8 Matroid Intersection

Transcription

1 8 Matroid Intersection 8.1 Definition and examples 8.2 Matroid Intersection Algorithm 8.1 Definitions Given two matroids M 1 = (X, I 1 ) and M 2 = (X, I 2 ) on the same set X, their intersection is M 1 M 2 = (X, I 1 I 2 ). Note that the intersection of two matroids is an independence system, but not typically a matroid. We can similarly define the intersection of any finite number of matroids. Below are some examples of graph-theoretical objects that can be viewed as elements of a matroid intersection. Keep in mind that since these will (typically) not be matroids, we know that the greedy algorithm cannot work for them (in general). In the next section, we ll see an algorithm that works for any intersection of two matroids. We ll use some shorter notation. The forest matroid of an undirected graph G will be written as F (G); the partition matroid of a set X, corresponding to a partition {X 1,..., X m } of X, will be written as P ({X i }) (see Problem Set 7), so P ({X i }) = {Y X : Y X i 1 i}. Simultaneous forests Given two graphs G 1 and G 2, and a bijection ϕ : E(G 1 ) E(G 2 ) between their edges, we could look for a (maximum) forest F 1 in G 1 such that ϕ(f 1 ) is a forest in G 2. This is the same as looking for a maximum forest in the matroid intersection M 1 M 2, where M 1 = F (G 1 ), M 2 = F (G 2 ). Here we should identify E 1 and E 2 via ϕ, and call both X. Bipartite matchings We saw in the last lecture that matchings (bipartite or not) do not form a matroid, at least not as sets of edges. But in the problem set we saw that the edge sets of bipartite matchings do form an intersection of the following two partition matroids (let A and B be the partite sets): M 1 = P ({δ(a)} a A ), M 2 = P ({δ(b)} b B ) 1

2 Directed forests Given a directed graph D, let G D be the undirected graph with the same edges, so G D = {{a, b} : ab E(D)}. Recall that a directed forest in D is an F E(D) that forms a forest in G D and satisfies F δ in (v) 1 for all v V (D). In the lecture on trees we saw that maximum directed forests cannot always be found by a greedy approach, which indirectly implies that they do not form a matroid. Actually, we saw this for spanning directed trees, but we also showed that these problems are equivalent. More directly, in the directed graph d c a b the directed forests are not a matroid, since {a, b, c} and {c, d} are bases of different size. But directed forests are the intersection of the two matroids M 1 = F (G D ), M 2 = P ({δ in (v)}). This should be clear from the definition, since the two conditions in the definition of directed forest (being a forest, single-entry) correspond exactly to these two matroids. Colorful forests The two previous examples were objects that we have already seen before, and in fact we already have polynomial algorithms for them. But many more less standard combinations are possible. For instance, given a graph G and a partition E = E i of its edges, take M 1 = F (G), M 2 = P ({E i }). Then M 1 M 2 can be interpreted as follows. The partition can be seen as a coloring of the edges, giving all the edges in E i the same color. Then a maximum edge set in M 1 M 2 is a colorful forest: it uses every color at most once. Intersections of three or more matroids Intersecting more than two matroids can also give meaningful objects, like colorful directed forests or colorful matchings. More significantly, you ll see in the problem set that subsets of Hamilton paths form an intersection of 3 matroids. Since finding Hamilton paths is NP-hard (as we ll see next time), this right away shows that finding a maximum independent set in an intersection of 3 (or more) matroids is NP-hard, so there is no polynomial algorithm for it unless P = NP. So to summarize: matroids can be solved easily by the greedy algorithm, intersections of 2 matroids are not as easy but can be solved by the polynomial algorithm in the next section, and for intersections of 3 or more matroids there is no general polynomial algorithm. Below we will show that every independence system, which includes many combinatorial optimization problems, is an intersection of matroids. However, for some intersections of 3 or more matroids there may be a polynomial algorithm. For instance, one can intersect one of the previous intersection of 2 matroids with another matroid in a trivial way, like with the matroid of all subsets. Or less trivially, one can intersect with a uniform matroid of sets of size k, which gives a problem that can be solved just as easily. More interestingly, we have a polynomial algorithm for shortest paths, and the set of subsets 2

3 of shortest paths forms an independence system. By the theorem below, this is an intersection of matroids, but it does not seem to be an intersection of 2 matroids. Theorem 8.1. Any independence system (X, I) with finite X is an intersection of a finite number of matroids, i.e. there are matroids M i = (X, I i ) with I = I i. Proof. Recall that a circuit in an independence system is a minimal dependent set (in a forest matroid, a circuit is the same as a cycle in the graph). If C is a circuit of (X, I), then (X, I C ) with I C = {Y X : C Y } is a matroid. Indeed, it clearly satisfies M1, and it satisfies M2 since the bases are exactly the sets X\{c} for some c C, so they all have the same size. There are finitely many circuits, and I is the intersection of the I C for all circuits, because a set is independent if and only if it does not contain a dependent set, which is equivalent to not containing a circuit. 3

