21-301, Spring 2019 Homework 4 Solutions

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1 21-301, Spring 2019 Homework 4 Solutions Michael Anastos Not to be published without written consent of the author or the instructors of the course. (Please contact me if you find any errors!) Problem 1 (vl&w 5A) A perfect matching in a graph G (not necessarily bipartite) is a matching so that each vertex of G is incident with one edge of the matching. (i) Show that a finite regular bipartite graph (regular of degree d > 0) has a perfect matching. (ii) Find a trivalent (regular of degree 3) simple graph which does not have a perfect matching. (iii) Suppose G is bipartite with vertices X Y (every edge having one end in X and one in Y ). Further assume that every vertex in X has the same degree s > 0 and every vertex in Y has the same degree t. (This condition is called semiregularity.) Prove: If X Y (equivalently, if s t), then there is a complete matching M of X into Y. Proof. (i) Let G be an r-regular bipartite graph, V (G) = A B and r > 0. Since G is r- regular and bipartite we have that r A = v A d(v) = v B d(v) = r B and therefore A = B. For v V and Z V let d Z (v) be the numbers of edges with an endpoint in each of {v} and Z. Clearly d Z (v) d(v). In addition for X A we have that r X = d(v) = d X (v) d(v) = r N(X). v X v N(X) v N(X) That is X N(X). Hall s Theorem implies that there exists a complete matching M from A into B. Finally A = B implies that M is a perfect matching. (ii)

2 v Let M be a matching of the above graph G Removing v creates 3 connected components. M may have at most 2 edges spanned by each of those components and at most 1 edge incident to v. Thus M = 7 < 8 = V (G) /2. Therefore M is not a perfect matching of G. (iii)for A X we have that s A = v A d(v) = v N(A) d A (v) v N(A) d(v) = t N(A). Thus N(A) s A A. Hall s Theorem implies that there exists a complete matching t M from X into Y. Problem 2 (vl&w 5C) In the hypothesis of Theorem 5.5, we replace integers by reals. Show that in this case, A is a non-negative linear combination of permutation matrices. (Equivalently, every doubly stochastic matrix see Chapter 11 is a convex combination of permutation matrices.) Proof. Let A (R 0 ) n n. We proceed by induction on the number of nonzero entries of A, denoted by nz(a). Observe that if A 0 then, since each column sums to a strictly positive value, A must have at least n non-zero entries. Base Case: nz(a) = n. Then each column/row has exactly one non-zero entry. Since every column/row sums to the same number all of the non-zero entries should be equal, say equal to m. Let A = A/m. Then A satisfies the conditions of Theorem 5.5 with l = 1 and therefore it is a permutation matrix. A = ma implies that the inductive statement is true for the base case. Let k > n and assume that every matrix A (R 0 ) n n such that (i) every column and every row has sum l > 0 and (ii) n nz(a ) < k, can be written as a non-negative linear 2

3 combination of permutation matrices. Let A (R 0 ) n n be such that (i) every column and every row has sum l > 0 and (ii) nz(a) = k. For i i n define A i by A i := {j : a ij > 0}. For 1 k n and any k-tuple {A i1,..., A ik }, since every column/row of A sums to l, we have that k k k n a ih j = a ih j = a ih j = kl j A i1... A ik j=1 and k a ih j h=1 h=1 n a ih j = A i1... A ik l. Therefore A i1... A ik k. Thus the A i s satisfy property H. An SDR of the A i s corresponds to a permutation matrix P = (p ij ) such that a ij > 0 if p ij = 1. Let a = min{a ij : p ij = 1} and set A := A ap. Since P is a permutation matrix every column/row of A sums to l = l a. In addition nz(a ) k 1 since the entries with value equal to a in A have value 0 in A while the rest of the entries are non-negative. Finally nz(a) > n implies that A has at least nz(a) n 1 nonzero entries thus, as discussed in the beginning, it has at least n nonzero entries i.e. n nz(a ) < k. Thus we can apply the induction hypothesis to the matrix A. That gives, A = a 1 P 1 + a 2 P a z P z where a 1, a 2,..., a z > 0 and P 1, P 2,..., P z are permutation matrices. Therefore completing the induction. A = ap + a 1 P 1 + a 2 P a z P z Problem 3 (vl&w 5D) Let S be the set {1, 2,..., mn}. We partition S into m sets A 1,..., A m of size n. Let a second partitioning into m sets of size n be B 1,..., B m. Show that the sets A i can be renumbered in such a way that A i B i. Proof. Consider the bipartite graph G = (V, E) where V = A B with A = {a 1,..., a m }, B = {b 1,..., b m } and there is an edge from a j to b i iff A j B i =. Then G is an n-regular bipartite graph. Problem 1 implies that G has a perfect matching M. We define the new sets A 1,..., A n and the permutation π on {1, 2,..., n} as follows: for 1 i, j n we set A i = A j and π(j) = i (equivalenlty j = π 1 (i)) iff the edge a j b i belongs to the matching M. Since M is a perfect matching all the sets A 1,..., A n as well as the permutation π are well defined. Moreover A i B i = A π 1 (i) B i. The last equality follows from the construction of G, M and π. 3

