+Samah Awad If you have 8 bits, 12345678, then apply IP you get an output X = 26314857. Then using input X as input to IP-1 must produce the original 8 bits as output, i.e. 12345678. Therefore IP-1 must be the permutation such that the 1st input bit moves to the 2nd position, the 2nd input bit moves to the 6th position, ..., the 8th input bit moves to the 7th position.
some tutorials make LS-2 by the values that obtained after p10 (LS-2 for 10000|01100) not for the values after LS-1(not 00001|11000) . what can i use??
First, here are my notes on the simplified DES example: ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf. I checked the Stallings textbook (Appendix G of 5th edition - you may find that chapter online) and it applies LS-2 for the values after LS-1. (If the textbook is wrong, then so am I).
After applying P10 we have 10000|01100. Then a left shift (by 1 position) is applied separately to both the left half and the right half. 10000 becomes 00001 (because the left most bit wraps around to become the right most bit). 01100 becomes 11000. A PDF showing the steps is available at ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf
If you mean the 2nd round of S-DES, then yes, the input it S1 is 1000 and the output is 11 (although I don't remember doing that in the lecture). For a more detailed write up of the example see: ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf
fk2 starts with the output of the SWap and then applies EP, XOR, S0/S1, P4 and XOR in the same steps as was previously done except it uses subkey 2 as input to the XOR. Look at slide 22 in ict.siit.tu.ac.th/~sgordon/css322y13s2/slides/css322y13s2l03-block-ciphers-and-des.pdf to see fk is the steps enclosed in the dark grey box. IP-1 is the inverse of IP, which means IP-1(IP(X)) = X. So if you take some input X and apply the arrangement specified by IP and get Y, then if you apply IP-1 on Y you must get the original X back. Since IP is [2 6 3 1 4 8 5 7], it means the 1st bit of X goes to the 4th position, and the 2nd bit of X goes to the 1st position in Y. So IP-1 must take the 4th bit of Y and put it in the 1st position, and take the 1st bit of Y and put it in the 2nd position. The start of IP-1 is [4 1 ...]. Try to find the rest.
This was done 10 years ago, but it is still very valuable today. Thank you.
The best explanation I have watched so far. Thank you.
you are amazing Prof Steven , Thank you very much
Fantastic explanation of DES ,, visually very effective. Thanks Mr Steven Gordon
Thanks a lot!!!! Helped me understand the SDES key generation and ciphering.
thanks Mr. Steven for sharing your knowledge. this video really helps me a lot to understand the concept.
Thank you professor, this helped me a lot in understanding DES & now I can easily use S-DES.
very good explanation, thanks a lot for helping me understanding the S DES Algo and making it clear to my students
sir i'm from india and it really helped me... on my clg project
Can we use 3DES in EED mode with K1,K1,K2 keys? how will the strength differ in finding the keys in terms of EDE(k1,k2,k1) and EED mode(k1,k1,k2?)?
Nice explanation and useful links. Thanks a lot.
thnx alot mr. Steven Gordon
Would there be much of a change of i had a 9 bit key? input message 100010110101 and the key 111000111?
understood to the very best level..thanks a lot
please How made S0 box and S1 box 4×4 matrix...is it rule .
Very good explanation... finally understand thanks
+butterfly4Christ plese can u tell me how we find IP-1 ??
+Samah Awad If you have 8 bits, 12345678, then apply IP you get an output X = 26314857. Then using input X as input to IP-1 must produce the original 8 bits as output, i.e. 12345678. Therefore IP-1 must be the permutation such that the 1st input bit moves to the 2nd position, the 2nd input bit moves to the 6th position, ..., the 8th input bit moves to the 7th position.
+Steven Gordon thanks i understood
Please give some exercise of DES Algorithm.
some tutorials make LS-2 by the values that obtained after p10 (LS-2 for 10000|01100) not for the values after LS-1(not 00001|11000) . what can i use??
First, here are my notes on the simplified DES example: ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf. I checked the Stallings textbook (Appendix G of 5th edition - you may find that chapter online) and it applies LS-2 for the values after LS-1. (If the textbook is wrong, then so am I).
agree.. thanks for the lesson!!!
it was a very helpful for me
Regards
Thanks a lot. Great tutorial.
Awesome class sir.. (y)
Very helpful video....Thank you very much.
Thank you Sir.
شرح رائع ..شكراً جزيلاً :)
in the LS1 isn't it supposed to be 00001|00110 ? I don't get it
After applying P10 we have 10000|01100. Then a left shift (by 1 position) is applied separately to both the left half and the right half. 10000 becomes 00001 (because the left most bit wraps around to become the right most bit). 01100 becomes 11000. A PDF showing the steps is available at ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf
you made a mistake on s-box 2(the right side). its supposed to be 1000
great tutorial though
explain please
If you mean the 2nd round of S-DES, then yes, the input it S1 is 1000 and the output is 11 (although I don't remember doing that in the lecture). For a more detailed write up of the example see: ict.siit.tu.ac.th/~sgordon/reports/simplified-des-example.pdf
Very helpful thank you
真是非常感谢！救了我一命啊！
Thank u very much
thank you!
Also i can't understand the last part at Fk2 and IP-1 ??
fk2 starts with the output of the SWap and then applies EP, XOR, S0/S1, P4 and XOR in the same steps as was previously done except it uses subkey 2 as input to the XOR. Look at slide 22 in ict.siit.tu.ac.th/~sgordon/css322y13s2/slides/css322y13s2l03-block-ciphers-and-des.pdf to see fk is the steps enclosed in the dark grey box.
IP-1 is the inverse of IP, which means IP-1(IP(X)) = X. So if you take some input X and apply the arrangement specified by IP and get Y, then if you apply IP-1 on Y you must get the original X back. Since IP is [2 6 3 1 4 8 5 7], it means the 1st bit of X goes to the 4th position, and the 2nd bit of X goes to the 1st position in Y. So IP-1 must take the 4th bit of Y and put it in the 1st position, and take the 1st bit of Y and put it in the 2nd position. The start of IP-1 is [4 1 ...]. Try to find the rest.
Steven Gordon
thank you professor Steven :)
Boring talk faster ad humor