Our model will certainly not work very well because we are missing important car attributes impacting the price, but the goal is to use real data while keeping things simple.

The first step is to normalize our data (known as feature scaling). We have:

- Number of kilometers: quantitative, number between 0 and 350k.
- Type of fuel: binary data diesel/gasoline.
- Age: quantitative, number between 0 and 40.
- Price: quantitative, number between 0 and 40k.

The number of kilometers and age (both quantitative) will be normalized using the means and standard deviation. The goal is to bring all the data to the same scale, generally between $[-6, 6]$. Else the number of kilometers will have a value up to 350k and the age up to 40 years, and a change in a weight value will not impact the age or number of kilometers in the same way. The formula for the normalization is:

$$x' = {x-\bar{x}\over \sigma}$$
Where $x$ is the original feature dataset (all the cars), $\bar{x}$ is the mean of that feature dataset, and $\sigma$ is the standard deviation of that feature dataset. We will do it in python later.

The type of fuel (binary) will be encoded with $-1$ for one value and $1$ for the other. There is no categorical data here in order to keep the number of features light because each categorical feature would become a sparse matrix of the size of the number of classes. If you have categorical data in your features (i.e: red, green, pink and blue for a feature) you must use effect encoding or dummy encoding. Our number of features will not change, only the values after the encoding.

We will normalize the price so that all the values are between $[0,1]$. This is necessary because our neural network will produce results between $[0,1]$. If the neural network guesses a price of 0.45 for a car whereas our initial price is 17k, the error between the guess and the real price will be high, 16 999.55€. As the guess will be at most 1, the error will be at best 16 999€ and will never improve. We will normalize the price using the formula:

$$\frac{x_i-\min(x)}{\max(x)-\min(x)}$$

$x_i$ is one car price whereas $x$ is the whole price dataset. It means that each car price will be processed using the maximum available price and the minimum available price.

Our 17K€ price will become something like 0.43 and if our network predict 0.45 it will be a good prediction. We will do the exact opposite from the above formula to find the price back from 0.45.

So we have three inputs (features), one binary, two quantitatives and one quantitative output. As we will predict a quantitative variable and we will use past data to train our network, it is called a supervised regression problem.

One input is a car.

Number of kms | Fuel | Age | Price |
---|---|---|---|

38 000 | Gasoline | 3 | 17 000 |

Once the data has been normalized and encoded as previously described, we will have.

Number of kms | Fuel | Age | Price |
---|---|---|---|

1.4 | -1 | 0.4 | 0.45 |

As we will have more than one car, our array will have more lines.

Number of kms | Fuel | Age | Price |
---|---|---|---|

1.4 | -1 | 0.4 | 0.45 |

0.4 | -1 | 0.1 | 0.52 |

5.4 | -1 | 4 | 0.25 |

1.5 | -1 | 1 | 0.31 |

... | ... | ... | ... |

We can see this array as matrix.

$$
X = \begin{bmatrix}

1.4 & -1 & 0.4 & 0.45 \\

0.4 & -1 & 0.1 & 0.52 \\

5.4 & -1 & 4 & 0.25 \\

1.5 & -1 & 1 & 0.31 \\

... & ... & ... & ...
\end{bmatrix}
$$

The price is not a feature and will not be available for a new car. We separate inputs (features) and outputs (prediction).

$$
X = \begin{bmatrix}

1.4 & -1 & 0.4 \\

0.4 & -1 & 0.1 \\

5.4 & -1 & 4 \\

1.5 & -1 & 1 \\

... & ... & ... & ...
\end{bmatrix}
y = \begin{bmatrix}

0.45 \\

0.52 \\

0.25 \\

0.31 \\

...
\end{bmatrix}
$$

For now we have two matrices $X$ and $y$, it's time to complete our network. We choose to have two hidden layers between our inputs and our output, the first layer will have three neurons and the second one two. The number of hidden layers and the number of neurons by layer is up to you. These are two hyper parameters that you have to define before running your neural network.

Each of our input neuron will reach all the neurons in the next layer because we are using a fully connected network. Each link from one neuron to another is called a synapse and comes with a weight. A weight on a synapse is of the form $W_{jk}^l$ where $l$ denotes the number of the layer, $j$ the number of the neuron from the $l^{th}$ layer and $k$ the number of the neuron from the $(l+1)^{th}$ layer.

We can view the weights as a matrix.

$$
W^1 = \begin{bmatrix}

W_{11}^1 & W_{12}^1 & W_{13}^1 \\

W_{21}^1 & W_{22}^1 & W_{23}^1 \\

W_{31}^1 & W_{32}^1 & W_{33}^1

\end{bmatrix}
$$

The number of rows equals the number of features (inputs) and the number of columns equals the number of neurons in the layer number $2$. Here we have three features and the first layer has three neurons so our $W^1$ matrix is of the size $3 \times 3$. For now I don't talk about the bias unit to keep things simple.

Now we compute the values of the neurons of the first hidden layer (the first column of green circles). For each car defined by three features, we will have three neurons computed for the first hidden layer. We do that using matrix calculus.

Let's say that we have only one car in our dataset. We don't need $y$ for now.

$$
X = \begin{bmatrix}

1.4 & -1 & 0.4 \\

\end{bmatrix}
$$

Our $W^1$ matrix is randomly initialized the first time so we can use random values for the weights.

$$
W^1 = \begin{bmatrix}

0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13

\end{bmatrix}
$$

The values of the neurons of the first hidden layer are called the activities of the first hidden layer. They will be saved into a matrix called $a^{(2)}$. We begin by computing $Z^{(2)}$ using the formula:

$$Z^{(2)} = X.W^1$$
$$Z^{(2)} = \begin{bmatrix}
1.4 & -1 & 0.4 \\

\end{bmatrix} . \begin{bmatrix}
0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13

\end{bmatrix}$$
$$Z^{(2)} = \begin{bmatrix}
-0.17 & 0.253 & 0.04 \\
\end{bmatrix}$$
We can view it on our network.

We are almost done with the first hidden layer, we just have to apply an activation function on $Z^{(2)}$. We can view it like that.

Once we apply an activation function element wise to $Z^{(2)}$ we obtain $a^{(2)}$, the activities of our second layer.

$$a^{(2)} = \sigma(Z^{(2)})$$

Where $\sigma(x)$ is our activation function. The activation function is applied element wise and is non linear, allowing the network to compute complicated problems using only a small number of nodes. The common ones are Sigmoid, Tanh, ReLU, Leaky ReLU, Maxout. The list is actually bigger than that. Which one to choose? Andrej Karpathy says the following:

TLDR: “What neuron type should I use?” Use the ReLU non-linearity, be careful with your learning rates and possibly monitor the fraction of “dead” units in a network. If this concerns you, give Leaky ReLU or Maxout a try. Never use sigmoid. Try tanh, but expect it to work worse than ReLU/Maxout.

So we should use ReLU as an activation function, it is short for rectified linear unit and is actually quite simple. If $x$ is greater than 0 we take it, else we take 0.

$$\sigma(x) = \max(0, x)$$
Nonetheless we have very little neurons and one or two dying neuron will undermine the results of our network (when the value of the neuron becomes zero). Instead we will use the Tanh as an activation function.

$$ \sigma(x) = tanh(x) = {e^x-e^{-x} \over e^x+e^{-x}}$$

So the first neuron of our first hidden layer will become:

$$ \sigma(-0.17) = tanh(-0.17) = {e^{-0.17}-e^{0.17} \over e^{-0.17}+e^{0.17}} = -0.168$$

The second one:

$$ \sigma(0.253) = tanh(0.253) = {e^{0.253}-e^{-0.253} \over e^{0.253}+e^{-0.253}} = 0,248$$

So our first hidden layer with only one car looks like:

Thanks to the matrix calculus property, each neuron of the first hidden layer will receive a weighted sum of the inputs. We have:

$$Z^{(2)} = \begin{bmatrix}
1.4 & -1 & 0.4 \\

\end{bmatrix} . \begin{bmatrix}
0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13

\end{bmatrix}$$

$$Z^{(2)} = \begin{bmatrix} -0.17 & 0.253 & 0.04 \\ \end{bmatrix}$$

$Z^{(2)}_1$ the first neuron of the layer two, for which we found $-0.17$ is the result of $1.4\times0.01 + -1\times0.2 + 0.4\times0.04$ (cf Matrix Multiplication). If you compare this to the neural network drawing, you see that in fact the first neuron of the layer two is the input 1 (number of kms) times the weight on the synapse plus the input 2 (type of fuel) times the weight on the synapse plus the input 3 (age) times the weight on the synapse. It is exactly what matrix calculus does for us.

For our example we assumed that only one car was in our dataset, that's why $X = \begin{bmatrix}

1.4 & -1 & 0.4 \\

\end{bmatrix}$. In reality we will have several cars, let's say five. So our inputs are:

$$X = \begin{bmatrix}
1.4 & -1 & 0.4 \\

0.4 & -1 & 0.1 \\

5.4 & -1 & 4 \\

1.5 & -1 & 1 \\

1.8 & 1 & 1 \\

\end{bmatrix}$$

And our $Z^{(2)}$ calculation will be:

$$Z^{(2)} = \begin{bmatrix}
1.4 & -1 & 0.4 \\

0.4 & -1 & 0.1 \\

5.4 & -1 & 4 \\

1.5 & -1 & 1 \\

1.8 & 1 & 1 \\

\end{bmatrix} . \begin{bmatrix}
0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13

\end{bmatrix}$$

$$Z^{(2)} = \begin{bmatrix}
-0.17 & 0.253 & 0.04 \\
-0.192 & 0.035 & -0.069 \\
0.014 & 2.469 & 0.788 \\

-0.145 & 0.594 & 0.125 \\
0.258 & 0.691 & 0.366

\end{bmatrix}$$

With the tanh applied element wise:

$$a^{(2)} = \sigma(Z^{(2)}) = tanh(Z^{(2)}) = \begin{bmatrix}
tanh(-0.17) & tanh(0.253) & tanh(0.04) \\

tanh(-0.192) & tanh(0.035) & tanh(-0.069) \\

tanh(0.014) & tanh(2.469) & tanh(0.788) \\

tanh(-0.145) & tanh(0.594) & tanh(0.125) \\

tanh(0.258) & tanh(0.691) & tanh(0.366)

\end{bmatrix}$$

$$a^{(2)} = \begin{bmatrix}
-0.16838105 & 0.24773663 & 0.03997868 \\
-0.18967498 & 0.03498572 & -0.06889071 \\
0.01399909 & 0.98576421 & 0.65727455 \\

-0.14399227 & 0.53276635 & 0.124353 \\
0.25242392 & 0.59862403 & 0.35048801

\end{bmatrix}$$

$Z^{(2)}$ and $a^{(2)}$ are of size $5\times3$, one row for each car and one column for each hidden unit (neuron) of the layer 2.

To summarise, for now we have three matrices, our input $X$, our weights $W^1$ between the layer 1 and 2 and our first hidden layer $a^{(2)}=tanh(X.W^1)$ ($Z^{(2)}$ is an intermediary matrix that holds the value of $X.W^1$).

We will repeat the exact same steps but instead of using our matrix $X$ as inputs, we will now use $a^{(2)}$. We add synapses from the layer 2 to the layer 3.

