没有合适的资源？快使用搜索试试~ 我知道了~

首页《密码学原理与实践》第三版部分章节的部分习题答案

《密码学原理与实践》第三版部分章节的部分习题答案

4星 · 超过85%的资源需积分: 31 327 下载量 16 浏览量更新于2023-03-03 评论 19 收藏 149KB PDF 举报

身份认证购VIP最低享 7 折!

领优惠券(最高得80元）

试读

21页

Douglas R. Stinson "Cryptography: theory and Pratice" third edition 《密码学原理与实践》第三版部分章节的部分习题答案

资源详情

资源评论

资源推荐

Homework of Cryptography

Tony Zhu

December 20, 2010

1 Classical Cryptography

1.21

Solution: Since in the aﬃne cryptosystem only φ(26) × 26 = 312 cases for key

K = (a, b), list all the cases brutally with the help of computer and scan quickly, I

found only one case seems meaningful:

ocanadaterredenosaieuxtonfrontestceintdeﬂeuronsglorieuxcartonbrassaitporterlepeeils

aitporterlacroixtonhistoireestuneepopeedesplusbrillan tsexploitsettavaleurdefoitrem-

peeprotegeranosfoyersetnosdroits

After breaking the long sentence, I found it should be the French version of

Canada national anthem:

O Canada

Terre de nos a

ıeux

Ton front est ceint de ﬂeurons glorieux!

Car ton bras sait porter l’

Il sait porter la croix!

Ton histoire est une

epop

Des plus brillants exploits

Et ta valeur, de foi tremp

Prot

egera nos foyers et nos droits

1.22

Proof. (a) Obviously in the case p

≥ p

≥ 0, q

≥ q

≥ 0, the sum p

+ p

is maximized.

Suppose S =

i=1

is maximized, and if there exists i > j such that q

< q

permute the positions of the two numbers and, according to the statement of last

paragraph, we get S

0≤h≤n

h,i, j

+ p

> S , contradicts. So the only

arrangement for S is q

≥ . . . ≥ q

. v

2 Shannon’s Theory

2.2

Proof. P = C = K = {1, . . . , n}. ∀x ∈ P, ∀y ∈ C

Pr[Y = y] =

k∈K

Pr[K = k]Pr[x = d

(y)] = 1/n

Pr[x|y] =

Pr[x]Pr[y|x]

Pr[y]

Pr[x]

= Pr[x]

that is this Latin Square Cryptosystem achieves perfect secrecy provided that ev-

ery key is used with equal probability. v

2.3

Proof. Similarly as (2.2). v

2.5

Proof. According to Theorem 2.4, this cryptosystem provides perfect secrecy iﬀ

every key is used with equal probability 1/|K |, and for every x ∈ P and y ∈ C ,

there is unique key k ∈ K , such that y = e

(x).

Due to perfect secrecy, we have

Pr[x|y] = Pr[x]

=⇒

Pr[x]Pr[y|x]

Pr[y]

= Pr[x]

=⇒ Pr[y] = Pr[y|x ] = Pr

(x)=y

[K = k] = 1/|K |.

that is every ciphertext is equally probable. v

2.11

Proof. (Perfect secrecy ⇒ H(P|C)=H(P)).

H(P|C) = −

y∈C

x∈P

Pr[y]Pr[x|y] log

Pr[x|y]

= −

y∈C

x∈P

Pr[y]Pr[x] log

Pr[x]

= H(P)

y∈C

Pr[y]

= H(P).

(H(P|C) = H(P) ⇒ Perfect secrecy).

According to Theorem 2.7,

H(P, C) ≤ H(P) + H(C)

with equality iﬀ P and C are independent random variables.

H(P|C) = H(P)

=⇒ H(P, C) = H(P) + H(C)

=⇒ P and C are independent random variables

=⇒ ∀x ∈ P, ∀y ∈ C, Pr[x|y] = Pr[x],

that is the cryptosystem achieves perfect secrecy. v

2.12

Proof. We have

H(K, P, C) = H(P|K, C) + H(K, C) = H(K, C),

H(K, P, C) ≥ H(P, C)

=⇒ H(K, C) ≥ H(P, C)

=⇒ H(K|C) = H(K, C) − H(C) ≥ H(P, C) − H(C) = H(P|C).

