Page 34 - 《应用声学》2024年第6期
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1210 2024 年 11 月
ρ
(t+1)
步骤4 更新辅助变量z + (1 + µ g ||v|| 2 − z (t+1) + τ (t) 2
)
2 r 1
(t)
z (t+1) + λ (β (t+1) + µ g ||v|| 2 − η + τ (t) )
2 2
R ρ (t+1) (t) 2
∑ (t) (t) + (β + µ g ||v|| 2 − η + τ 2 ) . (52)
= arg min λ (1 + µ g ||v (t) || 2 − z r + τ ) 2
z 1,r 1
2
r=1 式(52)可转化为ℓ 2 + ℓ 结构的优化问题:
ρ (t) (t) 2 2
+ (1 + µ g ||v || 2 − z r + τ 1 ) (t+1) (t+1) ρ (t+1) 2
2 v =argmin ψ ||v|| 2 + ||v−Γ v || , (53)
2
v 2
R
∑
(t)
+ κ (z r − G T ˜ w (t+1) ) (t+1) (t) (t) (t)
r 0,r 式 (53) 中,ψ = µ g [2 + λ + ρ(Rτ + τ +
2 1 2
r=1 R
ρ T (t+1) 2 (t+1) ∑ (t) (t+1) )]/[1+µ (1+R)],
2
+ (z r − G 0,r ˜ w ) . (46) β +R−η)+ (λ 1,r − ρz r g
2 r=1
2
对式(46)做进一步化简: Γ v (t+1) = ( ˜ w (t+1) − γ (t) /ρ)/[1 + µ (1 + R)]。根据
g
附录A,通过微分运算可求得其最优解:
R
∑
(t+1) ˆ (t) 2 ¯ (t+1) 2
z = arg min (z r − Γ z,r ) + (z r − Γ z,r ) , (t+1) (t+1)
ˆ
z Γ v [Υ] + , ||Γ v || 2 ̸= 0,
r=1 v (t+1) = (54)
(47) 0 U , ||Γ v (t+1) || 2 = 0,
(t)
ˆ
式 (47) 中,Γ z,r = 1 + µ g ||v (t) || 2 + τ (t) + λ (t) /ρ, ˆ (t+1) (t+1) (t+1) || 2 。
1 1,r 其中,Υ = (||Γ v || 2 − ψ /ρ)/||Γ v
(t)
¯ (t+1) T (t+1) − κ r /ρ,r = 1, 2, · · · , R。将 (t+1)
Γ z,r = G ˜ w 步骤7 更新辅助变量τ
0,r 1
式(47) 拆分为 R 个独立子问题,容易求得每个子问 R
(t+1) ∑ (t) (t+1)
题的解: τ 1 = arg min λ 1,r (1 + µ g ||v || 2
τ 1 >0
r=1
ρ
z r (t+1) = (Γ ˆ (t) + Γ ¯ (t+1) )/2, r = 1, 2, · · · , R. (48) − z (t+1) + τ 1 ) + (1 + µ g ||v (t+1)
z,r
z,r
r
2 || 2
步骤5 更新辅助变量Y (t+1) − z (t+1) + τ 1 ) . (55)
2
r
T (t+1)
˜
(t)
(t+1)
Y = arg min δ||Y || 1 + σ (Y − b ˜ w ) 易求得其最优解为
Y
ρ (t+1) 2 (t+1) [
˜
+ ||Y − b ˜ w || . (49) τ = − 1 − µ g ||v (t+1) || 2
2
2 1
等价于求解如下问题: R ]
∑ (t+1) (t)
+ (z − λ /ρ)/R . (56)
ρ (t+1) 2 r 1,r
(t+1)
Y =arg min δ||Y || 1 + ||Y − Γ Y || , (50) r=1 +
2
Y 2
(t+1)
(t+1) ˜ (t+1) (t) 步骤8 更新辅助变量τ 2
式 (50) 中,Γ = b ˜ w − σ /ρ。式 (50) 为
Y
(t)
典型的 ℓ 1 + ℓ 范数凸优化问题,可求得其软阈值 τ (t+1) = arg min λ (β (t+1) + µ g ||v (t+1) || 2 − η
2
2 2 2
τ 2 >0
解 [17] : ρ
2
+ τ 2 ) + (β (t+1) + µ g ||v (t+1) || 2 − η + τ 2 ) . (57)
2
(t+1)
(t+1)
Y u (t+1) = sign(Γ Y,u )[|Γ Y,u | − δ/ρ] + , 类似步骤7,可求得最优解为
u = 1, 2, · · · , U, (51) (t+1) (t)
τ = [η − β (t+1) − µ g ||v (t+1) || 2 − λ /ρ] + .
2 2
式(51)中,sign(·)表示符号函数。 (58)
步骤6 更新辅助变量v (t+1) (t+1) (t+1)
步骤9 更新拉格朗日乘子变量{λ 1 , λ 2 ,
v (t+1) = ξ (t+1) , ς (t+1) , κ (t+1) , σ (t+1) , γ (t+1) }
T (t+1)
(t)
arg min 2µ g ||v|| 2 + γ (v − ˜ w ) (t+1) (t) (t+1) (t+1)
v λ 1,r = λ 1,r + ρ(1 + µ g ||v || 2 − z r
ρ (t+1) 2 (t+1)
+ ||v − ˜ w || 2 + τ 1 ), r = 1, 2, · · · , R, (59a)
2
(t+1) (t) (t+1) (t+1)
R λ = λ + ρ(β + µ g ||v
∑ (t) (t+1) (t) 2 2 || 2
+ λ 1,r (1 + µ g ||v|| 2 − z r + τ 1 ) (t+1)
− η + τ ), (59b)
r=1 2