Let Π = (Gen,Enc,Dec) be a private-key encryption scheme that has indistinguishable en- cryptions in the presence of an eavesdropper. Which of the following encryption schemes are also necessarily secure against an eavesdropper? If you think a scheme is secure, sketch a proof, if not, provide a counterexample. Here, for a bit string s, parity(s) is 1 if the number of 1’s in s is odd, and 0 otherwise. The || symbol stands for concatenation. So, for strings if x = 00 and y = 11, x||y = 0011. (a) Enc1k(m) = 0||Enck(m) (b) Enc2k(m) = Enck(m)||parity(m) (c) Enc3k(m) = Enck(m)||Enck(m) (d) Enc4k(m) = Enck(m)||Enck(m + 1). Here think of m as an integer.      Please type answer note write by hend.

A B C D 0 1 N 13 111 0 P Q 14 15 16 17 F5 E 4 D 3 S U225 E R F G H I 19 V 6 7 8 T U 18 19 20 21 22 23 24 25 1932 J K L M 10 11 12 W ✗ Y Ꮓ (Show your work.) Decrypt the ciphertext "tyvq" using Affine Cipher in English alphabet with a = 7; b =22 encryption functions is f(p) = ☐ A) hens (7p+22) mod 26 B) hect C) zine D) helo

1. Please apply columnar transposition cipher to encrypt the message "a digital computer is a machine that can solve problems for" ignoring the space in between. If the probable word is "network", please give the corresponding ciphertext (no spaces). State the result: Provide Proof: Solution: Proof is as the following, N (3) E(1) T(6) W(7) O(4) R(5) K(2)

RSA cryptosystem Let p = 29 and q = 31. Then the RSA modulus n = pq = 899. We set the length of plaintext blocks to be 2 and the length of ciphertext blocks 3. Let the encryption exponent b = 29. If, for example, we want to encrypt the word live, we first divide the word into blocks of length 2, li and ve, and then find its corresponding number blocks 11 8 and 21 4. We consider these number blocks as numbers in base 26. Then we find the corresponding integers in base 10, that is 11 8 = 11 × 26 + 8 = 294 and 21 4 = 21 × 26 + 4 = 550. The integers 294 and 550 are regarded as modular numbers in Z899 and encrypted as 29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492. Now write 91 and 492 in base 26 of length 3 we get 91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 +18 × 26 +24 and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Hen the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.