《计算机网络 自顶向下方法》（第7版）答案（第八章）（二）
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