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P5

中文依然是屑翻译，按英文版做

a) There are 8 tables. Each table has 28 entries. Each entry has 8 bits.

number of tables * size of each table * size

of each entry = 828 8= 214 bits

b) There are 264 entries. Each entry has 64 bits. 271

bits

P6

a) 011011011

b) 该密文对应的明文由三个相同bit组成 c) c(i) = KS(m(i) XOR c(i-1))

c(1) =KS(100 XOR 111) = KS (011) = 100

c(2) =KS(100 XOR 100) = KS (000) = 110

c(1) =KS(100 XOR 110) = KS (010) = 101

P7

a) We are given p=3 and q=11

We thus have n=33 and q=11

Choose e=9 (it might be a good idea to give students a hint that 9 is a good value to choose, since the resulting calculations are less likely to run into numerical stability problems than other choices for e) since 3 and (p-1)(q-1)=20 have no common factors. Choose d=9 also so that ed=81 and thus e*d-1=80 is exactly divisible by 20. We can now perform the RSA encryption and decryption using n=33,e=9 and d=9

<🐎>

letter m

m ** e ciphertext = m**e mod 33

d     4   262144       25      o     15

38443359375 3

g     7

40353607 19

ciphertext cd m = cd mod n letter

25 38146972265625 4 d

3         19683               15             o 19     322687697779        7              g

</🐎>

We first consider each letter as a 5-bit number: 00100, 01111, 00111. Now we concatenate each letter to get 001000111100111 and encrypt the resulting decimal number m=4583. The concatenated decimal number m (= 4583) is larger than current n (= 33). We need m < n. So we use p= 43, q = 107, n = p*q = 4601, z = (p-1)(q-1) = 4452. e = 61, d = 73 ciphertext = m**e mod 4601

m**e= 21386577601828057804089602156530567188611499869029788733808438804302864595620613956725840720949764845640956118784875246785033236197777129730258961756918400292048632806197527785447791567255101894492820972508185769802881718983

ciphertext

= m**e mod 4601 = 402

c**d

= 12838133136197716341957121325397932876435331474825362093284052627930271588610123920532872496335709674931222802214538150129342413705402045814598714979387232141014703227794586499817945633390592

ciphertext = m**e mod 4601 = 4583

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