Tidy up

docs/lectures/cryptography/11_diffie-hellman.md

@@ -2,7 +2,7 @@
 
 - Two parties can jointly agree a *shared secret* over an *insecure channel*
 - Mathematically, what we are doing is both calculating the same value, mod a prime $p$
-  - Remember $p$ is $\times 10^{600}$
+    - Remember $p$ is $\times 10^{600}$
 - The parties separately compute the same key, rather than share it
 
 ### $\mathbb{Z}_n^*$
@@ -12,7 +12,7 @@
 > This set forms an *abelian* group under multiplication modulo $n$. The identity element is 1
 
 - In the majority of cases, we use a prime number as the modulus:
-  - $\mathbb{Z}_p^* = \{1,2,...,p-1\}$
+    - $\mathbb{Z}_p^* = \{1,2,...,p-1\}$
 
 **Group Cardinality** - The number of elements in that group
 
@@ -21,11 +21,11 @@ $$
 |\mathbb{Z}_m^*| = \Phi(n) \\
 $$
 
-- The security of ciphers often depend on the cardinality of the group
+- The security of ciphers often depends on the cardinality of the group
 
 #### Cyclic Groups
 
-- Lets consider group $\mathbb{Z}_{11}^*$
+- Let's consider group $\mathbb{Z}_{11}^*$
 - Consider calculating powers of 3 in this group
 
 $$
@@ -71,28 +71,30 @@ $$
 
 - A group that contains an element $g$ of maximum order is called a cyclic group
 - Any element of maximum order is called a primitive root, or a generator
-  - $2$ is a generator of $\mathbb{Z}_{11}^* \quad ord(2)=10$
-  - 3 is not a generator $\mathbb{Z}_{11}^* \quad ord(3)=5$
+    - $2$ is a generator of $\mathbb{Z}_{11}^* \quad ord(2)=10$
+    - 3 is not a generator $\mathbb{Z}_{11}^* \quad ord(3)=5$
 
 ##### Cyclic Subgroups
 
 - For all primes, $(\mathbb{Z}_{11}^*, \cdot)$ is an *abelian finite cyclic group*
-  - Let $g \in G$ where $G$ is a cyclic group:
-    1. $g^{|G|}=1$
-    2. $ord(g)$ divides $|G|$
-  - These are called **cyclic subgroups** 
+    - Let $g \in G$ where $G$ is a cyclic group:
+        1. $g^{|G|}=1$
+        2. $ord(g)$ divides $|G|$
+    - These are called **cyclic subgroups**
 - Orders of $\mathbb{Z}_{11}^*$
-  - ![1647359471.png](img/1647359471.png)
-  - Note the neutral element generates an order of $1$
+
+    - ![1647359471.png](img/1647359471.png)
+
+    - Note the neutral element generates an order of $1$
 
 ## Diffie-Hellman
 
 1. Alice and Bob agree on a large prime $p$, and a generator $g$ that is a primitive root of $p$
 2. Alice and Bob choose private numbers $a$ and $b$ at random in $\mathbb{Z}_p^*$
-   - Where $a\in \{1,2,...,p-1\}$
-   - and $b\in \{1,2,...,p-1\}$
+    - Where $a\in \{1,2,...,p-1\}$
+    - and $b\in \{1,2,...,p-1\}$
 3. Alice calculates $A=g^a\space mod \space p$ and sends $A$ publicly to Bob
-4. Bob calculates $B=g^b\space mod \space p$ and sends $B$ pubicly to Alice
+4. Bob calculates $B=g^b\space mod \space p$ and sends $B$ publicly to Alice
 5. Alice computes $k_{ab}=B^a\space mod \space p$
 6. Bob computes $k_{ab}=A^b\space mod \space p$
 
@@ -105,13 +107,13 @@ $$
 
 - Why is Diffie-Hellman so hard to break
 - Consider $\mathbb{Z}^*_{10000079},\space g=3$
-  - Alice calculates $A=3^a\space mod \space 10000079 = 4675535$
-  - What is $a$?
+    - Alice calculates $A=3^a\space mod \space 10000079 = 4675535$
+    - What is $a$?
 - This is the discrete logarithm problem
 
 **Brute Force** requires $O(|G|)$
 
-**Shank’s Baby-Step Giant-Step** requires $O(\sqrt{|G|})$ and $\sim \sqrt{|G|}$ space
+**Shanks’ Baby-Step Giant-Step** requires $O(\sqrt{|G|})$ and $\sim \sqrt{|G|}$ space
 
 - Using 128 bits, this is $2^{64}$, which would need a cluster
 
@@ -121,15 +123,13 @@ $$
 
 - The discrete log problem is solved mod each prime factor and the results combined using the Chinese remainder theorem
 
-**Index calculus** directly attacks $\mathbb{Z}_p^*$ and is the reason Elliptic Curves is so much more efficient
+**Index calculus** directly attacks $\mathbb{Z}_p^*$ and is the reason elliptic curves are so much more efficient
 
 ##### Choosing Primes
 
 - To avoid any unexpected small subgroup attacks, commonly used DH primes are **safe primes**
 - A safe prime is a prime $p$ where $\frac{(p-1)}{2}$ is also a prime
 - Consider the order of $\mathbb{Z}_p^*$ for a safe prime
-  - This will have two subgroups of order $p-1$ and $2$
-  - By choosing a generator of the **subgroup of large prime order**, we avoid attacks on small factors of the group order
-  - Basically this ensures the prime factorisation has one massive prime in it
-
-
+    - This will have two subgroups of order $p-1$ and $2$
+    - By choosing a generator of the **subgroup of large prime order**, we avoid attacks on small factors of the group order
+    - Basically this ensures the prime factorisation has one massive prime in it