sel=: {each <
Note also though:
q: 75600
2 2 2 2 3 3 3 5 5 7
# q: 75600
10
$N=:5 parRuskeyE 10
42525 5
$~. /:"1~ */@> N sel q:75600
798 5
The number of factorizations using factors s greater than 1 of an
integer will often be different than the number of partitions as we
had defined them here.
FYI,
--
Raul
On Sat, Nov 4, 2017 at 1:00 PM, 'Skip Cave' via Programming
<[email protected]> wrote:
> So, given the full parution set of say 3 par 4
>
> ]a =. 3 par 4
>
> ┌───┬───┬───┐
>
> │0 1│2 │3 │
>
> ├───┼───┼───┤
>
> │0 2│1 │3 │
>
> ├───┼───┼───┤
>
> │0 │1 2│3 │
>
> ├───┼───┼───┤
>
> │0 3│1 │2 │
>
> ├───┼───┼───┤
>
> │0 │1 3│2 │
>
> ├───┼───┼───┤
>
> │0 │1 │2 3│
>
> └───┴───┴───┘
>
>
> What would the verb 'sel' look like that would use those indices to select
> from a different set of objects
>
>
> a sel 'abcd'
>
> ┌───┬───┬───┐
>
> │a b│c │d │
>
> ├───┼───┼───┤
>
> │a c│b │d │
>
> ├───┼───┼───┤
>
> │a │b c│d │
>
> ├───┼───┼───┤
>
> │a d│b │c │
>
> ├───┼───┼───┤
>
> │a │b d│c │
>
> ├───┼───┼───┤
>
> │a │b │c d│
>
> └───┴───┴───┘
>
>
> Skip
>
>
>
> Skip Cave
> Cave Consulting LLC
>
> On Sat, Nov 4, 2017 at 11:09 AM, Skip Cave <[email protected]> wrote:
>
>> Raul,
>> Yes, the original Quora question specified positive factors only, but i
>> forgot to include that in the specification.
>>
>> Skip
>>
>> Skip Cave
>> Cave Consulting LLC
>>
>> On Sat, Nov 4, 2017 at 3:52 AM, Raul Miller <[email protected]> wrote:
>>
>>> Well, ok, though that was not a part of your re-specification this time.
>>>
>>> Actually, though, re-reading your spec, i left out a factor of 16 of
>>> the solutions: integers can be negative and as long as we include an
>>> even number of negatives they cancel out in a product.
>>>
>>> Thanks,
>>>
>>> --
>>> Raul
>>>
>>>
>>> On Sat, Nov 4, 2017 at 2:28 AM, 'Skip Cave' via Programming
>>> <[email protected]> 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
>>> >>
>>> >>
>>> >>
>>> > ----------------------------------------------------------------------
>>> > For information about J forums see http://www.jsoftware.com/forums.htm
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>
>>
>>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
