This is known as the 'Hamming Weight', 'popcount' or 'sideways addition'.

Some CPUs have a single built-in instruction to do it and others have parallel instructions which act on bit vectors. Instructions like x86's `popcnt`

(on CPUs where it's supported) will almost certainly be fastest for a single integer. Some other architectures may have a slow instruction implemented with a microcoded loop that tests a bit per cycle (*citation needed* - hardware popcount is normally fast if it exists at all.).

The 'best' algorithm really depends on which CPU you are on and what your usage pattern is.

Your compiler may know how to do something that's good for the specific CPU you're compiling for, e.g. C++20 `std::popcount()`

, or C++ `std::bitset<32>::count()`

, as a portable way to access builtin / intrinsic functions (see another answer on this question). But your compiler's choice of fallback for target CPUs that don't have hardware popcnt might not be optimal for your use-case. Or your language (e.g. C) might not expose any portable function that could use a CPU-specific popcount when there is one.

### Portable algorithms that don't need (or benefit from) any HW support

A pre-populated table lookup method can be very fast if your CPU has a large cache and you are doing lots of these operations in a tight loop. However it can suffer because of the expense of a 'cache miss', where the CPU has to fetch some of the table from main memory. (Look up each byte separately to keep the table small.) If you want popcount for a contiguous range of numbers, only the low byte is changing for groups of 256 numbers, making this very good.

If you know that your bytes will be mostly 0's or mostly 1's then there are efficient algorithms for these scenarios, e.g. clearing the lowest set with a bithack in a loop until it becomes zero.

I believe a very good general purpose algorithm is the following, known as 'parallel' or 'variable-precision SWAR algorithm'. I have expressed this in a C-like pseudo language, you may need to adjust it to work for a particular language (e.g. using uint32_t for C++ and >>> in Java):

GCC10 and clang 10.0 can recognize this pattern / idiom and compile it to a hardware popcnt or equivalent instruction when available, giving you the best of both worlds. (https://godbolt.org/z/qGdh1dvKK)

```
int numberOfSetBits(uint32_t i)
{
// Java: use int, and use >>> instead of >>. Or use Integer.bitCount()
// C or C++: use uint32_t
i = i - ((i >> 1) & 0x55555555); // add pairs of bits
i = (i & 0x33333333) + ((i >> 2) & 0x33333333); // quads
i = (i + (i >> 4)) & 0x0F0F0F0F; // groups of 8
return (i * 0x01010101) >> 24; // horizontal sum of bytes
}
```

For JavaScript: coerce to integer with `|0`

for performance: change the first line to `i = (i|0) - ((i >> 1) & 0x55555555);`

This has the best worst-case behaviour of any of the algorithms discussed, so will efficiently deal with any usage pattern or values you throw at it. (Its performance is not data-dependent on normal CPUs where all integer operations including multiply are constant-time. It doesn't get any faster with "simple" inputs, but it's still pretty decent.)

References:

### How this SWAR bithack works:

```
i = i - ((i >> 1) & 0x55555555);
```

The first step is an optimized version of masking to isolate the odd / even bits, shifting to line them up, and adding. This effectively does 16 separate additions in 2-bit accumulators (SWAR = SIMD Within A Register). Like `(i & 0x55555555) + ((i>>1) & 0x55555555)`

.

The next step takes the odd/even eight of those 16x 2-bit accumulators and adds again, producing 8x 4-bit sums. The `i - ...`

optimization isn't possible this time so it does just mask before / after shifting. Using the same `0x33...`

constant both times instead of `0xccc...`

before shifting is a good thing when compiling for ISAs that need to construct 32-bit constants in registers separately.

The final shift-and-add step of `(i + (i >> 4)) & 0x0F0F0F0F`

widens to 4x 8-bit accumulators. It masks *after* adding instead of before, because the maximum value in any 4-bit accumulator is `4`

, if all 4 bits of the corresponding input bits were set. 4+4 = 8 which still fits in 4 bits, so carry between nibble elements is impossible in `i + (i >> 4)`

.

So far this is just fairly normal SIMD using SWAR techniques with a few clever optimizations. Continuing on with the same pattern for 2 more steps can widen to 2x 16-bit then 1x 32-bit counts. But there is a more efficient way on machines with fast hardware multiply:

Once we have few enough "elements", **a multiply with a magic constant can sum all the elements into the top element**. In this case byte elements. Multiply is done by left-shifting and adding, so **a multiply of **`x * 0x01010101`

results in `x + (x<<8) + (x<<16) + (x<<24)`

. Our 8-bit elements are wide enough (and holding small enough counts) that this doesn't produce carry *into* that top 8 bits.