We also view the weights as a matrix.

$$W^2 = \begin{bmatrix}
W_{11}^2 & W_{12}^2 \\

W_{21}^2 & W_{22}^2 \\

W_{31}^2 & W_{32}^2

\end{bmatrix}$$

The number of rows equals the number of neurons in the layer 2 and the number of columns equals the number of neurons in the layer 3. We will compute the values of our second hidden layer into the matrix $Z^{(3)}$ as we did previously. For a given layer, the input is always the output of the previous layer. For the layer 2, the output of the previous layer are the data from the layer 1, meaning $X$, for the layer 3 the output of the previous layer are the data from the layer 2, meaning $a^{(2)}$.

$$Z^{(3)} = a^{(2)}.W^2$$

And then we apply the activation function, we are keeping $tanh(x)$ as activation function because it is unusual to use different activation functions for each layer.

$$a^{(3)} = tanh(Z^{(3)})$$

Previously we found that:

$$a^{(2)} = \begin{bmatrix}
-0.16838105 & 0.24773663 & 0.03997868 \\
-0.18967498 & 0.03498572 & -0.06889071 \\
0.01399909 & 0.98576421 & 0.65727455 \\

-0.14399227 & 0.53276635 & 0.124353 \\
0.25242392 & 0.59862403 & 0.35048801

\end{bmatrix}$$

Our $W^2$ matrix is randomly initialized the first time so we can use random values for the weights.

$$W^2 = \begin{bmatrix}
0.04 & 0.78 \\

0.40 & 0.45 \\

0.65 & 0.23

\end{bmatrix}$$

We calculate $Z^{(3)}$:

$$Z^{(3)} = a^{(2)}.W^2$$

$$Z^{(3)} = \begin{bmatrix}
-0.16838105 & 0.24773663 & 0.03997868 \\
-0.18967498 & 0.03498572 & -0.06889071 \\
0.01399909 & 0.98576421 & 0.65727455 \\

-0.14399227 & 0.53276635 & 0.124353 \\
0.25242392 & 0.59862403 & 0.35048801

\end{bmatrix} . \begin{bmatrix}
0.04 & 0.78 \\

0.40 & 0.45 \\

0.65 & 0.23

\end{bmatrix}$$

$$Z^{(3)} = \begin{bmatrix}
0.11834555 & -0.01066064 \\

-0.03837167 & -0.14804778 \\
0.8220941 & 0.60568633 \\

0.2881763 & 0.15603208 \\

0.47736378 & 0.54688371 \\

\end{bmatrix}$$

We then apply our activation function:

$$a^{(3)} = tanh(Z^{(3)})$$

$$a^{(3)} = \begin{bmatrix}
tanh(0.11834555) & tanh(-0.01066064) \\

tanh(-0.03837167) & tanh(-0.14804778) \\

tanh(0.8220941) & tanh(0.60568633) \\

tanh(0.2881763) & tanh(0.15603208) \\

tanh(0.47736378) & tanh(0.54688371) \\

\end{bmatrix}$$

$$a^{(3)} = \begin{bmatrix}
0.11779613 & -0.01066023 \\

-0.03835285 & -0.14697553 \\
0.67620804 & 0.54108347 \\

0.28045542 & 0.15477804 \\

0.44412987 & 0.49818098 \\

\end{bmatrix}$$

$Z^{(3)}$ and $a^{(3)}$ are of size $5\times2$, one row for each car and one column for each hidden unit (neuron) of the layer 3.

Our network looks like:

We will now connect the last layer, using the same technique as previously seen.

As for the two previous layers, we have:

$$Z^{(4)} = a^{(3)}.W^3$$ $$a^{(4)} = tanh(Z^{(4)})$$

Where:

$$W^3 = \begin{bmatrix}
W^3_{11} \\

W^3_{21}

\end{bmatrix}$$

The number of rows equals the number of neurons in the layer 3 and the number of columns equals the number of neurons in the layer 4.

Our $W^3$ matrix is randomly initialized the first time so we can use random values for the weights. For instance:

$$W^3 = \begin{bmatrix}
0.04 \\

0.41

\end{bmatrix}$$

We then calculate $Z^{(4)}$ using $a^{(3)}$ which is the result of the previous layer.

$$Z^{(4)} = a^{(3)}.W^3$$
$$Z^{(4)} = \begin{bmatrix}
0.11779613 & -0.01066023 \\

-0.03835285 & -0.14697553 \\
0.67620804 & 0.54108347 \\

0.28045542 & 0.15477804 \\

0.44412987 & 0.49818098 \\

\end{bmatrix} . \begin{bmatrix}
0.04 \\

0.41

\end{bmatrix}$$
$$Z^{(4)} = \begin{bmatrix}
0.00034115 \\

-0.06179408 \\
0.24889254 \\

0.07467721 \\

0.22201939 \\

\end{bmatrix}$$

We then apply the activation function:

$$a^{(4)} = tanh(Z^{(4)}) = \begin{bmatrix}
tanh(0.00034115) \\

tanh(-0.06179408) \\

tanh(0.24889254) \\

tanh(0.07467721) \\

tanh(0.22201939) \\

\end{bmatrix}$$
$$a^{(4)} = \begin{bmatrix}
0.000341156 \\

-0.0617156 \\
0.243877 \\
0.0745387 \\
0.218442 \\
\end{bmatrix}$$

Finally our network looks like the following:

What we did is called the forward propagation, we have an input and we propagate it through the network. Each time we have an hidden layer, we compute the values of its neurons using:

$$a^{(l)} = \sigma(a^{(l - 1)} \times W^{(l - 1)})$$

If $l = 1$, $a^{(1)} = X$ (our inputs), $\sigma(x)$ in our case is $tanh(x)$. Each time you see the above formula, you should think "neural network layer".

We didn't used a bias unit to keep things simple, we will add it now. Why do we need a bias? On the above picture we saw that $Z^{(2)}_1 = X_1 \times W^1_{11} + X_2 \times W^1_{21} + X_3 \times W^1_{31}$ where $X_1$, $X_2$ and $X_3$ are the attributes of our car. This calculation will produce a number an then our activation function $tanh(Z^{(2)}_1)$ will be applied. We have a 3D input ($X_1$, $X_2$, $X_3$), to explain the bias usefulness, we will keep only one feature, for instance the number of kilometers and assume that only the number of kilometers of a car describes its price. We now have $a^{(2)}_1 = X_1 \times W^1_{11}$ where $X_1$ is the car's number of kilometers. And once the activation function is applied, we have $a^{(2)}_1 = tanh(Z^{(2)}_1) = tanh(X_1 \times W^1_{11})$.

After the training part described later, our $W^1_{11}$ will be a learned value. For instance 1.8. Our $X_1$ is unknown and will be different for each car, $a^{(2)}_1$ will be the result of $tanh(1.8 \times X_1)$. We can draw this function:

As you can see this function is centered in zero, as our $W^1_{11}$ is a learned value, our network will be able to tweak it. We assumed that our network learned $1.8$ but let's draw the graph when $W^1_{11}$ is larger, let's say $7$.

And when $W^1_{11} = 0.4$:

As we can see, changing $W^1_{11}$ changes the stiffness of the graph, as our network can only tweak the weights, it will only be able to change the stiffness but will stay centered in zero. Is this a problem ?

Firstly the number of kilometers of a car will always be positive, so all the left part of the graph will be useless, $a^{(2)}_1$ will never be negative, is that a good thing ? We don't really know, but even if it was a good thing, we would have no choice.

Secondly, let's say that we have two cars, one with 30k kilometers and one with 170k kilometers, once normalized, we will have 0.5 and 2.5. Let's say that the impact of the number of kilometers is clearly determined, like if the car is less than 50k kilometers the price is high and if the car is more than 50k kilometers the price is low, the limit is clear. Once normalized 50k will be around 0.7, so we need a steep graph that produce -1 when x < 0.7 and produce 1 when x > 0.7. Like this one:

That would be perfect but unfortunately we can't move the graph to the right, so we are stuck with a graph centered in zero (the first image). We can only make the graph steep using a high value for $W^1_{11}$. The above picture would be a good solution for us, it is the graph of the function:

$$tanh(10x - 8)$$
Where 10 is the value for our $W^1_{11}$ and 8 is a constant that comes handy, the bias. Instead of $a^{(2)}_1 = tanh(X_1 \times W^1_{11})$ we will have $a^{(2)}_1 = tanh(X_1 \times W^1_{11} + b)$ and this little $b$ will greatly improve our network performances because it will move the graph of our activation function to the left or the right and the result produced will be more representative of our problem. In our example (the above picture) $b = -8$.

We used only one feature to easily draw graphs and display the impact of the bias but it is the same thing with more dimensions, our original example with a bias would be $a^{(2)}_1 = tanh(X_1 \times W^1_{11} + X_2 \times W^1_{21} + X_3 \times W^1_{31} + b)$.

This value $b$ also needs to be learned, because regarding the problem, the bias will be different. The same bias must be added to each car. We saw previously the origin of the weights and their matrix representation, how can we add a bias?

We will have a bias for each neuron. For instance for the first neuron of the first hidden layer we have:

$a^{(2)}_1 = tanh(X_1 \times W^1_{11} + X_2 \times W^1_{21} + X_3 \times W^1_{31} + b)$

As we have three neurons in the first hidden layer, we will need three biases, we could use a bias matrix.

Nonetheless managing the biases and the weights in separate matrices can be cumbersome. As the biases are learned, there are not different from the weights nonetheless they should not depend of a car attributes because all the car will have different attributes whereas the biases should be the same for all the cars. The solution is to add an additional feature to each car with a value $1$. So that this feature when multiplied with the bias will not change its value.

Note that $X_4$ is equal to $1$ so the calculations of the two previous pictures are exactly the same but using the bias trick we only have one weights matrix.

Adding a bias means adding a $1$ feature to all our inputs. We add biases to our input layer and each one of our hidden layers. Now our network looks like.

The link between the bias and the neurons are dotted to keep the network readable but they are normal weights. Our cars have a new feature with the value $1$. Each hidden layer has also a new neuron $1$ because the problem that the bias solve appears each time we use the activation function (so each layer). There is no link between the bias and the previous layer because the bias is added after the calculations, it is not the result of a matrix multiplication.

We will do again all the calculus with the biases. We began by adding the bias unit to our input data, this means adding a new column of $1$.

$$X = \begin{bmatrix}
1.4 & -1 & 0.4 & 1\\

0.4 & -1 & 0.1 & 1 \\

5.4 & -1 & 4 & 1 \\

1.5 & -1 & 1 & 1 \\

1.8 & 1 & 1 & 1 \\

\end{bmatrix}$$

And our $W_1$ matrix has a new row $

\begin{bmatrix}
W_{41}^1 & W_{42}^1 & W_{43}^1

\end{bmatrix}
$, because adding $1$ to our input created new links. This row is the biases, we init them at $0.1$ so that we will see the differences with the previous values (they should be initialised at $0$ or around $0$).

$$
W^1 = \begin{bmatrix}

0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13 \\

0.1 & 0.1 & 0.1

\end{bmatrix}
$$

Then we are doing the same calculations as before.

$$Z^{(2)} = \begin{bmatrix}
1.4 & -1 & 0.4 & 1\\

0.4 & -1 & 0.1 & 1 \\

5.4 & -1 & 4 & 1 \\

1.5 & -1 & 1 & 1 \\

1.8 & 1 & 1 & 1

\end{bmatrix} . \begin{bmatrix}
0.01 & 0.05 & 0.07 \\

0.20 & 0.041 & 0.11 \\

0.04 & 0.56 & 0.13 \\

0.1 & 0.1 & 0.1

\end{bmatrix}$$

$$Z^{(2)} = \begin{bmatrix}
-0.07 & 0.353 & 0.14 \\
-0.092 & 0.135 & 0.031\\
0.114 & 2.569 & 0.888\\

-0.045 & 0.694 & 0.225\\
0.358 & 0.791 & 0.466\\

\end{bmatrix}$$

If you compare these $Z^{(2)}$ results to the previous ones where we were not using biases, you see that each $Z^{(2)}$ has $+0.1$. As before we then apply the activation function.

$$a^{(2)} = tanh(Z^{(2)})$$ $$a^{(2)} = \begin{bmatrix} -0.06988589 & 0.33903341 & 0.13909245\\ -0.09174131 & 0.13418581 & 0.03099007\\ 0.11350871 & 0.98832966 & 0.71040449\\ -0.04496965 & 0.60054553 & 0.22127847\\ 0.34345116 & 0.65897516 & 0.43496173\\ \end{bmatrix}$$

We also add a bias to our first hidden layer. As we did with the inputs we must add $1$ as a fourth neuron to $a^{(2)}$. As a reminder, each row contains data for one car.

