I'd like to propose a different kind of golfing challenge to this community:

(Artificial) Neural Networks are very popular machine learning models that can be designed and trained to approximate any given (usually unknown) function. They're often used to solve highly complex problems that we don't know how to solve algorithmically like speech recognition, certain kinds of image classifications, various tasks in autonomous driving systems, ... For a primer on neural networks, consider this excellent Wikipedia article.

As this is the first in what I hope to be a series of machine learning golf challenges, I'd like to keep things as simple as possible:

In the language and framework of your choice, design and train a neural network that, given \$(x_1, x_2)\$ calculates their product \$x_1 \cdot x_2\$ for all integers \$x_1, x_2\$ between (and including) \$-10\$ and \$10\$.

Performance Goal

To qualify, your model may not deviate by more than \$0.5\$ from the correct result on any of those entries.


Your model

  • must be a 'traditional' neural network (a node's value is calculated as a weighted linear combination of some of the nodes in a previous layer followed by an activation function),
  • may only use the following standard activation functions:
    1. \$\textrm{linear}(x) = x\$,
    2. \$\textrm{softmax}(\vec{x})_i = \frac{e^{x_i}}{\sum_j e^{x_j}}\$,
    3. \$\textrm{selu}_{\alpha, \beta}(x) = \begin{cases} \beta \cdot x & \text{, if } x > 0 \\ \alpha \cdot \beta (e^x -1 ) & \text{, otherwise} \end{cases}\$,
    4. \$\textrm{softplus}(x) = \ln(e^x+1)\$,
    5. \$\textrm{leaky-relu}_\alpha(x) = \begin{cases} x & \text{, if } x < 0 \\ \alpha \cdot x & \text{, otherwise} \end{cases}\$,
    6. \$\tanh(x)\$,
    7. \$\textrm{sigmoid}(x) = \frac{e^x}{e^x+1}\$,
    8. \$\textrm{hard-sigmoid}(x) = \begin{cases} 0 & \text{, if } x < -2.5 \\ 1 & \text{, if } x > 2.5 \\ 0.2 \cdot x + 0.5 & \text{, otherwise} \end{cases}\$,
    9. \$e^x\$
  • must take \$(x_1, x_2)\$ either as a tupel/vector/list/... of integers or floats as its only input,
  • return the answer as an integer, float (or a suitable container, e.g. a vector or list, that contains this answer).

Your answer must include (or link to) all code necessary to check your results -- including the trained weights of your model.


The neural network with the smallest number of weights (including bias weights) wins.


  • 9
    \$\begingroup\$ Welcome to the site! I think this challenge could benefit a good deal from a more robust definition of a neural network. There are a couple of things here 1) It would be very nice for you to state it in language that does not already imply knowledge of NNs 2) You really should list the activation functions in your post rather than link out to an external source (outside links can change or disappear). \$\endgroup\$
    – Wheat Wizard
    Commented Jul 2, 2019 at 12:29
  • 4
    \$\begingroup\$ Can we reuse weights/use convolutional layers? (I recommend removing the bonus, as it doesn't add anything to the challenge and just distracts from the main goal.) Are the weights supposed to be real or can they be complex? \$\endgroup\$
    – flawr
    Commented Jul 2, 2019 at 14:41
  • 4
    \$\begingroup\$ Your wording implies nodes from layer 3 can't use inputs from layer 1. Does it cost a weight to have a layer 2 node simply doing f(x) = x to forward its input? \$\endgroup\$
    – Grimmy
    Commented Jul 2, 2019 at 14:46
  • 4
    \$\begingroup\$ There should be a link in the right column to the Sandbox, which was created expressly to fix these kind of issues before the question is even posted to the main site. And the network philosophy is that it's better to close a question, fix it, and reopen it than to get a bunch of answers which either will make no sense after the question is fixed or will tightly constrain the changes which can be made to the question. \$\endgroup\$ Commented Jul 3, 2019 at 16:22
  • 7
    \$\begingroup\$ Not at all. These kinds of issues are detected by many years' experience of seeing other people make the same kind of mistakes. Some ambiguities slip past the sandbox, but many more are caught there. And this would definitely have been caught, because as indicated in my first comment we had exactly the same problems with a neural net question two months ago. \$\endgroup\$ Commented Jul 3, 2019 at 16:29

5 Answers 5


21 13 11 9 weights

This is based on the polarization identity of bilinear forms which in the one dimensional real case reduces to the polynomial identity:

$$ x\cdot y = \frac{(x+y)^2 - (x-y)^2}{4}$$

So y1 just computes [x+y, x-y] using a linear transformation, and y3 is just the absolute value of y1 as a preprocessing step for the next one: Then the "hard" part is computing the squares which I will explain below, and after that just computing a difference and scaling which is again a linear operation.

