Lecture 8

Chapter 2: Computational Hardness

Computational number theory/arithmetic

We want to have a easy-to-use one-way functions for cryptography.

How to find a^x\mod N quickly. a,x,N are positive integers. We want to reduce [a\mod N]

Example: 129^{39}\mod 41\equiv (129\mod 41)^{39}\mod 41=6^{39}\mod 41

Find the binary representation of x. e.g. express as sums of powers of 2.

x=39=bin(1,0,0,1,1,1)

Repeatedly square floor(\log_2(x)) times.


\begin{aligned}
    6^{39}\mod 41&=6^{32+4+2+1}\mod 41\\
    &=(6^{32}\mod 41)(6^{4}\mod 41)(6^{2}\mod 41)(6^{1}\mod 41)\mod 41\\
    &=(-4)(25)(-5)(6)\mod 41\\
    &=7
\end{aligned}

The total multiplication steps is floor(\log_2(x))

looks like fast exponentiation right?

Goal: f_{g,p}(x)=g^x\mod p is a one-way function, for certain choice of p,g (and assumptions)

A group (Nice day one for MODERN ALGEBRA)

A group G is a set with, a binary operation \oplus. and \forall a,b\in G, a \oplus b\to c

a,b\in G,a\oplus b\in G (closure)
(a\oplus b)\oplus c=a\oplus(b\oplus c) (associativity)
\exists e such that \forall a\in G, e\oplus g=g=g\oplus e (identity element)
\exists g^{-1}\in G such that g\oplus g^{-1}=e (inverse element)

Example:

\mathbb{Z}_N=\{0,1,2,3,...,N-1\} with addition \mod N, with identity element 0. a\in \mathbb{Z}_N, a^{-1}=N-a.
A even simpler group is \Z with addition.
\mathbb{Z}_N^*=\{x:x\in \mathbb{Z},1 \leq x\leq N: gcd(x,N)=1\} with multiplication \mod N (we can do division here! yeah...).
- If N=p is prime, then \mathbb{Z}_p^*=\{1,2,3,...,p-1\}
- If N=24, then \mathbb{Z}_{24}^*=\{1,5,7,11,13,17,19,23\}
  - Identity is 1.
  - Let a\in \mathbb{Z}_N^*, by Euclidean algorithm, gcd(a,N)=1,$\exists x,y \in Z$ such that ax+Ny=1,ax\equiv 1\mod N,x=a^{-1}
  - a,b\in \mathbb{Z}_N^*. Want to show gcd(ab,N)=1. If gcd(ab,N)=d>1, then some prime p|d. so p|(a,b), which means p|a or p|b. In either case, gcd(a,N)>d or gcd(b,N)>d, which contradicts that a,b\in \mathbb{C}_N^*

Euler's totient function

\phi:\mathbb{Z}^+\to \mathbb{Z}^+,\phi(N)=|\mathbb{Z}_N^*|=|\{1\leq x\leq N:gcd(x,N)=1\}|

Example: \phi(1)=1, \phi(24)=8, \phi (p)=p-1, \phi(p\cdot q)=(p-1)(q-1)

Euler's Theorem

For any a\in \mathbb{Z}_N^*, a^{\phi(N)}\equiv 1\mod N

Consequence: a^x\mod N, x=K\cdot \phi(N)+r,0\leq r\leq \phi(N)


a^x\equiv a^{K \cdot \phi (N) +r}\equiv ( a^{\phi(n)} )^K \cdot a^r \mod N$

So computing a^x\mod N is polynomial in \log (N) by reducing a\mod N and x\mod \phi(N)<N

Corollary: Fermat's little theorem:

1\leq a\leq p-1,a^{p-1}\equiv 1 \mod p

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