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Chapter 29 Cryptography and N k Network Security Security 1

Chapter 29 Crypyp g p ytography and NkNetwork Securityggn.dronacharya.info/.../QuestionBank/Mtech/SemIII/cryptography.pdf · To show how secret keys in symmetric-key cryptography

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Page 1: Chapter 29 Crypyp g p ytography and NkNetwork Securityggn.dronacharya.info/.../QuestionBank/Mtech/SemIII/cryptography.pdf · To show how secret keys in symmetric-key cryptography

Chapter 29

Cryptographyyp g p yand

N kNetworkSecuritySecurity

1

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OBJECTIVES:OBJECTIVES:T i d i l d di h f kTo introduce security goals and to discuss the types of attacks that threaten these goals.

To introduce traditional ciphers as symmetric key ciphers toTo introduce traditional ciphers as symmetric-key ciphers to create the background for understanding modern symmetric-key ciphers.

To introduce the elements of modern block ciphers and show an example of a modern block cipher in which these elements are

dused.

To discuss the general idea behind asymmetric-key ciphers and introduce one common cipher in this categoryintroduce one common cipher in this category.

To discuss message integrity and show how to use a cryptographic hash function to create a message digest

2

cryptographic hash function to create a message digest.

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OBJECTIVES (OBJECTIVES (continuedcontinued):):T i d h id f h i i d h hTo introduce the idea of message authentication and show how a message digest combined with a secret can authenticate the sender.

To show how the idea of digital signatures can be used to authenticate a message using a pair of private-public keys.

To introduce the idea of entity authentication and show some simple schemes using either a secret key or a pair of private-

bli kpublic keys.

To show how secret keys in symmetric-key cryptography and how public keys in asymmetric key cryptography can behow public keys in asymmetric-key cryptography can be distributed and managed using KDCs or certificate authorities (CAs).

3

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Chapter Chapter O tliO tli

29.1 Introduction29.1 IntroductionOutlineOutline 29.2 Traditional Ciphers29.2 Traditional Ciphers

29 3 Modern Ciphers29 3 Modern Ciphers29.3 Modern Ciphers29.3 Modern Ciphers

29.4 Asymmetric29.4 Asymmetric--Key CiphersKey Ciphers

29.5 Message Integrity29.5 Message Integrity

29 6 Message Authentication29 6 Message Authentication29.6 Message Authentication29.6 Message Authentication

29.7 Digital Signature29.7 Digital Signatureg gg g

29.8 Entity Authentication29.8 Entity Authentication

4

29.9 Key Management29.9 Key Management

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29-1 INTRODUCTION

We are living in the information age. We need tog gkeep information about every aspect of our lives. Inother words, information is an asset that has a valuelik h A i f i dlike any other asset. As an asset, information needsto be secured from attacks. To be secured,information needs to be hidden from unauthorizedinformation needs to be hidden from unauthorizedaccess (confidentiality), protected from unauthorizedchange (integrity), and available to an authorizedchange (integrity), and available to an authorizedentity when it is needed (availability).

5

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Topics Discussed in the SectionTopics Discussed in the SectionSecurity GoalsAttacksS iServicesTechniques

6

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Figure 29.1 Taxonomy of attacks with relation to security goals

7

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29-2 TRADITIONAL CIPHERS

We now look at the first goal of security,g y,confidentiality. Confidentiality can be achieved usingciphers. Traditional ciphers are called symmetric-keyi h ( k i h ) b hciphers (or secret-key ciphers) because the samekey is used for encryption and decryption and thekey can be used for bidirectional communicationkey can be used for bidirectional communication.Figure 29.2 shows the general idea behind asymmetric-key cipher.symmetric key cipher.

8

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Topics Discussed in the SectionTopics Discussed in the SectionKeySubstitution CiphersT iti Ci hTransposition CiphersStream and Block Ciphers

9

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Figure 29.2 General idea of traditional cipher

10

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Note

A substitution cipher replaces one symbol with another.

11

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Figure 29.3 Symmetric-key: locking and unlocking with the same key

12

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Note

A substitution cipher replaces one symbol with another.

13

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Figure 29.4 Representation of characters in modulo 26

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Note

In additive cipher, the plaintext, ciphertext, and key are integers in

26modulo 26.

15

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Use the additive cipher with key = 15 to encrypt the message

ExampleExample 29.1Use the additive cipher with key 15 to encrypt the message“hello”.