To compute the squares I use an exponential series \$s\$ which should be accurate for all integers \$\{0,1,2,\ldots,20\}\$ within around \$0.5\$. This series is of the form

$$ \text{approx_square}(x) = \sum_{i=0}^2 w_i \exp(0.0001 \cdot i \cdot x)$$

where I just optimized for the weights W2 (\$=(w_i)_i\$). This whole approximation comprises again just two linear transformations with an exponential activation sandwiched in between. This approach results in a maximal deviation of about 0.02.

function p = net(x)
% 9 weights
one = 1; 
mone =-1;
zero = 0;
fourth = 0.25;
W1 = [1e-4, 2e-4];
W2  = [-199400468.100687;99700353.6313757];
b2 = 99700114.4299316;
leaky_relu = @(a,x)max(a*x,x); 

% Linear
y0 = [one, one; one, mone] * x;

% Linear + ReLU
y1 = mone * y0;
y2 = [leaky_relu(zero, y0), leaky_relu(zero, y1)];

% Linear
y3 = y2 * [one; one];

% Linear + exp
y4 = exp(y3 * W1); 

% Linear + Bias
y5 =  y4 * W2 + b2;

% Linear
y6 = [one, mone]*y5;
p = y6 * fourth;


Try it online!

  • \$\begingroup\$ I think your checking code in the footer of the TIO link misses an application of abs. But everything is fine anyway. \$\endgroup\$ Commented Jul 2, 2019 at 22:27
  • \$\begingroup\$ @ChristianSievers Thanks, I updated the TIO link! \$\endgroup\$
    – flawr
    Commented Jul 3, 2019 at 5:32
  • \$\begingroup\$ I'm not an expert on NN, out of curiosity, how is the weight count done? y0 needs 4, y1 needs 2, y3 needs 2, y4 needs 1, y5 needs 1 and y6 needs 2. That's 12? \$\endgroup\$ Commented Jul 3, 2019 at 19:27
  • 3
    \$\begingroup\$ @MargaretBloom Yes this is indeed a little bit unusual, but the OP said in the comments that we can reuse weights and only ever have to count them once, even if we use the same weight multiple times. So all the weights I'm using are defined in the first part of the function. \$\endgroup\$
    – flawr
    Commented Jul 3, 2019 at 19:50

7 weights

eps = 1e-6
c = 1 / (2 * eps * eps)

def f(A, B):
	e_s = exp(eps * A + eps * B)  # 2 weights, exp activation
	e_d = exp(eps * A - eps * B)  # 2 weights, exp activation
	return c * e_s + (-c) * e_d + (-1 / eps) * B  # 3 weights, linear activation

Try it online!

Uses the following approximate equality for small \$\epsilon\$ based on the Taylor expansion \$ e^x \approx 1 + x + \frac{x^2}{2}\$:

$$ AB \approx \frac{e^{\epsilon A+\epsilon B} - e^{\epsilon A-\epsilon B}}{2 \epsilon^2} - \frac{B}{\epsilon} $$

Picking \$\epsilon\$ small enough gets us within the required error bounds. Note that eps and c are constant weights in the code.

  • 1
    \$\begingroup\$ Not sure that this counts as a 'traditional neural network' (rule #1) but it's obvious that it can be reformatted into one, so I see no issue with that. Nice solution! \$\endgroup\$ Commented Jul 3, 2019 at 3:55
  • 1
    \$\begingroup\$ You could define C = -B (1 weight) and then have [e_s, e_d] = conv([A,B,C], [eps, eps]) (2 weights) to save one weight:) (BTW: Very clever approach!) \$\endgroup\$
    – flawr
    Commented Jul 3, 2019 at 5:39
  • \$\begingroup\$ (I forgot to add the exp) \$\endgroup\$
    – flawr
    Commented Jul 3, 2019 at 6:21
  • 4
    \$\begingroup\$ You can even get a lot lower by reusing the weights - you don't have to count the same weight multiple times. \$\endgroup\$
    – flawr
    Commented Jul 3, 2019 at 6:54
  • 2
    \$\begingroup\$ @flawr That's a nice trick, but I think the allowances for convolution and reusing weights in the comments make this so much a different challenge that I'm going to keep this answer as is. \$\endgroup\$
    – xnor
    Commented Jul 4, 2019 at 2:10

