The J video here - https://www.youtube.com/watch?v=OzaIi8BnHIc - walks you
through building the J code for Project Euler problem 1.  It ends with a
variant that leads into Pascal's extremely terse version.


On Tue, Apr 8, 2014 at 11:56 AM, Pascal Jasmin <[email protected]>wrote:

>    3 5 (0=|)"0 1 i.10
> 1 0 0 1 0 0 1 0 0 1
> 1 0 0 0 0 1 0 0 0 0
>
>    3 5 (0  ([: +./ =) |"0 1) i.10
> 1 0 0 1 0 1 1 0 0 1
>
>    3 5 (] #~   0  ([: +./ =) |"0 1) i.10
> 0 3 5 6 9
>
>    +/ 3 5 (] #~  0 ([: +./ =) |"0 1)  i.1000
> 233168
>
>
>
> ----- Original Message -----
> From: Jon Hough <[email protected]>
> To: "[email protected]" <[email protected]>
> Cc:
> Sent: Tuesday, April 8, 2014 11:30:22 AM
> Subject: [Jprogramming] Euler Project Problem 1
>
> Euler Project  #1 is:If we list all the natural numbers below 10 that are
> multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is
> 23. Find the sum of all the multiples of 3 or 5 below 1000.
> In C/Java I could easily do something like:
> int total =0;
> for (int counter = 0; counter < 1000; counter ++ ){if(counter %3 = 0 ||
> counter %5 = 0){    total += counter;}}
> return counter;
>
> The main features here being the if statement and the for-loop.
> Obviously in J things need to be done differently. Here is my first
> attempt:
> div3 =. 0&(="0)@:( 3&|) NB. is residue zerodiv5 =. 0&(="0)@:( 5&|)
> div3or5 =. div3"0 +."0 div5"0
> dummy =. 1&*"0 NB. You'll see why I need this...
> result =.  dummy * div3or5
> NB. final result is:
> +/result i.1000
>
> I used dummy because I wanted to do something like  multiply the value of
> y by "div3or5 y" so only multiples of 3 or 5 would be nonzero.
> Before attempting this problem I assumed it could be done in only one line
> and am pretty unhappy by my long solution. Any help explaining where I have
> gone wrong, or why my attempt at a fork is silly would be appreciated.
> Thanks in advance,
> Jon
>
>
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-- 
Devon McCormick, CFA
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