How does Haskell evaluate this function which undoes list intercalation?

Question

I'm trying to understand how Haskell evalutes sep [1, 2, 3, 4, 5] to get ([1, 3], [2, 4, 5]) where:

sep [ ] = ([ ], [ ])
sep [x] = ([ ], [x])
sep (x1:x2:xs) = let (is, ps) = sep xs in (x1:is, x2:ps)

I start like this:

sep [1, 2, 3, 4, 5] = let (is, ps) = sep [3, 4, 5] in (1:is, 2:ps)

but then?

next step is to match `sep [3, 4, 5]` with `sep (x1:x2:xs) = ...` — ymonad, Apr 22 '14 at 23:45
Ok, so. 1) sep [1, 2, 3, 4, 5] = let (is, ps) = sep [3, 4, 5] in (1:is, 2:ps) 2) sep [3, 4, 5] = let (is, ps) = sep [5] in (3:is, 4:ps) 3) sep [5] = ([], [5]) In 2) sep [3, 4, 5] = let (is, ps) = ([], [5]) in (3:is, 4:ps). Any idea in how to continue? — Fof, Apr 22 '14 at 23:58

score 4 · Accepted Answer · answered Apr 23 '14 at 00:15

Finally I understood.

1) sep [1, 2, 3, 4, 5] = let (is, ps) = sep [3, 4, 5] in (1:is, 2:ps)

2) sep [3, 4, 5] = let (is, ps) = sep [5] in (3:is, 4:ps)

3) sep [5] = ([], [5])

In 2) sep [3, 4, 5] = let (is, ps) = ([], [5]) in (3:is, 4:ps) = ([3], [4, 5])

In 1) sep [1, 2, 3, 4, 5] = let (is, ps) = ([3], [4, 5]) in (1:is, 2:ps) = ([1, 3], [2, 4, 5])

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