To do what you ask would require the result vector to have 145, not 144, elements, unless the vector is considered circular with the first element regarded as following the last element. In the following code you can have your choice since the last element in B always equals the first element. You can either leave B as is with its 145 elements or you can delete the last element, depending whether the vector is to be considered circular.
n = 12;
b = true;
while b
A = zeros(n);
B = zeros(n^2+1,1);
j = ceil(n*rand);
B(1) = j;
for k = 1:n^2
f = find(A(:,j)==0);
m = length(f);
if m == 0
break;
else
i = f(ceil(m*rand));
A(i,j) = k;
j = i;
B(k+1) = j;
end
end
b = any(any(A==0));
end
This is not highly efficient code since it continues to repeat the while loop until it succeeds in completing the vector B with the requested property. This usually takes a number of trials. When it succeeds, the vector B will have one occurrence of each possible pairing of successive elements, and since the 'rand' function is used in this placement, B can be considered random in a sense.
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