Given a random number generator r() which produces a (pseudo-)random double in the interval [0,1] with uniform density, ie p(x) = 1 for 0 <= x <= 1 and p(x) = 0 elsewhere, create a random number generator r(a,b) which generates a double in the interval [a,b] with density p(x) = 1/(b-a) for a <= x <= b.
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bountiful
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possible duplicate of [Generating Uniform Random Deviates within a given range](http://stackoverflow.com/questions/446153/generating-uniform-random-deviates-within-a-given-range) – Paul R Nov 15 '12 at 22:04
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The linked question is only considering integers. – bountiful Nov 16 '12 at 09:17
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Why the elaborate answers? Isn't it as simple as `a + (b - a) * r()`. What am I missing? – Mark Dickinson Nov 16 '12 at 20:02
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I believe (and have tested) that this produces a uniform distribution.
r(a,b) = ((r * b) mod (b-a)) + a
but is there another more obvious way?
Yes:
r(a, b) = r*(b-a) + a
bountiful
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