Is the XOR problem solvable with a 2x2x1 neural network without bias?

Question

Is a neural network with two input nodes, two hidden nodes and an output that should be able to solve the XOR problem provided there is no bias? Or can he get stuck?

Answer 1

Leave the prejudice. He does not see the meaning of your inputs.

In terms of a one-to-one analogy, I like to think of displacement as the displacement of a c value in a straightforward equation: y = mx + c ; it adds an independent degree of freedom to your system that is not affected by the entrances to your network.

Answer 2

If I remember correctly, it is impossible to have XOR without bias.

Answer 3

I built a neural network without prejudice, and 2x2x1 architecture solves XOR in 280 eras. New to this, so I didn’t know anyway, but it works, so it’s possible.

Hi,

Answer 4

Yes, you can use an activation function such as Relu (f (x) = max (0, x))

An example of the weight of such a network:

 Layer1: [[-1, 1], [1, -1]] Layer2: [[1], [1]] 

For the first (hidden) layer:

• If the input is [0,0], both nodes will have activation 0: ReLU (-1 * 0 + 1 * 0) = 0, ReLU (1 * 0 + -1 * 0) = 0
• If the input is [1,0], one node will have activation 0 ReLU (-1 * 1 + 1 * 0) = 0, and the other activation 1 ReLU (1 * 1 + -1 * 0) = 1
• If the input is [0,1], one node will have activation 1 ReLu (-1 * 0 + 1 * 1) = 1, and the other activation 0 ReLU (1 * 0 + -1 * 1) = 0
• If the input is [1,1], both nodes will have activation 0: ReLU (-1 * 1 + 1 * 1 = 0) = 0, ReLU (1 * 1 + -1 * 1 = 0) = 0

For the second (output) level: Since the scales [[1], [1]] (and due to ReLU there can be no negative activations from the previous layer), the layer simply acts as a summation of the activations in layer 1

• If the input is [0,0], the sum of activations in the previous layer is 0
• If the input is [1,0], the sum of activations in the previous layer is 1
• If the input is [0,1], the sum of activations on the previous layer is 1
• If the input is [1,1], the sum of activations in the previous layer is 0

Although this method by coincidence works in the above example, it is limited to using the zero mark (0) for false examples of the XOR problem. If, for example, we used for False examples and deuces for True examples, this approach no longer worked.