Optimize fair teams when team size is can vary.

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Levi Shelton
Levi Shelton 2022년 11월 21일
편집: Matt J 2022년 11월 21일
I am writing a script to pick teams as fairly as possible from a list of names and associated player values. The teams must have 4 or 5 members per team. I have a good method of optimizing teams when the list is divisible by 4 or 5 directly so that each team has the same number of players. My hang up is the list quanity is unpredictable and may end up being something like 17, where I would want to optimize each team's total value with the option of some teams having 4 members and others 5, i.e. 3 teams of 4 and 1 team of 5. I'm open to modifications to my current picking method, or if it's not possible, having a seperate optimizing loop for just this scenario that could be selected when needed. Thanks!
Current code is:
minstd = 5;
%teamsz = 4;
teamsz = 5;
I = height(playerval)/teamsz;
for i = 1:100000
idx = kron(eye(I),ones(teamsz,1));
idx = idx(randperm(end),randperm(end));
teamtry = playerval.*idx;
teamsums = sum(teamtry);
stdev = std(teamsums);
if stdev < minstd
minstd = stdev;
finalstd = stdev;
finalteams = idx;
finalteamscore = teamsums;
end
end
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Matt J
Matt J 2022년 11월 21일
편집: Matt J 2022년 11월 21일
@Levi Shelton Rather than optimizing standard deviation of teamsums, you should consider instead optimizing the mean absolute deviation about the mean. This can be posed as a binary linear program and would probably be more reliable than random trial and error.
Levi Shelton
Levi Shelton 2022년 11월 21일
@Matt J Thanks for the advice, I'll look into this method. I would certainly prefer a discrete solution as opposed to hammering randperm and hoping for the best.

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Matt J
Matt J 2022년 11월 21일
편집: Matt J 2022년 11월 21일
Instead of using kron to set up your initial matrix idx, use blkdiag:
nPlayers=17;
minTeams=floor(nPlayers/5); %minimum number of teams
maxTeams=ceil(nPlayers/4); %minimum number of teams
nTeams=minTeams:maxTeams;
M=numel(nTeams);
for i=1:M
N=nTeams(i);
c=diophantine(ones(1,N),nPlayers,[4,5]);
if ~isempty(c);
memberships=unique(sort(c,2,'descend'),'rows'); break;
end
end
memberships %number of people on each team
memberships = 1×4
5 4 4 4
args=arrayfun(@(z) ones(z,1), memberships,'uni',0);
idx=blkdiag(args{:})
idx = 17×4
1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0
  댓글 수: 1
Matt J
Matt J 2022년 11월 21일
편집: Matt J 2022년 11월 21일
Thanks for the advice, I'll look into this method. I would certainly prefer a discrete solution as opposed to hammering randperm and hoping for the best.
After membership is computed as above, we can continue as below, using the FEX download minl1intlin,
to compute the mean absolute deviation from mean solution:
M=sum(membership); Im=speye(M); em=ones(1,M); %number of players
N=numel(membership); In=speye(N); en=ones(1,N); %number of teams
V=kron(In,playerval(:)');
C=full(V-mean(V,1))/N; %difference-from-mean operator
Aeq_row=kron(en,Im); beq_row=em(:); %each player belongs to 1 team
Aeq_col=kron(In,em); beq_col=membership(:); %total players per team = membership
Aeq=[Aeq_row;Aeq_col]; %total equality constraints
beq=[beq_row;beq_col];
lb=zeros(M*N,1); ub=lb+1; intcon=1:M*N; %Force solution to be binary
X=minL1intlin(C,zeros(N,1),intcon, [],[],Aeq,beq,lb,ub); %solve
X=reshape(round(X),M,N) %X(i,j) assigns i-th player to j-th team

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