Matlab: construct a binary matrix satisfying horizontal and vertical constraints - matlab

Question

I want to initiate a mxn binary matrix. The summation of each column of the matrix equals to a given value s(j), for j=1..n. In addition, the summation of each row of the matrix should be within the range of given bounds: lhs is dl(i) and rhs is du(i), for i=1..m.
Now I can only randomly generate binary columns, in each of which the sum of ones equals to s(j) of that column, such as the following codes.
xij = zeros(m,n);
for j=1:n
randRows=randperm(m); %a row vector containing a random permutation of the integers from 1 to m inclusive.
rowsWithOne=randRows(1:sj(j)); %row indices having 1, and sum up to sj
xij(rowsWithOne,j)=1;
end
However, xij usually doesn't satisfy the horizontal constraints. I was thinking I should create a matrix first meets the row constraints lhs (lower bound) dl(i), then, column constraint s(j), and finally fill the offsets to meet rhs du(i), but I don't know how to implement it in Matlab. Is there any ideas to create xij?
Thanks in advance.

Answer 1

There are a couple things to first take into account, mainly if the problem can even be solved with the given constraints. First, you must check that the sum of s(j), sum(s) is greater than the sum of the lower bound constraints, sum(dl), and also less than the sum of the upper bound constraints, sum(du).
A simple if statement should be able to check this. After this has been done, the following code should be one solution to the problem, but given the nature of the problem, there will probably not be a unique solution.
%initialize counters
x = 1;
y = 1;
%make sure the lower bounds are first satisfied
for k = 1:nrows
while (dl(k) > 0)
if (s(y) > 0)
xij(k,y) = 1;
dl(k) = dl(k)-1;
du(k) = du(k)-1;
s(y) = s(y)-1;
end
y = y+1;
end
y = 1;
end
%make sure the columns all add to their specified values
for k = 1:ncols
while (s(k) > 0)
if (xij(x,k) == 0 && du(x) > 0)
xij(x,k) = 1;
du(x) = du(x)-1;
s(k) = s(k)-1;
end
x = x+1;
end
x = 1;
end
The first for loop adds 1s along the rows such that each row satisfies the minimum constraint. The next for loop adds 1s along the columns such that each column adds up to the desired value. The if statement in the first for loop ensures that a 1 isn't added if that column has already reached its desired value, and the if statement in the second for loop ensures that a 1 isn't added to a row that would put it over the maximum value of the row.
In order to keep track of the amount of ones, the du is lowered by 1 every time a 1 is added to the corresponding row, and likewise the desired value s(j) is subtracted by 1 every time a 1 is added to that column of the xij matrix.

Answer 2

%Create an initial xij matrix, each column of which consists of a random binary xij sequence, but sum of ones in each column equals to sj
Init_xij = zeros(nShift,nCombo);
for j=1:nCombo
randRows=randperm(nShift); %a row vector containing a random permutation of the integers from 1 to nShift inclusive.
rowsWithOne=randRows(1:sj(j)); %row indices having 1, and sum up to sj
Init_xij(rowsWithOne,j)=1;
end
%Adjust the initial xij matrix to make it feasible, satisfying horizontal
%LHS (dli) and RHS (dui) constraints
Init_xij_Feasible=Init_xij;
k=1;
while k
RowSum=sum(Init_xij_Feasible,2); %create a column vector containing the sum of each row
CheckLB=lt(RowSum,dli); %if RowSum <dli, true
CheckUB=gt(RowSum,dui); %if RowSum >dui, true
if ~any(CheckLB)&&~any(CheckUB) %if any element in CheckLB and CheckUB is zero
break,
else
[~,RowIdxMin]=min(RowSum);
[~,RowIdxMax]=max(RowSum);
ColIdx=find(Init_xij_Feasible(RowIdxMax,:) & ~Init_xij_Feasible(RowIdxMin,:),1); % returns the first 1 column index corresponding to the nonzero elements in row RowIdxMax and zero elements in row RowIdxMin.
%swap the min and max elements
[Init_xij_Feasible(RowIdxMin,ColIdx),Init_xij_Feasible(RowIdxMax,ColIdx)]=deal(Init_xij_Feasible(RowIdxMax,ColIdx),Init_xij_Feasible(RowIdxMin,ColIdx));
end
k=k+1;
end