Matroid Basics

Worker 2013 (10-11.04.2013) Scribe: Tomasz Kociumaka Lecturers: Stefan Kratsch, Saket Saurabh, Magnus Wahlstr¨om Matroid Theory and Kernelization

Part I Matroid basics

1 Matroid deﬁnition

Definition 1. Let E be a finite set and I ⊆ 2E a collection of its subsets. We say that M = (E, I) is a matroid if the following conditions are satisfied

(I1) I= 6 ∅,

(I2) if A0 ⊆ A and A ∈ I, then A0 ∈ I,

(I3) if A, B ∈ I and |A| < |B|, then there exists e ∈ B \ A such that A + e ∈ I.

Set E is called the ground set of M and sets A ∈ I are called independent sets. Property (I2) is called hereditary property.

Deﬁnition 2. Let M = (E, I) be a matroid. A set B ∈ I is called a basis of I if no proper superset of B belongs to I.

Observe that all bases of a matroid are of the same size. This number is called the rank of the matroid and denoted rank(M). Consider the following optimization problem: Maximum weight independent set Input: A matroid M = (E, I) represented by an independence oracle, a weight function w : E → R≥0 Problem: An independent set A ∈ I maximizing the total weight. and the following greedy algorithm: Algorithm 1: Greedy algorithm for matroids

(y1, . . . , yn) := the elements of E sorted by non-increasing w(yi); X := ∅; for i := 1 to n do if X + yi ∈ I then X := X + yi; return X;

Theorem 3 ([9]). A set family M = (E, I) is a matroid if and only if the greedy algorithm correctly solves the maximum weight independent set problem for any weight function w : E → R≥0.

1 2 Examples of matroids

Example 4. Let E be an n-element ground set, k ∈ {0, . . . , n} and I = {A ⊆ E : |A| ≤ k}. Then M = (E, I) forms a matroid, which is called a uniform matroid and denoted Un,k.

Example 5. Let E1,...,E` be a partition of a ﬁnite set E. Moreover, let k1, . . . , k` be non-negative integers. Then the family ` I = {X ⊆ E : ∀i=1 |X ∩ Ei| ≤ ki} satisﬁes the independence axioms. The corresponding matroid M = (E, I) is called a partition matroid.

Example 6. Let G = (V,E) be a graph and

I = {F ⊆ E : F forms a forest}.

Then M = (E, I) is called a graphic matroid.

Example 7. Let G = (V,E) be a connected graph and

I = {S ⊆ E : G \ S is connected}.

Then M = (E, I) is called a co-graphic matroid.

Example 8. Let G = (S, T, E) be a bipartite graph and

I = {X ⊆ S : there exists a matching that covers X}.

Then M = (S, I) is called a transversal matroid.

Deﬁnition 9. Let D = (V,A) be a digraph and let S, T ⊆ V . We say that T is linked to S if there exist |T | vertex-disjoint paths from S to T .

Note that we require the paths to be fully disjoint, in particular they cannot share endpoints. We allow zero-length paths if S ∩ T 6= ∅.

Example 10. Let G = (V,A) be a digraph and S, T ⊆ V . Then

I = {X ⊆ T : X is linked to S} satisﬁes the independence axioms and the corresponding matroid is called a gammoid. If T = V , we call it a strict gammoid.

3 Alternative axiom systems

The independence axioms are just one among many axiomatizations of matroids. Here, we present another axiomatic system, where bases are the primitive notion.

Deﬁnition 11. Let E be a ﬁnite set and B ⊆ 2E. The following properties are called the basis axioms:

2 (B1) B 6= ∅,

(B2) if B,B0 ∈ B, then |B| = |B0|,

(B3) if B,B0 ∈ B and x ∈ B \ B0, then there exists y ∈ B0 \ B such that B − x + y ∈ B. Fact 12. The family B of bases of a matroid satisﬁes the basis axioms. Proof. The only axiom that requires a proof is (B3). Let B,B0 ∈ B and x ∈ B \B0. By (I2) we have B−x ∈ I. Now, it suﬃces to apply (I3) for B−x and B0 to see that for some y ∈ B0\(B−x) = B0\B we have B − x + y ∈ I. Finally, by (B2) B − x + y must be a base since otherwise this set would extend to a base of size strictly greater that |B|.

Theorem 13 ([9]). If (E, B) satisﬁes the basis axioms, then for

I = I(B): {I ⊆ E : ∃B∈B I ⊆ B}

M = (E, I) is a matroid with B being the family of its bases.

4 Operations on matroids

Deﬁnition 14. Let M = (E, I) be a matroid and X ⊆ E. Deleting X from M gives a matroid M \ X = (E \ X, I0) where I0 = {I ∈ I : I ⊆ E \ X}. Deﬁnition 15. Let M = (E, I) be a matroid and X ⊆ E. Contracting X from M gives a matroid M/X = (E \ X, I0) where

0 0 I = {I ⊆ E \ X : ∀I∈I : I⊆X I ∪ I ∈ I}.

In other words the independent sets of M \ X are the independent sets of M disjoint from X while independent sets of M/X are the independent sets of M, which span X (i.e. cannot be extended by an element e ∈ X), with members from X removed.

Deﬁnition 16. Let M1 = (E1, I1),...,Mt = (Et, It) be a family of t matroids with pairwise disjoint ground sets. Then the direct sum of these matroids is a matroid M1 ⊕ · · · ⊕ Mt = (E, I) St such that E = i=1 Ei and t I = {I ⊆ E : ∀i=1 X ∩ Ei ∈ Ii}. Example 17. Any partition matroid can be obtained as a direct sum of several uniform matroids. Deﬁnition 18. Let M = (E, I) be a matroid and t be a nonnegative integer. A matroid M 0 = (E, I0) is the t-truncation of M if

I0 = {I ∈ I : |I| ≤ t}.

