Advanced Graph Algorithms Jan-Apr 2014 Lecture 14 | February, 1 2014 Lecturer: Saket Saurabh Scribe: Sanjukta Roy 1 Overview In the last lecture we learn about matroids and saw some examples of it. Continuing with an example we move to solve some interesting problems of graph theory using matroid intersection and matroid parity problem. 2 Problems on Matroid and Using Them 2.1 Transversal Matroids Recall from the last lecture, transversal matroid M = (E; I) of G has E = A as its ground set, where G is a bipartite graph with the vertex set V (G) being partitioned as A and B and I = fX j X ⊆ A; there is a matching that covers Xg The following shows the incidence matrix of a bipartite graph. columns are indexed by vertices ofA 0 1 rows are x11 x12 x13 ··· x1m indexed byB x x x ··· x C B 21 22 23 2m C verticesB x x x ··· x C T = B 31 32 33 3m C jAj = m and jBj = ` B . .. C of @ . A B x x x ··· x `1 `2 `3 `m `×m ( xi;j if(ai; bj) 2 E(G) ti;j = 0 otherwise xi;j are indeterminants. All of them are independent variables. Next we show that this matrix indeed gives the represention for transversal matroids. 2.1.1 Transversal Matroid is Representable Transversal Matroid is Representable over F[~z] 1 Proof. Forward direction: Suppose some subset X = fa1; :::; aqg is independent. Then there is a matching that saturates X. Let Y = fb1; b2; :::; bqg be the endpoints of this matching and aibi are the matching edges. Consider T [Y; X] a submatrix with rows in Y and columns in X. Consider the determinant of T [Y, X] then q Q we have a term zii which can not be cancelled by any other term. So these columns are linearly i=1 independent. Reverse direction: Suppose some subset X = fa1; ··· ; aqg of columns is independent in T . Then there is a submatrix of T [∗;X] that is nonsingular, say T [Y; X]. Consider the determinant of T [Y; X] q P Q det(T [Y; X]) = sgn(π) ziπ(i) 6= 0 π2S(Y ) i=1 q Q This implies that we have a term ziπ(i) 6= 0 and this gives us that there is a matching that i=1 saturates X in and hence X is independent. In this direction we remove the indeterminants xijs. 2.1.2 Finding the Representation Removing xij To remove the xij we assign xij from values in finite field F of size P uniformly at random. What should be the upper bound on P ? What is the probability that the randomly obtained T is a representation matrix for the transversal matroid. To answer this we use Schwartz-Zippel Lemma. Schwartz-Zippel Lemma Theorem 1. Let p(x1; x2; ··· ; xn) be a non-zero polynomial of degree d over some field F and let S be an N element subset of F. If each xi is independently assigned a value from S with uniform d probability, then p(x1; x2; ··· ; xn) = 0 with probability at most N We think det(T [Y; X]) as polynomial in xij's of degree at most n = jAj. n Probability ( det(T [Y; X]) = 0) < P . There are 2n independent sets in A and thus by union bound, probability that not all of them 2nn are independent in the matroid represented by T is at most P . Thus probability that T is the 2nn representation is at least 1 − P . Take P to be some field with at least 2nn2n elements. Size of this representation with be nO(1) bits. This way we can find the representation in polynomial time using randomisation. Open Problem Can we find the representation of transversal matroids in polynomial time using deterministic algorithm? 2.2 Two Problem Definitions n-Matroid Intersection Given matroids M1 = (E; I1); M2 = (E; I2); ··· ; Mn = (E; In) on the same ground set E, and a number k, does there exist a set X ⊆ E such that jXj > k and X 2 Ii for 1 ≤ i ≤ n? 2 Matroid Pairity Given a matroid M = (E; I), whose elements are partitioned into m pairs. (hence, jEj is even). Objective: Find the maximum cardinality collection of pairs, X ⊆ E such that X 2 I ? Matroid Parity (Linear) Given a matrix whose elements are partitioned into m pairs. z}|{ z}|{ ··· z}|{ 0 1 a11 b11 a12 b12 ··· a1m b1m B a b a b ··· a b C B 21 21 22 22 2m 2m C B a b a b ··· a b C A = B 31 31 32 32 3m 3m C r = rank of the matroid B . .. C @ . A a b a b ··· a b r1 r1 r2 r2 rm rm r×2m Objective: Find the maximum cardinality collection of pairs, such that unoin of columns of these pairs are linearly independent. ? 