S l tiSolutionWe apply the encryption algorithm to the plaintext, character bycharacter. The result is “WTAAD”. Note that the cipher ismonoalphabetic because two instances of the same plaintextcharacter (ls) are encrypted as the same character (A).

16

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Use the additive cipher with key = 15 to decrypt the message

ExampleExample 29.2Use the additive cipher with key 15 to decrypt the message“WTAAD”.

S l tiSolutionWe apply the decryption algorithm to the plaintext character bycharacter. The result is “hello”. Note that the operation is inmodulo 26, which means that we need to add 26 to a negativeresult (for example −15 becomes 11).

17

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Figure 29.5 An example key for mono-alphabetic substitution cipher

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We can use the key in Figure 29 5 to encrypt the message

ExampleExample 29.3We can use the key in Figure 29.5 to encrypt the message

The ciphertext is

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Note

A transposition cipher reorders symbols.

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Figure 29.6 Transposition cipher

Write row by row

e ht tm ta oy zc ne ia kn gt sPlaintext

Read row by row

Plaintexte ht tm ta oy zc ne ia kn gt s

e

h

tt

m

t

ao

y

z

cn

e

i

ak

n

g

ts

e

h

tt

m

t

ao

y

z

cn

e

i

ak

n

g

ts

ET

MA

YC

EA

NT

ET

MA

YC

EA

NT

HT

TO

ZN

IK

GS

E HT T M TAO Y ZCNE IAK N GT S

Read column by column

HT

TO

ZN

IK

GS

E HT T M TAO Y ZCNE IAK N GT S

Write column by column

21

E HT T M TAO Y ZCNE IAK N GT SCiphertext

E HT T M TAO Y ZCNE IAK N GT SCiphertextTransmission

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29-3 MODERN CIPHERS

The traditional symmetric-key ciphers that we havey y pstudied so far are character-oriented ciphers. Withthe advent of the computer, we need bit-orientedi h Thi i b h i f i bciphers. This is because the information to beencrypted is not just text; it can also consist ofnumbers graphics audio and video data It isnumbers, graphics, audio, and video data. It isconvenient to convert these types of data into astream of bits, to encrypt the stream, and then tostream of bits, to encrypt the stream, and then tosend the encrypted stream. A modern block ciphercan be either a block cipher or a stream cipher.

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Topics Discussed in the SectionTopics Discussed in the SectionModern Block CiphersData Encryption Standard (DES)M d St Ci hModern Stream Ciphers

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Figure 29.7 A modern block cipher

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Figure 29.8 Components of a modern block cipher

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Figure 29.9 General structure of DES

LI 1 RI 132 bits 32 bits

xer

KI

LI–1 RI–1

f ( RI–1, KI )

pper

Mix KI( I–1, I )

Swap

LI RI

Each round

32 bits 32 bits

26

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Figure 29.10 DES function

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Figure 29.11 Key generation

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We choose a random plaintext block a random key and a

ExampleExample 29.4We choose a random plaintext block, a random key, and acomputer program to determine what the ciphertext block wouldbe (all in hexadecimal):

29

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To check the effectiveness of DES when a single bit is

ExampleExample 29.5To check the effectiveness of DES, when a single bit ischanged in the input, let us use two different plaintexts with onlyone single bit difference. The two ciphertexts are completelydiff t ith t h i th kdifferent without even changing the key:

Alth h th t l i t t bl k diff l i th i ht t bitAlthough the two plaintext blocks differ only in the rightmost bit,the ciphertext blocks differ in 29 bits.

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Figure 29.12 One-time pad

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29-4 ASYMMETRIC-KEY CIPHERS

In previous sections we discussed symmetric-keyp y yciphers. In this chapter, we start the discussion ofasymmetric-key ciphers. Symmetric-key and

i k i h ill i i ll l dasymmetric-key ciphers will exist in parallel andcontinue to serve the community. We actuallybelieve that they are complements of each other; thebelieve that they are complements of each other; theadvantages of one can compensate for thedisadvantages of the other.disadvantages of the other.

32

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Topics Discussed in the SectionTopics Discussed in the SectionKeysGeneral IdeaRSA C t tRSA CryptosystemApplications

33

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S

Note

Symmetric-key cryptography is based on sharing secrecy;

asymmetric-key cryptography is based on personal secrecy.

34

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Note

In symmetric-key cryptography, symbols are permuted or substituted;

in asymmetric-key cryptography, numbers are manipulated.

35

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Figure 29.13 Locking and unlocking in asymmetric-key cryptosystem

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Note

Asymmetric-key ciphers are sometimes called public-key ciphers.

