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+# Lecture 3
+
+All algorithms $C(x)\to y$, $x,y\in \{0,1\}^*$ 
+
+P.P.T= Probabilistic Polynomial-time Turing Machine.
+
+## Turing Machine: Mathematical model for a computer program
+
+A machine that can:
+
+1. Read in put
+2. Read/Write working tape move left/right
+3. Can change state
+
+### Assumptions
+
+Anything can be accomplished by a real computer program can be accomplished by a "sufficiently complicated" Turing Machine (TM).
+
+## Polynomial time
+
+We say $C(x),|x|=n,n\to \infty$ runs in polynomial time if it uses at most $T(n)$ operations bounded by some polynomials. $\exist c>0$ such that $T(n)=O(n^c)$
+
+If we can argue that algorithm runs in polynomially-many constant-time operations, then this is true for the T.M.
+
+$p,q$ are polynomials in $n$,
+
+$p(n)+q(n),p(n)q(n),p(q(n))$ are polynomial of $n$.
+
+Polynomial-time $\approx$ "efficient" for this course.
+
+## Probabilistic
+
+Our algorithm's have access to random "coin-flips" we can produce poly(n) random bits.
+
+$P[C(x)$ takes at most $T(n)$ steps $]=1$
+
+Our adversary $a(x)$ will be a P.P.T which is non-uniform (n.u.) (programs description size can grow polynomially in n)
+
+## Efficient private key encryption scheme 
+
+$m=\{0,1\}^n$
+
+$Gen(1^n)$ p.p.t output $k\in \mathcal{K}$
+
+$Enc_k(m)$ p.p.t outputs $c$
+
+$Dec_k(c')$ p.p.t outputs $m$ or "null"
+
+$P_k[Dec_k(Enc_k(m))=m]=1$
+
+## Negligible function
+
+$\varepsilon:\mathbb{N}\to \mathbb{R}$ is a negligible function if $\forall c>0$, $\exists N\in\mathbb{N}$ such that $\forall n\geq N, \varepsilon(n)<\frac{1}{n^c}$
+
+Idea: for any polynomial, even $n^{100}$, in the long run $\varepsilon(n)\leq \frac{1}{n^{100}}$
+
+Example: $\varepsilon (n)=\frac{1}{2^n}$, $\varepsilon (n)=\frac{1}{n^{\log (n)}}$
+
+Non-example: $\varepsilon (n)=O(\frac{1}{n^c})\forall c$
+
+## One-way function
+
+Idea: We are always okay with our chance of failure being negligible.
+
+Foundational concept of cryptography
+
+Goal: making $Enc_k(m),Dec_k(c')$ easy and $Dec^{-1}(c')$ hard.
+
+### Strong one-way function
+
+#### Definition: Strong one-way function
+
+$$
+f:\{0,1\}^n\to \{0,1\}^*(n\to \infty)
+$$
+
+There is a negligible function $\varepsilon (n)$ such that for any adversary $a$ (n.u.p.p.t)
+
+$$
+P[x\gets\{0,1\}^n;y=f(x):f(a(y))=y,a(y)=x']\leq\varepsilon(n)
+$$
+
+_Probability of guessing correct message is negligible_
+
+and
+
+there is a p.p.t which computes $f(x)$ for any $x$.
+
+- Hard to go back from output
+- Easy to find output
+
+$a$ sees output y, they wan to find some $x'$ such that $f(x')=y$.
+
+Example: Suppose $f$ is one-to-one, then $a$ must find our $x$, $P[x'=x]=\frac{1}{2^n}$, which is negligible.
+
+Why do we allow $a$ to get a different $x'$?
+
+> Suppose the definition is $P[x\gets\{0,1\}^n;y=f(x):a(y)=x]\neq\varepsilon(n)$, then a trivial function $f(x)=x$ would also satisfy the definition.
+
+To be technically fair, $a(y)=a(y,1^n)$, size of input $\approx n$, let them use $poly(n)$ operations.
+
+### Do one-way function exists?
+
+Unknown, actually...
+
+But we think so!
+
+We will need to use various assumptions. one that we believe very strongly based on evidence/experience
+
+Ex. $p,q$ are large random primes
+
+$N=p\cdot q$
+
+Factoring $N$ is hard. (without knowing $p,q$)