1 Approximation Algorithms for Demand- Robust and Stochastic Min-Cut Problems Vineet Goyal Carnegie...

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Approximation Algorithms for Demand-Robust and Stochastic Min-Cut Problems

Vineet Goyal

Carnegie Mellon University

Based on,

[Golovin, G, Ravi] (STACS’06)

[G, Gupta] (Manuscript’06)

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Why Robust?

Uncertainty in problem data and constraints for optimization.

Optimize for the worst case realization of uncertainties.

Stochastic optimization models uncertainty but does not address worst-case future.

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Demand-Robust Model

Two stage model motivated from two-stage stochastic optimization (IKMM’04, RS’04, GPRS’04, SS’04).

Models uncertainty in problem constraints or demands as well as the problem data.

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Demand-Robust Min-Cut Problem Given an undirected graph G, a cost function c

on edges, a root vertex r and a set of k future scenarios

Scenario i specifies a terminal ti, and

an inflation factor i, i.e.

cost in scenario i, ci(e)= i¢c(e)

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Demand-Robust Min-Cut Problem Goal :

first stage edges Ef and

second stage edges Esi, 8 scenario i

s.t. Ef[Esi separates r from ti

Objective : min [c(Ef) + maxi i¢c(Esi)]

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An Example

Root vertex r

Four possible scenarios terminal in scenario i is ti

Inflation factor = 3 for all scenarios

1.5 1

1.5

r

t1 t2

t3 t4

1

2 2

v1 v2

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A Feasible Solution

Buy edges (r,v1) and (r,v2) in the first stage.

First stage cost = 1.5+1.5 = 3 1.5 1

1.5

r

t1 t2

t3 t4

1

2 2

v1 v2

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A Feasible Solution

Buy edges (r,v1) and (r,v2) in the first stage.

First stage cost = 1.5+1.5 = 31

r

t1 t2

t3 t4

1

2 2

v1 v2

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A Feasible Solution

Buy edges (r,v1) and (r,v2) in the first stage.

First stage cost = 1.5+1.5 = 3

Maximum second stage cost is 3.

Total cost = 3 + 3 = 6

1

r

t1 t2

t3 t4

1

2 2

v1 v2

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Another Feasible Solution

Buy nothing today.

First stage cost = 0

Maximum second stage cost is 4.5.

Total cost = 0 + 4.5 = 4.5

1.5 1

1.5

r

t1 t2

t3 t4

1

2 2

v1 v2

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Previous Work Two stage model of demand-robustness was introduced

in [Dhamdhere, G, Ravi, Singh] (FOCS’05).

Prove a structural result for the first stage solution for a general covering problem.

Give approximation algorithms for min-cut, multicut, Steiner tree and facility location problems in the demand-robust model.

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Structural Theorem [DGRS’05]

There exists a first stage solution Ef which is a minimal solution for a subset of scenarios and it can be extended in the second stage to obtain a 2-approximate solution to the problem.

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Results [DGRS’05]

O(log n)-approximation for robust min-cut,

O(log n¢ loglog n)-approximation for robust multicut,

O(1)-approximation for Steiner tree and facility location problems.

Algorithms for robust min-cut (multicut) extend to stochastic min-cut (multicut) to give similar guarantees

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Improved Results

2-approximation for the robust min-cut problem [Golovin, G, Ravi] (STACS’06).

4-approximation for the stochastic min-cut problem [G, Gupta] (Manuscript’06).

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Robust Min-Cut: 2-approximationFor simplicity, we assume same inflation factor for all

scenarios.

Basic Idea: Suppose the maximum second stage cost in some optimal solution is C

Scenarios for which the individual best solution costs at most C/ (with respect to cost c) can be ignored: “they do not need any first stage help”.

Rest of the scenarios, OPT must help in the first stage.

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Warm-up (Exact Algorithm for Trees)

Suppose the maximum second stage cost in some optimal solution is C.

Ignore terminals for which the min-cut costs less than C/i (we can cut in second stage).

Separate the remaining terminals from the root by a minimum cost cut in the first stage.

2

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Algorithm (General case) Guess the second stage cost (say C) of some

optimal solution.

Ignore terminals for which the individual min-cut costs at most 2C/ (with respect to cost c).

Separate the rest of the terminals (say R) from root in the first stage by a minimum cost cut.

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Analysis

Clearly, maximum second stage cost incurred by our algorithm is at most 2C.

We need to prove that the minimum cut separating root from all the terminals in R is at most twice the first stage cost in OPT.

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Analysis

Fix an optimal solution, say edges Ef, Es1, Es

2,…, Es

k such that the maximum second stage cost is C.

We will construct a cut separating root from the terminals in R such that its cost is at most 2c(Ef).

R: set of terminals whose individual min-cut cost is more than 2C/

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Analysis

Remove edges Ef from G and consider the Gomory-Hu tree H on GnEf .

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Analysis Root H at the root

vertex r.

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Analysis Root H at the root

vertex r.

