The following verb handily produces the possible distributions given a set of
weights. Good luck finding which combinations gave you those sums!
NB. x is (#boxes),(upper bound), y is weights to use
NB. finds the unique (sorted) sums of groupings of y so that no sum exceeds the
upper bound.
sums=.4 :0
'nbox bound'=.x
y=.\:~ y
s=.,: nbox$0
for_e. y do.
s=.(#~ bound >: {."1) ~. \:~"1 ,/ s+"1/ e* =i.nbox
end.
s
)
4 124 sums m
124 124 124 121
124 124 123 122
124 123 123 123
Marshall
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Viktor Cerovski
Sent: Sunday, April 10, 2011 6:45 PM
To: [email protected]
Subject: Re: [Jprogramming] Weight distribution problem
Skip Cave-3 wrote:
>
> [...]
> To make the problem more concrete, Paul gives a specific example of
> the problem. There are four buckets, and 20 masses. The masses are 23,
> 43, 12, 54, 7, 3, 5, 10, 54, 55, 26, 9, 9, 43, 54, 1, 8, 6, 38, 33
> respectively. What distribution of the 20 masses gives the smallest
> mass difference between the four buckets?
>
> Paul gives a link to his proposed solution, which I have not examined
> as yet, since I want to see how far I can get with a J solution.
> [...]
>
After some quick tries with J, I got a bunch of 123 123 123 124 distributions,
and they are all mutually different.
Here is one of them:
conf
0 0 2 3 1 0 3 3 0 2 1 1 2 1 3 1 2 2 1 2
23 43 12 54 7 3 5 10 54 55 26 9 9 43 54 1 8 6 38 33
+//./conf
123 123 123 124
</./conf
┌──────────┬──────────────┬──────────┬──────────────┐
│23 43 3 54│12 55 9 8 6 33│54 5 10 54│7 26 9 43 1 38│
└──────────┴──────────────┴──────────┴──────────────┘
--
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http://old.nabble.com/Weight-distribution-problem-tp31365679s24193p31366193.html
Sent from the J Programming mailing list archive at Nabble.com.
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