Enumerating all 3-tuples

[Ben Bacarisse]

bartc <bc at freeuk.com> writes:

> [repost as original seems to have gone to email; my newsreader has
> somehow acquired a 'Reply' button where 'Followup' normally goes.]

[I thought it was intended but my reply bounced.]

> On 10/03/2018 14:23, Ben Bacarisse wrote:
>> Ben Bacarisse <ben.usenet at bsb.me.uk> writes:
>
>> Off topic: I knocked up this Haskell version as a proof-of-concept:
>>
>>    import Data.List
>>
>>    pn n l = pn' n (map (:[]) l)
>>             where pn' n lists | n == 1 = lists
>>                               | otherwise = diag (pn' (n-1) lists) lists
>>                   diag l1 l2 = zipWith (++) (concat (inits l1))
>>                                             (concat (map reverse (inits l2)))
>>
<snip>
> What's the output? (And what's the input; how do you invoke pn, if
> that's how it's done?)

You pass a number and a list which should probably be infinite like
[1..].  You'd better take only a few of the resulting elements then:

*Main> let triples = pn 3 [1..]
*Main> take 20 triples
[[1,1,1],[1,1,2],[1,2,1],[1,1,3],[1,2,2],[2,1,1],[1,1,4],[1,2,3],[2,1,2],[1,3,1],[1,1,5],[1,2,4],[2,1,3],[1,3,2],[2,2,1],[1,1,6],[1,2,5],[2,1,4],[1,3,3],[2,2,2]]

or you can index the list to look at particular elements:

*Main> triples !! 10000000
[70,6,1628]

but, as I've said, the order of the results is not the usual one (except
for pairs).

-- 
Ben.