33 31 weights

# Activation functions
sub hard { $_[0] < -2.5 ? 0 : $_[0] > 2.5 ? 1 : 0.2 * $_[0] + 0.5 }
sub linear { $_[0] }

# Layer 0
sub inputA() { $a }
sub inputB() { $b }

# Layer 1
sub a15() { hard(5*inputA) }

# Layer 2
sub a8()  { hard(-5*inputA + 75*a15 - 37.5) }

# Layer 3
sub aa()  { linear(-5*inputA + 75*a15 - 40*a8) }

# Layer 4
sub a4()  { hard(aa - 17.5) }

# Layer 5
sub a2()  { hard(aa - 20*a4 - 7.5) }

# Layer 6
sub a1()  { linear(0.2*aa - 4*a4 - 2*a2) }

# Layer 7
sub b15() { hard(0.25*inputB - 5*a15) }
sub b8()  { hard(0.25*inputB - 5*a8) }
sub b4()  { hard(0.25*inputB - 5*a4) }
sub b2()  { hard(0.25*inputB - 5*a2) }
sub b1()  { hard(0.25*inputB - 5*a1) }

# Layer 8
sub output() { linear(-300*b15 + 160*b8 + 80*b4 + 40*b2 + 20*b1 - 10*inputA) }

# Test
for $a (-10..10) {
        for $b (-10..10) {
                die if abs($a * $b - output) >= 0.5;

print "All OK";

Try it online!

This does long multiplication in (sorta) binary, and thus returns the exact result. It should be possible to take advantage of the 0.5 error window to golf this some more, but I'm not sure how.

Layers 1 to 6 decompose the first input in 5 "bits". For golfing reasons, we don't use actual binary. The most significant "bit" has weight -15 instead of 16, and when the input is 0, all the "bits" are 0.5 (which still works out fine, since it preserves the identity inputA = -15*a15 + 8*a8 + 4*a4 + 2*a2 + 1*a1).

  • 1
    \$\begingroup\$ I did expect someone to come up with a hard-coded, ANN-ified multiplication algorithm. But I didn't think it would be the first response. Well done! (I'm also eager to see whether you'll be able to pull something like this off with the MNIST dataset or some other, more relastic ML problem :D.) \$\endgroup\$ Commented Jul 2, 2019 at 15:55

43 weights

The two solutions posted so far have been very clever but their approaches likely won't work for more traditional tasks in machine learning (like OCR). Hence I'd like to submit a 'generic' (no clever tricks) solution to this task that hopefully inspires other people to improve on it and get sucked into the world of machine learning:

My model is a very simple neural network with 2 hidden layers built in TensorFlow 2.0 (but any other framework would work as well):

model = tf.keras.models.Sequential([
tf.keras.layers.Dense(6, activation='tanh', input_shape=(2,)),
tf.keras.layers.Dense(3, activation='tanh'),
tf.keras.layers.Dense(1, activation='linear')

As you can see, all layers are dense (which most certainly isn't optimal), the activation function is tanh (which might actually be okay for this task), except for the output layer that, due to the nature of this task, has a linear activation function.

There are 43 weights:

  • \$(2+1) \cdot 6 = 18\$ between the input and first hidden layer,
  • \$(6+1) \cdot 3 = 21\$ between the hidden layers and
  • \$(3+1) \cdot 1 = 4\$ connecting the last hidden and the output layer.

The weights have been trained (with an adam optimizer) by a layered fitting approach: First they've been fitted to minimize the mean squarred error not only on integer multiplication between \$-10\$ and \$10\$ but actually on inputs in a certain neighborhood around these values. This results in much better convergence due to the nature of gradient descent. And it accounted for 400 epochs worth of training on 57,600 training samples each, using a batch size of 32.

Next, I've fine-tuned them -- optimizing for the maximum deviation on any of the integer multiplication tasks. Unfortunately, my notes don't show much fine tuning I ended up doing, but it was very minor. In the neighborhood of 100 epochs on those 441 training samples, with a batch size of 441.

