Topics in Combinatorial Optimization Orlando Lee Unicamp 18 de - - PowerPoint PPT Presentation

Topics in Combinatorial Optimization

Orlando Lee – Unicamp 18 de junho de 2014

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Agradecimentos

Este conjunto de slides foram preparados originalmente para o curso T´

• picos de Otimiza¸

c˜ ao Combinat´

• ria no primeiro

semestre de 2014 no Instituto de Computa¸ c˜ ao da Unicamp. Preparei os slides em inglˆ es simplesmente porque me deu vontade, mas as aulas ser˜ ao em portuguˆ es (do Brasil)! Agradecimentos especiais ao Prof. M´ ario Leston Rey. Sem sua ajuda, certamente estes slides nunca ficariam prontos a tempo. Qualquer erro encontrado nestes slide ´ e de minha inteira responsabilidade (Orlando Lee, 2014).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Edmonds-Giles theorem

We will describe a general framework involving digraphs and submodular functions that generalizes several combinatorial results. This framework includes MaxFlow-MinCut theorem, Lucchesi-Yonger theorem, minimum cost orientations of graphs and weighted matroid intersection theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Crossing families

Let C be a collection of subsets of a ground set V . A pair X, Y of subsets of V is crossing or cross if X − Y = ∅, Y − X = ∅, X ∩ Y = ∅ and X ∪ Y = ∅. We say that C is crossing if X, Y ∈ C, X ∩ Y = ∅, X ∪ Y = ∅ ⇒ X ∩ Y , X ∪ Y ∈ C. This is equivalent to require that X, Y ∈ C, X, Y cross ⇒ X ∩ Y , X ∪ Y ∈ C.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Crossing families

Some simple examples of crossing families are: 2V , 2V − {∅, V }, {{v} : v ∈ V } or any collection of disjoint sets. A more interesting and important example is the following. Let D = (V , A) be a digraph. Then {X : X ⊆ V , r ∈ X} = 2V −r for some r ∈ V and {X : X ⊆ 2V − {∅, V }, din(X) = 0} are a crossing families.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Crossing submodular functions

Let C be a crossing familly on V . A function b : C → R is called submodular on crossing pairs or crossing submodular if b(X) + b(Y ) b(X ∩ Y ) + b(X ∪ Y ) for every X, Y ∈ C with X ∩ Y = ∅ and X ∪ Y = ∅. Note that this is equivalent to require submodular on crossing pairs

• f C.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flows

Let D = (V , A) be a digraph, let C be a crossing family on V and let b : C → R be a crossing submodular function. A submodular flow is a vector (function) x ∈ RA such that x(δin(U)) − x(δout(U)) b(U) for each U ∈ C. (∗) The set of vectors in RA that satisfy (∗) is called submodular flow

• polyhedron. We will show that if b is integral then this polyhedron

is integral. More precisely, we will show that the system above is TDI.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Incidence matrix of a family in a digraph

Let D = (V , A) be a digraph and let C be a family of subsets of V . Let N be the C × A-matrix defined by: NX,a =    1 if a enters X, −1 if a leaves X,

• therwise,

for each X ∈ C and each a ∈ A. We say that N is the incidence matrix of D, C. So x ∈ RA is a submodular flow if satisfies Nx b.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free families

A family C ⊆ 2V is cross-free if for all X, Y ∈ C we have that X ⊆ Y or Y ⊆ X or X ∩ Y = ∅ or X ∪ Y = V , that is, no distinct members of C cross. The family C is laminar if for all X, Y ∈ C we have that X ⊆ Y or Y ⊆ X or X ∩ Y = ∅, that is, no distinct members of C intersect. So a laminar family is also a cross-free family. Two members of a cross-free family may intersect as long as their union is V .

