polyhedral optimization lecture 5 – part 2 m. pawan kumar [email protected] slides available...
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Polyhedral OptimizationLecture 5 – Part 2
M. Pawan Kumar
Slides available online http://cvn.ecp.fr/personnel/pawan/
• (Extended) Polymatroid
• Optimization over Polymatroids
• Minimizing Submodular Functions
Outline
Polymatroid
Set S
Real vector x of size |S|x1
Submodular function f
Pf = {x ≥ 0, x(U) ≤ f(U) for all U S}⊆
EPf = {x(U) ≤ f(U) for all U S}⊆
Polymatroid
Extended Polymatroid
Tight Sets
Set S Submodular function f
EPf = {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 wTx
x EP∈ f
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 wTx
x EP∈ f
Assume w ≥ 0
Otherwise the optimal solution is infinity Why?
Let us first try to find a feasible solution
Greedy Algorithm
max wTx
x EP∈ f
Order s1,s2,…,sn S such that w(s∈ i) ≥ w(si+1)
Define Ui = {s1,s2,..,si}
xGi = f(Ui) – f(Ui-1) xG EP∈ f Proof?
Proof Sketch
We have to show that xG(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 sk T∈
Clearly |T| ≤ |Uk|
xG(T) = xG(T\{sk}) + xGk
= f(T\{sk}) + f(Uk) - f(Uk-1)
≤ f(T)
≤ f(T\{sk}) + xGk Why?Induction
Why?
Why?Submodularity
Definition
Dual Problem
max wTx
x EP∈ f
min ∑T yT f(T)
yT ≥ 0, for all T S⊆
∑T yTvT = w
Let us first try to find a feasible dual solution
Greedy Algorithm
Order s1,s2,…,sn S such that w(s∈ i) ≥ w(si+1)
Define Ui = {s1,s2,..,si}
yGUi = w(si) - w(si+1) yG is feasible
yGS = w(sn)
yGT = 0, for all other T
Proof?
Proof Sketch
Trivially, yG ≥ 0
Consider si S∈
∑T s∋ i yG
T = ∑j≥i yGUj = w(si)
∑T yTvT = w
Optimality
Primal feasible solution xG
Dual feasible solution yG
Primal value at xG = Dual value at yG
Proof?
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)
= ∑T yGT f(T)
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
• (Extended) Polymatroid
• Optimization over Polymatroids
• Minimizing Submodular Functions
Outline
Submodular Function Minimization
minT S⊆ f(T)
We will assume f(null set) = 0
If not, we can add a constant
Submodular Function Minimization
minT 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 = vU
Proof?
f is a submodular function over S
Property 2
f is a submodular function over S
Define f’(U) = minT U⊆ f(T)
f’ is also submodular Proof?
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) = minT U⊆ f(T)
f’ is also submodular
EPf’ = { 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 xs ≤ f(null set) = 0
Submodular Function Minimization
minT S⊆ f(T) = f’(S) f’(U) = minT U⊆ f(T)
Optimization over EPf is easy
= max{x(S) | x EP∈ f’}
= max{x(S) | x EP∈ f, x ≤ 0}
Separation over EPf is easy
Separation over EPf’ is easy
Optimization over EPf’ is easy
HenceProved