Number recognition

This repository contains a neural network written from scratch in rust using the nalgebra crate. It is part of a blog series currently in progress here. You may find it useful to read the blogs if you hope to understand the code, otherwise I hope it is readable enough to be understood without them.

Running 🏃

This project can currently only be run by using cargo:

git clone https://github.com/max-amb/number_recognition.git
cd number_recognition
cargo run

Nix

If you are using nix there is no need to try install dependencies as they are all contained in the flake.nix. You can enter this dev shell as follows:

git clone https://github.com/max-amb/number_recognition.git
cd number_recognition
nix develop

Speed comparison with pytorch

The data for training runs can be found in ./results/. The tests were done utilising hyperfine utilising the following commands:

hyperfine --runs 100 "python3 main.py" --export-json ./py_results.json && \
cd ../ && \
hyperfine --runs 100 "./target/release/number_recognition" --export-json ./rust_results.json

The network that the models were tested on followed a $[728] \to [256] \to [10]$ architecure and utilised stochastic gradient descent and He initialisation. The program exited when the accuracy on test data exceeded $96%$. The test data was checked every epoch. I have attempted to ensure all parameters in the models are the same but if you spot any disparity please email me or raise an issue. The results in green are results from the parallelised rust version, orange is the non-parallelised rust version and blue is the pytorch version (also non-parallelised). Below is a graph displaying the results

Now, when parallelised, using the same conditions as above, the parallelised version was about $10\times$ faster than the non parallelised training

Testing data

This section hopes to detail how I obtained my testing data. If you are unsure of anything I recommend you read my blog post which walks through the mathematics!

Initial conditions

We start with a network with 3 layers (one input, one hidden and one output). They have sizes: 3, 2, and 3. The weights look like this:

$$\omega^{[1]} = \begin{bmatrix} -0.75 & 0.5 & -0.25 \\ 0.25 & -0.5 & 0.75 \end{bmatrix}$$

$$\omega^{[2]} = \begin{bmatrix}-0.75 & 0.75 \\ 0.5 & -0.5 \\ -0.25 & 0.25 \end{bmatrix}$$

Then we will have biases:

$$b^{[1]} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

$$b^{[2]} = \begin{bmatrix} 0.75 \\ 0.5 \\ 0.25 \end{bmatrix}$$

We will have mock data:

$$\begin{bmatrix} 0.25 \\ 0.5 \\ 0.75 \end{bmatrix}$$

and expect an output of

$$\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$

For the activation functions, we use a leaky relu ($\alpha = 0.2$), denoted $ReLU$, for the hidden layers and a logistical sigmoid function, denoted $\sigma$, for the output layer.

Everything is shortened to 3 decimal points for conciseness, but the full output can be found in the wolfram outputs.

Forward pass

Layer 1:

$$ \begin{aligned} a^{[1]} & = ReLU((\omega^{[1]} \times a^{[0]}) + b^{[1]}) \\ & = ReLU(\begin{bmatrix}-0.75 & 0.5 & -0.25 \\ 0.25 & -0.5 & 0.75 \end{bmatrix} \times \begin{bmatrix} 0.25 \\ 0.5 \\ 0.75 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix})\\ & = \begin{bmatrix} -0.025 \\ 1.375 \end{bmatrix} \end{aligned} $$

Layer 2:

$$ \begin{aligned} a^{[2]} & = \sigma((\omega^{[2]} \times a^{[1]}) + b^{[2]}) \\ & = \sigma(\begin{bmatrix}-0.75 & 0.75 \\ 0.5 & -0.5 \\ -0.25 & 0.25 \end{bmatrix} \times \begin{bmatrix} -0.025 \\ 1.375 \end{bmatrix} + \begin{bmatrix} 0.75 \\ 0.5 \\ 0.25 \end{bmatrix})\\ & = \begin{bmatrix} 0.858 \\ 0.450 \\ 0.646 \end{bmatrix} \end{aligned} $$