Fact 19 ([9]). Assume (E, B) satisfies the basis axioms and let B∗ = {E \ B : B ∈ B}. Then (E, B?) also satisfies the basis axioms. Definition 20. Let M be a matroid and B be the family of its bases. Then the matroid defined by B∗ as the family of bases is called the dual of M and denoted by M ∗.

3 5 Matroid representation

Example 21. Let V be a vector space and E = {v1, . . . , vn} ⊆ V . Then M = (E,I) where

I = {{vi1 , . . . , vik } : vectors vi1 , . . . , vik are linearly independent} is called a linear matroid. Observe that it suffices to consider finitely-dimensional spaces since we could exchange V with n the subspace spanned by v1, . . . , vn. In particular this allows to consider V = F only where F is a field, so that one can represent M with a matrix A over F, where vectors vi are the columns of A. Observe that for linear matroids we can give the independence oracle which performs polyno- mially many operations on F. Definition 22. Matroids M = (E, I), M 0 = (E0, I0) are called isomorphic if there is a bijection φ : E → E0 such that I0 = {φ(I): I ∈ I}.

Definition 23. A matroid M is called representable over a field F if M is isomorphic to a linear 0 d 0 matroid M over F for some finite dimension d. The matrix corresponding to M is called a representation of M. Note that the representation gives an efficient independence oracle for matroids which were not a priori defined as linear matroids. In particular we will be interested in representability over small finite fields, since the operations on these fields can be implemented efficiently. Fact 24. Let M be a linear matroid of rank d over a ground set of size m. Then M can be represented by a matrix of the following form

I D d×d d×m where Id×d is the identity matrix of rank d. Proof idea. Use the Gaussian elimination.

5.1 Representability vs operations on matroids

Fact 25. If M1,...,Mt are linear matroids representable over F, then so is M = M1 ⊕ · · · Mt.

Proof. Let Ai be the representation of Mi. Then the following matrix is a representation of M   A1 0 ··· 0  0 A2 ··· 0     . . .. .   . . . .  0 0 ··· At

Fact 26. Let M be a matroid and X a subset of the ground set. If M is representable over F, so is M \ X. Proof. It suﬃces to remove columns corresponding to X in the underlying matrix.

4 ∗ Fact 27. Let M be a matroid representable over F. Then its dual M is also representable over F. Proof sketch. By Fact 24 we may assume that the representation of M is

I D d×d d×m for some d × (m − d) matrix D. Then, it is easy to check that the following matrix represents M ∗ with the correspondence between columns and elements of the ground set preserved.

−DT I (m−d)×(m−d) (m−d)×m

5.2 Representability of common matroid classes

Fact 28. Consider a uniform matroid Un,k and a ﬁeld F with at least n elements. Then Un,k is representable over F.

Proof. Let x1, . . . , xn be distinct elements of F. Consider the following k × n matrix  1 1 ··· 1   x1 x2 ··· xn     x2 x2 ··· x2   1 2 n   . . .. .   . . . .  k−1 k−1 k−1 x1 x2 ··· xn

Observe that any k columns (corresponding to xi1 , . . . , xi ) of this matrix form a Vandermonde Q k matrix, whose determinant is 1≤j

Note that this means that a representation of the uniform matroid Un,k can be stored using O(nk log n) bits.

Fact 29. Any partition matroid on a ground set of n elements is representable over any ﬁeld with at least n elements.

Proof. Follows directly from Facts 25 and 28.

Fact 30. Any graphic matroid is representable over any (non-zero) ﬁeld F. Proof. Let G = (V,E) be a graph and let G~ = (V,A) be an arbitrary orientation of G. Consider a |V | vector space F with unit vectors xi. Let (v1, . . . , vn) be an arbitrary ordering of V . With each arc e = vivj we relate a vector ve = xi − xj. It is easy to check that the linear matroid given by these vectors is isomorphic to the graphic matroid of G.

Theorem 31. Let G = (S, T, E) be a bipartite graph with n vertices in total. Then the corresponding transversal matroid M can be represented over any suﬃciently large ﬁeld. In particular, a representation can be encoded using nO(1)-bit integers. Moreover, such a representation can be computed in polynomial time by a randomized algorithm with inverse exponential failure probability.

5 Proof. First, let us give a representation over F[~z], a ring of multivariable polynomial over an arbitrary ﬁeld. Let S = {s1, . . . , sk},T = {t1, . . . , t`} and consider a k × ` matrix A with its entries aij deﬁned as follows: ( zij if sitj ∈ E aij = 0 otherwise

Claim 32. Matrix A represents M over F[~z]. Proof. Recall that for any matrix B of dimension n × n we have

n X Y det(B) = sgn(π) biπ(i). π∈Sn i=1

Observe that if each bij is either a distinct variable or zero, then no two terms may cancel. In particular, B is non-singular if and only if there is a permutation π such that biπ(i) 6= 0 for i = 1, . . . , n. This means that a submatrix A[X,Y ] is non-singular if and only if an induced subgraph G[X,Y ] has a perfect matching. In particular vectors corresponding to X are independent if and only if there exists some Y ⊆ S for which G[X,Y ] has a perfect matching, i.e. X can be covered by a matching.

Now, let us see that a ring of multivariate polynomials over any field can be replaced with a sufficiently large field. Recall the famous Zippel-Schwartz Lemma:

Lemma 33 (Zippel-Schwartz). Let p(x1, . . . , xn) be a non-zero polynomial of degree d over a ﬁeld F. Let us draw zi uniformly at random from a subset X ⊆ F of size N. Then p(z1, . . . , zn) = 0 d with probability at most N .

