Assume a single bit of the 30th byte in the encrypted file got corrupted. You ca

Question

Assume a single bit of the 30th byte in the encrypted file got corrupted. You can achieve this corruption by modifying the 30th byte of the encrypted file using GHex. (The first row has 18 bytes. Note, two hexadecimal digits represent a byte, such as, the hexadecimal number B6 is equivalent to the 8-bit binary number 10110110, which is a byte. )

Why modes of operation are needed? How much information can you recover by decrypting the corrupted file, if the encryption mode is CBC, CFB, and OFB, respectively? This tells the error propagation in different modes of operation. Please do a little research online and explain your observations.

Explanation / Answer

CBC

in the original encrypted file,the position 30th byte is 1e correspond to the balue !B,i changed the single bit of this value fron 1 to A so the value of 30tth byte now is AB then saved the file .

CFB

in the original encrypted file the position 30th byte is 1E corresponds to the value 70, where i change the single bit of the value from 7 to A so the value of 30th bytes now is A0 and save the file.

OFB

in the original encrypted file, the position 30th byte is 1E corresponds to the value 0C where i replace the value from A to 0 so the value of 30 byte is now AC the save the file.

In CBC mode two blocks was affected

In OFB the feedback is key generation system.the single digit of 30th byte is corrupted.

In CFB there is problen in n/r numbers of block, hte error propagation is needful.

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