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Please suggest some solution to this code if any mistakes

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function region=eregiongrowing(f)
region=cell(size(f)); %Region matrix with same size of image,storing the labels of grown region
[m,n]=size(f);
seed=[m/2,n/2];%position of seed(x,y)
rcount=1;%counter to keep track of current region being grown
i=1;
j=1;
pg=seed;%pg is stack to sore pixels to grow
x1=im2single(f);
%Adaptive Threshold code start;
Err = 1;
T0=(max(max(x1))+min(min(x1)))/2;
while Err > 0.0001,
u1=0;
u2=0;
cnt1=0;
cnt2=0;
for i=1:m
for j=1:n
if x1(i,j)<= T0
u1=u1+x1(i,j);
cnt1=cnt1+1;
else
u2=u2+x1(i,j);
cnt2=cnt2+1;
end
end
end
u1=u1/cnt1;
u2=u2/cnt2;
T=(u1+u2)/2;
Err=abs(T-T0);
if Err > 0.0001
T0=T;
end
end
y=im2bw(f,T0);
%threshold end
L=bwlabel(y,8);% labelling of region
h=f(:,:,1);
s=f(:,:,2);
v=f(:,:,3);
while ~isempty(pg)%while pg is not empty
cp=pg;% cp -8 neighbours
double(cp);
i=i-1;
for k=1:8
if region(cp(k))~=L;%if region(cp(k)is not labelled
%code to find similarity using Euclidean distance
for i=1:8
for j=1:8
Dh=(h(x+i,y+j,1)-h(x,y,1)).^2;
Ds=(s(x+i,y+j,2)-s(x,y,2)).^2;
Dv=(v(x+i,y+j,1)-v(x,y,1)).^2;
end
end
d=sqrt(Dh+Ds+Dv);
if(d<T)% threshold value founded in adaptive threshold
region(cp(k))=rcount;
i=i+1;
pg=cp(k);
else
j=j+1;
bp=cp(k);% bp is stack to store boundary pixels of grown region
end
end
end
end
while ~isempty(bp);
j=j-1;
rcount=rcount-1;
i=1;
pg(i)=seed;
end
Getting Following message
??? Subscript indices must either be real positive integers or logicals.
Error in ==> eregiongrowing at 48 if region(cp(k))~=L;%if region(cp(k)is not labelled
Please Check the code and tell
me whether I am doing some mistakes anywhere in the code,if so please suggest
the solution

Accepted Answer

Jan
Jan on 19 Mar 2013
Edited: Jan on 19 Mar 2013
It is a very general question to ask the forum for searching a mistake anywhere else in the code. Because the code does not contain a suitable number of comments, we cann only guess, what the lines should achieve. And this guessing is based on the code only. So how could we find the difference between what the code does and what you want it to do?!
At first I would mark the code and hit Cmd-I for a clean indentation, before I post it in the forum.
Then note, that the line double(cp); converts the variable cp to a double temporarily, but does not store the result anywhere, such that this is a waste of time. Do you want:
cp = double(cp);
This looks strange:
for i=1:8
for j=1:8
Dh=(h(x+i,y+j,1)-h(x,y,1)).^2;
Ds=(s(x+i,y+j,2)-s(x,y,2)).^2;
Dv=(v(x+i,y+j,1)-v(x,y,1)).^2;
end
end
Here Dh, Ds, Dv are overwritten in each iteration. Then only the value of i=8 and j=8 are stored, so why do you compute the rest?
pg(i)=seed; must fail, when seed is not a scalar.
The actual problem:
seed=[m/2,n/2];
...
pg=seed;
...
cp=pg;
...
if region(cp(k))~=L % ERROR
This must fail, if m or n is odd, because than cp contains non-integer values.
  2 Comments
Poonam
Poonam on 21 Mar 2013
Whta would be solution for these mistake you have mention,please reply soon
Jan
Jan on 21 Mar 2013
Edited: Jan on 21 Mar 2013
I cannot suggest fixes for all the bugs, because I cannot guess what the lines should produce. Without the necessary comments, a code with bugs is not usable and cannot be maintained. E.g. "pg(i)=seed;", which fails for a non-scalar seed, but how could I find out, what you actually want to achieve? Or if the Dh value should be accumulated by a sum or product, or if an vector or matrix should be created?
The last section must fail also:
while ~isempty(bp);
j=j-1;
rcount=rcount-1;
i=1;
pg(i)=seed;
end
Here the value of bp is not changed inside the loop. Therefore this loop is either not entered at all, or it is infinite.
I suggest not to try to fix this code, because it is contaions to many problems. It will be more efficient to restart from scratch. Dividie the problem into smaller parts, which can be tested more easily, and join the parts after they are working sufficiently.

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More Answers (1)

Walter Roberson
Walter Roberson on 19 Mar 2013
You start with
[m,n]=size(f);
seed=[m/2,n/2]
If the image is an odd size, then m/2 or n/2 will not be integers.
You initialize pg = seed, and then cp = pg, so when you index region(cp(k)) you would be attempting to index region at a non-integer.

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