I'm working on Project Euler, I'm on problem 8, and I'm trying a simple brute force: Multiply each consecutive 5 digit of the number, make a list with the results, and find the higher.
This is the code I'm currently trying to write in J:
n =: 731671765313x
NB. 'n' will be the complete 1000-digits number
itl =: (".@;"0@":)
NB. 'itl' transform an integer in a list of his digit
N =: itl n
NB. just for short writing
takeFive =: 5 {. ] }.~ 1 -~ [
NB. this is a dyad, I get this code thanks to '13 : '5{.(x-1)}.y'
NB. that take a starting index and it's applied to a list
How I can use takeFive for all the index of N? I tried:
(i.#N) takeFive N
|length error: takeFive
| (i.#N) takeFive N
but it doesn't work and I don't know why. Thank you all.
1. The reason that (i.#N) takeFive N
is not working is that you are essentially trying to run 5{. ((i.#N)-1) }. N
but you have to use x
not as a list but as an atom. You can do that by setting the appropriate left-right rank "
of the verb:
(i.#N) (takeFive"0 _) N
7 3 1 6 7
7 3 1 6 7
3 1 6 7 1
1 6 7 1 7
6 7 1 7 6
7 1 7 6 5
1 7 6 5 3
7 6 5 3 1
6 5 3 1 3
5 3 1 3 0
3 1 3 0 0
1 3 0 0 0
2. One other way is to bind (&
) your list (N
) to takeFive
and then run the binded-verb through every i.#N
. To do this, it's better to use the reverse version of takeFive: takeFive~
:
((N&(takeFive~))"0) i.#N
7 3 1 6 7
7 3 1 6 7
3 1 6 7 1
1 6 7 1 7
6 7 1 7 6
7 1 7 6 5
1 7 6 5 3
7 6 5 3 1
6 5 3 1 3
5 3 1 3 0
3 1 3 0 0
1 3 0 0 0
or (N&(takeFive~)) each i.#N
.
3. I think, though, that the infix dyad \
might serve you better:
5 >\N
7 3 1 6 7
3 1 6 7 1
1 6 7 1 7
6 7 1 7 6
7 1 7 6 5
1 7 6 5 3
7 6 5 3 1
6 5 3 1 3