MATLAB: How to write a better “if condition” to eliminate some unwanted matrices

conditionefficiencyifMATLAB

To be clear, I'm changing my question as follows with more details:
——
Here is the code segment I'm working on:
NphaseSteps = 4; 
% Assume that TM is a 10x10 random matrix that includes both real and complex numbers. 
% But it is fixed after created randomly.
TMsize = length(TM);
focusX = 5; focusY = 5;
function inFieldA = WSfunction(NphaseSteps, TM, TMsize, focusX, focusY)
% This function is supposed to return inFieldA that gives 1) the maximum value
% at the target position (focusX,focusY) inside the outField matrix 2) the 
% optimized outField such that all other elements other than (focusX,focusY) 
% inside outField have "at most" half of the target position value.
tic
fprintf('The number of possible random matrices: %s \n',NphaseSteps^(TMsize^2));
phases = exp( 2*pi*1i * (0:(NphaseSteps-1))/NphaseSteps );
nIterations = 5;
inField = cell(1,nIterations);
i = 1; j = 0;
while i <= nIterations %number of iterations
     ind = randi([1 NphaseSteps],TMsize,TMsize);
     inField{i} = phases(ind);
     j = j+1;
     outField{i} = TM * inField{i};
     outI = abs(outField{i}).^2;
     targetIafter(i) = abs(outField{i}(focusX,focusY)).^2;
     middleI = targetIafter(i) / 2;
     if max(max(outI)) == targetIafter(i)
        outI(focusX,focusY)=0;
           if (outI > middleI)  == zeros(TMsize,TMsize)
              i = i + 1;
           end
     end
     if mod(j,10^6) == 0
        fprintf('The number of random matrices tried: %d million \n',j/10^6)
     end         
end
fprintf('The number of random matrices tried in total: %d \n',j)
[maxInt, maxInd] = max(targetIafter);
inFieldA = inField(maxInd);
toc
end
This program basically creates a 10×10 random input matrix inField that only includes 4 elements (number of phase steps) and multiplies it with a 10×10 fixed matrix TM to obtain a 10×10 output matrix outField. Now, I want to fix two things about the program:
  1. Since the inField matrices are created randomly, the program will create equal matrices many times. I want to eliminate the equal inField matrices before they go into the "if statement". For this I need to change the cellfun part but I don't know how to do it in the most efficient way.
  2. In the outField matrix I want a specific matrix element to have the largest value. First part of the "if statement" deals with that. But, I additionally want the other elements (other than my target matrix element) to have values smaller than the half of the target value. In the second part of the "if statement" I wanted to handle this situation but I think it's too complicated and not very efficient. Can you offer a better way to do it?

Best Answer

Sahin: there is one thing obvious I didn't see until now, but the each column of the product
TM * inField
depends independenly of column of inField (since it matrix product), there for if you want to have a goal of outField(5,5) to stick out, all you need to do is solving two subproblems of (matrix x vectors) form.
TM*inGoal is proportional to e5:=[0,...0,1,0...]' (1 at 5th position)
TM*outgoal is as small as possible
where inGoal, outGoal are both (10x1) vectors.
An approximation of inGoal is very to find, all you do is small subsub-problem:
TM(:,j)*ingoal(j) maximizes a correlaton with e5
For outGoal, what you need is to minimize
argmin|TM*x|_inf / |x|_inf, where |x|_inf := max(abs(x(i)).
If we denote y:=TM*x, this is equivalent to minimizing
argmin |y|_inf / |inv[TM]*y|_inf
which is the same as maximizing
argmax |inv[TM]*y|_inf / |y|_inf.
This is nothing than find the matrix norm inv[TM]_inf
The solution of which is (just look at google):
argmax |inv[TM]*y|inf / |y|_inf = max_i sum_j (inv[TM]_i,j)
and the proof of that also provides the solution y that make the ratio reach the matrix norm.
One you have y, just compute x = TM \ y, and that gives pretty much the phase vector outGoal to make TM*outGoal small.
Put together to get
inField = [outGoal, ..., outGoal, inGoal, outGoal, ..., outGoal]
where the inGoal is at 5th column, there you are.
Related Question