Follow example 3, we need minimize
with ,
so we can compute
- $(X_1,C_1) = argmin
Let C = [c11c12
c21c22 ] ,X = [x11x12
x21x22 ] ,
we have
∥A − X∥F2 + 1
2ρ∥X − CR0 + U0∥F2 = ∥ [ 2 − x11 1 − x12
−1 − x212 − x22 ] ∥F2 + 1
2ρ∥ [x11 − 2c11 + 1 x12 − c12
x21 − 2c21 x22 − c22 + 1 ]∥F2
= (2 − x11)2 + (1 − x
12)2 + (−1 − x
21)2 + (2 − x
22)2 + 1
2ρ ((x11 − 2c11 + 1)2 + (x
12 − c12)2 + (x
21 − 2c21)2 + (x
22 − c22 + 1)2)
Now take derivatives w.r.t. to X,C,
we have
∂NMF
∂x11 = −2(2 − x11) + ρ(x11 − 2c11 + 1) = (ρ + 2)x11 − 2ρc11 + ρ − 4 = 0
∂NMF
∂x12 = −2(1 − x12) + ρ(x12 − c12) = (ρ + 2)x12 − ρc12 − 2 = 0
∂NMF
∂x21 = 2(1 + x21) + ρ(x21 − 2c21) = (ρ + 2)x21 − 2ρc21 + 2 = 0
∂NMF
∂x22 = −2(2 − x22) + ρ(x22 − c22 + 1) = (ρ + 2)x22 − ρc22 + ρ − 4 = 0
∂NMF
∂c11 = −2ρ(x11 − 2c11 + 1) = 0
∂NMF
∂c12 = −ρ(x12 − c12) = 0
∂NMF
∂c21 = −ρ(x21 − 2c21) = 0
∂NMF
∂c22 = −ρ(x22 − c22 + 1) = 0
Note, we also need satisfy X ≥ 0
and C ≥ 0, Solve the
equations we have X1 = [21 0 2 ]
and C1 = [1.51 0 3 ]
- Now solve for R1 = argmin∥X1 − C1R + U0∥F2
with R ≥ 0,
that is
∥X1 − C1R + U0∥F2 = ∥ [2 − 1.5R11 − R211 − 1.5R12 − R22
−3R21 2 − 3R22 ] ∥F2
= (2 − 1.5R11 − R21)2 + (1 − 1.5R
12 − R22)2 + (−3R
21)2 + (2 − 3R
22)2
Take derivatives w.r.t. R,
we have
∂
∂R11 = 2(−1.5)(2 − 1.5R11 − R21) = 0
∂
∂R12 = 2(−1.5)(1 − 1.5R12 − R22) = 0
∂
∂R21 = −2(2 − 1.5R11 − R21) + 18R21 = 0
∂
∂R22 = −2(1 − 1.5R12 − R22) − 6(2 − 3R22) = 0
Solve for the equations we have R = [4
32
9
02
3 ] ≥ 0
- Lastly, we have U1 = U0+X1−C1R1 = [51 0 5 ]
Note X1 = C1R1 = [21 0 2 ]
which is what we have solved in problem 5.