Approximation slides 1. Cut problems. The character flaw of of Michael Myers, Freddy Kruger and Jason Voorhees: Cut problems

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1 Approximation slides 1 Cut problems The character flaw of of Michael Myers, Freddy Kruger and Jason Voorhees: Cut problems

2 Approximation slides 2 The directed multicut problem S j T j S 1 Figure 1: The directed multicut problem A cut Sphere B can contain both s j,t j. This is a problem, but not the main one... The problems are not separated because of edges entering B. Thus an edge e from

3 Approximation slides 3 some s i to its i, enters B.

4 Approximation slides 4 The LP The LP: opt f = Minimize Subject to e c(e) x e e P x e 1, For all s i to t i path P x e 0 Remarks: The separation oracle is a shortest path problem. An r cut, r < 1 for s i is a cut of all e = (u,v) so that dist(s i,u) r and dist(s i,v) > r

5 Approximation slides 5 Middle cuts, and a new cut lemma t1 C s1 Figure 2: A middle cut An r > 1/3 cut: the edges in the cut are at distance between 1/3 and 2/3 from s. There exists (a polynomially computable) r cut for r > 1/3 of cost O(opt f (i)).

6 Approximation slides 6 Proof Let x be the LP optimum. A proof: Assume that x e is a multiple of ǫ Divide the radii into ǫ intervals. Let C i ǫ be the i ǫ cut for s i. 1/3 iǫ 2/3 ǫ cost(c i ǫ ) opt f. Thus, there is an i so that cost(c i ǫ ) 3 opt f. Remark: Back edges are ignored

7 Approximation slides 7 1. Solve the LP. The algorithm 2. Set to 1 every e so that x e 1/ n and add them to C 3. While C not feasible do: (a) pick i, s i can reach t i by a path. (b) Find an r cut C i, for r > 1/3, between s i and t i whose cost is O(opt f (i)). (c) C C C i 4. Return C

8 Approximation slides 8 Charging If e = (u,v) is relevant for i charge e O(x e ) cost. Because the cost is O(opt f (i)) the full cost is charged t i 2/3 v Figure3: R(v)becomessmallerby neachtime e = (v,u) is charged

9 Approximation slides 9 The ratio Each edge is charged O( n) times. The total charge is O( n) opt f Ratio O( n) is implied Also gives O(opt f ) ratio. Best known! Remark: O(n 2/3 /opt 1/3 ) is possible if all costs are 1 For units costs only opt = n is a bad case. If trying to break the n ratio, may assume opt = n

10 Approximation slides 10 Multicuts II: Multiway cuts We present here the work of: Calinescu Karloff and Rabani

11 Approximation slides 11 Problem definition G(V,E),w(e),S = {s 1,...,s k }. The goal is to delete min-cost E so that s i,s j will be be in different connected components in G (V,E \E ). s1 s2 s3 s4 Figure 4: A multiway cut. The optimum has value 3.

12 Approximation slides 12 Two facts on cuts 1. S = 2 solvable by min-cut max-flow Theorem, due to Ford Fulkerson. (Even in the directed case). 2. Separation s from S can also be solved exactly; Contract S into one vertex.

13 Approximation slides 13 A very simple 2 ratio approximation 1. For every i: compute C i = mincut(s i,s \{s i }) 2. Output C i

14 Approximation slides 14 A pictorial explanation OPT: s 1 e s 2 Figure 5: The optimum multiway cut The optimum defines, k, (s i,s \{s i }), cuts C i

15 Approximation slides 15 Ratio 2 Clearly, cost(c i) cost(c i ). So, cost(ci) i i cost(c i ) = cost(alg). But we only know that: i cost(opt) cost(c i ). 2 A ratio 2 results

16 Approximation slides 16 A geometric look (0, 0, 1 ) ( 0, 1, 0 ) (1, 0, 0 ) Figure 6: The i x i = 1, x i 0 simplex (polytope)

17 Approximation slides 17 How do we relate these concepts? Pick k +1 dimensions. The optimum gives a mapping: the vertices in the CC of s j to to e j But we need to define an optimization problem whose solution will tend to imitate a good multiway cut

