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Naive implementation of cartesian_power. #486
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For
pow==0
/Power::Empty
, shouldn't the iterator should yield a single element (namely the empty vector)? (Instead of yielding no elements at all.)This would ensure that
it.cartesian_power(n)
yieldsit.count().pow(n)
elements. (We had a similar case back then where (I think it was) combinations had a special case forn==0
, which lead to inconsitencies.)It would possibly also help in getting rid of some special casings (i.e. possible eliminate
Power
).There was a problem hiding this comment.
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@phimuemue Well, I had this vague feeling I was neglicting something sneaky here. Thank you for pointing it out.
The sneaky stuff is that
cartesian_product("abc", 0)
should not yield the same ascartesian_product("", 3)
. And I bet that your criterionit.count().pow(n)
is the right thing to consider. If we agree that:Then the adaptor should yield:
["aa", "ab", "ba", "bb"]
["a", "b"]
[""]
["aa"]
["a"]
[""]
[]
[]
[""]
?So there is not 1 degenerated case, but 2 or 3, depending on the convention we pick for the
0^0
situation. I propose we stick to the "combinatorics" convention that0^0=1
, so there are only 2 corner cases. The only other solution I can think of is erroring or panicking instead when encounteringcartesian_product("", 0)
.I'll update the PR to better accomodate these degenerated situations :)
[from the future] see 9027ef0