I saw this Java code that does a perfect 50% blend between two RGB888 colors extremely efficiently:
public static int blendRGB(int a, int b) {
return (a + b - ((a ^ b) & 0x00010101)) >> 1;
}
That's apparently equivalent to extracting and averaging the channels individually. Something like this:
public static int blendRGB_(int a, int b) {
int aR = a >> 16;
int bR = b >> 16;
int aG = (a >> 8) & 0xFF;
int bG = (b >> 8) & 0xFF;
int aB = a & 0xFF;
int bB = b & 0xFF;
int cR = (aR + bR) >> 1;
int cG = (aG + bG) >> 1;
int cB = (aB + bB) >> 1;
return (cR << 16) | (cG << 8) | cB;
}
But the first way is much more efficient. My questions are: How does this magic work? What else can I do with it? And are there more tricks similar to this?
(a ^ b) & 0x00010101 is what the least significant bits of the channels would have been in a + b if no carry had come from the right.
Subtracting it from the sum guarantees that the bit that is shifted into the most significant bit of the next channel is just the carry from that channel, untainted by this channel. Of course that also means that this channel is no longer effected by the carry from the next channel.
An other way to look this, not the way it does it but a way that may help you understand it, is that effectively the inputs are changed so that their sum is even for all channels. The carries then go nicely into the least significant bits (which are zero, because even), without disturbing anything. Of course what it actually does is sort of the other way around, first it just sums them, and only then does it ensure that the sums are even for all channels. But the order doesn't matter.
More concretely, there are 4 cases (before the carry from the next channel is applied):
the lsb of a channel is 0 and there is no carry from the next channel.
the lsb of a channel is 0 and there is a carry from the next channel.
the lsb of a channel is 1 and there is no carry from the next channel.
the lsb of a channel is 1 and there is a carry from the next channel.
The first two cases are trivial. The shift puts the carried bit back in channel it belongs to, it doesn't even matter whether it was 0 or 1.
Case 3 is more interesting. If the lsb is 1, that means the shift would shift that bit into the most significant bit of the next channel. That's bad. That bit has to be unset somehow - but you can't just mask it away because maybe you're in case 4.
Case 4 is the most interesting. If the lsb is 1 and there is a carry into that bit, it rolls over to a 0 and the carry is propagated. That can't be undone by masking, but it can be done by reversing the process, ie subtracting 1 from the lsb (which puts it back to 1 and undoes any damage done by the propagated carry).
As you can see, in both case 3 and case 4, the cure is subtracting 1 from the lsb, and those are also the cases in which the lsb really wanted to be 1 (though maybe it isn't any more, due to a carry from the next channel), and in both case 1 and 2, you don't have to anything (in other words, subtract 0). That exactly corresponds to subtracting "what the lsb would have been in a + b if no carry had come from the right".
Also, the blue channel can only fall into cases 1 or 3 (there is no next channel which could carry), and the shift would just discard that bit instead of putting it in the next channel (because there is none). So alternatively, you may write (note the mask has lost the least significant 1)
public static int blendRGB(int a, int b) {
return (a + b - ((a ^ b) & 0x00010100)) >> 1;
}
Doesn't really make any difference, though.
To make it work for ARGB8888, you can switch to the good old "SWAR average":
// channel-by-channel average, no alpha blending
public static int blendARGB(int a, int b) {
return (a & b) + (((a ^ b) & 0xFEFEFEFE) >>> 1);
}
Which is a variation on a recursive way to define addition: x + y = (x ^ y) + ((x & y) << 1) which computes the sum without carries, then adds the carries in separately. The base case is when one of the operands is zero.
Both halves are effectively shifted right by 1, in such a way that the carry out of the most significant bit is never lost. The mask ensures that bits don't move to a channel to the right, and simultaneously ensures that a carry won't propagate out of its channel.
I'm using Java and I'm coding a chess engine.
I'm trying to find the index of the first 1 bit and the index of the last 1 bit in a byte.
