Math778P Homework 2 Solution

Transcription

1 Math778P Homework Solutio Choose ay 5 problems to solve. 1. Let S = X i where X 1,..., X are idepedet uiform { 1, 1} radom variables. Prove that E( S = 1 ( 1 1 Proof by Day Rorabaugh: Let S = X i where X 1,..., X are idepedet uiform { 1, 1} radom variables. Prove that E( S = 1 ( 1. 1 E( S = = k= k Pr(S = k k (# of ways S = k k=0 = 1 k=0 = 1. ( k # of ways to have k 0 k k eve ( k k -1 s ad + k ( = 1 ( i i i=0 ( = 1! + (( i i 0 ( i!i! ( = 1! 1 + ( i 1!i!! ( i!(i 1! ( = 1 1 (( i ( 1 i 1 1 s. 1

2 [( ] 1 = 1 ( = 1 1 ( 1 ( 1 = Suppose that p > > 10m, with p prime, ad let 0 < a 1 < a < < a m < p be itegers. Prove that there is a iteger x, 0 < x < p for which the m umbers are pairwise distict. (xa i mod p mod, 1 i m Proof by Travis Johsto: Let x be chose uiformly at radom from []. First, ote that for each a i, xa i is ever 0 mod p sice x is ever 0 mod p; thus xa i mod p is oe of itegers. Cosider a umber j with 0 j <. Let us cout the maximum umber of values 0 i < p such that i j mod. Note that i j mod if i = j, or i = j +, or i = j +,..., or i = j + k where j + k < p ad j + (k + 1 p. Further, oe ca see that k = p ; hece there are at most p + 1 itegers i with 0 i < p such that i j mod. It follows that P((xa i mod p mod = (xa j mod p mod p + 1. Let X ij be the radom variable takig o the value 1 if (xa i mod p mod = (xa j mod p mod ad 0 otherwise. We the have that

3 the expected umber of o-distict pairs xa i, xa j is: E( i<j X ij = i<j E(X ij = P((xa i mod p mod = (xa j mod p mod i<j p + 1 i<j ( m p = + 1 ( p < = < p p 5 = p 5p 5 < 1 Thus, there exists some x such that the umber of o-distict pairs is smaller tha 1, i.e. all the pairs are distict. 3. Let H be a graph, ad let > V (H be a iteger. Suppose there is a graph o vertices ad t edges cotaiig o copy of H, ad suppose that tk > log e. Show that there is a colorig of edges of the complete graph o vertices by k colors with o moochromatic copy of H. Proof by Clifford Gaddy: Let G be the graph o vertices ad t edges cotaiig o copy of H. Create k copies of G, labeled G 1,..., G k. Color each with a differet color, c i for 1 i k. Now successively ad radomly lay G i o top of K. i.e. color a radom copy of G 1 i K with c 1. The idepedetly repeat the process for c. If a edge is radomly chose to be colored by c that has already bee colored by c 1, the recolor it with c. Repeat this process k times. If at the ed of the process we have maaged to color each edge, the we have colored K with k colors. Also ay moochromatic subgraph of K will be a subgraph of some G i, oe of which cotai a copy of 3

4 H. Thus there ca be o moochromatic copy of H i our colorig. So it suffices to show that with positive probability we have completely colored K, i.e. every edge has bee colored. Let e be a edge of K. The evet X e,i that e has NOT bee colored with c i has probability (1 ( t, sice t out of the ( edges get colored. Thus the evet that e does ot get colored at all is the itersectio k X e i, which has probability (1 ( t k. Now to get the probability that at least oe edge has ot bee colored, we take the uio of this probability over all edges: Let X be the evet that some edge is ot colored at the ed of the process. The X = e E(K( k X e,i, which has probability P r(x ( (1 t ( k tk (e ( 1 = 1/ < 1. Thus l( e with positive probability, all edges get colored. So there does exist a colorig of the edges of K with k colors that cotais o moochromatic copy of H. 4. Prove that there is a costat c > 0 such that for every eve 4 the followig holds: For every udirected complete graph K o vertices whose edges are colored red ad blue, the umber of alteratig Hamiltoia cycles i K is at most c!. Proof by Liyua Lu: If is odd, the there is o alteratig Hamiltoia cycle. We ca assume is eve. Write = k. Radomly partitio the vertex set ito two parts of size k: V = V 1 V. Defie a directed graph D o V as follows: a directed edge uv E(D if u V 1, v V, ad uv is red i G or u V, v V 1, ad uv is blue i G. O oe had, a directed Hamiltoia cycle of D is always a alteratig Hamiltoia cycle i G. Each alteratig Hamiltoia cycle i G has probability ( beig a directed Hamiltoia cycle of D. Let k AH(G be the umber of alteratig Hamiltoia cycles i G. There is a partitio of V so that the umber of directed Hamiltoia cycle is at least ( k AH(G. O the other had, the umber of directed Hamiltoia cycles ca be bouded by the per(a D, where A D is the adjacecy matrix of D. We have ( D per(a D (r i! kah(g 1/r i. 4