4 Problem 4 (vl&w 6A) Let a 1, a 2,..., a n 2 +1 be a permutation of the integers 1, 2,..., n Show that Dilworth s theorem implies that the sequence has a subsequence of length n + 1 that is monotone. Proof. Let C = {c i = (i, a i ) : 1 i n 2 + 1}. For 1 i, j n we say c i c j iff i j and a i a j. Clearly is reflexive, transitive and antisymetric, hence it defines a partial ordering over C i.e (C, ) is a poset. (I) Let S = {c i1, c i2,..., c ik } be a chain of elements of C. Wlog we may assume that i 1 i 2... i k. Since S is a chain, it must be the case that for 1 j k 1 the elements c ij and c ij+1 are comparable, that is c ij c ij+1 or c ij c ij+1. Since i j i j+1 the former holds i.e. c ij c ij+1 and therefore a ij a ij+1. Hence a i1, a i2,..., a ik is an increasing subsequence of a 1, a 2,..., a n (II) Let S = {c i1, c i2,..., c ik } be an antichain of elements of C. Wlog we may assume that i 1 i 2... i k. Since S is an antichain, c ij and c ij+1 are incomparable for 1 j k 1. Thus, since i j i j+1 it must be the case that a ij > a ij+1 holds. Therefore a i1, a i2,..., a ik is a decreasing subsequence of a 1, a 2,..., a n Let M be the size of the largest antichain in C, m be the minumum number of chains that C can be partitioned into, S 1, S 2,..., S m be such a partition and l be the maximum size of a set in {S 1, S 2,..., S m }. Dilworth s theorem states that M = m. Therefore, since l S i for i {1, 2,..., m} and m = M, we have that m lm = lm S i = P = n (1) i=1 If (C, ) has an antichain of length at least n + 1 (i.e. M n + 1) then, from (II) it follows that our original sequence has a decreasing subsequence of length at least n + 1. Otherwise M n. In this case (1) implies that l (n 2 + 1)/M (n 2 + 1)/n n + 1. Thus C has a chain of length n + 1. (I) states that such a chain corresponds to an increasing subsequence of the given sequence of the same length. Therefore a 1, a 2,..., a n 2 +1 has an increasing subsequence of length n + 1. Problem 5 (vl&w 6B) Let the sets A i, 1 i k, be distinct subsets of {1, 2,..., n}. Suppose A i A j for all i and j. Show that k 2 n 1 and give an example where equality holds. 4

5 Proof. Let A = {A i : 1 i k}. We partition the set of subsets of {1,..., n} into 2 n 1 pairs of the form {A, V \ A}. For every A {1,..., n} we have that A (V \ A) =. Therefore A contains at most one set from each such pair. Hence k = A 2 n 1. Let A be all the subsets of {1,...n 1, n} that contain n. Clearly there is a one to one correspondence between sets in A and subset of {1,..., n 1} (consider adding/subtracting form a given set the element n), thus A = 2 n 1. In addition for A i, A j A we have that {n} A i A j. Hence A is such an example. 5