4 8.2 Matroid Intersection Algorithm Here we will use the notation S + T = S T, S T = S\T, and also S + x = S {x} and S x = S\{x}. The algorithm for intersections of two matroids will again use augmenting paths, though not in a way that directly derives from the augmenting paths for matchings. Let M 1 = (X, I 1 ) and M 2 = (X, I 2 ). Given an independent set I I 1 I 2, we define a directed graph D I = (X, E(D I )) which is bipartite with partite sets I and X I. The edges are defined by, for each y I and z X I, putting yz E(D I ) if I y + z I 1, zy E(D I ) if I y + z I 2. In words, we put an edge from y to z if exchanging y in I for z gives an independent set in I 1, and an edge from z to y if it gives an independent set in I 2. Note that it is possible that there is no edge between y and z, or an edge in both directions. What we hope for is a set Y I and Z X I such that Z = Y +1 and I Y +Z I 1 I 2, so we can augment I to I Y + Z. This suggests taking a directed path in D I from X I to X I, but not any such path would work. It turns out that one should take a shortest path from X 1 X I to X 2 X I, where X 1 = {z X I : I + z I 1 }, X 2 = {z X I : I + z I 2 }. If Q = z 0 y 1 z 1 y 2 z 2 y m z m is such a path, then with Y = {y i } and Z = {z i } it is true that I V (Q) = I Y + Z I 1 I 2. This is the augmentation step in the algorithm. Algorithm for max independent set in matroid intersection M 1 M 2 on X 1. Set I = and V (D I ) = X; 2. Greedily add to I any z X I such that I + z I 1 I 2 ; 3. Set E(D I ) = {yz : y X, z X I, I y + z I 1 } {zy : y X, z X I, I y + z I 2 }; Let X 1 = {z X I : I + z I 1 }, X 2 = {z X I : I + z I 2 }; 4. Find a shortest X 1 X 2 -path Q in D I ; if none exists, go to 6; 5. Augment using Q: I := I V (Q); go back to 2; 6. Return I. The second step is not necessary, since such a z will be found as the shortest path Q = z, but when doing the algorithm by hand it is highly recommended, since it saves a lot of work in setting up D I. 4

5 Theorem 8.2. Given two matroids M 1 and M 2 on the same X, the algorithm above returns a maximum common independent set I M 1 M 2 in polynomial time. Proof. (partial) That it terminates and is polynomial should be clear. We will omit the proof that each I I 1 I 2, since this is somewhat long and technical; we will merely justify it at the end of the proof in the two simplest cases. The only other thing to prove is that the final I is indeed maximum. Given any J I 1 I 2 and U X, observe that J = J U + J (X U) r M1 (U) + r M2 (X U). (1) The inequality holds because J U is independent and contained in U, hence for any basis B U of U we have J U B U = r M1 (U); similarly J (X U) B X U = r M2 (X U). For the final I in the algorithm, we show that there is a U X for which we have equality, i.e. I = r M1 (U) + r M2 (X U). This proves that I is maximum. Consider U = {z X : a path in D I from z to X 2 }. Then X 2 U and X 1 U =. We claim that simply r M1 (U) = I U and r M2 (X U) = I (X U), which proves that we have equality in (1) for I. Suppose I U < r M1 (U). Then there is an M 1 -basis B U of U with I U < B U, so by M1 there is an x B U I such that (I U) + x I 1. Since X 1 U =, we know that I + x I 1, so (I U) + x I. If (I U) + x = I, then (I U) + x = I y + x for some y I U. If (I U) + x < I, by M1 there must be an x I U such that (I U) + x + x I 1, and we can repeat this until we also have a set of the form I y + x I 1, for some y I U. Either way we have a y I U such that I y + x I 1, which implies yx E(D I ) by definition of D I. But this would imply that there is a path from y via x to X 2, so y U, a contradiction. The proof that r M2 (X U) = I (X U) is very similar. As we said, we omit the proof that each I during the algorithm is in I 1 I 2, but we will show it for the two simplest cases. If z X 1 X 2, then Q = z is the shortest path, and indeed I V (Q) = I + z I 1 I 2. This is basically a greedy step. If Q = z 0 y 1 z 1, then we know that I + z 0 I 1 because z 0 X 1, I + z 1 I 2 because z 1 X 2, I y 1 + z 0 I 2 because z 0 y 1 E(D I ), and finally I y 1 + z 1 I 1 because y 1 z 1 E(D I ). Since I +z 0 > I y 1 +z 1, we have by M1 that there is a z I +z 0 that is not in I y 1 +z 1 such that (I y 1 + z 1 ) + z I 1. This z would have to be either y 1 or z 0, but if it was y 1, then we d have I + z 1 = (I y 1 + z 1 ) + y 1 I 1, which means z 1 X 1 X 2 and Q = z 1 would have been a shorter path. So z = z 0, and we have I V (Q) = (I y 1 + z 1 ) + z 0 I 1. An identical argument gives I V (Q) I 2. The inequality that we used in the proof in fact gives a min-max formula that is interesting in its own right, since one can use it to prove that a particular common independent set is maximal. Had we worked out the integer program for matroid intersection, and proven integrality, then we would have seen that this formula follows from our good old friend the LP duality theorem. Theorem 8.3 (Matroid Intersection Theorem). max J I 1 I 2 J = min U X (r M 1 (U) + r M2 (X U)). Proof. We saw in the proof above, and the algorithm gives us an I and a U such that equality holds. 5