Asked by Berkay Sen
on 3 Jun 2019

Hi to all,

I'm working on my graduating thesis what about cross kidney transplants matching algorithms.

I've a solution with genetic algorithm but i want to know 'what is the optimum?' (of course GA is %90-99 interval, not %100-optimum-)

So;

I have a binary matrix(also symetric)

for example=

1 2 3 4 5 6 7 8 9 10

1 [0 1 0 0 0 0 0 0 0 0

2 1 0 1 0 0 0 0 0 0 0

3 0 1 0 0 0 0 0 0 0 0

4 0 0 0 0 0 0 0 0 1 0

5 0 0 0 0 0 0 0 0 0 0

6 0 0 0 0 0 0 0 0 0 0

7 0 0 0 0 0 0 0 0 0 0

8 0 0 0 0 0 0 0 0 0 0

9 0 0 0 1 0 0 0 0 0 0

10 0 0 0 0 0 0 0 0 0 0]

I want to find optimum match number beetween column numbers. (How many matches can be made in this matrix?)

In this case; Possible matches are:

NO2 - NO1

NO2 - NO3

NO 4- NO9

But when NO2 was matched with NO1, it can no longer match with NO1. Its kinda monogamy.(If patient and donor NO2 exchanges kidneys with p and d NO1, of course NO2 can not exchange with NO3)

In this example, optimum match value is = 2.

Let's think about 1000*1000 matrix.

When i solve this matrix with genetic algortihm, i had 410 as a result. ( theoretical match number is 500 ((1000/2)), of course impossible.)

HOW CAN I SOLVE THIS MATRIX USING CPLEXLP OR MILP?(or something else solver)

I will be very pleased if anybody help,

Yours truly.

Answer by Stephan
on 6 Jun 2019

Edited by Stephan
on 6 Jun 2019

Accepted Answer

Hi,

you could think about using graphs for this job:

% your matrix

A = [0 1 0 0 0 0 0 0 0 0;

1 0 1 0 0 0 0 0 0 0;

0 1 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 1 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 1 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0];

% create a graph object

g = graph(A);

% plot graph

plot(g);

% remove nodes who have no matches / are unconnected

g = rmnode(g,find(degree(g)==0));

% count the number of different matches

result = numel(unique(conncomp(g)));

% plot resulting graph

hold on

plot(g)

hold off

Stephan
on 6 Jun 2019

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Answer by Alan Weiss
on 6 Jun 2019

Alan Weiss

MATLAB mathematical toolbox documentation

Berkay Sen
on 6 Jun 2019

Hi Alan,

thank you for your great suggest.

But matchpairs doesn't working for me because matchpairs lookin rows(1. row, 2.row, 3.row...) and choosing best choise.

Imagine that,

No1 can match with No4,No5,No6;

But No2 can only match with No4,

And No3 can only match with No5

If matchpairs choose no4 because placement, then No2 can no longer match with anyone.

Same for no3-no5.

So algorithm must know this and choose NO6 for NO1.

I hope i could explain that :)

Christine Tobler
on 6 Jun 2019

Thank you for thinking of matchpairs, Alan. Unfortunately it wouldn't work for this case: Matchpairs will match rows of A to columns, but it doesn't have an understanding that if row i of A is matched, then column i of A is also unavailable for another match.

Basically, in matchpairs the rows and columns represent two different sets of resources to be matched up, while in this problem both rows and columns represent the same set of resources.

Berkay Sen
on 6 Jun 2019

Hi Christine, That's completely true, have you got any idea? Thank you for your comment

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Answer by Christine Tobler
on 6 Jun 2019

You can solve this using intlinprog. This is probably not the most efficient way of solving the problem, but reliable

A = [0 1 0 0 0 0 0 0 0 0;

1 0 1 0 0 0 0 0 0 0;

0 1 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 1 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0;

0 0 0 1 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0];

% Define an optimization variable: x(i,j) == 1 if (i, j) are matched, otherwise x(i, j) == 0

x = optimvar("x", size(A), "Type", "integer", "LowerBound", 0, "UpperBound", 1);

% Define an optimization problem

problem = optimproblem;

% Maximize the number of indices (i, j) for which A(i, j) == 1 and x(i, j) == 1

% (change sign because Objective is minimized in solve)

problem.Objective = -sum(sum(x.*A));

% Every column of x can only contain 1 non-zero (otherwise that column is matched to more than one row)

problem.Constraints.noDuplicates = sum(x, 1) == 1;

% Require x(i, j) == x(j, i)

problem.Constraints.symmetric = x == x';

% Solve the problem

s = solve(problem);

% Find indices i, j for which A(i, j) == 1 and s.x(i, j) == 1

[i, j] = find(triu(A & s.x))

% This returns i = [2; 4] and j = [3; 9], so two pairs (2, 3) and (4, 9)

Berkay Sen
on 6 Jun 2019

Christine,

I don't know how to thank you. Algorithm is running smoothly. You're superb!

Thank you to all who spent their time!

I'm gonna share the project(cross kidney transplant algorithm) when i finish it.

Christine Tobler
on 6 Jun 2019

You're very welcome. I just realized you probably need an additional constraint:

% Require x(i, i) == 0

n = size(x, 1);

problem.Constraints.NonDiagonal = x(1:n+1:end) == 0;

This will prevent the solver from matching an index with itself.

Berkay Sen
on 7 Jun 2019

Thanks but i've already adjust that. My matrix's diagonal is full of 0. It's working perfectly 👍

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## 7 Comments

## KSSV (view profile)

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