I'm currently using Long.numberOfTrailingZeros() (or something like that) in Java, and would like to emulate that functionality, except with bytes.
Would it be something like:
byte b = 0b011000101;
int firstOneBit = bitCount ((b & -b) - 1);
If so, how would I implement bitCount relatively efficiently. I don't mind good explainations, please don't just give me code.
use a lookup tabel with 256 entries.
to create it:
unsigned int bitcount ( unsigned int i ) {
unsigned int r = 0;
while ( i ) { r+=i&1; i>>=1; } /* bit shift is >>> in java afair */
return r;
}
this of course does not need to be fast as you do it at most 256 times to init your tabel.
/* Count Leading Zeroes */
static uint8_t clzlut[256] = {
8,7,6,6,5,5,5,5,
4,4,4,4,4,4,4,4,
3,3,3,3,3,3,3,3,
3,3,3,3,3,3,3,3,
2,2,2,2,2,2,2,2,
2,2,2,2,2,2,2,2,
2,2,2,2,2,2,2,2,
2,2,2,2,2,2,2,2,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0
};
uint32_t clz(uint32_t val)
{
uint32_t accum = 0;
accum += clzlut[val >> 24];
accum += (accum == 8 ) ? clzlut[(val >> 16) & 0xFF] : 0;
accum += (accum == 16) ? clzlut[(val >> 8) & 0xFF] : 0;
accum += (accum == 24) ? clzlut[ val & 0xFF] : 0;
return accum;
}
Explanation:
This works by storing the number of leading zeroes for each permutation of a byte as a lookup table. You use the byte value to look up the count of leading zeroes for that value. Since the example does this for an unsigned int, you shift and mask the four individual bytes, and accumulate the lookups accordingly. The ternary statement is used to stop the accumulation as soon as we find a bit which is set. That the accumulated value is 8, 16 or 24 implies that no set bit is found so far.
Also, some architectures have hardware support for this (as an instruction). The assembly mnemonic is often called 'CLZ' or 'BSR'. They are abbreviations for "Count leading Zeroes" and "Bit Scan Reverse" respectively.
The correct answer is that most all processors have some special instructions to do this sort of thing (leading zeros, trailing zeros, number of ones, etc). x86 has bsf/bsr, powerpc has clz, and so on. Hopefully Integer.numberOfTrailingZeros is smart enough to use these, but that's probably the only way that has a chance of using this sort of platform-specific function in Java (if it even uses them).
The Aggregate Magic Algorithms is another place with some approaches to this sort of problem, ranging from the obvious (lookup tables), to some rather clever SWAR approaches. But I suspect they all lose to Integer(x).numberOfTrailingZeros() if the java runtime is smart about the latter; it ought to be possible to optimize out the boxing and use a platform-specific technique for numberOfTrailingZeros, and if it does both that'll win.
Just for completeness, the other classic archive of brilliant bit-whacking is the old MIT HAKMEM collection (there's also a semi-modernized C version if your PDP-6/10 assembler skills have gotten rusty).
If you assume that Long.numberOfTrailingZeros is fast (i.e. JIT compiled/optimized to use a single ASM instructions when available), then why can't you simply do something like this:
max(8,Long.numberOfTrailingZeros(val))
where val is your byte value converted to a Long. This is also assuming that max() is available and again optimizes to use asm select or max instructions.
Theoretically, on a machine that supports it, these operations could be JIT compiled to two assembler instructions.
int x = n / 3; // <-- make this faster
// for instance
int a = n * 3; // <-- normal integer multiplication
int b = (n << 1) + n; // <-- potentially faster multiplication
The guy who said "leave it to the compiler" was right, but I don't have the "reputation" to mod him up or comment. I asked gcc to compile int test(int a) { return a / 3; } for an ix86 and then disassembled the output. Just for academic interest, what it's doing is roughly multiplying by 0x55555556 and then taking the top 32 bits of the 64 bit result of that. You can demonstrate this to yourself with eg:
$ ruby -e 'puts(60000 * 0x55555556 >> 32)'
20000
$ ruby -e 'puts(72 * 0x55555556 >> 32)'
24
$
The wikipedia page on Montgomery division is hard to read but fortunately the compiler guys have done it so you don't have to.
