Find the error in the Decryption pseudo code for RC5 – for i = 1 to r do RDi-1 = ((RDi – S [2xi+1] >>> LDi ) ⊕ L
Posted: Thu Jul 14, 2022 8:10 am
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];
b) for i = 1 to r do
c) LDi-1 = ((LDi – S [2xi] >>> RDi-1) ⊕ RDi-1);
d) A = LD0 – S [0];