Hi all!
New list, which in my opinion does not answer the quora question.
list=:4 : 0
v=.q: y
if. (#v)<x do.
r=:(0,x)$0
else.
r1=.x parRuskeyE #v
r2=. >r1 (*/)@:{&.>"1 0 < v
r3=./:~"1 r2
r=. ~.r3
end.
r
)
Cheers,
Erling Hellenäs
On 2017-11-04 07:28, 'Skip Cave' via Programming wrote:
Raul, very nice!
Actually I prefer the solution that doesn't allow 1 as a factor of p. Of
course, that restricts the max number of partitions to the max number of
prime factors of any p. That also greatly reduces the number of partition
instances that will be generated. Then:
5 par 358258
┌─┬─┬──┬──┬───┐
│2│7│11│13│179│
└─┴─┴──┴──┴───┘
Skip
Skip Cave
Cave Consulting LLC
On Fri, Nov 3, 2017 at 2:40 AM, Raul Miller <[email protected]> wrote:
So... 358358 has five prime factors (32 integer factors). We want to
find all sorted sequences (not sets - values can repeat) of five of
those factors whose product is 358358.
To restrict our search, we can investigate only those sorted sequences
of "number of prime factors represented in the variable" whose sum is
five:
~./:~"1 (#~ 5=+/"1) 6 #.inv i.6^5
0 0 0 0 5
0 0 0 1 4
0 0 0 2 3
0 0 1 1 3
0 0 1 2 2
0 1 1 1 2
1 1 1 1 1
In other words, the results of these seven expressions (use
require'stats' first to get comb):
1 1 1 1
358358
(1 1 1,(358358%*/),*/)"1 (4 comb 5){q:358358
/:~"1 (1 1 1,(358358%*/),*/)"1 (3 comb 5){q:358358
/:~"1 (1 1,q:@(358358%*/),*/)"1 (3 comb 5){q:358358
~./:~"1 (1 1,({.,*/@}.)@q:@(358358%*/),*/)"1 (2 comb 5){q:358358
/:~"1 (1,q:@(358358%*/),*/)"1 (2 comb 5){q:358358
q:358358
That's 44 different solutions:
1 1 1 1 358358
1 1 1 179 2002
1 1 1 13 27566
1 1 1 11 32578
1 1 1 7 51194
1 1 1 2 179179
1 1 1 154 2327
1 1 1 182 1969
1 1 1 143 2506
1 1 1 286 1253
1 1 1 91 3938
1 1 1 77 4654
1 1 1 358 1001
1 1 1 26 13783
1 1 1 22 16289
1 1 1 14 25597
1 1 13 154 179
1 1 11 179 182
1 1 11 13 2506
1 1 7 179 286
1 1 7 13 3938
1 1 7 11 4654
1 1 2 179 1001
1 1 2 13 13783
1 1 2 11 16289
1 1 2 7 25597
1 1 11 14 2327
1 1 7 22 2327
1 1 7 26 1969
1 1 7 143 358
1 1 2 77 2327
1 1 2 91 1969
1 1 2 143 1253
1 11 13 14 179
1 7 13 22 179
1 7 11 26 179
1 7 11 13 358
1 2 13 77 179
1 2 11 91 179
1 2 11 13 1253
1 2 7 143 179
1 2 7 13 1969
1 2 7 11 2327
2 7 11 13 179
We could of course come up with a routine which does something similar
for other examples (but we will run into prohibitive resource
limitations if we allow large enough integers).
So... just to confirm... this is the problem we are trying to solve?
Thanks,
--
Raul
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