This is the fastest as the compiler will optimize it if it can depending on the output processor.
int a;
int b;
a = some value;
b = a / 3;
There is a faster way to do it if you know the ranges of the values, for example, if you are dividing a signed integer by 3 and you know the range of the value to be divided is 0 to 768, then you can multiply it by a factor and shift it to the left by a power of 2 to that factor divided by 3.
eg.
Range 0 -> 768
you could use shifting of 10 bits, which multiplying by 1024, you want to divide by 3 so your multiplier should be 1024 / 3 = 341,
so you can now use (x * 341) >> 10
(Make sure the shift is a signed shift if using signed integers), also make sure the shift is an actually shift and not a bit ROLL
This will effectively divide the value 3, and will run at about 1.6 times the speed as a natural divide by 3 on a standard x86 / x64 CPU.
Of course the only reason you can make this optimization when the compiler cant is because the compiler does not know the maximum range of X and therefore cannot make this determination, but you as the programmer can.
Sometime it may even be more beneficial to move the value into a larger value and then do the same thing, ie. if you have an int of full range you could make it an 64-bit value and then do the multiply and shift instead of dividing by 3.
I had to do this recently to speed up image processing, i needed to find the average of 3 color channels, each color channel with a byte range (0 - 255). red green and blue.
At first i just simply used:
avg = (r + g + b) / 3;
(So r + g + b has a maximum of 768 and a minimum of 0, because each channel is a byte 0 - 255)
After millions of iterations the entire operation took 36 milliseconds.
I changed the line to:
avg = (r + g + b) * 341 >> 10;
And that took it down to 22 milliseconds, its amazing what can be done with a little ingenuity.
This speed up occurred in C# even though I had optimisations turned on and was running the program natively without debugging info and not through the IDE.
See How To Divide By 3 for an extended discussion of more efficiently dividing by 3, focused on doing FPGA arithmetic operations.
Also relevant:
Optimizing integer divisions with Multiply Shift in C#
Depending on your platform and depending on your C compiler, a native solution like just using
y = x / 3
Can be fast or it can be awfully slow (even if division is done entirely in hardware, if it is done using a DIV instruction, this instruction is about 3 to 4 times slower than a multiplication on modern CPUs). Very good C compilers with optimization flags turned on may optimize this operation, but if you want to be sure, you are better off optimizing it yourself.
For optimization it is important to have integer numbers of a known size. In C int has no known size (it can vary by platform and compiler!), so you are better using C99 fixed-size integers. The code below assumes that you want to divide an unsigned 32-bit integer by three and that you C compiler knows about 64 bit integer numbers (NOTE: Even on a 32 bit CPU architecture most C compilers can handle 64 bit integers just fine):
static inline uint32_t divby3 (
uint32_t divideMe
) {
return (uint32_t)(((uint64_t)0xAAAAAAABULL * divideMe) >> 33);
}
As crazy as this might sound, but the method above indeed does divide by 3. All it needs for doing so is a single 64 bit multiplication and a shift (like I said, multiplications might be 3 to 4 times faster than divisions on your CPU). In a 64 bit application this code will be a lot faster than in a 32 bit application (in a 32 bit application multiplying two 64 bit numbers take 3 multiplications and 3 additions on 32 bit values) - however, it might be still faster than a division on a 32 bit machine.
On the other hand, if your compiler is a very good one and knows the trick how to optimize integer division by a constant (latest GCC does, I just checked), it will generate the code above anyway (GCC will create exactly this code for "/3" if you enable at least optimization level 1). For other compilers... you cannot rely or expect that it will use tricks like that, even though this method is very well documented and mentioned everywhere on the Internet.
