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Difference between revisions of "Euclidean algorithm"

 
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The Euclidean algorithm allows to find the [[greatest common divisor]] of any two non-negative integers. The main idea is very simple: if we have two non-negative integers <math>a,b</math> with <math>a\ge b</math> and <math>b=0</math>, then the greatest common divisor is <math>\displaystyle{a}</math>. If <math>a\ge b>0</math>, then the set of common divisors of <math>\displaystyle{a}</math> and <math>b</math> is the same as the set of common divisors of <math>b</math> and <math>r</math> where <math>r</math> is the [[remainder]] of division of <math>\displaystyle{a}</math> by <math>b</math>. Indeed, we have <math>a=mb+r</math> with some integer<math>m</math>, so, if <math>\displaystyle{d}</math> divides both <math>\displaystyle{a}</math> and <math>b</math>, it must divide both <math>\displaystyle{a}</math> and <math>mb</math> and, thereby, their difference <math>r</math>. Similarly, if <math>\displaystyle{d}</math> divides both <math>b</math> and <math>r</math>, it should divide <math>\displaystyle{a}</math> as well. Thus, the greatest common divisors of <math>\displaystyle{a}</math> and <math>b</math> and of <math>b</math> and <math>r</math> coincide: <math>GCD(a,b)=GCD(b,r)</math>. But the pair <math>(b,r)</math> consists of smaller numbers than the pair <math>(a,b)</math>! So, we reduced our task to a simpler one. And we can do this reduction again and again until the smaller number becomes <math>0</math>. To see how it works, just take an example <math>\displaystyle a=112</math>, <math>b=42</math>. We have <math>112=2\cdot 42+ 28</math>, so the pair <math>(112,42)</math> can be replaced by <math>(42,28)</math>. Similarly, <math>42=1\cdot 28+14</math>, so we get the pair <math>(28,14)</math>. Now, <math>28=2\cdot 14+0</math>, so we get the pair <math>(14,0)</math>, which has the greatest common divisor <math>14</math>. Thus <math>GCD(112,42)=14</math>.
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The '''Euclidean algorithm''' (also known as the '''Euclidean division algorithm''' or '''Euclid's algorithm''') is an algorithm that finds the [[greatest common divisor]] (GCD) of two elements of a [[Euclidean domain]], the most common of which is the [[nonnegative]] [[integer]]s <math>\mathbb{Z}{\geq 0}</math>, without [[factoring]] them.
 +
 
 +
==Main idea and Informal Description==
 +
The basic idea is to repeatedly use the fact that <math>\gcd({a,b}) \equiv \gcd({b,a - b})</math>
 +
 
 +
If we have two non-negative integers <math>a,b</math> with <math>a|b</math> and <math>b\ne0</math>, then the greatest common divisor is <math>{a}</math>. If <math>a\ge b>0</math>, then the set of common divisors of <math>{a}</math> and <math>b</math> is the same as the set of common divisors of <math>b</math> and <math>r</math> where <math>r</math> is the [[remainder]] of division of <math>{a}</math> by <math>b</math>. Indeed, we have <math>a=mb+r</math> with some integer <math>m</math>, so, if <math>{d}</math> divides both <math>{a}</math> and <math>b</math>, it must divide both <math>{a}</math> and <math>mb</math> and, thereby, their difference <math>r</math>. Similarly, if <math>{d}</math> divides both <math>b</math> and <math>r</math>, it should divide <math>{a}</math> as well. Thus, the greatest common divisors of <math>{a}</math> and <math>b</math> and of <math>b</math> and <math>r</math> coincide: <math>GCD(a,b)=GCD(b,r)</math>. But the pair <math>(b,r)</math> consists of smaller numbers than the pair <math>(a,b)</math>! So, we reduced our task to a simpler one. And we can do this reduction again and again until the smaller number becomes <math>0</math>.
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 +
== General Form ==
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Start with any two elements <math>a</math> and <math>b</math> of a [[Euclidean Domain]]
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 +
* If <math>b=0</math>, then <math>\gcd(a,b)=a</math>.
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* Otherwise take the remainder when <math>{a}</math> is divided by <math>a \pmod{b}</math>, and find <math>\gcd(a,a \bmod {b})</math>.
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* Repeat this until the remainder is 0.<br>
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 +
<math>a \pmod{b} \equiv r_1</math><br>
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<math>b \pmod {r_1} \equiv r_2</math><br>
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<math> \vdots</math> <br>
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<math>r_{n-1} \pmod {r_n} \equiv 0</math><br>
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Then <math>\gcd({a,b}) = r_n</math><br>
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 +
~The congruence sign above should be replaced by the normal equal sign. It's important to note that <math>a \pmod{b} = r</math><br>is the same as <math>a \equiv {r} \pmod{b}</math>.  
 +
 
 
Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:
 
Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:
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<br>
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<math>112=2\cdot 42+ 28</math>
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for <math>r_{k+1} < r_k < r_{k-1}</math><br>
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<br>
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<math>a = b \cdot q_1+r_1</math><br>
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<math>42=1\cdot 28+14</math>
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<math>b = r_1 \cdot q_2 + r_2</math><br>
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<br>
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<math>r_1 = r_2 \cdot q_3 + r_3</math><br>
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<math>28=2\cdot 14+0</math>
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<math>\vdots</math><br>
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<br>
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<math>r_{n-1} = r_n \cdot q_{n+1} +0</math><br>
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The Euclidean algorithm has one nice bonus: the so called linear representation of the greatest common divisor. To see how it works, let's use our equations backwards starting with the last but one:
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and so <math>\gcd({a,b}) = r_n</math><br>
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<br>
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<math>14=42-1\cdot 28=42-1\cdot(112-2\cdot 42)=3\cdot 42-1\cdot 112</math>
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==  Example ==
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<br>
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To see how it works, just take an example. Say <math>a = 93, b=42</math>. <br/>
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i.e., <math>14</math> is expressed as a sum of <math>112</math> and <math>42</math> with integer coefficients.
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We have <math>93 \equiv 9 \pmod{42}</math>, so <math>\gcd(93,42) = \gcd(42,9)</math>. <br/>  
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<br>
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Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/>
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How fast is the Euclidean algorithm? One can prove that if the larger of two numbers does not exceed the <math>n</math>-th [[Fibonacci number]], then the required number of steps does not exceed <math>n</math>, so the number of steps is logarithmic in the size of the initial numbers (i.e., the number of steps is not much more than the number of digits in the largest of the two numbers).
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Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. <br/>
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Then <math>6 \equiv 0 \pmod{3}</math>, so <math>\gcd(6,3) = \gcd(3,0) = 3</math>. <br/>
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Thus <math>\gcd(93,42) = 3</math>.
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* <math>93 = 2 \cdot 42 + 9 \qquad (1)</math>
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* <math>42 = 4 \cdot 9 + 6 \qquad (2)</math>
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* <math>9 = 1 \cdot 6 + 3 \qquad (3)</math>
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* <math>6 = 2 \cdot 3 + 0 \qquad (4)</math>
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 +
== Extended Euclidean Algorithm ==
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An added bonus of the Euclidean algorithm is the "linear representation" of the greatest common divisor. This allows us to write <math>\gcd(a,b)=ax+by</math>, where <math>x,y</math> are some elements from the same [[Euclidean Domain]] as <math>a</math> and <math>b</math> that can be determined using the algorithm. We can work backwards from whichever step is the most convenient.
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Continuing the previous example, our goal is to find <math>a</math> and <math>b</math> such that <math>93a + 42b = \gcd(93,42) = 3.</math>  We can work backwards from equation <math>(3)</math> since <math>3</math> appears there:
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<math>3 = 9 - 1 \cdot 6</math>
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We currently have <math>3</math> as a linear combination of <math>6</math> and <math>9</math>.  Our goal is to replace <math>6</math> and <math>9</math> so that we have a linear combination of <math>42</math> and <math>93</math> only.  We start by rearranging <math>(2)</math> to <math>6 = 42 - 4 \cdot 9</math> so we can substitute <math>6</math> to express <math>3</math> as a linear combination of <math>9</math> and <math>42</math>:
 +
 
 +
<math>3 = 9 - 1 \cdot (42 - 4 \cdot 9)</math> <br>
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<math>3 = -1 \cdot 42 + 5 \cdot 9.</math> <br>
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 +
Continuing, we rearrange <math>(1)</math> to substitute <math>9 = 93 - 2 \cdot 42</math>:
 +
 
 +
<math>3 = -1 \cdot 42 + 5 \cdot (93 - 2 \cdot 42)</math> <br>
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<math>3 = -11 \cdot 42 + 5 \cdot 93. \qquad (5)</math> <br>
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 +
We have found one linear combination. To find others, since <math>42 \cdot 93 - 93 \cdot 42 = 0</math>, dividing both sides by <math>\gcd(93,42) = 3</math> gives <math>14 \cdot 93 - 31 \cdot 42 = 0</math>. We can add <math>k</math> times this equation to <math>(5)</math>, so we can write <math>3</math> as a linear combination of <math>93</math> and <math>42</math>
 +
 
 +
<math>3 = (14k + 5) \cdot 93 + (-31k - 11) \cdot 42</math>
 +
 
 +
for any integer <math>k</math>.
 +
 
 +
 
 +
===Introductory===
 +
https://artofproblemsolving.com/community/c1677139h2442945p20256095
 +
 
 +
===Intermediate===
 +
* [[2020 AMC 10A Problems/Problem 24]]
 +
* [[1985 AIME Problems/Problem 13]]
 +
* [[1959 IMO Problems/Problem 1]] (Note: this problem is widely regarded as the easiest problem ever asked in the IMO)
 +
* [[2021 AIME I Problems/Problem 10]]
 +
 
 +
===Olympiad===
 +
 
 +
==See Also==
 +
*[[Divisibility]]
 +
 
 +
[[Category:Algorithms]]
 +
[[Category:Number theory]]

Latest revision as of 16:39, 30 September 2024

The Euclidean algorithm (also known as the Euclidean division algorithm or Euclid's algorithm) is an algorithm that finds the greatest common divisor (GCD) of two elements of a Euclidean domain, the most common of which is the nonnegative integers $\mathbb{Z}{\geq 0}$, without factoring them.

