solution for algorithm to link status?
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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
please to generate this
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Walter Roberson
2021-3-17
编辑：Walter Roberson
2021-3-17
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.)
31 个评论
ankanna
2021-3-17
Walter Roberson
2021-3-17
Yes, that is to be expected. Your code uses ni twice but does not define it at all. You need to define it as something appropriate to the situation.
I would, however, point out that if ni is not either 1 or a vector of all 1's, then the if ni = 1 would fail, and your code would not construct l(i,j) or l(j,i) . That would suggest that for all the interesting cases, ni = 1 would have to be in effect.
ankanna
2021-3-17
ankanna
2021-3-17
ankanna
2021-3-17
Walter Roberson
2021-3-17
Sorry, I looked at the document but it uses notation that I do not understand. I think that the authors would need to be asked for the meaning of part of it. In particular, I do not understand the semi-colon between the less-than-or-equal-to and the lambda.
nodes = 3; lambda = 0.7;
r = rand(1, nodes) * 0.3 + 0.7; %reliability
%4.2.1.1
%no loop needed in MATLAB
n = rand(1,nodes) <= r;
%4.2.1.2
%no loop needed in MATLAB
test = squareform(rand(1,nodes*(nodes-1)/2));
test(1:nodes+1:end) = 0; %node to itself is fully reliable
L = double(test <= lambda & n & n.');
L(1:nodes+1:end) = nan; %but algorithm leaves node to itself undefined
L
ankanna
2021-3-21
format long g
nodes = 10;
lamda = 0.7;
bits = dec2bin(0:2^nodes-1)-'0';
nl = sum(bits,2);
nu = nodes-nl;
P = lamda.^nl .* (1-lamda).^nu;
P(1:20)
[minP, minidx] = min(P)
bits(minidx,:)
[maxP, maxidx] = max(P)
bits(maxidx,:)
histogram(P)
ankanna
2021-3-21
ankanna
2021-3-21
Walter Roberson
2021-3-21
I do not understand the question. You want each matrix to have 3 rows and NN columns ? What should be in the columns?
ankanna
2021-3-22
Walter Roberson
2021-3-22
I have no idea what you are asking, if you are not asking something that I solved for you days ago.
ankanna
2021-4-7
ankanna
2021-4-7
ankanna
2021-4-7
Walter Roberson
2021-4-7
that code have an error. the error is undefined variable l
Please post the URL of the code that I posted that had the undefined variable l ?
Walter Roberson
2021-4-7
this is the actual procedure to generate link status
Sorry, I looked at the document but it uses notation that I do not understand. I think that the authors would need to be asked for the meaning of part of it. In particular, I do not understand the semi-colon between the less-than-or-equal-to and the lambda.
If the semi-colon were not there, then the test
if (test<=;λ∩ni = 1∩nj = 1) then Li,j = 1
else; Li,j = 0
would translate as
L(i,j) = test <= lambda && ni == 1 && nj == 1;
However it is not clear in the document whether ni and nj are intended to be distinct variables or are intended to be a vector n indexed at locations i and j -- but that would be a problem because n occurs as a limit in the for statement, implying that it is a scalar.
When I looked at the paper, it seemed most likely that the equation was wrong, that the operator in the test should be ∪ instead of ∩
ankanna
2021-4-7
ankanna
2021-4-7
You seem to be referring to the earlier posting,
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
No undefined variable l because it never gets that far due to ni and nj being undefined.
So what happens if we define them?
ni = randi([0 1])
nj = randi([0 1])
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
Okay, that starts to make sense. You only assign to l if ni == 1 . So, we will have to initialize l
ni = randi([0 1])
nj = randi([0 1])
nodes = 3; m = 0.7;
l = zeros(nodes, nodes);
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
l
ankanna
2021-4-8
ankanna
2021-4-8
ankanna
2021-4-8
ankanna
2021-4-8
Walter Roberson
2021-4-8
You cannot get that output with that algorithm. The paper published the wrong algorithm.
ankanna
2021-4-8
ankanna
2021-4-8
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