Lecture 5. 1 Goermans-Williamson Algorithm for the maxcut problem

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1 Math 280 Geometric and Algebraic Ideas in Optimization April 26, 2010 Lecture 5 Lecturer: Jesús A De Loera Scribe: Huy-Dung Han, Fabio Lapiccirella 1 Goermans-Williamson Algorithm for the maxcut problem This is a probabilistic approach to choosing a CUT Step 1: Solve the relax problem Let the result be X and it can be factorized as X = V T V (1) where V = [ v 1 v 2 v n ] Step 2: Generate a random uniform vector r on the unit n-sphere S n Step 3: A CUT which approximates the MAXCUT is y i = sign ( r T v ) = { 1 if r T v 0 1 if r T v < 0 (2) Since this CUT is randomly generated, we are interested in its expected value Lemma 1 Pr [ sign ( ) ( )] r T v i sign r T v j = arccos ( vi T v ) j (3) π Theorem 2 Expected value of CUT is not less than the MAXCUT The expected value of CUT is E (CUT) = 1 w ij Pr [1 y i y j = 2] 2 = 1 2 = 1 2 = 1 2 w ij Pr [ sign ( r T v i ) sign ( r T v j )] arccos ( vi T w v ) j ij π w ij arccos (X ij ) π = 1 w ij (1 X ij ) 2 arccos (X ij ) 4 π (1 X ij ) 1 ( ) 2 arccos (t) w ij (1 X ij ) min 4 1 t 1π (1 t) ( ) 2 θ = RELAX min 0 θ π π (1 cos (θ)) RELAX (4) 5-1

2 2 Global optimization using SDP So far we have seen linear programs and Semidefinite programs are very useful and can be solved efficiently They will be used quite a bit later We will consider now situations where min/max f (x) (5) subject to g 1 (x) = 0 h 1 (x) 0 g 2 (x) = 0 h 2 (x) 0 g k (x) = 0 h s (x) 0 There is no integrality restriction except f, g i, h j are all polynomials 21 Univariate polynomial Let us start understanding such problem for one variable Let f (x) be a polynomial of variable x as We wish to find the global minimum of f (x) f (x) = p n x n + p n 1 x n 1 + p 1 x + p 0 (6) min f (x) (7) This problem is equivalent to find max γ (8) st f (x) γ 0 for all x First note: f (x) γ must be of even degree Lemma 3 A univariate polynomial p (x) is non-negative iff p (x) is a sum of square (SOS) It is obvious to see if p (x) is a SOS then p (x) is non-negative Now, given non-negative polynomial p (x) of order n and coefficients p j, (j = 1n) we proof that p(x) is also SOS According to the fundamental theorem of algebra, p (x) can be factorized as p (x) = p n (x r j ) nj n j m k (x a k + ib k ) mk (x a k ib k ) mk (9) 5-2

3 where r j and a k ± ib k are the real and complex roots of p (x), respectively Because p 0, n j are even,ie n j = 2s j, and p n 0 Moreover, (x a k + ib k ) (x a k ib k ) = (x a k ) 2 + b 2 k Therefore ( p (x) = p n (x r j ) 2sj (x a k ) 2 + bk) 2 mk (10) s j m k The first product of p (x) in (9) is a product of square terms Note that we prove that p (x) is a SOS (a 2 + b 2 )(c 2 + d 2 ) = (ac bd) 2 + (ad + bc) 2, (11) Corollary 4 Any non-negative polynomial in R is a sum of 2 squares The next question is how to check quickly whether p (x) is a SOS p (x) = q 2 1 (x) + q 2 2 (x) (12) Theorem 5 A univariate polynomial p (x) of even degree 2d is a SOS iff there exits a positive semidefinite matrix Q such that where x d = [ 1 x x d ] T p (x) = x T d Qx d (13) We write p (x) in SOS form p (x) = m qk 2 (x) (14) k=1 Define q = [ q 1 (x) q 2 (x) q m (x) ] T, we have where k-th row of V is made of coefficients of q m (x) Therefore q = Vx d, (15) where Q = V T V By definition Q 0 p (x) = q T q = x T d V T Vx d = x T d Qx d (16) The coefficients of p (x) has the following relation with elements of Q p i = Q jk (17) j+k=i 5-3

4 We have p (x) = x T d Qx d = d d Q jk x j+k = = Conclusion: SDP can solve (7) globally j=0 j=0 2d i=0 j+k=i 2d i=0 Q jk x i p i x i (18) 22 Multivariate polynomial We wish to find a global minimum of a multivariate polynomial f (x) This problem is equivalent to find min f (x) (19) x R n max γ (20) subject to f (x) γ 0 for all x R n The question is whether we can apply the same procedure as in the univariate polynomial case The answer is no because we do not know if f (x) is a SOS for x R In fact Motzkin and Robinson (1950) has shown that the polynomial x 4 y 2 + x 2 y x 2 y 2 can not be written as a SOS 3 Motzkin function The multivariate polynomial f(x), x R n of even degree deg(f(x)) = 2d is a non-negative polynomial if and only if it satisfies: Clearly, a sufficient condition in order for f(x) to be non-negative is: f(x) 0, x R n (21) f(x) = i q 2 i (x), (22) that is, if f(x) is reducible to a sum of squares (SOS), it is a non-negative polynomial In general, condition (22) is only sufficient Hilbert proved that it becomes also necessary if: n = 1 (23) d = 1 (24) n = 2 and d = 2 (25) 5-4

