is this what you want? not sure if i've understood, and not sure it's efficient. andrew grow :: Int -> [[Int]] -> [[Int]] grow n (x:xs) = x:(grow n (xs ++ [x++[i] | i <- [1..n]])) seqn :: Int -> [[Int]] seqn n = grow n [[i] | i <- [1..n]] main :: IO () main = do print $ take 100 (seqn 3) Ron de Bruijn said:
Hi there,
I have written this little function:
f :: (Num a, Enum a) => a -> [[a]] f n = [[a]|a<-fu n] ++ [a:[b]|a<-fu n,b<-fu n] ++ [a:b:[c]|a<-fu n,b<-fu n,c<-fu n] fu n = [1..n]
This is an example of the function in action:
*Mod> f 4 [[1],[2],[3],[4],[1,1],[1,2],[1,3],[1,4],[2,1],[2,2],[2,3],[2,4],[3,1],[3,2],[3, 3],[3,4],[4,1],[4,2],[4,3],[4,4],[1,1,1],[1,1,2],[1,1,3],[1,1,4],[1,2,1],[1,2,2] ,[1,2,3],[1,2,4],[1,3,1],[1,3,2],[1,3,3],[1,3,4],[1,4,1],[1,4,2],[1,4,3],[1,4,4] ,[2,1,1],[2,1,2],[2,1,3],[2,1,4],[2,2,1],[2,2,2],[2,2,3],[2,2,4],[2,3,1],[2,3,2] ,[2,3,3],[2,3,4],[2,4,1],[2,4,2],[2,4,3],[2,4,4],[3,1,1],[3,1,2],[3,1,3],[3,1,4] ,[3,2,1],[3,2,2],[3,2,3],[3,2,4],[3,3,1],[3,3,2],[3,3,3],[3,3,4],[3,4,1],[3,4,2] ,[3,4,3],[3,4,4],[4,1,1],[4,1,2],[4,1,3],[4,1,4],[4,2,1],[4,2,2],[4,2,3],[4,2,4] ,[4,3,1],[4,3,2],[4,3,3],[4,3,4],[4,4,1],[4,4,2],[4,4,3],[4,4,4]] *Mod>
This is ofcourse nice and all, but I need a function that does this same trick, but then doesn't only generate listelements of the maximum hard-coded length of three, but to length m, where m is an extra parameter. It's important that the elements are precisely in this order.
I tried rewriting the listcomprehension to map, concat and filter, but that only complicated things.
There is a possibility where I can write a function that gives the next of a list. So the next of [1,1,1] would be [1,1,2], but that would be somewhat inefficient when the list becomes large. Then I just call this function with (replicate n (head $ fu n)) and (last $ fu n). Then I just apply this function next to the rest, until I reach a state where all elements of the list equal the last value of the inputlist, so I have the required list of length n. Then to simply get the complete list as in function f. I need to map(\x->otherFunction x) [1..], but this "next"-function is almost identical to what the built-in listcomprehension does. I just don't like my solution.
Then I can probably use Template Haskell(which I have never used), but that seems overkill.
So does anyone has a better solution?
Greets Ron
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