37

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Figure 29.14 General idea of asymmetric-key cryptosystem

To public

Public-key distribution �channel

Key-generation�procedure

Public key Private key

Insecure channelPlaintext PlaintextCiphertextCiphertext

Encryption Decryption

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Figure 29.15 Encryption, decryption, and key Generation in RSA

Key calculation

To public(e, n) Select p, q

n = p × qTo public

Private (d)

Select e and d

(e, n)C: Ciphertext

E iC = Pe mod nP

PlaintextP

PlaintextP = Cd mod n

Encryption DecryptionPlaintext Plaintext

39

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For the sake of demonstration let Bob choose 7 and 11 as p

ExampleExample 29.6For the sake of demonstration, let Bob choose 7 and 11 as pand q and calculate n = 7 × 11 = 77. The value of φ(n) =(7 − 1)(11 − 1), or 60. If he chooses e to be 13, then d is 37.N t th t d d 60 1 N i i th t Ali t tNote that e × d mod 60 = 1. Now imagine that Alice wants tosend the plaintext 5 to Bob. She uses the public exponent 13 toencrypt 5. This system is not safe because p and q are small.

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Here is a more realistic example calculated with a computer

ExampleExample 29.7Here is a more realistic example calculated with a computer.We choose a 512-bit p and q, calculate n and φ(n), We thenchoose e and calculate d. Finally, we show the results of

ti d d ti Th i t i 159 di it bencryption and decryption. The integer p is a 159-digit number.

The integer q is a 160-digit number.

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The modulus n = p × q It has 309 digits

ExampleExample 29.7 ContinuedThe modulus n p × q. It has 309 digits.

φ(n) = (p − 1)(q − 1) has 309 digits.

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Bob chooses e = 35535 (the ideal is 65537) He then finds d

ExampleExample 29.7 ContinuedBob chooses e 35535 (the ideal is 65537). He then finds d.

Alice wants to send the message “THIS IS A TEST”, which canbe changed to a numeric value using the 00−26 encodingscheme (26 is the space character).( p )

43

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The ciphertext calculated by Alice is C = Pe which is

ExampleExample 29.7 ContinuedThe ciphertext calculated by Alice is C P , which is

Bob can recover the plaintext from the ciphertext using P = Cd,which is

The recovered plaintext is “THIS IS A TEST” after decoding.

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29-5 MESSAGE INTEGRITY

The cryptography systems that we have studied sofar provide secrecy, or confidentiality, but notintegrity. However, there are occasions where wemay not even need secrecy but instead must havemay not even need secrecy but instead must haveintegrity. For example, Alice may write a will todistribute her estate upon her death The will doesdistribute her estate upon her death. The will doesnot need to be encrypted. After her death, anyonecan examine the will. The integrity of the will,g yhowever, needs to be preserved. Alice does notwant the contents of the will to be changed.

45

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Topics Discussed in the SectionTopics Discussed in the SectionMessage and Message DigestHash Functions

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Figure 29.16 Message and digest

Message MessageI h l

Hashfunction

CreatedHashfunction Digest

M

S ?

Insecure channel

DigestSent

Keep

DiscardReceived

function

Digest

g

[Yes] [No]

Same?

Keepthe message

47

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f

Note

The message digest needs to be safe from change.

48

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29-6 MESSAGE AUTHENTICATION

A digest can be used to check the integrity of ag g ymessage: that the message has not been changed.To ensure the integrity of the message and the datai i h i i h Ali i h i i forigin authentication— that Alice is the originator of

the message, not somebody else—we need toinclude a secret held by Alice (that Eve does notinclude a secret held by Alice (that Eve does notpossess) in the process; we need to create amessage authentication code (MAC). Figure 29.17message authentication code (MAC). Figure 29.17shows the idea.

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Topics Discussed in the SectionTopics Discussed in the SectionMACHMAC

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Figure 29.17 Message authentication code

K

HashMAC

MK: A shared secret keyM: Message

MAC: Message authenticationcode

K

HashMAC

M

code

Discard[No]

Same?

M + MAC

M + MAC

Keepthe message

[Yes]M + MACInsecure channel

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C

Note

A MAC provides message integrity and message authentication using a

f fcombination of a hash function and a secret key.

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29-7 DIGITAL SIGNATURE

Another way to provide message integrity andy p g g ymessage authentication (and some more securityservices as we see shortly) is a digital signature. AMAC k h di di i lMAC uses a secret key to protect the digest; a digitalsignature uses a pair of private-public keys.

