Euclid’s Algorithm for GCD

Objectives

• Use Euclid’s algorithm to calculate the GCD of two numbers.
• Use the extended Euclid’s algorithm to solve Linear Diophantine equations.

Euclid’s Algorithm

• Reminder: the GCD of two numbers a and b is the largest factor g that divides both a and b.
• Notation: g|a means “g divides a.”
• To take the GCD of two numbers a and b you could just factor the numbers, but that is expensive.
• Euclid’s algorithm is faster!

Calculating GCD with simple subtraction.

• Let a be the largest number, let b be the smaller one. We want g=gcd(a,b)
• Mathematical fact: If g|a and g|b then also g|(a+b) and g|(a−b)
  • So, we can recursively define gcd(a,b)=gcd(b,a−b) (when a>b>0).
  • Also gcd(a,0)=a.

• How long would this take?

  int gcd(int a, int b) {
    if (b==0) return a;
    if (a<b) return gcd(b,a);
    return gcd(b,a-b);
  }
  

Example 1

• 28 = 2×2×7, 20 = 2×2×5
• gcd(28,20)=gcd(20,8)=gcd(12,8)=gcd(8,4)=gcd(4,0)=4

Euclid’s Algorithm

• Subtraction is slow. Let’s try subtracting multiples.
  • gcd(a,b)=gcd(b,a−nb) where n is an integer and a−nb≥0.
• For such n, notice r=a−nb=mod(a,b).
• We know g|a and g|nb.
• Therefore, we also know g|r.
• Therefore, gcd(a,b)=gcd(b,mod(a,b)).

int gcd(int a, int b) {
  if (b==0) return a;
  if (a<b) return gcd(b,a);
  return gcd(b,a % b);
}

Diophantine Equations

• A Diophantine Equation is a polynomial equation where we are only interested in integer solutions.
• Linear Diophantine equation: ax+by=g,
  • Let g=gcd(a,b) for this discussion.
• Running example: Suppose you go to the store. You buy x apples at 72 cents each and y oranges at 33 cents each. You spend $5.85. How many of each did you buy?

Calculating x and y.

• Given:
  • ax+by=g
  • g=gcd(a,b).
• We can “Euclid Algorithm” the equation: bx1+(a mod b)y1=g

• Get rid of modulus: a mod b=a−⌊ab⌋×b

  bx1+(a−⌊ab⌋y1)=g

• Rearrange to get coefficients of a and b:

  ay1+b(x1−⌊ab⌋y1)=g

• Therefore x=y1 and y=x1−⌊ab⌋y1

Source code

void extendedEuclid(int a, int b, int &x, int &y, int &d) {
   if (b == 0) {
     x = 1; y = 0; d = a; return;
   }
   extendedEuclid(b, a % b);

   int x1 = y;
   int y1 = x - (a / b) * y;

   x = x1;
   y = y1;
}

Example

Suppose you go to the store. You buy x apples at 72 cents each and y oranges at 33 cents each. You spend $5.85. How many of each did you buy?

Result

Suppose you go to the store. You buy x apples at 72 cents each and y oranges at 33 cents each. You spend $5.85. How many of each did you buy?

• 72×−5+33×11=3

• Multiply both sides by 195
  • 72×−975+33×2145=3
• We can add 72(333)n to the 72 term and subtract 33(723)n to the 33 term.
• Solve for n:
  • −975+11n>0
• n>88.6, so take n=89.
• Final equation:
  • 72×4+33×9=585