**A 64-bit version of this** can do 8x 8-bit elements in a 64-bit integer with a 0x0101010101010101 multiplier, and extract the high byte with `>>56`

. So it doesn't take any extra steps, just wider constants. This is what GCC uses for `__builtin_popcountll`

on x86 systems when the hardware `popcnt`

instruction isn't enabled. If you can use builtins or intrinsics for this, do so to give the compiler a chance to do target-specific optimizations.

### With full SIMD for wider vectors (e.g. counting a whole array)

This bitwise-SWAR algorithm could parallelize to be done in multiple vector elements at once, instead of in a single integer register, for a speedup on CPUs with SIMD but no usable popcount instruction. (e.g. x86-64 code that has to run on any CPU, not just Nehalem or later.)

However, the best way to use vector instructions for popcount is usually by using a variable-shuffle to do a table-lookup for 4 bits at a time of each byte in parallel. (The 4 bits index a 16 entry table held in a vector register).

On Intel CPUs, the hardware 64bit popcnt instruction can outperform an SSSE3 `PSHUFB`

bit-parallel implementation by about a factor of 2, but only if your compiler gets it just right. Otherwise SSE can come out significantly ahead. Newer compiler versions are aware of the popcnt false dependency problem on Intel.

```
if (RectA.Left < RectB.Right && RectA.Right > RectB.Left &&
RectA.Top > RectB.Bottom && RectA.Bottom < RectB.Top )
```

or, using Cartesian coordinates

(With X1 being left coord, X2 being right coord, **increasing from left to right** and Y1 being Top coord, and Y2 being Bottom coord, **increasing from bottom to top** -- if this is not how your coordinate system [e.g. most computers have the Y direction reversed], **swap the comparisons below**) ...

```
if (RectA.X1 < RectB.X2 && RectA.X2 > RectB.X1 &&
RectA.Y1 > RectB.Y2 && RectA.Y2 < RectB.Y1)
```

Say you have Rect A, and Rect B.
Proof is by contradiction. Any one of four conditions guarantees that **no overlap can exist**:

- Cond1. If A's left edge is to the right of the B's right edge,
- then A is Totally to right Of B
- Cond2. If A's right edge is to the left of the B's left edge,
- then A is Totally to left Of B
- Cond3. If A's top edge is below B's bottom edge,
- then A is Totally below B
- Cond4. If A's bottom edge is above B's top edge,
- then A is Totally above B

So condition for Non-Overlap is

NON-Overlap => Cond1 Or Cond2 Or Cond3 Or Cond4

Therefore, a sufficient condition for Overlap is the opposite.

Overlap => NOT (Cond1 Or Cond2 Or Cond3 Or Cond4)

De Morgan's law says

`Not (A or B or C or D)`

is the same as `Not A And Not B And Not C And Not D`

so using De Morgan, we have

Not Cond1 And Not Cond2 And Not Cond3 And Not Cond4

This is equivalent to:

- A's Left Edge to left of B's right edge, [
`RectA.Left < RectB.Right`

], and
- A's right edge to right of B's left edge, [
`RectA.Right > RectB.Left`

], and
- A's top above B's bottom, [
`RectA.Top > RectB.Bottom`

], and
- A's bottom below B's Top [
`RectA.Bottom < RectB.Top`

]

**Note 1**: It is fairly obvious this same principle can be extended to any number of dimensions.

**Note 2**: It should also be fairly obvious to count overlaps of just one pixel, change the `<`

and/or the `>`

on that boundary to a `<=`

or a `>=`

.

**Note 3**: This answer, when utilizing Cartesian coordinates (X, Y) is based on standard algebraic Cartesian coordinates (x increases left to right, and Y increases bottom to top). Obviously, where a computer system might mechanize screen coordinates differently, (e.g., increasing Y from top to bottom, or X From right to left), the syntax will need to be adjusted accordingly/

## Best Solution

Why not just compute the convex hull of the points? Depending on the algorithm you use, it takes either

`O(n)`

or`O(n log n)`

time and eliminates all the inner points from consideration. Then, only check these outermost points to find the two that are the farthest away.