Hi Roelof, I don't have a book, so I don't know the exercise exactly, but if I'm correct, you want the list [(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)] So the basic idea is to nest one list comprehension into another: You create a list [(x,4), (x,5), (x,6)] by the inner list comprehension with y <- [4..6] as generator. It should be [(x,y) | y <- [4..6]], where x is just free. The whole comprehension you stick as result into an outer comprehension whos generator creates the values for x (ie. 1,2 and 3). This will return the list of lists [[(1,4), (1,5), (1,6)],[(2,4), (2,5), (2,6)], [(3,4), (3,5), (3,6)]] which you can flatten by concat. I hope you manage the solution with the description above. I think it helps you more if I don't write the plain solution as working code. Cheers, Daniel. On Fri, 2011-07-22 at 16:48 +0000, Roelof Wobben wrote:
Hello,
I don't see the answer here. If I brake down the problem.
We have this :
[(x,y) | x <- [1..3], y <- [4,5,6]]
I have to use one generator with two nested list compreshession and make use of concat.
So the x generator is [x | x <- [1..3]] and the Y genarator is [y| y <- [4..6]]
So if I put it together it uses the numbers [1..6] so I could do something like this in pseudo code as guard. If generator smaller or equal 3 then its x else it's a y.
But if I do concat [x,y] then I get [1..6] and that not good. If I use zip [x,y] then I get [ (1,4) (2,5) (3,6)] which is also not good but better.
So im stuck now and im puzzeling the whole afternoon about this ?
Anyone who can give me a hint ?
Roelof
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