backpropagation Archives - Page 2 of 2

배치 정규화 는 심층 신경망에서 상당한 성능 향상으로 인정되었습니다. 인터넷에 많은 자료가 활성화별로이를 구현하는 방법을 보여줍니다. 나는 이미 행렬 대수를 사용하여 backprop를 구현했으며, 고밀도 언어 ( Rcpp고밀도 행렬 곱셈 에 (그리고 결국 GPU)에 의존하는 동안 )에서 모든 것을 추출하고 for-loops를 사용하면 코드가 느려질 것입니다 대단한 고통에 더해

배치 정규화 함수는

b (x_{p}) = γ (x_{p} - μ_{x_{p}}) σ_{x_{p}}^{- 1} + β

$b(x_p) = \gamma \left(x_p - \mu_{x_p}\right) \sigma^{-1}_{x_p} + \beta$
.

$x_{p}$
$γ$
$μ_{x_{p}}$

매트릭스 형태에서, 전체 층에 대한 배치 정규화는

b (X) = (γ \otimes 1_{p}) ⊙ (X - μ_{X}) ⊙ σ_{X}^{- 1} + (β \otimes 1_{p})

$b(\mathbf{X}) = \left(\gamma\otimes\mathbf{1}_p\right)\odot \left(\mathbf{X} - \mu_{\mathbf{X}}\right) \odot\sigma^{-1}_{\mathbf{X}} + \left(\beta\otimes\mathbf{1}_p\right)$
.

$X$
$1_{N}$
$γ$
및 는 행렬이며, 각 열은 열별평균 및 표준 편차의 벡터입니다. $μ_{X}$
는 Kronecker 제품이고 는 요소 별 (Hadamard) 제품입니다 $\otimes$

배치 정규화가없고 연속적인 결과가없는 매우 간단한 단층 신경망은

y = a ({X Γ}_{1}) Γ_{2} + ϵ

$y = a\left(\mathbf{X\Gamma}_1\right)\Gamma_2 + \epsilon$

어디

은 $Γ_{1}$
는 $Γ_{2}$
는 활성화 함수입니다 $a (.)$

손실의 경우 다음, 기울기는
$R = N^{- 1} \sum {(y - \hat{y})}^{2}$

R = N^{- 1} \sum {(y - \hat{y})}^{2}

$R = N^{-1}\displaystyle\sum\left(y - \hat{y}\right)^2$

\begin{array}{lr} \frac{\partial R}{\partial Γ_{1}} = - 2 V^{T} \hat{ϵ} \\ \frac{\partial R}{\partial Γ_{2}} = X^{T} (a^{'} (X Γ_{1}) ⊙ - 2 \hat{ϵ} Γ_{2}^{T}) \end{array}

$\begin{array}{lr} \frac{\partial R}{\partial \Gamma_1} = -2\mathbf{V}^T \hat\epsilon\\ \frac{\partial R}{\partial \Gamma_2} = \mathbf{X}^T \left(a'(\mathbf{X}\mathbf{\Gamma}_1) \odot -2\hat\epsilon \mathbf{\Gamma}_2^T\right) \\ \end{array}$

어디

$V = a (X Γ_{1})$
$\hat{ϵ} = y - \hat{y}$

배치 정규화 하에서,

또는

y = a (b (X Γ_{1})) Γ_{2}

$y = a\left(b\left(\mathbf{X}\Gamma_1\right)\right)\Gamma_2$

y = a ((γ \otimes 1_{N}) ⊙ (X Γ_{1} - μ_{X Γ_{1}}) ⊙ σ_{X Γ_{1}}^{- 1} + (β \otimes 1_{N})) Γ_{2}

$y = a\Big(\left(\gamma\otimes\mathbf{1}_N\right)\odot \left(\mathbf{X\Gamma_1} - \mu_{\mathbf{X\Gamma_1}}\right) \odot\sigma^{-1}_{\mathbf{X\Gamma_1}} + \left(\beta\otimes\mathbf{1}_N\right)\Big)\mathbf{\Gamma_2}$
Hadamard 및 Kronecker 제품의 파생 상품을 계산하는 방법을 모르겠습니다. 크로네 커 (Kronecker) 제품의 주제에 관한 문헌 은 상당히 비현실적이다.

