1 thinking the impossible “modern cryptography” jeremy r. johnson

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Thinking the Impossible“Modern Cryptography”

Jeremy R. Johnson

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Introduction

• Objective: To see how to securely communicate on the internet without giving up privacy. To understand what a public key cryptosystem is and how the RSA algorithm works. To do impossible things.

– Modern cryptography– Solutions to some “impossible problems”– Public Key Cryptosystems– Modular Arithmetic– RSA Algorithm

References: Rivest, Shamir, Adelman

CS Unplugged

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Importance of the Area

• Did you buy anything online recently? Use an ATM machine? If so, whether you know it or not, you used cryptography. Cryptography (in the guise of the SSL protocol) protects your credit card information as it whizzes across the Internet, and ensures that others can't withdraw money from your account.

• The ubiquitous use of tools such as SSL and SSH shows that cryptography, once an esoteric military concern, has now burst into the mainstream. Yet, this is only the beginning of a coming flood.

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“Impossible” Problem One

How can you determine the outcome of a vote on the intenet without revealing individual votes?

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Classical Cryptography

• Basic problem: Secure communication over an insecure channel

• Solution: private key encryption– m E(k,m) = c D(k,c) = m

• Shannon provided a rigorous theory of perfect secrecy based on information theory– Adversary has unlimited computational resources, but

key must be as long as message

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Substitution Cypher

HELLO

ALL HAIL CEASAR

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Substitution Cypher

KHOOR

DOO KDLO FHDVDU

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Frequency Analysis

en.wikipedia.org/wiki/Frequency_analysis_(cryptanalysis)

scottbryce.com/cryptograms

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One Time Pad

• Pad = b1 bn {0,1}* chosen randomly

• m = m1 mn

– E(Pad,m) = c = m Pad– D(Pad,c) = c Pad = (m Pad) Pad = m

m,c PrPad[E(Pad,m) = c] = 1/2n

– No information gained from seeing c– However, E(Pad,m) E(Pad,m’) = m m’

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“Impossible” Problem Two

How can you send a secret over the internet without previously sending a courier to distribute the secret key?

Is your method secure?

The answer comes from modern cryptography and relies on public key cryptography

Whitfield Diffie and Martin E. Hellman, "New Directions in Cryptography", IEEE Transactions on Information Theory, Vol. IT-22, No. 6, Nov. 1976.

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Public Key Cryptosystem

Let M be a message and let C be the encrypted message (ciphertext). A public key cryptosystem has a separate method E() for encrypting and D() decrypting. •D(E(M)) = M•Both E() and D() are easy to compute•Publicly revealing E() does not make it easy to determine D()•E(D(M)) = M - needed for signatures

The collection of E()’s are made publicly available but the D()’s remain secret. Called a one-way trap-door function (hard to invert, but easy if you have the secret information)

Public Key Encryption Map (From CS Unplugged)

Public Key Encryption Map

The Map

What To Do

Come up with 10 numbers that add up to your ASCII value.

Label your vertices with the values.

Take each vertex and its neighbors, compute the sum, and replace the vertex value with that sum.

Erase the old values!!!

ASCII Table

Private Key Encryption Map

Private Key Encryption Map

The Private Key

What To Do

• Just add up the values of each bold vertex from the public map you were given.

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Modern Cryptography

• Adversary’s resources are computationally bounded– Probabilistic polynomial time algorithm

• Impossibility of breaking the encryption system Infeasibility of breaking

• Rely on gap between efficient algorithms for encryption and computational infeasibility of decryption by adversary

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Dominating Sets & NP Completeness

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Dominating Sets & NP Completeness

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“Impossible” Problem Three

How can you flip a coin over the phone?

The answer comes from modern cryptography and is the key to secure communication over the internet, provides privacy, authentication and digital signatures

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“Impossible” Problem Four

How can you prove you know something to an adversary without revealing your secret?