Main idea and Informal Description

The basic idea is to repeatedly use the fact that $\gcd({a,b}) \equiv \gcd({b,a - b})$

If we have two non-negative integers $a,b$ with $a|b$ and $b\ne0$, then the greatest common divisor is ${a}$. If $a\ge b>0$, then the set of common divisors of ${a}$ and $b$ is the same as the set of common divisors of $b$ and $r$ where $r$ is the remainder of division of ${a}$ by $b$. Indeed, we have $a=mb+r$ with some integer $m$, so, if ${d}$ divides both ${a}$ and $b$, it must divide both ${a}$ and $mb$ and, thereby, their difference $r$. Similarly, if ${d}$ divides both $b$ and $r$, it should divide ${a}$ as well. Thus, the greatest common divisors of ${a}$ and $b$ and of $b$ and $r$ coincide: $GCD(a,b)=GCD(b,r)$. But the pair $(b,r)$ consists of smaller numbers than the pair $(a,b)$! So, we reduced our task to a simpler one. And we can do this reduction again and again until the smaller number becomes $0$.

General Form

Start with any two elements $a$ and $b$ of a Euclidean Domain

  • If $b=0$, then $\gcd(a,b)=a$.
  • Otherwise take the remainder when ${a}$ is divided by $a \pmod{b}$, and find $\gcd(a,a \bmod {b})$.
  • Repeat this until the remainder is 0.

$a \pmod{b} \equiv r_1$
$b \pmod {r_1} \equiv r_2$
$\vdots$
$r_{n-1} \pmod {r_n} \equiv 0$
Then $\gcd({a,b}) = r_n$

~The congruence sign above should be replaced by the normal equal sign. It's important to note that $a \pmod{b} = r$
is the same as $a \equiv {r} \pmod{b}$.

Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:

for $r_{k+1} < r_k < r_{k-1}$
$a = b \cdot q_1+r_1$
$b = r_1 \cdot q_2 + r_2$
$r_1 = r_2 \cdot q_3 + r_3$
$\vdots$
$r_{n-1} = r_n \cdot q_{n+1} +0$
and so $\gcd({a,b}) = r_n$

Example

To see how it works, just take an example. Say $a = 93, b=42$.
We have $93 \equiv 9 \pmod{42}$, so $\gcd(93,42) = \gcd(42,9)$.
Similarly, $42 \equiv 6 \pmod{9}$, so $\gcd(42,9) = \gcd(9,6)$.
Continuing, $9 \equiv 3 \pmod{6}$, so $\gcd(9,6) = \gcd(6,3)$.
Then $6 \equiv 0 \pmod{3}$, so $\gcd(6,3) = \gcd(3,0) = 3$.
Thus $\gcd(93,42) = 3$.

  • $93 = 2 \cdot 42 + 9 \qquad (1)$
  • $42 = 4 \cdot 9 + 6 \qquad (2)$
  • $9 = 1 \cdot 6 + 3 \qquad (3)$
  • $6 = 2 \cdot 3 + 0 \qquad (4)$

Extended Euclidean Algorithm

An added bonus of the Euclidean algorithm is the "linear representation" of the greatest common divisor. This allows us to write $\gcd(a,b)=ax+by$, where $x,y$ are some elements from the same Euclidean Domain as $a$ and $b$ that can be determined using the algorithm. We can work backwards from whichever step is the most convenient.

Continuing the previous example, our goal is to find $a$ and $b$ such that $93a + 42b = \gcd(93,42) = 3.$ We can work backwards from equation $(3)$ since $3$ appears there:

$3 = 9 - 1 \cdot 6$

We currently have $3$ as a linear combination of $6$ and $9$. Our goal is to replace $6$ and $9$ so that we have a linear combination of $42$ and $93$ only. We start by rearranging $(2)$ to $6 = 42 - 4 \cdot 9$ so we can substitute $6$ to express $3$ as a linear combination of $9$ and $42$:

$3 = 9 - 1 \cdot (42 - 4 \cdot 9)$
$3 = -1 \cdot 42 + 5 \cdot 9.$

Continuing, we rearrange $(1)$ to substitute $9 = 93 - 2 \cdot 42$:

$3 = -1 \cdot 42 + 5 \cdot (93 - 2 \cdot 42)$
$3 = -11 \cdot 42 + 5 \cdot 93. \qquad (5)$

We have found one linear combination. To find others, since $42 \cdot 93 - 93 \cdot 42 = 0$, dividing both sides by $\gcd(93,42) = 3$ gives $14 \cdot 93 - 31 \cdot 42 = 0$. We can add $k$ times this equation to $(5)$, so we can write $3$ as a linear combination of $93$ and $42$

$3 = (14k + 5) \cdot 93 + (-31k - 11) \cdot 42$

for any integer $k$.


Introductory

https://artofproblemsolving.com/community/c1677139h2442945p20256095

Intermediate

Olympiad

See Also

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