Problem is that it only works for constant numbers, not for variable ones. You always need to know the magic number (here 0xAAAAAAAB) and the correct operations after the multiplication (shifts and/or additions in most cases) and both is different depending on the number you want to divide by and both take too much CPU time to calculate them on the fly (that would be slower than hardware division). However, it's easy for a compiler to calculate these during compile time (where one second more or less compile time plays hardly a role).
For 64 bit numbers:
uint64_t divBy3(uint64_t x)
{
return x*12297829382473034411ULL;
}
However this isn't the truncating integer division you might expect.
It works correctly if the number is already divisible by 3, but it returns a huge number if it isn't.
For example if you run it on for example 11, it returns 6148914691236517209. This looks like a garbage but it's in fact the correct answer: multiply it by 3 and you get back the 11!
If you are looking for the truncating division, then just use the / operator. I highly doubt you can get much faster than that.
Theory:
64 bit unsigned arithmetic is a modulo 2^64 arithmetic.
This means for each integer which is coprime with the 2^64 modulus (essentially all odd numbers) there exists a multiplicative inverse which you can use to multiply with instead of division. This magic number can be obtained by solving the 3*x + 2^64*y = 1 equation using the Extended Euclidean Algorithm.
What if you really don't want to multiply or divide? Here is is an approximation I just invented. It works because (x/3) = (x/4) + (x/12). But since (x/12) = (x/4) / 3 we just have to repeat the process until its good enough.
#include <stdio.h>
void main()
{
int n = 1000;
int a,b;
a = n >> 2;
b = (a >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
printf("a=%d\n", a);
}
The result is 330. It could be made more accurate using b = ((b+2)>>2); to account for rounding.
If you are allowed to multiply, just pick a suitable approximation for (1/3), with a power-of-2 divisor. For example, n * (1/3) ~= n * 43 / 128 = (n * 43) >> 7.
This technique is most useful in Indiana.
I don't know if it's faster but if you want to use a bitwise operator to perform binary division you can use the shift and subtract method described at this page:
Set quotient to 0
Align leftmost digits in dividend and divisor
Repeat:
If that portion of the dividend above the divisor is greater than or equal to the divisor:
Then subtract divisor from that portion of the dividend and
Concatentate 1 to the right hand end of the quotient
Else concatentate 0 to the right hand end of the quotient
Shift the divisor one place right
Until dividend is less than the divisor:
quotient is correct, dividend is remainder
STOP
For really large integer division (e.g. numbers bigger than 64bit) you can represent your number as an int[] and perform division quite fast by taking two digits at a time and divide them by 3. The remainder will be part of the next two digits and so forth.
eg. 11004 / 3 you say
11/3 = 3, remaineder = 2 (from 11-3*3)
20/3 = 6, remainder = 2 (from 20-6*3)
20/3 = 6, remainder = 2 (from 20-6*3)
24/3 = 8, remainder = 0
hence the result 3668
internal static List<int> Div3(int[] a)
{
int remainder = 0;
var res = new List<int>();
for (int i = 0; i < a.Length; i++)
{
var val = remainder + a[i];
var div = val/3;
remainder = 10*(val%3);
if (div > 9)
{
res.Add(div/10);
res.Add(div%10);
}
else
res.Add(div);
}
if (res[0] == 0) res.RemoveAt(0);
return res;
}
If you really want to see this article on integer division, but it only has academic merit ... it would be an interesting application that actually needed to perform that benefited from that kind of trick.
Easy computation ... at most n iterations where n is your number of bits:
uint8_t divideby3(uint8_t x)
{
uint8_t answer =0;
do
{
x>>=1;
answer+=x;
x=-x;
}while(x);
return answer;
}
A lookup table approach would also be faster in some architectures.
uint8_t DivBy3LU(uint8_t u8Operand)
{
uint8_t ai8Div3 = [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ....];
return ai8Div3[u8Operand];
}