# What is the simplest neural network? One neuron

The basis of neural networks is the neuron. A neuron has some inputs, and based on those, either *fires* or doesn’t fire. We can interpret the decision to fire as a binary *classification* of the inputs into two classes.

To neuron whether to fire, the neuron follows a simple procedure: take each input, multiply it by a *weight*, sum those products, then fire if the sum is above a threshold. The specific behavior of a neuron is given by its weights and its threshold.

The following C code shows two neurons at work. One classifies numbers according to whether they are even or odd. Another neuron classifies numbers according to whether they are greater than five. Each is defined by specific weights and threshold.

The neuron only understands inputs as a vector of numbers, so we have to encode our inputs. To encode the natural numbers, I’ve encoded them in binary.

```
#include <stdbool.h>
#include <stdio.h>
bool neuron(float weights[4], float threshold, float input[4]) {
float sum = 0;
for (size_t i = 0; i < 4; i++) {
sum += input[i] * weights[i];
}
return threshold < sum;
}
float is_odd_weights[] = { 0, 0, 0, 1 };
float real_number_weights[] = { 8, 4, 2, 1 };
float four[] = {0, 1, 0, 0};
float five[] = {0, 1, 0, 1};
float six[] = {0, 1, 1, 0};
float seven[] = {0, 1, 1, 1};
int main() {
printf("4 is odd: %d\n", neuron(is_odd_weights, 0, four));
printf("5 is odd: %d\n", neuron(is_odd_weights, 0, five));
printf("6 is odd: %d\n", neuron(is_odd_weights, 0, six));
printf("7 is odd: %d\n", neuron(is_odd_weights, 0, seven));
printf("4 is greater than 5: %d\n", neuron(real_number_weights, 5, four));
printf("5 is greater than 5: %d\n", neuron(real_number_weights, 5, five));
printf("6 is greater than 5: %d\n", neuron(real_number_weights, 5, six));
printf("7 is greater than 5: %d\n", neuron(real_number_weights, 5, seven));
return 0;
}
```

What kinds of classifications can this pattern make? For example, it understands *divisibility by two*, but could we find weights which classify according to divisibility by *three*? Find out in a future episode.

With Vidrio

With generic competitor

### More by Jim

- Your syntax highlighter is wrong
- Granddad died today
- The Three Ts of Time, Thought and Typing: measuring cost on the web
- I hate telephones
- The sorry state of OpenSSL usability
- The dots do matter: how to scam a Gmail user
- My parents are Flat-Earthers
- How Hacker News stays interesting
- Project C-43: the lost origins of asymmetric crypto
- The hacker hype cycle
- The inception bar: a new phishing method
- Time is running out to catch COVID-19
- A probabilistic pub quiz for nerds
- Smear phishing: a new Android vulnerability

Tagged . All content copyright James Fisher 2017. This post is not associated with my employer. Found an error? Edit this page.