Discrete Optimization 2010 Lecture 6 Total Unimodularity ...

Total Unimodularity Matching Problem

Discrete Optimization 2010Lecture 6

Total Unimodularity & Matchings

Marc UetzUniversity of Twente

[email protected]

Lecture 6: sheet 1 / 29 Marc Uetz Discrete Optimization

mailto:[email protected]


Outline

1 Total Unimodularity

2 Matching Problem



Question

Under what conditions on A and b is it true that all vertices ofP = {x ∈ Rn | Ax ≤ b, x ≥ 0} happen to be integer?

Answer: If A is totally unimodular, and b integer



(Totally) Unimodular Matrices

Definition

1 An integer square matrix B ∈ Zn×n is unimodular ifdet(B) = ±1

2 An integer matrix A ∈ Zm×n is totally unimodular (TU) ifeach square submatrix B of A has det(B) ∈ {0,±1}.

Some easy facts:

entries of a TU matrix are {0,±1} by definition (1× 1submatrices)

if A is TU, adding or deleting a row or column vector(0, . . . , 1, . . . , 0), the result is again TU

if A is TU, multiplying any row/column by −1, the result isagain TU



Motivation

If all vertices of {x ∈ Rn | Ax ≤ b, x ≥ 0} are integer, for any cthe linear program

max ctx

s.t. Ax ≤ b, x ≥ 0

(if not infeasible or unbounded) has an integer optimal solutionas the set of optimal solutions of P is a face of {x | Ax ≤ b, x ≥ 0}

In that case, we can even solve integer linear programmingproblems by solving an LP only (say with Simplex), because anoptimum solution (if it exists) also occurs at a vertex whichhappens to be integer.



Integrality of Linear Equality Systems

Theorem

Let B be an integer square matrix that is nonsingular (that is,det(B) 6= 0), consider system Bx = b. Then x is integer for anyinteger right-hand-side b if and only if B is unimodular.

”if” x = B−1b, and by Cramer’s rule (Linear Algebra):

xj =det(B j)

det(B)

with B j = B, but with jth column of B replaced by b”only if” x = B−1b integer for all b, also for bt = (0, . . . , 1, . . . , 0)

such an x exactly equals a column of B−1

so all columns of B−1 are integer, and so is det(B−1)but det(B) det(B−1) = 1, both integer, so both are ±1



TU Matrices and Integrality

Define for A ∈ Rm×n, b ∈ Rm

P(A) = {x ∈ Rn | Ax ≤ b, x ≥ 0}

Theorem

If A is TU and b integer, then P(A) has integer vertices only.



Proof

P(A) = {x ∈ Rn | Ax ≤ b, x ≥ 0}

If P(A) = ∅ nothing to prove. We write

P(A) = {x |[

A−E

]x ≤

[b0

]}. As A is TU, so is

[A−E

].

A vertex x of P(A) exists (see Exercise), and is defined by nlinearly independent inequalities of this system, so by Aox = bo for

non-singular subsystem Ao , bo of

[A−E

]x ≤

[b0

].

As Ao non-singular and TU, det(Ao) = ±1, so x = (Ao)−1bo isinteger.



The whole story: Hoffman-Kruskal Theorem

P(A) = {x ∈ Rn | Ax ≤ b, x ≥ 0}

We proved sufficiency:

Theorem

If A is TU and b integer, then P(A) has integer vertices only.

It also holds necessity (see Literature, Theorem 6.25):

Theorem

If P(A) has integer vertices for all integer b, then A is totallyunimodular.



Sufficient Condition Total Unimodularity

Theorem

A matrix A is totally unimodular if no more than 2 nonzeros are ineach column, and if the rows can be partitioned into two sets I1and I2 such that

1 If a column has two entries of the same sign, their rows are indifferent sets of the partition

2 If a column has two entries of different sign, their rows are inthe same set of the partition

Proof: By induction on the size of the square submatrices



Sufficient Condition Total Unimodularity!"#$%&'

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Minimum Cost Flows

Theorem

The node-arc incidence matrix of any directed graph G = (V ,E ) istotally unimodular.

Proof: Rows have one +1 and one −1, so take I1 = V , I2 = ∅

Consequence

The linear program for Min-Cost Flow always has integer optimalsolutions, as long as capacities uij and balances b(i) are integer.

