Polyhedral Optimization Lecture 5 – Part 2 M. Pawan Kumar Slides available online

(Extended) Polymatroid Optimization over Polymatroids Minimizing Submodular Functions Outline

Polymatroid Set S Real vector x of size |S|x1 Submodular function f P f = {x ≥ 0, x(U) ≤ f(U) for all U ⊆ S} EP f = {x(U) ≤ f(U) for all U ⊆ S} Polymatroid Extended Polymatroid

Tight Sets Set SSubmodular function f EP f = {x(U) ≤ f(U) for all U ⊆ S} U is tight with respect to x ∈ EP f if x(U) = f(U) Tight sets are closed under union Tight sets are closed under intersection Proof?

Proof Sketch Let T and U be tight wrt x ∈ EP f ≥ x(T ∪ U) + x(T ∩ U) f(T) + f(U) ≥ f(T ∪ U) + f(T ∩ U) = x(T) + x(U) = f(T) + f(U) All inequalities must be equalities

(Extended) Polymatroid Optimization over Polymatroids Minimizing Submodular Functions Outline

Primal Problem max w T x x ∈ EP f Assume f(null set) ≥ 0 Otherwise EP f is empty f(null set) can be set to 0Why? Decreasing f(null set) maintains submodularity

Primal Problem max w T x x ∈ EP f Assume w ≥ 0 Otherwise the optimal solution is infinityWhy? Let us first try to find a feasible solution

Greedy Algorithm max w T x x ∈ EP f Order s 1,s 2,…,s n ∈ S such that w(s i ) ≥ w(s i+1 ) Define U i = {s 1,s 2,..,s i } x G i = f(U i ) – f(U i-1 ) x G ∈ EP f Proof?

Proof Sketch We have to show that x G (T) ≤ f(T) for all T ⊆ S Trivial when T = null set Mathematical induction on |T|

Proof Sketch Let k be the largest index such that s k ∈ T Clearly |T| ≤ |U k | x G (T) = x G (T\{s k }) + x G k = f(T\{s k }) + f(U k ) - f(U k-1 ) ≤ f(T) ≤ f(T\{s k }) + x G k Why? Induction Why? Submodularity Definition

Dual Problem max w T x x ∈ EP f min ∑ T y T f(T) y T ≥ 0, for all T ⊆ S ∑ T y T v T = w Let us first try to find a feasible dual solution

Greedy Algorithm Order s 1,s 2,…,s n ∈ S such that w(s i ) ≥ w(s i+1 ) Define U i = {s 1,s 2,..,s i } y G U i = w(s i ) - w(s i+1 ) y G is feasible y G S = w(s n ) y G T = 0, for all other T Proof?

Proof Sketch Trivially, y G ≥ 0 Consider s i ∈ S ∑ T ∋ s i y G T = ∑ j≥i y G U j = w(s i ) ∑ T y T v T = w

Optimality Primal feasible solution x G Dual feasible solution y G Primal value at x G = Dual value at y G Proof?

Proof Sketch w T x G = ∑ s ∈ S w(s)x G s = ∑ i ∈ {1,2,…,n} w(s i )(f(U i ) - f(U i-1 )) = ∑ i ∈ {1,2,…,n-1} f(U i )(w(s i ) - w(s i+1 )) + f(S)w(s n ) = ∑ T y G T f(T)

Optimality Primal feasible solution x G Dual feasible solution y G Primal value at x G = Dual value at y G Therefore, x G is an optimal primal solution And, y G is an optimal dual solution

(Extended) Polymatroid Optimization over Polymatroids Minimizing Submodular Functions Outline

Submodular Function Minimization min T ⊆ S f(T) We will assume f(null set) = 0 If not, we can add a constant

Submodular Function Minimization min T ⊆ S f(T) Brute force search is exponential in |S| We will prove that SFM is easy First, we need two properties

Property 1 f(U) = max {x(U) | x ∈ EP f } Set w = v U Proof? f is a submodular function over S

Property 2 f is a submodular function over S Define f’(U) = min T ⊆ U f(T) f’ is also submodularProof?

Proof Sketch We have to prove the following f’(T) + f’(U) ≥ f’(T ∪ U) + f’(T ∩ U) for all T, U ⊆ S f’(T) = f(X) for some X ⊆ T f’(U) = f(Y) for some Y ⊆ S

Proof Sketch f’(T) + f’(U)= f(X) + f(Y) ≥ f(X ∪ Y) + f(X ∩ Y) ≥ f’(T ∪ U) + f’(T ∩ U)

Property 2 Continued f is a submodular function over S Define f’(U) = min T ⊆ U f(T) f’ is also submodular EP f’ = { x ∈ EP f, x ≤ 0} Proof?

Proof Sketch If x ∈ P, then x(U) ≤ f(U) for all U ⊆ S x(T\U) + x(U) ≤ f(U), for all U ⊆ T ⊆ S Why? Because x ∈ EP f Why? Because x ≤ 0 We can show that { x ∈ EP f, x ≤ 0} = P ⊆ EP f’

Proof Sketch We can show that { x ∈ EP f, x ≤ 0} = P ⊇ EP f’ If x ∈ EP f’ then x ∈ EP f For any s ∈ S, x s ≤ 0 Why? Because x(U) ≤ f(U) for all U ⊆ S Why? Because x s ≤ f(null set) = 0

Submodular Function Minimization min T ⊆ S f(T) = f’(S) f’(U) = min T ⊆ U f(T) Optimization over EP f is easy = max{x(S) | x ∈ EP f’ } = max{x(S) | x ∈ EP f, x ≤ 0} Separation over EP f is easy Separation over EP f’ is easy Optimization over EP f’ is easy Hence Proved