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First interval size is less than the following #93

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If we have a population of 10, the first interval will be ]0, 0.1[ (size: 0.1). The second one will be ]0.1, 0.3[ (size: 0.2), greater than the first, the third one will be ]0.3, 0.48[ (size: 0.18)...

weigth = 1/10 = 0.1

i = 0
2*weight * ((pop.length-i)/pop.length) = 2 * 0.1 * ((10-0)/10) = 2 * 0.1 * 10/10 = 0.2

i = 1
2*weight * ((pop.length-i)/pop.length) = 2 * 0.1 * ((10-1)/10) = 2 * 0.1 * 9/10 = 0.18

So the first individual has less chances to be selected than the following.

And Math.random() returns a float in the [0, 1[ interval, so 0 can be returned but will never be picked because of the condition:

rand > lower && rand < upper

So we need to replace it by:

rand >= lower && rand < upper

If we have a population of 10, the first interval will be ]0, 0.1[ (size: 0.1). The second one will be ]0.1, 0.3[ (size: 0.2), greater than the first, the third one will be ]0.3, 0.48[ (size: 0.18)...

weigth = 1/10 = 0.1

i = 0
2*weight * ((pop.length-i)/pop.length) = 2 * 0.1 * ((10-0)/10) = 2 * 0.1 * 10/10 = 0.2

i = 1
2*weight * ((pop.length-i)/pop.length) = 2 * 0.1 * ((10-1)/10) = 2 * 0.1 * 9/10 = 0.18

So the first individual has less chances to be selected than the following.

And Math.random() returns a float in the [0, 1[ interval, so 0 can be returned but will never be picked because of the condition:

rand > lower && rand < upper

So we need to replace it by:

rand >= lower && rand < upper
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