[Math] Fibonacci Loop Invariants

algorithmsinductioninvariance

I've taking an Algorithms course. This is non-graded homework. The concept of loop invariants are new to me and it's taking some time to sink in. This was my first attempt at a proof of correctness today for the iterative Fibonacci algorithm. I don't feel good about it. I know this is not the right candidate invariant, because I believe it doesn't say anything about whether or not the algorithm works. It appears something like this is the correct one, though maybe not exactly for my loop. It seems my proof only shows the number increased each iteration, not that it computed any Fibonacci numbers from the sequence at a given index. I need more practice, but are there any general tips about seeing the invariant more easily?

Also aren't invariant suppose to be true before we start the loop, after each iteration, and after the loop completes?

Here was my first attempt:

Logical Implication

If $N$ is a natural number then $F(N)$ will calculate:

$
F(N) =
\begin{cases}
0, & N=0 \\
1, & N=1 \\
F(N-1) + F(N-2), & otherwise
\end{cases}
$

F(n):    
    l, r = 0, 1
    for i in [0,n):
        l, r = l+r, l
    return l

Note, l=left branch in recursive definition, r=right branch:
Fibonacci drawing

Invariant

Let's see, we're given $i,l,r,n \in \mathbb{N}$.
We want to add $r$ to $l$ every $i^\text{th}$ iteration and set $r$ to the original value of $l$.

Consider candidate invariant, $P = l + r \ge l$

Proof by Mathematical Induction

  • Basis step.
    Since we have base cases when $n\le1$, I'll consider both of these.
    When $n=0$, before and after the loop (which we don't enter) $l=0, r=1,$ and $ 0+1 \ge 0$. $P$ is true in this case.
    When $n=1$, before we enter the loop, $l=0, r=1,$ and $ 0+1 \ge 0$. $P$ holds true. After entering and terminating the only iteration, $i=0$; $l=1, r=0,$ and $ 1+0 \ge 1$. $P$ continues to hold true.
  • Inductive Hypothesis. $\forall k\text{ iterations}, 1 \le k \lt i$ where $k \in \mathbb{N}$, and $n\ge2$, suppose $P$ is true.
  • Inductive Step. Consider the $k+1$ iteration. P held true for the iteration $k$ according to the inductive hypothesis. Thus, by the time the $k+1$ iteration terminates it must be the case that P holds true because $l$ gets replaced with $l+r$ and $r$ gets replaced with the original value for $l$, and it follows that $(l+r) + (l) \ge l$.

And my second attempt:

Invariant

Let's see, we're given $i,l,r,n \in \mathbb{N}$.We want to compute $l$ to be the $i^{th}$ Fibonacci number. Consider candidate invariant $P$:

For $\forall i$ iterations,
$
l,r =
\begin{cases}
f_0,1 & ,n=0 |i=0\text{ loop hasn't terminated} \\
f_{i+1}, f_{i} & ,otherwise
\end{cases}
$
from the sequence $\{f_i\}_{i=0}^\infty = \{0, 1, 1, 2, 3, 5, 8, 13,\dots\}$.

Proof by Mathematical Induction

  • Basis step.
    Since we have base cases from the logical implication when $n\le1$, I'll consider both of these.
    When $n=0$ and before entering the loop when $n=1$, $l=f_0$ and $r=1$. $P$ is true and the algorithm returns the correct answer.
    When $n=1$, and after terminating the only loop iteration $i=0$, $l=f_{1}$ and $r=f_{0}$. $P$ continues to hold true, and the algorithm returns the correct answer.
  • Inductive Hypothesis. $\forall k \text{ iterations}, 1 \le k \lt i = n-1$ where $k \in \mathbb{N}$ and $n\ge2$, suppose $P$ is true if $k$ is substituted for $i$ in the invariant.
  • Inductive Step. Consider the $k+1$ iteration. $P$ is true for the iteration $k$ according to the inductive hypothesis. Thus, by the time the $k+1$ iteration terminates $l$ gets replaced with $l+r$, and $r$ gets replaced with the original value for $l$. Thus $P$ remains true because:
    \begin{align}
    l = l+r &= f_{k+1} + f_{k} &&\dots\text{by definition of inductive hypothesis}\\
    & = f_{k+2} &&\dots\text{by definition of Fibonacci sequence} \\
    \text{and} \\
    r = l &= f_{k+1} &&\dots\text{by definition of inductive hypothesis}
    \end{align}

UPDATE My third attempt after reading comments and answers:

F(n):    
    l, r = 0, 1
    for i in [0,n):
        l, r = l+r, l
    return l

For clarity (so I can formulate the invariant clearly):

for i in [0,n) is equivalent to i=0; while (i<n){ stuff; i++;}

Note, l=left branch in recursive definition, r=right branch:
Fibonacci drawing

Invariant

Let's see, we're given $i,l,r,n \in \mathbb{N}$.

We want to compute $l$ to be the $i^{th}$ Fibonacci number.

Consider candidate invariant $P$ (not sure if I need $n=0$ in the first case, is it redundant?):

  • For $\forall i$ iterations,
    $l,r =
    \begin{cases}
    f_{i},1 & ,n=0 |i=0\\
    f_{i}, f_{i-1} & ,otherwise
    \end{cases}$
    from the sequence $\{f_i\}_{i=0}^\infty = \{0, 1, 1, 2, 3, 5, 8, 13,\dots\}$.

