[R] A combinatorial assignment problem

[Ravi Varadhan]

Thanks, Bert.   
I have written this simple code, which is crude, but seems to do a decent job.  It works perfectly when M is a factor of R. Otherwise, it gives decent balance (of course, balance is not guaranteed).  I guess it is possible to take the results, that are somewhat unbalanced and then reshuffle it semi-randomly to get better balance.  Any improvements are appreciated! 

assign <- function(K, R, M, iseed=1234) {
  assignment <- matrix(NA, K, M)
  N <- R %/% M
  Nrem <- R %% M
  iseq <- seq(1,K,N)
  for (i in iseq){
  	size <- ifelse(K-i >= N, R-Nrem, M*(K-i+1))
  	sel <- sample(1:R, size=size, replace=FALSE)
  	end <- min((i+N-1),K)
  	assignment[i:end, ] <- sel
  	}
assignment
}

sol <- assign(40,16,3)
table(sol)

Thanks,
Ravi

-----Original Message-----
From: Bert Gunter [mailto:gunter.berton at gene.com] 
Sent: Thursday, May 01, 2014 10:46 AM
To: Ravi Varadhan
Cc: r-help at r-project.org; djnordlund at frontier.com
Subject: Re: A combinatorial assignment problem

Ravi:

You cannot simultaneously have balance and guarantee random mixing.
That is, you would need to specify precisely what you mean by balance and random mixing in this context, as these terms are now subjective and undefined.

You could, of course, randomize the initial assignment of referees to positions and note also that some mixing of referee groups would occur if m does not divide r evenly. More explicitly, note that a very fast way to generate the groups I described is:

rmk <- function(nrefs,size,ncands){
  n <- ncands * size
  matrix(rep(seq_len(nrefs), n %/% nrefs +1)[seq_len(n)],nrow=
ncands,byrow=TRUE)
}
## corner case checks and adjustments need to be made to this, of course.

>  rmk(7,3,10)
      [,1] [,2] [,3]
 [1,]    1    2    3
 [2,]    4    5    6
 [3,]    7    1    2
 [4,]    3    4    5
 [5,]    6    7    1
 [6,]    2    3    4
 [7,]    5    6    7
 [8,]    1    2    3
 [9,]    4    5    6
[10,]    7    1    2

You could then modify this by randomly reordering the referees every time a complete cycle of the groupings occurred, i.e. after each
nrefs/gcd(nrefs,size) candidates = rows. This would give variable groups and even assignment. This algorithm could be further fiddled with by choosing nrefs and size to assure they are relatively prime and then adding and removing further refs randomly during the cycling.
etc.

Cheers,
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 Thu, May 1, 2014 at 6:09 AM, Ravi Varadhan <ravi.varadhan at jhu.edu> wrote:
> 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 at 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
>
>