# 8.7. Backpropagation Through Time¶ Open the notebook in Colab Open the notebook in Colab Open the notebook in Colab

So far we repeatedly alluded to things like exploding gradients, vanishing gradients, truncating backprop, and the need to detach the computational graph. For instance, in the previous section we invoked s.detach() on the sequence. None of this was really fully explained, in the interest of being able to build a model quickly and to see how it works. In this section we will delve a bit more deeply into the details of backpropagation for sequence models and why (and how) the math works. For a more detailed discussion about randomization and backpropagation also see the paper by [Tallec & Ollivier, 2017].

We encountered some of the effects of gradient explosion when we first implemented recurrent neural networks (Section 8.5). In particular, if you solved the problems in the problem set, you would have seen that gradient clipping is vital to ensure proper convergence. To provide a better understanding of this issue, this section will review how gradients are computed for sequence models. Note that there is nothing conceptually new in how it works. After all, we are still merely applying the chain rule to compute gradients. Nonetheless, it is worth while reviewing backpropagation (Section 4.7) again.

Forward propagation in a recurrent neural network is relatively straightforward. Backpropagation through time is actually a specific application of back propagation in recurrent neural networks. It requires us to expand the recurrent neural network one timestep at a time to obtain the dependencies between model variables and parameters. Then, based on the chain rule, we apply backpropagation to compute and store gradients. Since sequences can be rather long, the dependency can be rather lengthy. For instance, for a sequence of 1000 characters, the first symbol could potentially have significant influence on the symbol at position 1000. This is not really computationally feasible (it takes too long and requires too much memory) and it requires over 1000 matrix-vector products before we would arrive at that very elusive gradient. This is a process fraught with computational and statistical uncertainty. In the following we will elucidate what happens and how to address this in practice.

## 8.7.1. A Simplified Recurrent Network¶

We start with a simplified model of how an RNN works. This model ignores details about the specifics of the hidden state and how it is updated. These details are immaterial to the analysis and would only serve to clutter the notation, but make it look more intimidating. In this simplified model, we denote $$h_t$$ as the hidden state, $$x_t$$ as the input, and $$o_t$$ as the output at timestep $$t$$. In addition, $$w_h$$ and $$w_o$$ indicate the weights of hidden states and the output layer, respectively. As a result, the hidden states and outputs at each timesteps can be explained as

(8.7.1)$h_t = f(x_t, h_{t-1}, w_h) \text{ and } o_t = g(h_t, w_o).$

Hence, we have a chain of values $$\{\ldots, (h_{t-1}, x_{t-1}, o_{t-1}), (h_{t}, x_{t}, o_t), \ldots\}$$ that depend on each other via recursive computation. The forward pass is fairly straightforward. All we need is to loop through the $$(x_t, h_t, o_t)$$ triples one step at a time. The discrepancy between outputs $$o_t$$ and the desired targets $$y_t$$ is then evaluated by an objective function as

(8.7.2)$L(x, y, w_h, w_o) = \sum_{t=1}^T l(y_t, o_t).$

For backpropagation, matters are a bit more tricky, especially when we compute the gradients with regard to the parameters $$w_h$$ of the objective function $$L$$. To be specific, by the chain rule,

(8.7.3)\begin{split}\begin{aligned} \partial_{w_h} L & = \sum_{t=1}^T \partial_{w_h} l(y_t, o_t) \\ & = \sum_{t=1}^T \partial_{o_t} l(y_t, o_t) \partial_{h_t} g(h_t, w_h) \left[ \partial_{w_h} h_t\right]. \end{aligned}\end{split}

The first and the second part of the derivative is easy to compute. The third part $$\partial_{w_h} h_t$$ is where things get tricky, since we need to compute the effect of the parameters on $$h_t$$.

To derive the above gradient, assume that we have three sequences $$\{a_{t}\},\{b_{t}\},\{c_{t}\}$$ satisfying $$a_{0}=0, a_{1}=b_{1}$$, and $$a_{t}=b_{t}+c_{t}a_{t-1}$$ for $$t=1, 2,\ldots$$. Then for $$t\geq 1$$, it is easy to show

(8.7.4)$a_{t}=b_{t}+\sum_{i=1}^{t-1}\left(\prod_{j=i+1}^{t}c_{j}\right)b_{i}.$

Now let us apply (8.7.4) with

(8.7.5)$a_t = \partial_{w_h}h_{t},$
(8.7.6)$b_t = \partial_{w_h}f(x_{t},h_{t-1},w_h),$
(8.7.7)$c_t = \partial_{h_{t-1}}f(x_{t},h_{t-1},w_h).$