$$a^{(2)} = \begin{bmatrix} -0.06988589 & 0.33903341 & 0.13909245 & 1\\ -0.09174131 & 0.13418581 & 0.03099007 & 1\\ 0.11350871 & 0.98832966 & 0.71040449 & 1\\ -0.04496965 & 0.60054553 & 0.22127847 & 1\\ 0.34345116 & 0.65897516 & 0.43496173 & 1\\ \end{bmatrix}$$

As we added a $1$ neuron to $a^{(2)}$ new links are created between the layer 2 and 3, as for $W^1$, $W^2$ will gain a row of biases that we init at $0.1$.

$$W^2 = \begin{bmatrix}
0.04 & 0.78 \\

0.40 & 0.45 \\

0.65 & 0.23 \\

0.1 & 0.1

\end{bmatrix}$$

We do the same calculations as before:

$$Z^{(3)} = a^{(2)}.W^2$$

$$Z^{(3)} = \begin{bmatrix}
-0.06988589 & 0.33903341 & 0.13909245 & 1\\
-0.09174131 & 0.13418581 & 0.03099007 & 1\\
0.11350871 & 0.98832966 & 0.71040449 & 1\\
-0.04496965 & 0.60054553 & 0.22127847 & 1\\
0.34345116 & 0.65897516 & 0.43496173 & 1\\
\end{bmatrix} . \begin{bmatrix}
0.04 & 0.78 \\

0.40 & 0.45 \\

0.65 & 0.23 \\

0.1 & 0.1

\end{bmatrix}$$

$$Z^{(3)} = \begin{bmatrix}
0.32322802 & 0.2300453 \\

0.17014822 & 0.09595311\\

0.96163513 & 0.79667817\\

0.48225043 & 0.38606321\\

0.66005324 & 0.76447193\\

\end{bmatrix}$$

$$a^{(3)} = tanh(Z^{(3)})$$

$$a^{(3)} = \begin{bmatrix}
0.31242279 & 0.22607134 \\

0.16852506 & 0.09565971 \\

0.74500533 & 0.66217559 \\

0.44804409 & 0.3679614 \\

0.57839884 & 0.64370347 \\

\end{bmatrix}$$

Finally we also add a bias neuron to $a^{(3)}$, as before new links are created between the layer 3 and 4, as for $W^2$, $W^3$ will gain a row of biases that we init at $0.1$.

$$Z^{(4)} = a^{(3)}.W^3$$
$$Z^{(4)} = \begin{bmatrix}
0.31242279 & 0.22607134 & 1\\

0.16852506 & 0.09565971 & 1\\

0.74500533 & 0.66217559 & 1\\

0.44804409 & 0.3679614 & 1\\

0.57839884 & 0.64370347 & 1\\

\end{bmatrix} . \begin{bmatrix}
0.04 \\

0.41 \\

0.1

\end{bmatrix}$$
$$Z^{(4)} = \begin{bmatrix}
0.20518616 \\

0.14596148 \\

0.4012922 \\

0.26878594 \\

0.38705438 \\

\end{bmatrix}$$

Then we apply the activation function.

$$a^{(4)} = tanh(Z^{(4)}) = \begin{bmatrix}
0.2023543 \\

0.14493368 \\

0.38105408 \\

0.26249479 \\

0.36881806 \\

\end{bmatrix}$$

The values obtained on the output layer are our network predictions. We will save them into the matrix $\hat{y} = a^{(4)}$. It is the price of the cars predicted by our network whereas $y$ is the real price of the cars (given by the dataset). For instance our network thinks that the car number one is less expensive than the car number five.

Adding the biases is really simple (just add a neuron $1$) and will greatly improve our results. The whole calculation that we did, from $X$ to $a^{(4)}$ is called forward propagation. This is how we will infer the price of a car in the future. We will take the car attributes, make the same calculations and get a price in the $a^{(4)}$ matrix.

Right now our results are pretty bad because we randomly initialised our weights. We will train our network step by step in order to tweak the weights until it outputs good predictions.

In our original dataset we have the car attributes and the car price. The features (attributes) of the first car with the bias was $X = \begin{bmatrix}

1.4 & -1 & 0.4 & 1\end{bmatrix}$ and its price was $y = 0.45$. Our network output was $\hat{y} = 0.2023543$ given in $a^{(4)}_1$. We clearly see that we are missing around 0.25 by doing $y - \hat{y}$. This is called the cost, how bad our network predicted the price compared to the actual price. Actually we will define a cost function in order to mesure how bad our predictions are.

$$ J(W) = \frac{1}{2}(y-\hat{y})^2$$

This gives us the cost for one example. For instance for our first car, its real price is 0.45, our network outputted 0.2023543, using the above formula we have an error of $ J(W) = \frac{1}{2}(0.45-0.2023543)^2 = 0,031$

This gives us the error for the first car, we will do it for all the cars then sum up the errors, so our formula become:

$$ J(W) = \sum_{1}^{n}\frac{1}{2}(y-\hat{y})^2$$

Where $n$ is the number of cars. We squared the error to get its absolute value but also because the quadratic function $x^2$ is convex, it means that the function has only one minimum. We introduced $\frac{1}{2}$ for later convenience.

The function $J(W)$ gives us the error of our network regarding our inputs $X$ and the weights of our network. If we replace $\hat{y}$ by its calculations, our function is:

$$ J(W) = \sum_{1}^{n}\frac{1}{2}(y- \tanh(\tanh(\tanh(X.W_1).W_2).W_3))^2$$

$J(W)$ is a function that gives us the cost regarding our examples (the cars) and the weights ($W_1$, $W_2$ and $W_3$). The minimum the cost is, the better our network predicts. Our goal is to minimize the function $J(W)$, i.e: find its minimum. This is an optimization problem. We can't touch our examples $X$ so we will minimize our function $J(W)$ by tweaking the weights. We will use the batch gradient descent algorithm (with a non convex cost function it is better to use the stochastic gradient descent). We choose the gradient descent as the optimization algorithm but other alternatives could be used. Let's see what a gradient is.

In mathematics, a function is a rule explaining how to process an input to give an output. The input is noted as \(x\) and the output as \(y\), the function is generally written as \(y = f(x)\). It is possible to have multiple inputs and outputs, multiple inputs is common and looks like \(z = f(x, y)\), multiple outputs is a vector valued function that produces a vector instead of only \(y\).

Strictly speaking the inputs are called independent variables and the outputs dependent variables. The function explains how the dependent variables depend on the independent variables.

The derivative of a function is a key tool in machine learning, it is leveraged among others by the gradient descent algorithm. The derivative measures how a change in the independent variables impact the dependent variables (how changing \(x\) impacts \(y\)). The process of finding the derivative is called differentiation.

There are two cases useful to us.

The first case is when there is only one independent variable and one dependent variable (\(y = f(x)\) where \(x \in \mathbb{R}\) and \(y \in \mathbb{R}\)). In that case the derivative of a function at a given input is the slope of the tangent line to the graph of the function at the given input. It means that at the given input, the function is well-approximated by a straight line, and the derivative is the slope of this straight line. Let's take an example. For instance if we take the sigmoïd function:

$$\sigma(x) = \frac{1}{1+e^{-x}}$$
The derivative of this function is:

$$\dfrac{\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}$$
You can compute the derivative detail here. Let's use this derivative to calculate the slope of the tangent at \(x=2\), we have:

$$\dfrac{\mathrm{e}^2}{\left(\mathrm{e}^2+1\right)^2} \approx{0,105}$$
If we draw the sigmoid function with its tangent at \(x=2\) we see that the slope of the tangent is \(\approx{0,105}\):

Our function does not looks like the sigmoid function because we used the same scale for the $x$ and $y$, it is easier to see that the slope of the tangent, the blue line, is indeed $\approx{0,105}$.

The tangent is the best linear approximation of a function at a given value, it shows how the independent variable impacts the dependent variable (small or big slope).

The second case is when there are several independent variables (\(y = f(x)\) where \(x \in \mathbb{R}^n\) and \(y \in \mathbb{R}\)). As the function takes as input several variables, we will now compute the partial derivative of the function with respect to each input variable.

The partial derivative of a function w.r.t (with respect to) one of the input variable is the derivative of the function where others variables held constant. It indicates the rate of change of a function with respect to that variable surrounding an infinitesimally small region near a particular point.

If we take for example, \(z = x^2 + y^3\), we have two inputs \(x\) and \(y\) so the derivative of the function will be a vector containing two partial derivatives:

$$ F = \Bigg[\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\Bigg]$$
To calculate the partial derivative \(\frac{\partial f}{\partial x}\) we held \(y^3\) as a constant so we have:

$$\frac{\partial f}{\partial x} = 2x + 0 = 2x$$
To calculate the partial derivative \(\frac{\partial f}{\partial y}\) we held \(x^2\) as a constant so we have:

$$\frac{\partial f}{\partial y} = 3y^2 + 0 = 3y^2$$
So the derivative of our initial function is:

$$ F = [2x, 3y^2]$$
This vector of partial derivatives is the gradient. It represents the slope of the tangent of the graph of the function, it means that the gradient points in the direction of the greatest rate of increase of the function and its magnitude is the slope of the graph in that direction.

For instance, for the previous function, if we are at \(x=4\) and \(y=2\), the corresponding gradient is \((8, 12)\) (we are using \((2x, 3y^2)\)). It means that the function increases more in the direction of \(y\) than \(x\). Using the two informations, it gives us a scalar (a direction) for which Z will increase the most.

If we draw the previous function in a 3D space, it looks like the following:

If we forget the \(z\) (because we can't tune its value) we have a 2D surface on which we can move on the \(x\) and \(y\) axis, it is like watching the previous 3D drawing from above, the \(z\) axis does not appear.

We can place our original point (\(x=4\) and \(y=2\)) in red, its derivative \((8, 12)\) in blue and draw an arrow between them (orange).

The orange arrow is the gradient, it gives us the direction of the steepest increase of \(z\) and the length of the arrow gives us how \(z\) changes, a long arrow means a big slope. If it is always unclear, here is a 5 minutes video that can help you understand.

Gradient descent is an optimization algorithm, it allows to find a local minimum of a function.

The algorithm comes from the observation that if one goes in the direction of the negative gradient of a function, the function decreases faster.

As we saw previously, the gradient gives us the direction of the maximum increase of the function, we use the negation of this gradient to update the coordinates of our position in the space.

By doing it repeatedly we are sure to find a local minimum of the function.

Let's say that our function is \(z = x^2 + y^3\), its gradient is \(F = [2x, 3y^2]\) so if we are at \(x=4\) and \(y=2\), the corresponding gradient is \((8, 12)\).

The basic idea is to substract this gradient from our original coordinates, but directly substracting the gradient would produce a big jump from our original coordinates. For our example we are at $(4,2)$, if we directly substract the gradient, we will be at $(4 - 8, 2 - 12) = (-4, -10)$. What a big jump. So before to substract it we multiply it by a coefficient so that its value will be reduced. We have $(4 - 0.01 * 8, 2 - 0.01 * 12) = (3.92, 1.88)$. We did a small move but at least we will not miss the minima.

This coefficient $0.01$ is called the learning rate, how much the gradient will impact our current position. A big learning rate means bigger steps but we could jump over the minimum of the function. A small learning rate means a little but precise step and more steps needed to find the minimum. This value must be small and the perfect value is determined by a try/fail process.

So if we summarize, we are at a point in our space and we want to find the minimum of a function. If we find the minimum we will have found the weights that give us the lowest error. So for the given coordinates, we compute the gradient, we multiply the gradient by a tiny coefficient and then we substract this updated gradient from our coordinates. This gives us our new coordinates and we start again.

This repeated process is the gradient descent algorithm. If the function that we are trying to minimize is convex, all local minimas are also global minimas so gradient descent descent will give us the global solution (but it can take an eternity). A convex cost function is not mandatory.

If we complete the image showing several iterations from wikipedia, we have:

The red cross is our minima, the red arrow is the negative gradient that we substract from our coordinates and the orange arrows are the actual gradients. As you can see the red arrow is smaller than the orange arrow because the gradient has been multiplied by the learning rate (so reduced). The green dots are the new coordinates produced after each iteration. If $X_{0}$ is our original point, we see four iterations on the picture.

