One of the questions in the problem sets that I'm struggling in is this specific number that asks me to prove an iterative Fibonacci algorithm. The algorithm is written below:

function fib(n)
    if n = 0 then
        a = 0; b = 1; i = 2;
        while i <= n do
            c = a + b
            a = b
            b = c
            i = i + 1

The way too prove correctness, according to my professor was to make sure that there are these three steps:

  1. Initialization - the loop invariant must hold true prior to the first iteration
  2. Maintenance - the loop invariant must hold true after an iteration
  3. Termination - the loop invariant must hold true when the loop terminates

The loop invariant I've chosen is a <= b since I find this to be true for steps 1 through 3. First of all, I'm not sure if this is a valid loop invariant and this is the only observation I saw since i <= n isn't always true for inputs n that are natural numbers.

Assuming that I've chosen the correct loop invariant, I need to answer the proof by doing three steps so for this number I plan to answer it this way

  1. Initialization - before the start of the loop a is assigned a value of 0 while b is assigned a value of 1 which starts the Fibonacci sequence. a <= b holds true prior to the start of the loop
  2. Maintenance - during the loop, another variable c is added such that it is equal to the sum of a and b. After which, b is assigned to variable a while c is assigned to b thus making the invariant a <= b true during the iteration.
  3. Termination - the loop ends when i > n. Before i is incremented, the procedures in the maintenance step is still done, thus the loop invariant still holds true

For my questions, is my loop invariant a <= b correct? And are the three statements I mentioned above sufficient to prove the correctness of an iterative Fibonacci algorithm?

  • $\begingroup$ Although I'm not a computer scientist, I stongly suspect that your loop invariant must be related to Fibonacci numbers. Something along the lines of b being the (i-1)th Fobinacci number. Couldn't you else "prove" that your code computes anything correctly? $\endgroup$
    – tobwin
    Feb 15, 2018 at 18:31
  • $\begingroup$ This is the part where I am a bit confused. Because, from what I get from the proof definition, to prove that this is correct, the loop invariant must hold true for the three cases. I am really not sure if it has something to do with correct computation. $\endgroup$
    – Jessie
    Feb 15, 2018 at 18:38
  • $\begingroup$ What exactly is your question? If it is "Is a<=b a loop invariant in my algorithm?" then the answer is yes. If your question is "Does this loop invariant help me prove the algorithm correct?" then the answer is no. $\endgroup$
    – tobwin
    Feb 15, 2018 at 18:52

1 Answer 1


The condition $a \leq b$ is a loop invariant, but it captures very little of the operation of the loop. A better loop invariant is $$ a = F_{i-2} \text{ and } b = F_{i-1}. $$ This clearly holds just before the loop. The loop sets $a' = b = F_{i-1}$, $b' = a+b = F_{i-2}+F_{i-1}=F_i$, $i' = i+1$, from which it is easy to prove that it maintains the loop invariant. When the loop has finished, we must have $i = n+1$, and so $a = F_{i-2} = F_{n-1}$ and $b = F_{i-1} = F_n$.

There is a weak point in this argument: How do we know that $i = n+1$ at the end of the loop? We can fix this by strengthening the loop invariant to $$ a = F_{i-2} \text{ and } b = F_{i-1} \text{ and } i \leq n+1. $$ This holds just before the loop since $n \geq 1$ (assuming $n \geq 0$, a precondition that needs to be added to the entire function). It is maintained during the loop since at the beginning of each iteration $i \leq n$. Finally, after the loop we have $i \leq n+1$ and $i > n$, which forces $i = n+1$ since $i$ is an integer.

  • $\begingroup$ Now I see the part where the computation of the Fibonacci takes place. I was thinking of something like this but I just couldn't properly sew it in. I guess the part of determining the loop invariant is the most important part for proving iterative algorithms $\endgroup$
    – Jessie
    Feb 15, 2018 at 19:44

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