My original approach was even more naive than Marc's:
NB. from Roger Hui's Combinations essay on the J website:
https://goo.gl/WL4nXn
c=: ((= +/"1) |.@:I.@# ]) #:@i.@(2&^) NB. Roger called this 'comb2'
NB. I used this definition because I could cut & paste one line, and not
require an editor
a =. 2 2 5 5
~.(*/"1(4 c 4){a),(*/"1(3 c 4){a),(*/"1(2 c 4){a),, |:(1 c 4){a
100 20 50 4 10 25 2 5
I like to sort it:
/:~~.(*/"1(4 c 4){a),(*/"1(3 c 4){a),(*/"1(2 c 4){a),, |:(1 c 4){a
2 4 5 10 20 25 50 100
The final goal of this exercise was to find all the divisors of an integer
(not just the prime divisors).
So you need to find the prime factors of the integer, for example 100:
]a =. q:100
2 2 5 5
/:~~.(*/"1(4 c 4){a),(*/"1(3 c 4){a),(*/"1(2 c 4){a),, |:(1 c 4){a
2 4 5 10 20 25 50 100
So these are all the divisors of 100.
To use Raul's verb, it needs some mods. I can't do the unique until the
end, because prime factors of an integer are often duplicated.
F1=:1 :'u@#~ #:@i.@(2^#)' NB. Here's Raul's verb minus the initial
unique
*/F1 q:100 NB. we take the product
1 5 5 25 2 10 10 50 2 10 10 50 4 20 20 100
~.*/F1 q:100 NB. Now we take the unique
1 5 25 2 10 50 4 20 100
/:~~.*/F1 q:100 NB. Sort it to make it pretty
1 2 4 5 10 20 25 50 100
Didn't really need the 1, but Raul likes the empty combination.
Now we put it all in one verb:
F2=. /:~~.*/F1 q:
F2 100
|length error: F2
| F2 100
F2=. /:~~.*/F1 q:]
F2 100
|domain error: F2
| F2 100
So this is above my pay grade. I'll have to stick with my inline code:
/:~~.*/F1 q:110
1 2 5 10 11 22 55 110
/:~~.*/F1 q:43
1 43
/:~~.*/F1 q:45
1 3 5 9 15 45
/:~~.*/F1 q:444
1 2 3 4 6 12 37 74 111 148 222 444
So I can find all the divisors of an integer.
Skip
Skip Cave
Cave Consulting LLC
On Mon, Oct 2, 2017 at 12:15 PM, Marc Simpson <[email protected]> wrote:
> More naive than Raul's approach, first pass using the 'stats' lib:
>
> a=.2 5 7
> require'stats'
> combv=: ] {~ (comb #@])
> 3 combv a
> 2 5 7
> 2 combv a
> 2 5
> 2 7
> 5 7
> 1 combv a
> 2
> 5
> 7
> combos=: (1 + i.@#) <@combv"0 1 ]
> combos a
> ┌─┬───┬─────┐
> │2│2 5│2 5 7│
> │5│2 7│ │
> │7│5 7│ │
> └─┴───┴─────┘
> f=: 1 : ';u/"1 each combos y'
> +f a
> 2 5 7 7 9 12 14
> *f a
> 2 5 7 10 14 35 70
>
> /M
>
> On Mon, Oct 2, 2017 at 10:06 AM, Raul Miller <[email protected]>
> wrote:
> > For a first effort, I would go with
> >
> > F=:1 :'(u@#~ #:@i.@(2^#))@~.'
> > f=: /F
> >
> > Hopefully that makes the issues obvious - the specification here calls
> > for a result which grows exponentially with the size of the argument
> > set.
> >
> > Also:
> >
> > The ~. might be extra work, but for typical cases the effort of
> > ensuring that the argument is a set is trivial compared to the effort
> > of constructing the result.
> >
> > You did not include the empty combination in your example results, but
> > given your specification my initial inclination is to treat that as an
> > oversight.
> >
> > I defined F instead of going straight for f because for testing
> > purposes I want to be able to do (<F a), and perhaps similar things.
> >
> > Thanks,
> >
> > --
> > Raul
> >
> >
> > On Mon, Oct 2, 2017 at 12:49 PM, Skip Cave <[email protected]>
> wrote:
> >> Given a set of integers, what is the most concise and or efficient way
> to
> >> list the numbers along with the sum of all combinations of the numbers?
> the
> >> products of all combinations?
> >>
> >> for example:
> >>
> >> a =. 2 5 7
> >> + f a NB. 2, 5, 7, (2+5), (2+7), (5+7), (2+5+7)
> >> 2 5 7 7 9 12 14
> >>
> >> * f a NB. 2, 5, 7, (2*5), (2*7), (5*7), (2*5*7)
> >> 2 5 7 10 14 35 70
> >>
> >> The function 'f' should work for any verb and any size right argument
> noun
> >> vector.
> >>
> >> Skip
> >>
> >> Skip Cave
> >> Cave Consulting LLC
> >> ----------------------------------------------------------------------
> >> For information about J forums see http://www.jsoftware.com/forums.htm
> > ----------------------------------------------------------------------
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> ----------------------------------------------------------------------
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