Our goal is therefore to find the gradients of our function $J(W)$ and use them to update the weights of our network. Our cost function computes three inputs that are the network weights. We have to find the partial derivatives (gradients) with regards to its weights.

$$\nabla (J(W)) = \Bigg[\frac{\partial J(W)}{\partial W_1}, \frac{\partial J(W)}{\partial W_2}, \frac{\partial J(W)}{\partial W_3}\Bigg]$$

Then we will update the weights using the gradients, for instance $W_1$ will be updated using the following rule:

$$W_1 = W_1 - \alpha \frac{1}{n} \frac{\partial J(W)}{\partial W_1}$$

Where $\alpha$ is our learning rate, we divide by $n$, which is the number of inputs (our cars) because the gradients for each car will be summed and we are using the averaged gradient to update our weights $W_1$.

We could directly apply the backpropagation algorithm to find the gradients of $J(W)$ but we will compute the derivative of our cost function to get a better understanding of the algorithm.

As a reminder:

$$J(W) = \sum_{1}^{n}\frac{1}{2}(y - \tanh(\tanh(\tanh(X.W_1).W_2).W_3))^2$$

Our cost function is a composition of several functions, you can see a $tanh$ into a $tanh$ into a $tanh$. To derive it we will use the chain rule. It is a formula for computing the derivative of the composition of two or more functions.

The usual formula is:

$$(f \circ g)' = (f' \circ g).g'$$

For instance if we take the function $f(x) = (2x^2 + 8)^3$ we see a composition. The result of the first function $g(x) = 2x^2 + 8$ is used by the second function $f(g(x)) = (g(x))^3$. The derivative of $g(x)$ is $g'(x) = 4x$ and the derivative of $f(g(x))$ is $f'(g(x)) = 3g(x)^2$. We apply the above formula:

$$f'(x) = f'(g(x)).g'(x) = 3(2x^2 + 8)^2 \cdot 4x$$

The chain rule can also be written in the following way:

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x}$$

Meaning that if $y$ depends on $x$ and $z$ depends on $y$, $z$ also depends on $x$. If we use our previous example. We want $\frac{\partial f}{\partial x}$. We know that $f$ depends on $g$ and $g$ depends on $x$ because $f(g(x)) = g(x)^3$ and $g(x) = 2x^2 + 8$ so we can write:

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial x}$$

Then we compute the derivative:

$$\frac{\partial f}{\partial g} = 3g(x)^2$$

$$\frac{\partial g}{\partial x} = 4x$$

$$\frac{\partial f}{\partial x} = 3(2x^2 + 8)^2 \cdot 4x$$

We will use the same way to compute the gradients of our cost function. There is a sum in our cost function, meaning that we have to add all together the cost of each input (our cars) in order to obtain the overall cost. We will forget it for now and talk about it later, the sum rule allows us to ignore it for now. The derivative of the sums equals the sum of the derivatives. At the end, once we have the derivative for one example we will just sum up the derivatives of all examples. So we have:

$$J(W) = \frac{1}{2}(y-\hat{y})^2$$

And we have to find:

$$\nabla (J(W)) = \Bigg[\frac{\partial J(W)}{\partial W_1}, \frac{\partial J(W)}{\partial W_2}, \frac{\partial J(W)}{\partial W_3}\Bigg]$$

We begin from the output of our network, so let's find $\frac{\partial J(W)}{\partial W_3}$ first. We want the derivative of $J(W)$ with regards to $W_3$, it means that other variables, meaning $W_1$ and $W_2$ are held constants.

We have:

$$\frac{\partial J(W)}{\partial W_3} = \frac{1}{2}(y-\hat{y})^2$$

We can see a first composition. $J(W) = \frac{1}{2}(g(x))^2$ where $g(x) = y-\hat{y}$ so we have:

$$\frac{\partial J(W)}{\partial W_3} = \frac{\partial J(W)}{\partial g} \cdot \frac{\partial g}{\partial W_3}$$

Using the power rule, we have:

$$\frac{\partial J(W)}{\partial g} = 2 \times \frac{1}{2}(g(x)) = g(x)$$

Thats why we put $\frac{1}{2}$ in our cost function, so that when differentiating things go smoothly and the term $2 \times \frac{1}{2}$ disappear. So:

$$\frac{\partial J(W)}{\partial W_3} = g(x) \cdot \frac{\partial g}{\partial W_3} = (y-\hat{y}) \cdot \frac{\partial g}{\partial W_3}$$

Now let's find the $\frac{\partial g}{\partial W_3}$ term, as a reminder $g(x) = y-\hat{y}$, we know that $y$ is a constant that does not depend on $W_3$ whereas $\hat{y}$ does depend on it. We have:

$$\frac{\partial g}{\partial W_3} = 0 - \frac{\partial \hat{y}}{\partial W_3}$$

And:

$$\frac{\partial J(W)}{\partial W_3} = (y-\hat{y}) \cdot -\frac{\partial \hat{y}}{\partial W_3}$$

As we said before, $\hat{y}$ is our predictions, meaning the output of our network, meaning $a^{(4)}$. We know that $\hat{y} = a^{(4)} = tanh(Z^{(4)})$ so our $\hat{y}$ depends on $Z^{(4)}$ and $Z^{(4)}$ depends on $W_3$. So we can write:

$$\frac{\partial J(W)}{\partial W_3} = (y-\hat{y}) \cdot -\frac{\partial \hat{y}}{\partial Z^{(4)}} \cdot \frac{\partial Z^{(4)}}{\partial W_3}$$

We can find $\frac{\partial \hat{y}}{\partial Z^{(4)}}$ directly, we have: $\hat{y} = tanh(Z^{(4)})$ so our derivative with regards to $Z^{(4)}$ is the following:

$$\frac{\partial \hat{y}}{\partial Z^{(4)}} = tanh'(Z^{(4)}) = 1-\tanh(Z^{(4)})^2$$

We can replace its value in our initial formula:

$$\frac{\partial J(W)}{\partial W_3} = (y-\hat{y}) \cdot - (1-\tanh(Z^{(4)})^2) \cdot \frac{\partial Z^{(4)}}{\partial W_3}$$

Now we have one final term to compute $\frac{\partial Z^{(4)}}{\partial W_3}$ and we know that $Z^{(4)} = a^{(3)} \cdot W^3$. Finally $Z^{(4)}$ depends on $W_3$ directly so no more chain rule needed for this first gradient. We keep $a^{(3)}$ as a constant and $W^3$ becomes one because we are differentiating with regard to $W^3$. We have:

$$\frac{\partial Z^{(4)}}{\partial W_3} = a^{(3)} \times 1 = a^{(3)}$$

We can replace it in our initial formula:

$$\frac{\partial J(W)}{\partial W_3} = (y-\hat{y}) \cdot - (1-\tanh(Z^{(4)})^2) \cdot a^{(3)}$$

We found our first gradient! We will use it during the training process to update our weights $W_3$.

We will introduce $\delta^{(4)}$ equals to:

$$\delta^{(4)} = (y-\hat{y}) \cdot - (1-\tanh(Z^{(4)})^2)$$

So our previous gradient is in fact:

$$\frac{\partial J(W)}{\partial W_3} = \delta^{(4)} \cdot a^{(3)}$$

Now we need our second gradient, $\frac{\partial J(W)}{\partial W_2}$, we will use the same steps as before for the beginning. We begin from:

$$J(W) = \frac{1}{2}(y-\hat{y})^2$$

Using the exact same steps as for $W_3$ we will arrive at:

$$\frac{\partial J(W)}{\partial W_2} = (y-\hat{y}) \cdot - (1-\tanh(Z^{(4)})^2) \cdot \frac{\partial Z^{(4)}}{\partial W_2}$$

You can see above that we have the same term as before, that's why we introduced $\delta^{(4)}$, we can replace it in our formula:

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(4)} \cdot \frac{\partial Z^{(4)}}{\partial W_2}$$

Before we were searching the derivative of $Z^{(4)}$ with regards to $W_3$ so the derivative was $a^{(3)} \times 1$, as a reminder $Z^{(4)} = a^{(3)} \cdot W_3$ but this time we are searching the derivative with regards to $W_2$. $W_3$ does not depend on $W_2$ so it becomes a constant, meanwhile $a^{(3)}$ depends on $W_2$ so we have to find its derivative with regards to $W_2$. This gives us:

$$\frac{\partial Z^{(4)}}{\partial W_2} = W_3 \cdot \frac{\partial a^{(3)}}{\partial W_2}$$

We can replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(4)} \cdot W_3 \cdot \frac{\partial a^{(3)}}{\partial W_2}$$

Now we have to compute $\frac{\partial a^{(3)}}{\partial W_2}$, we know that $a^{(3)}$ depends on $z^{(3)}$ (because $a^{(3)} = tanh(z^{(3)})$) which itself depends on $W_2$ (because $z^{(3)} = a^{(2)} \cdot W_2$). Using the chain rule we can write:

$$\frac{\partial a^{(3)}}{\partial W_2} = \frac{\partial a^{(3)}}{\partial z^{(3)}} \cdot \frac{\partial z^{(3)}}{\partial W_2}$$

We can replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(4)} \cdot W_3 \cdot \frac{\partial a^{(3)}}{\partial z^{(3)}} \cdot \frac{\partial z^{(3)}}{\partial W_2}$$

We differentiate $\frac{\partial a^{(3)}}{\partial z^{(3)}}$, we have:

$$\frac{\partial a^{(3)}}{\partial z^{(3)}} = tanh'(z^{(3)}) = 1-\tanh(Z^{(3)})^2$$

We can replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(4)} \cdot W_3 \cdot 1-\tanh(Z^{(3)})^2 \cdot \frac{\partial z^{(3)}}{\partial W_2}$$

And we differentiate the last missing term $\frac{\partial z^{(3)}}{\partial W_2}$:

$$\frac{\partial z^{(3)}}{\partial W_2} = a^{(2)} \cdot 1 = a^{(2)}$$

We can replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(4)} \cdot W_3 \cdot 1-\tanh(Z^{(3)})^2 \cdot a^{(2)}$$

We found the second gradient of our function $J(W)$. As you can see, the more you go toward the beginning of the network, the more the differentiation will be long. That's why we introduced the $\delta^{(l)}$ terms where $l$ is the layer number. So that we don't have to differentiate again the first part of the function but directly use $\delta^{(l)}$. For the second gradient we introduce:

$$\delta^{(3)} = \delta^{(4)} \cdot W_3 \cdot 1-\tanh(Z^{(3)})^2$$

We now have to find our last gradient $\frac{\partial J(W)}{\partial W_1}$, we will use the same steps as before for the beginning. We begin from:

$$J(W) = \frac{1}{2}(y-\hat{y})^2$$

Using the exact same steps as for $W_2$ we will arrive at:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(4)} \cdot W_3 \cdot 1-\tanh(Z^{(3)})^2 \cdot \frac{\partial z^{(3)}}{\partial W_1}$$

As we introduced $\delta^{(3)}$ we can use it:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(3)} \cdot \frac{\partial z^{(3)}}{\partial W_1}$$

Before we were searching the derivative of $z^{(3)}$ with regards to $W_2$ and so the derivative was equal to $a^{(2)}$. As a reminder $z^{(3)} = a^{(2)} \cdot W_2 $. This time we are searching the derivative of $z^{(3)}$ with regards to $W_1$ and so $W_2$ is only a constant, we have:

$$\frac{\partial z^{(3)}}{\partial W_1} = W_2 \cdot \frac{\partial a^{(2)}}{\partial W_1}$$

We can replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(3)} \cdot W_2 \cdot \frac{\partial a^{(2)}}{\partial W_1}$$

We now have to find the derivative of $\frac{\partial a^{(2)}}{\partial W_1}$, we know that $a^{(2)}$ depends on $z^{(2)}$ which itself depends on $W_1$. As before we can use the chain rule here. We have:

$$\frac{\partial a^{(2)}}{\partial W_1} = \frac{\partial a^{(2)}}{\partial z^{(2)}} \cdot \frac{\partial z^{(2)}}{\partial W_1}$$

