Add 'GCD of multiple numbers' section in euclid-algorithm.md (#1355)

mdeapedro · adamant-pwn · web-flow · commit aeeac9217aec · 2024-10-15T12:56:28.000+02:00
* Update euclid-algorithm.md

Add 'GCD of multiple numbers' section.

* Update euclid-algorithm.md

* Update euclid-algorithm.md

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Co-authored-by: Oleksandr Kulkov &lt;adamant.pwn@gmail.com&gt;
diff --git a/src/algebra/euclid-algorithm.md b/src/algebra/euclid-algorithm.md
@@ -15,7 +15,7 @@ $$\gcd(a, b) = \max \{k > 0 : (k \mid a) \text{ and } (k \mid b) \}$$
 
 When one of the numbers is zero, while the other is non-zero, their greatest common divisor, by definition, is the second number. When both numbers are zero, their greatest common divisor is undefined (it can be any arbitrarily large number), but it is convenient to define it as zero as well to preserve the associativity of $\gcd$. Which gives us a simple rule: if one of the numbers is zero, the greatest common divisor is the other number.
 
-The Euclidean algorithm, discussed below, allows to find the greatest common divisor of two numbers $a$ and $b$ in $O(\log \min(a, b))$.
+The Euclidean algorithm, discussed below, allows to find the greatest common divisor of two numbers $a$ and $b$ in $O(\log \min(a, b))$. Since the function is **associative**, to find the GCD of **more than two numbers**, we can do $\gcd(a, b, c) = \gcd(a, \gcd(b, c))$ and so forth.
 
 The algorithm was first described in Euclid's "Elements" (circa 300 BC), but it is possible that the algorithm has even earlier origins.
 
@@ -70,7 +70,7 @@ Moreover, it is possible to show that the upper bound of this theorem is optimal
 
 Given that Fibonacci numbers grow exponentially, we get that the Euclidean algorithm works in $O(\log \min(a, b))$.
 
-Another way to estimate the complexity is to notice that $a \bmod b$ for the case $a \geq b$ is at least $2$ times smaller than $a$, so the larger number is reduced at least in half on each iteration of the algorithm.
+Another way to estimate the complexity is to notice that $a \bmod b$ for the case $a \geq b$ is at least $2$ times smaller than $a$, so the larger number is reduced at least in half on each iteration of the algorithm. Applying this reasoning to the case when we compute the GCD of the set of numbers $a_1,\dots,a_n \leq C$, this also allows us to estimate the total runtime as $O(n + \log C)$, rather than $O(n \log C)$, since every non-trivial iteration of the algorithm reduces the current GCD candidate by at least a factor of $2$.
 
 ## Least common multiple
 
