1. Submodularity Reading Group Polymatroid
M. Pawan Kumar

2. 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

3. 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

4. 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

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

6. Primal Problem max wTx 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

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

8. Greedy Algorithm max wTx x ∈ EPf
Order s1,s2,…,sn ∈ S such that w(si) ≥ w(si+1)
Define Ui = {s1,s2,..,si}
xGi = f(Ui) – f(Ui-1)
xG ∈ EPf
Proof?

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

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

11. Greedy Algorithm max wTx x ∈ EPf
Order s1,s2,…,sn ∈ S such that w(si) ≥ w(si+1)
Define Ui = {s1,s2,..,si}
xGi = f(Ui) – f(Ui-1)
xG is optimal
Proof?

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

13. Greedy Algorithm Order s1,s2,…,sn ∈ S such that w(si) ≥ w(si+1)
Define Ui = {s1,s2,..,si}
yGUi = w(si) - w(si+1)
yG is feasible
yGS = w(sn)
Proof?
yGA = 0, for all other A

14. Proof Sketch Trivially, yG ≥ 0 Consider si ∈ S
∑A∋si yGA = ∑j≥i yGUj = w(si)
∑A yAvA = w

15. Optimality Primal feasible solution xG Dual feasible solution yG
Primal value at xG = Dual value at yG
Proof?

16. Proof Sketch wTxG = ∑s∈S w(s)xGs = ∑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 yGA f(A)

17. Optimality Primal feasible solution xG Dual feasible solution yG
Primal value at xG = Dual value at yG
Therefore, xG is an optimal primal solution
And, yG is an optimal dual solution