Op 9/04/2022 om 02:01 schreef duncan smith:
On 08/04/2022 22:08, Antoon Pardon wrote:
Well my first thought is that a bitset makes it less obvious to calulate
the size of the set or to iterate over its elements. But it is an idea
worth exploring.
def popcount(n):
"""
Returns the number of set bits in n
"""
cnt = 0
while n:
n &= n - 1
cnt += 1
return cnt
and not tested,
def iterinds(n):
"""
Returns a generator of the indices of the set bits of n
"""
i = 0
while n:
if n & 1:
yield i
n = n >> 1
i += 1
Sure but these seem rather naive implementation with a time complexity of
O(n) where n is the maximum number of possible elements. Using these would
turn my O(n) algorithm in a O(n^2) algorithm.
--
Antoon Pardon.
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