ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

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1 1. Graph Theory Prove that there exist o simple plaar triagulatio T ad two distict adjacet vertices x, y V (T ) such that x ad y are the oly vertices of T of odd degree. Do ot use the Four-Color Theorem. (Simple meas o loops or parallel edges.) Solutio: Let G be obtaied from T by deletig the edge xy. The every vertex of G has eve degree, ad hece the dual is bipartite. Thus the faces of G may be colored usig colors 1 ad 2 such that every edge is icidet with a face of either color. Let a be the umber of faces colored 1, ad let b be the umber of faces colored 2. From the symmetry we may assume that the uique face of G of size four is colored 1. Sice every face of G colored 2 is a triagle, we deduce that E(G) = 3b. Sice every face of G colored 1 is a triagle, except for precisely oe face of size four, we deduce that E(G) = 3(a 1) + 4. But the equatio 3a + 1 = 3b has o itegral solutio, ad hece o such triagulatio exists.

2 2. Probability Let X, X 1, X 2,... be idepedet idetically distributed radom variables. Deote S := X X. (a) If X 0 a.s. ad EX = +, the (b) If E X = +, the S + as a.s. S = + a.s. Solutio: (a) Deote X c j = X ji(x j c), c > 0. The, by the SLLN, for all c > 0, Hece implyig (a). (b) We have Hece, for all c > 0, S Xc Xc lim if + = E X = S sup c>0 + P{ X c} = =0 0 EX c 1 EX c 1 as a.s. P{ X t}dt c = EX = + a.s., P{ X c}. =0 P{ X c} = +. By Borel-Catelli lemma, this implies that for all c > 0 with probability 1 Therefore, =0 X c i.o.. X = + a.s. ( ) It follows from the zero-oe law that the probability of the evet { } S = + is either 0 or 1. If it is equal to 0, the with probability 1 the sequece {S /} is bouded implyig that the same is true for the sequece {S 1 /}, ad, as a cosequece, for {X /}. This cotradicts ( ). Therefore with probability 1. S = +

3 3. Aalysis of Algorithms Suppose we would like to fid a collectio of matchigs such that every edge of the graph is a member of (at least) oe of the matchigs we selected. The goal is to pick the miimum umber of matchigs for our collectio. Give a O(log )-approximatio for this problem that rus i polyomial time, where is the umber of vertices of the iput graph. Solutio: Let the m edges of the iput graph be the elemets of our uiverse, ad let each matchig of the graph be a subset of this uiverse. The miimum umber of matchigs for our collectio is the just a miimum set cover. The greedy algorithm gives a (log m + 1)-approximatio algorithm for set cover that proceeds by iteratively choosig the set that covers the maximum umber of elemets that have ot yet bee covered. We ca implemet this efficietly by fidig a maximum matchig o the thus far uvisited edges, takig O( 2 m) time to fid the matchig. Sice we remove at least oe edge i each step, the total ruig time is at most O( 2 m 2 ) ad this yields a O(log )-approximatio algorithm.

4 4. Theory of Liear Iequalities Let C R be a fiitely-geerated coe of full dimesio ad let H be a itegral Hilbert basis for C. Suppose w T x 0 is a facet-defiig iequality for C such that the compoets of w are relatively prime itegers. Show that there exists a vector h H such that w T h = 1. Solutio available upo request.

5 Cosider a assigmet problem (AP) 5. Combiatorial Optimizatio max i N j N c ijx ij j N x ij = 1 for i N i N x ij = 1 for j N x ij = 0 or 1 for i, j N where N = {1,..., }. Note: You should be able to aswer (1)-(5) quickly, which is prelimiary to (6). (6) is the part of this questio that couts the most. (1) State a property of the costrait matrix from which it ca be deduced that all of the extreme poits of the LP relaxatio are itegral. j J x ij where I, J N. Suppose I + J = + k, k 1. Prove that (2) Let x(i, J) = i I if x is a feasible solutio, the x(i, J) k. (3) Now cosider a costraied AP called CAP where we require x 2k 1,2k 1 x 2k,2k = 0 for k = 1,..., m. Suppose we have a fractioal poit say x = 1/2 0 1/2 1/2 1/ /2 1/2 for = 3, m = 1. What is required of such a poit to coclude that the LP relaxatio of CAP is ot itegral? Does this poit x suffice? (4) Let = 5, m = 2 i CAP. Prove that x 11 (x 34 + x 43 + x 44 ) 0 is a valid iequality. (5) Give I, J N with I + J = 1, let K = {2r 1 : {2r 1, 2r} I J, r m} ˆK = {2r 1 : {2r 1, 2r} (N\I) (N\J), r m}. Show that the iequality i (4) is of the form i K ˆK x ii x(i, J) 0 with I = J = {3, 4}. (6) Prove i geeral that ( ) is a valid iequality for CAP if I + J = 1 ad ˆK 1. (7) Describe how you would prove that ( ) with the coditio give i (6) is a facet of the covex hull of CAP. A actual proof is ot required. (8) Give the form of ( ) with the coditio give i (6) for m = 1. I this case, the assigmet costraits, side costraits, oegativity ad ( ) give the covex hull. Describe how you could prove this. A actual proof is ot required. (9) Suggest a idea for separatig ( ) with m = 1. Solutio ot yet available. ( )