Therefore, $$a_{t}=b_{t}+c_{t}a_{t-1}$$ becomes the following recursion

(8.7.8)$\partial_{w_h}h_{t}=\partial_{w_h}f(x_{t},h_{t-1},w)+\partial_{h}f(x_{t},h_{t-1},w_h)\partial_{w_h}h_{t-1}.$

By (8.7.4), the third part will be

(8.7.9)$\partial_{w_h}h_{t}=\partial_{w_h}f(x_{t},h_{t-1},w_h)+\sum_{i=1}^{t-1}\left(\prod_{j=i+1}^{t}\partial_{h_{j-1}}f(x_{j},h_{j-1},w_h)\right)\partial_{w_h}f(x_{i},h_{i-1},w_h).$

While we can use the chain rule to compute $$\partial_w h_t$$ recursively, this chain can get very long whenever $$t$$ is large. Let us discuss a number of strategies for dealing with this problem.

• Compute the full sum. This is very slow and gradients can blow up, since subtle changes in the initial conditions can potentially affect the outcome a lot. That is, we could see things similar to the butterfly effect where minimal changes in the initial conditions lead to disproportionate changes in the outcome. This is actually quite undesirable in terms of the model that we want to estimate. After all, we are looking for robust estimators that generalize well. Hence this strategy is almost never used in practice.

• Truncate the sum after $$\tau$$ steps. This is what we have been discussing so far. This leads to an approximation of the true gradient, simply by terminating the sum above at $$\partial_w h_{t-\tau}$$. The approximation error is thus given by $$\partial_h f(x_t, h_{t-1}, w) \partial_w h_{t-1}$$ (multiplied by a product of gradients involving $$\partial_h f$$). In practice this works quite well. It is what is commonly referred to as truncated BPTT (backpropgation through time). One of the consequences of this is that the model focuses primarily on short-term influence rather than long-term consequences. This is actually desirable, since it biases the estimate towards simpler and more stable models.

• Randomized Truncation. Last we can replace $$\partial_{w_h} h_t$$ by a random variable which is correct in expectation but which truncates the sequence. This is achieved by using a sequence of $$\xi_t$$ where $$E[\xi_t] = 1$$ and $$P(\xi_t = 0) = 1-\pi$$ and furthermore $$P(\xi_t = \pi^{-1}) = \pi$$. We use this to replace the gradient:

(8.7.10)$z_t = \partial_w f(x_t, h_{t-1}, w) + \xi_t \partial_h f(x_t, h_{t-1}, w) \partial_w h_{t-1}.$

It follows from the definition of $$\xi_t$$ that $$E[z_t] = \partial_w h_t$$. Whenever $$\xi_t = 0$$ the expansion terminates at that point. This leads to a weighted sum of sequences of varying lengths where long sequences are rare but appropriately overweighted. [Tallec & Ollivier, 2017] proposed this in their paper. Unfortunately, while appealing in theory, the model does not work much better than simple truncation, most likely due to a number of factors. First, the effect of an observation after a number of backpropagation steps into the past is quite sufficient to capture dependencies in practice. Second, the increased variance counteracts the fact that the gradient is more accurate. Third, we actually want models that have only a short range of interaction. Hence, BPTT has a slight regularizing effect which can be desirable.

Fig. 8.7.1 From top to bottom: randomized BPTT, regularly truncated BPTT and full BPTT

Fig. 8.7.1 illustrates the three cases when analyzing the first few words of The Time Machine: * The first row is the randomized truncation which partitions the text into segments of varying length. * The second row is the regular truncated BPTT which breaks it into sequences of the same length. * The third row is the full BPTT that leads to a computationally infeasible expression.

## 8.7.2. The Computational Graph¶

In order to visualize the dependencies between model variables and parameters during computation in a recurrent neural network, we can draw a computational graph for the model, as shown in Fig. 8.7.2. For example, the computation of the hidden states of timestep 3, $$\mathbf{h}_3$$, depends on the model parameters $$\mathbf{W}_{hx}$$ and $$\mathbf{W}_{hh}$$, the hidden state of the last timestep $$\mathbf{h}_2$$, and the input of the current timestep $$\mathbf{x}_3$$.

Fig. 8.7.2 Computational dependencies for a recurrent neural network model with three timesteps. Boxes represent variables (not shaded) or parameters (shaded) and circles represent operators.

