# Lecture 20

## Construction of CRHF (Compression Resistant Hash Function)

Let $h: \{0, 1\}^{n+1} \to \{0, 1\}^n$ be a CRHF.

Base on the discrete log assumption, we can construct a CRHF $H: \{0, 1\}^{n+1} \to \{0, 1\}^n$ as follows:

$Gen(1^n):(g,p,y)$

$p\in \tilde{\Pi}_n(p=2q+1)$

$g$ generator for group of sequence $\mod p$ (G_q)

$y$ is a random element in $G_q$

$h_{g,p,y}(x,b)=y^bg^x\mod p$, $y^bg^x\mod p \in \{0,1\}^n$

$g^x\mod p$ if $b=0$, $y\cdot g^x\mod p$ if $b=1$.

Under the discrete log assumption, $H$ is a CRHF.

- It is easy to sample $(g,p,y)$
- It is easy to compute
- Compressing by 1 bit

Proof it is a CRHF:

Suppose there exists an adversary $\mathcal{A}$ that can break $h$ with non-negligible probability $\mu$.

$$
P[(p,g,y)\gets Gen(1^n);(x_1,b_1),(x_2,b_2)\gets \mathcal{A}(p,g,y):y^{b_1}g^{x_1}\equiv y^{b_2}g^{x_2}\mod p\land (x_1,b_1)\neq (x_2,b_2)]=\mu(n)>\frac{1}{p(n)}
$$

Where $y^{b_1}g^{x_1}=y^{b_2}g^{x_2}\mod p$ is the collision of $H$.

Suppose $b_1=b_2$.

Then $y^{b_1}g^{x_1}\equiv y^{b_2}g^{x_2}\mod p$ implies $g^{x_1}\equiv g^{x_2}\mod p$.

So $x_1=x_2$ and $(x_1,b_1)=(x_2,b_2)$.

So $b_1\neq b_2$, Without loss of generality, say $b_1=1$ and $b_2=0$.

$y\cdot g^{x_1}\equiv g^{x_2}\mod p$ implies $y\equiv g^{x_2-x_1}\mod p$.

We can create a adversary $\mathcal{B}$ that can break the discrete log assumption with non-negligible probability $\mu(n)$ using $\mathcal{A}$.

Let $g,p$ be chosen and set random $x$ such that $y=g^x\mod p$.

Let the algorithm $\mathcal{B}$ defined as follows:

```pseudocode
function B(p,g,y):
    (x_1,b_1),(x_2,b_2)\gets \mathcal{A}(p,g,y)
    If (x_1,1) and (x_2,0) and there is a collision:
        y=g^{x_2-x_1}\mod p
        return x_2-x_1 for b=1
    Else:
        return "Failed"
```

$$
P[B\text{ succeeds}]\geq P[A\text{ succeeds}]-\frac{1}{p(n)}>\frac{1}{p(n)}
$$

So $\mathcal{B}$ can break the discrete log assumption with non-negligible probability $\mu(n)$, which contradicts the discrete log assumption.

So $h$ is a CRHF.