Complexity of gcd

#1

how the complexity of gcd is logn ?please explain and also n here represents which of the two numbers the graeter one or the smaller one?

#2

// Function to return

// gcd of a and b

int gcd( int a, int b)

{

if (a == 0)

return b;

return gcd(b % a, a);

} Time Complexity: O(Log min(a, b))

#3

but how is it log n please also tell this?

#4

At every step, there are two cases

  • b >= a / 2, then a, b = b, a % b will make b at most half of its previous value
  • b < a / 2, then a, b = b, a % b will make a at most half of its previous value, since b is less than a / 2

So at every step, the algorithm will reduce at least one number to at least half less. In at most O(log2a)+O(log2b)O(log2a)+O(log2b) step, this will be reduced to the simple cases. Which yield an O(log2n)O(log2n) algorithm, where n is the upper limit of a and b.

#5

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