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1
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)
2
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.
3
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.
4
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)
5
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)]
6
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
7
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
8
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
9
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
10
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
11
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.
12
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.
13
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
14
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).
15
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.
19
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/
20
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
24
Analysis
t1
t2
r
t3
Es1
ti
v
Gn Ef
25
Analysis
t1
t2
r
t3
Es1
ti
v
G
26
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/
27
Analysis
t1
t2
r
t3
Es1
ti
v
G• Thus, we can buy Es1
and charge against the blue edges.
28
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
29
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
30
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
31
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
32
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.
33
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
35
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.
36
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).
37
Analysis Remove edges Ef from G
and consider the Gomory-Hu tree H on GnEf .
Root H at the root vertex r.
38
Analysis
t1
t2
r
t3
Es1
ti
v
Gn Ef
39
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.
40
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
41
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.
43
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.
44
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.
45
Thank You