Transcript Survivable Network Design with Degree or Order Constraints

Approximation Algorithms:
Bristol Summer School 2008
Seffi Naor
Computer Science Dept.
Technion
Haifa, Israel
Outline
1. Basics
2. Network design – the primal-dual method
3. Network design – iterative rounding and iterative relaxation
4. Competitive analysis via the primal-dial method
Outline
1. Basics
2. Network design – the primal-dual method
3. Network design – iterative rounding and iterative relaxation
4. Competitive analysis via the primal-dial method.
What is an Approximation Algorithm?
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Returns a sub-optimal solution for an optimization problem
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For a minimization problem: for every input I, ALG(I ) · ®
OPT(I )
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Running time: polynomial
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Then, ALG is an ®-approximation
Why are Approximation Algorithms needed?
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Many interesting/useful optimization problems are NP-hard
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Optimal solution cannot be computed
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How well can they be approximated?
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Leads to beautiful algorithmic theory.
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And, the problems still need to be solved in practice …
Simple Example: Vertex Cover
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Input: undirected graph G=(V,E)
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Output: a minimum cardinality V’ µ V such that
V’ Å e  ;
for each e 2 E
Classic NP-hard problem
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Approximation Algorithm:
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Find a maximal (inclusion-wise) matching M
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Vertex Cover V’ = V(M) , vertices adjacent to edges in M
Analysis
Feasibility of cover: V(M) covers all edges,
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Every edge must have an end point in V(M)
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Otherwise, M is not a maximal matching!
Approximation factor:
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Optimal solution must pick a representative from each
edge e 2 M
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Therefore, OPT ¸ V(M)/2
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Approximation factor = V(M)/OPT · 2
Question: Can this factor be improved? Probably not …
Linear Programming
Optimize a linear function over a set of linear constraints:
Xn
minimize
cj x j
j=1
minimize c ¢ x
subject to:
Ax ¸ 0
subject t o:
X¸0
Xn
ai j x j ¸ bj
1· j · n
j=1
x1; : : : ; xn ¸ 0
Linear programming is solvable
in polynomial time [Khachiyan ’79]
Back to Vertex Cover
Vertex cover can be formulated as integer/linear program:
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wv – weight of vertex v
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xv - indicator variable for choosing vertex v
X
minimize
wv ¢x v
v2 V
8 edge e = (u; v): x u + x v ¸ 1
x v 2 f 0; 1g
Relaxation: 0 · xi · 1
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LP can be solved in poly time
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LP provides a lower bound on
optimal integral solution!!!
Rounding Algorithm:
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if xv ¸ ½, then xv à 1, else xv à 0
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for each edge (u,v), at least one of xv, xu ¸ ½
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Approximation factor = 2
Duality in Linear Programming
Xm
Xn
minimize
cj x j
maximize
j=1
j=1
subject t o:
subject t o:
Xn
Xm
ai j x j ¸ bj
bi yi
1· i · m
ai j yi · cj
j=1
i= 1
x1; : : : ; xn ¸ 0
y1 ; : : : ; ym ¸ 0
Primal
1· j · n
Dual
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primal constraint $ dual variable
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dual constraint $ primal variable
Weak Duality Theorem: for u feasible dual solution
and w feasible primal solution, b·u · c·w
Set Cover
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Elements: U ={1,2, … ,n}
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Sets: S1,…,Sm
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Each set Si has cost ci
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Goal: find a min cost collection of sets that cover U
(each Si µ {1,2, … ,n})
set cover can be formulated as integer/linear program:
xi - indicator variable for choosing set Si
Xm
minimize
Relaxation: 0 · xi · 1
ci x i
i= 1
X
for every element j :
xi ¸ 1
i j j 2 Si
x i 2 f 0; 1g
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LP can be solved in poly time
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LP provides a lower bound on
optimal integral solution!!!
Rounding a Fractional Solution (1)
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For each Si: 0 · xi · 1
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For each element j: i xi ¸ 1
(summed over xi, j 2 Si)
Randomized Rounding:
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For each set Si: pick it to the cover with probability xi
Analysis:
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Exp[cost of cover] = i ci xi = LP cost
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Pr[element j is not covered] =
µ
Y
=
(1 ¡ x ` ) ·
` j j 2 S`
1¡
P
` x`
k
¶k
µ
·
1
1¡
k
¶k
1
·
e
Conclusion: probability of covering element j is at least a constant!
Rounding a Fractional Solution (2)
Amplify probability of success:
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Repeat experiment clogn times so that
µ ¶ c l og n
1
1
Pr[element j is not covered] ·
·
e
2n
Analysis:
1
1
·
2n
2
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Pr[some element is not covered] · n ¢
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Exp[cost of cover] = O(logn) i ci xi = O(logn)(LP cost)
Conclusion: approximation factor is O(logn)
Set Cover: Greedy Algorithm
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Initially: C is empty
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While there is an uncovered element:
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Add to C the set Si minimizing
ci/(# new elements covered)
Analysis: via dual fitting
Xm
minimize
Xn
maximize
ci x i
yi
j=1
i= 1
X
for every element j :
xi ¸ 1
X
for every set Si :
j 2 Si
i j j 2 Si
yj ¸ 0
xi ¸ 0
Primal: covering
¸
yj · ci
dual: packing
Dual Fitting (1)
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Primal solution is feasible
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Define a dual solution: if element j is covered by set Si then
yj = ci/(# new elements covered by Si)
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In the iteration in which Si is picked:
¢ primal = ¢ dual = ci
since cost of Si is “shared” between new elements covered by it
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Thus, cost of primal solution = cost of dual solution
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But is the dual solution feasible? Almost, but not quite …
Dual Fitting (2)
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Consider set S
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Suppose the elements in S are covered in the order e1, … , ek
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c(S)
When element ei is covered, yei ·
k¡ i+ 1
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Thus,
Xk
Xk
yei ·
i= 1
i= 1
Xk 1
c(S)
· c(S) ¢
· c(S) ¢H (k)
k¡ i + 1
i
i= 1
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Dividing dual variables by H(n) ¼ logn yields a feasible solution
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Greedy algorithm is O(logn)-approximation: primal · dual £ H(n)
Summary
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Linear programming: useful method for relaxing combinatorial
problems
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Randomized rounding: going from a fractional solution to an integral
solution without incurring too much damage
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Dual programs:
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useful for generating lower bounds
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useful for analyzing “primal” algorithms