18 Approximation slides 18 The multiway to L 1 norm relation Using the L 1 norm relates to the multiway cut problem, and leads to a polynomially computable optimum The L 1 distance is defined as: dist L1 (ū, v) = k ū i v i. i=1 Let e = (u,v) and say that f(u) ū and v v. Define: L(e,f) = dist L1 (f(u),f(v)) = dist L1 (ū, v)

19 Approximation slides 19 The value of the optimum as a mapping Now if we use the optimum as mapping then: 1. A cut edge will have L(e,opt) = 2 2. A non cut edge will have L(e,opt) = 0 3. L(e,opt) = 2 opt. e

20 Approximation slides 20 A geometric way of defining the problem 1. Map s i to e i 2. Map V \S via some f to the points (vertices!) of the simplex i x i = 1, x i Minimize the objective function e L(e,f). 2 The optimum value of the above is exactly opt.

21 Approximation slides 21 What CAN be computed in poly time? 1. Map s i to e i 2. Map v s i, for all i to v in the simplex 3. Minimize e L(e,f)

22 Approximation slides 22 This is an LP Create a variable v i for the i th coordinate of v. You can set d i (u,v) = v i u i /2 by d i (u,v) (u i v i )/2,(v i u i )/2 The program is thus: k Minimize e=(u,v) i=1 d i(u,v) Subject to d i (u,v) u i v i 2, v i u i 2 k i=1 v i = 1, for all v s i s i = e i x i 0

23 Approximation slides 23 Assumption: Property of an optimum solution Let e = (v,u), v = (v 1,...,v k ) and ū = (u 1,...,u k ) be the solution of the LP for u and v. Note that k i=1 u i = k i=1 v i = 1. WE MAY ASSUME: v and ū differ by at most 2 coordinates

24 Approximation slides 24 A simple modification Let ū = (x 1,x 2,x 3,...), and v = (x 1 +ρ,y 2,y 3...). ρ > 0 Say that ρ is the minimum (nonzero) difference. Hence, say that x 2 y 2 +ρ Add in the middle of the edge a vertex w with: w = (x 1,y 2 +ρ,y 3,y 4...). Clearly, the sum does not change. And w is by one coordinate more, equal to ū. And, the cut size does not change (removal of (u,w) or (v,w) is equivalent to (v,u) removal).

25 Approximation slides 25 Two vectors ū, v Say that ū = (x 1,x 2,z 1,z 2,...), and v = (y 1,y 2,z 1,z 2,...), As x 1 +x 2 = y 1 +y 2 = 1 i 1 z i, It follow that: x 1 y 1 = y 2 x 2 = d(e) d(e) d(e) x 1 y 1 x 2 y 2 Figure 7: Two segments

26 Approximation slides 26 A simple algorithm 1. Draw ρ R [0,1). 2. Let µ = (1,2...,k) and ν = (k,k 1,...,1) 3. Choose µ or ν at random. Say that µ was chosen, then: (a) For i = 1 to k 1 do: Insert ū V j,j < i so that the i coordinate u i ρ into V i (b) Insert the rest of the vertices into V k Key: If i is before j in µ, it is after j in ν.

27 Approximation slides 27 Analysis rho_3 rho_4 rho_1 rho_5 rho_2 x 1 y 1 x 2 y 2 Figure 8: Several possibilities for ρ 1. The choices ρ 1,ρ 2,ρ 3 do not cut the edge. 2. The choice ρ 5, cuts the edge 3. The choice of ρ 4 : It depends on the order of the permutation 4. If the x 1,y 1 coordinate comes first, the edge is cut. Otherwise, not cut

28 Approximation slides 28 Summary Pr(e is cut ) d(e)+ d(e) 2 = 3 d(e). 2 And since: opt f = e d(e), We can analyze the ratio

29 Approximation slides 29 Ratio Analysis Put X(e) = 1, if e is cut and X(e) = 0 otherwise. Clearly, And, Outsize = e X(e). E(Outsize) = e = e E(X(e)) Pr(e Cut).

30 Approximation slides 30 Ratio, cont. E( cut size ) = e A 3/2 ratio results 3d(e) 2 = 3 2 opt f 3 opt 2

31 Approximation slides 31 Other cases The directed case: Naor and Zosin: ratio 2. Complicated. Via fractional multi commodity flow. Solves an LP. The k cut problem: Delete few edges to get k parts. A 2 ratio is very simple. Improving ratio 2 is a big open problem.