Discrete Hopfield Network

Colab Open in Kaggle

In this article, we describe core ideas behind discrete hopfield networks and try to understand how it works. In addition, we explore main problems related to this algorithm. And finally, we take a look into simple example that aims to memorize digit patterns and reconstruct them from corrupted samples.

In this article we are going to learn about Discrete Hopfield Network algorithm.

Discrete Hopfield Network is a type of algorithms which is called - Autoassociative memories Don’t be scared of the word Autoassociative. The idea behind this type of algorithms is very simple. It can store useful information in memory and later it is able to reproduce this information from partially broken patterns. You can perceive it as human memory. For instance, imagine that you look at an old picture of a place where you were long time ago, but this picture is of very bad quality and very blurry. By looking at the picture you manage to recognize a few objects or places that make sense to you and form some objects even though they are blurry. It can be a house, a lake or anything that can add up to the whole picture and bring out some associations about this place. With these details that you got from your memory so far other parts of picture start to make even more sense. Though you don’t clearly see all objects in the picture, you start to remember things and withdraw from your memory some images, that cannot be seen in the picture, just because of those very familiarly-shaped details that you’ve got so far. That’s what it is all about. Autoassociative memory networks is a possibly to interpret functions of memory into neural network model.

Don’t worry if you have only basic knowledge in Linear Algebra; in this article I’ll try to explain the idea as simple as possible.


Discrete Hopfield Network is an easy algorithm. It’s simple because you don’t need a lot of background knowledge in Maths for using it. Everything you need to know is how to make a basic Linear Algebra operations, like outer product or sum of two matrices.

Let’s begin with a basic thing. What do we know about this neural network so far? Just the name and the type. From the name we can identify one useful thing about the network. It’s Discrete. It means that network only works with binary vectors. But for this network we wouldn’t use binary numbers in a typical form. Instead, we will use bipolar numbers. They are almost the same, but instead of 0 we are going to use -1 to decode a negative state. We can’t use zeros. And there are two main reasons for it. The first one is that zeros reduce information from the network weight, later in this article you are going to see it. The second one is more complex, it depends on the nature of bipolar vectors. Basically they are more likely to be orthogonal to each other which is a critical moment for the Discrete Hopfield Network. But as I mentioned before we won’t talk about proofs or anything not related to basic understanding of Linear Algebra operations.

So, let’s look at how we can train and use the Discrete Hopfield Network.

Training procedure

We can’t use memory without any patterns stored in it. So first of all we are going to learn how to train the network. For the Discrete Hopfield Network train procedure doesn’t require any iterations. It includes just an outer product between input vector and transposed input vector.

\[\begin{split}\begin{aligned} \begin{align*} W = x \cdot x^T = \left[ \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right] \cdot \left[ \begin{array}{c} x_1 & x_2 & \cdots & x_n \end{array} \right] \end{align*} \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} \begin{align*} = \left[ \begin{array}{c} x_1^2 & x_1 x_2 & \cdots & x_1 x_n \\ x_2 x_1 & x_2^2 & \cdots & x_2 x_n \\ \vdots\\ x_n x_1 & x_n x_2 & \cdots & x_n^2 \\ \end{array} \right] \end{align*} \end{aligned}\end{split}\]

\(W\) is a weight matrix and \(x\) is an input vector. Each value \(x_i\) in the input vector can only be -1 or 1. So on the matrix diagonal we only have squared values and it means we will always see 1s at those places. Think about it, every time, in any case, values on the diagonal can take just one possible state. We can’t use this information, because it doesn’t say anything useful about patterns that are stored in the memory and even can make incorrect contribution into the output result. For this reason we need to set up all the diagonal values equal to zero. The final weight formula should look like this one below.

\[\begin{split}\begin{aligned} \begin{align*} W = x x^T - I = \left[ \begin{array}{c} 0 & x_1 x_2 & \cdots & x_1 x_n \\ x_2 x_1 & 0 & \cdots & x_2 x_n \\ \vdots\\ x_n x_1 & x_n x_2 & \cdots & 0 \\ \end{array} \right] \end{align*} \end{aligned}\end{split}\]

Where \(I\) is an identity matrix.

But usually we need to store more values in memory. For another pattern we have to do exactly the same procedure as before and then just add the generated weight matrix to the old one.

\[W = W_{old} + W_{new}\]

And this procedure generates us a new weight that would be valid for both previously stored patterns. Later you can add other patterns using the same algorithm.

