Plagiarising Raul and Aai's brilliant insights (and presumably RE Boss's),
as I got nowhere near those efficiencies, it seems you can get a slight
improvement in time and space using a boolean array until termination:
1 ts'#(>:# ((1 0$~#)))^:(40) 1 '
1.62761 4.69765e8
1 ts'(# 2 1$~#)^:(40) 2 ' NB. slower laptop here!
1.84354 8.05309e8
Mike
On 15/01/2015 14:07, Aai wrote:
Probably identical to what you have:
(# 2 1$~#)^:(40) 2
R.E. Boss schreef op 15-01-15 om 13:03:
K=. 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1 2 1
2 2 1
2 2 1 1 2 1 2 2 1 2 1 1 2 1 1 2 2 1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2
1 2 2
1 2 1 1 2 2 1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 1 2 1 2 2 1 2
2 1 1
is the start of a sequence where each i-th number gives the length of
the
i-th substring of equal numbers.
Or in J
(}:#;.(2)~ 2~:/\]) K
2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1 2 1 2 2
1 2 2
1 1 2 1 2 2 1 2 1 1 2 1 1 2 2 1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 1 2
(Partial) equality from
((]-: ({.~ #)) }:#;.(2)~ 2~:/\]) K
1
I do have a way to generate them (not by hand), but what are other ways?
Elegant and lean svp.
The first 21717701 items I generated in
ts 'genREB 2'
0.9922127 8.0530995e8
R.E. Boss
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