Question

Find the error in the Decryption pseudo code for RC5 –

for i = 1 to r do

RDi-1 = ((RDi – S [2xi+1] >>> LDi ) ⊕ LDi);

LDi-1 = ((LDi – S [2xi] >>> RDi-1) ⊕ RDi-1);

B= RD0 + S [1];

A = LD0 – S [0];

(a) B= RD0 + S [1];

(b) for  i = 1  to  r    do

(c) LDi-1 = ((LDi – S [2xi] >>> RDi-1) ⊕ RDi-1);

(d) A = LD0 – S [0];

The question was posed to me during an online interview.

This interesting question is from RC4 and RC5 in section Pseudorandom Number Generation and Stream Ciphers and More Block Ciphers of Cryptograph & Network Security

Answer 1

Correct choice is (a) B= RD0 + S [1];

Easy explanation: B= RD0 + S [1]; is incorrect as there should be a ‘-’ instead.