But if you need to store multiple vectors inside the network at the same time you don’t need to compute the weight for each vector and then sum them up. If you have a matrix \(X \in \Bbb R^{m\times n}\) where each row is the input vector, then you can just make product matrix between transposed input matrix and input matrix.

\[W = X^T X - m I\]

Where \(I\) is an identity matrix (\(I \in \Bbb R^{n\times n}\)), \(n\) is a number of features in the input vector and \(m\) is a number of input patterns inside the matrix \(X\). Term \(m I\) removes all values from the diagonal. Basically we remove 1s for each stored pattern and since we have \(m\) of them, we should do it \(m\) times. Practically, it’s not very good to create an identity matrix just to set up zeros on the diagonal, especially when dimension on the matrix is very big. Usually linear algebra libraries give you a possibility to set up diagonal values without creating an additional matrix and this solution would be more efficient. For example in NumPy library it’s a numpy.fill_diagonal function.

Let’s check an example just to make sure that everything is clear. Let’s pretend we have a vector \(u\).

\[\begin{split}\begin{aligned} u = \left[\begin{align*}1 \\ -1 \\ 1 \\ -1\end{align*}\right] \end{aligned}\end{split}\]

Assume that network doesn’t have patterns inside of it, so the vector \(u\) would be the first one. Let’s compute weights for the network.

\[\begin{split}\begin{aligned} \begin{align*} U = u u^T = \left[ \begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array} \right] \left[ \begin{array}{c} 1 & -1 & 1 & -1 \end{array} \right] = \left[ \begin{array}{cccc} 1 & -1 & 1 & -1\\ -1 & 1 & -1 & 1\\ 1 & -1 & 1 & -1\\ -1 & 1 & -1 & 1 \end{array} \right] \end{align*} \end{aligned}\end{split}\]

Look closer to the matrix \(U\) that we got. Outer product just repeats vector 4 times with the same or inversed values. First and third columns (or rows, it doesn’t matter, because matrix is symmetrical) are exactly the same as the input vector. The second and fourth are also the same, but with an opposite sign. That’s because in the vector \(u\) we have 1 on the first and third places and -1 on the other.

To make weight from the \(U\) matrix, we need to remove ones from the diagonal.

\[\begin{split}\begin{aligned} W = U - I = \left[ \begin{array}{cccc} 1 & -1 & 1 & -1\\ -1 & 1 & -1 & 1\\ 1 & -1 & 1 & -1\\ -1 & 1 & -1 & 1 \end{array} \right] - \left[ \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right] \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} = \left[ \begin{array}{cccc} 0 & -1 & 1 & -1\\ -1 & 0 & -1 & 1\\ 1 & -1 & 0 & -1\\ -1 & 1 & -1 & 0 \end{array} \right] \end{aligned}\end{split}\]

\(I\) is the identity matrix and \(I \in \Bbb R^{n \times n}\), where \(n\) is a number of features in the input vector.

When we have one stored vector inside the weights we don’t really need to remove 1s from the diagonal. The main problem would appear when we have more than one vector stored in the weights. Each value on the diagonal would be equal to the number of stored vectors in it.

Recovery from memory

The main advantage of Autoassociative network is that it is able to recover pattern from the memory using just a partial information about the pattern. There are already two main approaches to this situation, synchronous and asynchronous. We are going to master both of them.


Synchronous approach is much more easier for understanding, so we are going to look at it firstly. To recover your pattern from memory you just need to multiply the weight matrix by the input vector.

\[\begin{split}\begin{aligned} \begin{align*} s = {W}\cdot{x}= \left[ \begin{array}{cccc} w_{11} & w_{12} & \ldots & w_{1n}\\ w_{21} & w_{22} & \ldots & w_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ w_{n1} & w_{n2} & \ldots & w_{nn} \end{array} \right] \left[ \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right] \end{align*} \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} \begin{align*} = \left[ \begin{array}{c} w_{11}x_1+w_{12}x_2 + \cdots + w_{1n} x_n\\ w_{21}x_1+w_{22}x_2 + \cdots + w_{2n} x_n\\ \vdots\\ w_{n1}x_1+w_{n2}x_2 + \cdots + w_{nn} x_n\\ \end{array} \right] \end{align*} \end{aligned}\end{split}\]

Let’s analyze the result. We summed up all information from the weights where each value can be any integer with an absolute value equal to or smaller than the number of patterns inside the network. It’s clear that total sum value for \(s_i\) is not necessary equal to -1 or 1, so we have to make additional operations that will make bipolar vector from the vector \(s\).

