Matlab - Not getting the expected result - matlab

Question

I have written a function that is supposed to do the following:
Take as an input two sets
Take the distance between the two sets using pdist2 which code is shown here.
It will take the distance between the two sets at the beginning. Then, for the second set, at each iteration, it will set the (i,j) location to 0 and calculate the distance with this change. An, when it goes to the next iteration, it should change the next location value to '0' while at the same time return the preceding value which was set to '0' to its original value.
Note that the result from pdist2 originally return as a matrix, but for the comparison, I sum up the matrix values to use them for comparison.
Based on that, I have written the following function (note that you can use the pdist2.m function from the link here):
function m = pixel_minimize_distance(x,y)
sum1=0;
sum2=0;
[r c] = size(y);
d1 = pdist2(x,y);
[r1 c1] = size(d1);
for i=1:r1
for j=1:c1
sum1=sum1+d1(i,j);
end
end
maximum = sum1;
for i=1:r
for j=1:c
o = y(i,j)
y(i,j) = 0;
d2 = pdist2(x,y);
[r2 c2] = size(d2);
for i=1:r2
for j=1:c2
sum2=sum2+d2(i,j);
end
end
if sum2 >= maximum
if o ~= 0
maximum = sum2;
m = o;
end
end
if sum2 <= maximum
maximum = maximum;
end
y(i,j)=o;
end
end
end
Now, this is what I have run as a test:
>> A=[1 2 3; 6 5 4];
>> B=[4 5 3; 7 8 1];
>> pixel_minimize_distance(A,B)
o =
4
o =
4
o =
1
o =
7
o =
7
o =
0
ans =
7
See the the answer here is 7 (scroll down if you cannot see it), while the expected value when I calculate this manually should be 3, as since when we set it to 0 the sum of the distance will be 142.
Any idea what could be wrong in the code? I think it would be in the location in the code of setting o = y(i,j) where o denotes original value, but really couldn't figure a way of solving that.
Thanks.

Answer 1

I think you have many redundant commands in your code. I just removed them, nothing else. I am getting value of m as 3. I used MATLAB's pdist2 function with squared euclidean distance (since that is the default in the function you provided). I did not get 142 as the distance.
Here is the code:
function m = pixel_minimize_distance(x,y)
[r c] = size(y);
maximum = (sum(sum(pdist2(x,y)))).^2; %explained below
for i=1:r
for j=1:c
o = y(i,j);
y(i,j) = 0
sum2 = (sum(sum(pdist2(x,y)))).^2;
if sum2 >= maximum
if o ~= 0
maximum = sum2;
m = o;
end
end
y(i,j)=o;
end
end
end
and output is:
y =
0 5 3
7 8 1
y =
4 0 3
7 8 1
y =
4 5 0
7 8 1
y =
4 5 3
0 8 1
y =
4 5 3
7 0 1
y =
4 5 3
7 8 0
m =
3
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Explanation:
You have written the following snippet of code:
d2 = pdist2(x,y);
[r2 c2] = size(d2);
for i=1:r2
for j=1:c2
sum2=sum2+d2(i,j);
end
end
what this simply does is calculates the distance between two sets using pdist2 and sums up the entire distance matrix to come up with one value stored in sum2 in your case.
Lets look at the my code:
sum2 = (sum(sum(pdist2(x,y)))).^2;
pdist2 will give the distance. First sum command will sum along the rows and then the second one will sum along the columns to give you a total of all values in the matrix (This is what you did with two for loops). Now, the reason behind .^2 is:
In the original pdist2 function in the link which you have provided, you can see from the following snippet of code:
if( nargin<3 || isempty(metric) ); metric=0; end;
switch metric
case {0,'sqeuclidean'}
that squared Euclidean is the default distance, whereas in MATLAB, Euclidean distance is default. Therefore, I have squared the term.
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