In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.

In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private

1.	In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.
A.	True
B.	False
Answer» C.

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Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).
In an RSA system the public key of a given user is e = 31, n = 3599. What is the private key of this user?
For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.
The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.
In Merkle-Hellman Cryptosystem, the public key can be used to decrypt messages, but cannot be used to decrypt messages. The private key encrypts the messages.
In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.
Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.
THE_PLAINTEXT_MESSAGE_CONSIST_OF_SINGLE_LETTERS_WITH_5-BIT_NUMERICAL_EQUIVALENTS_FROM_(00000)2_TO_(11001)2._THE_SECRET_DECIPHERING_KEY_IS_THE_SUPERINCREASING_5-TUPLE_(2,_3,_7,_15,_31),_M_=_61_AND_A_=_17._FIND_THE_CIPHERTEXT_FOR_THE_MESSAGE_‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖ‚À´WHY‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖŒÄ.?$#
IN_MERKLE-HELLMAN_CRYPTOSYSTEM,_THE_PUBLIC_KEY_CAN_BE_USED_TO_DECRYPT_MESSAGES,_BUT_CANNOT_BE_USED_TO_DECRYPT_MESSAGES._THE_PRIVATE_KEY_ENCRYPTS_THE_MESSAGES.?$
For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.$

In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.

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