solution for algorithm to link status?

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

Answer 1

[Walter Roberson]

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]

undefined function or variable ni is obtained

[Walter Roberson]

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]

for i = 1,2,...,n

for j = i+1,...,n

test - select random number from uniform distribution between (0,1)

if (test<=;λ∩ni = 1∩nj = 1) then Li,j = 1

else; Li,j = 0

Lj,i = Li,j

L - load Li,j and Lj,i into the matrix L

this is the question actual

[ankanna]

the document of 2tr

[ankanna]

please help to solve this algorithm

[Walter Roberson]

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.

[Walter Roberson]

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

L = 3×3

NaN 1 1 1 NaN 0 1 0 NaN

[ankanna]

probability of existence of a topology is given by

P(alfak=1) = lamda^nl*(1-lamda)^nu

Where

lamda = probability of link existence = 0.7

nl = linked nodes and

nu = unlinked nodes

now we take 0 0 0

p(alfak=1) =(0.7)^0*(1-0.7)^3

=0.027

0 0 1

p(alfak=1)=(0.7)^1*(1-0.7)^2

= 0.063

0 1 0 and 1 0 0 also sme because only one linked

p(alfak=1)=(0.7)^1*(1-0.7)^2

= 0.063

0 11, 1 0 1, 1 1 0

p(alfak=1)=(0.7)^2*(1-0.7)^1

= 0.147

1 1 1

p(alfak=1)=(0.7)^3*(1-0.7)^0

= 0.343

output will

L12 L13 L23 p(alfak=1)

___ ___ ___ _______

0 0 0 0.027

0 0 1 0.063

0 1 0 0.063

0 1 1 0.147

1 0 0 0.063

1 0 1 0.147

1 1 0 0.147

1 1 1 0.343

i need to generate above matrix

please help me to generate the code

[Walter Roberson]

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)

ans = 20×1

5.90490000000001e-06 1.37781e-05 1.37781e-05 3.21489e-05 1.37781e-05 3.21489e-05 3.21489e-05 7.50141000000001e-05 1.37781e-05 3.21489e-05

[minP, minidx] = min(P)

minP = 
      5.90490000000001e-06

minidx = 
     1

bits(minidx,:)

ans = 1×10

0 0 0 0 0 0 0 0 0 0

[maxP, maxidx] = max(P)

maxP = 
              0.0282475249

maxidx = 
        1024

bits(maxidx,:)

ans = 1×10

1 1 1 1 1 1 1 1 1 1

histogram(P)

[ankanna]

how to take 3 by matrix in the order NN will

NN = 1 2 3

1

2

3

for example,

0 0 1

in the matrix

[ 0 1 1

1 0 0

1 0 0 ]

how to generate these type of matrix

[ankanna]

up to n

[Walter Roberson]

I do not understand the question. You want each matrix to have 3 rows and NN columns ? What should be in the columns?

[ankanna]

columns will be in logic 123

and row logic should also 123

we take

001

1 2 3

1 0 1 1

2 1 0 0

3 1 0 0

here 1 2 3 are three nodes n =3

number of links will be 8 (000 to 111)

how to generate above logic matrix ?

[Walter Roberson]

I have no idea what you are asking, if you are not asking something that I solved for you days ago.

[ankanna]

that code have an error. the error is undefined variable l

[ankanna]

please help me to generate error free output

[ankanna]

for i = 1,2,...,n

for j = i+1,...,n

test - select random number from uniform distribution between (0,1)

if (test<=;λ∩ni = 1∩nj = 1) then Li,j = 1

else; Li,j = 0

Lj,i = Li,j

L - load Li,j and Lj,i into the matrix L

this is the actual procedure to generate link status

[Walter Roberson]

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]

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]

L = [l(i,j);l(j,i)];

obtain error only in this part

[ankanna]

how to change this . if any logic is available to change this

[Walter Roberson]

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

Unrecognized function or variable 'ni'.

No undefined variable l because it never gets that far due to ni and nj being undefined.

[Walter Roberson]

So what happens if we define them?

ni = randi([0 1])

ni = 0

nj = randi([0 1])

nj = 0

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

Unrecognized function or variable 'l'.

Okay, that starts to make sense. You only assign to l if ni == 1 . So, we will have to initialize l

[Walter Roberson]

ni = randi([0 1])

ni = 0

nj = randi([0 1])

nj = 0

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 = 2×1

0 0

L = 2×1

0 0

L = 2×1

0 0

l

l = 3×3

0 0 0 0 0 0 0 0 0

[ankanna]

i dont want l is 000

i need only this

l = 3×3

0 1 1

1 0 1

1 1 0

[ankanna]

please help to generate this code

[ankanna]

nodes=3; ri=0.9, lamda=0.7;

l = zeros(nodes, nodes);

for i=1: nodes

test=unifrnd (0,1)

if test<=ri

ni=1

else

ni=0

end

ns(i)=[ni]

end

for i=1:nodes

for j=i+3:nodes

test=unifrnd (0,1)

if test<=(lamda/ns(i)==1/ns(j)==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

this will be generated but the output is

L[ ]

[ankanna]

i want output L will be

0 1 1

1 0 1

1 1 0

please help me to generate this output

[Walter Roberson]

You cannot get that output with that algorithm. The paper published the wrong algorithm.

[ankanna]

if any possible to change this algarithm.

[ankanna]

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