Random Boolean array with 10 elements where 3 out of 10 are True

Question

I have been searching all over the forum and google but I think I am using the wrong key words or I do not understand the solution. What I am looking for is to create a list or array with 10 random Boolean elements where I can decide that 1 or 2 or 3 or 4 etc. elements are True. The order must be random, in the end I want a result like this e.g. (a 0.3 chance):

False
True
False
False
True
False
False
False
True
False

Thnx.

You can use the code below. I have not run it but should work. Please import System.Random. boolean[] randomBooleanArray(int len) { boolean[] arr = new boolean[len]; Random rand = new Random(); for(int i = 0; i < len; i++) { arr[i] = rand.NextDouble() > 0.5; } return arr; } — Egalitarian, Jul 06 '17 at 18:58
@Egalitarian this does not ensure that `true` appears only three times. — Max Play, Jul 06 '17 at 19:21

score 5 · Accepted Answer · answered Jul 06 '17 at 18:51

5

Why not create them in a know order (true, true, true, false, ....) and then use a good algorithm to shuffle them?

An option is the good old Fisher-Yates shuffle

answered Jul 06 '17 at 18:51

oerkelens

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The ask is different. – Egalitarian Jul 06 '17 at 18:59
2

@Egalitarian How is the question different from what I suggest? The result is exactly what the OP asks: a randomised list with N `true` values. I just suggest taking two steps in creating it. – oerkelens Jul 06 '17 at 19:01

Random Boolean array with 10 elements where 3 out of 10 are True

1 Answers1