finding minimum value in Quanterion matrix

I have a 386x514 quaternion array,in this please tell how to find the minimum value of that matrix

[Walter Roberson]

What does each element look like? Is there a reason you cannot just use min(TheMatrix(:)) ?

[FIR]

My form is 384x512 pure quaternion array with uint8 components

if i use min i get error

Undefined function or method 'min' for input arguments of type 'quaternion'.

[Matt J]

FIR Commented:

i downloaded [the quaternion] tool box from here

http://sourceforge.net/projects/qtfm/

FIR Commented:

Sorry Matt my code [which doesn't work] is

    A=imreadq('peppers.png');
    T=65;
    for i=2:6
        for j=2:6
            q= A(i-1:i+1,j-1:j+1);
            [minsum, minidx] = min( sum(abs(bsxfun(@minus, q(:), q(:).'))) );
            qVFM = q(minidx);
              V1=an equation
              V2=an equation
              S=min([V1 V1]);
              if S>T
                  THE CENTRE PIXEL IS REPLACED BY qVFM %%%%3x3 matrix
              else
                  THE CENTRE PIXEL IS not REPLACED BY qVFM%%%%3x3 matrix
              end
          end
      end

Answer 1

[Matt J]

BSXFUN isn't overloaded well in the quaternion toolbox that you're using. Here is a workaround

 q= convert(A(i-1:i+1,j-1:j+1),'single');
[minsum, minidx] = min( sum(abs(   repmat(q(:),1,9)-repmat(q(:).',9,1)  ) ));

[Walter Roberson]

Odd, a lot seemed to disappear there. As I know that I had commented that

min( sum(abs(bsxfun(@minus, q(:), q(:).')))

is code that was developed for numeric arrays, not for quanterion arrays, and that we need to be shown the formula that is being implemented by this code so that we can re-examine it with the quanterion toolbox in mind.

[Matt J]

Well, I think you rightly deduced that the desired formula is the same as in FIR's previous post

http://www.mathworks.com/matlabcentral/answers/56973-implementing-equation-doe-3x3-matrix

except that now we know the q_i in each 3x3 block are not scalars, but rather quaternions. My proposed workaround operates on this assumption.

[FIR]

Thanks a lot Matt it is working fine,

q= convert(A(i-1:i+1,j-1:j+1),'single');

q gives 3x3 matrix

final question, i have to add 2nd and 8th pixel and divide by 2 from q, please help

[Walter Roberson]

2nd and 8th pixel, or 2nd and 8th quaternion ?

[FIR]

q gives 3x3 quanterion,in that 3rd and 8th quanterion

sorry for mistake

[Matt J]

(q(3)+q(8))/2  ?

[FIR]

i have found 4 values using above equation

V1,V2,V3,V4

now i want to apply in code

A=imreadq('peppers.png');
    T=65;
    for i=2:6
        for j=2:6
            q= convert(A(i-1:i+1,j-1:j+1),'single');
[minsum, minidx] = min( sum(abs(   repmat(q(:),1,9)-repmat(q(:).',9,1)  ) ));

                S=min([V1 V2 V3 V4]);
                if S>T
                    THE CENTRE PIXEL OF QUANTERION ARRAY IS REPLACED BY minsum

                else
                    THE CENTRE PIXEL IS not REPLACED BY minsum
                end
            end
        end

please help,and original image must be added with noise,please assist

[Matt J]

Isn't the center pixel A(i,j)? If so, do A(i,j)=whatever.

[Walter Roberson]

Or, since that would affect the computations as you slid the window, create a second array and set the pixels in it -- as I showed you in one of your previous questions.