행렬 프레임 워크 내에 , 및 을 계산하는 실용적인 방법이 있습니까? 노드 별 계산에 의존하지 않고 간단한 표현? $\partial R / \partial γ$

\partial R / \partial γ

$\partial R/\partial \gamma$ $\partial R / \partial β$

\partial R / \partial β

$\partial R/\partial \beta$ $\partial R / \partial Γ_{1}$

\partial R / \partial Γ_{1}

$\partial R/\partial \mathbf{\Gamma_1}$

업데이트 1 :

$\partial R / \partial β$

\partial R / \partial β

$\partial R/\partial \beta$

1_{N}^{T} (a^{'} (X Γ_{1}) ⊙ - 2 \hat{ϵ} Γ_{2}^{T})

$\mathbf{1}_{N}^T \left(a'(\mathbf{X}\mathbf{\Gamma}_1) \odot -2\hat\epsilon \mathbf{\Gamma}_2^T\right)$

set.seed(1)
library(dplyr)
library(foreach)

#numbers of obs, variables, and hidden layers
N <- 10
p1 <- 7
p2 <- 4
a <- function (v) {
  v[v < 0] <- 0
  v
}
ap <- function (v) {
  v[v < 0] <- 0
  v[v >= 0] <- 1
  v
}

# parameters
G1 <- matrix(rnorm(p1*p2), nrow = p1)
G2 <- rnorm(p2)
gamma <- 1:p2+1
beta <- (1:p2+1)*-1
# error
u <- rnorm(10)

# matrix batch norm function
b <- function(x, bet = beta, gam = gamma){
  xs <- scale(x)
  gk <- t(matrix(gam)) %x% matrix(rep(1, N))
  bk <- t(matrix(bet)) %x% matrix(rep(1, N))
  gk*xs+bk
}
# activation-wise batch norm function
bi <- function(x, i){
  xs <- scale(x)
  gk <- t(matrix(gamma[i]))
  bk <- t(matrix(beta[i]))
  suppressWarnings(gk*xs[,i]+bk)
}

X <- round(runif(N*p1, -5, 5)) %>% matrix(nrow = N)
# the neural net
y <- a(b(X %*% G1)) %*% G2 + u

그런 다음 파생 상품을 계산하십시오.

# drdbeta -- the matrix way
drdb <- matrix(rep(1, N*1), nrow = 1) %*% (-2*u %*% t(G2) * ap(b(X%*%G1)))
drdb
           [,1]      [,2]    [,3]        [,4]
[1,] -0.4460901 0.3899186 1.26758 -0.09589582
# the looping way
foreach(i = 1:4, .combine = c) %do%{
  sum(-2*u*matrix(ap(bi(X[,i, drop = FALSE]%*%G1[i,], i)))*G2[i])
}
[1] -0.44609015  0.38991862  1.26758024 -0.09589582

$β \otimes 1_{N}$

β \otimes 1_{N}

$\beta \otimes \mathbf{1}_N$

\frac{\partial A \otimes B}{\partial A} = (I_{n q} \otimes T_{m p}) (I_{n} \otimes v e c (B) \otimes I_{m})

$\frac{\partial A \otimes B}{\partial A} = \left(I_{nq} \otimes T_{mp}\right)\left(I_n\otimes vec(B) \otimes I_m\right)$ $m$

m

$m$ $n$

n

$n$ $p$

p

$p$ $q$

q

$q$ $A$

A

$A$ $B$

B

$B$ $T$

T

$T$

# playing with the kroneker derivative rule
A <- t(matrix(beta))
B <- matrix(rep(1, N))
diag(rep(1, ncol(A) *ncol(B))) %*% diag(rep(1, ncol(A))) %x% (B) %x% diag(nrow(A))
     [,1] [,2] [,3] [,4]
 [1,]    1    0    0    0
 [2,]    1    0    0    0
 snip
[13,]    0    1    0    0
[14,]    0    1    0    0
snip
[28,]    0    0    1    0
[29,]    0    0    1    0
[snip
[39,]    0    0    0    1
[40,]    0    0    0    1