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free families

• Theorem. Let D = (V , A) be a digraph and let C a cross-free

family on V . Then N, the incidence matrix of D, C, is totally unimodular. Idea of the proof (see Schrijver’s book, Vol. A, p. 213-216) (a) Prove that certain matrices, known as network matrices, are totally unimodular. (b) Prove that if C is cross-free, then N is a network matrix, and hence N is totally unimodular. I have included the proofs in these slides but have not discussed in class.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

LP formulation

Consider the following linear programming (LP). max

• a∈A c(a)x(a)

s.t. x(δin(X)) − x(δout(X)) b(X) for every X ∈ C, l(a) x(a) u(a) for every a ∈ A. This is equivalent to: max cx s.t. Nx b l x u

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Dual problem

The dual problem (DP) is the following. min

• X∈C

yXb(X) −

• a∈A

u(a)w(a) +

• a∈A

l(a)z(a) s.t.

• X∈C:a∈δin(X)

yX −

• X∈C:a∈δout(X )

y X − w(a) + z(a) = c(a) a ∈ A, yX 0 X ∈ C, w(a), z(a) 0 a ∈ A. Or equivalently, min yb − 1w + 1z s.t yN − wI + zI = c y, w, z 0

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some applications

Let us describe how to model some well-known combinatorial

• ptimization problems as a submodular flow problem.

(a) Minimum Cost Circulation. Set C := {{v} : v ∈ V } and b = 0. (b) Minimum Cost Dijoin (Lucchesi-Younger). Set C := {X ⊆ V − {∅, V } : din(X) = 0} and b = −1.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some applications

(c) Minimum Cost k-Arc-Connected Orientation. Suppose we are given a digraph D = (V , A) and a cost function c : A → R+. The value c(a) represents the cost of reversing the orientation of a. Suppose we have a target arc-connectivity k, that is, we want to reorient D so that the resulting digraph is k-arc-connected (that is, dout(X) k for every X ⊆ 2V − {∅, V }). Set C := 2V − {∅, V } and b(X) := din(X) − k. You can check that a submodular flow for this pair C, b corresponds to a k-arc-connected reorientation.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some applications

(d) Weighted matroid intersection theorem. Let M1 := (E, r 1) and M2 := (E, r 2) be two matroids on the same ground set given by their respective independence oracles, and let c : E → R be a cost

• function. We want to find a common independent set I of both

matroids which maximizes c(I). Note that if I is a common independent then χI must be a feasible solution of the following LP: x(x(U)) r1(U) for every U ⊆ E, x(x(U)) r2(U) for every U ⊆ E, x(e) 0 for every e ∈ E. Actually, this defines the polyhedron of the intersection of two

• matroids. We derive this from Edmonds-Giles submodular flow

theeorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some applications

Construct two copies E ′ and E ′′ of the ground set E. For an element u ∈ E, let u′ and u′′ denote the corresponding element in the respective copy. Analogously, define U′ and U′′ for any U ⊆ E. Let D be a digraph with vertex set E ′ ∪ E ′′ and arc-set {(u′′, u′) : u ∈ E}, that is, an arc connects corresponding elements and goes from E ′′ to E ′.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some applications

Let C := {U′ : U ⊆ E} ∪ {E ′ ∪ U′′ : U ⊆ E}. This is a crossing family. Now define b : C → Z+ as follows: b(U′) = r 1(U) for U ⊆ E, U = E b(E ′ ∪ U′′) = r 2(U) for U ⊆ E, U = ∅, b(E ′) = min{r 1(E), r 2(E)}. In our model we let x(u′′, u′) = 1 meaning that we choose u to be in I, our desired common independent set. You can check that a {0, 1}-vector x is the incidence vector of a common independent set if and only if belongs to the submodular flow polyhedron defined above.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flow polyhedron

• Theorem. (Edmonds-Giles, 1977) Let D = (V , A) be a digraph, let

C be a crossing family, let b a crossing submodular function defined on C, let c : A → Z be an arc cost function, and let l, u ∈ ZA arc capacities. Then the dual LP problem (DP) has an

• ptimum solution (if it exists) which is integral.
• Proof. Suppose the dual has an optimum solution. Let y be an
• ptimum solution of the dual which minimizes
• X∈C

yX|X||X|, where X := V \ X.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flow polyhedron

Let C>0 := y + = {X ∈ C : y X > 0}. We will show that C>0 is cross-free. We need the following result.

