Peter Otten <[EMAIL PROTECTED]> wrote:
> I guess I have to move the goal posts to beat you:
>
>>>> set([-1,-2]), set([-2,-1])
> (set([-2, -1]), set([-1, -2]))
>
> For that one the number of slots doesn't matter because
>
>>>> hash(-1), hash(-2)
> (-2, -2)
>
Neat.
>>> last = []
>>> for i in range(0,10000,5):
pair = list(set(range(-2,i)))[-2:]
if pair != last:
print "%4d(%4d): %s" % (i, i*3/2, pair)
last = pair
0( 0): [-1, -2]
5( 7): [-2, -1]
20( 30): [-1, -2]
85( 127): [-2, -1]
340( 510): [-1, -2]
1365(2047): [-2, -1]
5460(8190): [-1, -2]
The order of those two values reverses each time the hash table is
resized, but they will always appear at the end of a set of consecutive
integers from -2 upwards. (Highly implementation dependant of course.)
I think this demonstrates quite nicely how the hash table always
contains some power of 2 slots and is resized whenever it is about
2/3rds full. Also it shows that even when you construct a set from a
list, the set is grown one element at a time rather than pre-allocated
(otherwise you wouldn't get the alternating order).
--
Duncan Booth http://kupuguy.blogspot.com
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