The answer comes from the area of zero knowledge proofs

Where’s Waldo

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Open Sesame

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Jean-Jacques Quisquater, Louis C. Guillou, Thomas A. Berson. How to Explain Zero-Knowledge Protocols to Your Children. Advances in Cryptology - CRYPTO '89:

Proceedings, v.435, p.628-631, 1990.

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Zero Knowledge Proof

1. Completeness: if the statement is true, the honest verifier (that is, one following the protocol properly) will be convinced of this fact by an honest prover.

2. Soundness: if the statement is false, no cheating prover can convince the honest verifier that it is true, except with some small probability.

3. Zero-knowledge: if the statement is true, no cheating verifier learns anything other than this fact. This is formalized by showing that every cheating verifier has some simulator that, given only the statement to be proven (and no access to the prover), can produce a transcript that "looks like" an interaction between the honest prover and the cheating verifier.

Secure Passwords

• Every users stores a statement of a theorem in a publicly readable directory

• Upon login, the user engages in a zero-knowledge proof of the correctness of the theorem

• If the proof is convincing access is granted• Guarantees that an adversary who overhears the

proof can not learn enough to gain access

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RSA Public Key Cryptosystem

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Clock Arithmetic

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3

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56

7

8

9

10

110

7+6 = ?

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Clock Arithmetic

1

2

3

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56

7

8

9

10

110

7 + 1

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Clock Arithmetic

1

2

3

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56

7

8

9

10

110

7 + 2

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Clock Arithmetic

1

2

3

4

56

7

8

9

10

110

7 + 3

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Clock Arithmetic

1

2

3

4

56

7

8

9

10

110

7 + 4

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Clock Arithmetic

1

2

3

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56

7

8

9

10

110

7 + 5

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Clock Arithmetic

1

2

3

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56

7

8

9

10

110

7 + 6 = 1 (mod 12)

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Clock Arithmetic

1

2

3

4

56

7

8

9

10

110

5 5 = ?

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Clock Arithmetic

1

2

3

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56

7

8

9

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110

5 2

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Clock Arithmetic

1

2

3

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56

7

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110

5 3

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Clock Arithmetic

1

2

3

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56

7

8

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110

5 4

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Clock Arithmetic

1

2

3

4

56

7

8

9

10

110

5 5 = 1 (mod 12)

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Multiplication Table mod 5

0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

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Multiplication Table mod 6

0 1 2 3 4 5

0 0 0 0 0 0 0

1 0 1 2 3 4 5

2 0 2 4 0 2 4

3 0 3 0 3 0 3

4 0 4 2 0 4 2

5 0 5 4 3 2 1

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Modular Arithmetic (Zn)

Definition: a b (mod n) n | (b - a)

Alternatively, a = qn + b

Properties (equivalence relation)– a a (mod n) [Reflexive]– a b (mod n) b a (mod n) [Symmetric]– a b (mod n) and b c (mod n) a c (mod n) [Transitive]

Definition: An equivalence class mod n

[a] = { x: x a (mod n)} = { a + qn | q }

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Modular Arithmetic (Zn)It is possible to perform arithmetic with equivalence classes mod n.

– [a] + [b] = [a+b]– [a] * [b] = [a*b]

In order for this to make sense, you must get the same answer (equivalence) class independent of the choice of a and b. In other words, if you replace a and b by numbers equivalent to a or b mod n you end of with the sum/product being in the same equivalence class.

a1 a2 (mod n) and b1 b2 (mod n) a1+ b1 a2 + b2 (mod n)

a1* b1 a2 * b2 (mod n)

(a + q1n) + (b + q2n) = a + b + (q1 + q2)n

(a + q1n) * (b + q2n) = a * b + (b*q1 + a*q2 + q1* q2)n

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Representation of Zn

The equivalence classes [a] mod n, are typically represented by the representatives a.

• Positive Representation: Choose the smallest positive integer in the class [a] then the representation is {0,1,…,n-1}.

