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The other day I saw this Hybrid Cryptographic Algorithm in the Google and felt curious about it. I am trying to understand this Hybrid Algorithm. And hopefully its not so hard to understand whats going on until AES portion comes up. Here in this research paper the author gives 48bytes input to AES and gets 88byte output from AES.

I hope the author won't mind that I copied a part of the image and gave it in here. enter image description here

My question is, How is this possible? AES always gives cipher size that is divisible by 16. But 88byte is wired! Can anyone help me by explaining or giving a hint what might be the way to achieve this?

-Thanks.

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Obviously I'm not using their schema. I'm just curious to know how it happens---thats all. –  Giliweed Jul 4 at 11:28
4  
Wow. That paper is....... poor. Like, really poor. –  Reid Jul 4 at 16:22
    
The paper considers that a block cipher's output is wider than its input, likely because the authors confused with some mode of operation of a block cipher complete with IV and padding; that's both for DES and AES. –  fgrieu Jul 4 at 17:25
    
OK, looked into it. This was a paper to get their Bachelors degree. If that is the level of education on the Mumbai university, I will put question marks on any students that graduated over there. I wonder why they let them put this online, even the spelling is horrible - although from a science point of view, that may be the least of their worries. Then again, this is from students, not full blown cryptographers. –  owlstead Jul 5 at 16:28
    
Removed first comment because it went beyond calling the paper poor. The advice about not using the paper stands of course. –  owlstead Jul 13 at 23:35

1 Answer 1

I don't think it's useful spending time on trying to understand that paper, but if you look at the screenshots and compare to their "character counts", you see that they are counting base64 characters and including the padding characters in the count.

That means "88 characters" could be one IV + three blocks of AES output (512 bits in base 64 = 85.33 + 2.67 padding).

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