# solution for algorithm to link status?

2 views (last 30 days)
ankanna on 17 Mar 2021
Commented: ankanna on 8 Apr 2021
nodes = 3; m = 0.7
for i = 1:nodes
for j = i+1:nodes
test = unifrnd(0,1)
if ni = 1
nj = 1:nodes
if (test<=m/ni=1/nj=1)
li,j = 1
else
li,j = 0
end
lj,i = li,j
end
L = li,j;lj,i
end

Walter Roberson on 17 Mar 2021
Edited: Walter Roberson on 17 Mar 2021
nodes = 3; m = 0.7;
for i = 1:nodes
for j = i+1:nodes
test = unifrnd(0,1);
if ni == 1
nj = 1:nodes;
if (test<=m/ni==1./nj==1)
l(i,j) = 1;
else
l(i,j) = 0;
end
l(j,i) = l(i,j);
end
L = [l(i,j);l(j,i)];
end
end
However, we can prove that the if (test<=m/ni==1./nj==1) will be false except when nodes = 1 (in which case the code is just rather strange but could be true sometimes.)
ankanna on 8 Apr 2021
please to generate link status i.e., the output will be contain this matrix. how to generate this code
0 1 1
1 0 1
1 1 0

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