Let us take a matrix A over F[~z] and uniformly at random assign each zij a value from a finite field F of size P . This cannot make any independent set dependent, so it suffices to make sure that the independent sets are still independent. As we have seen proving the Claim, any independent set corresponds to a non-singular submatrix of A. Singularity of a submatrix can be expressed by its determinant being zero and determinants of such submatrices of A are non-zero polynomials with some zij as variables. There are up to 2n determinants we need to control (one per independent set) and they are all polynomials of degree less than n. Combining Zippel-Schwartz Lemma with the union bound this n2n n limits the failure probability by P . We take P of size at least n4 , so that the failure probability n is inverse exponential. If P is not much larger than n4 , the arithmetic in F works in polynomial time and the whole representation requires polynomial space.

Theorem 34. Strict gammoids are representable over sufficiently large finite fields (of size 2O(n)). Moreover, such a representation can be computed in polynomial time (with inverse exponential error probability).

Proof. We shall prove that strict gammoids are duals of transversal matroids. Then, the Theorem 31 and Fact 27 imply the ﬁrst part of the statement. For the construction algorithm one just needs to see that all the steps we perform are constructive. Lemma 35. Any strict gammoid is a dual of a transversal matroid.

6 Proof. Take a strict gammoid for a digraph D = (V,A) and S ⊆ V = {v1, . . . , vn}. Let us build a ∗ bipartite graph G with bipartition U, W where U = {ui : vi ∈ V } and W = {wi : vi ∈ V \ S}. For each vi ∈ V \ S we add an edge uiwi and for each arc vivj – an edge uiwj. We need to show that for any V0 ⊆ V of size |S| the set V0 is linked to S in D if and only if ∗ there is a matching covering U \ U0 in G . Let us start with the ⇒ direction. Take a family of |S| disjoint paths going from S to V0 and consider the following matching covering U \ U0:

• if vivj is an arc on some path, then ui is matched to wj,

• if vi does not lie on any path, then ui is matched to wi. Observe that, by node-disjointness of the paths, this is indeed a matching which covers all vertices ∗ except from U0 in G . Now, let us proceed to the ⇐ direction. Assume that U \ U0 is matched with W by a matching M. Observe that for any vi ∈ S \ V0 we can grow a path to V0 using arcs vivj for uiwj ∈ M. The construction above can be performed in polynomial time, so together with Theorem 31 this gives the algorithm computing a representation of a strict gammoid.

Gammoids are obtained from strict gammoids by a deletion, so the theorem above proves representability of all gammoids. Nevertheless the representation uses nO(1)-bit integers. This can be further improved: Theorem 36 ([4]). Let G = (V,A) be a digraph and let S, T ⊆ V . The corresponding gammoid has 1 a representation using O(min(|T |, |S|log|T |) + log( ε ) + log|V |)-bit integers. Such a representation 1 O(1) can be found in n + log( ε ) time with error probability ε.

Part II Applications to kernelization

6 Gammoid as a minimum cut ‘data structure’

Fact 37. Let M be a gammoid on a digraph G = (V,A) with sources S and sinks T . The value of a maximum S, T -flow is equal to rank(M). Consequently, it can be found using just a representation of M, without knowledge of the whole graph G. Actually, gammoids can answer much more involved queries on maximum flow values. Theorem 38. Let G = (V,A) be a digraph and X ⊆ V be a set of terminals. There exists a gammoid with ground set of size 2|X|, which gives maximum S, T -flow values in G \ R for any partition X = S ∪ T ∪ R ∪ U. Moreover, such a gammoid can be constructed in polynomial time. Proof. Consider a digraph G0 obtained from G by adding a vertex x0 with an arc x0x for any x ∈ X and a gammoid M on G0 with sources X0 and sinks X0 ∪ X. We claim that an S, T -flow of value |T | exists if and only if there is an independent set I satisfying the following constraints. • x, x0 ∈ I for each x ∈ T ; forces a zero-length path x0 and a path y0, y, . . . , x.

7 • x, x0 ∈/ I for each x ∈ S; allows a path x0, x, . . . , y. • x ∈ I and x0 ∈/ I for each x ∈ R; blocks x from being used as an internal vertex, allows a path (x0, x). • x0 ∈ I for each x ∈ U; forces a zero-length path x0 preventing a x0, x, . . . , y path. Any independent set corresponds to the following set of paths: • for each x ∈ T ∪ U a zero-length path x0, • for each x ∈ T a path y0, y, . . . , x for some y ∈ S, • for each x ∈ R a path x0, x or y0, y, . . . , x for some y ∈ S. • possibly some x0, x, . . . , y paths for x ∈ S and y ∈ U. In particular, this implies |T | disjoint S, T -paths. A converse is also true, it suffices extend the original S, T -paths so that they start in S0 and to use x0, x paths for x ∈ R and x0 paths for x ∈ T ∪ U. Note that a greedy approach can be used to get the value of a max S, T flow, it suffices to greedily add conditions x ∈ I for x ∈ T starting with conditions for T as for U. The tools we have just developed is useful for the following problem. Odd Cycle Transversal Input: A graph G = (V,E), a positive integer k Problem: Does there exist X ⊆ V of size ≤ k such that G \ X is bipartite? Corollary 39. There exists a randomized polynomial compression for the Odd Cycle Transver- sal problem. Proof. See Lecture 6.