• Symmetric Representation: Choose the integer with the smallest absolute value in the class [a]. The representation is {-(n-1)/2 ,…, n/2 }. When n is even, choose the positive representative with absolute value n/2.

• E.G. Z6 = {-2,-1,0,1,2,3}, Z5 = {-2,-1,0,1,2}

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Euclidean Algorithm

gcd(a,b)

if b = 0 then

return a

else

return gcd(b, a mod b)

Example: gcd(30,12)

gcd(12,6)

gcd(6,0)

Efficient!!! O(log N), a, b N

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Modular Inverses

Definition: x is the inverse of a mod n, if ax 1 (mod n)

The equation ax 1 (mod n) has a solution iff gcd(a,n) = 1.

Extended Euclidean Algorithm, there exist x and y such that

ax + ny = gcd(a,n).

When gcd(a,n) = 1, ax + ny = 1 ax 1 (mod n)

Example

gcd(5,12) = 1, 5 5 + -2 12 = 1

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Euler phi function

• Definition: phi(n) = #{a: 0 < a < n and gcd(a,n) = 1}• Properties:

(p) = p-1, for prime p. (p^e) = (p-1)*p^(e-1) (m*n) = (m)* (n) for gcd(m,n) = 1. (p*q) = (p-1)*(q-1)

• Examples:

(15) = (3)* (5) = 2*4 = 8. = #{1,2,4,7,8,11,13,14} (9) = (3-1)*3^(2-1) = 2*3 = 6 = #{1,2,4,5,7,8}

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Euler’s Identity

• The number of elements in Zn that have multiplicative inverses is equal to phi(n).

• Theorem: Let (Zn)* be the elements of Zn with inverses (called units). If a (Zn)*, then a(n) 1 (mod n).

Proof. The same proof presented for Fermat’s theorem can be used to prove this theorem.

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Chinese Remainder Theorem

Theorem: If gcd(m,n) = 1, then given a and b there exist an integer solution to the system:

x a (mod m) and x = b (mod n).

Proof:

Consider the map x (x mod m, x mod n).

This map is a 1-1 map from Zmn to Zm Zn, since if x and y map to the same pair, then x y (mod m) and x y (mod n). Since gcd(m,n) = 1, this implies that x y (mod mn).

Since there are mn elements in both Zmn and Zm Zn, the map is also onto. This means that for every pair (a,b) we can find the desired x.

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Let M be a message and let C be the encrypted message (ciphertext). A public key cryptosystem has a separate method E() for encrypting and D() decrypting.

• D(E(M)) = M• Both E() and D() are easy to compute• Publicly revealing E() does not make it easy to determine

D()• E(D(M)) = M - needed for signatures


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RSA Public Key Cryptosystem

Based on the idea that it is hard to factor large numbers.

First encode M as an integer (e.g. use ASCII). Large messages will need to be blocked.

• Choose n = p*q, the product of two large prime numbers.• Choose e such that gcd(e,phi(n)) = 1. • Choose d such that de 1 (mod (n))

• E = (e,n) and E(M) = Me mod n• D = (d,n) and D(M) = Md mod n

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Correctness of the RSA Algorithm

Theorem: D(E(M)) = E(D(M)) = M.

Proof. D(E(M)) = (Me)d (mod n) = Med (mod n).

Since ed 1 (mod (n)), ed = k* (n) + 1, for some integer k.

Mk* (n)+1 (Mk* (n)+1 mod p, Mk* (n)+1 mod q)

= (Mk* (n) * M mod p, Mk* (n) * M mod q)

= (M(p-1)*(q-1)*k * M mod p, M(q-1)*(p-1)*k * M mod q) [since n = pq]

= ((M(p-1))(q-1)*k * M mod p, (M(q-1))(p-1)*k * M mod q)

= (M mod p, M mod q) [By Fermat’s theorem]

Therefore, by the CRT, Mk* (n)+1 M (mod n).

1 thinking the impossible “modern cryptography” jeremy r. johnson

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