• Theorem. If X, Y are subsets of V such that X ⊆ Y and Y ⊆ X,

then |X||X| + |Y ||Y | > |X ∩ Y ||X ∩ Y | + |X ∪ Y ||X ∪ Y |.

• Proof. See Theorem 2.1 in Schrijver’s book.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flow polyhedron

Suppose for a contradiction that C>0 is not cross-free. Let X, Y two members of C>0 which cross. Let α := min{y X, y Y } > 0. Define y′ on C by: y ′

S :=

   y S − α if S = X or S = Y , y S + α if S = X ∩ Y or S = X ∪ Y , y S

• thewise.

Then y ′ is a feasible dual solution (Exercise).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flow polyhedron

Moreover, by submodularity we have that

• X∈C

y ′

Xb(X) =

• X∈C

y Xb(X) + α(b(X∩) + b(X ∪ Y ) − b(X) − b(Y ))

• X∈C

y Xb(X), and hence y′ is optimum. However, by the previous theorem, we have that

• X∈C

y ′

X|X||X| =

• X∈C

y X|X||X| + α(|X ∩ Y ||X ∩ Y | + |X ∪ Y ||X ∪ Y | − |X||X| − |Y ||Y |) <

• X∈C

y X|X||X|, which contradicts the choice of y.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Submodular flow polyhedron

Thus C>0 is cross-free. Let N′ the submatrix of N indexed by the rows in C>0. So the dual program is equivalent to the following. min yb − 1w + 1z s.t yN′ − wI + zI = c y, w, z 0 Since N′ is totally unimodular, so is [N′ − I I] and hence y can be chosen integral.

• Corollary. The system described in the LP is TDI and hence

defines an integral polyhedron.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Network matrices

Let D = (V , A) be a digraph and let T := (V , A′) be a directed tree (it needs not be a subdigraph of D). Let M be a A × A-matrix defined as follows. For each a′ ∈ A′ and each a = (u, v) ∈ A define Mf ,a =    +1 if the uv-path in T goes through a′ forwardly, −1 if the uv-path in T goes through a′ backwardly, if the uv-path in T does not go through a′. Matrix M is called a network matrix generated by T and D. A matrix is a network matrix if it is the network matrix generated by some directed tree and some digraph.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Network matrices

• Theorem. Any submatrix of a matrix network is a network matrix.
• Proof. Suppose that M is a network matrix generated by

T = (V , A′) and D = (V , A). Deleting a column indexed by a ∈ A corresponds to deleting a from D. Deleting the row indexed by a′ = (u, v) ∈ A′ corresponds to contracting a′ in T and identifying u and v in D. Recall that a matrix M is totally unimodular (TU) if every square submatrix of M has determinant 0, +1 or −1.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Network matrices

• Theorem. (Tutte, 1965) A network matrix is TU.
• Proof. By the previous theorem, it suffices to show that any square

network matrix B has determinant 0, −1 or +1. We prove this by induction on the order of B. If B has order 1 then the result is

• trivial. So assume that B has order at least 2. Suppose that B is

generated by a directed tree T := (V , A′) and a digraph D = (V , A). Assume that det B = 0. Let u be an end vertex of T and let a′ be the arc of T incident to u. By reversing the orientation (this only changes the sign of det B) we may assume that every arc in A and A′, incident to u, leaves u. Then by definition of B, the row a′ contains only 0’s and 1’s.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Network matrices

Consider two 1’s in row a′. That is, consider two columns indexed by arcs a1 = (u, v 1) and a2 = (u, v 2) in A. Subtracting column a1 from column a2, is equivalent to resetting a2 to (v 1, v 2). So after this operation, column a2 has 0 in position a′. Since this operation does not change the determinant, we can assume that there exists exactly one arc in A incident to u; so the row a′ has exactly one

• nonzero. Then by expanding the determinant by row a′ and using

induction, we obtain that det B = ±1.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free families

A family C ⊆ 2S is cross-free if for all X, Y ∈ C we have that X ⊆ Y or Y ⊆ X or X ∩ Y = ∅ or X ∪ Y = S, that is, no distinct members of C cross. The family C is laminar if for all X, Y ∈ C we have that X ⊆ Y or Y ⊆ X or X ∩ Y = ∅, that is, no distinct members of C intersect. So a laminar family is also a cross-free family.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free families

Note that if C ⊆ 2S is a cross-free family, then adding, for each X ∈ C, the complement S − X to C mantains its cross-freeness. Moreover, for a fixed s ∈ S, the family {X ∈ C : s ∈ X} is laminar. Finally, for a fixed s ∈ S, the family obtained from C by replacing each set of C containing s by its complement is laminar. (We will use this result later.)

