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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.

  • Just a minor note: your `itl` could be replaced with a simple use of the inverse of Base (#.) `N =: 10 #.inv n` – Dane Oct 30 '13 at 15:12

1 Answers1

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1. The reason that (i.#N) takeFive N is not working is that you are essentially trying to run 5{. ((i.#N)-1) }. Nbut 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
Eelvex
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