Spectral Method for Modularity Maximization Yunqi Guo January 24, 2017 Problem . Maximize modularity Q : � A i j − k i k j � Q = 1 � δ g i g j , (1) 2 m 2 m i j where δ g i g j = 1 � � s i s j + 1 (2) 2 and + 1 , if vertex i belongs to group 1 s i = (3) − 1 , if vertex i belongs to group 2 Solution. � � Take δ g i g j = 1 s i s j + 1 to equation (1). Then 2 � � � Q = 1 A i j − k i k j � � s i s j + 1 . (4) 4 m 2 m i j Define the quantity B , called modularity matrix B i j = A i j − k i k j 2 m . (5) The sums of all its rows and columns are zero: k i k j � � � A i j − (6) B i j = 2 m j j j = k i − 2 m k i (7) 2 m = 0 (8) 1
Use substitute B i j for A i j : Q = 1 � � � B i j s i s j + 1 . (9) 4 m i j = 1 B i j s i s j + 1 � � B i j (10) 4 m 4 m i j i j = 1 � B i j s i s j (11) 4 m i j Our task is to maximize Q over the possible choices of the s i . Relax s i to any real value: s i ∈ R Use constrain: � k i s 2 i = 2 m , (12) i where � i k i = 2 m . The original vector s is mapped to the boundary of a hyper ellipsoid. To make the problem simple, we define Q ′ as the objective function: � Q ′ = B i j s i s j (13) i j Based on the theorem of the Lagrange multiplier [P5. Theorem Lagrange multiplier], we define Lagrange function: � � k i s 2 F = B i j s i s j + λ 2 m − (14) i i j i when all s i and λ satisfy ∂ F ∂ s i = 0 (15) ∂ F ∂λ = 0 vector s = ( s 1 , s 2 , ... ) will be the stationary point. ∂ F = ∂ � � k i s 2 B i j s i s j + λ 2 m − = 0 (16) i ∂ s i s i i j i � 2 B i j s j − 2 λ k i s i = 0 (17) j 2
Then, � B i j s j = λ k i s i (18) j or, in matrix notation, Bs = λ Ds (19) where D is the diagonal matrix with elements equal to the vertex degrees D ii = k i Use the adjacency matrix A to substitute modularity matrix B . Then, according to equation (5), � � A i j − k i k j � s j = λ k i s i (20) 2 , j k j � � A i j s j = k i λ s i + 2 m s j (21) j j or, in matrix notation, λ s + k T s � � As = D 2 m 1 (22) where 1 = (1 , 1 , ... ). It can be obvious observed that A1 = D1 = k (23) Also, for that A , D are symmetric matrices, we have: A = A T D = D T (24) And for that � i k i = 2 m , we have: k T 1 = 2 m (25) So, equation (22) can be deformed step by step. Let both the left side and right side multiply 1 T . λ s + k T s � � 1 T As = 1 T D 2 m 1 (26) λ s + k T s � � ( A1 ) T s = ( D1 ) T 2 m 1 (27) 3
s + k T s � � k T s = λ k T 2 m 1 (28) k T s = λ k T s + k T s 2 m k T 1 (29) k T s = λ k T s + k T s 2 m × 2 m (30) k T s = λ k T s + k T s (31) λ k T s = 0 (32) Since we are assuming there exists a nontrivial eigenvalue value λ > 0, we know that λ � 0. Hence k T s = 0 (33) Equation (22) simplifies to As = λ Ds (34) Obviously, λ = 1 when s = 1 is a solution to this function. But it does not satisfy constrain equation (33). By Perron-Frobenius theorem, λ = 1 is the most positive eigenvalue. So to maxi- mize Q , we need to use the second positive eigenvalue. To make it more simple, define: u = D 1 / 2 s (35) and use the normalized Laplacian: L = D − 1 / 2 AD − 1 / 2 (36) equation (34‘) can be deformed as: L u = λ u (37) Because the elements of s and u have the same sign correspondingly. So the solution of the original maximization problem is the sign of eigenvector u , when the eigenvalue is the second positive one. � 4
Theorem Lagrange multiplier. Maximize f ( x , y ), subject to g ( x , y ) = c . We need both f and g to have continuous first partial derivatives. We introduce a new variable λ called a Lagrange multiplier and study the Lagrange function (or Lagrangian) defined by L = f ( x , y ) − λ ( g ( x , y ) − c ) where the λ term may be either added or subtracted. If f ( x 0 , y 0 ) is a maximum of f ( x , y ) for the original constrained problem, then there exists λ 0 such that ( x 0 , y 0 , λ 0 ) is a stationary point for the Lagrange function (stationary points are those points where the partial derivatives of L are zero). 5
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