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cnn突破六（四层bpnet网络公式）

news 2025/4/28 12:43:13

四层bpnet网络反向传播公式推导：

X【196】-》HI【128】/HO【128】-》H2I【60】/H2O【60】-》YI【10】/YO【10】，

期望是d【10】

X，HI之间用w1【196，128】

HO，H2I之间用w12【128,60】

H2O，YI之间用w2【60,10】

k=10;m=60；j=128;i=196

$\frac{\partial E}{\partial w2[m,k]}$ = $\frac{\partial E}{\partial yo[k]}$ $\frac{\partial yo[k]}{\partial yi[k]}$ $\frac{\partial yi[k]]}{\partial w2[m,k]}$ =(yo[k]-d[k])*ds(yo[k])*ho[m];

$\frac{\partial E}{\partial w12[j,m]}$ = $\sum_{k=0}^{9}$ $\frac{\partial E}{\partial yo[k]}$ $\frac{\partial yo[k]}{\partial yi[k]}$ $\frac{\partial yi[k]]}{\partial h2o[m]}$ $\frac{\partial h2o[m]}{\partial h2i[m]}$ $\frac{\partial h2i[m]}{\partial w12[j,m]}$

= $\sum_{k=0}^{9}$ (yo[k]-d[k])*ds(yo[k])*w2【m，k】*ds(h2o[m])*ho【j】

$\frac{\partial E}{\partial w1[i,j]}$ = $\sum_{k=0}^{9}$ $\frac{\partial E}{\partial yo[k]}$ $\frac{\partial yo[k]}{\partial yi[k]}$ $\frac{\partial yi[k]]}{\partial h2o[m]}$ $\frac{\partial h2o[m]}{\partial h2i[m]}$ $\frac{\partial h2i[m]}{\partial ho[j]}$ $\frac{\partial ho[j]}{\partial hi[j]}$ $\frac{\partial hi[j]}{\partial w1[i,j]}$