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docs/lectures/cryptography/15_hash_functions.md

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+# Hash Functions
+
+#### Multiple Signatures
+
+- Could we simply split up a message and sign parts?
+
+![1649192960.png](img/1649192960.png)\
+
+A lot of faff for signing large files
+
+- An attacker can remove $s_{n-1}$ (or $s_{any}$) and it would still be valid
+
+### Properties of Hash Functions
+
+1. Any input length
+2. Fixed output length
+3. Pre-image resistance (one way)
+4. Second pre-image resistance
+   - If we have a hashed message, we cannot find another message with the same hash
+5. Collision resistance
+
+#### Pre-image Resistance
+
+- Hash functions must be one-way
+- Given a hash of a message $H(x)$ it must be infeasible to calculate $x$
+- Less applicable to digital signatures
+  - Crucial to password storage and key derivation
+
+#### Second Pre-image Resistance
+
+- Weak collision resistance
+- Given a message $x_1$ and a hash of that message $H(x_1)$ it should be infeasible to find a second message $x_2$ such that $H(x_1)=H(x_2)$
+
+![1649193613.png](img/1649193613.png)
+
+##### Second pre-image attack
+
+![1649193645.png](img/1649193645.png)
+
+Oscar finds a weak message (one of the messages is known ahead of time), he replaces the message $x_1$ with $x_2$. Now Oscar can send a signed message to Alice
+
+#### Collision Resistance
+
+- Strong collision resistance
+- It is not possible to find **any** message pair $x_1, x_2$ such that $H(x_1)=H(x_2)$
+- In practice, this is *much easier than finding a weak collision*
+
+![1649193812.png](img/1649193812.png)
+
+### Preventing Collisions
+
+![1649193848.png](img/1649193848.png)
+
+#### Collision Attack
+
+![1649193973.png](img/1649193973.png)
+
+##### How Likely
+
+**Second pre-image attacks**
+
+- For a 256 bit hash with good random properties we might expect $2^{256}$ bit brute force before we find a collision with $x_1$
+
+**Collision Attacks**
+
+- There are many other possible collisions beyond those simply with $x_1$
+
+### The Birthday Paradox
+
+> What is the probability two people in this room share a birthday
+
+- It is easier to first calculate the probability $P(n)$ that $n$ people do not share any birthdays:
+
+$$
+\begin{align*}
+P(2)&=(1-\frac{1}{365}) \\
+P(3)&=(1-\frac{1}{365})\cdot (1-\frac{2}{365}) \\
+P(n)&=(1-\frac{1}{365})\cdot (1-\frac{2}{365})\dots (1-\frac{n-1}{365})
+\end{align*}
+$$
+
+- The probability of at least one collision is $1 – P(\textrm{no collision})$.
+  - The probability of a collision with only 23 people is ~50%!
+  - For 40 people it’s ~90%
+- The same principle applies to hash functions, the more hashes computed, the more likely a collision becomes
+
+![1649194470.png](img/1649194470.png)
+
+#### The Birthday Attack
+
+- The output of the hash must be long enough to avoid a birthday attack
+- Given a hash function outputs $n$ bit hashes
+- You will find a collision after approx $\sqrt{(2^n)}=2^{\frac n2}$ random attempts
+- This means that your bit length needs to be double the size of your desired security margin
+- `SHA-256` therefore offers equivalent security to `AES 128`
+- left at `25:55`