随机算法课件ch06

Ch6 - Randomized Rounding of Semidefinite Programs
Peng Zhang
∗
May 13, 2012
∗ School
of Computer ቤተ መጻሕፍቲ ባይዱcience and Technology, Shandong University, Ji’nan 250101,
0
. Clearly vi · vi = 1 and vi · vj = yi yj .
• So {vi } is a feasible solution to (V P ) with the same objective value.
0-15
Algorithm Random Hyperplane [Goemans, Williamson, STOC’94 / JACM’95] 1 Solve (V P ) for MAX CUT. Let {vi } be an optimal solution. 2 Pick a random vector r = (r1 , r2 , · · · , rn )T by drawing each component from N (0, 1). ¯). 3 Let S = {i | vi · r ≥ 0}. return (S, S Fact 3.1 The normalization of r, r/ r , is uniformly distributed over the n-dimensional unit sphere Sn−1 . Proof: • Each component ri of vector r has density function 1 − x2 i 2 f (xi ) = √ e , 2π 0-16 −∞ < x < ∞.
• “•” is the matrix inner product, a.k.a. Frobenius multiplication. C •X =
i,j cij xij .
• Mn is the cone of all symmetric real matrices.
0-11
A vector program (V P ) which is equivalent to (SDP ). max or min
i,j
cij (vi · vj ) aij (vi · vj ) = bk ,
i,j (k )
(V P ) ∀k
s.t.
vi ∈ Rn Lemma 2.1 (V P ) is equivalent to (SDP ). Proof: • Given a solution X to (SDP ), compute V such that X = V T V by Cholesky decomposition. • Let vi the the ith column of V . Then xij = vi · vj and one can verify {vi } satisfies the constraint in (V P ). 0-12
China. Email: algzhang@.
0-0
1
The MAX CUT Problem
• Instance. A undirected graph G = (V, E ), and nonnegative weights wij for each edge (i, j ) ∈ E . • Query. Partition V into (U, W ) so as to maximize the total weight of edges in the cut. A simple randomized
0-8
1 −1 2 √ 3 One can verify that V T V = 0 2 0 0 1 1 −2 −1 2 1 1 − 2 1 − 2 = X. 1 −1 − 1 2 2
1 √ 1 − 23 − 2 0 −1 2 −1 2 0
0-4
• Now, define matrix √ λ 1 D= Then Λ = DDT . • So, we have X = QDDT QT = QD(QD)T . Now, (3) follows by letting V = (QD)T . • (3 ⇒ 1). For any y ∈ Rn , y T Xy = y T V T V y = (V y )T (V y ) ≥ 0. 0-5 √ λ2 ··· √ λn .
So the corresponding to eigenvalue λ2 = λ3 = 0 are eigenvectors −1 −1 v2 = 1 and v3 = 0 . 1 0 1 0 0 √ 1 3 • By performing Cholesky decomposition, we have V = − 2 . 0 2 √ −1 − 23 0 2
• (1 ⇒ 2). Let λ be an eigenvalue of X , and v be the corresponding eigenvector. Then, Xv = λv . • By pre-multiplying v T , we get v T Xv = λv T v. • Since X is positive semidefinite, v T Xv ≥ 0. Since v T v > 0 (v is non-zero), λ ≥ 0. • (2 ⇒ 3). Let λ1 , λ2 , · · · , λn be the n eigenvalues of X , and v1 , v2 , · · · , vn be the corresponding complete collection of orthonormal eigenvectors. • Definte matrix Q = v1 v2 ··· vn
• Given a solution {vi } to (V P ). Let V be the matrix whose ith column is vi . • Then X = V T V is symmetric and positive semidefinite, and satisfies the constraint in (SDP ). Moreover, xij = vi · vj .
3 2 √ − 23 √
0 0 = 0
0-9
2.2
Semidefinite Programming
max or min
i,j
cij xij aij xij = bk ,
i,j (k )
(SDP ) ∀k ∀i, j 0
s.t.
xij = xji , X = (xij )
0-10
SDP in matrix form. max or min s.t. C •X Ak • X = bk , X ∈ Mn X 0 (SDP ) ∀k
λ−1
1 2
=
λ−1
• The 3 eigenvalues are λ1 = 0, λ2 = λ2 = 3 2. • The eigenvector for λ1 = 0.
0-6
y 1 Solve (λ1 I − X )Y = 0, where Y = y2 . y3 1 1 1 1 1 −1 1 − 1 − 1 2 2 2 2 2 2 1 3 3 λ1 I −X = 1 → → −1 2 0 −4 0 1 −1 . 2 4 1 1 3 3 − 1 0 − 0 0 0 2 4 4 2 y − 1y − 1y = 0 y − 1y = 1y 1 1 2 3 2 2 2 2 2 3 , i.e., . So we have y =y y −y =0 2 3 2 3 1 So the eigenvector corresponding to eigenvalue λ1 = 0 is v1 = 1. 1 • The eigenvector for λ2 = λ3 = 3 2. 0-7
Solve (λ2 I − X )Y = 0.
1 2 1 2 1 2
1
0 . 0 = −y2 − y3 .
1 1 1 1 λ2 I − X = 1 → 0 0 2 2 2 1 1 1 0 0 2 2 2 So we have y1 + y2 + y3 = 0, i.e., y1
0-13
3
Approximation Algorithm for MAX CUT
An integer quadratic program for MAX CUT. max s.t. 1 2 wij (1 − yi yj )
(i,j )∈E
(IP ) ∀i
yi ∈ {−1, 1},
A vector program relaxation of (IP ). max s.t. 1 2 wij (1 − vi · vj )
Definition 2.1 A matrix X ∈ Rn×n is positive semidefinite if and only if for all y ∈ Rn , y T Xy ≥ 0. Fact 2.1 If X ∈ Rn×n is a symmetric matrix, then the following statements are equivalent: 1. X is positive semidefinite; 2. X has non-negative eigenvalues; 3. X = V T V for some V ∈ Rm×n where m ≤ n. Proof: 0-2