I used modulo outer product |/ found which ones equal zero and then anded the
rows of the matrix together to find the numbers that have both 3 and 5 as
divisors
I. (*./ (0 = 3 5 |/i.1000))
0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 . . .
to generalize into a tacit I cheat and use 13 : '' for automated tacit
geneuler1 =: 13 : 'I. (*./ (0 = x |/i.y))'
which produces the following tacit form:
geneuler1
[: I. [: *./ 0 = [ |/ [: i. ]
3 5 geneuler1 1000
0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 . . .
2 6 geneuler1 1000
0 6 12 18 24 30 36 42 48 54 60 66 72 . . .
3 5 7 geneuler1 1000
0 105 210 315 420 525 630 735 840 945
That's the fun of these problems in J there are so many ways to do them. I
didn't do a manual check on these to ensure they are right they look like
correct answers at a glance
Tom McGuire
On May 4, 2016, at 10:35 PM, Geoff Canyon <gcan...@gmail.com> wrote:
> So I tried to write code to solve the general case of Project Euler problem
> 1. The problem given is to find the sum of all the positive integers less
> than 1000 that are divisible by 3 or 5. Obviously the specific case is
> highly optimizable. But I wanted to solve the general, with any number of
> divisors and any upper limit.
>
> So I started by enumerating the integers from 0 to the limit, and finding
> the remainder for each list when divided by each of the divisors.
>
> Then I transpose and multiply, so the list is now 0 for each integer
> divisible by any of the divisors.
>
> Then I search for 0 in the list, so I end up with 1 for numbers that are
> divisible by any of the divisors, and 2 for numbers that aren't.
>
> Then I subtract that list from 1, so divisible numbers are 1, and
> non-divisible numbers are 0.
>
> Then I enumerate from 0 to the limit again, and multiply the two lists, so
> I have a list with 0 for the numbers that are non-divisible, and the
> numbers when they are divisible.
>
> Then I sum that list, and that's the result.
>
> Here's the code. As always, I suck at J, so improvements/suggestions are
> welcome.
>
> pe1 =: +/@(([:i.]) * 1&-@(0&i.)@*/"1@|:@(|"0 1 i.))
>
>
> 1 pe1 10 NB. sum of 1..9
>
> 45
>
>
> 2 pe1 101 NB. sum of the first 50 even numbers
>
> 5050
>
>
> 2 3 5 7 11 13 17 19 23 29 pe1 1000 NB. difficult to do the "optimized" way
>
> 305863
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