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]

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];

Answer 1

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

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