Rob's analysis assumes that each bucket must contain a fixed number of objects. In my problem, each bucket can hold any number of objects. However, in each trial, every bucket must hold at least one object. Of course, that means that for a specific x & y, the most that any bucket can hold is y-x-1 where y is the number of objects, and x is the number of buckets.
Skip Cave Cave Consulting LLC On Sat, Oct 21, 2017 at 7:29 PM, 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 > >>>> > >>>> > >>>> > >>>> > >>>>> > >>>>> > >>> ---------------------------------------------------------------------- > >>> 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 > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