$γ$

γ

$\gamma$ $Γ_{1}$

Γ_{1}

$\mathbf{\Gamma_1}$ $β \otimes 1$

β \otimes 1

$\beta \otimes \mathbf{1}$

업데이트 2

$\partial R / \partial Γ_{1}$

\partial R / \partial Γ_{1}

$\partial R/\partial \Gamma_1$ $\partial R / \partial γ$

\partial R / \partial γ

$\partial R/\partial \gamma$ vec() $\partial R / \partial Γ_{1}$

\partial R / \partial Γ_{1}

$\partial R/\partial \Gamma_1$ $w ⊙ X Γ_{1}$

w ⊙ X Γ_{1}

$w\odot\mathbf{X\Gamma_1}$ $Γ_{1}$

Γ_{1}

$\mathbf{\Gamma_1}$ $w \equiv (γ \otimes 1) ⊙ σ_{X Γ_{1}}^{- 1}$

w \equiv (γ \otimes 1) ⊙ σ_{X Γ_{1}}^{- 1}

$w \equiv (\gamma \otimes \mathbf{1}) \odot \sigma_{\mathbf{X\Gamma_1}}^{-1}$

$w ⊙ X$

w ⊙ X

$w\odot\mathbf{X}$ $w$

w

$w$ $X$

X

$\mathbf{X}$

\partial (A ⊙ B) = \partial A ⊙ B + A ⊙ \partial B

$\partial(A \odot B) = \partial A \odot B + A \odot \partial B$

과에서 이 , 그

\frac{\partial v e c (w ⊙ X Γ_{1})}{\partial v e c (Γ_{1})^{T}} = v e c (X Γ_{1}) I \frac{\partial v e c (w)}{\partial v e c (Γ_{1})^{T}} + v e c (w) I \frac{\partial v e c (X Γ_{1})}{\partial v e c (Γ_{1})^{T}}

$\frac{\partial vec(w \odot \mathbf{X\Gamma_1})}{\partial vec(\mathbf{\Gamma_1})^T} = vec(\mathbf{X\Gamma_1})I\frac{\partial vec(w)}{\partial vec(\mathbf{\Gamma_1})^T} + vec(w)I\frac{\partial vec(\mathbf{X\Gamma_1})}{\partial vec(\mathbf{\Gamma_1})^T}$

업데이트 3

여기에서 진보하고 있습니다. 나는 지난 밤 2시에이 아이디어로 일어났다. 수학은 수면에 좋지 않습니다.

$\partial R / \partial Γ_{1}$

\partial R / \partial Γ_{1}

$\partial R/\partial \mathbf{\Gamma_1}$

$w \equiv (γ \otimes 1) ⊙ σ_{X Γ_{1}}^{- 1}$
$"stub" \equiv a^{'} (b ({X Γ}_{1})) ⊙ - 2 \hat{ϵ} Γ_{2}^{T}$

\frac{\partial R}{\partial Γ_{1}} = \frac{\partial w ⊙ {X Γ}_{1}}{\partial Γ_{1}} (“stub”)

$\frac{\partial R}{\partial \Gamma_1} = \frac{\partial w \odot \mathbf{X\Gamma}_1}{\partial \Gamma_1}\left(\text{"stub"}\right)$ $i$

i

$i$ $j$

j

$j$ $I$

I

$\mathbf{I}$

\frac{\partial R}{\partial Γ_{i j}} = {(w_{i} ⊙ X_{i})}^{T} ({“stub”}_{j})

$\frac{\partial R}{\partial \Gamma_{ij}} = \left(w_i \odot \mathbf{X_i}\right)^T\left(\text{"stub"}_j\right)$

\frac{\partial R}{\partial Γ_{i j}} = {(I w_{i} X_{i})}^{T} ({“stub”}_{j})

$\frac{\partial R}{\partial \Gamma_{ij}} = \left(\mathbf{I} w_i \mathbf{X_i}\right)^T\left(\text{"stub"}_j\right)$

\frac{\partial R}{\partial Γ_{i j}} = {X_{i}}^{T} I w_{i} ({“stub”}_{j})

$\frac{\partial R}{\partial \Gamma_{ij}} = \mathbf{X_i}^T\mathbf{I} w_i\left(\text{"stub"}_j\right)$

\frac{\partial R}{\partial Γ} = X^{T} (“stub” ⊙ w)

$\frac{\partial R}{\partial \Gamma} = \mathbf{X}^T\left(\text{"stub"}\odot w\right)$

그리고 실제로는 :

stub <- (-2*u %*% t(G2) * ap(b(X%*%G1)))
w <- t(matrix(gamma)) %x% matrix(rep(1, N)) * (apply(X%*%G1, 2, sd) %>% t %x% matrix(rep(1, N)))
drdG1 <- t(X) %*% (stub*w)

loop_drdG1 <- drdG1*NA
for (i in 1:7){
  for (j in 1:4){
    loop_drdG1[i,j] <- t(X[,i]) %*% diag(w[,j]) %*% (stub[,j])
  }
}