Where:

$$\frac{\partial a^{(2)}}{\partial z^{(2)}} = tanh'(z^{(2)}) = 1-\tanh(Z^{(2)})^2$$

We replace it in our original formula:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(3)} \cdot W_2 \cdot 1-\tanh(Z^{(2)})^2 \cdot \frac{\partial z^{(2)}}{\partial W_1}$$

We have one last term to differentiate, if you remember $z^{(2)} = X \cdot W_1$ as we differentiate with regards to $W_1$ we have:

$$\frac{\partial z^{(2)}}{\partial W_1} = X \cdot W_1 = X$$

So our last gradient is:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(3)} \cdot W_2 \cdot 1-\tanh(Z^{(2)})^2 \cdot X$$

We also introduce the term $\delta^{(2)}$, we have:

$$\delta^{(2)} = \delta^{(3)} \cdot W_2 \cdot 1-\tanh(Z^{(2)})^2$$

And:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(2)} \cdot X$$

Here we are, we found the gradient of $J(W)$ with regards to its weights. As a reminder we found:

$$\frac{\partial J(W)}{\partial W_1} = \delta^{(2)} \cdot X$$

$$\delta^{(2)} = \delta^{(3)} \cdot W_2 \cdot 1-\tanh(Z^{(2)})^2$$

$$\frac{\partial J(W)}{\partial W_2} = \delta^{(3)} \cdot a^{(2)}$$

$$\delta^{(3)} = \delta^{(4)} \cdot W_3 \cdot 1-\tanh(Z^{(3)})^2$$

$$\frac{\partial J(W)}{\partial W_3} = \delta^{(4)} \cdot a^{(3)}$$

$$\delta^{(4)} = (y-\hat{y}) \cdot - (1-\tanh(Z^{(4)})^2)$$

If you remember, each gradient will be used to update a weight matrix during one gradient descent iteration. This means that our gradients should have the same size as the weights matrix that will use it. For instance $W_3$ is a $2 \times 1$ matrix so $\frac{\partial J(W)}{\partial W_3}$ must be a $2 \times 1$ matrix.

We will detail the dimensions of the matrices used to calculate our gradients so that it appears clearly that the gradients are summed up. It will also be useful to know when were are using element wise or matrix multiplication. Let's say that we have five cars, for $\frac{\partial J(W)}{\partial W_3}$ we found:

The $\odot$ means an element wise multiplication, the $\cdot$ means a matrix multiplication. From now on we will distinguish between the two. Our $\delta^{(4)}$ term has a dimension of $5 \times 1$ and $a^{(3)}$ has a dimension of $5 \times 2$. As we want the same dimension as $W_3$ meaning $2 \times 1$ there is only one way to achieve that:

By inverting $\delta^{(4)}$ and $a^{(3)}$ and using the transpose of $a^{(3)}$ we are able to get the desired result. $a^{(3)}$ contains the neurons values of the the third layer, two for each car and $\delta^{(4)}$ contains the error for each car. By doing the matrix multiplication we are actually summing the neurons of all the cars where each car neuron value is multiplied by the error for the given car. If you remember we removed the sum from our cost function, this step takes care of the summation of the errors using the matrix multiplication.

So we modify our third gradient so that the summation is handled and the dimension of the matrices are compatible (the little $\intercal$ means transpose):

$$\frac{\partial J(W)}{\partial W_3} = a^{(3)\intercal} \cdot \delta^{(4)}$$

$$\delta^{(4)} = (y-\hat{y}) \odot - (1-\tanh(Z^{(4)})^2)$$

Our $\delta^{(l)}$ will always have the same size as $a^{(l)}$ because it mesures how much a neuron is responsible for any error in our output. This remark is important because during the forward propagation we added bias units to each of our layer, having the effect of increasing the $a^{(l)}$ dimension by one column. $\delta^{(l)}$ will also contain the error of the bias units. These bias units are not linked to the previous layers so when we backpropagate our $\delta^{(l)}$ into our $(l - 1)$ layer we will have to remove the bias error. We will detail the dimensions of the matrices, we also have five cars. Using the same reasoning as before (transpose + inversion) we find that:

$$\frac{\partial J(W)}{\partial W_2} = a^{(2)\intercal} \cdot \delta^{(3)}$$

$$\delta^{(3)} = (\delta^{(4)} \cdot W_3^{\intercal}) \odot 1-\tanh(Z^{(3)})^2$$

As you can see the order of the calculations follows the output toward input flow. We have:

The third element of $W_3$ is the bias value, as we also multiply it with $\delta^{(4)}$ the result of our product $\delta^{(4)} \cdot W_3^{\intercal}$ is of size $5 \times 3$ whereas $Z^{(3)}$ is of size $5 \times 2$. To allow a smooth element wise multiplication we add a column of $1$ to $Z^{(3)}$. The result is unchanged and $\delta^{(3)}$ is of size $5 \times 3$ like $a^{(3)}$.

Nonetheless the bias is not linked to the previous layer, this means that when backpropagating the error $\delta^{(3)}$ we must remove the part about the bias in the error matrix. We have:

The greyed circles are removed from the $\delta^{(3)}$ before the multiplication, $\frac{\partial J(W)}{\partial W_2}$ has the same size as $W_2$.

We use the same tricks for $\frac{\partial J(W)}{\partial W_1}$.

$$\frac{\partial J(W)}{\partial W_1} = X^{\intercal} \cdot \delta^{(2)}$$

$$\delta^{(2)} = (\delta^{(3)} \cdot W_2^{\intercal}) \odot 1-\tanh(Z^{(2)})^2$$

We have:

And:

$\frac{\partial J(W)}{\partial W_1}$ has the same size as $W_1$.

This may seem difficult but using the dimension analysis and knowing when to remove the bias error and add a $1$ column, there is only one way to obtain the same dimension as the weight matrix.

As you may have noticed the partial derivatives follow the same pattern, we did the differentiation manually to see the foundations but we can apply the backpropagation algorithm, for the last layer ($l$ is the index of the last layer) we have:

$$ \delta^{(l)} = -(y - a^{(l)})\odot\sigma'(z^{(l)})$$

For non output layer ($l$ is not the index of the last layer):

$$ \delta^{(l)} = ((\delta^{(l + 1)} \cdot W_{(l)}^{\intercal})\odot\sigma'(z^{(l)}))$$

Then we compute the gradients:

$$\frac{\partial J(W)}{\partial W_{(l)}} = a^{(l)\intercal} \cdot \delta^{(l + 1)}$$

You can stack a bunch of layers and use the backpropagation formula to easily compute the gradients. Let's use our previous examples to compute our gradients and do a gradient descent iteration.

$$\delta^{(4)} = (y-\hat{y}) \odot - (1-\tanh(Z^{(4)})^2)$$

$$\delta^{(4)} = \begin{bmatrix} 0.45 \\ 0.8 \\ 0.2 \\ 0.5 \\ 0.55 \\ \end{bmatrix}-\begin{bmatrix} 0.202354302599 \\ 0.144933682554 \\ 0.381054078721 \\ 0.262494787219 \\ 0.368818057375 \\ \end{bmatrix} \odot - \begin{bmatrix} 0.95905273622 \\ 0.978994227661 \\ 0.85479778909 \\ 0.931096486683 \\ 0.863973240554 \\ \end{bmatrix}$$

We took care of applying $1-\tanh(x)^2$ element wise to $Z^{(4)}$ values.

$$\delta^{(4)} = \begin{bmatrix} 0.247645697401 \\ 0.655066317446 \\ -0.181054078721 \\ 0.237505212781 \\ 0.181181942625 \\ \end{bmatrix} \odot - \begin{bmatrix} 0.95905273622 \\ 0.978994227661 \\ 0.85479778909 \\ 0.931096486683 \\ 0.863973240554 \\ \end{bmatrix}$$

$$\delta^{(4)} = \begin{bmatrix} -0.237505283705 \\ -0.641306143515 \\ 0.154764626197 \\ -0.221140269189 \\ -0.156536350099 \\ \end{bmatrix}$$

Our gradient for W3 is:

$$\frac{\partial J(W)}{\partial W_3} = a^{(3)\intercal} \cdot \delta^{(4)}$$

$$\frac{\partial J(W)}{\partial W_3} = \begin{bmatrix} 0.312422790277 & 0.16852505915 & 0.745005333554 & 0.448044091981 & 0.578398840625 \\ 0.226071339877 & 0.0956597075832 & 0.662175586399 & 0.367961400586 & 0.643703471035 \\ 1.0 & 1.0 & 1.0 & 1.0 & 1.0 \\ \end{bmatrix} \cdot \begin{bmatrix} -0.237505283705 \\ -0.641306143515 \\ 0.154764626197 \\ -0.221140269189 \\ -0.156536350099 \\ \end{bmatrix}$$

$$\frac{\partial J(W)}{\partial W_3} = \begin{bmatrix} -0.25659878177 \\ -0.194693013848 \\ -1.10172342031 \\ \end{bmatrix}$$

For $W_2$ we have:

$$\delta^{(3)} = (\delta^{(4)} \cdot W_3^{\intercal}) \odot 1-\tanh(Z^{(3)})^2$$

$$\delta^{(3)} = \begin{bmatrix} -0.237505283705 \\ -0.641306143515 \\ 0.154764626197 \\ -0.221140269189 \\ -0.156536350099 \\ \end{bmatrix} \cdot \begin{bmatrix} 0.04 & 0.41 & 0.1 \\ \end{bmatrix} \odot \begin{bmatrix} 0.902392000116 & 0.948891749286 & 0.419974341614 \\ 0.971599304439 & 0.990849220345 & 0.419974341614 \\ 0.444967052977 & 0.561523492777 & 0.419974341614 \\ 0.799256491641 & 0.864604407679 & 0.419974341614 \\ 0.665454781164 & 0.585645841377 & 0.419974341614 \\ \end{bmatrix}$$

As before we took care of applying $1-\tanh(x)^2$ element wise to $Z^{(3)}$ values. We also concatenated a column of 1 to $Z^{(3)}$ as described before. The 1s became 0.419974341614 when applying $1-\tanh(x)^2$.

$$\delta^{(3)} = \begin{bmatrix} -0.0095002113482 & -0.0973771663191 & -0.0237505283705 \\ -0.0256522457406 & -0.262935518841 & -0.0641306143515 \\ 0.00619058504786 & 0.0634534967406 & 0.0154764626197 \\ -0.00884561076757 & -0.0906675103676 & -0.0221140269189 \\ -0.00626145400397 & -0.0641799035407 & -0.0156536350099 \\ \end{bmatrix} \odot \begin{bmatrix} 0.902392000116 & 0.948891749286 & 0.419974341614 \\ 0.971599304439 & 0.990849220345 & 0.419974341614 \\ 0.444967052977 & 0.561523492777 & 0.419974341614 \\ 0.799256491641 & 0.864604407679 & 0.419974341614 \\ 0.665454781164 & 0.585645841377 & 0.419974341614 \\ \end{bmatrix}$$

$$\delta^{(3)} = \begin{bmatrix} -0.00857291472002 & -0.092400389689 & -0.00997461251539 \\ -0.0249237041189 & -0.260529453845 & -0.0269332125396 \\ 0.00275460638495 & 0.0356306291187 & 0.0064997171992 \\ -0.0070699118285 & -0.078391529097 & -0.00928732389571 \\ -0.00416671450398 & -0.0375866936086 & -0.00657412505716 \\ \end{bmatrix}$$

Our gradient for W2 is:

$$\frac{\partial J(W)}{\partial W_2} = a^{(2)\intercal} \cdot \delta^{(3)}$$

$$\frac{\partial J(W)}{\partial W_2} = \begin{bmatrix} -0.0698858903164 & -0.0917413131084 & 0.113508705786 & -0.0449696495836 & 0.34345116481 \\ 0.339033408721 & 0.134185809931 & 0.988329664432 & 0.600545525169 & 0.658975160566 \\ 0.139092447878 & 0.0309900734824 & 0.710404487737 & 0.221278467898 & 0.434961731831 \\ 1.0 & 1.0 & 1.0 & 1.0 & 1.0 \\ \end{bmatrix} \cdot \begin{bmatrix} -0.00857291472002 & -0.092400389689 \\ -0.0249237041189 & -0.260529453845 \\ 0.00275460638495 & 0.0356306291187 \\ -0.0070699118285 & -0.078391529097 \\ -0.00416671450398 & -0.0375866936086 \\ \end{bmatrix}$$

