEEJ: Mach 2014. All Rights Reseved. Japan s tading losses each JPY20 tillion Enegy accounts fo moe than half of the tading losses YANAGISAWA Akia Senio Economist Enegy Demand, Supply and Foecast Goup

More information

Comparing Availability of Various Rack Power Redundancy Configurations

Comparing Availability of Various Rack Power Redundancy Configurations Compaing Availability of Vaious Rack Powe Redundancy Configuations By Victo Avela White Pape #48 Executive Summay Tansfe switches and dual-path powe distibution to IT equipment ae used to enhance the availability

More information

Multiple choice questions [70 points]

Multiple choice questions [70 points] Multiple choice questions [70 points] Answe all of the following questions. Read each question caefull. Fill the coect bubble on ou scanton sheet. Each question has exactl one coect answe. All questions

More information

Explicit, analytical solution of scaling quantum graphs. Abstract

Explicit, analytical solution of scaling quantum graphs. Abstract Explicit, analytical solution of scaling quantum gaphs Yu. Dabaghian and R. Blümel Depatment of Physics, Wesleyan Univesity, Middletown, CT 06459-0155, USA E-mail: ydabaghian@wesleyan.edu (Januay 6, 2003)

More information

An Efficient Group Key Agreement Protocol for Ad hoc Networks

An Efficient Group Key Agreement Protocol for Ad hoc Networks An Efficient Goup Key Ageement Potocol fo Ad hoc Netwoks Daniel Augot, Raghav haska, Valéie Issany and Daniele Sacchetti INRIA Rocquencout 78153 Le Chesnay Fance {Daniel.Augot, Raghav.haska, Valéie.Issany,

More information

Approximation Algorithms for Data Management in Networks

Approximation Algorithms for Data Management in Networks Appoximation Algoithms fo Data Management in Netwoks Chistof Kick Heinz Nixdof Institute and Depatment of Mathematics & Compute Science adebon Univesity Gemany kueke@upb.de Haald Räcke Heinz Nixdof Institute

More information

Problem Set # 9 Solutions

Problem Set # 9 Solutions Poblem Set # 9 Solutions Chapte 12 #2 a. The invention of the new high-speed chip inceases investment demand, which shifts the cuve out. That is, at evey inteest ate, fims want to invest moe. The incease

More information

CHAT Pre-Calculus Section 10.7. Polar Coordinates

CHAT Pre-Calculus Section 10.7. Polar Coordinates CHAT Pe-Calculus Pola Coodinates Familia: Repesenting gaphs of equations as collections of points (, ) on the ectangula coodinate sstem, whee and epesent the diected distances fom the coodinate aes to

More information

Top K Nearest Keyword Search on Large Graphs

Top K Nearest Keyword Search on Large Graphs Top K Neaest Keywod Seach on Lage Gaphs Miao Qiao, Lu Qin, Hong Cheng, Jeffey Xu Yu, Wentao Tian The Chinese Univesity of Hong Kong, Hong Kong, China {mqiao,lqin,hcheng,yu,wttian}@se.cuhk.edu.hk ABSTRACT

More information

Physics 235 Chapter 5. Chapter 5 Gravitation

Physics 235 Chapter 5. Chapter 5 Gravitation Chapte 5 Gavitation In this Chapte we will eview the popeties of the gavitational foce. The gavitational foce has been discussed in geat detail in you intoductoy physics couses, and we will pimaily focus

More information

Determining solar characteristics using planetary data

Determining solar characteristics using planetary data Detemining sola chaacteistics using planetay data Intoduction The Sun is a G type main sequence sta at the cente of the Sola System aound which the planets, including ou Eath, obit. In this inestigation

More information

Deflection of Electrons by Electric and Magnetic Fields

Deflection of Electrons by Electric and Magnetic Fields Physics 233 Expeiment 42 Deflection of Electons by Electic and Magnetic Fields Refeences Loain, P. and D.R. Coson, Electomagnetism, Pinciples and Applications, 2nd ed., W.H. Feeman, 199. Intoduction An

More information

Supplementary Material for EpiDiff

Supplementary Material for EpiDiff Supplementay Mateial fo EpiDiff Supplementay Text S1. Pocessing of aw chomatin modification data In ode to obtain the chomatin modification levels in each of the egions submitted by the use QDCMR module

More information

Do Vibrations Make Sound?

