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
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