traditional private/secret/single key cryptography uses one key shared by both sender and receiver if this key is disclosed communications are compromised also is symmetric , parties are equal hence does not protect sender from receiver forging a message

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Cryptography and

Network Security

Chapter 9

Fifth Edition

by William Stallings

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Chapter 9 –

Public Key

Cryptography and RSA

Every Egyptian received two names, which were Every Egyptian received two names, which were

known respectively as the true name and the known respectively as the true name and the

good name, or the great name and the little good name, or the great name and the little

name; and while the good or little name was name; and while the good or little name was

made public, the true or great name appears to made public, the true or great name appears to

have been carefully concealed. have been carefully concealed.

—

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Private-Key Cryptography



traditional

_traditional

private/secret/single key

_{private/secret/single key}

cryptography uses

one

key



shared by both sender and receiver

_{shared by both sender and receiver}



if this key is disclosed communications are

_{if this key is disclosed communications are}

compromised



also is

_{also is}

symmetric

_symmetric

, parties are equal

_{, parties are equal}



hence does not protect sender from

_{hence does not protect sender from}

receiver forging a message & claiming is

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Public-Key Cryptography



probably most significant advance in the

_{probably most significant advance in the}

3000 year history of cryptography



uses

_uses

two

_two

keys – a public & a private key

_{keys – a public & a private key}



asymmetric

_asymmetric

since parties are

_{since parties are}

not

_not

equal

_equal



uses clever application of number

_{uses clever application of number}

theoretic concepts to function



complements

_complements

rather than

_{rather than}

replaces private

_{replaces private}

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Why Public-Key

Cryptography?



developed to address two key issues:

_{developed to address two key issues:}

 key distributionkey distribution – how to have secure – how to have secure

communications in general without having to communications in general without having to

trust a KDC with your key trust a KDC with your key

 digital signaturesdigital signatures – how to verify a message – how to verify a message

comes intact from the claimed sender comes intact from the claimed sender



public invention due to Whitfield Diffie &

_{public invention due to Whitfield Diffie &}

Martin Hellman at Stanford Uni in 1976

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Public-Key Cryptography

 public-key/two-key/asymmetric_{public-key/two-key/asymmetric} cryptography _cryptography

involves the use of

involves the use of twotwo keys: keys:

 a a public-keypublic-key, which may be known by anybody, and can , which may be known by anybody, and can

be used to

be used to encrypt messagesencrypt messages, and , and verify signaturesverify signatures

 a related a related private-keyprivate-key, known only to the recipient, used , known only to the recipient, used

to

to decrypt messagesdecrypt messages, and , and signsign (create) (create) signatures signatures

 infeasible to determine private key from public_{infeasible to determine private key from public}  is _isasymmetric_asymmetric because_because

 those who encrypt messages or verify signatures those who encrypt messages or verify signatures cannotcannot

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Public-Key Applications



can classify uses into 3 categories:

_{can classify uses into 3 categories:}

 encryption/decryptionencryption/decryption (provide secrecy) (provide secrecy)

 digital signaturesdigital signatures (provide authentication) (provide authentication)

 key exchangekey exchange (of session keys) (of session keys)



some algorithms are suitable for all uses,

_{some algorithms are suitable for all uses,}

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Public-Key Requirements

 Public-Key algorithms rely on two keys where:_{Public-Key algorithms rely on two keys where:}

 it is computationally infeasible to find decryption key it is computationally infeasible to find decryption key

knowing only algorithm & encryption key knowing only algorithm & encryption key

 it is computationally easy to en/decrypt messages it is computationally easy to en/decrypt messages

when the relevant (en/decrypt) key is known when the relevant (en/decrypt) key is known

 either of the two related keys can be used for either of the two related keys can be used for

encryption, with the other used for decryption (for encryption, with the other used for decryption (for some algorithms)

some algorithms)



_these

_{are formidable requirements which}

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Public-Key Requirements



need a trapdoor one-way function

_{need a trapdoor one-way function}



one-way function has

_{one-way function has}

 Y = f(X) easy Y = f(X) easy

 X = fX = f–1–1(Y) infeasible(Y) infeasible



a trap-door one-way function has

_{a trap-door one-way function has}

 Y = fY = f

k

k(X) easy, if k and X are known(X) easy, if k and X are known

 X = fX = f

k

k–1–1(Y) easy, if k and Y are known(Y) easy, if k and Y are known

 X = fX = f

k

k–1–1(Y) infeasible, if Y known but k not known(Y) infeasible, if Y known but k not known



