Submodularity Reading Group Polymatroid M Pawan Kumar http

Submodularity Reading Group Polymatroid M. Pawan Kumar http: //www. robots. ox. ac. uk/~oval/

Submodular Function Ground set S Function f over power set of S f(T) + f(U) ≥ f(T ∪ U) + f(T ∩ U) for all T, U ⊆ S

Diminishing Returns Define df(s|T) = f(T ∪{s}) - f(T) Gain by adding s to T If f is submodular, df (s|T) is non-increasing df (s|T) ≥ df(s|U), for all T ⊆ U

A Clarification on Notation Vector: x (in bold font) An element of the vector: xi (non-bold x) Consider ground set S = {1, 2, …, n} x(U) where U ⊆ S: ∑i∈U xi

Polymatroid Set S Submodular function f Real vector x of size |S|x 1 Pf = {x ≥ 0, x(U) ≤ f(U) for all U ⊆ S} Polymatroid EPf = {x(U) ≤ f(U) for all U ⊆ S} Extended Polymatroid

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

Primal Problem max w. Tx x ∈ EPf Assume w ≥ 0 Otherwise the optimal solution is infinity Why?

Greedy Algorithm max w. Tx x ∈ EPf Order s 1, s 2, …, sn ∈ S such that w(si) ≥ w(si+1) Define Ui = {s 1, s 2, . . , si} x. Gi = f(Ui) – f(Ui-1) x. G ∈ EPf Proof?

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

Proof Sketch Let k be the largest index such that sk ∈ A Clearly |A| ≤ |Uk| x. G(A) = x. G(A{sk}) + x. Gk ≤ f(A{sk}) + x. Gk Induction = f(A{sk}) + f(Uk) - f(Uk-1) Definition ≤ f(A) Submodularity Why?

Greedy Algorithm max w. Tx x ∈ EPf Order s 1, s 2, …, sn ∈ S such that w(si) ≥ w(si+1) Define Ui = {s 1, s 2, . . , si} x. Gi = f(Ui) – f(Ui-1) x. G is optimal Proof?

Dual Problem max w. Tx x ∈ EPf min ∑A y. A f(A) y. A ≥ 0, for all A ⊆ S ∑A y. Av. A = w Let us first try to find a feasible dual solution

Greedy Algorithm Order s 1, s 2, …, sn ∈ S such that w(si) ≥ w(si+1) Define Ui = {s 1, s 2, . . , si} y. GUi = w(si) - w(si+1) y. GS = w(sn) y. GA = 0, for all other A y. G is feasible Proof?

Proof Sketch Trivially, y. G ≥ 0 Consider si ∈ S ∑A∋si y. GA = ∑j≥i y. GUj = w(si) ∑A y. Av. A = 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. Tx. G = ∑s∈S w(s)x. Gs = ∑i∈{1, 2, …, n} w(si)(f(Ui) - f(Ui-1)) = ∑i∈{1, 2, …, n-1} f(Ui)(w(si) - w(si+1)) + f(S)w(sn) = ∑A y. GA f(A)

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