In general this is not true. m=. 1 1 1 100 nb=. 2
The solution is obvious, but you get [bm=: nb %~ +/ m 51.5 ]'a b'=.(,~nb&-) nb * (-<.) x: bm 1 1 R.E. Boss > -----Oorspronkelijk bericht----- > Van: [email protected] [mailto:programming- > [email protected]] Namens Aai > Verzonden: dinsdag 12 april 2011 12:22 > Aan: Programming forum > Onderwerp: Re: [Jprogramming] Weight distribution problem > > I'm not sure if the following observation scales down the problem, but may > be it's of some help. > > One of the problems here is finding a (nice) partitioning of bucket sums that > are equal or close to the mean bucketsum. Looking at the two presented > examples I found the following: > > We have > bm : mean bucket sum > nb : # buckets > > then you may say > (,>:) <. bm > are two possible sums very close to bm. > > Then we have to find a and b in the following equation: > > (+/m) = +/ (a, b) * (,>:) <. bm > > Finding a and b: > b=. nb * (-<.) x: bm > > and a follows from > > a =. nb - b > > For the low scale example we have: > > nb=. 4 > > ]bm=.nb%~+/m > 123.25 > > ]'a b'=.(,~nb&-) nb * (-<.) x: bm > 3 1 > > (+/m) = +/ (a, b) * (,>:) <. bm > 1 > > For the large scale ex. we have: > nb=.20 > bm=.5273.4 > > ]'a b'=.(,~nb&-) nb * (-<.) x: bm > 12 8 > > Giving bucket sums: > > 12 8 # (,>:)<. bm > > > I'm not sure of course if this particular partitioning (a,b) of integer bucket > sums is possible. > > -- > Met vriendelijke groet, > =@@i > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
