-----Original Message-----
From: Programming [mailto:[email protected]]
On Behalf Of Rob B
Sent: zondag 22 oktober 2017 11:40
To: [email protected]
Subject: Re: [Jprogramming] Partitions
In the twelvefold way at https://www.johndcook.com/TwelvefoldWay.pdf I
think this is case 6, labelled, unlabelled, >=1.
Regards, Rob.
On 22 Oct 2017, at 01:29, Rob Hodgkinson <[email protected]> wrote:
Skip, not sure the status of a solution for this yet (Raul’s was last closest I
believe ?).
I thought through the following analysis over the weekend, more around
combinatorics.
Problem is to split a list of objects o, into y “buckets” and you want to all
the
ways possible given the constraints below.
A combinatoric view is to consider filling each bucket (without
replacement) until no more objects, for example; Suppose 7 objects (e.g.
‘abcdefg’) split into 3 buckets of size 2 3 2.
There are 2C7 (or 2!7) ways to fill the first bucket, leaving 5 left.
From these 5 there are then 3!5 ways to fill the second bucket, leaving 2
left.
From these 2 there are then 2!2 ways to fill the last bucket
The combinations (with decreasing sample sets) can be calculated as
follows (I pasted in Courier New, hope this appears OK).
smoutput bset=:(+/bsize)- +/\ 0,}:bsize=:2 3 2
7 5 2
So from the reducing sample from which to fill each successive bucket, the
possible combinations are:
bsize ! bset=:(+/bsize)- +/\ 0,}:bsize=:2 3 2
21 10 1
And the combinations in each bucket are then “indices” within each bucket
from the available objects remaining;
smoutput bcombs=:bsize comb each bset=:(+/bsize)- +/\ 0,}:bsize=:2 3
2 ┌───┬─────┬───┐
│0 1│0 1 2│0 1│
│0 2│0 1 3│ │
│0 3│0 1 4│ │
│0 4│0 2 3│ │
│0 5│0 2 4│ │
│0 6│0 3 4│ │
│1 2│1 2 3│ │
│1 3│1 2 4│ │
│1 4│1 3 4│ │
│1 5│2 3 4│ │
│1 6│ │ │
│2 3│ │ │
│2 4│ │ │
│2 5│ │ │
│2 6│ │ │
│3 4│ │ │
│3 5│ │ │
│3 6│ │ │
│4 5│ │ │
│4 6│ │ │
│5 6│ │ │
└───┴─────┴───┘
So for example the first row in each bucket only says e.g. from ‘abcdefg’,
choose (‘ab’;’cde’;’fg’), the first “trial”.
$ each bcombs
┌────┬────┬───┐
│21 2│10 3│1 2│
└────┴────┴───┘
This shows there would be 210 possible combinations (trials), although I am
unclear on your rule 5 below, to remove “like” combinations, but this could
be achieved post the algorithm.
To get the actual 210 samples you could then index into each cell, but the
“samples remaining” would then change for each combination.
For example, bucket1 is straightforward to calculate, along with the
remainder samples available to fill the remaining buckets …
smoutput bucket1=:'abcdefg' {~ >{.bcombs ab ac ad ae af ag bc bd be
bf bg cd ce cf cg de df dg ef eg fg
bucket1;'abcdefg'-."1 bucket1
┌──┬─────┐
│ab│cdefg│
│ac│bdefg│
│ad│bcefg│
│ae│bcdfg│
│af│bcdeg│
│ag│bcdef│
│bc│adefg│
│bd│acefg│
│be│acdfg│
│bf│acdeg│
│bg│acdef│
│cd│abefg│
│ce│abdfg│
│cf│abdeg│
│cg│abdef│
│de│abcfg│
│df│abceg│
│dg│abcef│
│ef│abcdg│
│eg│abcdf│
│fg│abcde│
└──┴─────┘
The steps would be repeated for each product, at each stage “expanding”
the combinations within each bucket by the successive indices (avoiding
selecting any items already in existing buckets).
