how to do an enumeration in haskell

Question

I have some chars, for example ["ab"], and I just have two number, zero and one. And I want to get an answer like

[[('a', 1), ('b', 1)],
 [('a', 0), ('b', 1)],
 [('a', 1), ('b', 0)],
 [('a', 0), ('b', 0)]]

I am a beginner, could someone help me?

Thankyou very much.

possible duplicate of [Cartesian product](http://stackoverflow.com/questions/4119730/cartesian-product) — Daniel Wagner, May 07 '12 at 06:32
Well, I don't thing so, he's looking for a special order and pairing, as it seems. — phipsgabler, May 07 '12 at 06:35
@phg This is not actual homework, and as a beginner, it is difficult for me. — lpy, May 07 '12 at 06:39
@Laurent no problem. That was a reply to Daniel, since you're actually not searching for a cartesian product, like he thought first. — phipsgabler, May 07 '12 at 06:42

score 4 · Accepted Answer · answered May 07 '12 at 06:37

4

The most interesting part is generating the list of all possible sequences of 0 and 1; one way is via replicateM:

> replicateM 2 [1,0]
[[1,1],[1,0],[0,1],[0,0]]

You can use map and zip from there:

> map (zip "ab") (replicateM 2 [1,0])
[[('a',1),('b',1)],[('a',1),('b',0)],[('a',0),('b',1)],[('a',0),('b',0)]]

To match your answer exactly, you'd have to do a tiny bit of extra munging:

> map (zip "ab" . reverse) (replicateM 2 [1,0])
[[('a',1),('b',1)],[('a',0),('b',1)],[('a',1),('b',0)],[('a',0),('b',0)]]

answered May 07 '12 at 06:37

Daniel Wagner

Thankyou, but I still can't understand, what if it is "abcd"? Does it still work? – lpy May 07 '12 at 06:42
1

@Laurent You may use [Hoogle](http://www.haskell.org/hoogle/) to find functions by name or type. If you want to match up with different strings, you will need some minor changes: the `2` argument to `replicateM` should be the length of the string. – Daniel Wagner May 07 '12 at 06:57
Thankyou all of you! I will remember this website, it works well, thankyou for your help. – lpy May 07 '12 at 07:09

1 Answers1