Thank you, Dan and Bert.


Bert - Your approach provides a solution.  However, it has the undesired property of referees lumping together (I apologize that I did not state this as a condition).  In other words, it does not "mix" the referees in some random fashion.

Dan - your approach attempts to have the desired properties, but is not guaranteed to work.  Here is a counterexample:

> set.seed(1234)
> a <- assignment(40,15,3)
> table(a)
a
1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
10  7 12  7  4 10  8  6  8 13  7  7 11  3  7

Notice that the difference between maximum and minimum candidates for referees is 13 - 3 = 10.  Of course, I have to increase the # iters to get a better solution, but for large  K and R this may not converge at all.

Best regards,
Ravi

From: Ravi Varadhan
Sent: Wednesday, April 30, 2014 1:49 PM
To: r-help@r-project.org
Subject: A combinatorial assignment problem

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


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