7 Representative sets lemma

Definition 40. Let M = (E, I) be a matroid and let X,Y ∈ I. Then X extends Y if X ∩ Y = ∅ and X ∪ Y ∈ I. Definition 41. Let M = (E, I) and let Y∗ ⊆ Y ⊆ I. We say that Y∗ is an r-representative for Y if for every X ∈ I of size at most r, some Y ∈ Y extends X if and only if some Y ∗ ∈ Y∗ extends X. Example 42. Let G = (V,E) be a graph, k a positive integer, and M = (V, 2V ) the uniform V V ∗ matroid of rank |V |. We assume E ⊆ 2 ⊆ 2 . Let E be a k-representative of E. Observe that G = (V,E) has a vertex cover of size ≤ k if and only if G∗ = (V,E∗) has. The representative sets lemma shows that small representative sets exist and can be efficiently constructed. Together with a trivial isolated vertex removal reduction rule, this gives an alternative O(k2)-edge kernel for Vertex Cover. Lemma 43 (Representative sets lemma, [8, 6]). Let M be a matroid of rank r + s represented by a matrix A and let Y a collection of independent sets, each of size s. An r-representative Y∗ of Y r+s O of size at most s can be found in (|Y| + kAk) (1) time.

8 8 Cuts vs independent sets of gammoids

Deﬁnition 44. Let G = (V,A) be a digraph. A set X is closest to S if X is the unique minimum S, X-cut.

Fact 45. Let G = (V,A) be a digraph and let X,S ⊆ V . There is a unique S, X-cut X0 that is closest to X, it can be found in polynomial time.

Proof idea. Use submodularity of cuts.

We say that v0 is a sink-copy of v ∈ G if for each arc or edge uv there is an arc uv0 and these are the only arcs incident to v0. Below this point we assume that the gammoids are build for (di)graphs extended with suﬃciently many sink-copies of each vertex, denoted by v0, v00,.... The following lemma characterizes the closest sets in terms of the gammoid.

Lemma 46. Let G = (V,A) be a digraph, and let M be a strict gammoid for S ⊆ V . Then X is closest to S if and only X ∪ {x0} is independent in M for each x ∈ X \ S.

Proof. ⇐ Assume Z 6= X is an S, X-cut of size at most |X| and let x ∈ (X \ S) \ Z. Such a vertex exists since X ∩ S ⊆ Z. There are |X| + 1 vertex-disjoint paths from S to X ∪ {x0}, one of them must avoid Z. However, this path either is an S, X-path or an S, x0 path. In the latter case the last vertex can be changed from x0 to x (since x∈ / Z), so in both cases we have an S, X-path avoiding Z, a contradiction. ⇒ Assume there is x ∈ X \ S such that X ∪ {x0} is not independent. This gives an S, (X ∪ {x0}) cut Z of size at most |X|. If x0 ∈ Z then Z −x0 is an S, X-cut smaller than X, a contradiction. Thus x0 ∈/ Z. If x ∈ Z, by minimality of Z there exists an S, x-path avoiding Z − x. However, changing the last vertex on this path to x0 gives an S, x0-path avoiding Z, a contradiction. Consequently x∈ / Z and Z 6= X is a diﬀerent S, X-cut of size at most |X|, a contradiction.

Lemma 47. Let X be a set closest to S and let v ∈ V \ X. Then X ∪ {v} is independent if and only if v is reachable from S in G \ X.

Proof. ⇒ If X ∪ {v} is independent, then there are |X| + 1 disjoint paths from S to X ∪ {v}, so an S, v-path avoids X, i.e. v is reachable from S in G \ X. ⇐ If X ∪ {v} is not independent, there is an S, (X ∪ {v}) cut Z of size at most |X|. This is in particular an S, X-cut, so Z = X. Consequently X is an S, v-cut.

9 Randomized compression for digraph pair cut problem

Digraph Pair Cut Input: A digraph G = (V,A), source vertices S ⊆ V , a collection of pairs of sink vertices V P ⊆ 2 , an integer k Problem: Is there a set X ⊆ V of size at most k such that there is no pair {u, v} ∈ P with both u and v reachable from S in G \ X?

Example 48. Digraph Pair Cut generalizes Vertex Cover. An instance (G = (V,E), k) of Vertex Cover is equivalent to an instance of Digraph Pair Cut with G0 = (V, ∅), sources S = V and pairs P = E.

9 We shall generalize the approach used for Vertex Cover, i.e. ﬁnd a smaller representative set of pairs P ∗ ⊆ P such that shrinking P to P ∗ does not change the answer. Then, the problem can be encoded by a gammoid with sources S and sinks V (P ∗).

Fact 49. There is an optimal solution X of the Digraph Pair Cut problem such that X is closest to S.

Proof. Assume X is an optimal solution, which is not closest to S. Then, there is a set X0 of size at most |X| which is closest to S and an S, X-cut. Observe that any vertex not reachable from S in G \ X is not reachable in G \ X0 since any S, v-path would need to intersect X and X0 is an S, X-cut.

Lemma 50. Consider an instance of the Digraph Pair Cut problem. In randomized polynomial time one can obtain an equivalent instance with |P | = O(k2).

Proof. In order to use the representative sets lemma, we need to build an appropriate matroid. Let M be a strict gammoid constructed as follows. The underlying digraph G0 consists of two disjoint copies of G with the copies of v denoted as v1 and v2. The set of sources is {s1, s2 : s ∈ S}. We apply the representative sets lemma for a family Y = {{u1, v2} : {u, v} ∈ P } to obtain a ∗ 2k+2 2 ∗ ∗ 2k-representative subset Y of size 2 = O(k ). Let P = {{u, v} : {u1, v2} ∈ Y }. Now, it remains to prove the correctness of this construction, i.e. the following claim. Note that Fact 49 allows to assume that X is closest. Claim 51. Let X ⊆ V be closest to S of size at most k. Then S can reach a pair {p, q} ∈ P ∗ if and only if S can reach a pair {u, v} ∈ P , where reaching a pair means reaching both its members.

