ClassW04ApproxAlgs/QiFu

Tutorial: Multiway Cut Problem

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1. Problem description:
Given:

• an undirected graph G=(V,E)
• a nonnegative cost function c(u,v)
• a set of terminals S ⊆ V : {s₁, s₂, ... s_k}

Ask for:

• a multiway cut C ⊆ E such that the removal of all edges in C from E disconnects each terminal from all the other terminals.
• cost of C: ∑_{e∈ C} c(e) should be minimized.

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2. Example:

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3. NP-hardness of the problem.

The problem of finding a minimum weight multiway cut is NP-hard for any fixed k ≥ 3.
When k=2, it is the minimum s-t cut problem which can be solved in polynomial time using Maximum Flow algorithm.

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4. A 2-2/k approximation algorithm:

1. For each i = 1, 2, ..., k, compute a minimum weight isolating cut for s_i , say C_i .
   
2. C = C₁ ∪ C₂ ∪ ... ∪ C_k - C_j in which C_j is the cut with the maximum weight.
   
3. Output C.
   

What's isolating cut?

• An isolating cut for s_i is a set of edges whose removal disconnects s_i from the rest of the terminals.
  

This can be done by identifying the terminals in S-{s_i} into a single node, and finding a minimum cut separating this node from s_i, this takes one max-flow computation.
Notice that C contains k-1 sets of isolating cut and the removal of C from the graph G disconnects every pair of terminals, so C is a multiway cut.

Analysis of the approximation ratio: 2-2/k

• Let A be the set of edges in the optimal solution.
  
• Remove all edges in A from the graph:
  
  • create k connected components, no more, no less;
    
  • every terminal s_i is in a different connected component;
    
  • let A_i ⊆ A be the isolating cut for s_i.
    
• Every edge appears in exactly two sets, so:

          ∑i=1k c(Ai) = 2c(A) = 2 • OPT 

• C_i is the minimum isolating cut for s_i, so:
  

          c(Ci) ≤ c(Ai) 

• The isolating cut with the heaviest cost C_j is discarded, so:
  

          c(Cj) ≥ 1/k ∑ c(Ci) 

• Finally we get:
  

          c(C) ≤ (1-1/k) ∑i=1k c(Ci) ≤ (1-1/k) ∑i=1k c(Ai) ≤ (2-2/k) OPT 

An example showing this ratio is tight:


The blue cut is the approximation solution, c(C) = 6-3ε
The red cut is the OPT solution, c(A) = 4

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5. An 3/2-1/k approximation algorithm:
This is an interesting relaxation of linear program by Gruia Calinescu, et al.

• Minimize ∑_{(u,v)∈ E} c(u,v)d(u,v) subject to:
  
  • d(u,v) = 1/2 ∑_i=1^k | x_uⁱ - x_vⁱ |     (u,v) ∈ E
    
  • x_v ∈ Δ_k                                     ∀ v ∈ V
    
  • x_t = e_t                                        ∀ t ∈ S
    

Explainations:

• Δ_k denotes the k-1 simplex, a k-1 dimensional convex polytope in R^k defined by:

            {x ∈ Rk | x ≥ 0 and ∑i=1k xi = 1} .

• For x ∈ R^k , ||x|| is its L₁ norm, ||x|| = ∑_i=1^k ||x_i|| .
  
• e_t denotes the unit vector given by (e_j)^j = 1, (e_j)ⁱ = 0 for all i¬= j.
  
• Vertices of G are mapped to the points in Δ_k, terminals are mapped to simplex corners.
• Every vertex v is associated with a k-dimensional vector x_v, with each component between 0 and 1. For k = 7: x_v = (x_v¹, x_v², x_v³, x_v⁴, x_v⁵, x_v⁶, x_v⁷)
• An integral solution maps each vertex of G to a vertex of the simplex, respectively. Each edge (u,v) has length 0 or 1. Edges of length 1 form a multiway cut.Optimal integral solution corresponds to an optimal multiway cut.
  
• Terminals have forms:

           x3 = (0,0,1,0,0,0,0); x7 = (0,0,0,0,0,0,1).

• Non-terminals:
  • Integer solution:
    • x_v^t = 1 iff v and t in the same partition.
    • x_v^t = 0 iff v and t in the different partitions.
    • Exactly one coordinate is 1, all others are 0.
  • Fractional solution:
    • 0 ≤ x_v^t ≤ 1
    • ∑_t=1^k x_v^t = 1
    • x_v = (0, 0.2, 0, 0, 0.4, 0, 0.4)
• The distance between two vertices is defined as:
  

              d(u,v)=1/2 ∑t=1k | xut - xvt | 

• d(u,v)=0 ⇒ u and v in the same partition;
• d(u,v)=1 ⇒ u and v in different partitions;e.g. d(s₁,s₂) = 1/2(|1-0|+|0-1|+|0-0|) = 1 in the following figure.
• The fractional solution will give us 0 ≤ d(u,v) ≤ 1 , we need some technique to decide how to partition those vertices.

             e.g.  d(u,s1) = 1/2(|0.5-1|+|0.5-0|+|0-0|) = 0.5  in the following figure.

Example of G with 3 terminals {s₁, s₂, s₃} and 3 non-terminals {u, v, w}:

In this example, the optimal fractional solution has cost 7.5:

• 6 edges with weight 2, 3 edges with weight 1, all edges has distance 1/2, so,
  
• ∑_{(u,v)∈ E} c(u,v)d(u,v) = (2*1/2) * 6 + (1*1/2) * 3 = 7.5 .