Backpropagation

Deltas for layer 2

$$ \begin{aligned} \delta^{[2]} & = \nabla_a C \ \odot\ f'(z^{[2]}) \\ & = 2 \ (\begin{bmatrix} 0.858 \\ 0.450 \\ 0.646 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}) \odot \begin{bmatrix} 0.858 (1-0.858) \\ 0.450 (1-0.450) \\ 0.646(1-0.646) \end{bmatrix} \\ & = \begin{bmatrix} 0.209 \\ -0.272 \\ 0.295 \end{bmatrix} \end{aligned} $$

Wolfram input: 2 * ({(1/(1+e^{-1.8})),(1/(1+e^{0.2})),(1/(1+e^{-0.6}))} - {0,1,0}) * ({(1/(1+e^{-1.8}))*(1-(1/(1+e^{-1.8}))), (1/(1+e^{0.2}))*(1-(1/(1+e^{0.2}))), (1/(1+e^{-0.6}))*(1-(1/(1+e^{-0.6})))})

Output: {{0.208924}, {-0.272186}, {0.295432}}

Biase derivatives for layer 2

$$ \frac{\partial C}{\partial b^{[2]}} = \delta^{[2]} = \begin{bmatrix} 0.209 \\ -0.272 \\ 0.295 \end{bmatrix} $$

Weight derivatives for layer 2

$$ \begin{aligned} \frac{\partial C}{\partial \omega^{[2]}} & = \delta^{[2]} {a^{[1]}}^T \\ & = \begin{bmatrix} 0.209 \\ -0.272 \\ 0.295 \end{bmatrix} \begin{bmatrix} -0.025 & 1.375 \end{bmatrix} \\ & = \begin{bmatrix} -0.005 & 0.287 \\ 0.007 & -0.374 \\ -0.007 & 0.406 \end{bmatrix} \end{aligned} $$

Wolfram input: {{0.208924}, {-0.272186}, {0.295432}} * {{-0.025, 1.375}}

Output: {{-0.0052231, 0.287271}, {0.00680465, -0.374256}, {-0.0073858, 0.406219}}

Deltas for layer 1

$$ \begin{aligned} \delta^{[1]} & = ({\omega^{[2]}}^T \delta^{[2]}) \odot f'(z^{[1]}) \\ & = (\begin{bmatrix} -0.75 & 0.5 & -0.25 \\ 0.75 & -0.5 & 0.25 \end{bmatrix} \begin{bmatrix} 0.209 \\ -0.272 \\ 0.295 \end{bmatrix}) \otimes \begin{bmatrix} 0.2 \\ 1 \end{bmatrix} \\ & = \begin{bmatrix} -0.073 \\ 0.367 \end{bmatrix} \end{aligned} $$

Wolfram input: ({{-0.75, 0.5, -0.25}, {0.75, -0.5, 0.25}} * {{0.208924}, {-0.272186}, {0.295432}}) * {{0.2}, {1}}

Output: {{-0.0733288}, {0.366644}}

Biase derivatives for layer 1

$$ \frac{\partial C}{\partial b^{[1]}} = \delta^{[1]} = \begin{bmatrix} -0.073 \\ 0.367 \end{bmatrix} $$

Weight derivatives for layer 1

$$ \begin{aligned} \frac{\partial C}{\partial \omega^{[1]}} & = \delta^{[1]} {a^{[0]}}^T \\ & = \begin{bmatrix} -0.073 \\ 0.367 \end{bmatrix} \begin{bmatrix} 0.25 & 0.5 & 0.75 \end{bmatrix} \\ & = \begin{bmatrix} -0.018 & -0.037 & -0.055 \\ 0.092 & 0.183 & 0.275 \end{bmatrix} \end{aligned} $$

Wolfram input: {{-0.0733288}, {0.366644}} {{0.25, 0.5, 0.75}}

Output: {{-0.0183322, -0.0366644, -0.0549966}, {0.091661, 0.183322, 0.274983}}

FAQ

• Is it good? Yes