We took care of removing the last column of $\delta^{(3)}$ as the error on the bias is not backpropagated.

$$\frac{\partial J(W)}{\partial W_2} = \begin{bmatrix} 0.0020851994346 & 0.0250192301883 \\ -0.010520017991 & -0.102917746604 \\ -0.00338471099243 & -0.0293089952796 \\ -0.0419786387864 & -0.433277437121 \\ \end{bmatrix}$$

Finally for $W_1$ we have:

$$\delta^{(2)} = (\delta^{(3)} \cdot W_2^{\intercal}) \odot 1-\tanh(Z^{(2)})^2$$

$$\delta^{(2)} = \begin{bmatrix} -0.00857291472002 & -0.092400389689 \\ -0.0249237041189 & -0.260529453845 \\ 0.00275460638495 & 0.0356306291187 \\ -0.0070699118285 & -0.078391529097 \\ -0.00416671450398 & -0.0375866936086 \\ \end{bmatrix} \cdot \begin{bmatrix} 0.04 & 0.4 & 0.65 & 0.1 \\ 0.78 & 0.45 & 0.23 & 0.1 \\ \end{bmatrix} \odot \begin{bmatrix} 0.995115962335 & 0.885056347771 & 0.980653290943 & 0.419974341614 \\ 0.991583531469 & 0.981994168413 & 0.999039615346 & 0.419974341614 \\ 0.987115773711 & 0.0232044744039 & 0.495325463803 & 0.419974341614 \\ 0.997977730616 & 0.6393450722 & 0.951035839645 & 0.419974341614 \\ 0.88204129739 & 0.565751737757 & 0.810808291842 & 0.419974341614 \\ \end{bmatrix}$$

We took care of removing the last column of $\delta^{(3)}$, we also concatenated a column of 1 to $Z^{(2)}$ and applied $1-\tanh(x)^2$ element wise to $Z^{(2)}$ values.

$$\delta^{(2)} = \begin{bmatrix} -0.0724152205462 & -0.0450093412481 & -0.0268244841965 & -0.0100973304409 \\ -0.204209922164 & -0.127207735878 & -0.0761221820616 & -0.0285453157964 \\ 0.0279020749679 & 0.0171356256574 & 0.00998553884751 & 0.00383852355036 \\ -0.0614281891688 & -0.0381041528251 & -0.0226254943808 & -0.00854614409256 \\ -0.0294842895948 & -0.0185806979255 & -0.0113533039576 & -0.00417534081126 \\ \end{bmatrix} \odot \begin{bmatrix} 0.995115962335 & 0.885056347771 & 0.980653290943 & 0.419974341614 \\ 0.991583531469 & 0.981994168413 & 0.999039615346 & 0.419974341614 \\ 0.987115773711 & 0.0232044744039 & 0.495325463803 & 0.419974341614 \\ 0.997977730616 & 0.6393450722 & 0.951035839645 & 0.419974341614 \\ 0.88204129739 & 0.565751737757 & 0.810808291842 & 0.419974341614 \\ \end{bmatrix}$$

$$\delta^{(2)} = \begin{bmatrix} -0.0720615418815 & -0.0398358031806 & -0.0263055187051 & -0.00424061970398 \\ -0.20249119578 & -0.124917254809 & -0.0760490754861 & -0.0119883002078 \\ 0.0275425783201 & 0.000397623186961 & 0.00494609166096 & 0.00161208140083 \\ -0.0613039648226 & -0.0243617023391 & -0.0215176560459 & -0.00358916123861 \\ -0.0260063610469 & -0.0105120621401 & -0.00920535298859 & -0.00175353600822 \\ \end{bmatrix}$$

Our gradient for $W_1$ is:

$$\frac{\partial J(W)}{\partial W_1} = X^{\intercal} \cdot \delta^{(2)}$$

$$\frac{\partial J(W)}{\partial W_1} = \begin{bmatrix} 1.4 & 0.4 & 5.4 & 1.5 & 1.8 \\ -1.0 & -1.0 & -1.0 & -1.0 & 1.0 \\ 0.4 & 0.1 & 4.0 & 1.0 & 1.0 \\ 1.0 & 1.0 & 1.0 & 1.0 & 1.0 \\ \end{bmatrix} \cdot \begin{bmatrix} -0.0720615418815 & -0.0398358031806 & -0.0263055187051 \\ -0.20249119578 & -0.124917254809 & -0.0760490754861 \\ 0.0275425783201 & 0.000397623186961 & 0.00494609166096 \\ -0.0613039648226 & -0.0243617023391 & -0.0215176560459 \\ -0.0260063610469 & -0.0105120621401 & -0.00920535298859 \\ \end{bmatrix}$$

We took care of removing the last column of $\delta^{(2)}$ as the error on the bias is not backpropagated.

$$\frac{\partial J(W)}{\partial W_1} = \begin{bmatrix} -0.171920111136 & -0.159054126528 & -0.0893845808607 \\ 0.282307763117 & 0.178205075002 & 0.109720805588 \\ -0.0262137489196 & -0.0617093184844 & -0.0290657574213 \\ -0.334320485211 & -0.199229199282 & -0.128131511565 \\ \end{bmatrix}$$

Now that we have our gradients, we can apply one iteration of the gradient descent algorithm to update our weights. We saw previously the update formula:

$$W_{(l)} = W_{(l)} - \alpha \frac{1}{n} \frac{\partial J(W)}{\partial W_{(l)}}$$

We have five cars in our dataset so $n = 5$, we choose a learning rate of $\alpha = 0.1$.

$$W_1 = W_1 - \alpha \frac{1}{n} \frac{\partial J(W)}{\partial W_1}$$

$$W_1 = \begin{bmatrix} 0.01 & 0.05 & 0.07 \\ 0.2 & 0.041 & 0.11 \\ 0.04 & 0.56 & 0.13 \\ 0.1 & 0.1 & 0.1 \\ \end{bmatrix} - 0.1 \odot \frac{1}{5} \odot \begin{bmatrix} -0.171920111136 & -0.159054126528 & -0.0893845808607 \\ 0.282307763117 & 0.178205075002 & 0.109720805588 \\ -0.0262137489196 & -0.0617093184844 & -0.0290657574213 \\ -0.334320485211 & -0.199229199282 & -0.128131511565 \\ \end{bmatrix}$$

$$W_1 = \begin{bmatrix} 0.0134384022227 & 0.0531810825306 & 0.0717876916172 \\ 0.194353844738 & 0.0374358985 & 0.107805583888 \\ 0.0405242749784 & 0.56123418637 & 0.130581315148 \\ 0.106686409704 & 0.103984583986 & 0.102562630231 \\ \end{bmatrix}$$

As you can see our weights matrix has been updated, some values are bigger, others smaller, we are moving in our space of solutions toward the best set of weights.

We do the exact same thing for $W_2$ and $W_3$:

$$W_2 = W_2 - \alpha \frac{1}{n} \frac{\partial J(W)}{\partial W_2}$$

$$W_2 = \begin{bmatrix} 0.04 & 0.78 \\ 0.4 & 0.45 \\ 0.65 & 0.23 \\ 0.1 & 0.1 \\ \end{bmatrix} - 0.1 \odot \frac{1}{5} \odot \begin{bmatrix} 0.0020851994346 & 0.0250192301883 \\ -0.010520017991 & -0.102917746604 \\ -0.00338471099243 & -0.0293089952796 \\ -0.0419786387864 & -0.433277437121 \\ \end{bmatrix}$$

$$W_2 = \begin{bmatrix} 0.0399582960113 & 0.779499615396 \\ 0.40021040036 & 0.452058354932 \\ 0.65006769422 & 0.230586179906 \\ 0.100839572776 & 0.108665548742 \\ \end{bmatrix}$$

$$W_3 = W_3 - \alpha \frac{1}{n} \frac{\partial J(W)}{\partial W_3}$$

$$W_3 = \begin{bmatrix} 0.04 \\ 0.41 \\ 0.1 \\ \end{bmatrix} - 0.1 \odot \frac{1}{5} \odot \begin{bmatrix} -0.25659878177 \\ -0.194693013848 \\ -1.10172342031 \\ \end{bmatrix}$$

$$W_3 = \begin{bmatrix} 0.0451319756354 \\ 0.413893860277 \\ 0.122034468406 \\ \end{bmatrix}$$

By doing the forward propagation, backward propagation and weights update in a loop you have an algorithm that learn.

To do the backpropagation we found the derivative of our cost function. It is easy to do a mistake during the differentiation and while working ostensibly fine our neural network will perform poorly. We used the chain rule to find our gradients, this is called analytical differentiation.

Nonetheless the original definition of a derivative is the following:

$$\frac{df(x)}{dx} = \lim_{h\ \to 0} \frac{f(x + h) - f(x)}{h}$$
The idea is that for a small enough \(h\) value the previous formula approximates correctly the value of the derivative of the function \(f\). Because if you zoom enough on a tiny part of the graph of the function, it will appear linear, and so approximate the tangent of the graph. This is called numerical differentiation. As you can see we are computing twice $f(x)$, that's why the numerical differentiation is slower and not used (even if it is less error prone).

The idea is to use the numerical differentiation to check that our analytical differentiation implementation is correct. We will not use the strict definition but rather the centered formula (works better):

$$\frac{df(x)}{dx} = \frac{f(x + h) - f(x - h)}{2h} \hspace{0.1in}$$

In total we have three matrices containing the weights. We will test one weight at a time. We will proceed as follow.

- We do a forward and backward propagation.
- We save our gradients into a vector $V_1$.
- For each weight:
- We compute f(weight + h).
- We compute f(weight - h).
- We compute the centered formula and save the value into a vector $V_2$.

- The difference between $V_1$ and $V_2$ should be less than $10^-8$.

The function that we differentiated during the backpropagation is our cost function:

$$ J(W) = \sum_{1}^{n} \frac{1}{2}(y-\hat{y})^2$$

Where $\hat{y}$ is in reality $a^{(4)}$.

During the backpropagation we found our gradients equal to:

$$\frac{\partial J(W)}{\partial W_1} = \begin{bmatrix} -0.171920111136 & -0.159054126528 & -0.0893845808607 \\ 0.282307763117 & 0.178205075002 & 0.109720805588 \\ -0.0262137489196 & -0.0617093184844 & -0.0290657574213 \\ -0.334320485211 & -0.199229199282 & -0.128131511565 \\ \end{bmatrix}$$

$$\frac{\partial J(W)}{\partial W_2} = \begin{bmatrix} 0.0020851994346 & 0.0250192301883 \\ -0.010520017991 & -0.102917746604 \\ -0.00338471099243 & -0.0293089952796 \\ -0.0419786387864 & -0.433277437121 \\ \end{bmatrix}$$

$$\frac{\partial J(W)}{\partial W_3} = \begin{bmatrix} -0.25659878177 \\ -0.194693013848 \\ -1.10172342031 \\ \end{bmatrix}$$

We save them into a vector $V^{(1)}$ of size $23 \times 1$. These gradients have been computed during the forward pass with the analytical differentiation.

As you will see later in the code I am using a perturbation vector, I will keep the explanation simple and don't talk about it.