This will become a cartesian product of samples at each iteration to
generate the final 210 samples.
This would likely involve recursion or some form of (f ,/), but wanted to
check my understanding is on track. It may make the solution clearer.
I hope this makes sense, Rob
On 21 Oct 2017, at 10:06 am, 'Skip Cave' via Programming
<[email protected]> wrote:
Perhaps here is a better way to describe the problem I was trying to solve:
1. I have y unique objects. I have x containers. I want to find all
the different ways I can put those y unique objects into x containers.
2. A single trial is defined as distributing *all* y of the objects
into all of the x containers. After the trial, the objects are
removed from the containers and redistributed in the containers for
the next trial. Objects can not be duplicated.
3. Empty containers are not allowed in any trial. There must be at
least one object in every container in every trial.
4. There is no order requirement for objects in a container. If two
trials are identical except for the order of the objects in any of
the containers, those two trials are considered as one trial.
5. There is no order requirement for the containers. If two trials
are identical in that both trials have the same number of objects in
each of the x containers, but the order of the containers is
different, those two trials are considered as one trial.
Skip
Skip Cave
Cave Consulting LLC
On Fri, Oct 20, 2017 at 3:54 AM, Raul Miller <[email protected]>
wrote:
With that description, I think I might do something like this:
nparts=: ~.@([: /:~"1 i.@] </."1~ [ #.^:_1 [ i.@^ ])
2 nparts 3
┌───┬─────┐
│ │0 1 2│
├───┼─────┤
│0 1│2 │
├───┼─────┤
│0 2│1 │
├───┼─────┤
│0 │1 2 │
└───┴─────┘
I should be able to do something more efficient, but before I
attempt that, I would like to clarify something:
In your examples, you do not have any empty partitions, so is that a
part of the specification also?
I am unsure if I should be paying close attention to your examples
because you said "The order of the objects in each partition is not
important" but your examples also omit partitions which contain the
objects out of order.
Actually... there's several kinds of order here which we could be
discussing:
(1) the order of objects within a partition.
(2) the order of objects across partitions.
(3) the order of partitions.
In other words:
NB. (1) the order of objects within a partition \:~each 2 nparts 3
┌───┬─────┐
│ │2 1 0│
├───┼─────┤
│1 0│2 │
├───┼─────┤
│2 0│1 │
├───┼─────┤
│0 │2 1 │
└───┴─────┘
NB. (2) the order of objects across partitions
2 ~.@([: \:~"1 i.@] </."1~ [ #.^:_1 [ i.@^ ]) 3 ┌─────┬───┐
│0 1 2│ │
├─────┼───┤
│2 │0 1│
├─────┼───┤
│1 │0 2│
├─────┼───┤
│1 2 │0 │
└─────┴───┘
NB. (3) the order of partitions
2 ((i.@[ ,"1 [ #.^:_1 i.@^) <@}./."1 {. , i.@]) 3 ┌─────┬─────┐
│0 1 2│ │
├─────┼─────┤
│0 1 │2 │
├─────┼─────┤
│0 2 │1 │
├─────┼─────┤
│0 │1 2 │
├─────┼─────┤
│1 2 │0 │
├─────┼─────┤
│1 │0 2 │
├─────┼─────┤
│2 │0 1 │
├─────┼─────┤
│ │0 1 2│
└─────┴─────┘
I have presumed that you are thinking of both the partition contents
and the partitions themselves as sets. In other words, I think that
none of these orders matter. But... this kind of thing is worth
verifying?
Thanks,
--
Raul
On Fri, Oct 20, 2017 at 12:19 AM, 'Skip Cave' via Programming
<[email protected]> wrote:
Mike,
I wasn't very thorough in my definition of the original problem. I
thought
that the example I gave was enough to clarify the requirements, but
looking
back, more definition would have been good.
The original problem I posted was to develop a dyadic verb that
would
take
y objects and show all the ways that those y objects could be
partitioned into x groups. Each partition set must include all objects
exactly once.
