There are various type of algorithm is used for the purpose of the key scheduling such as AES .in the AES algorithm we used same key for encryption and decryption of text .The main objective of the AES algorithm it is used by Various round of the similar plain text encryption to reinforce the cipher text.

The Option (a) is wrong because In the key scheduling the creating keys are not being used one after just another in the various communication cycles.

The Option (b) is wrong because In the key scheduling we do not used the the random key for the encryption process .

The Option (c) is wrong because we will never arbitrarily subdivided into groups of public and private key.

3 0

3 years ago

Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th

Evgen [1.6K]

Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)

=> P(f(A), A, g(A)), P(f(A), A, g(A))

Thus , mgu = {x/y, z/y , y/A }.

Explanation:

Unification: Any substitution that makes two expressions equal is called a unifier.

a) P(B,A,B), P(x,y,z)

Use { x/B}

=> P(B,A,B) , P(B,y,z)

Now use {y/A}

=> P(B,A,B) , P(B,A,z)

Now, use {z/B}

=> P(B,A,B) , P(B,A,B}

Hence, most general unifier = {x/B , y/A , z/B }

b. P(x,x), Q(A,A)

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q

c. Older(Father(y),y), Older(Father(x),John)

Use {y/x}

=> Older(Father(x),x), Older(Father(x),John)

Now use { x/John }

=> Older(Father(John), John), Older(Father(John), John)

Thus , mgu ={ y/x , x/John }

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

Use { y/x }

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))

This is not unifiable as x cannot be bound for both A and B

e) P(f(x), x, g(x)), P(f(y), A, z)

Use {x/y}

=> P(f(y), y, g(y)), P(f(y), A, z)

Now use {z/g(y)}

P(f(y), y, g(y)), P(f(y), A, g(y))

Now use {y/A}

=> P(f(A), A, g(A)), P(f(A), A, g(A))

Thus , mgu = {x/y, z/y , y/A }.

7 0

3 years ago