Nice solution. Not only faster, but also leaner.
rank '+/"1 {&f 10&#.^:_1 i.1e6';'([:,f&(+/))^:5 f'
+----+-----+-----+----+
|rank|tm*sz|time |size|
+----+-----+-----+----+
| 1 |94.60|14.89|6.35|
+----+-----+-----+----+
| 0 | 1.00| 1.00|1.00|
+----+-----+-----+----+
-:/".&.> '+/"1 {&f 10&#.^:_1 i.1e6';'([:,f&(+/))^:5 f'
1
However, only applicable if the universe is i.10^n.
R.E. Boss
> -----Oorspronkelijk bericht-----
> Van: [email protected] [mailto:programming-
> [email protected]] Namens Marshall Lochbaum
> Verzonden: zondag 5 september 2010 23:58
> Aan: Programming forum
> Onderwerp: [Jprogramming] splitting stuff into digits the way where you
> don't split it into digits
>
> Thought I would let everyone know about the alternate approach I just
> found
> to the sums of factorials of digits problem:
>
>
>
> f=.(+!)i.10
>
> r=.([:,f&(+/))^:6 f
>
> r=.r - 6,(9*10^i.7)#(i._7)
>
>
>
> This generates the sum of the digit plus the factorial for each of i.1e6
>
> This way, you don't have to actually work starting with i.1e6. By
> modifying
> the 6s and 7s, you can generate any list of the form 10^n, although 1e7 is
> about the limit in terms of memory.
>
>
>
> It is also faster than the other method:
>
>
>
> 6!:2 '([:,f&(+/))^:6 f=.(+!)i.10'
>
> 0.205724
>
> 6!:2 '+/"1 {&f 10&#.^:_1 i.1e6'
>
> 0.378998
>
>
>
> Where each has an error based on the number of leading zeros (corrected
> for
> in the last line of the script above).
>
>
>
> You can probably also generate length 10^2^n by taking ([:,+/~)^:n f, but
> that is a bit ridiculous.
>
>
>
> Marshall
>
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