Add some more encryption fun

Cryptography/IntroToCryptography/Lecture2.md

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+# Modular Arithmetic and Historical Ciphers
+
+**GOAL:** Computation in finite sets 
+- i.e. infinite sets being real numbers, integers, etc... 
+- modern crypto is dealt with FINITE sets, though... you need special computational rules for this
+
+**Example:** for a finite set in everyday life...
+- If you go to the bakery and order 20 loaves of bread, and then order 12 more, you'll have 32 loaves of bread
+- But, 20 hours from 12 o'clock isn't 32 hours, it's 8 o'clock
+- This is the basis of modular/modulo arithmetic. It goes around "in circles." When the finite set goes up to 24, 20+12=8, 32 % 24 = 8
+
+**_(Def)_: modulo operator:** Let `a`, `r`, `m` ∈ `ℤ` and `m > 0`, we write `a` ≡ `r mod m`. If `m` divides `(a-r)`, i.e. `m / (a-r)`.
+
+**Example:** a = 13, m = 9, r = ?
+
+9 ≡ (13 - r), r = 4
+
+## 1a. Computation of the remainder 
+Given: `a, m ∈ ℤ`, write `a = q * m` (`q` being quotient)
+
+**Example:** a = 42, m = 9
+
+42 = q*9, q = 4, so r = 6
+
+but also...
+
+42 = 3 * 9 + 15, and in this case r = 15. **BUT, does this hold in the actual definition?**
+
+Check (42-15) = 27, 9 / 27 checks out
+
+but also......
+
+42 = 5 * 9 + (-3), and in this case r = -3 **BUT, does this hold in the actual definition?**
+
+Check (42 + 3) = 45, 9 / 45 checks out
+
+### LESSON: The remainder is not unique!!
+
+## 1b. Equivalence Classes
+**Example:** a = 12, m = 5, 12 ≡ 2 mod 5 (check 5 / (12-2)), 12 ≡ 7 mod 5 (check 5 / (12-7))
+
+Def'n the set `{..., -8, -3, 2, 7, 12, ...}` forms an **equivalence class** modulo 5. All members of the class behave equivalent modulo 5
+
+Let's look at all equivalence classes modulo 5...
+
+`{..., -10, -5, 0, 5, 10, ...}` A
+
+`{..., -9, -4, 1, 6, 11, ...}` B
+
+`{..., -8, -3, 2, 7, 12, ...}` C
+
+`{..., -7, -2, 3, 8, 13, ...}` D
+
+`{..., -6, -1, 4, 9, 14, ...}` E
+
+Let's say you have to do a computation: `13 * 16 - 8` = `208-8` = `200 ≡ 0 mod 5`. But, we can think of this as `D * B - D`... you can use **ANY OTHER MEMBERS OF THE EQUIVALENCE CLASS**. This is the same thing as `3 (i.e. 13 mod 5) * 1 (i.e. 16 mod 5) - 3 (i.e. 13 mod 5)` = `3-3` ≡ 0 mod 5
+
+**Example:** $3^8$ mod 7 ≡ ? - this is what HTTPS does (but the numbers are 200 bits long)
+- This could be done as $3^8$ = 6561 ≡ 2 mod 7
+- This could done more easily as $3^8$  = $3^4$ * $3^4$ = 81 * 81 **BUT YOU CAN MAKE EQUIVALENCE CLASSES FOR MODULO 7 AS ABOVE TO GET A MUCH SMALLER NUMBER!!** So you get `4 (i.e. 81 mod 7) * 4 = 16 ≡ 2 mod 7`
+
+**By convention, you'll usually be multiplying by a maximum of 4, as the smallest members of equivalence classes will usually be `{1, 2, 3, 4}`**
+
+### LESSON: equivalence classes are interchangeable!!
+
+## 2. Rings: An algebraic view on modular arithmetic 
+The "integer ring" $ℤ_m$ consists of:
+1. The set $ℤ_m$ = {0, 1, ..., m-1}
+2. Two operators `+` and `*` such that for all a, b, c, d ∈ $ℤ_m$
+
+             i. a + b ≡ c mod m
+
+             ii. a * b ≡ d mod m
+
+**Example for multiplicative inverses:** m = 9 (i.e. $ℤ_m$ = {0, 1, 2, 3, 4, 5, 6, 7, 8}), a = 2, $a^{-1}$ = ?
+- 2 * $2^{-1}$ ≡ 1 mod 9. So, 2 * 5 ≡ 1 mod 9, so $2^{-1}$ ≡ 5 mod 9
+- Or, you could do something random... 6 * ? ≡ 1 mod 9... compute _gcd_(6,9) = 3 != 1, so the inverse does not exist
+
+i.e. the nature of rings is:
+- you can always add and subtract
+- you can always multiply 
+- taking the inverse (or division), only works sometimes
+
+## 3. Shift (or Caesar) Cipher
+**Idea:** Shift letters in alphabet 
+
+**Example:** k = 3
+- A -> d
+- B -> e
+
+...
+
+- W -> z
+- X -> a
+- Y -> b
+- Z -> c
+
+Wrapping around at the end of the alphabet is done nicely with modulo 26!
+
+**Encryption:** $e_k$(x) = x + k mod 26
+
+**Decryption:** $d_k$(y) = y + k mod 26
+
+**EXAMPLE:** ATTACK = $x_1$,$x_2$,...$x_6$ = 0,19,19,0,2,10. This becomes 17,10,10,17,19,1 or rkkrtb
+
+Two attacks possible:
+1. Frequency analysis 
+2. Brute force
+
+## 4. Affine Cipher
+
+k = (a, b) -> key space becomes larger
+y ≡ a * x + b mod 26
+
+y - b ≡ a * x mod 26
+
+x ≡ $a^{-1}$(y-b) mod 26
+
+#k = ?, #b = 26, condition for a = _gcd_(a, 26)
+
+#k values = (#a values) * (#b values)
+
+**NOTE:** Write s$t^{-1}$ instead of s/t