# Chapter 6: Random rounding of semidefinite programs

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1 Chapter 6: Random rounding of semidefinite programs Section 6.1: Semidefinite Programming Section 6.: Finding large cuts Extra (not in book): The max -sat problem. Section 6.5: Coloring 3-colorabale graphs. Section 6.1: Semidefinite Programming The book gives some properties of semidefinite matrices but not all are used in this chapter. Here we list only the ones that are used, and additionally give a definition of a strict quadratic program. Definition 1. A matrix X R n n y R n, y T Xy 0. Notation: X 0. is positive semidefinite (psd) iff for all Fact 1. If X R n n is symmetric then the following equivalence holds X 0 X = V T V for some V R m n with m n. We may assume that m = n since if m < n we can extend V with zero-rows. Definition. A semidefinite program (SDP) is a linear program with an additional constraint of the following form: X is positive semidefinite, where X is a square symmetric matrix containing all the variables. Example: The following LP min x + y + z s.t. x + y z = x, y, z 0 [ ] x y with the additional constraint: 0, is called a semidefinite program. y z The general form with m linear constraints is min or max s.t. ij c ij x ij a ijk x ij = b k ij x ij = x ji X = (x ij ) 0. for k = 1,..., m for all i, j Fact. Semidefinite programs can be solved in polynomial time up to an arbitrarily small additive constant ɛ > 0. (1) 1

2 A quadratic program (QP) is a linear program in which, additionally, also the product of two variables is allowed, and is called strict if it only contains constant terms and products of two variables. For example, the constraint y 1 + y 1 y + y 3 = 1 is not allowed in a strict QP because of the term y 3. Definition 3. A vector program is obtained from a strict quadratic program by replacing all variables y i by vectors v i R n (where n is the number of variables y i ) and by replacing the scalar products of variables by vector inproducts. See for an example Section 6.. The general form of a vector program is min or max s.t. n n c ij v i v j i=1 j=1 n i=1 j=1 n a ijk v i v j = b k, k = 1,..., m v i R n i = 1... n. () Theorem 1. Semidefinite programs are equivalent with vector programs. Consequently, vector programs can be solved in polynomial time up to an arbitrarily small additive constant ɛ > 0. Proof. Let X R n n be a feasible solution for SDP (1). Then X = V T V for some n n matrix V. Let v i be the i-th column of V. Then v 1, v,..., v n is feasible for vector program () and the value stays the same. Conversely, if v 1, v,..., v n is feasible for VP, then define V = [v 1 v v n ] and X = V T V. Then, X = (x ij ) is feasible for SDP and the value stays the same. Section 6.: Finding large cuts. In the weighted version of max cut problem, we may assume without loss of generality that the graph is complete since we can define w ij = 0 for any missing edge. To simplify notation we write (i,j) to denote the summation over all edges of the complete graph. The following quadratic program is an exact formulation of the max-cut problem. 1 (QP) max (1 y i y j )w ij (i,j) s.t. y i { 1, 1} i = 1,..., n.

3 Figure 1: The maximum cut in this graph on 3 vertices has value. Note that the constraint y i { 1, 1} can also be written as yi = 1. It is easy to see that this is indeed an exact formulation of the max cut problem: Given a feasible solution for (QP) we get a cut with the same value by taking i U if y i = 1 and taking i W if y i = 1. Conversely, given a cut (U, W ) we get a feasible solution for (QP ) with the same value by setting y i = 1 if i U and taking y i = 1 if i W. The quadratic program (QP) is strict and we can formulate its vector program relaxation. 1 (VP) max (1 v i v j )w ij (i,j) s.t. v i v i = 1, v i R n i = 1,..., n. It is easy to see that (VP) really is a relaxation of (QP): Let y 1, y,..., y n be feasible for (QP), then v i = (y i, 0,..., 0) for all i defines a feasible solution for (V P ) with the same value. In the graph of Figure 1, an optimal solution to it quadratic program is (y 1, y, y 3 ) = (1, 1, 1) and the corresponding solution for (V P ) is v 1 = (1, 0, 0), v = v 3 = ( 1, 0, 0). However, this is not an optimal solution to (V P ). An optimal solution is given in Figure. Figure : An optimal solution for the vector program (VP) for the graph of Figure 1. The inproduct v i v j is 0.5 for all pairs i, j. Hence, the optimal value is =.5. The max-cut algorithm. Goemans and Williamson gave the following algorithm for the max-cut problem: 3

4 Step 1. Solve the VP-relaxation v1, v,..., vn, ZV P. Step. (Randomized rounding) Take a vector r R n of length 1 uniformly at random. Add vertex i to U if vi r 0 and add it to W otherwise. Analysis: Let ϕ ij be the angle between vi and vj. Then, v i v j and we can express the optimal value of the VP in terms of ϕ ij : = cos ϕ ij ZV P = 1 (1 vi vj )w ij = 1 (1 cos ϕ ij )w ij. (3) On the other hand, (i,j) (i,j) Pr{ edge (i, j) is in the cut } = Pr{v i and v j are separated by the hyperplane perpendicular to r.} = ϕ ij π Let W be the weight of the cut. Then E[W ] = (i,j) ϕ ij π w ij. () Let [ ] ϕ α = min 0 ϕ π π. 1 cos ϕ One can show that the minimum α of this function is well-defined and satisfies: α > Consequently, for any ϕ ij we have ϕ ij π α1 cos ϕ ij. Now we combine (3) and (): E[W ] = ϕ ij π w ij α (1 cos ϕ ij )w ij = αzv P αopt > 0.878Opt. (ij) (i,j) The algorithm can be derandomized using a sophisticated application of the method of conditional expectations (See section 5.) but this is not easy. The algorithm above is best possible if P N P and if the so called unique games conjecture holds. (See Section 13.3 for this unique games conjecture.)

