how to create a matrix in matlab?

2021 3 16
3 个回答
回答已采纳
更新时间：2021 4 20
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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

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 采纳的回答

Veronica Taurino
Veronica Taurino 2021-3-16
编辑：Veronica Taurino 2021-3-16

1 个投票

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

更多回答（2 个）

Stephen23
Stephen23 2021-3-16
Ran in:

1 个投票

Ran in:
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
Walter Roberson
Walter Roberson 2021-3-16
Ran in:

0 个投票

Ran in:
[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

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ankanna
ankanna 2021-3-16
how to generate (S - T) path?
ankanna
ankanna 2021-3-16
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
Walter Roberson 2021-3-16
I do not understand your pseudocode.
Are you trying to create an algorithm for link assignment between communicating users?
Walter Roberson
Walter Roberson 2021-3-16
编辑：Walter Roberson 2021-3-16
%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
Walter Roberson 2021-3-16
Ran in:
Ran in:
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
ankanna 2021-3-20
how to put this topologies in matrix
Walter Roberson
Walter Roberson 2021-3-20
编辑：Walter Roberson 2021-3-20
for n = 0:2^nodes-1
out(:,:,n+1)= bitget(n, NN);
end
ankanna
ankanna 2021-3-20
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
ankanna 2021-3-20
编辑：Walter Roberson 2021-3-24
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
ankanna 2021-3-20
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
Walter Roberson 2021-3-24
Ran in:
Ran in:
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
ankanna 2021-4-2
how to give reliability that is ri = 0.9?
please help me to generate code
ankanna
ankanna 2021-4-11
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
Walter Roberson 2021-4-11
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
ankanna 2021-4-20
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.

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