How to rearrange random binary input for each iteration?

I am having 7 randomnly generated binary input. With this input I have performed some set of mathematical calculations. Now i need to repeat the mathematical calculations with changed order of binary input and this has to be followed for each iteration until a certain condition is met.

Example:

A = [011; 101; 010; 001; 100; 110; 111]

Expected output: Input order of A has to be varied for each iteration.

Answer 1

[Bjorn Gustavsson]

Have a look at the nextperm tool on the file exchange. It will allow you to cycle through all permutations.

HTH

[ASHA PON]

This is not working. Can you suggest some other method.

[Image Analyst]

This does not give a binary vector

A = [011; 101; 010; 001; 100; 110; 111]

A = 7×1

11 101 10 1 100 110 111

Those are decimal, base 10 numbers, not base 2 numbers.

Please attach your actual code.

[ASHA PON]

I have used the following code to generate binary numbers and this is stored in a text file and it is called in main program.

A= randperm(7);
b= dec2bin(r);

Unrecognized function or variable 'r'.

c = convertCharsToStrings(b)
mat = reshape (b,[7,3]);
matstng=string(mat);

[Walter Roberson]

A = randperm(7)

A = 1×7

5 2 7 3 6 4 1

for col = (dec2bin(A) - '0').'
    row = col.'
    result = row(1)*11 + row(2).^2 - row(3)   %some calculation
end

row = 1×3

1 0 1

result = 10

row = 1×3

0 1 0

result = 1

row = 1×3

1 1 1

result = 11

row = 1×3

0 1 1

result = 0

row = 1×3

1 1 0

result = 12

row = 1×3

1 0 0

result = 11

row = 1×3

0 0 1

result = -1

[ASHA PON]

Actually this numbers has to vary for each iteration. Can you please suggest the mehod.

[Walter Roberson]

The number does vary for each iteration. Each row = output line shows the input that was received, and you can see that no two of the inputs were the same.

[ASHA PON]

Actually i need to apply it for higher number of bits, this is my example problem.

result = row(1)*11 + row(2).^2 - row(3), this means if i need to generate 12 bit number, then how should i write this code.

[Walter Roberson]

Just write the expression using indexing.

If you have a linear expression, you could use multiplication such as [3 -1 5 11 4 2 -5 ...] * col

A = randi([0 2^12-1], 1, 5)
for col = (dec2bin(A) - '0').'
    row = col.'
    result = row(1)*11 + row(2).^2 - row(3) + 7*row(4).^2 - row(5).^9 + row(6)*2 - row(7)*5 + row(8).^3 - row(9)*15 + row(10)*3 + row(11)*8 - row(9)*4  %some calculation
end

row = 1×12

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

result = -6

row = 1×12

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

result = 1

row = 1×12

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

result = 17

row = 1×12

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

result = -6

row = 1×12

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

result = -3

[ASHA PON]

Thank you for the reply