Two of the possibilities for finding all divisors
of a number:
] p=: q: 24
2 2 2 3
] b=: #: i.2^#p
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
b */ .(^~) p
1 3 2 6 2 6 4 12 2 6 4 12 4 12 8 24
~. b */ .(^~) p
1 3 2 6 4 12 8 24
__ q: 24
2 3
3 1
*/&> , { (^i.@>:)&.>/__ q: 24
1 3 2 6 4 12 8 24
----- Original Message -----
From: Arved Sandstrom <[EMAIL PROTECTED]>
Date: Wednesday, November 22, 2006 4:03 pm
Subject: [Jprogramming] Avoiding fills
> Another novice question. I trying to get the factors of a number
> N. I use q:
> to get the prime factors. As many of those as there are, say M, I
> generatethe 2^M binary numbers using
>
> #: i.(2^M)
>
> For M=3 we therefore have
>
> 0 0 0
> 0 0 1
> 0 1 0
> 0 1 1
> 1 0 0
> 1 0 1
> 1 1 0
> 1 1 1
>
> Now, any given item of this, when used as the left operand to #,
> with q: N
> being the right, gives me a combination of prime factors that I will
> multiply with */ to get a possible factor. I'd just finish off
> with ~. to
> clean up the list of factors.
>
> My problem is in taking that list of binary numbers and using each
> item to #
> the prime factors. In the above example the longest copy will
> result in a
> list of shape 3, and so every result item has shape 3.
>
> I had a really ugly work around that replaced all the zeros with
> one's,after doing the #, and seeing as how I followed with */ that
> worked. But I
> didn't like it.
>
> Is there a more elegant solution? Or I should say, what might it
> be? Could
> it involve boxes?
>
> Thanks.
>
> AHS
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