Proof by Mathematical Induction

  • Basis step(s).
    Since we have base cases from the logical implication when $n\le1$, I'll consider both of these.
    When $n=0$ we don't enter the loop so $l=f_0$ and $r=1$. $P$ is true and the algorithm returns the correct answer.
    When $n=1$, before the first iteration terminates, $i=0$ so $l=f_0$ and $r=1$. After terminating the only loop iteration, $i=1$, and so $l=f_{1}$ and $r=f_{0}$. $P$ continues to hold true, and the algorithm returns the correct answer.
  • Inductive Hypothesis. $\forall k \text{ iterations}, 2 \le k \lt n$ and $k \in \mathbb{N}$, suppose $P$ is true if $k$ is substituted for $n$ in the invariant.
  • Inductive Step. Consider the $k+1$ iteration. $P$ is true for the end of iteration $k$ according to the inductive hypothesis. Thus, at the start of the $k+1$ iteration, $P$ holds true for the current loop variable $i$. By the time this iteration terminates $l$ gets replaced with $l+r$, and $r$ gets replaced with the original value for $l$, and the loop variable becomes $i+1$. Thus $P$ remains true because:
    \begin{align}
    l = l+r &= f_{i} + f_{i-1} &&\dots\text{by definition of inductive hypothesis}\\
    & = f_{i+1} &&\dots\text{by definition of Fibonacci sequence} \\
    \text{and} \\
    r = l &= f_{i} &&\dots\text{by definition of inductive hypothesis}
    \end{align}
    Thus $r$ is the previous Fibonacci number prior to $l$ as required by $P$, and $l$ is the correct answer for $F(k+1)=f_{i+1}$.

Best Answer

First of all, yes, the loop invariant must be true before we start the loop, after each iteration, and after the loop completes. See https://en.wikipedia.org/wiki/Loop_invariant, http://www.cs.miami.edu/home/burt/learning/Math120.1/Notes/LoopInvar.html, any of several other pages from university course notes that you'll find if you search for "loop invariant", or (presumably) your textbook.

In the most general sense, finding a loop invariant is about constructing a mathematical proof, and there is not a simple set of instructions that will give a good invariant whenever one exists. But if you have an inductive definition of the result of the algorithm, as you do here, your invariant might also look like an inductive definition.

In this case you have that for all $i>1$, $F(i) = F(i-1) + F(i-2)$. You also know the values of $F(0)$ and $F(1)$. This suggests an algorithm in which before the first iteration of a loop, you store the values $F(0)$ and $F(1)$ in two variables, $A$ and $B$, respectively, and you maintain the invariant that $A=F(k)$ and $B=F(k+1)$ where $k$ is the number of iterations of the loop that have completed. You can prove that the invariant is maintained by showing that at the end of each loop, the value in $A$ is the value that previously was in $B$, and the value in $B$ is the previous value of $A+B$. That is, if at the end of $k$ iterations you had $F(k)$ stored in $A$ and $F(k+1)$ stored in $B$, at the end of the next iteration (that is, after $k+1$ iterations) you will have $F(k+1)$ stored in $A$ and $F(k+2) = F(k+1) + F(k)$ stored in $B$.

My $A$ and $B$ look suspiciously like your r and l, respectively. But note that in the algorithm I described, $B$ will store $F(n)$ after only $n-1$ iterations, while your algorithm is written to do $n$ iterations, and to return the larger number at the end of the $n$th iteration. Your algorithm resolves this by doing an extra iteration at the beginning, that is, it starts by setting l to $F(0)$ and r to $F(-1)$, respectively, where $F(-1)=1$ in order that $F(1) = F(0) + F(-1)$. So after $k$ iterations, where $k\geq 0$, your algorithm always has $F(k-1)$ stored in r and $F(k)$ stored in l.

Alternatively, rather than defining $F(-1) = 1$, you can write a special case into your loop invariant for the case when you have completed $0$ iterations. That is what your proposed invariant does, which is fine.

One of the potential difficulties of writing loop invariants is notational. The invariant as I stated it depends on knowing how many iterations of the loop have been completed. You have a variable i whose value during each loop is clearly defined, but whose values at the beginning and end of each loop are not clear to me. At the end of the first loop, does i store the value $0$ or the value $1$? You should have a clearly defined convention for some kind of variable associated with a loop control such as for i in [0,n), with clearly-defined values of this variable before the first iteration of the loop and at the end of each iteration of the loop. Your notes or textbook should have established a convention for how you construct such a variable for such a loop. Without a clearly established convention, I'm not sure whether l at the end of the last iteration is $f_n$ (which it is if $i$ in your invariant has the value $k-1$ at the end of $k$ iterations) or $f_{n+1}$ (which is what you get if $i$ denotes the number of iterations that have been completed).

EDIT: "How I would have done it," not necessarily how you should do it:

If I were writing this for a formal methods course (or possibly even in an ordinary program), I probably would have started by laying out the base cases in code like this (using = for an equality test, not assignment):

if n=0
  return 0
else if n=1
  return 1
else if n>1
  ... do loop ...
  return result

Note that the return value is undefined if $n<0$. For $n>1$, the loop would do $n-1$ iterations of the $A,B$ algorithm I described above, and the result would be the last value of $B$.

I might then notice that the outcomes would be the same if I wrote

if n=0
  return 0
else if n>0
  ... do loop ...
  return result

That is, one of the "if" conditions was superfluous. But I cannot get rid of the special handling of the case $n=0$ with the loop as written, because the loop depends on doing $n-1$ iterations to produce $F(0)$, and you can't do $-1$ iterations of a loop.

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