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Matroids Graphs Give Isomorphic Matroid Structure?

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Matroids Graphs Give Isomorphic Matroid Structure? EECS 495: Combinatorial Optimization Lecture 7 Matroid Representation, Matroid Optimization Reading: Schrijver, Chapters 39 and 40 graph and S = E. A set F ⊆ E is indepen- dent if it is acyclic. Food for thought: can two non-isomorphic Matroids graphs give isomorphic matroid structure? Recap Representation Def: A matroid M = (S; I) is a finite ground Def: For a field F , a matroid M is repre- set S together with a collection of indepen- sentable over F if it is isomorphic to a linear dent sets I ⊆ 2S satisfying: matroid with matrix A and linear indepen- dence taken over F . • downward closed: if I 2 I and J ⊆ I, 2 then J 2 I, and Example: Is uniform matroid U4 binary? Need: matrix A with entries in f0; 1g s.t. no • exchange property: if I;J 2 I and jJj > column is the zero vector, no two rows sum jIj, then there exists an element z 2 J nI to zero over GF(2), any three rows sum to s.t. I [ fzg 2 I. GF(2). Def: A basis is a maximal independent set. • if so, can assume A is 2×4 with columns The cardinality of a basis is the rank of the 1/2 being (0; 1) and (1; 0) and remaining matroid. two vectors with entries in 0; 1 neither all k Def: Uniform matroids Un are given by jSj = zero. n, I = fI ⊆ S : jIj ≤ kg. • only three such non-zero vectors, so can't Def: Linear matroids: Let F be a field, A 2 have all pairs indep. F m×n an m×n matrix over F , S = f1; : : : ; ng be index set of columns of A. Then I ⊆ S is 2 independent if the corresponding columns are Question: representation of U4 ? linearly independent. (1; 0); (0; 1); (1; −1); (1; 1) in <. Note: WLOG any linear matroids can be Def: A binary matroid is a matroid repre- sentable over GF (2). written as A = [ImjB] where m is rank of matroid and B is an (n−m)×m matrix over Def: A regular matroid is representable over F . any field. Def: Graphic matroids: Let G = (V; E) be a Example: Graphic matroids are regular. 1 Proof: Take A to be vertex/edge incidence Conjecture (Rota, 1971): Matroids repre- matrix with +1= − 1 in each column in any sentable over a finite field can be character- order. ized by a finite list of excluded minors. Much like planar graphs are those with no • Minimally dependent sets sum to zero K or K as a minor. perhaps with multiplying by −1. 3;3 5 • Works over any field with +1 as multi- Matroid Optimization plicative identity and −1 additive inverse of +1. Given: Matroid M = (S; I) and weights c : S ! R Note: Have graphic ⊂ binary ⊂ regular ⊂ linear. Find: max-weight (or min-weight) basis ""Recall Kruskal's Alg for min spanning## Note: There are matroids that are not linear tree: select edges in increasing order of (MacLane, 1936; Lazarson, 1958). weight Algorithm: Greedy Matroid Operations • Set J = ;. Def: (from last lecture): The dual M ∗ of ma- • Order S s.t. c ≥ ::: ≥ c . troid M = (S; I) is the matroid with ground 1 n set S whose independent sets I are such that • For i = 1 to n, if J [ fig is independent, S n I contains a basis of M. J := J [ fig Def: The deletion M n Z of matroid M = 22If weights are non-neg, this is max-weight33 (S; I) and subset Z ⊂ S is the matroid with indep set; otherwise stop selecting elts ground set S n Z and independent sets fI ⊆ 66 77 44when ci becomes negative for max-weight55 S n Z : I 2 Ig. indep set. Example: Take graph, delete edges, take Claim: Greedy finds maximal-weight basis. acyclic subsets of remaining edges. [[First rephrase second axiom. ]] Def: The contraction M=Z of ::: is ::: (M ∗ n Proof: Clearly a basis. Suppose not max- Z)∗. weight, i.e., for greedy set J and opt J 0, ""So for X ⊆ Z maximal independent set## c(J) < c(J 0). of M, I independent in M=Z if I [ X independent in M. • Let J = fe1; : : : ; elg be greedy set la- 0 Def: If a matroid M arises from M by a beled according to chosen order so ce1 ≥ 0 series of deletions and contractions, then M ::: ≥ cel . is a minor of M. 0 • Let J = fq1; : : : ; qkg be max-weight ba- Claim: (Tutte, 1958) A matroid is binary if sis labeled s.t. cq ≥ ::: ≥ cq . 2 1 k and only if it has no U4 minor. • Let i be smallest index s.t. c > c (if ""Similar characterization of ternary ma-## qi ei no such index, must have k > l so let troids as those that exclude the so-called i = l + 1). Fano matroid and its dual as a minor. 2 • Consider independent sets I = Let OP ;OD be primal/dual value. To prove 0 n fe1; : : : ; ei−1g and I = fq1; : : : ; qig. TDI need for any w 2 Z exists opt dual soln that's integral. • since jI0j > jIj exchange property says 22Recall TDI means for integral cost vector33 9z 2 I0 s.t. I + z independent 66c s.t. primal soln finite, there exists in-77 66 77 • but each elt in I0 has greater weight than 66tegral opt dual. Furthermore if polytope77 I and z was available to greedy at step i 44is TDI and b is integral, then polytope is55 integral. by above, so greedy can't have chosen ei over z. • WLOG w non-negative (else discard neg In fact, matroids are precisely set systems elts and note dual constraint satisfied on which greedy works, see book. since y ≥ 0. 22What about running time? Depends on33 • Let J be independent set found by 66matroid representation to test if I + z in-77 66 77 greedy. 66dependent. Want poly in jSj given indep77 66 77 66set oracle, or sometimes given sucinct77 • Note w(J) ≤ maxI2I w(I) ≤ OP = OD. 66 77 66representation of M like in graphs (note77 66 77 • Find integral y s.t. dual value equals 66listing all indep sets is exponential in77 66 77 w(J) hence proving both claims. Label 66jSj). Question, is there a matroid with77 44a sucinct rep in which checking indepen-55 elts in decreasing order of weight and let dence is hard? Ui = fs1; : : : ; sig. y = w(s ) − w(s ) Matroid Polytopes Ui i i+1 yUn = w(sn) Variables: xs for each s 2 S Constraints: yU = 0; otherwise xS ≥ 0; 8s 2 S { feasible: for any si 2 S, P Pn X yU = yU x ≤ r(U); 8U ⊆ S U:si2U j=i j s Pn−1 s2U = j=i (w(si)+w(si+1))+w(sn) = w(si). Claim: Greedy is optimal. { optimal: Claim: Matroid polytope integral. n−1 X X Proof: Consider primal objective r(U)y = r(U )(w(s ) − w(s )) max P w(s)x . Dual is: U i i i+1 s2S S U⊆S i=1 X +r(Un)w(sn) min r(U)yU = w(s1)r(U1) n U⊆S X + w(si)(r(Ui) − r(Ui−1)) X s:t: y ≥ w(s); 8s 2 S i=2 U = w(J) U:s2U yU ≥ 0; 8U ⊆ S 3.
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