Tidy up

docs/lectures/cryptography/10_RSA.md

@@ -21,19 +21,17 @@
 ### Euler Totient Function
 
 - Integers $a$ and $m$ are *relatively prime* if they do not share a divisor (except 1)
-  - $gcd(a,m) = 1$
+    - $gcd(a,m) = 1$
 - The **Euler totient** $\Phi$ is the number of integers in $\mathbb{Z}_m = \{0,1,...m-1\}$ for which $gcd(a,m)=1$
-  - For example $\Phi(9)=6$ as:
-    - $gcd(1,9)=1$ :white_check_mark:
-    - $gcd(2,9)=1$ :white_check_mark:
-    - $gcd(3,9)=3$ ❌
-    - $gcd(4,9)=1$ :white_check_mark:
-    - $gcd(5,9)=1$ :white_check_mark:
-    - $gcd(6,9)=3$ ❌
-    - $gcd(7,9)=1$ :white_check_mark:
-    - $gcd(8,9)=1$ :white_check_mark:
-    
-      ### 
+    - For example $\Phi(9)=6$ as:
+        - $gcd(1,9)=1$ :white_check_mark:
+        - $gcd(2,9)=1$ :white_check_mark:
+        - $gcd(3,9)=3$ ❌
+        - $gcd(4,9)=1$ :white_check_mark:
+        - $gcd(5,9)=1$ :white_check_mark:
+        - $gcd(6,9)=3$ ❌
+        - $gcd(7,9)=1$ :white_check_mark:
+        - $gcd(8,9)=1$ :white_check_mark:
 
 #### Integer Factorisation
 
@@ -64,19 +62,19 @@ $$
 #### Fermat’s Little Theorem
 
 - Fermat’s little theorem states that for some prime $p$, and any integer $a$:
-  - $a^{p-1} \equiv 1 \space (mod \space p)$
-    - Also note that $a^{p-1} = a\cdot a^{p-2} \equiv 1 \space (mod \space p)$
-    - Therefore $a^{p-2}$ is actually the inverse of $a\space (mod \space p)$
-  - It follows that $a^p \equiv p \space (mod \space p)$
+    - $a^{p-1} \equiv 1 \space (mod \space p)$
+        - Also note that $a^{p-1} = a\cdot a^{p-2} \equiv 1 \space (mod \space p)$
+        - Therefore $a^{p-2}$ is actually the inverse of $a\space (mod \space p)$
+    - It follows that $a^p \equiv p \space (mod \space p)$
 
 #### Euler’s Theorem
 
 - Generalisation of Fermat’s little theorem, not exclusive to primes
-  - $a^{\Phi(m)} \equiv 1 \space (mod \space m)$
-  - If $gcd(a,m)=1$
+    - $a^{\Phi(m)} \equiv 1 \space (mod \space m)$
+    - If $gcd(a,m)=1$
 - This works for any integer ring $\mathbb{Z}_m$
-  - We can see that FLT is a special case of this
-  - $\Phi(p) = (p-1) \therefore a^{\Phi(p)} = a^{p-1} \equiv 1 \space (mod \space p)$
+    - We can see that FLT is a special case of this
+    - $\Phi(p) = (p-1) \therefore a^{\Phi(p)} = a^{p-1} \equiv 1 \space (mod \space p)$
 
 ## RSA Key Generation
 
@@ -86,7 +84,7 @@ $$
 4. Choose a value $e\in \{2, ..., \Phi(n) -1\}$ where $gcd(\Phi(n),e)=1$
 5. Compute $d$ where $d\cdot e \equiv 1 \space (mod \space \Phi(n))$
 
- ![1647285406.png](img/1647285406.png)
+![1647285406.png](img/1647285406.png)
 
 $d$ is very easy to calculate if you know $p$ and $q$
 
@@ -97,29 +95,29 @@ $d$ is very easy to calculate if you know $p$ and $q$
 ##### Encryption
 
 - Now we have a public key $(3, 187)$ and private key $107$
-- Encryption and decryption is performed by:
-  - $x^e \equiv y \space (mod \space n)$
-  - $y^d \equiv x \space (mod \space n)$
+- Encryption and decryption are performed by:
+    - $x^e \equiv y \space (mod \space n)$
+    - $y^d \equiv x \space (mod \space n)$
 
 ![1647285650.png](img/1647285650.png)
 
 #### Proof
 
 - We want to show that $(x^e)^d = x^{ed} \equiv x \space (mod \space n)$
-- Let’s assume $gcd(x,n)=1$ So Euler’s theorem applies
-  - $e\cdot d=1\space (mod \space \Phi(n))$
-  - $\therefore e\cdot d = 1 + k\cdot \Phi(n)$
-  - $x^{e\cdot d} = x^{1+k\cdot \Phi(n)} = x\cdot x^{k+\Phi(n)}$
-  - $x\cdot (x^{\Phi(n)})^k=x\cdot(1)^k=x$
+- Let’s assume $gcd(x,n)=1$, so Euler’s theorem applies
+    - $e\cdot d=1\space (mod \space \Phi(n))$
+    - $\therefore e\cdot d = 1 + k\cdot \Phi(n)$
+    - $x^{e\cdot d} = x^{1+k\cdot \Phi(n)} = x\cdot x^{k+\Phi(n)}$
+    - $x\cdot (x^{\Phi(n)})^k=x\cdot(1)^k=x$
 
 ### Why is RSA Secure
 
 - We’d like the message $x$ based on some ciphertext $y$, given the public key $e$:
-  - $y \equiv ?^d \space (mod \space n)$
-  - $x \equiv y^? \space (mod \space n)$
+    - $y \equiv ?^d \space (mod \space n)$
+    - $x \equiv y^? \space (mod \space n)$
 - It can be fairly easy to calculate $d$:
-  - $e\cdot d \equiv q \space (mod \space \Phi(n))$
-  - $\Phi(n) = (p-1)(q-1)$
+    - $e\cdot d \equiv q \space (mod \space \Phi(n))$
+    - $\Phi(n) = (p-1)(q-1)$
 - As an attacker we only have access to $e$ and $d$
 
 ### Exponentiation
@@ -129,7 +127,7 @@ x^4 = x^2 \cdot x^2 \\
 x^8 = x^4 \cdot x^4
 $$
 
-When calculating a exponent raised to a power of two, we can use previously calculated values.
+When calculating an exponent raised to a power of two, we can use previously calculated values.
 
 ##### Binary Exponentiation
 
@@ -158,8 +156,7 @@ $$
 ##### Computational Complexity
 
 - What is the computational complexity of exponentiation?
-  - For a 2048 key:
-    - $X^{2^{2048}}$ - A ridiculously big number
-  - Where as using square and multiply
-    - $2048=T$ we need $\frac{3T}{2}$ calculations
-
+    - For a 2048 key:
+        - $X^{2^{2048}}$ - A ridiculously big number
+    - Whereas using square and multiply
+        - $2048=T$ we need $\frac{3T}{2}$ calculations