Table 1 (a–f) are cipher text blocks, cipher text sub-blocks, Brain disease cover image blocks, assignment of random number to cover image blocks, cover image block into bits and diagonal queues, respectively

Suppose we have the following data

  a. N blocks of permuted cipher text: 1x16

b. Permuted 4 sub-blocks of each N block: 1x4

c. M block of 8 pixels each, from cover image: 1x8

d) Now, assign pseudo-random number to each pixel for converting the above block into bits: As example the pseudo-random number generated as \(01100110\)

e) M block of \(64\) bits each. It is obtained by converting the above block into bits as follows:

If the generated random value is \(0\) for particular pixel, then the \(8\) bits will be distributed from right to left direction.

If the generated random value is \(1\) for particular pixel, then the \(8\) bits will be distributed from left to right direction example as above.

Using the above matrix, various diagonal queues from left to right inserted bits are created on the basis of above example.

The above shown bold bits in the example are used to swap with the cipher text bits using FIFO property of queue.

Now, we will select one of these eligible diagonal queues, using chaos theory of random number generation.

We will also select one of the N blocks and sub-blocks using chaos theory of random number generation.

We will then put the selected ciphertext bits, in selected diagonal queue at 5th to 8th bit LSB position.

Example for selection of hiding position in diagonal queues using pseudo-random number is as follows:

If \(L5(1)\) is having value \(1\) then \(4 + L5\left( 1 \right)\) is \(5\), i.e., secret cipher data bit will be hidden at 5th LSB position in selected diagonal.

If \(L5(1)\) is having value \(2\) then \(4 + L5\left( 1 \right)\) is \(6\), i.e., secret cipher data bit will be hidden at 6th LSB position in selected diagonal.

If \(L5(1)\) is having value \(3\) then \(4 + L5\left( 1 \right)\) is \(7\), i.e., secret cipher data bit will be hidden at \(7\) th LSB position in selected diagonal.

If \(L5(1)\) is having value \(4\) then \(4 + L5\left( 1 \right)\) is \(8\), i.e., secret cipher data bit will be hidden at 8th LSB position in selected diagonal.