Let’s think about this product operation. What does it actualy do? Basically after training procedure we saved our pattern dublicated \(n\) times (where \(n\) is a number of input vector features) inside the weight. When we store more patterns we get interception between them (it’s called a crosstalk) and each pattern add some noise to other patterns. So, after perfoming product matrix between \(W\) and \(x\) for each value from the vector \(x\) we’ll get a recovered vector with a little bit of noise. For \(x_1\) we get a first column from the matrix \(W\), for the \(x_2\) a second column, and so on. Then we sum up all vectors together. This operation can remind you of voting. For example we have 3 vectors. If the first two vectors have 1 in the first position and the third one has -1 at the same position, the winner should be 1. We can perform the same procedure with \(sign\) function. So the output value should be 1 if total value is greater then zero and -1 otherwise.

\[\begin{split}\begin{aligned} sign(x) = \left\{ \begin{array}{lr} &1 && : x \ge 0\\ &-1 && : x < 0 \end{array} \right.\\ \end{aligned}\end{split}\]

\[y = sign(s)\]

That’s it. Now \(y\) store the recovered pattern from the input vector \(x\).

Maybe now you can see why we can’t use zeros in the input vectors. In voting procedure we use each row that was multiplied by bipolar number, but if values had been zeros they would have ignored columns from the weight matrix and we would have used only values related to ones in the input pattern.

Of course you can use 0 and 1 values and sometime you will get the correct result, but this approach give you much worse results than explained above.


Previous approach is good, but it has some limitations. If you change one value in the input vector it can change your output result and value won’t converge to the known pattern. Another popular approach is an asynchronous. This approach is more likely to remind you of real memory. At the same time in network activates just one random neuron instead of all of them. In terms of neural networks we say that neuron fires. We iteratively repeat this operation multiple times and after some point network will converge to some pattern.

Let’s look at this example: Consider that we already have a weight matrix \(W\) with one pattern \(x\) inside of it.

\[\begin{split}\begin{aligned} \begin{align*} W = \left[ \begin{array}{cccc} 0 & 1 & -1 \\ 1 & 0 & -1 \\ -1 & -1 & 0 \end{array} \right] \end{align*} \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} \begin{align*} x = \left[ \begin{array}{c} 1\\ 1\\ -1 \end{array} \right] \end{align*} \end{aligned}\end{split}\]

Let’s assume that we have a vector \(x^{'}\) from which we want to recover the pattern.

\[\begin{split}\begin{aligned} \begin{align*} x^{'} = \left[ \begin{array}{c} 1\\ -1\\ -1 \end{array} \right] \end{align*} \end{aligned}\end{split}\]

In first iteration one neuron fires. Let it be the second one. So we multiply the first column by this selected value.

\[\begin{split}\begin{aligned} \begin{align*} x^{'}_2 = sign(\left[ \begin{array}{c} 1 & -1 & -1 \end{array} \right] \cdot \left[ \begin{array}{c} 1\\ 0\\ -1 \end{array} \right]) = sign(2) = 1 \end{align*} \end{aligned}\end{split}\]

And after this operation we set up a new value into the input vector \(x\).

\[\begin{split}\begin{aligned} \begin{align*} x^{'} = \left[ \begin{array}{c} 1\\ 1\\ -1 \end{array} \right] \end{align*} \end{aligned}\end{split}\]

As you can see, after first iteration value is exactly the same as \(x\) but we can keep going. In second iteration random neuron fires again. Let’s pretend that this time it was the third neuron.

\[\begin{split}\begin{aligned} \begin{align*} x^{'}_3 = sign(\left[ \begin{array}{c} 1 & 1 & -1 \end{array} \right] \cdot \left[ \begin{array}{c} -1\\ -1\\ 0 \end{array} \right]) = sign(-2) = -1 \end{align*} \end{aligned}\end{split}\]

\(x^{'}_3\) is exactly the same as in the \(x^{'}\) vector so we don’t need to update it. We can repeat it as many times as we want, but we will be getting the same value.

Memory limit

Obviously, you can’t store infinite number of vectors inside the network. There are two good rules of thumb.

Consider that \(n\) is the dimension (number of features) of your input vector and \(m\) is the number of patterns that you want to store in the network. The first rule gives us a simple ration between \(m\) and \(n\).

\[m \approx 0.18 n\]

The main problem with this rule is that proof assumes that stored vectors inside the weight are completely random with an equal probability. Unfortunately, that is not always true. Let’s suppose we save some images of numbers from 0 to 9. Pictures are black and white, so we can encode them in bipolar vectors. Will the probabilities be the same for seeing as many white pixels as black ones? Usually no. More likely that number of white pixels would be greater than number of black ones. Before use this rule you have to think about type of your input patterns.

The second rule uses a logarithmic proportion.

\[m = \left \lfloor \frac{n}{2 \cdot log(n)} \right \rfloor\]

Both of these rules are good assumptions about the nature of data and its possible limits in memory. Of course, you can find situations when these rules will fail.