Proof. ⇐ Clearly P ∗ ⊆ P , so this implication is trivial. 0 ⇒ Observe that being closest to S1 ∪S2 the set X1 ∪X2 is independent in D . Assume that X leaves a pair {u, v} ∈ P connected to S. This means that u1 can be reached from S1 in D1 \ X1 and v2 can be reached from S2 in D2 \ X2. Consequently, by Lemma 47, X1 ∪ X2 ∪ {u1, v2} is independent ∗ in M, so {u1, v2} extends X1 ∪ X2. However, {u1, v2} ∈ Y, so there is a pair {p1, q2} ∈ Y such that {p1, q2} also extends X1 ∪ X2. Again, by Lemma 47, this means that p1 is reachable from S1 in D1 \ X1 and q2 is reachable from S2 in D2 \ X2, so both p and q are reachable from S in D \ X, a contradiction.

This completes the proof of Lemma 50.

This way we have reduced the size of P to kO(1). Nevertheless the whole graph still has n vertices, so we need to compress the instance. It turns out that, similar to OCT, it suﬃces to store a gammoid instead of the whole graph.

Lemma 52. A gammoid M on G with sources S and sinks V (P ∗) is sufficient to solve the Digraph Pair Cut problem. Proof. Consider the following NP-algorithm. Guess a choice T ⊆ V (P ∗) of vertices to disconnect from S such that T contains at least one element of each pair in P ∗. Observe that the cost of disconnecting S from T is equal to the maximum value of an S, T -flow and Fact 37 can be used to find its value just using M.

Theorem 53 ([5]). The Digraph Pair Cut admits a randomized polynomial kernel.

10 Proof. First, observe that the problem admits an O?(2k) FPT algorithm, which works in polynomial time if k ≤ log n. This lets us assume k ≥ log n. Let us show that |S| can be without loss of generality limited to k + 1. Create k + 1 auxiliary source vertices and connect them to the original source vertices. These vertices will be new sources. Observe that no optimal solution uses these vertices, so there is a direct correspondence between optimal solutions in the original and modiﬁed instances. Now, the construction we have given above gives a polynomial compression. Indeed |S|+|P ∗|+ log |V | = O(k2), so by Theorem 36 the gammoid has kO(1) representation, which gives the desired compression to the problem described in the proof of Lemma 52. This problem belongs to NP, so it can be reduced back to any NP-complete problem, in particular to Digraph Pair Cut, or even its special case: Vertex Cover.

10 Kernel for multiway cut with deletable terminals

Multiway Cut with Deletable Terminals (DTMWC) Input: A graph G = (V,E), a set T ⊆ V of terminals, an integer k Problem: Is there X ⊆ V of size at most k such that each connected component of G \ X contains at most one terminal? Note that this is also a generalization of the Vertex Cover problem, which corresponds to instances with T = V . LP rounding arguments [3] allow to reduce T so that |T | ≤ 2k. Therefore we can focus on limiting the number of non-terminal vertices only.

Fact 54. There exists an optimal solution X of DTMWC such that X is closest to T .

Proof. Assume that X is an optimal solution and X0 6= X is a minimum T,X-cut. Observe that 0 0 X is an optimal solution. Indeed, if there is a path joining different terminals t1 and t2 in G \ X , then this path must contain a vertex from X, since it didn’t exist in G \ X. However, X0 is an 0 T,X-cut, so no vertex from X is reachable from neither t1 nor t2 in G \ X . Our current aim is to identify vertices that separate many terminals with respect to some optimal solution. We would like to formalize this notion in terms of uniquely extending some independent set. Then, such vertices would survive after applying the representative sets lemma, so we can get a small set containing all these vertices. This is because if Y ∈ Y is a unique extension of X ∈ I, then Y belongs to Y∗. We fix M to be a gammoid on G = (V,E) with sources T and sufficiently many sink-copies of each vertex.

Attempt 1. Let Y = {{v0} : v ∈ V \ T } and let Y∗ be its k-representative subset. Which vertices survive? Let us ﬁx a minimum solution X of size ≤ k and x ∈ X \ T . Observe that {x0} extends X since X is closest to T , but other sets {v0} also do, so there is no guarantee that {x0} survives in Y∗.

Attempt 2. Let Y = {{v, v0} : v ∈ V \ T } and let Y∗ be its k − 1-representative subset. Again let us ﬁx a minimum solution X of size ≤ k and x ∈ X \ T . Clearly {x, x0} extends X − x, but again other pairs {v, v0} also do, so we still need a stronger property.

11 Final attempt. Observation 55. Let X be an optimal solution minimizing X \ T . Then for each x ∈ X \ T in G0 = G − (X − x) at least three terminals are reachable from x. Proof. If at most one terminal is reachable from x, then X − x is a smaller solution. If two terminals t1, t2 are reachable from x, then (X − x) ∪ {t1} is a solution with the same size and less non-terminals.

Once we have the intuition from Observation 55, let us prove a more general result. Lemma 56. Let X be an optimal closest solution minimizing |X \ T |. Then X ∪ {x0, x00} is independent for each x ∈ X \ T . Proof. Let us build an auxiliary bipartite graph H = ((X \ T ) ∪ {x0, x00},T \ X,F ) with yt ∈ F if and only if t reaches y is G \ ((X ∪ {x0, x00}) − y). We shall see that there is a matching saturating (X \ T ) ∪ {x0, x00}. For a proof by contradiction 0 00 assume otherwise. By Hall’s theorem there is Q ⊆ (X \ T ) ∪ {x , x } with |NH (Q)| < |Q|. Let 0 0 00 0 0 0 0 Q = Q \{x , x }. Observe that |NH (Q )| = |NH (Q)| < |Q| ≤ |Q | + 2, i.e. |NH (Q )| ≤ |Q | + 1. Now, as in the proof of Observation 55 we argue that instead of including Q0 in the solution, 0 one could rather include all but one terminal from NH (Q ). This preserves the solution size and decreases the number of non-terminals included. Moreover, the new solution can be made closest without changing these parameters. The matching gives each y ∈ (X \ T ) ∪ {x0, x00} a private terminal t ∈ T reachable without passing through other vertices of X ∪ {x0, x00}. Since X is a solution, each of these terminals lies in a diﬀerent component of G \ X, so the paths are pairwise disjoint. Counting also zero-length paths for x ∈ X ∩ T this gives |X| + 2 pairwise disjoint paths from T to X ∪ {x0, x00}, which proves that X ∪ {x0, x00} is indeed independent.