> loop_drdG1
           [,1]       [,2]       [,3]       [,4]
[1,] -61.531877  122.66157  360.08132 -51.666215
[2,]   7.047767  -14.04947  -41.24316   5.917769
[3,] 124.157678 -247.50384 -726.56422 104.250961
[4,]  44.151682  -88.01478 -258.37333  37.072659
[5,]  22.478082  -44.80924 -131.54056  18.874078
[6,]  22.098857  -44.05327 -129.32135  18.555655
[7,]  79.617345 -158.71430 -465.91653  66.851965
> drdG1
           [,1]       [,2]       [,3]       [,4]
[1,] -61.531877  122.66157  360.08132 -51.666215
[2,]   7.047767  -14.04947  -41.24316   5.917769
[3,] 124.157678 -247.50384 -726.56422 104.250961
[4,]  44.151682  -88.01478 -258.37333  37.072659
[5,]  22.478082  -44.80924 -131.54056  18.874078
[6,]  22.098857  -44.05327 -129.32135  18.555655
[7,]  79.617345 -158.71430 -465.91653  66.851965

업데이트 4

$\partial R / \partial γ$

\partial R / \partial γ

$\partial R / \partial \gamma$

$\tilde{X Γ} \equiv (X Γ - μ_{X Γ}) ⊙ σ_{X Γ}^{- 1}$
$\tilde{γ} \equiv γ \otimes 1_{N}$

이전과 마찬가지로 체인 규칙은 까지

\frac{\partial R}{\partial \tilde{γ}} = \frac{\partial \tilde{γ} ⊙ \tilde{X Γ}}{\partial \tilde{γ}} (“stub”)

$\frac{\partial R}{\partial \tilde\gamma} = \frac{\partial \tilde\gamma \odot \widetilde{\mathbf{X\Gamma}}}{\partial \tilde\gamma}\left(\text{"stub"}\right)$

\frac{\partial R}{\partial {\tilde{γ}}_{i}} = (\tilde{X Γ})_{i}^{T} I {\tilde{γ}}_{i} ({“stub”}_{i})

$\frac{\partial R}{\partial \tilde\gamma_i} = (\widetilde{\mathbf{X\Gamma}})_i^T \mathbf{I}\tilde\gamma_i \left(\text{"stub"}_i\right)$

\frac{\partial R}{\partial \tilde{γ}} = (\tilde{X Γ})^{T} (“stub” ⊙ \tilde{γ})

$\frac{\partial R}{\partial \tilde\gamma} = (\widetilde{\mathbf{X\Gamma}})^T \left(\text{"stub"} \odot \tilde\gamma \right)$

그것은 일종의 일치합니다.

drdg <- t(scale(X %*% G1)) %*% (stub * t(matrix(gamma)) %x% matrix(rep(1, N)))

loop_drdg <- foreach(i = 1:4, .combine = c) %do% {
  t(scale(X %*% G1)[,i]) %*% (stub[,i, drop = F] * gamma[i])
}

> drdg
           [,1]      [,2]       [,3]       [,4]
[1,]  0.8580574 -1.125017  -4.876398  0.4611406
[2,] -4.5463304  5.960787  25.837103 -2.4433071
[3,]  2.0706860 -2.714919 -11.767849  1.1128364
[4,] -8.5641868 11.228681  48.670853 -4.6025996
> loop_drdg
[1]   0.8580574   5.9607870 -11.7678486  -4.6025996

$γ$

γ

$\gamma$

본인의 질문에 답변 한 것 같지만 본인이
맞는지 잘 모르겠습니다. 이 시점에서 나는 내가 함께 해킹 한 것을 엄격하게 증명 (또는 반증)하는 대답을 받아 들일 것이다.

while(not_answered){
  print("Bueller?")
  Sys.sleep(1)
}

답변

b (X) = (X - e_{N} μ_{X}^{T}) Γ Σ_{X}^{- 1 / 2} + e_{N} β^{T}

$b(X)=(X−e_N\mu_X^T)ΓΣ_X^{-1/2}+e_N\beta^T$ $Γ = d i a g (γ)$

Γ = d i a g (γ)

$\Gamma=\mathop{\mathrm{diag}}(\gamma)$ $Σ_{X}^{- 1 / 2} = d i a g (σ_{X_{1}}^{- 1}, σ_{X_{2}}^{- 1}, \dots)$

Σ_{X}^{- 1 / 2} = d i a g (σ_{X_{1}}^{- 1}, σ_{X_{2}}^{- 1}, \dots)

$\Sigma_X^{-1/2}=\mathop{\mathrm{diag}}(\sigma_{X_1}^{-1},\sigma_{X_2}^{-1},\dots)$ $e_{N}$

e_{N}

$e_N$

\nabla_{β} R = [- 2 \hat{ϵ} (Γ_{2}^{T} \otimes I) J_{X} (a) (I \otimes e_{N})]^{T}

