 Hello everyone, this is Alice Gao. In this video, I am going to talk about why we cannot use a perceptron to represent the exclusive OR function. In the main lecture video, I gave an intuitive argument by drawing these pictures for all the logical functions, and intuitively you can see that with AND and OR, you can draw a line to separate the positive and the negative examples, but you are not able to do that with the XOR function. However, this is now the rigorous proof, and some of you are asking for rigorous proofs in this course, so let's look at a rigorous proof to show this. I've drawn the examples for the XOR function again. Now, this proof starts by assuming that we can represent the XOR function using a perceptron. We're going to assume that the activation function is a step function, just to make our argument simpler. Given our assumption, the weights w01, w11, and w21 need to satisfy the following inequalities. For example, the example 1, 0 is a positive example. That means if we plug in x1 is equal to 1, x2 is equal to 0, the weighted sum should be positive. I've written the first inequality, but I realized that I should probably remind you what the perceptron looks like, so I've drawn a little picture of the perceptron as well. So the first inequality says, well, if we plug in 1 and 0, the weighted sum should be positive. And similarly, we can come up with three other inequalities for the three other data points. So these are the four inequalities. The first two are for the two positive examples, and the last two are for the two negative examples. Let's simplify these so that we can get rid of the terms where the value plug in 0. Here are the four simplified inequalities. Well, next, our goal is to eventually derive a contradiction using these inequalities. So we need to arrive at two inequalities that cannot be true at the same time. Well, all of these inequalities have w01 in there, so perhaps we can use that to make a connection and derive the contradiction. Let me do some manipulations with these inequalities. So I rewrote the first two inequalities and named them 1 and 2, and then added those two together to get this one. So w21 plus w11 is greater than negative 2 times w01. And then I also took inequality 3 and then rewrote a little bit. So on the left-hand side, we also have w21 plus w11. So comparing 4 and 5, you can see that their left-hand sides are the same, but their right-hand sides both only involve w01. So let's look at the right-hand side and see how do they compare with each other. Well, this inequality 6 says that w01 is a non-positive value, either 0 or negative. So either those two are the same or minus 2 w01, which would be a positive value, is greater than minus w01. Which means the right-hand side of inequality 4 is greater than or equal to the right-hand side of inequality 5. So combining inequalities 4 and 5, I get the following inequality, and then taking out the intermediate steps, I get a contradiction, right? Because the same number cannot be strictly greater than itself. That's it for this proof. It's basically a mathematical exercise, right? If we can represent this function, then the weights have to satisfy these inequalities. Well, turns out it's impossible for the weights to satisfy these inequalities. That's it. So, were you able to come up with this proof yourself? That's everything for this video. I will see you in the next one. Bye for now.