The Backpropagation Algorithm

1.

Propagates inputs forward in the usual way, i.e.

• All outputs are computed using sigmoid thresholding of the inner product of the corresponding weight and input vectors.
• All outputs at stage n are connected to all the inputs at stage n+1

2.

Propagates the errors backwards by apportioning them to each unit according to the amount of this error the unit is responsible for.

We now derive the stochastic Backpropagation algorithm for the general case. The derivation is simple, but unfortunately the book-keeping is a little messy.

• $\vec{x_j} =$ input vector for unit j (x_ji = ith input to the jth unit)
• $\vec{w_j} =$ weight vector for unit j (w_ji = weight on x_ji)
• $z_j = \vec{w_j}\cdot \vec{x_j}$, the weighted sum of inputs for unit j
• o_j = output of unit j ( $o_j = \sigma(z_j)$)
• t_j = target for unit j
• Downstream(j) = set of units whose immediate inputs include the output of j
• Outputs = set of output units in the final layer

Since we update after each training example, we can simplify the notation somewhat by imagining that the training set consists of exactly one example and so the error can simply be denoted by E.

We want to calculate $\frac{\partial E}{\partial w_{ji}}$ for each input weight w_ji for each output unit j. Note first that since z_j is a function of w_ji regardless of where in the network unit j is located,

$$\begin{eqnarray*}\frac{\partial E}{\partial w_{ji}} &=& \frac{\partial E}{\parti... ...rtial w_{ji}} \\ &=& \frac{\partial E}{\partial z_j} x_{ji}\\ \end{eqnarray*}$$

Furthermore, $\frac{\partial E}{\partial z_j}$ is the same regardless of which input weight of unit j we are trying to update. So we denote this quantity by $\delta_j$.

Consider the case when $j \in Outputs$. We know

$$E = \frac{1}{2}\sum_{k \in Outputs} (t_k - \sigma(z_k))^2$$

Since the outputs of all units $k \ne j$ are independent of w_ji, we can drop the summation and consider just the contribution to E by j.

$$\begin{eqnarray*}\delta_j = \frac{\partial E}{\partial z_j} &=& \frac{\partial }... ...o_j)(1-\sigma(z_j))\sigma(z_j)\\ &=& -(t_j - o_j)(1-o_j)o_j\\ \end{eqnarray*}$$

Thus

$$\Delta w_{ji} = -\eta \frac{\partial E}{\partial w_ij} = \eta \delta_j x_{ji}$$ (17)

Now consider the case when j is a hidden unit. Like before, we make the following two important observations.

1.

For each unit k downstream from j, z_k is a function of z_j

2.

The contribution to error by all units $l \ne j$ in the same layer as j is independent of w_ji

We want to calculate $\frac{\partial E}{\partial w_{ji}}$ for each input weight w_ji for each hidden unit j. Note that w_ji influences just z_j which influences o_j which influences $z_k \forall k \in Downstream(j)$ each of which influence E. So we can write

$$\begin{eqnarray*}\frac{\partial E}{\partial w_{ji}} &=& \sum_{k \in Downstream(j... ...al o_j} \cdot \frac{\partial o_j}{\partial z_j} \cdot x_{ji}\\ \end{eqnarray*}$$

Again note that all the terms except x_ji in the above product are the same regardless of which input weight of unit j we are trying to update. Like before, we denote this common quantity by $\delta_j$. Also note that $\frac{\partial E}{\partial z_k} = \delta_k$, $\frac{\partial z_k}{\partial o_j} = w_{kj}$ and $\frac{\partial o_j}{\partial z_j} = o_j (1-o_j)$. Substituting,

$$\begin{eqnarray*}\delta_j &=& \sum_{k \in Downstream(j)} \frac{\partial E}{\par... ... &=& \sum_{k \in Downstream(j)} \delta_k w_{kj} o_j (1-o_j)\\ \end{eqnarray*}$$

Thus,

$$\delta_k = o_j (1-o_j) \sum_{k \in Downstream(j)} \delta_k w_{kj}$$ (18)

We are now in a position to state the Backpropagation algorithm formally.

Formal statement of the algorithm:

Stochastic Backpropagation(training examples, $\eta$, n_i, n_h, n_o)

Each training example is of the form $\langle \vec{x}, \vec{t} \rangle$ where $\vec{x}$ is the input vector and $\vec{t}$ is the target vector. $\eta$ is the learning rate (e.g., .05). n_i, n_h and n_o are the number of input, hidden and output nodes respectively. Input from unit i to unit j is denoted x_ji and its weight is denoted by w_ji.

• Create a feed-forward network with n_i inputs, n_h hidden units, and n_o output units.
• Initialize all the weights to small random values (e.g., between -.05 and .05)
• Until termination condition is met, Do
  • For each training example $\langle \vec{x}, \vec{t} \rangle$, Do

    1.

    Input the instance $\vec{x}$ and compute the output o_u of every unit.

    2.

    For each output unit k, calculate
    

    $$\delta_k = o_k(1-o_k)(t_k - o_k)$$

    
    

    3.

    For each hidden unit h, calculate
    

    $$\delta_h = o_h(1-o_h)\sum_{k \in Downstream(h)} w_{kh}\delta_k$$

    
    

    4.

    Update each network weight w_ji as follows:
    

    $$\begin{eqnarray*}w_{ji} &\leftarrow& w_{ji} + \Delta w_{ji}\\ {\rm where} \quad \Delta w_{ji} &=& \eta \delta_j x_ji\\ \end{eqnarray*}$$