Hallucinations is one of the main problems in the Discrete Hopfield Network. Sometimes network output can be something that we hasn’t taught it.

To understand this phenomena we should firstly define the Hopfield energy function.

\[E = -\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} w_{ij} x_i x_j + \sum_{i=1}^{n} \theta_i x_i\]

Where \(w_{ij}\) is a weight value on the \(i\)-th row and \(j\)-th column. \(x_i\) is a \(i\)-th values from the input vector \(x\). \(\theta\) is a threshold. Threshold defines the bound to the sign function. For this reason \(\theta\) is equal to 0 for the Discrete Hopfield Network. In terms of a linear algebra we can write formula for the Discrete Hopfield Network energy Function more simpler.

\[E = -\frac{1}{2} x^T W x\]

But linear algebra notation works only with the \(x\) vector, we can’t use matrix \(X\) with multiple input patterns instead of the \(x\) in this formula. For the energy function we’re always interested in finding a minimum value, for this reason it has minus sign at the beginning.

Let’s try to visualize it. Assume that values for vector \(x\) can be continous in order and we can visualize them using two parameters. Let’s pretend that we have two vectors [1, -1] and [-1, 1] stored inside the network. Below you can see the plot that visualizes energy function for this situation.

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

def energy(input_vector):
    input_vector = np.array(input_vector)
    X = np.array([[1, -1], [-1, 1]])
    weight = X.T.dot(X) - 2 * np.eye(2)
    return -0.5 * input_vector.dot(weight).dot(input_vector)

fig = plt.figure(figsize=(9, 9))
ax = fig.add_subplot(111, projection='3d')

x = y = np.arange(-1.0, 1.0, 0.01)
X, Y = np.meshgrid(x, y)
energies = map(energy, zip(np.ravel(X), np.ravel(Y)))
zs = np.array(list(energies))
Z = zs.reshape(X.shape)

ax.view_init(elev=6, azim=-72)
ax.plot_surface(X, Y, Z, cmap='Reds')


As you can see we have two minimum values at the same points as those patterns that are already stored inside the network. But between these two patterns function creates a saddle point somewhere at the point with coordinates \((0, 0)\). In this case we can’t stick to the points \((0, 0)\). But in situation with more dimensions this saddle points can be at the level of available values and they could be hallucination. Unfortunately, we are very limited in terms of numbers of dimensions we could plot, but the problem is still the same.


import brainpy as bp
import brainpy.math as bm

class HopfieldNet(bp.DynamicalSystem):
  def __init__(self, num):
    self.num = num
    self.weight = bm.Variable(bm.zeros([num, num]))

  # Train function for the Hopfield net. (By updating the weights)
  def store_patterns(self, samples):
    # data: An array with [d, N]. 'd' is the number of data samples used to train. (In this case 2)
    assert samples.ndim == 2
    assert samples.shape[1] == self.num

    # Loop through all data samples.
    bm.for_loop(self.store, samples)

    # Hopfield nets are a form of RNNs, albeit without self-connections, and so,
    # we need to make sure that the diagonal elements of the final weight matrix are zero.
    bm.fill_diagonal(self.weight, 0)

  # Storing one sample pattern
  def store(self, sample):
    # sample is an array with the shape of (N,)
    assert self.num == sample.shape[0]

    # Data cross-product gives neural hopfield update rule.
    w_update = bm.outer(sample, sample)

    # Sum all pattern cross-products.
    self.weight += w_update

  def async_recover(self, sample, n, energy=False):
    # n: the number of iterations to recover
    # energy: calculate the energy function
    idxs = bm.random.randint(0, self.num, n)  # the sampled positions
    # JIT compilation requires to label the value to be changed as Variable
    sample = bm.Variable(sample)

    def recover(i):
      # i: the position to update
      # update
      sample[i] = bm.sign(bm.inner(self.weight[i], sample))
      # return energy
      if energy:
        return self.energy(sample)

    r = bm.for_loop(recover, idxs) # for loop JIT
    return (sample, r) if energy else sample

  def sync_recover(self, sample, n, energy=False):
    # n: the number of iterations to recover
    # energy: calculate the energy function
    idxs = bm.arange(n)  # the update times
    # JIT compilation requires to label the value to be changed as Variable
    sample = bm.Variable(sample)

    def recover(i):
      # i: the position to update
      # update
      sample.value = bm.sign(self.weight @ sample)
      # return energy
      if energy:
        return self.energy(sample)

    r = bm.for_loop(recover, idxs)  # for loop JIT
    return (sample, r) if energy else sample

  # This function computes the Hopfield nets' energy.
  def energy(self, x):
    # x: [N] data vector
    return bm.inner(- x @ self.weight, x)

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