Let us therefore consider Y = {{v, v0, v00} : v ∈ V \ T } and let Y∗ be its k − 1-representative set. We have seen that any non-terminal x present in some solution X extends X − x. In order to make sure that {{x, x0, x00}} survives in Y∗ we shall also prove the following result. Lemma 57. Let X be any solution and let v ∈ V \ X, v 6= x. Then {v, v0, v00} does not extend X − x. Proof. For a proof by contradiction assume that {v, v0, v00} does extend X − x. This means that in G \ (X − x) there are disjoint paths to v from three terminals t1, t2, t3. At most one of these paths contains x, so in G \ X two of them remain. This contradicts X being a solution.

We can combine lemmas stated above with the representative sets lemma to obtain the following corollary. Corollary 58. In polynomial time we can (with inverse polynomial error probability) ﬁnd a set Y of size O(k3) such that for some optimal solution X we have X ⊆ T ∪ Y . 0 00 k+2 3 Proof. Set Y = {{v, v , v } : v ∈ V \T }, which has k −1-representative subset of size 3 = O(k ) and Y = {v : {v, v0, v00} ∈ Y∗}. By Lemmas 56 and 57 for any optimal solution minimizing |X \ T | and any x ∈ X \ T the triple {x, x0, x00} is the only triple from Y extending X − x. Consequently, it belongs to Y∗, i.e. x ∈ Y . The error comes from computing the representation of the underlying gammoid.

12 Note that here we do not use Theorem 36, its simpler variant — Theorem 34 is suﬃcient. Now, we can make any vertex v ∈ V \ (T ∪ Y ) undeletable. This is done by applying a torso operation, i.e. adding an edge between any vertices u, v ∈ T ∪Y which can be joined by a path with all inner vertices out of T ∪Y . This way we obtain an equivalent instance with O(k3) non-terminals. LP rounding arguments allow to reduce the number of terminals to 2k beforehand, so we obtain the following theorem. Theorem 59 ([5]). The Multiway Cut with Deletable Terminals problem admits a randomized (Monte Carlo, inverse exponential error probability) kernel with O(k3) vertices.

Part III Irrelevant vertex rules and capturing all solutions

11 Multiway cut and related problems

11.1 s-Multiway cut

Multiway Cut Input: A graph G = (V,E), a set of terminals T , an integer k Problem: Does there exist a set X ⊆ V \ T of at most k vertices such that each connected component of G \ X contains at most one terminal? We study the s-Multiway Cut problem, where s = |T | is assumed to be constant. We shall apply the irrelevant vertex technique, originally introduced by Robertson and Seymour in the Graph Minors project. The applications in kernelization are of significantly different nature, but the technique itself is still very useful. Definition 60. We say that a vertex v ∈ V is essential, if v ∈ X for every optimal solution X and irrelevant otherwise. Note that in Multiway Cut problem we cannot just remove an irrelevant vertex, instead we can make it undeletable by joining its neighbours with a clique and the removing the vertex; this is actually a torso operation on V −v. It is easy to see that this is equivalent to making v undeletable and allows to reduce |V |. This way, if we can find an irrelevant vertex in polynomial time provided that |V | ≥ f(k), we get an f(k)-kernel. Definition 61. Let G = (V,A) be a digraph and S, X ⊆ V Lemma 62 ([1, 7]). Let X be a minimum multiway cut, and t ∈ T a terminal. Let X0 = Push(X,T − t) the X, (T − t)-cut closest to (T − t). Then X0 is also a minimum multiway cut. Proof. By definition X0 is not larger that X. We shall see that X0 is a valid multicut. For a proof 0 by contradiction assume that a t1, t2 path exists in G \ X . Since X was a multiway cut such a path must intersect X, i.e. can be decomposed into a t1,X and X, t2-paths. However, at least one 0 of the terminals t1, t2 is in T − t, so this implies an X, (T − t)-path in G \ X . This a contradiction with X0 being an X, (T − t) cut.

13 Deﬁnition 63. A vertex v ∈ V \ T is stable in X if v ∈ X and v ∈ Push(X,T − t) for every t ∈ T . Observe that essential vertices are stable, so if v is not stable in X, it is irrelevant. The following lemma characterizes stable vertices. Lemma 64. Let X be a multiway cut for (G, T ). Then v is stable in X if and only if X + v0 is linked to T − t for every t ∈ T . Proof. Recall that X +v0 is linked to T −t if and only if v0 is reachable from T −t in G\X0(t) where X0(t) = Push(X,T − t). Observe that for v ∈ X this holds if and only if v ∈ X0. Consequently, the lemma holds for every vertex v ∈ X. For v∈ / X, v is reachable from at most one terminal, so X + v0 is not linked to T − t for some t ∈ T .

