On Jul 22, 2011, at 12:48 PM, 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 liOt 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.

You won't need a guard to do this.

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 ?

I think that this exercise is very artificial. It is very easy to solve if you know how list comprehensions are implemented using map and other library functions. Personally, Roelof, I think you will be better off to forget about list comprehensions for the present and focus on understanding basic functions on lists like map, filter and concat. I'll give you the answer to the question because, I can't think of any hints. Don't look if you want to figure it out yourself.

concat [[(x,y) | y <- [4,5,6]] | x <- [1..3]]

Roelof

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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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