2.13

Solution: H(P) =

log

3 +

log

6 ≈ 1.45915.

H(K) = log

3 ≈ 1.58496.

Pr[1] = 2/9

Pr[2] = 5/18

Pr[3] = 1/3

Pr[4] = 1/6

H(C) ≈ 1.95469.

H(K|C) = H(K) + H(P) − H(C) ≈ 1.08942.

Pr[K

|1] = 3/4 Pr[K

|1] = 0 Pr[K

|1] = 1/4

Pr[K

|2] = 2/5 Pr[K

|2] = 3/5 Pr[K

|2] = 0

Pr[K

|3] = 1/6 Pr[K

|3] = 1/3 Pr[K

|3] = 1/2

Pr[K

|4] = 0 Pr[K

|4] = 1/3 Pr[K

|4] = 2/3

H(P|C) ≈ 1.08942.

2.14

Solution: H(K) = log

216, H(P) = log

26, H(C) = log

26,

H(K|C) = H(K) + H(P) − H(C) = log

216 ≈ 7.75489.

It is easy to evaluate:

H(K|P, C) = log

12 ≈ 3.58496.

2.19

Proof. A key in the product cipher S

×S

has the form (s

, s

), where s

, s

∈ Z

, s

)

(x) = x + (s

+ s

But this is precisely the deﬁnition of a key in the Shift Cipher. Further, the proba-

bility of a key in S

× S

equals

i=0

, which is also the probability

of a key in the Shift Cipher. Thus S

× S

is the Shift Cipher. v

2.20

Proof. (a) Similarly as (2.19).

(b) The number of keys in S

equals 26

lcm(m

, m

)

. Form the following fact we

can see that the keys in S

× S

are lesser when m

. 0 (mod m

Notice

< m

, m

. 0 (mod m

)

=⇒ m

+ m

< 2m

≤

gcd(m

, m

)

= l, where l , lcm(m

, m

)

the equation below, which essentially gives the representation of the keys of prod-

uct cryptosystem S

×S

, would’t always has a solution (K

, K

) = (x

, . . . , x

, y

, . . . , y

) ∈

× K

, for any selected key K = (z

, . . . , z

) ∈ K





































because the coeﬃcient matrix’s column number is less than the row number. Con-

clusively, there must exist key K ∈ K

that not in K

× K

. v

3 The RSA Cryptosystem and Factoring Integers

5.2

剩余20页未读，继续阅读

zzcpp

2011-10-24

还可以只有部分答案需要的同学谨慎

llkkop

粉丝: 3
资源: 4

上传资源快速赚钱

我的内容管理收起

我的资源快来上传第一个资源

我的收益

登录查看自己的收益

我的积分登录查看自己的积分

我的C币登录后查看C币余额

我的收藏

我的下载

下载帮助

会员权益专享

《密码学原理与实践》第三版部分章节的部分习题答案

评论28

会员权益专享

最新资源

《密码学原理与实践》第三版 部分章节的部分习题答案

评论28

密码学原理与实践课后习题答案.7z

（珍贵资料）哈尔滨工业大学密码学原理与实践五次作业答案（李秋豪学长版）

《密码编码学与网络安全：原理与实践 （第四版）》答案

密码学原理与实践冯登国pdf

密码学原理与实践答案第四章hash函数

"密码编码学与网络安全 原理与实践第六版课后答案 -site:\"baidu.com"

现代密码学 原理与协议 pdf

现代密码学理论与实践 pdf

计算机安全原理与实践第四版pdf

哈工大密码学原理期末

密码学实践 pdf弗格森

现代密码学理论与实践.caj

密码学课后习题stinson

现代密码学第五版 杨波pdf

ustc 密码学 答案

密码学引论 第二版 csdn

现代密码学教程pdf版

密码学工程实践java

现代密码学 陈鲁生 [第3章] 课后答案.pdf

密码编码学与网络安全第六版pdf

会员权益专享

最新资源

《密码学原理与实践》第三版部分章节的部分习题答案

《密码编码学与网络安全：原理与实践（第四版）》答案

"密码编码学与网络安全原理与实践第六版课后答案 -site:\"baidu.com"

现代密码学原理与协议 pdf

现代密码学第五版杨波pdf

ustc 密码学答案

密码学引论第二版 csdn

现代密码学陈鲁生 [第3章] 课后答案.pdf