2.3 Problems Question:1 Phrase bipartite matching problem as intersection of two matroids Let M1 = (E; I1) where I1 = fX ⊂ E j if (i1; j1) 2 X and (i2; j2) 2 X and i1; i2 2 A and j1; j2 2 B and i1 6= i2g that is 8xi 2 X every xi are incident on different vertex on A and M2 = (E, I2) where I2 = fX ⊂ E j if (i1; j1) 2 X and (i2; j2) 2 X and i1; i2 2 B and j1; j2 2 A and i1 6= i2g We want to show that set family M1 and M2 are matroids. Proof. M1 satisfies (I2), since if all the edges in X are incident on different vertices of A then all its subsets are also incident on different vertex of A. Suppose (I3) does not hold. That is there is set X1 and X2 2 I. jX1j < jX2j. For all edges e 0 0 2 X2 nX1. e is incident on a vertex v such that 9 an edge e 2 X1 nX2. e is also incident on v. But since jX2j > jX1j and all edges in X2 are incident on differet vertices then 9 atleast one vertex u such that there is no edge (u; v) 2 X1. Then we can extend X1 using that edge, Hence (I3) holds. Therefore M1 is a matroid. Using the same reason M2 is also a matroid. Therefore M1 \ M2 gives the matching in the bipartite graph. Question:2 Phrase matchng in general graph as matroid parity problem 3 We represent the graph as the following matrix. z }| { z }| { z }| { e11 e12 e21 e22 ··· ejEj ejEj 0 1 2 1 1 0 0 ··· . B . C 8 . 1 if i = u; j = e (u; v) = e 2 E B C > l1 l u B 1 0 ··· C < A = B C a = 1 if i = v; j = e ; (u; v) = e 2 E B C i;j l2 l v B 0 1 ··· C > . B . C :0 otherwise . @ . A jV j · · · jV |×|2Ej that is we take a jV j×j2Ej matrix, we take two columns for each edge. This partitions the columns of the matrix in E pairs. th If two coulms are linearly dependent then they are same. So the i row have only one 1 if ek1 and ek2 are amongst the linearly independent columns and (i; j) = ek 2 E. Hence every vertex in the subgraph formed from these pair of linearly independent columns has degree atmost one, so it corresponds to a matching. Hence the union of columns of these pairs which are linearly independent gives the matching. Question:3 How do you test if a graph has two edge disjoint spanning trees Let G(V,E) be the graph. M1 = (E; I1) be the graphic matroid of G and M2 = (E; I2) be the co-graphic matroid of G. If M1 and M2 has an intersection of size jV j − 1. Call this set of edges S. Then S 2 I1 i.e., S forms spanning trees of G. S 2 I2 i.e., G n S is connected. Therefore 9 another spanning tree of G in E n S. Hence there are atleast two edge disjoint spanning trees. Question:4 3 Matroid intersection Problem is NPC Proof. To prove this we show, directed Hamiltonian path ≤p 3 matroid intersection. Directed Hamiltonian Path Problem: Given directed graph D = (V; A) and s; t 2 V Is there a path from s to t that goes through each vertex exactly once. M1 = (E; I1) is the graphic matroid of underlying undirected graph i.e., E is the same set of edges as A ignoring direction. M2 = (A; I2), I2 = fX 2 E j each v 2 V has at most one incoming edge in X, except s which has no edge entering itg. M3 = (A; I3), I3 = fY 2 E j each v 2 V has at most one outgoing edge in Y , except t which has no edge leaving itg. Now we prove that M2 and M3 are matroids. In question 6 we prove that M2 is a matroid. For M3 heriditary property holds trivially. Suppose A; B 2 I3 and jAj < jBj but @x 2 B n A such that A [ fxg 2 I3. But each edge in B is outgoing edge for exactly one vertex and since no of edges in B is greater than no of edges in A then there is atleast one vertex, v with outdegree 0 in A but has outdegree 4 1 is B. Therefore we can add this edge comming out of v to A.