I have f(x) = rand(1,k), for an integer k >= 2, that generates uniformly a random integer between 1 and k.
How can I use f to compute g = rand(1,n), for an integer n >= k, that generates uniformly a random integer between 1 and n?
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Fakher Mokadem
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How about `rand(1,k) * rand(1,k)`? – goodvibration Sep 15 '20 at 11:25
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1what about math? `(rand(1,k) - 1) / 4 * n` – Alberto Sinigaglia Sep 15 '20 at 11:26
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@Berto99: That would give `0` 80% of the time and `n` in the remaining cases. – goodvibration Sep 15 '20 at 11:27
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1@goodvibration why? from the question f(x) = `random number between 1 and 5`, so with `-1` you go to `[0,4]`, then you normalize and so you go to `[0,1]`, and at the and you multiply by `n`, so you go to `[0,n]` – Alberto Sinigaglia Sep 15 '20 at 11:29
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@goodvibration, that would make it impossible to get *k+1* as output. – trincot Sep 15 '20 at 11:45