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Commit 1d8e2c4a authored by toto's avatar toto
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last results of mlp

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......@@ -43,32 +43,6 @@ Ces fonctions permettent de calculer la précision de l'algorithme k-NN pour dif
### Backpropagation in a Neural Network
#### Partial Derivatives with Chain Rule
1. **\(\frac{\partial C}{\partial A^{(2)}}\)**:
\(\frac{\partial C}{\partial A^{(2)}} = \frac{2}{N_{out}}(\hat{Y} - Y)\)
2. **\(\frac{\partial C}{\partial Z^{(2)}}\)**:
\(\frac{\partial C}{\partial Z^{(2)}} = \frac{2}{N_{out}}(\hat{Y} - Y) \cdot \sigma(Z^{(2)}) \cdot (1 - \sigma(Z^{(2)}))\)
3. **\(\frac{\partial C}{\partial W^{(2)}}\)**:
\(\frac{\partial C}{\partial W^{(2)}} = \frac{\partial C}{\partial Z^{(2)}} \cdot A^{(1)T}\)
4. **\(\frac{\partial C}{\partial B^{(2)}}\)**:
\(\frac{\partial C}{\partial B^{(2)}} = \frac{\partial C}{\partial Z^{(2)}}\)
5. **\(\frac{\partial C}{\partial A^{(1)}}\)**:
\(\frac{\partial C}{\partial A^{(1)}} = (W^{(2)T} \cdot \frac{\partial C}{\partial Z^{(2)}})\)
6. **\(\frac{\partial C}{\partial Z^{(1)}}\)**:
\(\frac{\partial C}{\partial Z^{(1)}} = \frac{\partial C}{\partial A^{(1)}} \cdot \sigma'(Z^{(1)})\)
7. **\(\frac{\partial C}{\partial W^{(1)}}\)**:
\(\frac{\partial C}{\partial W^{(1)}} = \frac{\partial C}{\partial Z^{(1)}} \cdot A^{(0)T}\)
8. **\(\frac{\partial C}{\partial B^{(1)}}\)**:
\(\frac{\partial C}{\partial B^{(1)}} = \frac{\partial C}{\partial Z^{(1)}\)
# Neural Network Training and Testing Overview
......
from read_cifar import read_cifar, split_dataset
from knn import evaluate_knn_for_k, plot_accuracy_versus_k
import matplotlib.pyplot as plt
from mlp2 import run_mlp_training, plot_accuracy_versus_epoch
from mlp import run_mlp_training, plot_accuracy_versus_epoch
......
results/mlp1.png

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