Lecture 3.2: Public Key Cryptography II

CS 436/636/736 Spring 2012

Nitesh Saxena

Today’s Informative/Fun Bit – Acoustic Emanations

• http://www.google.com/search?source=ig&hl=en&rlz=&q=keyboard+acoustic+emanations&btnG=Google+Search

• http://tau.ac.il/~tromer/acoustic/

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Course Administration

• HW2 – due at 11am on Feb 06• Any questions, or help needed?

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Outline of Today’s Lecture

• Number Theory• Modular Arithmetic

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Modular Arithmetic

• Definition: x is congruent to y mod m, if m divides (x-y). Equivalently, x and y have the same remainder when divided by m.

Notation: Example: • We work in Zm = {0, 1, 2, …, m-1}, the group of

integers modulo m• Example: Z9 ={0,1,2,3,4,5,6,7,8}• We abuse notation and often write = instead

of 5

)(modmyx 14 5(mod9)

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Addition in Zm :

• Addition is well-defined:

– 3 + 4 = 7 mod 9.– 3 + 8 = 2 mod 9.

6

)(mod''

)(mod'

)(mod'

myxyx

then

myy

mxx

if

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Additive inverses in Zm

• 0 is the additive identity in Zm

• Additive inverse of a is -a mod m = (m-a)– Every element has unique additive inverse. – 4 + 5= 0 mod 9. – 4 is additive inverse of 5.

7

)(mod0)(mod0 mxmxx

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Multiplication in Zm :

• Multiplication is well-defined:

– 3 * 4 = 3 mod 9.– 3 * 8 = 6 mod 9.– 3 * 3 = 0 mod 9.

8

)(mod''

)(mod'

)(mod'

myxyx

then

myy

mxx

if

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Multiplicative inverses in Zm

• 1 is the multiplicative identity in Zm

• Multiplicative inverse (x*x-1=1 mod m)– SOME, but not ALL elements have unique

multiplicative inverse. – In Z9 : 3*0=0, 3*1=3, 3*2=6, 3*3=0, 3*4=3, 3*5=6, …,

so 3 does not have a multiplicative inverse (mod 9)– On the other hand, 4*2=8, 4*3=3, 4*4=7, 4*5=2,

4*6=6, 4*7=1, so 4-1=7, (mod 9)

9

)(mod1)(mod1 mxmxx

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Which numbers have inverses?

• In Zm, x has a multiplicative inverse if and only if x and m are relatively prime or gcd(x,m)=1– E.g., 4 in Z9

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Extended Euclidian: a-1 mod n

• Main Idea: Looking for inverse of a mod n means looking for x such that x*a – y*n = 1.

• To compute inverse of a mod n, do the following:– Compute gcd(a, n) using Euclidean algorithm.– Since a is relatively prime to m (else there will be no inverse) gcd(a, n)

= 1.– So you can obtain linear combination of rm and rm-1 that yields 1.

– Work backwards getting linear combination of ri and ri-1 that yields 1.

– When you get to linear combination of r0 and r1 you are done as r0=n and r1= a.

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Example – 15-1 mod 37

• 37 = 2 * 15 + 7• 15 = 2 * 7 + 1• 7 = 7 * 1 + 0Now,• 15 – 2 * 7 = 1• 15 – 2 (37 – 2 * 15) = 1• 5 * 15 – 2 * 37 = 1So, 15-1 mod 37 is 5.

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Modular Exponentiation:Square and Multiply method

• Usual approach to computing xc mod n is inefficient when c is large.

• Instead, represent c as bit string bk-1 … b0 and use the following algorithm:

z = 1For i = k-1 downto 0 doz = z2 mod n

if bi = 1 then z = z* x mod n

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Example: 3037 mod 77

14

z = z2 mod n

if bi = 1 then z = z* x mod n

i b z

5 1 30 =1*1*30 mod 77

4 0 53 =30*30 mod 77

3 0 37 =53*53 mod 77

2 1 29 =37*37*30 mod 77

1 0 71 =29*29 mod 77

0 1 2 =71*71*30 mod 77

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Other Definitions• An element g in G is said to be a generator of a

group if a = gi for every a in G, for a certain integer i– A group which has a generator is called a cyclic group

• The number of elements in a group is called the order of the group

• Order of an element a is the lowest i (>0) such that ai = e

• A subgroup is a subset of a group that itself is a group

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Lagrange’s Theorem

• Order of an element in a group divides the order of the group

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Euler’s totient function

• Given positive integer n, Euler’s totient function is the number of positive numbers less than n that are relatively prime to n

• Fact: If p is prime then – {1,2,3,…,p-1} are relatively prime to p.

17

( ) 1p p

)(n

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Euler’s totient function

• Fact: If p and q are prime and n=pq then

• Each number that is not divisible by p or by q is relatively prime to pq.– E.g. p=5, q=7:

{1,2,3,4,-,6,-,8,9,-,11,12,13,-,-,16,17,18,19,-,-,22,23,24,-,26,27,-,29,-,31,32,33,34,-}

– pq-p-(q-1) = (p-1)(q-1)

18

)1)(1()( qpn

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Euler’s Theorem and Fermat’s Theorem

• If a is relatively prime to n then

• If a is relatively prime to p then ap-1 = 1 mod p

Proof : follows from Lagrange’s Theorem

19

na n mod1)(

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Euler’s Theorem and Fermat’s Theorem

EG: Compute 9100 mod 17:

p =17, so p-1 = 16. 100 = 6·16+4. Therefore, 9100=96·16+4=(916)6(9)4 . So mod 17 we have 9100

(916)6(9)4 (mod 17) (1)6(9)4 (mod 17) (81)2 (mod 17) 16

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Some questions

• 2-1 mod 4 =?• What is the complexity of

– (a+b) mod m– (a*b) mod m– a-1 mod (m)– xc mod (n)

• Order of a group is 5. What can be the order of an element in this group?

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Further Reading

• Chapter 4 of Stallings• Chapter 2.4 of HAC

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