EUCLID (a, b)
1. if b=0
2. then return (a)
3. else return(EUCLID(b, a mod b))

Lemma: If a > b ≥ 1 and the invocation EUCLID(a, b) performs k ≥ 1 calls
then a ≥ F_k+2 and b ≥ F_k+1

Proof: (By induction)

Basis: Let k=1, we know a > b ≥ 1
⇒ b ≥ F₂ = 1 (here k+1=2)

Since a > b ⇒ a ≥ 2 ⇒ a ≥ F₃ (here k+2=3)

If a > b initially then this property a > b is maintained at each recursive invocation in EUCLID ( a , b ) algorithm, since b > a mod b always.

NOTE: Since a mod b < b ⇒ The invariant 1^st argument > 2^nd argument of EUCLID's algorithm is maintained during each iteration.

Inductive Hypothesis: Assume the result holds for # of invocations ≤ k -1

Inductive proof: Let EUCLID ( a , b ) makes k invocations

                        ⇒ EUCLID ( b , a mod b ) makes ( k -1) invocation

From our inductive hypothesis:

b ≥ F _{( k -1)+2} , a mod b ≥ F _{( k -1)+1}

Therefore b ≥ F _{k +1} , a mod b ≥ F _k

We know, a = * b + a mod b (where = Floor ( a ))

≥ 1 ⇒ a ≥ b + a mod b .

Since a mod b ≥ F _k we have a ≥ F _{k +1} + F_k ⇒ a ≥ F _{k +2} .

Lame’s Theorem: For any integer k ≥ 1 if a > b ≥ 1 and if b < F_{k +1} then EUCLID (a, b) makes fewer than k recursive calls gcd ( F_{k +1}, F_k ) = gcd (F_k, F_{k -1}) = … = gcd (1,0) = 1

Therefore # of recursive invocation = k-1

This shows that the bound k-1 is tight.

F_k / F_{k -1} Φ[Golden Ratio Φ =]

To represent F_k, # of bits required = k

Therefore for two β bit numbers running time complexity of EUCLID is O (β)