Consider terminals which do not have any parent terminals.

t1

t3

t2

r

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Analysis Consider min-cuts

from r corresponding to these maximal terminals

t1

t2

r

t3

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Analysis

t1

t2

r

t3

Es1

ti

v

Gn Ef

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Analysis

t1

t2

r

t3

Es1

ti

v

G

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Blue edges are edges of Ef

• We know r-t1 min-cut in G has cost more than 2C/

• Also, c(Es1) · C/

• Thus, c(blue edges) >C/

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Thus, we can buy Es1

and charge against the blue edges.

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Similarily, we can buy edges Es

2 and Es3 and

charge against corresponding blue edges.

Es2

Es3

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Similarily, we can buy edges Es

2 and Es3 and

charge against corresponding blue edges.

• Red edges and blue edges form the final first stage cut.

• Note that each blue edge is charged at most twice.

Es2

Es3

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Thus, c(red edges) · 2 c(Ef)

• Hence, we have a cut separating all terminals in R from root and has cost at most 3c(Ef) Es

2

Es3

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Analysis

t1

t2

r

t3

Es1

ti

v

G• In fact, we can construct a separating cut of cost at most 2c(Ef).

• Thus, we get a 2-approximation for the problem.

Es2

Es3

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Stochastic Min-Cut

Each scenario i has an associated probability p i and the objective is to minimize the expected cost of the solution (instead of maximum over all scenarios).

We obtain a 4-approximation for the stochastic min-cut problem.

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Natural IP formulation

Consider the following IP formulation, xf(e) is the first stage variable and

xsi(e) is the second stage variable for scenario i

min e2 E c(e)¢ xf(e) + i=1k pii¢ e2 E c(e)¢ xs

i(e)

(xf+xsi)(P) ¸ 1, 8 r-ti paths P, 8 i

xf, xsi 2 {0,1} 8 e, i

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MIP Formulation

Let i be the cost of root to ti min-cut in G with respect to cost c. Consider the following MIP: x(e) is the first stage variable and yi denotes whether terminal i is cut in first stage or not.

min e2 E c(e)¢ x(e) + i=1k pii¢(1-yi)¢i

dx(e)(r,ti) ¸ yi, 8 i=1,…,k

yi 2 {0,1}, 8 i=1,…,k

0 · x(e) · 1, 8 e2 E

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MIP Formulation

min e2 E c(e)¢ x(e) + i=1k piI ¢ (1-yi) ¢ i

dx(e)(r,ti) ¸ yi, 8 i=1,…,k

yi 2 {0,1}, 8 i=1,…,k

0 · x(e) · 1, 8 e2 E

Note that this is an approximate MIP. We can show a feasible solution of cost at most 2OPT.

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MIP Feasible Solution Consider an optimal solution, say Ef, Es

1,…,Esk. Thus,

OPT = c(Ef) + i=1k pii ¢ c(Es

i).

Let Efi be the edges that scenario i uses from first stage.

If c(Esi) ¸ c(Ef

i), then let yi=0 (we can buy a min-cut for ti in the second stage for at most twice the cost that OPT pays for Es

i).

For the remaining terminals, let yi=1. We show that all the terminals in R can be separated from root by a cut of cost at most 2c(Ef).

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Analysis Remove edges Ef from G

and consider the Gomory-Hu tree H on GnEf .

Root H at the root vertex r.

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Analysis

t1

t2

r

t3

Es1

ti

v

Gn Ef

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Analysis

t1

t2

r

t3

Es1

ti

v

G• Blue edges are edges of Ef

1

• We know c(Ef1) > c(Es

1)

• Thus, we can buy red edges in the first charging against the blue edges.

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Analysis

t1

t2

r

t3

Es1

ti

v

G

• Red edges and blue edges form the final cut.

• From the previous argument, we know that the cost of the final first stage cut · 2c(Ef).

Es2

Es3

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MIP is 2-approximate

min e2 E c(e)¢ x(e) + i=1k pii¢(1-yi)¢i

dx(e)(r,ti) ¸ yi, 8 i=1,…,k

yi 2 {0,1}, 8 i=1,…,k

0 · x(e) · 1, 8 e2 E

Thus, opt(MIP) is at most 2OPT.

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LP Relaxation : Rounding

min e2 E c(e)¢ x(e) + i=1k pii¢(1-yi)¢i

dx(e)(r,ti) ¸ yi, 8 i=1,…,k

0 · yi · 1, 8 i=1,…,k

0 · x(e) · 1, 8 e2 E

If yi < ½ , then ignore ti in the first stage. Separate rest of the terminals from root in the first

stage by a minimum cut.

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Rounding the LP

We can round the LP within a factor of 2. Therefore, we obtain a 4-approximation for the stochastic min-cut problem.

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Conclusions and Open Problems Our technique “guess and plan” crucially exploits the

structure of the demand-robust problem: every scenario in the second stage can pay up to the maximum second stage cost of OPT without worsening the solution cost.

Can we use these ideas for the multi-cut and Steiner tree problems in the demand-robust model?

Hardness of demand-robust and stochastic min-cut on undirected graphs is unknown. Natural LP relaxation has an integrality gap. Directed versions are NP-hard.

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Thank You