## 8.7.3. BPTT in Detail¶

After discussing the general principle, let us discuss BPTT in detail. By decomposing $$\mathbf{W}$$ into different sets of weight matrices ($$\mathbf{W}_{hx}, \mathbf{W}_{hh}$$ and $$\mathbf{W}_{oh}$$), we will get a simple linear latent variable model:

(8.7.11)$\mathbf{h}_t = \mathbf{W}_{hx} \mathbf{x}_t + \mathbf{W}_{hh} \mathbf{h}_{t-1} \text{ and } \mathbf{o}_t = \mathbf{W}_{oh} \mathbf{h}_t.$

Following the discussion in Section 4.7, we compute the gradients $$\frac{\partial L}{\partial \mathbf{W}_{hx}}$$, $$\frac{\partial L}{\partial \mathbf{W}_{hh}}$$, $$\frac{\partial L}{\partial \mathbf{W}_{oh}}$$ for

(8.7.12)$L(\mathbf{x}, \mathbf{y}, \mathbf{W}) = \sum_{t=1}^T l(\mathbf{o}_t, y_t),$

where $$l(\cdot)$$ denotes the chosen loss function. Taking the derivatives with respect to $$W_{oh}$$ is fairly straightforward and we obtain

(8.7.13)$\partial_{\mathbf{W}_{oh}} L = \sum_{t=1}^T \mathrm{prod} \left(\partial_{\mathbf{o}_t} l(\mathbf{o}_t, y_t), \mathbf{h}_t\right),$

where $$\mathrm{prod} (\cdot)$$ indicates the product of two or more matrices.

The dependency on $$\mathbf{W}_{hx}$$ and $$\mathbf{W}_{hh}$$ is a bit more tricky since it involves a chain of derivatives. We begin with

(8.7.14)\begin{split}\begin{aligned} \partial_{\mathbf{W}_{hh}} L & = \sum_{t=1}^T \mathrm{prod} \left(\partial_{\mathbf{o}_t} l(\mathbf{o}_t, y_t), \mathbf{W}_{oh}, \partial_{\mathbf{W}_{hh}} \mathbf{h}_t\right), \\ \partial_{\mathbf{W}_{hx}} L & = \sum_{t=1}^T \mathrm{prod} \left(\partial_{\mathbf{o}_t} l(\mathbf{o}_t, y_t), \mathbf{W}_{oh}, \partial_{\mathbf{W}_{hx}} \mathbf{h}_t\right). \end{aligned}\end{split}

After all, hidden states depend on each other and on past inputs. The key quantity is how past hidden states affect future hidden states.

(8.7.15)$\partial_{\mathbf{h}_t} \mathbf{h}_{t+1} = \mathbf{W}_{hh}^\top \text{ and thus } \partial_{\mathbf{h}_t} \mathbf{h}_T = \left(\mathbf{W}_{hh}^\top\right)^{T-t}.$

Chaining terms together yields

(8.7.16)\begin{split}\begin{aligned} \partial_{\mathbf{W}_{hh}} \mathbf{h}_t & = \sum_{j=1}^t \left(\mathbf{W}_{hh}^\top\right)^{t-j} \mathbf{h}_j \\ \partial_{\mathbf{W}_{hx}} \mathbf{h}_t & = \sum_{j=1}^t \left(\mathbf{W}_{hh}^\top\right)^{t-j} \mathbf{x}_j. \end{aligned}\end{split}

A number of things follow from this potentially very intimidating expression. First, it pays to store intermediate results, i.e., powers of $$\mathbf{W}_{hh}$$ as we work our way through the terms of the loss function $$L$$. Second, this simple linear example already exhibits some key problems of long sequence models: it involves potentially very large powers $$\mathbf{W}_{hh}^j$$. In it, eigenvalues smaller than $$1$$ vanish for large $$j$$ and eigenvalues larger than $$1$$ diverge. This is numerically unstable and gives undue importance to potentially irrelevant past detail. One way to address this is to truncate the sum at a computationally convenient size. Later on in Section 9 we will see how more sophisticated sequence models such as LSTMs can alleviate this further. In practice, this truncation is effected by detaching the gradient after a given number of steps.

## 8.7.4. Summary¶

• Backpropagation through time is merely an application of backprop to sequence models with a hidden state.

• Truncation is needed for computational convenience and numerical stability.

• High powers of matrices can lead to divergent and vanishing eigenvalues. This manifests itself in the form of exploding or vanishing gradients.

• For efficient computation, intermediate values are cached.

## 8.7.5. Exercises¶

1. Assume that we have a symmetric matrix $$\mathbf{M} \in \mathbb{R}^{n \times n}$$ with eigenvalues $$\lambda_i$$. Without loss of generality, assume that they are ordered in ascending order $$\lambda_i \leq \lambda_{i+1}$$. Show that $$\mathbf{M}^k$$ has eigenvalues $$\lambda_i^k$$.

2. Prove that for a random vector $$\mathbf{x} \in \mathbb{R}^n$$, with high probability $$\mathbf{M}^k \mathbf{x}$$ will be very much aligned with the largest eigenvector $$\mathbf{v}_n$$ of $$\mathbf{M}$$. Formalize this statement.

3. What does the above result mean for gradients in a recurrent neural network?

4. Besides gradient clipping, can you think of any other methods to cope with gradient explosion in recurrent neural networks?

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