Re: Enumerating all 3-tuples

[bartc]

On 10/03/2018 20:06, Ben Bacarisse wrote:
bartc <b...@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.use...@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).

OK. I ran it like this:

 main = print (take 20 (pn 3 [1..]))

But I'm trying to understand the patterns in the sequence. If I use:

 main = print (take 50 (pn 2 [1..]))

then group the results into sets of 1, 2, 3, etc pairs, showing each 
group on a new line, then this gives sequences which are equivalent to 
the diagonals of the OP's 2D grid. (Except they don't alternate in 
direction; is that what you mean?)

I'll have to study the pn 3 version some more. (pn 10 gives an 
interesting view of it too.)

--
bartc
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