$\nabla_\beta R=[-2\hat{\epsilon}(\Gamma_2^T\otimes I)J_X(a)(I\otimes e_N)]^T$ $- 2 \hat{ϵ} (Γ_{2}^{T} \otimes I) = v e c (- 2 \hat{ϵ} Γ_{2}^{T})^{T}$

- 2 \hat{ϵ} (Γ_{2}^{T} \otimes I) = v e c (- 2 \hat{ϵ} Γ_{2}^{T})^{T}

$-2\hat{\epsilon}(\Gamma_2^T\otimes I)=\mathop{\mathrm{vec}}(-2\hat{\epsilon}\Gamma_2^T)^T$ $J_{X} (a) = d i a g (v e c (a^{'} (b (X Γ_{1}))))$

J_{X} (a) = d i a g (v e c (a^{'} (b (X Γ_{1}))))

$J_X(a)=\mathop{\mathrm{diag}}(\mathop{\mathrm{vec}}(a^\prime(b(X\Gamma_1))))$

\nabla_{β} R = (I \otimes e_{N}^{T}) v e c (a^{'} (b (X Γ_{1})) ⊙ - 2 \hat{ϵ} Γ_{2}^{T}) = e_{N}^{T} (a^{'} (b (X Γ_{1})) ⊙ - 2 \hat{ϵ} Γ_{2}^{T})

$\nabla_\beta R=(I\otimes e_N^T)\mathop{\mathrm{vec}}(a^\prime(b(X\Gamma_1))\odot-2\hat{\epsilon}\Gamma_2^T)=e_N^T(a^\prime(b(X\Gamma_1))\odot-2\hat{\epsilon}\Gamma_2^T)$ $v e c (A X B) = (B^{T} \otimes A) v e c (X)$

v e c (A X B) = (B^{T} \otimes A) v e c (X)

$\mathop{\mathrm{vec}}(AXB)=(B^T\otimes A)\mathop{\mathrm{vec}}(X)$

\begin{aligned} \nabla_{γ} R & = [- 2 \hat{ϵ} (Γ_{2}^{T} \otimes I) J_{X} (a) (Σ_{X Γ_{1}}^{- 1 / 2} \otimes (X Γ_{1} - e_{N} μ_{X Γ_{1}}^{T})) K]^{T} \\ = K^{T} v e c ((X Γ_{1} - e_{N} μ_{X Γ_{1}}^{T})^{T} W Σ_{X Γ_{1}}^{- 1 / 2}) \\ = d i a g ((X Γ_{1} - e_{N} μ_{X Γ_{1}}^{T})^{T} W Σ_{X Γ_{1}}^{- 1 / 2}) \end{aligned}

$\begin{align}\nabla_\gamma R&=[-2\hat{\epsilon}(\Gamma_2^T\otimes I)J_X(a)(\Sigma_{X\Gamma_1}^{-1/2}\otimes (X\Gamma_1-e_N\mu_{X\Gamma_1}^T))K]^T\\&=K^T\mathop{\mathrm{vec}}((X\Gamma_1-e_N\mu_{X\Gamma_1}^T)^TW\Sigma^{-1/2}_{X\Gamma_1})\\&=\mathop{\mathrm{diag}}((X\Gamma_1-e_N\mu_{X\Gamma_1}^T)^TW\Sigma^{-1/2}_{X\Gamma_1})\end{align}$ $W = a^{'} (b (X Γ_{1})) ⊙ - 2 \hat{ϵ} Γ_{2}^{T}$

W = a^{'} (b (X Γ_{1})) ⊙ - 2 \hat{ϵ} Γ_{2}^{T}

$W=a^\prime(b(X\Gamma_1))\odot-2\hat{\epsilon}\Gamma_2^T$ $K$

K

$K$ $N p \times p$

N p \times p

$Np\times p$ $d Γ_{i \neq j} = 0$

d Γ_{i \neq j} = 0

$d\Gamma_{i\neq j}=0$ $b$

b

$b$ $γ_{i}$

γ_{i}

$\gamma_i$ $γ_{i}$

γ_{i}

$\gamma_i$ $Γ_{1}$

Γ_{1}

$\Gamma_1$ $w$

w

$w$ $Σ_{X}$

Σ_{X}

$\Sigma_X$ $μ_{X}$

μ_{X}

$\mu_X$ $X$

X

$X$

How IT

언제든지 물어보세요.

태그 보관물: backpropagation

배치 정규화를 통한 역 전파의 매트릭스 형태 = gamma){ xs <- scale(x)

답변

답변