I'm working on a biometric matching system and I would like to have few suggestions regarding vectorizing the following code:

% tpol and ipol 
% column 1: x co-ordinate of the feature
% column 2: y co-ordinate of the feature
% column 3: angle orientation
% column 4: type of feature

ref_points=size(tpol,1); %number of rows=number of reference image points
   in_points=size(ipol,1); %number of rows=number of input image points
   radsize=7*ones(ref_points,1); %Radius size
   angsize=11* ones(ref_points,1); %Angle size
   radlow=-radsize./2;  % Lower Radius 
   radhigh=radsize./2; % Upper Radius
   anglow=-angsize./2;  % Lower Angle
   anghigh=angsize./2; % Upper Angle
   mscore=0; % initializing the matching score

   for i=1:ref_points
      for j=1:in_points
        rdiff=tpol(i,1)-ipol(j,1);    % Difference between the x-coordinate
        ediff=tpol(i,2)-ipol(j,2);      % Difference between the y-coordinate
        thetadiff=tpol(i,3)-ipol(j,3);   %Difference between the orientation 
        if ((radlow(i) < rdiff) && (rdiff < radhigh(i)) && (anglow(i) < ediff) && ...
            (ediff < anghigh(i)) && (abs(thetadiff) < epsillon) && ...
            tpol(i,4)=3; %Change type of the feature to know that the line was used

ipol and tpol are the input image and reference image points respectively. They are both a matrix of size (A x 4) each, A being the number of features detected in each image.

I asked this question on Mathworks for which I received the following answer:

    r = [-3.5 3.5];  % Lower Radius and Upper Radius
    a = [-5.5 5.5];  % Lower Angle and Upper Angle

    % Vectorized computation of rdiff,ediff,thetadiff i.e. x,y-co-ordinate
    % and orientation
    pol = bsxfun(@minus,tpol,permute(ipol,[3 2 1])); 
    p = all([bsxfun(@gt,pol(:,1:2,:),[r(1),a(1)]) & bsxfun(@lt,pol(:,1:2,:),[r(2),a(2)]),...
     pol(:,3,:) < ep, pol(:,4,:) == 0],2);
    mscore = nnz(p);
    tpol(any(p,3),4) = 4;

I checked the mscore received from the original code and verified it with the solution provided by plugging in one "if condition" at a time. All the "if conditions" provided me the same mscore except for:

if ((tpol(i,4)==ipol(j,4))


p= all ([pol(:,4,:) == 0],2);

It would be very helpful if someone could help me out with this. The for loop takes a long time when the points detected are many.

The input data can be downloaded from Google Drive


Regarding the vectorized code:

Therefore, it's close to impossible to improve the performance of the vectorized code you've posted.

Now, the question is: Is it correct?

It's hard to tell without knowing the exact data, but just by reading the code and your comments, I would think it is.

The reason why I think it is correct, even though the if conditions didn't return the same result:

You're comparing floating point values using ==. Check out Is floating point math broken?

When checking for equality, you should always use some tolerance. Have a look at the example below, where we are comparing 0.3 with 0.1 + 0.2:

x = 0.1;
y = 0.2;
z = 0.3;
the_same = (z == (x + y))
the_same =
the_same = abs(z - (x + y)) < eps
the_same =
  • \$\begingroup\$ I agree with the explanation. But, the data on the fourth column is either type 1 feature or type 2 feature. So I do not have any floating values. Sorry, I forgot to upload the data set. The data set is made available now. Please take a look at it. \$\endgroup\$ – shreyas kamath Jun 17 '16 at 1:36

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