|
| 1 | +# Euclidean algorithm |
| 2 | + |
| 3 | +## Problem |
| 4 | + |
| 5 | +Find the Greatest Common Divisor (GCD) of two positive integers $a$ and $b,ドル which is defined as the largest number $g = gcd(a, b)$ such that $g$ divides both $a$ and $b$ without remainder. |
| 6 | + |
| 7 | +## Idea |
| 8 | + |
| 9 | +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: |
| 10 | + |
| 11 | +Let $a > b$. Let $g$ be the GCD of $a$ and $b$. |
| 12 | +Then $g$ divides $a$ and $b$. Thus $g$ also divides $a - b$. |
| 13 | + |
| 14 | +Let $g'$ be the GCD of $b$ and $a - b$. |
| 15 | + |
| 16 | +Proof by contradiction that $g' = g$: |
| 17 | + |
| 18 | +Assume $g' < g$ or $g' > g$. |
| 19 | + |
| 20 | +If $g' < g,ドル $g'$ would not be the *greatest* common divisor, |
| 21 | +since $g$ is also a common divisor of $a - b$ and $b$. |
| 22 | + |
| 23 | +If $g' > g,ドル $g'$ divides $b$ and $a - b$ - |
| 24 | +that is, there exist integers $n, m$ |
| 25 | +such that $g'n = b$ and $g'm = a - b$. |
| 26 | +Thus $g'm = a - g'n \iff g'm + g'n = a \iff g'(m + n) = a$. |
| 27 | +This imples that $g' > g$ also divides $a,ドル |
| 28 | +which contradicts the initial assumption that $g$ is the GCD of $a$ and $b$. |
| 29 | + |
| 30 | +## Implementation |
| 31 | + |
| 32 | +To speed matters up in practice, modulo division is used instead of repeated subtractions: |
| 33 | +$b$ can be subtracted from $a$ as long as $a >= b$. |
| 34 | +After these subtractions only the remainder of $a$ when divided by $b$ remains. |
| 35 | + |
| 36 | +A straightforward Lua implementation might look as follows: |
| 37 | + |
| 38 | +```lua |
| 39 | +function gcd(a, b) |
| 40 | + while b ~= 0 do |
| 41 | + a, b = b, a % b |
| 42 | + end |
| 43 | + return a |
| 44 | +end |
| 45 | +``` |
| 46 | + |
| 47 | +note that `%` is the modulo/remainder operator; |
| 48 | +`a` is assigned to the previous value of `b`, |
| 49 | +and `b` is assigned to the previous value of `a` |
| 50 | +modulo the previous value of `b` in each step. |
| 51 | + |
| 52 | +## Analysis |
| 53 | + |
| 54 | +### Space Complexity |
| 55 | + |
| 56 | +The space complexity can trivially be seen to be constant: |
| 57 | +Only two numbers (of assumed constant size), $a$ and $b,ドル need to be stored. |
| 58 | + |
| 59 | +### Time Complexity |
| 60 | + |
| 61 | +Each iteration of the while loop runs in constant time and at least halves $b$. |
| 62 | +Thus $O(log_2(n))$ is an upper bound for the runtime. |
| 63 | + |
| 64 | +## Walkthrough |
| 65 | + |
| 66 | +Finding the GCD of $a = 42$ and $b = 12$: |
| 67 | + |
| 68 | +1. 42ドル \mod 12 = 6$ |
| 69 | +2. 12ドル \mod 6 = 0$ |
| 70 | + |
| 71 | +The result is $gcd(42, 12) = 6$. |
| 72 | + |
| 73 | +Finding the GCD of $a = 633$ and $b = 142$ using the Euclidean algorithm: |
| 74 | + |
| 75 | +1. 633ドル \mod 142 = 65$ |
| 76 | +2. 142ドル \mod 65 = 12$ |
| 77 | +3. 65ドル \mod 12 = 5$ |
| 78 | +4. 12ドル \mod 5 = 2$ |
| 79 | +5. 5ドル \mod 2 = 1$ |
| 80 | +6. 2ドル \mod 1 = 0$ |
| 81 | + |
| 82 | +The result is $gcd(633, 142) = 1$: $a$ and $b$ are co-prime. |
| 83 | + |
| 84 | +## Applications |
| 85 | + |
| 86 | +* Shortening fractions |
| 87 | +* Finding the Least Common Multiple (LCM) |
| 88 | +* Efficiently checking whether two numbers are co-prime (needed e.g. for the RSA cryptosystem) |
| 89 | + |
| 90 | +## Resources |
| 91 | + |
| 92 | +* [Wikipedia Article](https://en.wikipedia.org/wiki/Euclidean_algorithm) |
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