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Lecture 5: Lovász extension and matroids
1.1
Lovász extension
Extension of an arbitrary (sub or non submodular) set function z : 2V 7→ R to
g : Rn+ 7→ R where n = |v| .
Denition 1. g 1A = z (A) ∀A ⊆ V .
For any w ∈ Rn+ let p1 > p2 > · · · > pn , be distinct values, where pn = 0.
Let vj = {i|w (i) ≥ pj }
Denition 2. The Lovász extension g is given by
g (w) =
n
X
(pj − pj+1 ) z (vj ) .
j=1
Example 1. If w = (0.75, 0.3, 0.2, 0.3, 0) we get
0.75 > |{z}
0.3 > |{z}
0.2 .
|{z}
p1
p2
p3
g (w) = 0.45z ({1}) + 0.1z ({1, 2, 4}) + 0.2z ({1, 2, 3, 4}) .
Example 2. For w ∈ [0, 1]n1 we get a positive linear combination of vertices.
Lemma 1. Let z : 2V 7→ R be a set function with z (∅) = 0 and g : Rn+ 7→ R
it's Lovász extension. If z is submodular, then the following equality holds:
g (w) = max wT x for all w ∈ [0, 1]
n
x∈Pz
where Pz is the submodular polyhedron for z.
Proof. Exercise 3.2.
Theorem 1. A set function
z : 2V 7→ R is submodular (or supermodular) if
and only if it's extensions g is convex (or concave).
Proof. (⇒) Let w1 , w2 ∈ Rn+ , λ ∈ (0, 1) and w = λw1 + (1 − λ) w2 . From
Lemma (1) let x? be such that
g (w) = wT x? = λw1T x? + (1 − λ) w2T x? .
Then g (λw1 ) ≥ λw1T x? and g ((1 − λ)) ≥ (1 − λ) w2T x? and thus
g (w) = g (λw1 + (1 − λ) w2 ) ≤ g (λw1 )+g ((1 − λ) w2 ) = λg (w1 )+(1 − λ) g (w2 )
(⇒) Let A, B ⊆ V from the denition
g 1A + 1B = g 1A∪B + g 1A∩B .
By convexity we get
g 1A + 1B ≤ g 1A + g 1B
⇒ z (A) + z (B) ≥ z (A ∪ B) + z (A ∩ B) .
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1.2
Matroids and greedy algorithms
Denition 3. A forest is a subset of edges without any cycles.
Example 3. Let G = (V, E) be a connected graph with edge
P weights we .
Let I = { forests of G}, nd a spanning tree while maximizing e∈J we where
J ⊆ E.
Greedy algorithm (kruskal) Set J = ∅ while ∃e ∈/ J such that J ∪ {e} ∈ I
choose e with maximum weight we . Replace J with J ∪ {e} .
Example 4. Let
G = (V, E) be a graph. M ⊆ E is matching if no vertex is
incident to more than one edge. Let I = {matchinigs of G}, and say we want
to nd the matching with maximum sum of the edge weights. If we try Kruskal
greedy algorithm on the graph in Figure (1), The result will be to rst pick the
edge pq then rs for a total sum of 12. The maximum matching however is to
pick ps and qr for a sum of 15. So for this example the greedy approach fails.
Figure 1: Graph for matching example.
Denition 4. Let V be a nite set and let I be a family of subsets of V . We
call V the grounds set and I the independent sets. The pair m = (V, I) is called
a matroid if
(i) ∅ ∈ I
(ii) If J ∈ I and J 0 ⊆ J then J 0 ∈ I . (Down-closed)
(iii) ∀J 0 , J ∈ I with |J 0 | < |J| ∃ j ∈ J\J 0 such that J 0 ∪ {j} ∈ I . (Extension
axiom).
Denition 5.
B is a base of A ⊆ V if B ∈ I and @ x ∈ A\B such that
B ∪ {x} \I.
Lemma 2.
m = (V, I) is a matroid if and only if I fullls axiom (i) and (ii)
and all bases B of A where A ⊆ I have the same cardinality.
Proof. (⇒)Let B1 and B2 be basis. If |B1 | < |B2 | then B1 can be extended
which is a contradiction of the denition of base.
(⇐) For J 0 , J ∈ I where |J 0 | < |J| let A = J ∪ J 0 . J 0 is not a basis of A
⇒∃j ∈ A\J 0 = J\J 0 such that J 0 ∪ (j) ∈ I.
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Example 5. Let
weights
m = (V, I) be a matroid. Take a subset A ⊆ V with the
(
wij =
1 j∈A
0 otherwise.
For J ∈ I let w (J) = |J ∩ A| . If J ∈ I ⇒ J ∩A ∈ I and w (J ∩ A) = |J ∩ A| .
A base of A will solve the maximum independent set.
Example 6. Continuation of the spanning tree example Let m = (E, I) with
I = { forests of G}. The basis are spanning trees with |B| = |V | − 1.
Example 7. Continuation of the matching example. Let A = {pq, qr, rs} basis
of A are {pq, rs} and {qr}.
Example 8. Let
A be a matrix. A linear matroid M = (V, I) is a matroid
where I = {linearly indepdent columns of A} .
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