Here is a different approach for the Project Euler #1 solution:
+/~.(3*i.>.1000%3),5*i.>.1000%5
How to refactor it?
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Did you check project euler forum for problem #1? http://projecteuler.net/index.php?section=forum&id=1 – Macarse Oct 12 '09 at 23:57
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I got this code from there. But the forum is locked. – Jader Dias Oct 13 '09 at 00:04
3 Answers
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+/(#~ ( (0= 3| ]) +. (0 = 5 |]) )) 1+i.999
0 = ( 3 | ]) uses (twice) the trick of verb train (fork) with n u v (discussed at the end of http://www.jsoftware.com/help/learning/09.htm)
A different way of writing it:
+/(#~ ( ((0&=) @ (3&|)) +. ((0&=) @ (5&|)))) 1+i.999
Zhe Hu
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[:+/@~.@,3 5([*i.@>.@%~)]
usage example:
f =: [:+/@~.@,3 5([*i.@>.@%~)]
f 1000
or
+/~.,3 5([*i.@>.@%~)1000
%~ = 4 : 'y % x'
i.@>.@%~ = 4 : 'i. >. y % x'
[*i.@>.@%~ = 4 : 'x * i. >. y % x'
3 5([*i.@>.@%~)] = 3 : '3 5 * i. >. y % 3 5'
[:+/@~.@,3 5([*i.@>.@%~)] = 3 : '+/ ~. , 3 5 * i. >. y % 3 5'
ephemient
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this is legible for you? I am still trying to figure out each step of the refactoring... – Jader Dias Oct 23 '09 at 22:07
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Here is another approach, using a simple, generic verb
multiplesbelow =: 4 : 'I. 0 = x | i.y'
+/ ~. ,3 5 multiplesbelow"0 [ 1000
MPelletier
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1Nice! How about `multiplesBelow =: 4 : '(#~ +./(0 = x | ])"0) i. y'`? Then you can say `+/ 3 5 multiplesBelow 1000`. – Gregory Higley Jun 29 '10 at 18:52