5 Here r i is the out degree of i i D. The above upper boud reaches the maximum whe all r i s are about equal. Sice r i = k, the average of r i is k. Whe k is eve, write k = r. The (r i! 1/r i (r! r = (r! 4. We have AH(G 1 ( (r! 4 = 1 k! = (1 + o(1π!. ( k r Whe k is odd, write k = r + 1. The (r i! 1/r i (r! r ((r + 1! (r+1 = (1 + o(1(r!(r + 1!. We have AH(G (1+o(1 1 ( (r!(r+1! = (1+o(1 1 k! = (1+o(1π!. ( k r 5. Let X be a collectio of pairwise orthogoal uit vectors i R ad suppose the projectio of each of these vectors o the first k coordiates is of Euclidea orm at least ɛ. Show that X k ɛ, ad this is tight for all ɛ = k/ r < 1. Proof by Heather Smith: Fix ɛ > 0 ad k []. Let X be a collectio vectors i R as described i the problem. Let l = X. We ca exted X to a ormal basis X for R. Let A be a square matrix with the vectors i X makig the rows. Sice the rows are pairwise orthogoal uit vectors, AA T = I. So A 1 = A T ad A T A = I which implies v i (s = 1 for ay s []. Hece l v i (s 1. By assumptio k s=1 v i (s ɛ. Puttig these facts together, lɛ l k k l vi (s = vi (s k 1 = k. s=1 s=1 5

6 Therefore X k as desired. For the tight case, fix r ad such ɛ that r. If we ca create r pairwise orthogoal uit vectors of the form ± 1 r/,..., ± 1 r/, 0..., 0 R where the first r coordiates are ± 1 r/ ad the remaiig r are 0. The the projectio oto the first k coordiates, for k r, has Euclidea orm ( k v (i 1/ = ( k 1/ ( 1 k 1/ r = r = ɛ ( k 1/ ad for k > r, v (i = 1 > ɛ.. So it suffices to determie a assigmet of ±1 s to the vectors ± 1,..., ± 1 R r so r/ r/ that we have r pairwise orthogoal vectors. This assigmet comes directly from the Hadamard matrices which exist for orders r. 6. Let G = (V, E be a bipartite graph with vertices ad a list S(v of more tha log colors associated with each vertex v V. Prove that there is a proper colorig of G assigig to each vertex v a color from its list S(v. Proof by Edward Boehlei: Let V = L R be a bipartitio. Defie S = v V S(v. The radomly distribute each elemet of S ito either S L (with probability 1 or ito S R (with probability 1. Now for each v L, defie S L (v = S L S ad likewise for each v R, defie S R (v = S R S. The, for all v L, Pr(S L (v = = ( 1 S(v < ( 1 log = log = 1 Similarly, for each v S R, Pr(S R (v = = 1. Ad so, Pr({ v L (S L (v = } { v R (S R (v = } = v L < v L = v V = 1 Pr(S L (v = + v R Pr(S R (v = v R 1 6

7 Therefore, there exists a vertex colorig where vertices from L oly use colors from S L, ad vertices of R use oly colors from S r. Sice L R was a bipartitio, such a colorig is proper. 7