Do Vibrations Make Sound? Do Vibations Make Sound? Gade 1: Sound Pobe Aligned with National Standads oveview Students will lean about sound and vibations. This activity will allow students to see and hea how vibations do in fact

More information

Experiment 6: Centripetal Force

Experiment 6: Centripetal Force Name Section Date Intoduction Expeiment 6: Centipetal oce This expeiment is concened with the foce necessay to keep an object moving in a constant cicula path. Accoding to Newton s fist law of motion thee

More information

Mechanics 1: Motion in a Central Force Field

Mechanics 1: Motion in a Central Force Field Mechanics : Motion in a Cental Foce Field We now stud the popeties of a paticle of (constant) ass oving in a paticula tpe of foce field, a cental foce field. Cental foces ae ve ipotant in phsics and engineeing.

More information

Comparing Availability of Various Rack Power Redundancy Configurations

Comparing Availability of Various Rack Power Redundancy Configurations Compaing Availability of Vaious Rack Powe Redundancy Configuations White Pape 48 Revision by Victo Avela > Executive summay Tansfe switches and dual-path powe distibution to IT equipment ae used to enhance

More information

12. Rolling, Torque, and Angular Momentum

12. Rolling, Torque, and Angular Momentum 12. olling, Toque, and Angula Momentum 1 olling Motion: A motion that is a combination of otational and tanslational motion, e.g. a wheel olling down the oad. Will only conside olling with out slipping.

More information

Theory and practise of the g-index

Theory and practise of the g-index Theoy and pactise of the g-index by L. Egghe (*), Univesiteit Hasselt (UHasselt), Campus Diepenbeek, Agoalaan, B-3590 Diepenbeek, Belgium Univesiteit Antwepen (UA), Campus Die Eiken, Univesiteitsplein,

More information

PAN STABILITY TESTING OF DC CIRCUITS USING VARIATIONAL METHODS XVIII - SPETO - 1995. pod patronatem. Summary

PAN STABILITY TESTING OF DC CIRCUITS USING VARIATIONAL METHODS XVIII - SPETO - 1995. pod patronatem. Summary PCE SEMINIUM Z PODSTW ELEKTOTECHNIKI I TEOII OBWODÓW 8 - TH SEMIN ON FUNDMENTLS OF ELECTOTECHNICS ND CICUIT THEOY ZDENĚK BIOLEK SPŠE OŽNO P.., CZECH EPUBLIC DLIBO BIOLEK MILITY CDEMY, BNO, CZECH EPUBLIC

More information

ESCAPE VELOCITY EXAMPLES

ESCAPE VELOCITY EXAMPLES ESCAPE VELOCITY EXAMPLES 1. Escape velocity is the speed that an object needs to be taveling to beak fee of planet o moon's gavity and ente obit. Fo example, a spacecaft leaving the suface of Eath needs

More information

Lesson 7 Gauss s Law and Electric Fields

Lesson 7 Gauss s Law and Electric Fields Lesson 7 Gauss s Law and Electic Fields Lawence B. Rees 7. You may make a single copy of this document fo pesonal use without witten pemission. 7. Intoduction While it is impotant to gain a solid conceptual

More information

Unit Vectors. the unit vector rˆ. Thus, in the case at hand, 5.00 rˆ, means 5.00 m/s at 36.0.

Unit Vectors. the unit vector rˆ. Thus, in the case at hand, 5.00 rˆ, means 5.00 m/s at 36.0. Unit Vectos What is pobabl the most common mistake involving unit vectos is simpl leaving thei hats off. While leaving the hat off a unit vecto is a nast communication eo in its own ight, it also leads

More information