Assalam Alikoum all,
I have this code in matlab - which I don't know anything about it - and need to transfer it to java, so could anybody please help on that ?
Many thanks in advance
function s = Skeleton(a, varargin)
% function s = Skeleton(a);
% function s = Skeleton(a, iterations);
%
% Skeletonizing a binary image using Zhang and Suen's method from
%
% "A fast parallel algorithm for thinning digital patterns"
% Comm ACM, Vol. 7, No. 23, pp. 326 -- 329, 1985.
%
% as described in Gonzales & Wintz.
%
% 'a' should contain binary data with 0 considered to be the
% background and 1 the foreground (or object to be thinned).
%
% 'iterations' is the number of iterations of the algorithm to
% perform. Each iteration can only strip away a one pixel wide
% boundary. If no 'iterations' argument is supplied, the function
% iterates until the result converges.
%
% This is a straightforward implementation and as such is very slow.
% There is much scope for optimisation.
%
% Gary Dickson <moc.liamtoh@noskcidyrag>
%
% Check arguments
%
if nargin == 1
iterations = -1; % Flag to iterate until convergence
elseif nargin == 2
iterations = varargin(1); % Specified number of iterations
iterations = iterations{1}; % Want plain old number, not a cell
else
disp('Too many arguments supplied.');
s = a;
return;
end
% Check input data
%
if (min(min(a)) < 0) | (max(max(a)) > 1)
disp('input not binary')
s = a;
return;
end
[h w] = size(a);
s = a;
it = 1;
prevsum = 0;
while 1
% Step 1 (thinning top and left sides)
%
m = ones([h w]);
for j = 2 : h-1
for i = 2 : w-1
if (s(j, i) == 1)
condA = sum(sum(s(j-1 : j+1, i-1 : i+1))) - s(j, i);
condB = Num01Transitions(s, j, i);
condC = s(j-1, i) * s(j, i+1) * s(j+1, i); % p2 * p4 * p6
condD = s(j, i+1) * s(j+1, i) * s(j, i-1); % p4 * p6 * p8
if (condA >= 2) & (condA <= 6) & (condB == 1) & (condC == 0) & (condD == 0)
m(j, i) = 0;
end
end
end % i
end % j
s = s .* m;
% Step 2 (thinning bottom and right sides)
%
m = ones([h w]);
for j = 2 : h-1
for i = 2 : w-1
if (s(j, i) == 1)
condA = sum(sum(s(j-1 : j+1, i-1 : i+1))) - s(j, i);
condB = Num01Transitions(s, j, i);
condC = s(j-1, i) * s(j, i+1) * s(j, i-1); % p2 * p4 * p8
condD = s(j-1, i) * s(j+1, i) * s(j, i-1); % p2 * p6 * p8
if (condA >= 2) & (condA <= 6) & (condB == 1) & (condC == 0) & (condD == 0)
m(j, i) = 0;
end
end
end % i
end % j
s = s .* m;
% Time to stop? As points are always being removed and never added, we can
% check if the method has converged by checking if the sum of the image
% values is the same for two successive iterations.
%
newsum = sum(sum(s));
if (newsum == prevsum) & (iterations == -1)
break;
end;
if (it >= iterations) & (iterations ~= -1)
break;
end
it = it + 1;
prevsum = newsum;
end % it
%------------------------------------------------------------------------------%
% %
%------------------------------------------------------------------------------%
% Count the number of 0 -> 1 transitions that occur while traversing the eight
% neighbours of the point of interest.
%
% p9 p2 p3
% p8 p1 p4
% p7 p6 p3
%
function Nt = Num01Transitions(c, j, i)
p = [c(j-1, i) c(j-1, i+1) c(j, i+1) c(j+1, i+1)];
p = [p c(j+1, i) c(j+1, i-1) c(j, i-1) c(j-1, i-1) c(j-1, i)];
pp = zeros(1, 8);
for k = 1 : 8
pp(k) = p(k+1) - p(k);
end
Nt = sum(pp == 1);