6 6. Algebra Suppose G is a group of order 255. Prove that G is cyclic. (Hit: First show G has a ormal subgroup of order 17 ad that G has a ormal cyclic subgroup of order 85.) Solutio: Let 17 be the umber of Sylow 17 subgroup of G. We kow 17 divides 15 ad is cogruet to 1 mod 17. Thus 17 = 1 ad there is a uique Sylow 17 subgroup, call it P. Now G/P is a group of order 15. If we look at the umber of Sylow 5 subgroups of G/P we see this umber must divide 3 ad be cogruet to 1 mod 5. Thus there is just oe such subgroup Q ad it must be ormal. By the fourth isomorphism theorem we kow there is a subgroup Q i G such that Q/P = Q i G/P. This subgroup will be ormal ad of order 85. (You could alterately argue that if P was ay Sylow 5 subgroup of G the P P is a subgroup of G of order 85 ad sice its idex is the smallest prime dividig the order of G we ca coclude it is ormal.) Clearly P Q is ormal ad sice 5 + 1, ad does ot divide 17 the Sylow 5 subgroup R of Q is ormal i Q. So Q = P R, P R = {1} ad P ad Q are both ormal i Q. Thus Q = P Q ad sice P ad Q have prime orders we kow they are cyclic. Thus Q = (Z/17Z) (Z/5Z) = Z/85Z is a cyclic subgroup of G. Now if 3 is the umber of Sylow 3 subgroups the 3 divides 85 ad is cogruet to 1 mod 3. The oly way for this to happe is for 3 = 1 (sice 4, 7, 10, 13, 16 ad 19 do ot divide 85). Thus the Sylow 3 subgroup S of G is ormal i G. Sice G = QS, Q S = {1} ad Q ad S are ormal subgroups of G we see G = Q S = Z/255Z.

7 7. Radomized Algorithms Recall for a pair of distributios µ ad ν o a fiite set Ω, their variatio distace is d TV (µ, ν) = 1 µ(z) ν(z) 2 z Ω Cosider a ergodic Markov chai o state space Ω, trasitio matrix P ad uique statioary distributio π. Let P t (x, ) deote the t-step distributio of the Markov chai startig from x Ω. Recall the mixig time is defied to be T (ɛ) = max x Ω T x(ɛ) where T x (ɛ) = mi { t : d TV (P t (x, ), π) ɛ } For the purposes of this problem we cosider the followig otio of itersectio time. For x, y Ω, defie their itersectio time as T x,y = mi { t : d TV ( P t (x, ), P t (y, ) ) 1/2 } ad let T = max x,y Ω T x,y Prove that T (ɛ) T log(1/ɛ), where the log is base 2. Solutio. We will prove the iequality usig the couplig techique. For x Ω, cosider 2 copies (X t ) ad (Y t ) of the Markov chai of iterest where the iitial states are X 0 = x ad Y 0 is chose accordig to π. Thus Y t is distributed accordig to π for all t 0. The couplig is defied i segmets of T steps. From X t, Y t, we couple the 2 chais as follows: 1. Ru X t for T steps. 2. If X t = Y t, the we use the idetity couplig for these T steps, i.e., let Y t+1 = X t+1,..., Y t+t = X t+t. 3. Otherwise, idepedetly of what happeed for X t, ru Y t for T steps. 4. Repeat the process from X t+t, Y t+t. By the defiitio of T, from ay 2 iitial states, if the 2 chais ru idepedetly for T steps, the with probability 1/2 the 2 chais reach the same state. Thus, for ay iteger i 1, for all x, y Ω, we have P r(x it Y it X (i 1)T = x, Y (i 1)T = y ) 1/2 The couplig is defied so that if the 2 chais reach the same state at the ed of a roud of T steps the they always stay i the same state, therefore we have that P r(x it Y it ) (1/2) i By the couplig lemma, sice Y t is distributed accordig to π we have for t = it : d TV (P t (x, ), π) (1/2) i Settig i = log(1/ɛ), ad sice the above holds for all iitial states x, this proves the iequality.