1 Lecture notes Classication. Denition 1. Let z be a set function on the nite set V i.e. z : 2 V R, then z is submodular if

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1 Lecture notes 3 Denition. Let z be a set function on the nite set V i.e z : 2 V R, then z is submodular if z A) + z B) z A B) + z A B) A, B V. Denition 2. If z is submodular then z is supermodular. Denition 3. z is modular if z is both submodular and supermodular that is z A) + z B) = z A B) + z A B). The modular functions can be characterized as z A) = z ) + z {a}) ) a A. Classication Denition 4. z is a monotone function if z A) z B) whenever A B. Denition 5. z is a symmetric function if z A) = z V \A) A V. Denition 6. ρ j S) = z S {j}) z S). Theorem 7. The following things are equivalent i) z is a submodular function. ii) ρ j S) ρ j T ) S T V, j V \T. iii) ρ j S) ρ j S {k}) S V, j V \S {k}.

2 Proof. i) ii)) Take S T, j / T, A = S {j} and B = T. And insert into dention of submodular function z S {j}) + z T ) z T {j}) + z S) z S {j}) z S) z T {j}) z T ) ρ j S) ρ j T ). i) ii)) Let A\B = {j i,..., j r } Start with ρ j A B {j i,..., j r }) ρ j B {j i,..., j r }) Summing allr inequalities, using telescopic sum this becomes ii) iii)) Obvious. ii) iii)) Left as exercise. z A) z A B) z A B) z B) Example 8. Let V be the set of columns of matrix M. For A V, let r A) be the rank function then r is submodular. Example 9 Cut functions). Let G V, E) be a directed graph where V are the vertices and E the edges. Dene also a weight function w e) e E. Denition 0. The cut is dened asδ T A) = {e = i, j) E i A, j / A} Denition. The global cut function z A) = w δ T A) ) = w e) e δ T A) Denition 2. Given s, t V such that s t. Dene V st = V \ {s, t}. The s t cut function is z st : 2 Vst R such that z st A) = z A {s}) A V st. Lemma 3. If all weights w 0 then these cut functions z and z st are submodular. 2

3 Proof. In Figure each arrow w i indicate the sum of all edges going out from the region into another region. From this we get z A) + w + w 2 + w 4 + w 6 z B) = w 2 + w 3 + w 5 + w 7 z A B) = w 2 + w 6 + w 7 z A B) = w + w 2 + w 3. Using all this we get z A) + z B) z A B) z A B) = w 4 + w 5 > 0 Figure : The w i arrows indicate the sum of all edges..2 Operations If z A) is submodular then z V \A) is too. z and z 2 submodular then z + z 2 is submodular. Neither min z, z 2 ) nor maxz, z 2 ) is submodular in general when z, z 2 is submodular. Proposition 4. Let z and z 2 be submodular such that z z 2 is monotone. Then min z, z 2 ) is submodular. Proof. Exercise..3 Pseudo-boolean functions Let f : B n R be a pseudo-boolean function and associate with it z : 2 V R. Denition 5. Characteristic vector B n for S V is dened by { S if j S j = 0 otherwise. 3

4 z S) = f ) S j, and hence z A B) = f A B) = f A B), z A B) = f A B) = f A B), where is elementwise min and is elementwise max. Example 6. z is submodular if and only if f is ane. Example 7. f : B 2 R is submodular if and only if f, 0) + f 0, ) f 0, 0) + f, ). Notation: f 2 x i x j means ordinary derivative as for, e.g, polynomials. Theorem 8. The following things are equivalent i) f is a submodular function ii) f 2 x i x j 0 x B n iii) Any restriction of f to 2 variables f ij x i, x j ) is submodular. Proof. We show that i) ii) V = {,..., n}. Let S V, i, j / S and i j. Dene f x) = x i x j ϕ ij x) + x i ψ i x) + x j ψ j x) + ψ 0 x) 2) where ϕ ij, ψ i, ψ j, ψ 0 are independent of x i, x j. ρ i s) ρ j S {i}) = z S {i}) z S) z S {i, j}) z S {i})) = ψ i S ) + ψ 0 S ) ψ 0 S ) ϕ ij S ) + ψ i S ) + ψ j S ) + ψ 0 S ) ψ j S ) ψ 0 ) ) = So ϕ ij S ) 0 and for ) we have ii) iii) left as exercise. f 2 x i x j = ϕ ij. ϕ ij S ) 0 4

5 Corollary 9. A quadratic pseudo boolean function is submodular if and only if it can be written in the form f x,..., x n ) = a 0 + n a i x i i= i<j n where a ij 0 i, j. Or, equivalently, as a posiform a ij x i x j φ = a 0 + u a u u + a ij x i x j. Remark 20. x i x j is not submodular but ȳ i = x i, gives ȳ i x j = y i x j + y i whic is submodular..4 Cubic functions A cubic pseudo boolean function can be written uniquely as L + a ij x i x j + a ijk u i u j u k 3) such that i<j n i<j<k n. L is a linear posiform contaning only one of x i and x i ). 2. For each i, j, k either u i u j u k = x i x j x k or u i u j u k = x i x j x k. 3. a ijk < 0 i, j, k 4. a ij may be negative or positive. Theorem 2. The cubic canonical form 3) is submodular if and only if a ij 0 i, j. Proof. ) Fix two indicies i, j. We now show that any restriction of 3) L + a ij x i x j + b ij x i x j + c ij x i x j to i, j. is submodular. b ij and c ij will be sums of a ijk and hence negative. Furthermre, x i x j = x i x j + linear term. by Collorary 9) the restriction is submodular. Since this holds for any restriction, we can use8 to conclude that 3) is submodular. ) Consider two indices i, j we want to show a ij 0. Look at the following restriction if a ijk x i x j x k then x k = 0 if a ijk x i x j x k then x k =, by this construction all a ijk terms vanishes so 3) becomes L + a ij x i x j a ij 0. 5

6 .5 Graph representation of cubics x i x j x k = min y B x i + x j + x k ) y = = min y x i + x j + x k ) y + ȳ. Replace each a ijk x i x j x k with a ijk > 0 by x i + x j + x k ) y + ȳ. Similary replace a ijk x i x j x k with a ijk x i ȳ + x j ȳ + x k ȳ + y). 6