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Edge-connectivity of undirected and directed hypergraphs

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40 Chapter 2. Submodular functions<br />

According to Claim 2.22 we can assume that all tight sets are in G. If X <strong>and</strong> Y are<br />

tight, then by Claim 2.20, 0 = p ∗ (Y ) + p ∗ (V − Y ) = p ∗ (X ∩ Y ) + p ∗ (Y − X) + p ∗ (V − Y ) ≤<br />

p ∗ (X ∩ Y ) + p ∗ (V − (X ∩ Y ), so X ∩ Y is tight, <strong>and</strong> similarly Y − X is tight. This means<br />

that the minimal tight sets form a partition W1, . . . , Wk. The conditions <strong>of</strong> the theorem<br />

are not changed if we add a modular set function m to p for which m(V ) = 0. Thus we<br />

can assume that p ∗ (W1) = · · · = p ∗ (Wk) = 0, so p ∗ (X) = 0 for every tight set X.<br />

We say that sets X, Y are crossing restricted to Wl if X ∩ Wl <strong>and</strong> Y ∩ Wl are crossing<br />

on the ground set Wl. For a given 1 ≤ l ≤ k we define a matrix Al. The rows <strong>of</strong><br />

Al correspond to the pairs X, Y ∈ G that are crossing restricted to Wl, <strong>and</strong> for which<br />

p ∗ (X, Y ) = p ∗ (X) + p ∗ (Y ). The columns <strong>of</strong> Al correspond to the sets in G. In the<br />

following we define the row <strong>of</strong> Al corresponding to a pair (X, Y ) <strong>of</strong> sets. There is a<br />

composition FX∩Y <strong>of</strong> X ∩ Y <strong>and</strong> a composition FX∪Y <strong>of</strong> X ∪ Y , both consisting <strong>of</strong> sets<br />

in G, for which p∗ (X, Y ) = p∗ (X ∩ Y ) + p∗ (X ∪ Y ) = <br />

Z∈FX∩Y p∗ (Z) + <br />

Z∈FX∪Y p∗ (Z).<br />

Observe that FX∩Y + FX∪Y contains at least 2 non-tight sets (because Wl is a minimal<br />

tight set), <strong>and</strong> if it contains exactly 2, then restricted to Wl these are (X ∩ Y ) ∩ Wl <strong>and</strong><br />

(X ∪ Y ) ∩ Wl. Using FX∩Y <strong>and</strong> FX∪Y , we define the row r <strong>of</strong> Al corresponding to X, Y :<br />

for a set Z ∈ G, let r(Z) := −χFX∩Y (Z) − χFX∪Y (Z) if Z = X <strong>and</strong> Z = Y , <strong>and</strong> let<br />

r(X) = r(Y ) := 1 − χFX∩Y − χFX∪Y (Z).<br />

Claim 2.23. The system {yAl ≤ 0, y > 0} has no solution.<br />

Pro<strong>of</strong>. Suppose indirectly that there exists such a y; we can assume that y is integral.<br />

The vector −yAl can be considered as a non-negative function whose domain <strong>of</strong> definition<br />

is G. Let F be the family defined by χF(Z) := −yAl(Z) (Z ∈ A), <strong>and</strong> let F ′ be the<br />

family obtained by leaving out the tight sets from F. By the definition <strong>of</strong> the rows <strong>of</strong> Al,<br />

<br />

Z∈F p∗ (Z) = 0, <strong>and</strong> F is regular. This is possible only if F ′ = ∅. But in a row <strong>of</strong> Al<br />

corresponding to a pair (X, Y ), exactly 2 columns corresponding to non-tight sets have<br />

value +1 (those corresponding to X <strong>and</strong> Y ), <strong>and</strong> at least 2 columns corresponding to non-<br />

tight sets have negative value. This is possible only if each row <strong>of</strong> Al for which y is positive<br />

contains exactly 2 columns corresponding to non-tight sets that have negative value (which<br />

is -1). We have seen that for a row corresponding to (X, Y ) the intersection <strong>of</strong> these non-<br />

tight sets with Wl must be (X ∩ Y ) ∩ Wl <strong>and</strong> (X ∪ Y ) ∩ Wl. But |X ∩ Wl| 2 + |Y ∩ Wl| 2 <<br />

|(X ∩ Y ) ∩ Wl| 2 + |(X ∪ Y ) ∩ Wl| 2 , so <br />

Z∈F ′ |Z ∩ Wl| 2 > 0, contradicting F ′ = ∅.<br />

According to the Lemma 1.16 the dual system {Alz ≫ 0, z ≥ 0} has a solution. Let zl<br />

be an arbitrary solution <strong>of</strong> the above system (for l = 1, . . . , k). This means that zl is a<br />

non-negative weight function on the sets <strong>of</strong> G, with the following property:

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