How can we prove that in binary search

$$\mathit{low} - \mathit{high} ≤ 1$$

Below is a sample algorithm for Binary Search.

while (low<=high) { 
  int mid = low + (high-low)/2; 
  if (data[mid]<value) low = mid+1; 
  else if (data[mid]>value) high = mid-1; 
  else return &data[mid]; 
return nullptr; 

A good stranger told me that

typically you would want to avoid unprotected mid-1 and mid+1 for fear of over- and underflow and have a special case for when the difference between high and low is less than 2.

What I still want to know is a formal proof that $\mathit{low} - \mathit{high} ≤ 1$ provided that you started off with $\mathit{low} ≤ \mathit{high}$.

My idea so far has been to adapt the explanation on loop invariant here but I can assure you I haven't made any progress, yet.


Let's denote the indices by $l,h,m$. If $l \leq h$ and $h-l$ is even then $m = \frac{l+h}{2}$ and so in the following iteration, the new values $l',h'$ will either be $\frac{l+h}{2}+1,h$ or $l,\frac{l+h}{2}-1$. In the first case $$ l'-h' = \frac{l-h}{2}+1 \leq 1, $$ and in the second case $$ l'-h' = \frac{l-h}{2}+1 \leq 1. $$

If $l \leq h$ and $h-l$ is odd then $m=\frac{l+h-1}{2}$ and so either $l',h' = \frac{l+h+1}{2},h$ or $l',h' = l,\frac{l+h-3}{2}$. In the first case $$ l' - h' = \frac{l-h+1}{2} \leq \frac{1}{2} $$ and so $l'-h' \leq 0$, since $l'-h'$ is an integer. In the second case $$ l' - h' = \frac{l-h+3}{2} \leq \frac{3}{2}, $$ and so $l'-h' \leq 1$, since $l'-h'$ is an integer.

Put together, we have shown:

If $l \leq h$ then $l'-h' \leq 1$.

Now let's use it to show that $l-h \leq 1$ always holds.

We first prove by induction on $t$ that if the loop executes at least $t$ times, then after $t$ iterations of the loop, $l-h \leq 1$. The base case $t = 0$ is by assumption. Now suppose that after $t$ iterations, we have $l-h \leq 1$. If $l > h$ then the loop terminates, and so there is nothing to prove. Otherwise, $l \leq h$, and so $l-h \leq 1$ at the end of the iteration, as shown above.

This shows that when the loop terminates (if it terminates — which has to be shown separately), $l-h \leq 1$ necessarily holds. In fact, we can say more: since the loop terminated, we must have $l > h$, and so $l-h = 1$.

  • $\begingroup$ Thanks. Very clear and concise $\endgroup$
    – Gilboot
    Apr 25 at 18:02
  • $\begingroup$ "In fact, we can say more: since the loop terminated, we must have $l\gt h$, and so $l-h=1$". Did you ignore the clause "else return &data[mid]; "? $\endgroup$
    – John L.
    Apr 26 at 6:26
  • $\begingroup$ In that case, the entire function terminates, and control never reaches beyond the loop. $\endgroup$ Apr 26 at 6:28

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