Now, let us characterize stable vertices in terms of a matroid. Let M = M1 ⊕ · · · ⊕ Ms, where Mi is a gammoid over G with NG(T − ti) as the source set. For each vertex v let vi be the copy of 0 0 v in Mi. We deﬁne t(v) = {v1, . . . , vs} and T = {t(v): v ∈ V \ T }. Lemma 65. Let X be an optimal solution of size ≤ k and let v ∈ X be stable. Then the only extensions of X = X1 ∪ · · · ∪ Xs among tuples t(u) are the tuples t(v) for v stable in X. 0 Proof. By Lemma 64, v is stable in X if and only if vi extends Xi for i = 1, . . . , s. Consequently, t(v) indeed extends X if and only if v is stable in X.

Theorem 66 ([5]). Multiway Cut with s terminals has a randomized polynomial kernel (with inverse exponential failure probability) with O(ks+1) vertices. Proof. By LP-based reductions (see [3]) we can assume |N(T )| ≤ 2k. Moreover the representative ∗ O(ks) s sets lemma can be used to find T , the (s + 1)k-representative set of T . Its size is s = O(k ). There are at most k stable vertices in any set X we care about and taking the representative we cover one, if we take a representative set k times, we cover all stable vertices. More precisely, ∗ ∗ ∗ first we find a representative set T1 of T , then a representative set T2 of T\T1 and so on. This gives a collection of O(ks+1) non-terminals containing all essential vertices. Thus, if there are more vertices, one can make any single one undeletable (and remove it). Note that making more vertices undeletable in a single step might lead to losing all solutions.

11.2 s-Multicut

s-Multicut Input: A graph G = (V,E), a set of s cut requests (a1, b1),..., (as, bs), an integer k Problem: Does there exists a set X ⊆ V \{a1, .., as, b1, . . . , bs} of size ≤ k that disconnects ai from bi for every i? Observe that each instance of s-Multiway cut is an instance of s-Multicut. We shall give √ 2 a kO( s)-vertex kernel for s-Multiway cut Consider a solution X and the connected components of G \ X. They deﬁne a partition V = (V1,...,V`) of V \ X. If there is no cut request within Vi ∪ Vj, the partition can be coarsened, i.e. 0 0 0 Vi and Vj can be joined to a single partition class. Let V = (V1,...,Vp) be a maximally coarse 0 version of V obtained this way and let P = V ∩ T = (T1,...,Tp) be the corresponding partition of the set of terminals T = {a1, . . . , as, b1, . . . , bs}. Observe that X is a multiway cut for P and any p multiway cut for P is a solution. Moreover 2 ≤ s, since the partition cannot be coarsened.

14 Now, we have a set T of up to 2s undeletable terminals and we would like to kernelize preserving √ an optimal multiway cut for any partition of P of T into O( s) parts. The number of possible partitions is a function of s, i.e. O(1), so each partition√ can be analyzed separately. In particular, for each partition we shall compute a set of kO( s) vertices including all essential vertices. If the graph is suﬃciently large, there is a vertex, which is irrelevant for all partitions. Again,√ since the number of partitions is O(1), an irrelevant vertex can be found unless there are kO( s) vertices.

Lemma 67. Let P = (T1,...,Tp) be a partition of the terminal set. Then in polynomial time one can ﬁnd a set of kp+1 vertices including vertices essential for any optimal multiway cut for P.

Proof. Note that this is a generalized version of Theorem 66. The approach is going to be similar; we omit most details. Again for any optimal X other optimal solutions are Push(X,T \ Ti) for any i, so a vertex v can be defined as stable in X if v ∈ X and v ∈ Push(X,T \ Ti) for i = 1, . . . , p. Any vertex which is not stable in some optimal X, is irrelevant. This time we consider M = M1 ⊕ · · · ⊕ Mp, where Mi is a gammoid over G with T as sources. 0 0 We define T = {{v1, . . . , vp} : v ∈ V \ T } and observe that t(v) is the unique extension of (X1 ∪ T1) ∪ · · · ∪ (Xs ∪ Ts). Here, adding Ti to Xi has an effect of ‘turning off’ the sources Ti in matroid Mi. √ Applying the representative sets lemma k times gives a representative set of T of size kO( s), which corresponds to the desired set of vertices.

Consequently, we obtain a kernel.

Theorem 68 ([5]). Multicut with s cut demands√ has a randomized polynomial kernel (with inverse exponential failure probability) with O(kO( s)) vertices.

12 Covering all solutions

d-Hitting Set V Input: A hypergraph H ⊆ d , an integer k Problem: Does there exist a set X ⊆ V of size at most k such that X intersects every edge E ∈ H? This is a classic generalization of the Vertex Cover problem with an O(kd)-size kernel. We shall recall a proof of slightly stronger result:

Lemma 69 ([2]). There exists a collection H0 ⊆ H of size O(kd) such that all minimal hitting sets (of size at most k) are contained in H0. In particular, for sets of size up to k, the hitting sets of H are exactly the hitting sets of H0.

Proof. Recall the definition of sunflower and the sunflower lemma (Lecture 1). Note that if |H| > d f(d)k , then there exists a sunflower E of size at least k + 2. We claim that an arbitrary member of E can be discarded and this preservers all hitting sets of size at most k. Let the core of E be C, the discarded edge be E, and X ⊆ V be any set of size at most k. If X hits C, then it hits E, so it is a hitting set of E if and only if it is a hitting set of H − E. Otherwise 0 it misses at least one other edge E ∈ E − E.

15 We shall see that the representative lemma can be used to give results of similar ﬂavour. Let us see an example for the s-Multiway Cut problem. The s-Multiway Cut kernel apart from an optimal solution captures every stable vertex of every solution to the problem, and even every stable vertex of every multiway cut of every re-partitioning of the terminals, see the formalization below

Lemma 70. Let T be a set of O(k) (deletable) terminals and let Z be the output of the s-Multiway Cut kernel (of size O(ks+1)). For every assignment T → {1, . . . , s} ∪ {deleted}, the set Z includes

• an optimal multiway cut of the resulting partition,

• all stable vertices of every alternative multiway cut of that partition.