a practical public-key scheme depends on

_{a practical public-key scheme depends on}

a suitable trap-door one-way function

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Security of Public Key Schemes

 like private key schemes brute force _{like private key schemes brute force}exhaustive _exhaustive

search

search attack is always theoretically possible attack is always theoretically possible

 but keys used are too large (>512bits) _{but keys used are too large (>512bits)}

 security relies on a _{security relies on a}large enough_{large enough} difference in _{difference in}

difficulty between

difficulty between easyeasy (en/decrypt) and (en/decrypt) and hardhard

(cryptanalyse) problems (cryptanalyse) problems

 more generally the _{more generally the}hard_hard problem is known, but _{problem is known, but}

is made hard enough to be impractical to break is made hard enough to be impractical to break

 requires the use of _{requires the use of}very large numbers_{very large numbers}

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RSA

 by Rivest, Shamir & Adleman of MIT in 1977 _{by Rivest, Shamir & Adleman of MIT in 1977}

 best known & widely used public-key scheme _{best known & widely used public-key scheme}  _{based on exponentiation in a finite (Galois) field}_{based on exponentiation in a finite (Galois) field}

over integers modulo a prime over integers modulo a prime

 nb. exponentiation takes O((log n)nb. exponentiation takes O((log n)33) operations (easy) ) operations (easy)  uses large integers (eg. 1024 bits)_{uses large integers (eg. 1024 bits)}

 _{security due to cost of factoring large numbers}_{security due to cost of factoring large numbers}

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RSA En/decryption



to encrypt a message M the sender:

_{to encrypt a message M the sender:}

 obtains obtains public keypublic key of recipient of recipient PU={e,n}PU={e,n}  computes: computes: C = MC = Mee mod n mod n, where , where 00≤≤MM<<nn



to decrypt the ciphertext C the owner:

_{to decrypt the ciphertext C the owner:}

 uses their private key uses their private key PR={d,n}PR={d,n}  computes: computes: M = CM = Cdd mod n mod n



note that the message M must be smaller

_{note that the message M must be smaller}

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RSA Key Setup

 each user generates a public/private key pair by: _{each user generates a public/private key pair by:}  selecting two large primes at random: _{selecting two large primes at random:}_{p, q}_{p, q}

 computing their system modulus _{computing their system modulus}_n=p.q_n=p.q  note note ø(n)=(p-1)(q-1)ø(n)=(p-1)(q-1)

 selecting at random the encryption key _{selecting at random the encryption key}_e_e

 where where 1<e<ø(n), gcd(e,ø(n))=1 1<e<ø(n), gcd(e,ø(n))=1

 solve following equation to find decryption key _{solve following equation to find decryption key}_d_d

 e.d=1 mod ø(n) and 0e.d=1 mod ø(n) and 0≤≤dd≤≤nn

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Why RSA Works

 because of Euler's Theorem:_{because of Euler's Theorem:}

 aaø(n)ø(n)mod n = 1 mod n = 1 where where gcd(a,n)=1gcd(a,n)=1  _{in RSA have:}_{in RSA have:}

 n=p.qn=p.q

 ø(n)=(p-1)(q-1)ø(n)=(p-1)(q-1)

 carefully chose carefully chose ee & & dd to be inverses to be inverses mod ø(n)mod ø(n)  hence hence e.d=1+k.ø(n)e.d=1+k.ø(n) for some for some kk

 hence :_{hence :}

C

Cdd_{= M}_{= M}e.d e.d _{= M}_{= M}1+k.ø(n)1+k.ø(n)_{= M}_{= M}11_.(M_.(Mø(n)ø(n)₎₎kk

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RSA Example - Key Setup

1.

1. Select primes: Select primes: p_p=17 & _{=17 &}q_q=11₌₁₁

2.

2. CalculateCalculate n n = = pq pq =17=17 x x 11=18711=187

3.

3. CalculateCalculate ø(ø(nn)=()=(p–p–1)(1)(q-q-1)=161)=16xx10=16010=160

4.

4. Select Select e_e:: gcd(e,160)=1; _{gcd(e,160)=1;}choose choose e_e=7₌₇

5.

5. Determine Determine dd:: de=de=1 mod 1601 mod 160 and and d d < 160< 160

Value is

Value is _d=23_d=23 since since ₂₃₂₃_x_x_{7=161= 10}_{7=161= 10}_x_x₁₆₀₊₁₁₆₀₊₁

6.

6. Publish public key Publish public key PU={7,187}PU={7,187}

7.

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RSA Example - En/Decryption



sample RSA encryption/decryption is:

_{sample RSA encryption/decryption is:}



given message

_{given message}

_{M = 88}

(nb.

