4 ms·
Slight variation: import itertools def below(n): return slice(n+1, None) def above(n): return slice(None, n) print([
by jhrmnn 7y ago
Slight variation:
import itertools
def below(n):
return slice(n+1, None)
def above(n):
return slice(None, n)
print([
q for q in itertools.product((True, False), repeat=6)
if q == tuple(
f(i) for i, f in enumerate([
lambda n: all(q[below(n)]), # 1. All of the below.
lambda n: not any(q[below(n)]), # 2. None of the below.
lambda n: all(q[above(n)]), # 3. All of the above.
lambda n: any(q[above(n)]), # 4. One of the above.
lambda n: not any(q[above(n)]), # 5. None of the above.
lambda n: not any(q[above(n)]), # 6. None of the above.
])
)
])
- lovasoa 7y agoI see what you are doing here! Pushing it one step further, one could even do this: https://repl.it/repls/CourteousCreepyAdaware https://repl.it/repls/CourteousCreepyAdaware