Perception is everything. Communication faculty students learn this in their early lessons. Machine learning practitioners learn this in their freshman days as well. perceptron is an early version of modern neural networks. Understanding the logic behind the classical single layer perceptron will help you to understand the idea behind deep learning as well. Because you can image deep neural networks as combination of nested perceptrons. You can also imagine single layer perceptron as legacy neural networks.
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Explaining perceptron with metaphors
Explaining perceptron with some metaphors might help you to understand the perceptron better.
Vlog
You can either watch the following video or read this blog post. They both cover the perceptron from scratch. Additionally, vlog explains perceptron in python.
Supportive problem
Suppose that we are going to work on AND Gate problem. The gate returns if and only if both inputs are true.
| X1 | X2 | Y |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
We are going to set weights randomly. Let’s say that w1 = 0.9 and w2 = 0.9
Round 1
We will apply 1st instance to the perceptron. x1 = 0 and x2 = 0.
Sum unit will be 0 as calculated below
Σ = x1 * w1 + x2 * w2 = 0 * 0.9 + 0 * 0.9 = 0
Activation unit checks sum unit is greater than a threshold. If this rule is satisfied, then it is fired and the unit will return 1, otherwise it will return 0. BTW, modern neural networks architectures do not use this kind of a step function as activation.
Activation threshold would be 0.5.
Sum unit was 0 for the 1st instance. So, activation unit would return 0 because it is less than 0.5. Similarly, its output should be 0 as well. We will not update weights because there is no error in this case.
Let’s focus on the 2nd instance. x1 = 0 and x2 = 1.
Sum unit: Σ = x1 * w1 + x2 * w2 = 0 * 0.9 + 1 * 0.9 = 0.9
What about errors?
Activation unit will return 1 because sum unit is greater than 0.5. However, output of this instance should be 0. This instance is not predicted correctly. That’s why, we will update weights based on the error.
ε = actual – prediction = 0 – 1 = -1
We will add error times learning rate value to the weights. Learning rate would be 0.5. BTW, we mostly set learning rate value between 0 and 1.
w1 = w1 + α * ε = 0.9 + 0.5 * (-1) = 0.9 – 0.5 = 0.4
w2 = w2 + α * ε = 0.9 + 0.5 * (-1) = 0.9 – 0.5 = 0.4
Focus on the 3rd instance. x1 = 1 and x2 = 0.
Sum unit: Σ = x1 * w1 + x2 * w2 = 1 * 0.4 + 0 * 0.4 = 0.4
Activation unit will return 0 this time because output of the sum unit is 0.5 and it is less than 0.5. We will not update weights.
Mention the 4rd instance. x1 = 1 and x2 = 1.
Sum unit: Σ = x1 * w1 + x2 * w2 = 1 * 0.4 + 1 * 0.4 = 0.8
Activation unit will return 1 because output of the sum unit is 0.8 and it is greater than the threshold value 0.5. Its actual value should 1 as well. This means that 4th instance is predicted correctly. We will not update anything.
Round 2
In previous round, we’ve used previous weight values for the 1st instance and it was classified correctly. Let’s apply feed forward for the new weight values.
Remember the 1st instance. x1 = 0 and x2 = 0.
Sum unit: Σ = x1 * w1 + x2 * w2 = 0 * 0.4 + 0 * 0.4 = 0.4
Activation unit will return 0 because sum unit is 0.4 and it is less than the threshold value 0.5. The output of the 1st instance should be 0 as well. This means that the instance is classified correctly. We will not update weights.
Feed forward for the 2nd instance. x1 = 0 and x2 = 1.
Sum unit: Σ = x1 * w1 + x2 * w2 = 0 * 0.4 + 1 * 0.4 = 0.4
Activation unit will return 0 because sum unit is less than the threshold 0.5. Its output should be 0 as well. This means that it is classified correctly and we will not update weights.
We’ve applied feed forward calculation for 3rd and 4th instances already for the current weight values in the previous round. They were classified correctly.
Learning term
We should continue this procedure until learning completed. We can terminate the learning procedure here. Luckily, we can find the best weights in 2 rounds.
Updating weights means learning in the perceptron. We set weights to 0.9 initially but it causes some errors. Then, we update the weight values to 0.4. In this way, we can predict all instances correctly.
I’ve written the logic of perceptron in python. You can find the source code here.
Homework
To reinforce the perceptron, you should apply learning procedure for OR Gate. The gate returns 0 if and only if both inputs are 0. Do not hesitate to change the initial weights and learning rate values.
| X1 | X2 | Y |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Basic perceptron can generalize any kind of linear problem. The both AND and OR Gate problems are linearly separable problems. On the other hand, this form cannot generalize non-linear problems such as XOR Gate. Perceptron evolved to multilayer perceptron to solve non-linear problems and deep neural networks were born.
Feature Importance
Explainable AI and machine learning interpretability are the hottest topics nowadays in the data world. Herein, perceptrons are naturally explainable algorithms. Similar to linear regression, coefficients are directly related to the feature importance values.
Multilayer perceptron
Multilayer perceptron or its more common name neural networks can solve non-linear problems. It is the evolved version of perceptron.
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Algebraic, Geometric, Statistical aspects of the weighted sum?
what do you mean exactly?
show hyperplane that created each iteration, will help me so much. thanks.
it would be clever! I will study on this.
thanks a lot
Hey, thx a lot for the post.
In Round 2 you have:
Sum unit:
Σ = x1 * w1 + x2 * w2 = 0 * 0.4 + 0 * 0.4 = 0.4
which is 0, not 0.4.
Although the conclusion still applies, you might want to correct this minor error, to avoid future confusion.
Cheers
Very helpful post to understand perceptron algorithm, thank you very much