How to partition node u, v and w?
1.We need normalization before present the algorithm:

• Let A be a feasible solution to the LP.
• It's possible to construct a graph G=(V,E) such that:
  • The value of A is at most A with same cost.
  • For all edges (u,v)∈ E the vectors x_u and x_v differ at most 2 coordinates.
• This can be done by dividing edges with adding dummy vertex.
• The transformation requires at most k|E| subdivisions.

2.We need some definitions before present the algorithm:

• x = optimal, normalized solution of LP.
• E_t = {(u,v) ∈ E | x_u^t ¬= x_v^t}
  • edges between vertices that differ in coordinate t.
  • each edge (u,v) appears in two of these sets.
• W_t = ∑_{e∈ Ei} c(e)d(e)
  • cost of all edges in E_t, weighted by the distances.
  • each edge with d(e) > 0 is counted twice.
• B(t,ρ) = {v ∈ V | x_v^t ≥ ρ}, for ρ ∈ [0,1]
  • set of all vertices that are with small distance to terminal t.
  • v is "close" to i ⇔ x_v^t ≥ ρ

3.The algorithm:

1. Compute the optimal, normalized fractional solution x of LP;
2. Relabel the terminals so that W_k is the largest among all W_i;
3. Pick random σ in two permutations: {(1,2,...,k-1,k), (k-1,k-2,...,1,k)};
4. Pick random ρ∈[0,1];
5. For i = 1 to k-1 do V_σ(i)=B(i,ρ)-∪_j<iV_σ(j);
6. V_k = V - ∪_i<k V_i;
7. Return C = {(u,v) ∈ E | u∈ V_i, v∈ V_j, i ¬= j} .

Analysis of the approximation ratio: 3/2 - 1/k
1. Intuition:

• The probability of an edge being cut = the probability of ρ falling in between the interval (x_uⁱ,x_vⁱ) = the distance of these two interval.
  

Every color represents a choice of ρ, only when the color line falls into interval (u,v), the edge (u,v) is cut by partition v into S1, partition u into S3.
This gives us probability d(e).

With the same ρ but different σ:

• In the left graph, set 1 is partitioned first, u and v are both picked, edge (u,v) is hidden to set 2.
• In the right graph, set 2 is partitioned first, with the same ρ, u is partitioned into set 2, v is partitioned into set 1. Edge (u,v) is made a cut.
• So σ increase the probability of an edge being cut.
• This gives us probability d(e)/2.

The summation of previous d(e) and d(e)/2 gives us the approximation ratio 3/2. And since the heaviest cut is discarded, we get 3/2-1/k.

2. Formal analysis:

   (1) If  e ∈ E - Ek , Prob[e∈ C]≤ 1.5• d(e) 

• Let e = (u,v), let i and j be the coordinates in which x_u and x_v differ.
• There are two cases: the intervals [x_uⁱ,x_vⁱ] and [x_v^j,x_u^j] either overlap or disjoint, but having the same length d(e):
  

• vertices u and v can end up in one of 3 sets, V_i , V_j , V_k. (other than these 3 sets make no difference)
• if ρ ∈ [0,1]-(α ∪ β) , u and v end up in the same set, ⇒ edge e will not be in the cut.
  • P₁ = Pr[ρ ∈ (α ∪ β)] = |α| + |β|
    
• special case:
  • ρ ∈ α and σ(j) < σ(i)
  • both u and v will end up in S_j
  • P₂ = Pr[(ρ ∈ α) and (σ(j) < σ(i))] = |α|/2
• Pr[e ∈ C] = P₁ - P₂ = |β| + |α|/2 ≤ 1.5d(e)

   (2) If  e ∈ Ek , Prob[e∈ C]≤ d(e) 

• Let e=(u,v) and i and k be the coordinates in which x_u and x_v differ.
  • σ(i) < σ(k) because set k is always the last set in the two permutations;
  • u and v will end up in different sets if ρ falls between interval x_u and x_v:

                      Pr[(u,v) ∈ C] = |xui - xvi| = d(e)

   (3) From the analysis of (1) and (2):

                      E[c(C)] = ∑e∈ E c(e)Pr[e∈ C] 
                              = ∑e∈ E-Ek c(e)Pr[e∈ C] + ∑e∈ Ek c(e)Pr[e∈ C] 
                              ≤ 1.5 ∑e∈ E-Ek c(e)d(e) + ∑e∈ Ek c(e)d(e) 
                              = 1.5 ∑e∈ E c(e)d(e) - 0.5 ∑e∈ Ek c(e)d(e) 
                              ≤ (1.5-1/k) • OPTf 

Discussion:

• The approximation ratio 3/2-1/k is not tight.
• According to Gruia Calinescu, et al, the worst case integrality ratio is 16/15 when k=3.
• According to David Karger, et al, the integrality ratio 12/11 with k=3 using a different relaxation: Rounding Algorithms for a Geometric Embedding of Minimum Multiway Cut

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Referrence:

[1] Vijay V. Vazirani Approximation Aglorithms, Springer, 2001.
[2] E. Dahlhaus, et al. The Complexity of Multiway Cuts, Extended abstract, 1983.
[3] Gruia Calinescu, et al. An Improved Approximation Algorithm for Multiway Cut, Journal of Computer and System Sciences, 60(3):564-574, 2000.