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Stack Gives Back Featured on Meta. New post summary designs on greatest hits now, everywhere else eventually. Related Hot Network Questions. As the diagram shows the extra two bit in each blocks comes from its neighbor.
It is consider as circular so that the first bit comes from the last bit and viceversa. Sub3 3 :So we get a 48 bit output from the P-box which is "bitwise xor" with the 48 bit key of that round. Sub4 3 :the xor output is of 48 bit which is then fitted in a compression S-box. The S-box consist of 8 small Sbox which takes 6 bit and results in 4 bit.
Each S-box has a different configuration which is fixed for each round. S-box 1 configuration:. Consider the first 6 bit, Find the decimal value of 1st and last bit which will give the row number and the remaining 4 bits are converted to decimal to get the column. The corresponding value is fetched from the above table and is again converted to binary which will be of 4 bit since every number in the matrix is less than Similarly you can do it for the rest of the bit.
The configuration of the other S -box as follows:. Let's Discuss Cancel reply. Iconic One Theme Powered by Wordpress. This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Close Privacy Overview This website uses cookies to improve your experience while you navigate through the website.
Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. And middle 4 numbers together represent column number. Since maximum number with 4 bits is 15, S box also contains columns 0 to 15 total of So for this input the number positioned at row 1 and column 13 will be picked.
As mentioned earlier S box only contains number in range 0 to All can be represented in 4 bits. So picked number 4 bits are output for the S box.
See the code for all S boxes. Permutation: After getting output from all S boxes, we are applying again permutation. Here also a matrix with different arrangements will be there, we have to arrange according to that. Final XOR: After this permutation, take the left half which initially divided 64bit text to two halves. Do XOR with this permutation output to left 32bit part.
This result is new Right part. And Right 32bit part which passed through all permutation will be come as new Left Part. These 2 parts will be the inputs for the second round. Same as keys also, the parts before left shift are next round input keys. All this explanation for a single round for a 62bit plain text. Like this, it passes through total 16 rounds.
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