Proof sketch. The crucial claims remain valid after appropriate modiﬁcations. In particular (X ∪ 0 Ti) + v is linked to T if and only if v ∈ Push(X,T \ Ti) or v is reachable from T \ Ti avoiding Push(X,T \ Ti). Lemma 70 has broad applications. In particular, it immediately implies Theorems 59 and 68. Below, we generalize the case of s = 2 to digraphs and analyze its broad applications.

12.1 Covering graph cuts Lemma 71 ([5]). Let G = (V,A) be a digraph, S, T ⊆ V and r be the size of the minimum S, T - cut. We can ﬁnd a set Z of size O(|S||T |r) such that for every P ⊆ S and Q ⊆ T , Z includes a minimum P,Q-cut.

Observe that this gives a kernel from Digraph Pair Cut. Indeed, as in the proof of Theorem 53 we can assume |S| = k + 1 and |P ∗| = O(k2), so |S| = k + 1, |T | = O(k2) and r ≤ |S| = k + 1. The lemma gives a kernel of size O(k4).

Lemma 72 ([5]). Let G = (V,A) be a digraph and let X ⊆ V . We can ﬁnd set Z of size O(|X|3) such that for any partition of X into

• sources S,

• sinks T ,

• deleted vertices R,

• other (standard) vertices X \ (S ∪ T ∪ R),

Z contains a minimum S, T -cut in D \ R.

Proof sketch. We present main claims in the directed case, but with R = ∅, i.e. with no forced deletions. Details are omitted. Claim 73. Let A ⊆ S and B ⊆ T and let C be a minimum A, B-cut. Then the vertices of all A, B-cuts (stable vertices for A, B) are exactly those vertices v such that C ∪ {v0} is linked both to ←− A in G and to B in G, the reversal of G.

16 Claim 74. Stable vertices for A, B are exactly those vertices v such that C + (S \ A) + v0 is independent in the gammoid over G with sources S and C + (T \ B) + v0 is independent in the ←− gammoid over G with sources T .

Again, we construct a matroid M = M1 ⊕ M2, where M1 and M2 are the gammoids mentioned 0 0 in the claim. We deﬁne t(v) = {v1, v2}, T = {t(v): v ∈ V } and notice that for C = (C1 ∪ (S \ A)) ∪ (C2 ∪ (T \ B)) only the stable vertices for A, B can extend C with t(v). A single representative set includes at least one stable vertex, so taking the representative set at least r times, where r is the size of the minimum S, T -cut, suﬃces to cover all vertices of C.

12.2 Applications Below we list several consequences of Lemma 72. Again, we refer to [5] for the details.

Replacing compression with kernelization. Due to similarity with Digraph pair cut, we get an O(k6)-variable kernel for Almost 2-SAT deﬁned as follows Almost 2-SAT Input: A 2-CNF-SAT formula φ, an integer k Problem: Does there exists a set X of at most k variables such that if we remove all clauses containing variables from X, we are left with a satisﬁable formula? There also exists an O(k3)-vertex kernel for a generalization of the Odd Cycle Transversal problem: Odd Cycle Transversal in Signed Graphs Input: A graph G :(V,E) and a function w : E → {0, 1}, an integer k Problem: Does there exists a set X ⊆ V of size at most k variables such G \ X contains no cycle with odd total weight? Another related problem also admits a polynomial kernel: Vertex Cover Above LP Input: A graph G :(V,E), an integer k Problem: Let x∗ be the value of the optimum solution of the standard LP relaxation for the vertex cover problem. Does there exists a vertex cover of size at least x∗ + k?

Other consequences. √ Corollary 75. Almost 2-SAT and Odd Cycle Transversal problems admit O( log OPT ) approximations.

Corollary 76. There is a 2k − k1/4-edge kernel for Vertex Cover.

References

[1] Elias Dahlhaus, David S. Johnson, Christos H. Papadimitriou, Paul D. Seymour, and Mihalis Yannakakis. The complexity of multiterminal cuts. SIAM J. Comput., 23(4):864–894, 1994.

17 [2] Fedor V. Fomin, Saket Saurabh, and Yngve Villanger. A polynomial kernel for proper interval vertex deletion. In Leah Epstein and Paolo Ferragina, editors, ESA, volume 7501 of Lecture Notes in Computer Science, pages 467–478. Springer, 2012.

[3] Sylvain Guillemot. Fpt algorithms for path-transversal and cycle-transversal problems. Discrete Optimization, 8(1):61–71, 2011.

[4] Stefan Kratsch and Magnus Wahlstr¨om.Compression via matroids: a randomized polynomial kernel for odd cycle transversal. In Yuval Rabani, editor, SODA, pages 94–103. SIAM, 2012.

[5] Stefan Kratsch and Magnus Wahlstr¨om.Representative sets and irrelevant vertices: New tools for kernelization. In FOCS, pages 450–459. IEEE Computer Society, 2012.

[6] LászlóLovász.Flats in matroids and geometric graphs. In Combinatorial surveys (Proc. Sixth British Combinatorial Conf., Royal Holloway Coll., Egham, 1977), pages 45–86. Academic Press London, 1977.

[7] D´anielMarx. Parameterized graph separation problems. Theor. Comput. Sci., 351(3):394–406, 2006.

[8] D´anielMarx. A parameterized view on matroid optimization problems. Theor. Comput. Sci., 410(44):4471–4479, 2009.

[9] J.G. Oxley. Matroid theory. Oxford Graduate Texts in Mathematics. Oxford University Press, 2011.