how to create a matrix in matlab?

L12 L13 L23

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

how to create a matrix

Answer 1

[Veronica Taurino]

Your question is not clear. If you have those 3 arrays:

L12    = [0 0 0 0 1 1 1 1];
L13   = [0 0 1 1 0 0 1 1];
L23 = [0 1 0 1 0 1 0 1];
Matrix = [ L12; L13; L23]

or

Matrix = [ L12; L13; L23 ]' 

depending or your needs

Answer 2

[Stephen23]

M = dec2bin(0:7)-'0'

M = 8×3

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

Answer 3

[Walter Roberson]

[L23, L13, L12] = ndgrid(0:1);
L12 = L12(:); L13 = L13(:); L23 = L23(:);
table(L12, L13, L23)

ans = 8×3 table
    L12    L13    L23
    ___    ___    ___

     0      0      0 
     0      0      1 
     0      1      0 
     0      1      1 
     1      0      0 
     1      0      1 
     1      1      0 
     1      1      1 

[ankanna]

how to generate (S - T) path?

[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

how to generate code by using this algorithm. please help me to generate code for link status

[Walter Roberson]

I do not understand your pseudocode.

Are you trying to create an algorithm for link assignment between communicating users?

[Walter Roberson]

%input is n, scalar integer starting from 0
[
  bitget(n,4), bitget(n,3), bitget(n,2);
  bitget(n,3), bitget(n,4), bitget(n,1);
  bitget(n,2), bitget(n,1), bitget(n,4);
]

Which can be written in terms of

NN = [4, 3, 2; 3 4 1; 2 1 4]
bitget(n,NN)

Why that particular NN? Well, for the case of node = 3, it is

NN = toeplitz(node+1:-1:2);
mask = logical(fliplr(diag(ones(1,node-1),-1)));
NN(mask) = 1;
out = bitget(n, NN)

It is likely that this would have to change for node = 4, but you would need to give more information about the pattern for 4 for me to predict.

[Walter Roberson]

node = 3;

NN = toeplitz(node+1:-1:2);
mask = logical(fliplr(diag(ones(1,node-1),-1)));
NN(mask) = 1;

for n = 0:7
  out = bitget(n, NN)
end

out = 3×3

0 0 0 0 0 0 0 0 0

out = 3×3

0 0 0 0 0 1 0 1 0

out = 3×3

0 0 1 0 0 0 1 0 0

out = 3×3

0 0 1 0 0 1 1 1 0

out = 3×3

0 1 0 1 0 0 0 0 0

out = 3×3

0 1 0 1 0 1 0 1 0

out = 3×3

0 1 1 1 0 0 1 0 0

out = 3×3

0 1 1 1 0 1 1 1 0

[ankanna]

how to put this topologies in matrix

[Walter Roberson]

for n = 0:2^nodes-1
  out(:,:,n+1)= bitget(n, NN);
end

[ankanna]

NN = toeplitz(node+1:-1:2);
mask = logical(fliplr(diag(ones(1,node-1),-1)));
NN(mask) = 1;

can you please explain this logic

and explain what the logic is used

clear explanation please

[ankanna]

n = 4;
L=(n-1)*n/2;
c = 2^L-1;
LN = 4
NN = toeplitz(LN+1:-1:2);
mask = logical(fliplr(diag(ones(1,LN-1),-1)));
NN(mask) = 1;
for n = 0:c
  out = bitget(n, NN)
end

output :

out =

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

but i want this diagnol matrix upto n nodes

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

[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 give reliability that is ri = 0.9?

please help me to generate code

[ankanna]

n = 3;ri=0.9; lamda = 0.7;
L=(n*(n-1))/2;
c=2^L;
l=zeros(n,n);
for i = 1:n
    for j = i+1:n
        test = unifrnd(0,1)
        if test<=ri
        ni=1
    else
        ni=0
    if ni == 1
        if  (test<=lamda/n(i)==1./n(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
    end
end

after this code how to add source-destination and two terminal reliability. please help to generate this code.

Source-destination procedure:

Initialize:

source = 1

destination = n

if ni = 1 then lamdasource = 1

for hop = 1,2,...,n-1

for i = 1; ... , n

if lamdai = 1

for j = 1; 2 ..., n

if lij = 1; then lamdaj = 1

else lamdaj = 0

if lamdadestination = 1; then path = 1

else; path = 0

These procedures are used to generate the following MAWN simulation approach, where Q is the number of runs in the simulation.

Calculation of 2TR m:

for q = 1,2,...,Q

Simulate Network --->Lq

Find Path->pathq

2TR m = (Summation with limits Q and q=1 (pathq))/Q

[Walter Roberson]

That code will never do what you want it to do, and it cannot be fixed.

Your source is a paper that is wrong. You need to get corrections from the authors of the paper.

[ankanna]

node = 3; ri=0.9;

L=(node*(node-1))/2;

configuration = dec2bin(0:(2^L-1))-'0';

alfak=configuration;

source node=1; destination node=3;

m = Limit on intermediate node;

2TR(alfak) == ri^m;

how to calculate two terminal reliability.

alfak Path 2TR(alfak)

1 r1r3 0.81

2 r1r3 0.81

3 r1r2r3 0.729

4 r1r3 0.81

5 None 0.00

6 r1r3 0.81

7 None 0.00

8 None 0.00

please help me to generate above and i want that 2terminal reliability at output.