Evolution of Simplicity for the xor problem :D (of my AIPuzzle app, level "Decision 3.0!")

The first try is three or one, not two,
The second is two xor gates for two inputs combined,
The third try is one, blocking two, unblocking three,
The 4th one uses a smaller weight for the blocking two, so an unblocking three is included
The 5th and final and probably simplest one possible is the same strategy, just reusing the output node, because the result is either 0 or 1 anyways (by the calculation, weight 1 for all white, weight -2 for the red one)


7363601s ago, by Antonio

and I'll maybe remove the black dot in future levels… because it's completely useless (it just looks nice XD)