The weights that we used are:

$$W_1 = \begin{bmatrix} 0.0134384022227 & 0.0531810825306 & 0.0717876916172 \\ 0.194353844738 & 0.0374358985 & 0.107805583888 \\ 0.0405242749784 & 0.56123418637 & 0.130581315148 \\ 0.106686409704 & 0.103984583986 & 0.102562630231 \\ \end{bmatrix}$$

$$W_2 = \begin{bmatrix} 0.0399582960113 & 0.779499615396 \\ 0.40021040036 & 0.452058354932 \\ 0.65006769422 & 0.230586179906 \\ 0.100839572776 & 0.108665548742 \\ \end{bmatrix}$$

$$W_3 = \begin{bmatrix} 0.0451319756354 \\ 0.413893860277 \\ 0.122034468406 \\ \end{bmatrix}$$

We disturb the first weight by adding a small value, we choose $h = 10^{-4}$.

$$W_1 = \begin{bmatrix} 0.0134384022227 + 0.0001 & 0.0531810825306 & 0.0717876916172 \\ 0.194353844738 & 0.0374358985 & 0.107805583888 \\ 0.0405242749784 & 0.56123418637 & 0.130581315148 \\ 0.106686409704 & 0.103984583986 & 0.102562630231 \\ \end{bmatrix}$$

All the other weights stay the same. We do a forward propagation using the weights disturbed (just one value has been disturbed). We obtain:

$$ a^{(4)} = \begin{bmatrix} 0.202395039754 \\ 0.144946047163 \\ 0.381136194829 \\ 0.262533502845 \\ 0.368843890153 \\ \end{bmatrix}$$

We compute the sum of the costs (square the cost before summing):

$$ J(W) = \frac{1}{2}\sum\left(\begin{bmatrix} 0.45 \\ 0.8 \\ 0.2 \\ 0.5 \\ 0.55 \\ \end{bmatrix}-\begin{bmatrix} 0.202395039754 \\ 0.144946047163 \\ 0.381136194829 \\ 0.262533502845 \\ 0.368843890153 \\ \end{bmatrix}\right)^2$$

$$ J(W) = loss2 = 0.30621105$$

We disturb again the first weight by removing the small value $h$.

$$W_1 = \begin{bmatrix} 0.0134384022227 - 0.0001 & 0.0531810825306 & 0.0717876916172 \\ 0.194353844738 & 0.0374358985 & 0.107805583888 \\ 0.0405242749784 & 0.56123418637 & 0.130581315148 \\ 0.106686409704 & 0.103984583986 & 0.102562630231 \\ \end{bmatrix}$$

All the other weights stay the same. We do a forward propagation using the weights disturbed (just one value has been disturbed). We obtain:

$$ a^{(4)} = \begin{bmatrix} 0.202313563549 \\ 0.144921317917 \\ 0.380971901463 \\ 0.262456067958 \\ 0.368792216736 \\ \end{bmatrix}$$

We also compute the sum of the costs:

$$ J(W) = \frac{1}{2}\sum\left(\begin{bmatrix} 0.45 \\ 0.8 \\ 0.2 \\ 0.5 \\ 0.55 \\ \end{bmatrix}-\begin{bmatrix} 0.202313563549 \\ 0.144921317917 \\ 0.380971901463 \\ 0.262456067958 \\ 0.368792216736 \\ \end{bmatrix}\right)^2$$

$$ J(W) = loss1 = 0.30624543$$

Then we compute the numerical gradient by doing:

$$V^{(2)}_1 = \frac{(loss2 - loss1)}{(2*h)}$$

$$V^{(2)}_1 = -0.17192014$$

This is the first weight computed using the numerical differentiation. Using the analytical differentiation we found $-0.171920111136$. As you can see they are almost identical.

We repeat the process for each weight until we have computed our 23 numerical gradients into the vector $V^{(2)}_1$. We then want to compare our analytical gradients with our numerical gradients. To quantify the difference we divide the norm of the difference by the norm of the sum of the vectors we would like to compare. Typical results should be on the order of $10^-8$ or less if we’ve computed our gradient correctly.

$$\frac{\left|V^{(1)} - V^{(2)}\right|}{\left|V^{(1)} + V^{(2)}\right|} = 1.15700817288e-08$$

As you can see the difference is on the order of $10^-8$ this means that our backward propagation computes correctly the gradients $\nabla (J(W))$. If it is not the case, it means that there is an error in the backpropagation algorithm and you have to debug it, good luck, as you will see, you sometimes lose a great amount of time because of a little mistake.

One last step is missing, the regularization. During the training our weights will evolve in order to predict correctly the price of a car from its attributes. Nonetheless it could happen that the weights are perfectly predicting the price for the training data but can't generalize for an unseen car. This problem is called overfitting. During the training steps our network will display a small loss but when testing it with unseen data it will display a larger loss. It means that our weights are well suited for the training data only.

Regularization is useful when the network has a lot of parameters and not enough data. Enough data is at least ten times the number of parameters. Our network has 23 parameters so we need at least 230 cars. As our dataset of cars has more than 9k cars, overfitting is not really a problem for us, nonetheless we will solve the overfitting problem as it regularly occurs.

One way to reduce overfitting is called regularization. The idea is to penalize large weights by modifying our cost function. There are several way to do regularization, the most used are L1, L2, max norm and dropout. We will use the L2 regularization because it is the most common. It is pretty simple, we only add one term to our cost function:

$$ J(W) = \sum_{1}^{n}\frac{1}{2}(y-\hat{y})^2 - \frac{1}{2} \lambda \sum_{1}^{3} W_{(n)}^2$$

The $\lambda$ is a hyper parameter that will change the impact of the regularization, $\lambda = 1$ means a strong regularization whereas $\lambda = 0.001$ means a light regularization. We sum all the weights squared and we add this sum times lambda to the cost function. If the weights are high, the regularization term will be high and the cost will increase, conversely if the weights are low the regularization term will be low.

Because of that, the weights will be smoothed over all the features. If a feature has a strong impact on the price, let's say the age, the weights corresponding to the age should be high. As we regularize it will not happen. As for the cost function, the regularization term contains $\frac{1}{2}$ to ease things during the differentiation. Previously we differentiated our cost function in order to find our gradients. We did not take into account the regularization term, but the differentiation is easy.

When we expand our regularization sum we have:

$$\frac{1}{2} \lambda \sum_{1}^{3} W_{(n)}^2 = \frac{1}{2} \lambda(W_{(1)}^2 + W_{(2)}^2 + W_{(3)}^2)$$

When differentiating our cost function w.r.t $W_{(3)}$, we found:

$$\frac{\partial J(W)}{\partial W_3} = a^{(3)\intercal} \cdot \delta^{(4)}$$

If we differentiate the regularization term w.r.t $W_{(3)}$ we have:

$$\frac{1}{2} \lambda(W_{(1)}^2 + W_{(2)}^2 + W_{(3)}^2)$$

$$\frac{1}{2} \lambda(0 + 0 + W_{(3)}^2)$$

$$2 \times \frac{1}{2} \lambda W_{(3)}$$

$$\lambda W_{(3)}$$

If we add the regularization, our gradient w.r.t $W_{(3)}$ is:

$$\frac{\partial J(W)}{\partial W_3} = a^{(3)\intercal} \cdot \delta^{(4)} - \lambda \cdot W_{(3)}$$

The process is the same for the gradient w.r.t each weight matrix. We remove from the gradient $\lambda \cdot W_{(n)}$. This means that during each gradient descent iteration, using the L2 regularization, each weight will linearly decay towards zero.

When building a model, here is the list of things you have to choose:

- The number of hidden layers
- The number of neurons for each hidden layer
- The activation function
- The cost function
- The optimization algorithm
- The learning rate
- The type of regularization
- The regularization rate
- The number of gradient descent steps
- The way of evaluating the accuracy of the network

We saw the theoritical part of a deep neural network, there are other types of neural network that will be the topic of other blog posts. To name a few: Convolutionnal Neural Network, Long Short Term Memory, Generative Adversarial Networks...

We are finally done with the theoritical part. We will now implement all the stuffs we described in python. We will make a first version using only numpy and a second one using TensorFlow. In a following blog post, we will build a version using CuDNN and TensorFlow C++.

All the following code is available on github.

The first step is to download the data from leboncoin.fr, I made a python script that loads each car page and save for each car its number of kilometers, type of fuel, age and price. The data are saved into a csv file. We have 8717 cars.

We have the raw input data, normalize_lbc_cars_data.py will normalize these raw data as we described previously. For the kilometers and the age, we substract the mean and divide by the standard deviation.

```
# normalize kilometers: (x - mean)/std
km = features[:, 0].astype("int")
mean_km = np.mean(km)
std_km = np.std(km)
km = (km - mean_km)/std_km
features[:, 0] = km
# normalize age: (x - mean)/std
age = features[:, 2].astype("int")
mean_age = np.mean(age)
std_age = np.std(age)
age = (age - mean_age)/std_age
features[:, 2] = age
```

For the type of fuel, we binary encode, diesel equals -1, essence equals 1:

```
# binary convert fuel: Diesel = -1, Essence = 1
features[:, 1] = [-1 if x == 'Diesel' else 1 for x in features[:,1]]
```

And finally for the price, we bring the data between [0, 1] because it will be the output of our network.

```
# normalize price: (x - min)/(max - min)
price = features[:, 3].astype("float")
min_price = np.min(price)
max_price = np.max(price)
features[:, 3] = (price - min_price)/(max_price - min_price)
```

Using the normalized data we create a new CSV file called normalized_car_features.csv. The first line contains mean_km, std_km, mean_age, std_age, min_price, max_price. We will use these data in the future when predicting. The user will input the raw car attributes, we will normalize them so that our network can use them and produce a result, than we will also process this result so that the user can see a price that looks like 13 456 euros and not 0.12789.

During the normalization process we skipped some lines that had a bad format or wrong data. Be careful if you are using the CSV file of raw data.

The first step is to implement our network using numpy. We implement each part of the neural network described previously. The corresponding code is dnn_from_scratch.py.

We load our normalized data.

```
reader = csv.reader(open("normalized_car_features.csv", "rb"), delimiter=",")
x = list(reader)
features = np.array(x[2:]).astype("float")
np.random.shuffle(features)
```

We shuffle the array because we will keep 80% for our data for the training set and 20% for the test set. The goal is to train our network using the training set and analyze the loss using the test set. If the loss is low with our test set, it means that our network generalizes well because it predicts correctly unseen data.

We split the car attributes and the car prices in two matrices. We also append 1 to each car for the bias unit.

```
data_x = np.concatenate((features[:, :3], np.ones((features.shape[0], 1))), axis=1)
data_y = features[:, 3:]
```

We save the dataset metadata for the prediction part of the network, it will help us to normalize the input that we will predict and convert the output of the network into a human readable price. All the metadata are on the first line of the dataset (mean_km, std_km, mean_age, std_age, min_price, max_price).

```
self.predict = util.Predict(float(x[0][0]), float(x[0][1]), float(x[0][2]), float(x[0][3]), float(x[0][4]), float(x[0][5]))
```

We calculate how many elements 80% of our network means and split the dataset into a train set and a test set.

```
self.m = float(features.shape[0])
self.m_train_set = int(self.m * 0.8)
self.x, self.x_test = data_x[:self.m_train_set, :], data_x[self.m_train_set:, :]
self.y, self.y_test = data_y[:self.m_train_set, :], data_y[self.m_train_set:, :]
```

We init all the matrices that we will use, also Lambda that is the regularization rate and the learning rate.

```
self.z2, self.a2, self.z3, self.a3, self.z4, self.a4 = (None,) * 6
self.delta2, self.delta3, self.delta4 = (None,) * 3
self.djdw1, self.djdw2, self.djdw3 = (None,) * 3
self.gradient, self.numericalGradient = (None,) * 2
self.Lambda = 0.01
self.learning_rate = 0.01
```

We init the weights with the values described in the blog post.

```
self.w1 = np.matrix([
[0.01, 0.05, 0.07],
[0.2, 0.041, 0.11],
[0.04, 0.56, 0.13],
[0.1, 0.1, 0.1]
])
self.w2 = np.matrix([
[0.04, 0.78],
[0.4, 0.45],
[0.65, 0.23],
[0.1, 0.1]
])
self.w3 = np.matrix([
[0.04],
[0.41],
[0.1]
])
```

