Can someone help me to optimize this matrix?

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David Mueller 2017-11-27

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编辑： David Mueller 2017-11-28

I have the following matrix

wall0 = [0 0 0 0 0 0 0 0 0 0;...
         0 0 0 0 0 0 0 0 0 0;...
         0 a 0 b 0 a 0 0 0 0;...
         0 0 a 0 b 0 a 0 0 0;...
         0 0 0 a 0 a 0 0 0 0;...
         0 a 0 b 0 a 0 0 0 0;...
         0 0 a 0 b 0 a 0 0 0;...
         0 0 0 a 0 a 0 0 0 0;...
         0 0 b 0 b 0 b 0 0 0;...
         0 0 0 0 0 0 0 0 0 0];

Where

a = 0.3;
b = 0.5;

The numbers in the matrix represent probabilities that a 'sound wave' would be absorbed, where 0 is a zero probability.

All this means is that a cell is chosen randomly in my code, and an 'if' statement determines whether or not the 'wave' is absorbed based on another random number between 0 and 1 that is attached to it.

An algorithm that randomizes the rows, as follows:

            [rows,cols] = size(wall0) ;
            for i=1:rows 
                idx = randperm(cols) ; 
                wallR(i,idx) = wall0(i,:);  
            end

An example of what I mean is:

wave1 hits wall0 (randomly) at [4,3] -> wall0(4,3) = a
if (different) random number for wave1 < a, wave1 is absorbed
number of absorptions += 1
repeat x number of times

The only constraints are that the matrix dimensions are fixed; and the number of a's and b's (and zeros) are fixed and can only move within the rows.

Basically, I would like to optimize this matrix such that the number of absorptions is maximized. I am hoping that someone may be able to help as I am a beginner in regards to optimization.

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David Mueller 2017-11-27

编辑：Walter Roberson 2017-11-27

在 MATLAB Online 中打开

Just FYI, the following is the full code for my project:

clc; clear variables; close;
    % sets the matrix which represents the wall to wall0. a and b both represent
    % the audio foam panels. Their numbers are the probability which will be
    % evaluated against a random number to determine if the sound wave is 
    % absorbed or not.
    a = 0.3;
    b = 0.5;
    n = 100;
    wall0 = [0 0 0 0 0 0 0 0 0 0;...
          0 0 0 0 0 0 0 0 0 0;...
          0 a 0 b 0 a 0 0 0 0;...
          0 0 a 0 b 0 a 0 0 0;...
          0 0 0 a 0 a 0 0 0 0;...
          0 a 0 b 0 a 0 0 0 0;...
          0 0 a 0 b 0 a 0 0 0;...
          0 0 0 a 0 a 0 0 0 0;...
          0 0 b 0 b 0 b 0 0 0;...
          0 0 0 0 0 0 0 0 0 0];
    % d0 is the perpendicular distance of the source from the wall. h0 and w0
    % are the starting height and width (corresponding indexes on the 
    % matrix)
    % AvgR will be used to average the number of absorbed waves per
    % configuration.
    d0 = 4;
    h0 = 4;
    w0 = 5; 
    AvgR = [];
    allAVec = [];
    %%begin for loop. This loop creates c number of configurations for the 
        for x = 1:10
            % JR is a placeholder vector that will concatenate the loop variable 
    % j, basically to create an 'x' variable to graph against number of 
    % absorptions. AbsorbVecR is another placeholder that will contain
    % the number of absorbed waves on the initial wall configuration,
    % per iteration.
        JR = []; AbsorbVecR = []; 
        j = 0; wallR = wall0;
        %The initial configuration is shuffled, row by row to get random 
        %configurations. This is stored in wallR, and is repeated on each loop. 
                [rows,cols] = size(wall0) ;
                for t=1:rows 
                    idx = randperm(cols) ; 
                    wallR(t,idx) = wall0(t,:);  
                end
                wallR
            while j < 1000
                absorbR = 0; %initializes number of absorptions to zero
        %initializes vectors that will contain the matrix indices
        %of where the wave hits. Height -> row, width -> column
                relHeightR = []; 
                relWidthR = [];
        %creates a vector of random numbers, each corresponding to the 
        %wave components. To be evaluated against a, b, or zero.
                rnjesusR = rand(1, n);
        %creates the azimuthal (thetaR) and longitudinal (phiR) angles 
        %between -pi/4 & pi/4 used in determining wave 'hit' position.
                thetaR = (pi/4) * 2*(rand(1, n) - 0.5);
                phiR = (pi/4) * 2 *(rand(1, n) - 0.5);
        %dHR and dWR are the vertical and horizontal distances the wave travels
        %from source.
                dHR = int8(d0.*tan(thetaR));
                dWR = int8(d0.*tan(phiR));
                %loops for each sound wave (elements of dHR and dWR)
                for i = 1:length(dHR)
                       %determines the actual hit indices on the matrix
                             relHeightR = [relHeightR, (dHR(i) + h0) + 1];
                             relWidthR = [relWidthR, (dWR(i) + w0) + 1];
                       %tests to see if wave is absorbed, adds to absorb total                        
                             if rnjesusR(i) < wallR(relHeightR(i) , relWidthR(i))
                                 absorbR = absorbR + 1;
                             end
                end
        %for each vector of random waves, the number of absorptions
        %are concatenated in AbsorbVecR, giving a vector which can be
        %used to find the average absorption for the configuration
        %loop variable j is incremented, iteration number added to JR.
                AbsorbVecR = [AbsorbVecR, absorbR];
                j = j + 1;
                JR = [JR, j];
            end
                AbsorbtotR = 0;
                for i = 1:length(AbsorbVecR)
                    AbsorbtotR = AbsorbtotR + AbsorbVecR(i);
                end
                AvgR = [AvgR,(AbsorbtotR)/length(AbsorbVecR)];
                plot(JR, AbsorbVecR)
                xlabel('Iteration Number', 'fontsize', 12);
                ylabel('Number of Absorbtions (per iteration)', 'fontsize'...
                    ,12);
                hold on
        end

Walter Roberson 2017-11-28

The a and b are fixed as to their rows and the number of each in a given row, but may move within the row. The matrix therefore cannot be taken fully randomly.

You could take a random matrix within those constraints, but the problem is doing meaningful optimization against that, short of just trying all of the possibilities.

David Mueller 2017-11-28

编辑：David Mueller 2017-11-28

Ren, thanks for your response. Wall0 and wallR are both 10x10 matrices, the difference being that wallR has its rows randomly shuffled. And yes, I want to optimize wallR for the greatest number of absorptions as you said

The wave is composed of two 100x1 vectors (I just called it wave1 in the pseudocode for simplicity, although now I see that may have been less helpful than intended). One is called relHeight which is just the row numbers where each wave hits, and the other is relWidth which is the column numbers. Together they give the contact index on the wall for each wave. I am open to suggestions regarding changing the organization if it would help.

Walter, yes that is correct regarding the a's and b's. Although, it doesn't necessarily have to be completely random shuffling. If there is a way to look at all the permutations and find the optimal arrangement using other than rng, I would be willing to try.

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