Perhaps use the distributed rounding function from the numeric script? For
example:
load 'numeric'
(+/,]) 1 rounddist a * 172 % +/a=. 42 23 5 20 41 40
172 42 23 5 21 41 40
(+/,]) 1 rounddist a * 172 % +/a=. 50 35 50 8 19 11
172 50 35 49 8 19 11
On Wed, Jun 27, 2018 at 3:47 PM, Skip Cave <s...@caveconsulting.com> wrote:
> Devon,
>
> That 0-1 fraction method is a really nice way to generate lots of sets of 6
> random integers 1-70 that sum to 172.
> That got me to thinking -- shouldn't we be able to make all the generated
> 6-integer sets sum to 172?
>
> f01=. 1e5 6?@$0 NB. Generating the fractions
>
> NB. Now if we instument your verb a bit more:
>
> itt1=. f01#~c=.172=+/"1 b=.([: <. 0.5 + ] * 172 % +/"1)"1 ] f01
>
> $itt1
>
> 55085 6
>
> NB. Not bad. Over half of the strings sum to 172.
>
>
> NB. What about the other half (invert the c selection vector)?
>
> 5{.b#~-.c
>
> 42 23 5 20 41 40
>
> 50 35 50 8 19 11
>
> 37 34 34 17 22 27
>
> 49 10 25 2 34 53
>
> 30 14 44 15 32 38
>
>
> How far off are the rest of these strings ?
>
> d=.172-+/"1 b#~-.-.c
>
> 20{.d
>
> 1 _1 1 _1 _1 _1 1 1 _1 1 1 _1 1 1 1 1 1 1 _1 1
>
> _20{.d
>
> 1 1 _1 _1 1 1 _1 _1 _1 _1 _1 1 _1 2 _2 1 1 _1 _1 1
>
>
> >./d
>
> 2
>
> <./d
>
> _2
>
> NB. Looks like the rest of the strings are off by as much as +/- 2
>
>
> So we should be able find the discrepancy in the remaining strings, and
> randomly pick one or more of the integers in each of the strings to add or
> subtract 1 or 2 to. Then all the strings will sum to 172.
>
>
> I'll leave that exercise to the reader...
>
>
> Skip
>
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