I noticed that in many books calculation of midpoint for binary search uses this:
int mid = left + (right - left) / 2;
Why not use
int mid = (left + right) / 2;
instead?
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3 Answers
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Because left + right may overflow. Which then means you get a result that is less than left. Or far into the negative if you are using signed integers.
So instead they take the distance between left and right and add half of that to left. This is only a single extra operation to make the algorithm more robust.
answered Aug 25, 2017 at 13:46
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Suppose your 'low' and 'high' are 16 bit unsigned integers. That means, they can only have a maximum value of 2^16=65536. Consider this, low = 65530 high = 65531
If we added them first, (low+high) would end up being junk since that big a number (131061) cannot be stored in a your 16-bit integer. And so, mid would be a wrong value.
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2$\begingroup$ Yeah david,, I found out a lot people saying overflow overflow on the different forums but couldn't understand what they really mean to say.. After i found out the example,, my confusion was cleared.. $\endgroup$ Commented Sep 24, 2018 at 18:06
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This answer gives a practical example of why the l + (r-l)/2 calculation is necessary.
In case you are curious how the two are equivalent mathematically, here is the proof. The key is adding 0 then splitting that into l/2 - l/2.
(l+r)/2 =
l/2 + r/2 =
l/2 + r/2 + 0 =
l/2 + r/2 + (l/2 - l/2) =
(l/2 + l/2) + (r/2 - l/2) =
l + (r-l)/2
right. $\endgroup$left=right, then(2 * right) / 2=right. $\endgroup$left + right >= right, intermediate values I mean. Kind of take actions against overflow. $\endgroup$