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Topics Discussed in the SectionTopics Discussed in the SectionComparisonProcessSi i th Di tSigning the DigestServicesRSA Di it l Si t S hRSA Digital Signature SchemeDigital Signature Standard (DSS)

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f

Note

A digital signature uses a pair of private-public keys.

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Figure 29.18 Digital signature process

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Note

A digital signature needs a public-key system.

The signer signs with her private key; the verifier verifies with the signer’s

bli kpublic key.

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Note

A cryptosystem uses the private and public keys of the receiver:

a digital signature uses the private and public keys of the sender.

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Figure 29.19 Signing the digest

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Figure 29.20 Using a trusted center for non-repudiation

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Figure 29.21 The RSA signature one the message digest

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29-8 ENTITY AUTHENTICATION

Entity authentication is a technique designed to lety q gone party prove the identity of another party. Anentity can be a person, a process, a client, or a

Th i h id i d bserver. The entity whose identity needs to be provenis called the claimant; the party that tries to prove theidentity of the claimant is called the verifieridentity of the claimant is called the verifier.

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Topics Discussed in the SectionTopics Discussed in the SectionEntity versus Message AuthenticationVerification CategoriesP dPasswords Challenge-Response

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Note

In challenge-response authentication, the claimant proves that she knows a

fsecret without sending it to the verifier.

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Figure 29.22 Unidirectional symmetric-key authentication

Alice1

Encrypted with Alice-Bob secret key

A nonce sent by BobRB

Alice1

RB 2

RB3

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Figure 29.23 Unidirectional asymmetric-key authentication

Encrypted with Alice’s public key

Alice1

yp p y

A nonce sent by BobRB

Alice1

Bob, RB 2

RB3

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Figure 29.24 Digital signature, unidirectional authentication

R

Alice1

A nonce sent by BobRB

Alice1

RB 2

B bBob, Sig (RB, Bob)Signed with�

Alice’s private key3

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29-9 KEY MANAGEMENT

We discussed symmetric-key and asymmetric-keyy y y ycryptography in the previous sections. However, wehave not yet discussed how secret keys in

i k h d bli k isymmetric-key cryptography, and public keys inasymmetric-key cryptography, are distributed andmaintained This section touches on these twomaintained. This section touches on these twoissues.

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Topics Discussed in the SectionTopics Discussed in the SectionSymmetric-Key Distribution Symmetric-Key AgreementP bli K Di t ib tiPublic-Key Distribution

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Figure 29.25 Multiple KDC’s

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Note

A session symmetric key between two parties is used only once.

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Figure 29.26 Creating a session key using KDC

KA Encrypted with Alice-KDC secret key

KB Encrypted with Bob-KDC secret key

Session key between Alice and Bob

Alice, Bob1

2

KAKB 2, Alice, Bob,

3KB

3 Alice, Bob,

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Figure 29.27 Diffie-Hellman method

The values ofp and g are public.

3R2 = gy mod p

2 R1

K = (R1)y mod p 5K = (R2)x mod p5

R2 4

K xy d

Shared secret key

73

K = gxy mod p

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( )

Note

The symmetric (shared) key in the Diffie-Hellman method is K = gxy mod p.

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Let us give a trivial example to make the procedure clear Our

ExampleExample 29.8Let us give a trivial example to make the procedure clear. Ourexample uses small numbers, but note that in a real situation,the numbers are very large. Assume that g = 7 and p = 23. Thet f llsteps are as follows:1. Alice chooses x = 3 and calculates R1 = 73 mod 23 = 21.2. Alice sends the number 21 to Bob.3. Bob chooses y = 6 and calculates R2 = 76 mod 23 = 4.4. Bob sends the number 4 to Alice.5 Alice calculates the symmetric key K = 43 mod 23 = 18 Bob5. Alice calculates the symmetric key K 4 mod 23 18. Bobcalculates the symmetric key K = 216 mod 23 = 18.

Th l f K i th f b th Ali d B bThe value of K is the same for both Alice and Bob;gxy mod p = 718 mod 35 = 18.

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Note

In public-key cryptography, everyone has access to everyone’s public key;

public keys are available to the public.

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Figure 29.28 Certification authority

DirectoryAlice KA

Bob KB

Recording2

Bob’spublic key

eco d g

public keyApplying1

Signed with

Distributingto publicIssuing

gCA’s private key

Bob’s certificate3

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