These are the weights I ended up with:

[<tf.Variable 'dense/kernel:0' shape=(2, 6) dtype=float32, numpy=
 array([[ 0.10697944,  0.05394982,  0.05479664, -0.04538541,  0.05369904,
         -0.0728976 ],
        [ 0.10571832,  0.05576797, -0.04670485, -0.04466859, -0.05855528,
         -0.07390639]], dtype=float32)>,
 <tf.Variable 'dense/bias:0' shape=(6,) dtype=float32, numpy=
 array([-3.4242163, -0.8875816, -1.7694025, -1.9409281,  1.7825342,
         1.1364107], dtype=float32)>,
 <tf.Variable 'dense_1/kernel:0' shape=(6, 3) dtype=float32, numpy=
 array([[-3.0665843 ,  0.64912266,  3.7107112 ],
        [ 0.4914808 ,  2.1569328 ,  0.65417236],
        [ 3.461693  ,  1.2072319 , -4.181983  ],
        [-2.8746269 , -4.9959164 ,  4.505049  ],
        [-2.920127  , -0.0665407 ,  4.1409926 ],
        [ 1.3777553 , -3.3750365 , -0.10507642]], dtype=float32)>,
 <tf.Variable 'dense_1/bias:0' shape=(3,) dtype=float32, numpy=array([-1.376577  ,  2.8885336 ,  0.19852689], dtype=float32)>,
 <tf.Variable 'dense_2/kernel:0' shape=(3, 1) dtype=float32, numpy=
 array([[-78.7569  ],
        [ 84.29587 ]], dtype=float32)>,
 <tf.Variable 'dense_2/bias:0' shape=(1,) dtype=float32, numpy=array([8.521169], dtype=float32)>]

which barely met the stated performance goal. The maximal deviation ended up being \$0.44350433\$ as witnessd by \$9 \cdot 10 = 90.443504\$.

My model can be found here and you can also Try it online! in a Google Colab environment.


2 weights

I was inspired by the other answers to approximate the polarization identity in a different way. For every small \$\epsilon>0\$, it holds that

$$ xy \approx \frac{e^{\epsilon x+\epsilon y}+e^{-\epsilon x-\epsilon y}-e^{\epsilon x-\epsilon y}-e^{-\epsilon x+\epsilon y}}{4\epsilon^2}.$$

It suffices to take \$\epsilon=0.01\$ for this challenge.

The obvious neural net implementation of this approximation takes weights in \$\{\pm\epsilon,\pm(4\epsilon^2)^{-1}\}\$. These four weights can be golfed down to three \$\{\pm\epsilon,(4\epsilon^3)^{-1}\}\$ by factoring \$\pm(4\epsilon^2)^{-1}=\pm\epsilon\cdot(4\epsilon^3)^{-1}\$. As I mentioned in a comment above, every neural net with weights in machine precision can be golfed to a (huge!) neural net with only two distinct weights. I applied this procedure to write the following MATLAB code:

function z=approxmultgolfed(x,y)

w1 = 0.1;   % first weight
w2 = -w1;   % second weight

k  = 250000;
v1 = w1*ones(k,1);
v2 = w2*ones(k,1);

L1 = w1*eye(2);
L2 = [ w1 w1; w2 w2; w1 w2; w2 w1 ];
L3 = [ v1 v1 v2 v2 ];
L4 = v1';

z = L4 * L3 * exp( L2 * L1 * [ x; y ] );

All told, this neural net consists of 1,250,010 weights, all of which reside in \$\{\pm0.1\}\$.

How to get away with just 1 weight (!)

It turns out you can simulate any neural net that has weights in \$\{\pm0.1\}\$ with a larger neural net that has only one weight, namely, \$-0.1\$. Indeed, multiplication by \$0.1\$ can be implemented as

$$ 0.1x = w^\top wx, $$

where \$w\$ is the column vector of \$10\$ entries, all equal to \$-0.1\$. For neural nets in which half of the weights are positive, this transformation produces a neural net that is \$10.5\$ times larger.

The obvious generalization of this procedure will transform any neural net with weights in \$\{\pm 10^{-k}\}\$ into a larger neural net with the single weight \$-10^{-k}\$. Combined with the procedure in my comment above, it therefore holds that every neural net with machine-precision weights can be transformed into a single-weight neural net.

(Perhaps we should modify how re-used weights are scored in future neural net golfing challenges.)


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