The first mistake. For AND, the example (1, 1) should give 1. The score is $0 + 0 + 0 = 0$ , not above zero, so the output is 0: wrong. The rule adds the inputs, giving $w_1 = 1$ , $w_2 = 1$ , and adds 1 to the bias, $b = 1$ .
The correction overshoots. On the next pass, (0, 0) now scores $b = 1 > 0$ , output 1, but should be 0. Subtract: the weights stay at 1 and 1 (the inputs were 0) and $b$ drops to 0.
It settles. Going round the four examples again and again, it makes its last correction on the fifth pass and ends at $w_1 = 2$ , $w_2 = 1$ , $b = -2$ . Check: (1, 1) scores $2 + 1 - 2 = 1$ , output 1; (1, 0) scores 0, (0, 1) scores $-1$ and (0, 0) scores $-2$ , all output 0. All four right.
XOR has no answer. XOR needs (0, 0) to score at most 0, so $b \le 0$ ; (1, 0) and (0, 1) to score above 0, so $w_1 + b > 0$ and $w_2 + b > 0$ ; and (1, 1) at most 0, so $w_1 + w_2 + b \le 0$ . Adding the middle two gives $w_1 + w_2 + 2b > 0$ , so $w_1 + w_2 + b > -b \ge 0$ , which contradicts the last condition. No weights work.
A hidden layer solves it. Use two neurons: one with score $x_1 + x_2 - 0.5$ (OR) and one with $x_1 + x_2 - 1.5$ (AND). Feed their outputs $h_1$ , $h_2$ to a third with score $h_1 - h_2 - 0.5$ . For (1, 0): $h_1 = 1$ , $h_2 = 0$ , score 0.5, output 1. For (1, 1): both are 1, score $-0.5$ , output 0. That is XOR.