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That's the problem

$$y=(x,w,\rho) = \begin{cases} 1 & \sum_{i=1}^3 w_ix_i >\rho\\ 0 & \text{otherwise} \end{cases},$$

where $x=\{x_1,x_2,x_3\}$ are inputs, $w=\{w_1,w_2,w_3\}$ are weights and $\rho$ is the threshold value.

The problem asks to find an opportune set of weights that can classify these inputs.

$$A = \{(1, 2, 0), (−1, 3, 0), (−2, −3, 0)\},$$ $$B = \{(0,1,2),(9,0,1),(−3,−3,3)\}.$$

The first step is to assign a random weights to all inputs

$w_1= 0.5$, $w_2= 0.7$, $w_3=0.3$.

For the first example $(1,2,0)$ that I know is part of the A class

$\sum_{i=1}^3 w_ix_i=1.9 < \rho~ \Rightarrow y=0$ is the result.

I need to update the weights but I can't understand how.

The formula is

$w_i'=w_i*n*x(t-y)$.

Correct?

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  • $\begingroup$ In your formula you use variables $t$ and $n$ that you never define. Also, should the final term of the formula be $x_i*(t-y)$? $\endgroup$ – FrankW Apr 9 '14 at 11:32
  • $\begingroup$ Thank you so much for suggestion. I don't know if the formula is correct... I think that t must be the target (value expected) and y is the value obtained. N is a constant called learning constant (typically a very small number like 0.1). what is the procedure to solve this kind of problem and obtain the weights to a correct classification of the inputs? thanks $\endgroup$ – Usi Usi Apr 9 '14 at 15:41
  • $\begingroup$ @UsiUsi you don't specify value of $\rho$. so it is not possible to solve the problem. $\endgroup$ – user13107 Sep 11 '17 at 9:28
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The biggest problem is your equation is wrong for how to update the weights. Your response that $t$ is the target value is correct. Furthermore, the $n$ should almost certainly be the greek letter eta, $\eta$. Anyway, sticking as close to your notation and example as possible, say we have some input/output pair $(x, t) \in \mathbb{R}^3 \times \{0,1\}$ and $y$ is the output of the perceptron on input $x$. Then the weight update should be

$$ w_i^\prime = w_i + \eta x_i (t - y), \; i = 1,2,3. $$

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  • $\begingroup$ Thanks for your answer. Can you show me one step for example to check the (1,2,0) input? How the algo classifies it? if the y>rho it means that it is in class A or B (for the algo obviously) $\endgroup$ – Usi Usi Apr 9 '14 at 16:12
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The first thing to do is to treat the bias ($\rho$) as a weight on an extra input whose activity is always 1 so that you won't need a separate learning rule for the bias and then the decision for your binary threshold outputs a 1 iff $\sum\limits_{i=1}^{3} w_i x_i \geq 0$

To learn the weights of perceptrons, you are right to start with random weights. The remainder of the procedure is to do the following: if for an input your weights correctly classify an output, you leave the weights alone. If, however, you incorrectly output a zero, add the input vector to the weight vector. And on the other hand, if you incorrectly output a 1, subtract the input vector from the weight vector. I would be able to show you an example run with your inputs and random weights above but you're missing the outputs which map to the inputs in order to correct the weights.

Also, note that the above procedure gives you a correct set of weights iff one exists. Otherwise, the learning procedure will fail and you will eventually find it impossible to classify one of the inputs.

EDIT: Here's another example to try and explain things. Let's use initial weights where $w = (w_1, w_2, w_3) = (1, 0, -2)$, letting $\rho = 0$, with:

Target: $t_1 = 1$, $t_2 = 1$, $t_3 = 0$, $t_4 = 1$, $t_5 = 0$

Input: $x_1 = (1,0,1)$, $x_2 = (0,1,1)$, $x_3 = (1, 0, 0)$, $x_4 = (1,0,1)$, $x_5 = (1, 0, 0)$

$y_1 = wx_1 = 1\times 1 + 0\times 0 + -2\times1 =-1 \leq \rho$ --> outputs a 0, which does not equal $t_1$. In this case, we add the input to the weight vector so $w\prime = w + x_1 = (1+1,0+0,-2+1) = (2,0,-1)$, $w = w\prime$

$y_2 = wx_2 = 0\times 2 + 1 \times 0 + 1\times -1 = -1 \leq \rho $ --> outputs a 0, which does not equal $t_2$. In this case, we add the input to the weight vector so $w\prime = w + x_2 = (2,1,0)$, $w = w\prime$

$y_3 = wx_3 = 1\times 2 + 0\times 1 + 0\times 0 = 2 > \rho$ --> outputs a 1, which does not equal $t_3$. In this case, we subtract the input from the weight vector so $w\prime = w - x_3 = (2-1,1-0,0-0) = (1,1,0)$, $w = w\prime$

$y_4 = wx_4 = 1\times 1 + 0\times 1 + 1\times 0 = 1 > \rho$ --> outputs a 1, which equals $t_4$ so the weight vector remains unchanged.

$y_5 = wx_5 = 1\times 1 + 0\times 1 + 0\times 0 = 1 > \rho$ --> outputs a 1, which does not equal $t_5$. In this case, we subtract the input from the weight vector so $w\prime = w - x_5 = (1-1,1-0,0-0) = (0,1,0)$, $w = w\prime$

Final weight vector $w= (0,1,0)$

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  • $\begingroup$ Thanks for your answer. The (1,2,0) is in class A... if the perceptron response is 0 it means that it doesn't classified the example... right? why it isn't the output? $\endgroup$ – Usi Usi Apr 12 '14 at 14:45
  • $\begingroup$ I had a bad choice of words; you can calculate the outputs, but what are the targets for those inputs? $\endgroup$ – Francesco Gramano Apr 13 '14 at 1:57
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    $\begingroup$ I think that 1 means ok is in class A. The exercise is not clear at all $\endgroup$ – Usi Usi Apr 13 '14 at 10:58
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    $\begingroup$ No, that's not what it means. Classification is correct if your target matches your output, but you haven't specified targets for the inputs in A. $\endgroup$ – Francesco Gramano Apr 14 '14 at 1:58
  • $\begingroup$ @FrancescoGramano can one choose target values for examples in A to be 1 (and target values for examples in B = 0)? (or vice versa) $\endgroup$ – user13107 Sep 11 '17 at 9:23

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