Once the network is initialized, we can begin the forward propagation. We do the calculations explained in the blog post. We put them into a forward method because we will call them repeatedly during the gradient descent.

```
def forward(self):
# first layer
self.z2 = np.dot(self.x, self.w1)
self.a2 = np.tanh(self.z2)
# we add the the 1 unit (bias) at the output of the first layer
ba2 = np.ones((self.x.shape[0], 1))
self.a2 = np.concatenate((self.a2, ba2), axis=1)
# second layer
self.z3 = np.dot(self.a2, self.w2)
self.a3 = np.tanh(self.z3)
# we add the the 1 unit (bias) at the output of the second layer
ba3 = np.ones((self.a3.shape[0], 1))
self.a3 = np.concatenate((self.a3, ba3), axis=1)
# output layer, prediction of our network
self.z4 = np.dot(self.a3, self.w3)
self.a4 = np.tanh(self.z4)
```

Then we do the backward propagation, we are using the data computed during the forward propagation to compute the gradients. As the backward propagation gives us the sum of the gradients we divide them by the size of our train set.

```
def backward(self):
# gradient of the cost function with regards to W3
self.delta4 = np.multiply(-(self.y - self.a4), tanh_prime(self.z4))
self.djdw3 = (self.a3.T * self.delta4) / self.m_train_set + self.Lambda * self.w3
# gradient of the cost function with regards to W2
self.delta3 = np.multiply(self.delta4 * self.w3.T, tanh_prime(np.concatenate((self.z3, np.ones((self.z3.shape[0], 1))), axis=1)))
self.djdw2 = (self.a2.T * np.delete(self.delta3, 2, axis=1)) / self.m_train_set + self.Lambda * self.w2
# gradient of the cost function with regards to W1
self.delta2 = np.multiply(np.delete(self.delta3, 2, axis=1) * self.w2.T, tanh_prime(np.concatenate((self.z2, np.ones((self.z2.shape[0], 1))), axis=1)))
self.djdw1 = (self.x.T * np.delete(self.delta2, 3, axis=1)) / self.m_train_set + self.Lambda * self.w1
```

Where tanh_prime is the derivative of the tanh function.

```
def tanh_prime(x):
return 1.0 - np.square(np.tanh(x))
```

The backward propagation gives us the gradients of our cost function, we can use them to update our weights.

```
def update_gradient(self):
self.w1 -= self.learning_rate * self.djdw1
self.w2 -= self.learning_rate * self.djdw2
self.w3 -= self.learning_rate * self.djdw3
```

By doing these three steps several times, our network is learning. We choose 5000 steps.

```
nb_it = 5000
for step in xrange(nb_it):
nn.forward()
nn.backward()
nn.update_gradient()
if step % 100 == 0:
nn.summary(step)
```

The summary method gives us the scores regarding our network. I don't detail here the R2 score method.

```
def summary(self, step):
print("Iteration: %d, Loss %f" % (step, self.cost_function()))
print("RMSE: " + str(np.sqrt(np.mean(np.square(self.a4 - self.y)))))
print("MAE: " + str(np.sum(np.absolute(self.a4 - self.y)) / self.m_train_set))
print("R2: " + str(self.r2()))
```

To be sure that our backpropagation algorithm computes the gradients (partial derivatives) of our cost function correctly we use gradient checking. We compute the gradients using the forward and backward propagation, then we compute them using the numerical gradient and we compare both results.

```
def check_gradients(self):
self.compute_gradients()
self.compute_numerical_gradients()
print("Gradient checked: " + str(np.linalg.norm(self.gradient - self.numericalGradient) / np.linalg.norm(
self.gradient + self.numericalGradient)))
```

The compute_gradient method is pretty simple, we save our three gradient matrices produced by the forward and backward prop into one [1 x 23] vector.

```
def compute_gradients(self):
nn.forward()
nn.backward()
self.gradient = np.concatenate((self.djdw1.ravel(), self.djdw2.ravel(), self.djdw3.ravel()), axis=1).T
```

The compute_numerical_gradients method is more complicated. To add or remove the perturbation as we described earlier our weights are flattened into one vector of size [1 x 23]. We test each weight separately. We have a perturbation vector of the same size [1 x 23] containing zeros everywhere except at the index of the weight that we are testing.

If we are testing the 5th weight, our perturbation vector will have 1e-4 at the 5th position. We add the perturbation vector to the weights, then we reconstruct our three weights matrices do a forward pass and compute the cost. Then we remove the perturbation vector from the weights, reconstruct our three weights matrices, do a forward pass and compute the cost. Our numerical gradient for the 5th weight is given by: `(loss2 - loss1) / (2 * e)`

. We do that for each weight. We obtain a [1 x 23] vector containing each numerical gradient for each weight.

```
def compute_numerical_gradients(self):
weights = np.concatenate((self.w1.ravel(), self.w2.ravel(), self.w3.ravel()), axis=1).T
self.numericalGradient = np.zeros(weights.shape)
perturbation = np.zeros(weights.shape)
e = 1e-4
for p in range(len(weights)):
# Set perturbation vector
perturbation[p] = e
self.set_weights(weights + perturbation)
self.forward()
loss2 = self.cost_function()
self.set_weights(weights - perturbation)
self.forward()
loss1 = self.cost_function()
self.numericalGradient[p] = (loss2 - loss1) / (2 * e)
perturbation[p] = 0
self.set_weights(weights)
```

The cost function method used in the code above is the implementation of our cost function. As the cost function gives us the sum of the cost we divide it by the size of our train set:

$$ J(W) = \sum_{1}^{n} \frac{1}{2}(y-\hat{y})^2 - \frac{1}{2} \lambda \sum_{1}^{3} W_{(n)}^2$$

```
def cost_function(self):
return 0.5 * sum(np.square((self.y - self.a4))) / self.m_train_set + (self.Lambda / 2) * (
np.sum(np.square(self.w1)) +
np.sum(np.square(self.w2)) +
np.sum(np.square(self.w3))
)
```

The only difference compared to the formula is that the cost is averaged for all the training examples (because the gradients are averaged too, remember the $\frac{1}{n}$ when we update the weights using the gradients). The set_weights method is useful to convert the weights from one [1 x 23] vector to three matrices back and forth.

```
def set_weights(self, weights):
self.w1 = np.reshape(weights[0:12], (4, 3))
self.w2 = np.reshape(weights[12:20], (4, 2))
self.w3 = np.reshape(weights[20:23], (3, 1))
```

For the regularization there is no explicit part in the code, you can see that we add the regularization term to the cost function `+ (self.Lambda / 2) * (np.sum(np.square(self.w1)) + np.sum(np.square(self.w2)) + np.sum(np.square(self.w3)))`

, also when computing the gradients we add the weights times lambda (due to the differentiation process) `+ self.Lambda * self.w3`

for each weight matrix.

If you remember we splitted our data between train set and test set. We want the network's summary (the metrics) with the test set, after 5000 iterations we have:

```
### Testing summary ###
Iteration: 5000, Loss 0.004642
RMSE: 0.0459143540873
MAE: 0.00770927707759
R2: 0.662649615919
```

Our R2 score could be better, but overall, given the number of features for a car (only 3) and the small size of our network, the results are pretty good.

Now let's predict the price of a car. We introduced a helper class called Predict available into predict.py. Our predict object is constructed during the init part of the neural network, it only saves the metadata of our dataset, meaning: mean_km, std_km, mean_age, std_age, min_price, max_price. We then use the predict object to convert real life car attributes into a normalized version, so that our network can use them. Our predict object is also useful to convert the output of the network into a human readable price.

To use our network to predict a price, we only need to do a forward pass where X, the input, contains the car's data for which we want to predict the price. This is called inference, and it will use our previously learned weights.

I took a random ads where a 5 years old car with 168000 kilometers and a Diesel engine was sold 16 000 euros.

Once the training steps are done, to predict the price we run the following:

```
print("### Predict ###")
nn.predict_price(168000, "Diesel", 5)
```

Where predict_price is:

```
def predict_price(self, km, fuel, age):
self.x = np.concatenate((self.predict.input(km, fuel, age), np.ones((1, 1))), axis=1)
nn.forward()
print("Predicted price: " + str(self.predict.output(self.a4[0])))
```

`self.predict.input(km, fuel, age)`

will took the raw car attributes and give us the normalized version to which we append 1 for the bias.

We then do a forward pass and the predicted price is `Predicted price: [[ 13484.89728828]]`

16 000 euros seems expensive according to our network.

We also implemented the same version of our network using TensorFlow. The code is easier because TF do things for you. Namely the whole optimization part. The corresponding code is dnn_from_scratch_tensorflow.py.

The idea of TF is that you build a graph in python and then once your graph is ready, you run it. The computation begins when the session is run, before it is only a graph definition. We are using the same normalized data as previously. Reading the CSV is the same code as before.

The Github code contains `with`

statements that are useful to group nodes under a same group when displaying the graph in TensorBoard, the name passed as parameter for each TF operation is also used by TensorBoard to name nodes beautifully. I will omit them here in order to keep the code clear.

We declare our variables.

```
x = tf.placeholder("float", name="cars")
y = tf.placeholder("float", name="prices")
w1 = tf.Variable(tf.random_normal([3, 3]), name="W1")
w2 = tf.Variable(tf.random_normal([3, 2]), name="W2")
w3 = tf.Variable(tf.random_normal([2, 1]), name="W3")
b1 = tf.Variable(tf.random_normal([1, 3]), name="b1")
b2 = tf.Variable(tf.random_normal([1, 2]), name="b2")
b3 = tf.Variable(tf.random_normal([1, 1]), name="b3")
```

A placeholder will be completed when the session is run, the variables are initialized by randomly choosing data from a normal distribution.

Then we declare our execution graph. We separate the weights and the bias whereas in our previous examples we merged them. TensorFlow manages them separately. We define our three layers.

```
layer_1 = tf.nn.tanh(tf.add(tf.matmul(x, w1), b1))
layer_2 = tf.nn.tanh(tf.add(tf.matmul(layer_1, w2), b2))
layer_3 = tf.nn.tanh(tf.add(tf.matmul(layer_2, w3), b3))
```

We compute the regularization. The sum of the L2 loss of each weight matrix.

```
regularization = tf.nn.l2_loss(w1) + tf.nn.l2_loss(w2) + tf.nn.l2_loss(w3)
```

We define our cost function (to which we add the regularization).

```
loss = tf.reduce_mean(tf.square(layer_3 - y)) + Lambda * regularization
```

We then define which algorithm we will use to minimize our cost function. We choose gradient descent and specify that the loss operation will be minimized. The loss operation is our cost function.

```
train_op = tf.train.GradientDescentOptimizer(learning_rate).minimize(loss)
```

We defined the learning rate to 0.01. We can then train the network for 5000 steps.

```
# launching the previously defined model begins here
init = tf.global_variables_initializer()
with tf.Session() as session:
session.run(init)
for i in range(5000):
session.run(train_op, feed_dict={x: x_data, y: y_data})
```

Our `feed_dict`

contains the data by which the placeholders will be replaced. x_data contains the training data, meaning 80% of our data set, these data will be used when the placeholder x is used in our graph. y_data contains the prices of our cars.

The great thing about TensorFlow is that it knows which operations we stacked during our graph definition and how to differentiate each one of them. As we called `GradientDescentOptimizer(learning_rate).minimize(loss)`

TF knows that the backward propagation will start from the `loss`

formula back to the input and that it has to differentiate the whole calculation graph.

When coding the network, the back propagation part is the most error prone. We don't check our gradient here, we assume that TF is unit tested and that the analytical differentiation of our cost function is correct.

We use our testing data to validate the performances of our network. At the 5000th iteration our training set gives a loss of: 0.024727896 and our test set gives a loss of: 0.024568973 which is good.

If we try to predict the same car as before, TensorFlow gives us a price of `Predicted price: [[ 13471.90332031]]`

. We also do a forward pass using one input:

```
feed_dict = {x: predict.input(168000, "Diesel", 5)}
print("Predicted price: " + str(predict.output(session.run(layer_3, feed_dict))))
```

As with our numpy network, the price is less than 16 000 euros. Definitely overpriced. Each launch will give a different price because we are learning the weights from scratch each time and they are randomly initialized.

Thanks for reading this post, if you have questions or see mistakes, do not hesitate to comment.

]]>