Hacker News Reader

Comment by arconis987 | Hacker News Reader

This is my go-to implementation, but I make one little tweak. It’s very unlikely that you will overflow a long, but you can you guarantee that you’ll avoid overflow like this:

mid = min + (max - min) / 2;

View on HN

In x86 (intel notation) you can also:

add rax, rbx

rcr rax, 1

It explicitly uses the carry flag from the addition as the top bit of the right-rotated result.

Not super useful for anything other than averaging two uints but that's x86 for you.

What I missed in my first pass was

   * Forgetting exactly which register was the low index and which was the high index
   * Underflow if array size was zero from setting the last index to (size - 1)

If the unsigned right shift is defined (java): "int mid = (min + max) >>> 1;"

edit: this is a bit longer version, if signed conversion/shift is not available: int mid = (min >> 1) + (max >> 1) + (min & max & 1)

That's my C# implementation, so long is 64 bits and int is 32 bits. 64 bits will never overflow in my lifetime. That's basically a Google scale amount of data.

can you explain to me how that fixes it? Thanks

There is an implicit invariant that 0 <= min, and min <= max, so max - min is between 0 and int_max inclusive, so you avoid overflowing with that calculation. To be sure that you avoid overflowing at the sum step, you need to prove that min + (max - min) / 2 <= max.

what if max is already overflown due to size of array (without even doing the lo + high calculation?) Then min + (max- min)/2 would be min + (smaller negative value) which would violate the min + (max-min)/2 <= max

Assuming the value of max passed in is "sane" (non-negative and within array bounds), it is strictly decreasing and should never overflow.

This makes sure that none of the quantities being calculated will overflow while still returning the same answer.

Where x is min and y is max, the midpoint is:

(x+y)/2 = x+(y-x)/2 x+y = 2x+y-x x+y=x+y

Reply on news.ycombinator.com