I would like to make a method where I could give it a list of lengths and it would return all combinations of cartesian coordinates up to those lengths. Easier to explain with an example:
cart [2,5]
Prelude> [ [0,0],[0,1],[0,2],[0,3],[0,4],[1,0],[1,1],[1,2],[1,3],[1,4] ]
cart [2,2,2]
Prelude> [ [0,0,0],[0,0,1],[0,1,0],[0,1,1],[1,0,0],[1,0,1],[1,1,0],[1,1,1] ]
A simple list comprehension won't work because I don't know how long the lists are going to be. While I love Haskell's simplicity for many problems, this is one that I could write procedurally (in C or something) in 5 minutes whereas Haskell gives me an aneurysm!
A solution to this specific problem would help me out a lot; I'd also love to hear about your thought processes when tackling stuff like this.
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This can be solved recursively. First, the Cartesian product of nothing is {∅}:
cart [] = [[]](Or define just the 1-element form if the empty product is invalid:
cart [x] = [[i] | i <- [0 .. x-1]])
Then, the Cartesian product of
x:xscan be written ascart (x:xs) = [i:rest | i <- [0 .. x-1], rest <- cart xs]
In general, if you what to write a function f that requires the list's length N, try to think of a way to make f(N) depends on smaller lists e.g. f(N - 1) only, then solve the base case f(0) or f(1) etc. This transforms the problem into a recursion that can be easily solved.
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I bet your procedural solution would involve recursion. Our Haskell solution will involve recursion too.
So, recursion. First the recursive case.
cart (c : cs) = [i : r | i <- [0 .. c-1], r <- rcart] where rcart = cart csHere we're just saying that for each possible initial co-ordinate, and each possible combination of cartesian co-ordinates from the remaining lengths, we do the obvious thing of combining the co-ordinate with the remaining co-ordinates.
Then the base case.
cart [] = [[]]You might think
cart [] = []. I did at first. But think about what the recursive case requires from the base case. -
Umm..
cart = sequence . map (enumFromTo 0 . subtract 1)It's reasonable to expect that a
[[a]] -> [[a]]function doing what we expect already exists in the library. So if one is not familiar withsequence, finding it is a simple matter of hoogling it.Edit: as newacct pointed out:
cart = mapM (enumFromTo 0 . subtract 1)This can also be found by feeding the previous solution to HLint.
newacct : `cart = mapM (enumFromTo 0 . subtract 1)`yairchu : @newacct: nice. btw hlint suggests this as well :)Reid Barton : Wow, it's funny how ingrained `sequence . map (...)` is as the answer to this kind of question, since it was my initial reaction too.
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