Could random.randint(1,10) ever return 11?

When researching for this question and reading the sourcecode in random.py, I started wondering whether randrange and randint really behave as "advertised". I am very much inclined to believe so, but the way I read it, randrange is essentially implemented as

start + int(random.random()*(stop-start))

(assuming integer values for start and stop), so randrange(1, 10) should return a random number between 1 and 9.

randint(start, stop) is calling randrange(start, stop+1), thereby returning a number between 1 and 10.

My question is now:

If random() were ever to return 1.0, then randint(1,10) would return 11, wouldn't it?

回答1:

From random.py and the docs:

"""Get the next random number in the range [0.0, 1.0)."""

The ) indicates that the interval is exclusive 1.0. That is, it will never return 1.0.

This is a general convention in mathematics, [ and ] is inclusive, while ( and ) is exclusive, and the two types of parenthesis can be mixed as (a, b] or [a, b). Have a look at wikipedia: Interval (mathematics) for a formal explanation.

回答2:

Other answers have pointed out that the result of random() is always strictly less than 1.0; however, that's only half the story.

If you're computing randrange(n) as int(random() * n), you also need to know that for any Python float x satisfying 0.0 <= x < 1.0, and any positive integer n, it's true that 0.0 <= x * n < n, so that int(x * n) is strictly less than n.

There are two things that could go wrong here: first, when we compute x * n, n is implicitly converted to a float. For large enough n, that conversion might alter the value. But if you look at the Python source, you'll see that it only uses the int(random() * n) method for n smaller than 2**53 (here and below I'm assuming that the platform uses IEEE 754 doubles), which is the range where the conversion of n to a float is guaranteed not to lose information (because n can be represented exactly as a float).

The second thing that could go wrong is that the result of the multiplication x * n (which is now being performed as a product of floats, remember) probably won't be exactly representable, so there will be some rounding involved. If x is close enough to 1.0, it's conceivable that the rounding will round the result up to n itself.

To see that this can't happen, we only need to consider the largest possible value for x, which is (on almost all machines that Python runs on) 1 - 2**-53. So we need to show that (1 - 2**-53) * n < n for our positive integer n, since it'll always be true that random() * n <= (1 - 2**-53) * n.

Proof (Sketch) Let k be the unique integer k such that 2**(k-1) < n <= 2**k. Then the next float down from n is n - 2**(k-53). We need to show that n*(1-2**53) (i.e., the actual, unrounded, value of the product) is closer to n - 2**(k-53) than to n, so that it'll always be rounded down. But a little arithmetic shows that the distance from n*(1-2**-53) to n is 2**-53 * n, while the distance from n*(1-2**-53) to n - 2**(k-53) is (2**k - n) * 2**-53. But 2**k - n < n (because we chose k so that 2**(k-1) < n), so the product is closer to n - 2**(k-53), so it will get rounded down (assuming, that is, that the platform is doing some form of round-to-nearest).

So we're safe. Phew!

Addendum (2015-07-04): The above assumes IEEE 754 binary64 arithmetic, with round-ties-to-even rounding mode. On many machines, that assumption is fairly safe. However, on x86 machines that use the x87 FPU for floating-point (for example, various flavours of 32-bit Linux), there's a possibility of double rounding in the multiplication, and that makes it possible for random() * n to round up to n in the case where random() returns the largest possible value. The smallest such n for which this can happen is n = 2049. See the discussion at http://bugs.python.org/issue24546 for more.

回答3:

From Python documentation: