In python the function random() generates a random float uniformly in the semi-open range [0.0, 1.0). In principle can it ever generate 0.0 (i.e. zero) and 1.0 (i.e. unity)? What is the scenario in practicality?
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3Not 1.0 surely. Apart from the fact that the range does not include 1.0, there also does not exist a float 1.0 (it can't be represented binary).extraneon– extraneon2010-04-12 09:48:18 +00:00Commented Apr 12, 2010 at 9:48
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9@extraneon - what? Of course there's a float 1.0. It's '0011111111110000000000000000000000000000000000000000000000000000' (as a double, but the float value is basically the same). The only numbers that don't have an exact float representations are decimals that don't have exact representations with a power-of-two denominator. en.wikipedia.org/wiki/…Daniel G– Daniel G2010-04-12 09:57:17 +00:00Commented Apr 12, 2010 at 9:57
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2 Answers
0.0 can be generated; 1.0 cannot (since it isn't within the range, hence the ) as opposed to [).
The probability of generating 0.0 is equal to the probability of generating any other number within that range, namely, 1/X where X is the number of different possible results. For a standard unsigned double-precision floating point, this usually means 53 bits of fractional component, for 2^53 possible combinations, leading to a 1/(2^53) chance of generating exactly 0.0.
So while it's possible for it to return exactly 0.0, it's unlikely that you'll see it any time soon - but it's just as unlikely that you'd see exactly any other particular value you might choose in advance.
7 Comments
Mark Dickinson
@Dav: Just a nitpick for what's already a nice answer, but I think you mean 53, not 52. With a perfect random source (which Mersenne Twister of course is not, but it's good enough for practical non-cryptographic purposes), random() generates a number of the form n/2^53 with n an integer uniformly and randomly chosen in the range [0, 2^53). If you look at the source, it takes two 32-bit words generated by the MT algorithm and combines the top 27 bits of the first with the top 26 bits of the second to create n. Incidentally, this means that most floats in [0.0, 1.0) can't ever be generated.
Amber
Ah, right - probably because I was thinking of signed floating point, and random() probably uses an unsigned format given its range. I'll change it.
Arkapravo
@Mark Dickinson : 2^52 is a troublesome number in MATLAB ! ..... does it also employ Mersenne Twister ?
Mark Dickinson
@Arkapravo: It looks like it, at least for not-too-old versions. The release notes for MATLAB 7.4 (released March 2007) say: "rand Function Uses the Mersenne Twister Algorithm as Default".
Arkapravo
@Mark Dickinson: I understand that Mersenne Twister is a significant improvement on Wichmann-Hill generator to generate random numbers ! ... anything more newer than the Twister ?
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The [ indicates that 0.0 is included in the range of valid outputs. The ) indicates 1.0 is not in the range of valid outputs.