5 Extra (not in book): The max -sat problem. Section 5.1 gives a 1/-approximation for the maximum satisfiability problem. In particular, it gives a 1 (1/) k -approximation if each clause contains at least k literals. For the max -sat problem, in which each clause contains exactly two literals, this yields a 3/-approximation. Also, the linear program of Section 5. yields a 3/-approximation when rounded as in Section 5.6. Here, we show that the approach used for the max-cut problem gives an α-approximation with the same factor α = for the max -sat problem. We shall restrict to the unweighted version but the algorithm works exactly the same in case each clause had a given weight and we want to maximize the total weight of the satisfied clauses. In the example below, Opt = 3. The examples also shows the four possible types of clauses. x 1 x, x 1 x, x x 1, x 1 x. Let us denote val(x i ) = 1 if x i is true and let it be zero otherwise. Similarly, for clause C let val(c) = 1 if the clause is true and let it be zero otherwise. An obvious choice for a quadratic program is to define binary variables y i {0, 1} with y i = val(x i ). Then the value of a clause x i x j can be expressed as val(x i x j ) = 1 (1 y i )(1 y j ) = y i + y j y i y j. We can do the same for the other 3 types, ( x i x j ), ( x i x j ), ( x i x j ), and formulate a quadratic program. However, its relaxation is not convex and nor is it a strict quadratic program. We need to try another approach. A strict quadratic program. We introduce one more binary variable, y 0 and make the following proposition: y i = y 0 x i = true y i { 1, 1} and y i y 0 x i = false Now, the value of a literal can be expressed as follows. val(x i ) = 1 + y 0y i and Then, the value of a clause x i x j becomes val( x i ) = 1 y 0y i. val(x i x j ) = 1 val( x i )val( x j ) ( ) ( ) 1 y0 y i 1 y0 y j = 1 = 1 + y 0y i y 0y j + 1 y iy j. 5

6 In the same way, we can write val( x i x j ) = 1 y 0y i val(x i x j ) = 1 + y 0y i val( x i x j ) = 1 y 0y i y 0y j 1 y 0y j + 1 y 0y j y iy j y iy j + 1 y iy j. We see that any instance of max -sat can be written as a strict quadratic program of the following form. Again, to simplify notation we write (i,j) to denote the summation over i, j with 0 i < j n. (QP) max (i,j) a ij (1 + y i y j ) + b ij (1 y i y j ) s.t. y i = 1, i = 0, 1,..., n. Here, a ij and b ij are constants depending on the instance. The vector program relaxation becomes (VP) max (i,j) a ij (1 + v i v j ) + b ij (1 v i v j ) Algorithm: s.t. v i v i = 1, v i R n+1 i = 0, 1,..., n. Step 1. Solve the VP-relaxation v 0, v 1,..., v n, Z V P. Step. Take a vector r R n of length 1 uniformly at random. For i = 0,..., n, let y i = 1 if if v i r 0 and let y i = 1 otherwise. Analysis: Let ϕ ij be the angle between v i and v j. Then, v i v j = cos ϕ ij. The optimal VP value can be written as follows. Z V P = (i,j) a ij (1+v i v j )+b ij (1 v i v j ) = (i,j) a ij (1+cos ϕ ij )+b ij (1 cos ϕ ij ). On the other hand, let W be the number of clauses satisfied by the algorithm. As we have seen for the max cut problem, Pr(y i y j ) = ϕ ij /π. Hence, E[W ] = (i,j) a ij Pr(y i = y j ) + b ij Pr(y i y j ) = (i,j) a ij (1 ϕ ij π ) + b ϕ ij ij π. (5) Let α be defined as in the max cut analysis. That means ϕ π α (1 cos ϕ), for all 0 ϕ π. 6

