Question: Consider the following protocol: A - > KDC :IDA | | IDB | | N 1 KDC - > A :E ( ka , [

Consider the following protocol:
A -> KDC :IDA||IDB||N1
KDC -> A :E(ka,[ks||IDB||N1||E(kb,[ks||IDA]))
A -> B :E(kb,[ks||IDA])
B -> A :E(ks, N2)
A -> B :E(ks, f(N2))
(1)
(a) Explain the protocol

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