Python 3: Generate not all permutations, but all non-repetitive combinations of length r?

Question

I'm trying to generate a lazily iterable collection of Vigenere cipher keys of length r. I'm aware of itertools and the permutations() method. However, this generates keys such as ABCD, ABCE, ABCF... but it will never do something like AABC.

So basically, I need tuples or strings of characters that aren't repetitive (that is, a repetitive key can be cut in half to get two identical halves), but can contain duplicate characters. Good example: AABABA, not AABAAB.

How can I create such a collection that won't generate keys like this, and is lazily iterated so I don't blow up my RAM when I want to explore keys longer than 3 characters?

You mean "*can't* cut the key in half to get two identical halves"? — Edwin, Jan 18 '12 at 00:24
Maybe it sounds ambiguous but I mean to say that "repetitive" means that you can do such a thing. — 2rs2ts, Jan 18 '12 at 00:25
It's not ambiguous, it's contradictory. First you say not repetitive (so `AABAAB` won't work), then you say you can cut the key in half and get two identical halves (so `AABAAB` *would* work). Your example supports the former. Just asking for clarification. — Edwin, Jan 18 '12 at 00:29
I think the OP means that `repetitive == can cut into equal halves` and that strings should not be repetitive, as the examples suggest. — Katriel, Jan 18 '12 at 00:35

Katriel · Accepted Answer · 2012-01-18T00:37:18.550

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("".join(s) for s in product(alphabet, repeat=n) if s[:n//2]!=s[n//2:])

_{EDIT: fixed up thanks to @PetrViktorin}

edited Jan 18 '12 at 00:37

answered Jan 18 '12 at 00:29

Katriel

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You can use `n` instead or `len(s)`. Also, use the `//` operator for integer division. – Petr Viktorin Jan 18 '12 at 00:35
I get a `TypeError: 'int' object is not iterable`. – 2rs2ts Jan 18 '12 at 01:05
Is `alphabet` what you think it is? It should be an iterable e.g. a string or a list of characters. – Katriel Jan 18 '12 at 08:27
I was using `string.ascii_uppercase` which I am pretty sure is an iterable. – 2rs2ts Jan 25 '12 at 13:12
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It works for me. One of your names is not defined right: either `alphabet` is not an iterable or `product` is not `itertools.product`. – Katriel Jan 25 '12 at 23:06

score 2 · Answer 2 · answered Jan 18 '12 at 00:25

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It sounds like you want to use itertools.combinations_with_replacement(). On top of that, you can write a generator around that to filter out the ones you don't want.

http://docs.python.org/library/itertools.html#itertools.combinations_with_replacement

answered Jan 18 '12 at 00:25

krousey

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+1. Very nice, never needed that although I use `itertools` frequently. Seems to be exactly what @agarett wants. – ChristopheD Jan 18 '12 at 00:28
I see how it creates all the combinations but I don't know how to write such a generator. While iterating through keys I could just check to see if the first half equals the second half, but if that could be magically avoided I would prefer that. – 2rs2ts Jan 18 '12 at 00:38

Python 3: Generate not all permutations, but all non-repetitive combinations of length r?

2 Answers2