7 On the other hand, when we substitute ϕ = π θ then, for all 0 θ π π θ π α (1 cos(π θ)) 1 θ π α (1 + cos θ). Now, apply the two inequalities above to (5): E[W ] = (i,j) a ij (1 ϕ ij π ) + b ϕ ij ij π α (i,j) a ij (1 + cos ϕ ij ) + b ij (1 cos ϕ ij ) = αz V P αopt. Section 6.5: Coloring 3-colorabale graphs. A graph coloring is a labeling of the vertices with labels 1,,... (the colors) such that any two adjacent vertices have a different label. The chromatic number χ(g) is the minimum number of colors needed for a feasible coloring of G. If χ(g) then finding a feasible -coloring is easy: Start with any vertex and give it color 1, then color all its neighbors with color, color all neighbors of neighbors with color 1, and so on. Hence, deciding if a graph is -colorable is easy. Also, any graph can be colored using at most + 1 colors if is the maximum degree of the graph. (Simply color the vertices one-by-one.) (i) Finding a -coloring is easy if the graph is -colorable. (ii) Finding a coloring with at most + 1 colors is easy, where is the maximum degree in the graph. On the other hand, deciding if χ(g) 3 is NP-complete. In fact, even when we know that the given graph is 3-colorable then finding a coloring with a small number of colors is difficult. First we give a simple algorithm to color the vertices of a 3-colorable graph G = (V, E) with at most n colors. Let σ = n. Algorithm 1: Step 1 As long as there is a vertex v with degree at least σ do the following: Use a new color for v and use two new colors to color the neighbors of v. Remove v and its neighbors from G. Step Color the remaining graph with at most σ colors. Clearly, Step 1 can be done in polynomial time since the set of neighboring points of any point v is -colorable. Also, Step is easy since the graph that remains after Step 1 has maximum degree σ 1. The total number of colors used is 3 n/(σ + 1) + σ < n. 7

8 Figure 3: Application of Algorithm 1 with σ =. We can reduce the number of colors by improving (ii). Note that in general (ii) is tight since for a complete graph on n vertices we need n = + 1 colors. However, if the graph is 3-colorable then we can do better. We use the notation Õ(.) to hide factors log n. For example, n log 3 n = Õ(n ). Theorem. If G is 3-colorable, then a coloring with Õ( log 3 ) colors in expectation can be found in polynomial time. Note, log Before proving Theorem, let us see how it leeds to a reduced approximation ratio. We apply the same algorithm but only need Õ( log 3 ) = Õ(σlog 3 ) colors in Step. That means, the total number of colors used is bounded by 3 n/(σ + 1) + Õ(σlog 3 ) Now we optimize over σ. Solving n/σ = σ log 3 gives σ = n log 6 3. Then, for both steps the number of colors used is Õ(n/σ) = Õ(nlog 6 ), which is better than the n we have seen earlier since log Proof of Theorem The idea is similar to that of the max-cut problem. There are two differences to notice. First, we do not formulate an exact quadratic program but only give a vector program relaxation for the 3-coloring problem. Second, the algorithm takes many random vectors in stead of just one. (VP) min λ s.t. v i v j λ, for all (i, j) E v i v i = 1, v i R n for all i V. If G = (V, E) is 3-colorable then the vector program above has optimal value at most λ 0.5. This can be seen from the solution of the max cut example in Figure. Given a feasible 3-coloring, let the three vectors correspond with 8

9 the three colors. If (i, j) E then i and j have different colors which implies that v i v j = 0.5. Remark: It is not true that λ = 0.5 for any graph with χ(g) = 3 as can be seen from Figure. Figure : A cycle on 5 vertices has chromatic number 3. The optimal solution to VP in this case has value λ = cos(π/5) < 0.5. For an odd cycle, λ 1 for n. The idea of the algorithm is as follows. First solve the VP. Then take t n unit vectors in R n at random. The corresponding hyperplanes partition R n in t regions. Let the regions correspond to colors. Say that a vertex is properly colored if all its neighbors have a different color. Remove all properly colored vertices and take a new set of t unit vectors in R n at random and use t new colors. Repeat until all vertices are properly colored. Algorithm: Step 1. Solve the VP-relaxation v 1,..., v n, λ. Step. Repeat until the graph is properly colored: Take vectors r 1,..., r t R n of length 1 uniformly at random. Each of the t regions defines a color. Give point i the color of the region containing vi. Say that a vertex is properly colored if all its neighbors have a different color. Remove all properly colored vertices from the graph. Consider an edge (i, j) E and assume i and j are not yet properly colored. Then, for the next iteration of Step : Pr{ i and j get the same color } = Pr{vi and vj are not separated by any of the r i s.} ( = 1 ϕ ) t ij π ( = 1 1 ) t π arccos(v i vj ( 1 1 ) t π arccos(λ ) ( 1 1 ) t ( π arccos( 1/) = 1 1 π 9 ) t π = 3 ( ) t 1 3

10 The first inequality follows from (VP) and from the fact that arccos is nonincreasing. The second inequality follows from λ 1/ and from arccos being non-increasing. Take t = 1 + log 3. Then, Pr{ i and j get the same color } 1 3. By the union bound, for any vertex i in an iteration of Step : Pr{ i is not properly colored } 1 3 = 1 3. Let n be the number of remaining vertices before some iteration of Step. Then, for this iteration By Markov s inequality: E[ # vertices not properly colored ] n 3. Pr{ # vertices not properly colored n 3 } 1. Hence, in each iteration of Step, the number of vertices still to be colored is reduced by a factor /3 with probability at least 1/. This implies that the expected number of iteration of Step is O(log n) and the expected number of colors used in total is O(log n) t = Õ(t ) = Õ(1+log 3 ) = Õ( log 3 ). 10

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