Add Euclidean algorithm (#201)

Co-authored-by: David Leal <[email protected]>

Author: appgurueu

en/Basic Math/Euclidean algorithm.md

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+# Euclidean algorithm
+
+## Problem
+
+Find the Greatest Common Divisor (GCD) of two positive integers $a$ and $b$, which is defined as the largest number $g = gcd(a, b)$ such that $g$ divides both $a$ and $b$ without remainder.
+
+## Idea
+
+The Euclidean algorithm is based on the simple observation that the GCD of two numbers doesn't change if the smaller number is subtracted from the larger number:
+
+Let $a > b$. Let $g$ be the GCD of $a$ and $b$.
+Then $g$ divides $a$ and $b$. Thus $g$ also divides $a - b$.
+
+Let $g'$ be the GCD of $b$ and $a - b$.
+
+Proof by contradiction that $g' = g$:
+
+Assume $g' < g$ or $g' > g$.
+
+If $g' < g$, $g'$ would not be the *greatest* common divisor,
+since $g$ is also a common divisor of $a - b$ and $b$.
+
+If $g' > g$, $g'$ divides $b$ and $a - b$ -
+that is, there exist integers $n, m$
+such that $g'n = b$ and $g'm = a - b$.
+Thus $g'm = a - g'n \iff g'm + g'n = a \iff g'(m + n) = a$.
+This imples that $g' > g$ also divides $a$,
+which contradicts the initial assumption that $g$ is the GCD of $a$ and $b$.
+
+## Implementation
+
+To speed matters up in practice, modulo division is used instead of repeated subtractions:
+$b$ can be subtracted from $a$ as long as $a >= b$.
+After these subtractions only the remainder of $a$ when divided by $b$ remains.
+
+A straightforward Lua implementation might look as follows:
+
+```lua
+function gcd(a, b)
+	while b ~= 0 do
+		a, b = b, a % b
+	end
+	return a
+end
+```
+
+note that `%` is the modulo/remainder operator;
+`a` is assigned to the previous value of `b`,
+and `b` is assigned to the previous value of `a`
+modulo the previous value of `b` in each step.
+
+## Analysis
+
+### Space Complexity
+
+The space complexity can trivially be seen to be constant:
+Only two numbers (of assumed constant size), $a$ and $b$, need to be stored.
+
+### Time Complexity
+
+Each iteration of the while loop runs in constant time and at least halves $b$.
+Thus $O(log_2(n))$ is an upper bound for the runtime.
+
+## Walkthrough
+
+Finding the GCD of $a = 42$ and $b = 12$:
+
+1. $42 \mod 12 = 6$
+2. $12 \mod 6 = 0$
+
+The result is $gcd(42, 12) = 6$.
+
+Finding the GCD of $a = 633$ and $b = 142$ using the Euclidean algorithm:
+
+1. $633 \mod 142 = 65$
+2. $142 \mod 65 = 12$
+3. $65 \mod 12 = 5$
+4. $12 \mod 5 = 2$
+5. $5 \mod 2 = 1$
+6. $2 \mod 1 = 0$
+
+The result is $gcd(633, 142) = 1$: $a$ and $b$ are co-prime.
+
+## Applications
+
+* Shortening fractions
+* Finding the Least Common Multiple (LCM)
+* Efficiently checking whether two numbers are co-prime (needed e.g. for the RSA cryptosystem)
+
+## Resources
+
+* [Wikipedia Article](https://en.wikipedia.org/wiki/Euclidean_algorithm)