gcd for fibonacci numbers number of steps proof

Question

I have to proof $gcd(F_{n+1},F_{n+2})$ need at least n steps to get gcd(we have to use version with subtraction).

My try

After first step we get $gcd(F_{n+1} - F_{n+2},F_{n+2})=gcd(F_{n+1},F_{n})$

After second step we get $gcd(F_{n+1}-F_{n},F_{n})=gcd(F_{n-1},F_{n})$

After n-th step we get $gcd(F_{3}-F_{2},F_{2})=gcd(F_{1},F_{2})$

So after n+1-th step we get $gcd(F_{1},F_{2}-F_{1})=gcd(F_{1},F_{0})$

That show us we need n+1 steps to get gcd of 2 fibonnaci numbers.

But it isn't formaly (I'm not sure that's proof) could someone help me how that should looks correctly

As for a proof that this is the minimum number of "steps", you'd have to define a step first. If it means either swapping the number or subtracting the smaller from the larger until you cannot anymore, then it should also be easy to prove with induction you need about n subtractions and n swaps. — TMM, Jun 06 '17 at 18:33
You can find your answer in the accepted answer of this post — Amin235, Jun 06 '17 at 18:34

score 2 · Accepted Answer · answered Jun 06 '17 at 18:41

First, you have a small mistake:

After first step we get $gcd(F_{n+1} - F_{n+2},F_{n+2})=gcd(F_{n+1},F_{n})$

This should be:

After first step we get $gcd(F_{n+1},F_{n+2}-F_{n+1})=gcd(F_{n+1},F_{n})$

since $F_{n+2} > F_{n+1}$ (for $n > 0$)

Second, here is what a more formal (inductive) proof would look like:

Claim: For any $n$, it takes at least $n$ steps to compute $gcd(F_{n+1},F_{n+2})$

Proof by (weak) induction:

Base: $n=0$: $gcd(F_1,F_2)= gcd(1,1)=1$. OK, so that took $0$ or $1$ steps, depending on how you count. If we count the number of subtractions, it is $0$. So yes, it takes (at least) $0$ steps to compute $gcd(F_1,F_2)$.

Step: Suppose that for some arbitrary $k$ we have that it takes at least $k$ steps to compute $gcd(F_{k+1},F_{k+2})$ . Now consider $k+1$, i.e. let's see how many steps it takes to compute $gcd(F_{(k+1)+1},F_{(k+1)+2})$. Well:

$$gcd(F_{(k+1)+1},F_{(k+1)+2}) =$$

$$gcd(F_{(k+1)+1},F_{(k+1)+2}-F_{(k+1)+1})=$$

$$gcd(F_{k+2},F_{k+1}) =$$

$$gcd(F_{k+1},F_{k+2})$$

Since by our inductive assumption this last one takes at least $k$ steps to compute, it will take at least $k+1$ steps to compute $gcd(F_{(k+1)+1},F_{(k+1)+2})$, because we added at least one more subtraction operation. Since this is what we needed to show, we have now completed the inductive step, and thereby the inductive proof.

score 0 · Answer 2 · answered Jun 06 '17 at 18:34

0

It looks acceptable to me, but to make it more formal you could start from the bottom (use induction). That means first show it for $n=0$ and then prove the claim for $n=k$ implies the claim for $n=k+1$, $k$ some integer.

answered Jun 06 '17 at 18:34

Akababa

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gcd for fibonacci numbers number of steps proof

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