_(nb.

_88<187

)

₎



encryption:

_encryption:

C = 88

C = 8877_{mod 187 = 11}_{mod 187 = 11}



decryption:

_decryption:

M = 11

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Exponentiation

 can use the Square and Multiply Algorithm_{can use the Square and Multiply Algorithm}  a fast, efficient algorithm for exponentiation _{a fast, efficient algorithm for exponentiation}

 _{concept is based on repeatedly squaring base}_{concept is based on repeatedly squaring base}  _{and multiplying in the ones that are needed to}_{and multiplying in the ones that are needed to}

compute the result

 _{look at binary representation of exponent}_{look at binary representation of exponent}

 only takes O(logonly takes O(log₂₂ n) multiples for number n n) multiples for number n

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Exponentiation

c = 0; f = 1

for i = k downto 0

do c = 2 x c

f = (f x f) mod n

if b

_{if b}

_i_i

== 1

_{== 1}

then

_then

c = c + 1

f = (f x a) mod n

return f

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Efficient Encryption



encryption uses exponentiation to power e

_{encryption uses exponentiation to power e}



hence if e small, this will be faster

_{hence if e small, this will be faster}

 often choose e=65537 (2often choose e=65537 (21616-1)-1)

 also see choices of e=3 or e=17also see choices of e=3 or e=17



but if e too small (eg e=3) can attack

_{but if e too small (eg e=3) can attack}

 using Chinese remainder theorem & 3 using Chinese remainder theorem & 3

messages with different modulii messages with different modulii



if e fixed must ensure

_{if e fixed must ensure}

_{gcd(e,ø(n))=1}

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Efficient Decryption



decryption uses exponentiation to power d

_{decryption uses exponentiation to power d}

 this is likely large, insecure if notthis is likely large, insecure if not



can use the Chinese Remainder Theorem

_{can use the Chinese Remainder Theorem}

(CRT) to compute mod p & q separately.

then combine to get desired answer

 approx 4 times faster than doing directlyapprox 4 times faster than doing directly



only owner of private key who knows

_{only owner of private key who knows}

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RSA Key Generation



users of RSA must:

_{users of RSA must:}

 determine two primes determine two primes at random - at random - p, qp, q  select either select either ee or or dd and compute the other and compute the other



primes

_primes

_p,q

must not be easily derived

_{must not be easily derived}

from modulus

_n=p.q

 means must be sufficiently largemeans must be sufficiently large

 typically guess and use probabilistic testtypically guess and use probabilistic test



exponents

_exponents

_e

,

_,

_d

are inverses, so use

_{are inverses, so use}

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RSA Security



possible approaches to attacking RSA are:

_{possible approaches to attacking RSA are:}

 brute force key search - infeasible given size brute force key search - infeasible given size

of numbers of numbers

 mathematical attacks - based on difficulty of mathematical attacks - based on difficulty of

computing ø(n), by factoring modulus n computing ø(n), by factoring modulus n

 timing attacks - on running of decryptiontiming attacks - on running of decryption

 chosen ciphertext attacks - given properties of chosen ciphertext attacks - given properties of

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Factoring Problem

 mathematical approach takes 3 forms:_{mathematical approach takes 3 forms:}

 factor factor n=p.qn=p.q, hence compute , hence compute ø(n)ø(n) and then and then dd  determine determine ø(n)ø(n) directly and directly and compute compute dd

 find d directlyfind d directly

 currently believe all equivalent to factoring_{currently believe all equivalent to factoring}  have seen slow improvements over the years have seen slow improvements over the years

• as of May-05 best is 200 decimal digits (663) bit with LS as of May-05 best is 200 decimal digits (663) bit with LS

 biggest improvement comes from improved algorithmbiggest improvement comes from improved algorithm

• cf QS to GHFS to LS_{cf QS to GHFS to LS}

 currently assume 1024-2048 bit RSA is securecurrently assume 1024-2048 bit RSA is secure

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Progress in

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Progress

in

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Timing Attacks

 developed by Paul Kocher in mid-1990’s_{developed by Paul Kocher in mid-1990’s}  exploit timing variations in operations_{exploit timing variations in operations}

 eg. multiplying by small vs large number eg. multiplying by small vs large number  or IF's varying which instructions executedor IF's varying which instructions executed  infer operand size based on time taken _{infer operand size based on time taken}  RSA exploits time taken in exponentiation_{RSA exploits time taken in exponentiation}  countermeasures_{countermeasures}

 use constant exponentiation timeuse constant exponentiation time  add random delaysadd random delays