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free familes and directed trees

Let T := (V , A) be a directed tree and let Φ : S → V be a function for some set S. We use T and Φ to define a family C of subsets of S as follows. For each arc a = (u, v) of T, let X a := {s ∈ S : Φ(s) is in the component of T − a containing v}. Let C(T, Φ) be the family of sets X a, that is, C(T, Φ) := {X a : a ∈ A}. If C = C(T, Φ) then we say that T, Φ is a tree-representation of C. If moreover T is a rooted tree, then T, Φ is a rooted tree-representation of C.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free familes and directed trees

It is easy to see that C(T, Φ) is cross-free. Moreover, if T is a rooted tree, then C(T, Φ) is laminar. Edmonds and Giles (1977) showed that the converse is true, that is, every cross-free family has a tree-representation and every laminar family has a rooted tree-representation.

• Theorem. (Edmonds and Giles, 1977) A family of subsets of S is

cross-free if and only if C has a tree-representation. Moreover, C is laminar if and only if C has a rooted tree representation.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free familes and directed trees

• Proof. We have seen that sufficiency holds. Let us prove necessity.

First, we prove that every laminar family C of subsets of S has a rooted tree-representation. We use induction on |C|, the case C = ∅ being trivial. So let X be a minimal set of C. By induction, the family C′ := C − {X} has a rooted tree-representation T ′ = (V , A′), Φ : S → V . If X = ∅, then we can add to T ′ a new arc from any vertex to a new vertex, to obtain a rooted tree-representation T, Φ′ of C. So we may assume that X = ∅. We claim that |Φ(X)| = 1. Suppose for a contradiction that Φ(x) = Φ(y) for some x, y ∈ X. Then there exists an arc a of T ′ separating Φ(x) and Φ(y). Hence the set X a ∈ C′ contains one of x and y. As C is laminar, this implies that X a ⊂ X, contradicting the choice of X. Thus |Φ(X)| = 1.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Cross-free familes and directed trees

Let v the vertex of T ′ for which Φ(X) = {v}. Add a new vertex w to T ′ and a new arc b = (v, w). Reset Φ(s) := w for each s ∈ X. Then the new tree and Φ form a rooted tree-representation of C. This shows that every laminar family has a rooted tree-representation. Now suppose C is a cross-free family and let s ∈ S. Let C′

• btained from C by replacing each set containing s by its
• complement. Then C′ is laminar and hence has a rooted

tree-representation. Reversing some arcs of the tree (if needed) we

• btain a tree-representation of C.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Total unimodularity of incidence matrices

Let D = (V , A) be a digraph and let C be a family of subsets of V . Let N be the C × A-matrix defined by: NX,a =    1 if a enters X, −1 if a leaves X,

• therwise,

for each X ∈ C and each a ∈ A.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Total unimodularity of a certain matrix

• Corollary. Let D = (V , A) be a digraph and let C be a cross-free

family on V . Then N is a network matrix and hence N is totally unimodular.

• Proof. Let T = (W , B), Φ : V → W be a tree-representation of C.

Let D′ := (W , A′) be a digraph with A′ := {(Φ(u), Φ(v)) : (u, v) ∈ A}. Let M the network matrix generated by T, D′. There exist a bijection between C and B (if X ∈ C then there exists exactly one arc b in T such that X = X b) and a bijection between A′ and A. It is easy to see that N is identical to M up to relabeling. Therefore N is TU.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

References

• A. Schrijver, Combinatorial Optimization, Vol. B, Springer.

(chapter 60)

Orlando Lee – Unicamp Topics in Combinatorial Optimization