factor-group function

[cameyo]

To solve a project euler problem i have written this factor-group function:

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(define (factor-group x)
  (letn (fattori (factor x)
         unici (unique fattori))
    (transpose (list unici (count unici fattori)))))

(factor-group 2000)
;-> ((2 4) (5 3))
(factor-group 232792560)
;-> ((2 4) (3 2) (5 1) (7 1) (11 1) (13 1) (17 1) (19 1))


Do you known a better/faster way?
Thanks.

cameyo

[ralph.ronnquist]

Nice.

The factorization is of course the "slow" part, whereas the result juggling is almost irrelevant time-wise. I wouldn't try to improve on your implementation, but as a hind-sight, I probably would have used

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(map list unici (count unici fattori))

instead of

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(transpose (list unici (count unici fattori)))

just because it intuitively would be less structure re-arrangement.
And I also wouldn't have known to use Italian named variables :)

[cameyo]

Thanks Ralph.
I am really enjoying newLisp :-)
cameyo

[cameyo]

I forgot to add the inverse function:

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(setq fg (factor-group 220))
;-> ((2 2) (5 1) (11 1))

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(setq num-fg (apply * (map (lambda (x) (pow (first x) (last x))) fg)))
;-> 220

cameyo