Lecture 7

Letter choosing experiment

For 100 letter tiles,

p_1,...,p_{27} (with oe blank)

(p_1)^2+\dots +(p_{27})^2\geq\frac{1}{27}

For any p_1,...,p_n, 0\leq p_i\leq 1.

\sum p_i=1

$P[$the same event twice in a row$]=p_1^2+p_2^2....+p_n^2$

By Cauchy-Schwarz: |u\cdot v|^2 \leq ||u||\cdot ||v||^2.

let \vec{u}=(p_1,...,p_n), \vec{v}=(1,..,1), so (p_1^2+p_2^2....+p_n)^2\leq (p_1^2+p_2^2....+p_n^2)\cdot n. So p_1^2+p_2^2....+p_n^2\geq \frac{1}{n}

So for an adversary A, who random choose x' and output f(x')=f(x) if matched. P[f(x)=f(x')]\geq\frac{1}{|Y|}

So P[x\gets f(x);y=f(x):f(a(y,1^n))=y]\geq \frac{1}{|Y|}

Modular arithmetic

For a,b\in \mathbb{Z}, N\in \mathbb{Z}^2

a\equiv b \mod N\iff N|(a-b)\iff \exists k\in \mathbb{Z}, a-b=kN,a=kN+b

Ex: N=23, -20\equiv 3\equiv 26\equiv 49\equiv 72\mod 23.

Equivalent relations for any `N` on `\mathbb{Z}`

a\equiv a\mod N

a\equiv b\mod N\iff b\equiv a\mod N

a\equiv b\mod N and b\equiv c\mod N\implies a\equiv c\mod N

Division Theorem

For any a\in \mathbb{Z}, and N\in\mathbb{Z}^+, \exists unique\ r,0\leq r<N.

\mathbb{Z}_N=\{0,1,2,...,N-1\} with modular arithmetic.

a+b\mod N,a\cdot b\mod N

Theorem: If a\equiv b\mod N and$c\equiv d\mod N$, then a\cdot c\equiv b\cdot d\mod N.

Definition: gcd(a,b)=d,a,b\in \mathbb{Z}^+, is the maximum number such that d|a and d|b.

Using normal factoring is slow... (Example: large p,q,r, N=p\cdot q,,M=p\cdot r)

Euclidean algorithm.

Recursively relying on fact that (a>b>0)

gcd(a,b)=gcd(b,a\mod b)

def euclidean_algorithm(a,b):
    if a<b: return euclidean_algorithm(b,a)
    if b==0: return a
    return euclidean_algorithm(b,a%b)

Proof:

We'll show d|a and d|b\iff d|b and d|(a\mod b)

\impliedby a=q\cdot b+r, r=a\mod b

\implies d|r, r=a\mod b

Runtime analysis:

Fact: b_{i+2}<\frac{1}{2}b_i

Proof:

Since a_i=q_i\cdot b_i+b_{i+1}, and b_1=q_2\cdot b_2+b_3, b_2>b_3, and q_2 in worst case is 1, so b_3<\frac{b_1}{2}

T(n)=2\Theta(\log b)=O(\log n) (linear in size of bits input)

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