Duplication of objects is not allowed. The order of the objects in
each partition is not important.
Erling got the right idea in his previous post:
par=: 4 : '(1,.2</\"1(i.x)#/~(y=+/"1 o)#o=.((x$v)#:i.v^x){1+i.v=.1+
y-x)<;.1[1+i.y'
2 par 3
┌───┬───┐
│1 │2 3│
├───┼───┤
│1 2│3 │
└───┴───┘
2 par 4
┌─────┬─────┐
│1 │2 3 4│
├─────┼─────┤
│1 2 │3 4 │
├─────┼─────┤
│1 2 3│4 │
└─────┴─────┘
2 par 5
┌───────┬───────┐
│1 │2 3 4 5│
├───────┼───────┤
│1 2 │3 4 5 │
├───────┼───────┤
│1 2 3 │4 5 │
├───────┼───────┤
│1 2 3 4│5 │
└───────┴───────┘
3 par 4
┌───┬───┬───┐
│1 │2 │3 4│
├───┼───┼───┤
│1 │2 3│4 │
├───┼───┼───┤
│1 2│3 │4 │
└───┴───┴───┘
3 par 5
┌─────┬─────┬─────┐
│1 │2 │3 4 5│
├─────┼─────┼─────┤
│1 │2 3 │4 5 │
├─────┼─────┼─────┤
│1 │2 3 4│5 │
├─────┼─────┼─────┤
│1 2 │3 │4 5 │
├─────┼─────┼─────┤
│1 2 │3 4 │5 │
├─────┼─────┼─────┤
│1 2 3│4 │5 │
└─────┴─────┴─────┘
3 par 6
┌───────┬───────┬───────┐
│1 │2 │3 4 5 6│
├───────┼───────┼───────┤
│1 │2 3 │4 5 6 │
├───────┼───────┼───────┤
│1 │2 3 4 │5 6 │
├───────┼───────┼───────┤
│1 │2 3 4 5│6 │
├───────┼───────┼───────┤
│1 2 │3 │4 5 6 │
├───────┼───────┼───────┤
│1 2 │3 4 │5 6 │
├───────┼───────┼───────┤
│1 2 │3 4 5 │6 │
├───────┼───────┼───────┤
│1 2 3 │4 │5 6 │
├───────┼───────┼───────┤
│1 2 3 │4 5 │6 │
├───────┼───────┼───────┤
│1 2 3 4│5 │6 │
└───────┴───────┴───────┘
Skip Cave
Cave Consulting LLC
On Thu, Oct 19, 2017 at 5:47 PM, 'Mike Day' via Programming <
[email protected]> wrote:
Skip, in your actual Quora Problem, why not include other triads,
such as 1 1 24, 2 3 4 etc; or, otherwise, why include both 2 4 3
and 4
2 3 ?
Anyway, this is (quite) short and brutish but not too nasty to
solve
your
Quora problem for quite small numbers and numbers of factors:
|: 24 ([ ( (= */"1)#]) [:>:[#.inv i.@^ ) 3 NB. transpose
gratuitous!
1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 4 4 4 4 6 6 6 8 8 12 12
24
1 2 3 4 6 8 12 24 1 2 3 4 6 12 1 2 4 8 1 2 3 6 1 2 4 1 3 1 2 1
24 12 8 6 4 3 2 1 12 6 4 3 2 1 8 4 2 1 6 3 2 1 4 2 1 3 1 2 1
1
It builds triads 1 1 1 , 1 1 2 ,...1 1 24, ... up to 24 24 24, and
keeps
just those whose product is 24.
No points for space or time, filtering 30 out of 13824 candidates,
but
it's
quite straightforward, and it does yield all 30 permutations,
which
some
of the Quora corresondents appear to consider the requirement.
NB - it's not clear to me what the problem actually is - is 30 the
required
answer (number of permutations of 3 suitable factors), or 6
(number of
combinations of same)?
Mike
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