The dual linear program always has integer optimal solution, aslong as the costs cij are integer.



Outline

1 Total Unimodularity

2 Matching Problem



The Matching Problem

Definition

A matching in an undirected graph G = (V ,E ) is a set M ⊆ E ofpairwise non-incident edges.

Given an undirected graph G = (V ,E ), a maximum matching isone with maximal cardinality. Perfect matching: |M| = |V |/2

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Special Case: Bipartite Matching

Special Case

Matching in bipartite graphs G = (V1,V2,E ), where E ⊆ V1 × V2.

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Bipartite Graphs

Theorem

Graph G = (V ,E ) is bipartite if and only if G contains no oddcycle.

Proof: (w.l.o.g. assume G connected)

Necessity: Trivially, a bipartite graph has no odd cycle.

Sufficiency: Given G with no odd cycle. Pick vo ∈ V . Define

V1 = {v ∈ V | dist(vo , v) even}V2 = {v ∈ V | dist(vo , v) odd}

Then V1 ∪ V2 = V is a bipartition with no edges within V1, V2.



Matching Algorithms: Basic Ideas

matchings are an independence system but no matroid

greedy algorithm yields maximal matching, but needn’t bemaximum

for M = maximal matching, and M∗ = maximum matching,|M| ≥ 1

2 |M∗| (Exercise)

Idea for matching algorithm: Start with maximal matching, lookfor augmenting paths

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Alternating Path Theorem

Observation: Let M be a matching of G = (V ,E ), and let P be anaugmenting path (for M), then M ⊕ P is a matching with oneedge more.

M ⊕ P = M ∪ P \M ∩ P

Theorem (Berge 1957)

M is a maximum matching if and only if there are no augmentingalternating paths (for M).

Proof: Necessity is clear, sufficiency: Assume matching M ′ largerthan M and consider M ⊕M ′. Show that this contains anaugmenting alternating path for M. (see page 129, reader)



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Maximum Bipartite Matching and Maximum Flow

Augmenting Path Algorithm for MaxFlow

Flow values are either 0 or 1

flow 1 on (v ,w) ⇔ edge {v ,w} ∈ M

any feasible flow yields matching (why?)

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Question

What are the flow augmenting paths?

red arc: flow 1, backward arc in residual graph

black arc: flow 0, forward arc in residual graph

So, flow augmentation = augmenting alt. path



Problem for Non-Bipartite Graphs

Odd cycles:

flow augm. pathno matching!



Edmonds’ Breakthrough

Theorem (Edmonds 1965)

There exists a polynomial time algorithm for finding a maximummatching in any (also non-bipartite) graph. O( n3 ) time

Paths, Trees, and Flowers, Canad. J. Math. 17 (1965)

Main idea: ‘Shrink the blossoms (odd cycles)’

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Matching: Integer Linear Programming Formulation

Recall δ(v) = edges incident with v

xe = 1 if edge e is in M

max∑e∈E

xe

s.t.∑

e∈δ(v)

xe ≤ 1 v ∈ V

xe ∈ {0, 1} e ∈ E

With A = node-edge incidence matrix, this is

max 1 · xs.t. Ax ≤ 1

x ≥ 0, integer



Matching: Linear Programming Formulation

Theorem

Node-edge incidence matrix A of an undirected graph is totallyunimodular if (and only if) G is bipartite. (Exercise)

Consequence?

The linear program (LP) for the matching problem

max 1 · xs.t. Ax ≤ 1

x ≥ 0

always has an integer optimal solution (a matching), if G bipartite.

And as before, the same holds for the dual LP . . . what is the dual?



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Matching & Node Cover in Bipartite Graphs

Theorem (Konig 1931)

In any bipartite graph G , the size of a maximum matching equalsthe size of a minimum node cover.

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Proof of Konig’s Theorem

The Linear Programming relaxations of Matching and Node Coverare the following primal dual pair:

Matchingmax 1 · xs.t. Ax ≤ 1

x ≥ 0

Node Covermin 1 · ys.t. Aty ≥ 1

y ≥ 0

Bipartite graph ⇒ matrix A is TU ⇒ ∃ optimal primal-dual pairthat is integer, and this is the desired matching/node-cover.



Square Submatrix

1 0 1

1

0

1

0

0

0

1

0

1

1 1

0 1

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