5 Motzkin showed an example of a non-negative polynomial with n = 2 and d = 3 that cannot be reduced to a sum of squares: It is easy to show that (x, y) R 2 f(x, y) 0: true, because: f(x, y) = 1 + x 4 y 2 + x 2 y 4 3x 2 y 2 (26) 1 + x 4 y 2 + x 2 y 4 3x 2 y 2 (27) and: a a n n n a 1 a 2 a n (28) 1 + x 4 y 2 + x 2 y x 4 y 2 (x 2 y 4 ) = 3x 2 y 2 (29) The proof that the non-negative polynomial f(x, y) cannot be reduced to a sum of squares is as follows: Suppose: f = s 2 i, s i R[x, y] (30) We note that each polynomial s i must have Newton polytope contained in the triangle (0, 0), (1, 2), (2, 1), therefore s i is a linear combination of 1, xy 2, x 2 y, xy, but s 2 i contains x2 y 2 and it must appear with a positive coefficient (s i = a + bx 2 y + cxy 2 + dxy implies that x 2 y 2 has coefficient d 2 0), but this is in contradiction with the fact that x 2 y 2 appears with coefficient 3 in f(x, y) Lemma 6 If p(x) = q 2 i (x), then Newton(q i) 1 2 Newton(p(x)) This reduces the size of the compounding PSD matrix Q In fact, recall that semi-definite programming can be used to prove whether a given polynomial is a sum of squares A polynomial f(x) = p α x α can be written as: f(x) = [x] T d Q[x] d (31) with [x] d vector of all monomials of n variables and degree deg([x i ] d ) d, if: 31 Example p α = Q 0, (32) Q βδ (33) α=β+δ f(x, y, z, ω) = (x 4 + 1)(y 4 + 1)(z 4 + 1)(ω 4 + 1) + 3x + 4y + 5z + 2ω (34) deg(f(x, y, z, ω)) = 16, the total number of variables necessary to solve this problem is: ( ) = 495 (35) 8 Such a large number of variables is not efficiently handled on most of current computers Recalling that Newton(f(x)) is a cube, we can solve the problem using only 3 4 = 81 monomials, which is duable on most of current computers 5-5

6 4 Optimization problem We wish to: maximize: f(x) subject to: g 1 (x) 0 h 1 (x) = 0, g m (x) 0 h k (x) = 0 (36) where functions f( ), g j ( ) (j {1,,m}), h i ( ) (i {1,,k}) are polynomials with coefficients in R In order to tackle the solution of problem 36, let us consider the set of all polynomials in n variables with coefficients on a field K {C, R, Z p }, K[x 1,, x n ] = K[x] and define an ideal I K[xx] such that: (a) if f 1, f 2 I f 1 + f 2 I, (b) if f I, g k[x] f g I Examples: (1) Let S C n, then: I(S) is defined as the vanishing ideal I(S) = {f(x) f(s) = 0, s S} (37) (2) Let g 1, g 2,, g s be a series of polynomials; the ideal generated by g 1,, g s is defined as: < g 1,, g s >= {f f = λ 1 g 1 + λ 2 g λ s g s, with λ i K[x]} (38) Exercise: show that in C[x 1 ] every ideal is principal Theorem: every polynomial ideal is finitely generated (HILBERT basis theorem) Given a set of polynomials h 1,, h k in K[x 1,, x n ], we define a variety as: V (h 1, h 2,, h k ) = {s K n h i (s) = 0, i = 1, 2,, k} (39) Example: V R 2(x 2 y 2 ) is the set of the x and y axis We define a basic semi-algebraic set as a subset of R n that satisfies: Examples: {x R n g 1 (x) 0,, g m (x) 0, h 1 (x) 0,, h k (x) 0} (40) (1) x [0, 1] is a semi-algebraic set of R that can be written as: {x R x + 1 0, x 0} (41) (2) The set of all symmetric n n matrices that are positive semidefinite is a basic semi-algebraic set: A 0 all 2 n 1 principal minors of A are non-negative Exercise: find fewer than 2 n 1 inequalities that define the same semi-algebraic set Given a set of polynomials g 1,, g m in K[x], a cone is defined as the set of all linear combination of all products of subsets of g i : cone(g 1,, g m ) = { s I π i I g i s I an SOS} (42) I {1,,m} 5-6

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