I had trouble with my email and it went before it should. Here's the
solution I meant to send:
Arrange the r referees in a circle.
start <- 0
Replicate k times{
end <- (start + m-1)%% r
output: c(start,end) +1
start <- (end+1)%% r
}
The start and end pairs give the subsets of referees around the
circle. The distributes the referees to the candidates as evenly as
possible. I leave it to you to translate this into code. It should be
very fast as no combinatorics are required.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
> On Wed, Apr 30, 2014 at 10:48 AM, Ravi Varadhan <ravi.varad...@jhu.edu> wrote:
>> Hi,
>>
>> I have this problem: K candidates apply for a job. There are R referees
>> available to review their resumes and provide feedback. Suppose that we
>> would like M referees to review each candidate (M < R). How would I assign
>> candidates to referees (or, conversely, referees to candidates)? There are
>> two important cases: (a) K > (R choose M) and (b) K < (R chooses M).
>>
>> Case (a) actually reduces to case (b), so we only have to consider case (b).
>> Without any other constraints, the problem is quite easy to solve. Here is
>> an example that shows this.
>>
>> require(gtools)
>> set.seed(12345)
>> K <- 10 # number of candidates
>> R <- 7 # number of referees
>> M <- 3 # overlap, number of referees reviewing each candidate
>>
>> allcombs <- combinations(R, M, set=TRUE, repeats.allowed=FALSE)
>> assignment <- allcombs[sample(1:nrow(allcombs), size=K, replace=FALSE), ]
>> assignment
>>> assignment
>> [,1] [,2] [,3]
>> [1,] 3 4 5
>> [2,] 3 5 7
>> [3,] 5 6 7
>> [4,] 3 5 6
>> [5,] 1 6 7
>> [6,] 1 2 7
>> [7,] 1 4 5
>> [8,] 3 6 7
>> [9,] 2 4 5
>> [10,] 4 5 7
>>>
>>
>> Here each row stands for a candidate and the column stands for the referees
>> who review that candidate.
>>
>> Of course, there are some constraints that make the assignment challenging.
>> We would like to have an even distribution of the number of candidates
>> reviewed by each referee. For example, the above code produces an
>> assignment where referee #2 gets only 2 candidates and referee #5 gets 7
>> candidates. We would like to avoid such uneven assignments.
>>
>>> table(assignment)
>> assignment
>> 1 2 3 4 5 6 7
>> 3 2 4 4 7 4 6
>>>
>>
>> Note that in this example there are 35 possible triplets of referees and 10
>> candidates. Therefore, a perfectly even assignment is not possible.
>>
>> I tried some obvious, greedy search methods but they are not guaranteed to
>> work. Any hints or suggestions would be greatly appreciated.
>>
>> Best,
>> Ravi
>>
>>
>> [[alternative HTML version deleted]]
>>
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>> and provide commented, minimal, self-contained, reproducible code.
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