Origin of Quake3's Fast InvSqrt() 402
geo writes "Beyond3D.com's Ryszard Sommefeldt dons his seersucker hunting jacket and meerschaum pipe to take on his secret identity as graphics code sleuth extraordinaire. In today's thrilling installment, the origins of one of the more famous snippets of graphics code in recent years is under the microscope — Quake3's Fast InvSqrt(), which has been known to cause strong geeks to go wobbly in the knees while contemplating its simple beauty and power." From the article: ""
Re:A famous quote (Score:5, Informative)
Re:A famous quote (Score:1, Informative)
To find the square root of say, N, you use InvSqrt(N) and then multiply the result by N to get the square root of N. Supposedly, it's faster than using the built-in sqrt() function.
Re:A famous quote (Score:4, Informative)
Re:A famous quote (Score:5, Informative)
Carmack quite graciously denied the code was his and helped direct the author closer to the true source.
This paper seems to have the info (Score:5, Informative)
Re:This paper seems to have the info (Score:3, Informative)
It was fast (Score:4, Informative)
That page compares the time it takes to calculate the sqrt various ways including Carmacks. Short version is that modern processors are significantly faster since it can be done in hardware. It may still be useful in cases where the processor doesn't have the sqrt function available.
His version took 428 cycles compared to 107 cycles doing it in hardware on the same system.
Re:What's with use of Pointers? (Score:5, Informative)
int i = (int)x;
Then C will simply convert the float value into an integer value (throwing away fractional part). But this isn't what we want. We want to operate on the bits of an IEEE floating point value directly, and integers are the best way to do that.
So first, we lie to the compiler by telling it we have a pointer to an int:
(int *) &f
And then we deference the pointer to get it into an operable int:
i = *(int *) &f
Note what's important here is to keep the compiler from modifying any part of the original 32-bit value.
Schwab
Re:And so why do we care? (Score:5, Informative)
Re:What's with use of Pointers? (Score:5, Informative)
*(int*) &x treats the bits as an integer, with no behind the scenes conversion to an actual int value.
Re:What's with use of Pointers? (Score:5, Informative)
If you do this:
int i = (int)3.0f;
You get i=3, like what you'd get from the floor() function.
If you do this:
float f = 3.0f;
int i = *(int*)
Then i contains a bit-for-bit copy of the IEEE floating-point representation of 3.0.
It's because C knows how to cast a float to an int by applying the floor function. However, if you do it the second way, you aren't casting a float to an int, you are casting a pointer-to-float to a pointer-to-int and then dereferencing it.
By the way, I just wanted to say... this is one of the most interesting things I've read on Slashdot in a while. Wow. That function is just amazing. I only wish I understood how it worked. I know nothing about what a "Newton-Raphson iteration" is.
Re:Poor function name (Score:1, Informative)
(-1/2) != (2).
x^(-1/2) != x^(2).
QED.
Re:It's Funny. Laugh. (Score:1, Informative)
Mirrordot the airticle cut-and-paste (Score:5, Informative)
Note!
This article is a republishing of something I had up on my personal website a year or so ago before I joined Beyond3D, which is itself the culmination of an investigation started in April 2004. So if timeframes appear a little wonky, it's entirely on purpose! One for the geeks, enjoy.
Origin of Quake3's Fast InvSqrt()
To most folks the following bit of C code, found in a few places in the recently released Quake3 source code, won't mean much. To the Beyond3D crowd it might ring a bell or two. It might even make some sense.
InvSqrt()
Finding the inverse square root of a number has many applications in 3D graphics, not least of all the normalisation of 3D vectors. Without something like the nrm instruction in a modern fragment processor where you can get normalisation of an fp16 3-channel vector for free on certain NVIDIA hardware if you're (or the compiler is!) careful, or if you need to do it outside of a shader program for whatever reason, inverse square root is your friend. Most of you will know that you can calculate a square root using Newton-Raphson iteration and essentially that's what the code above does, but with a twist.
How the code works
The magic of the code, even if you can't follow it, stands out as the i = 0x5f3759df - (i>>1); line. Simplified, Newton-Raphson is an approximation that starts off with a guess and refines it with iteration. Taking advantage of the nature of 32-bit x86 processors, i, an integer, is initially set to the value of the floating point number you want to take the inverse square of, using an integer cast. i is then set to 0x5f3759df, minus itself shifted one bit to the right. The right shift drops the least significant bit of i, essentially halving it.
Using the integer cast of the seeded value, i is reused and the initial guess for Newton is calculated using the magic seed value minus a free divide by 2 courtesy of the CPU.
But why that constant to start the guessing game? Chris Lomont wrote a paper analysing it while at Purdue in 2003. He'd seen the code on the gamedev.net forums and that's probably also where DemoCoder saw it before commenting in the first NV40 Doom3 thread on B3D. Chris's analysis for his paper explains it for those interested in the base math behind the implementation. Suffice to say the constant used to start the Newton iteration is a very clever one. The paper's summary wonders who wrote it and whether they got there by guessing or derivation.
So who did write it? John Carmack?
While discussing NV40's render path in the Doom3 engine as mentioned previously, the code was brought up and attributed to John Carmack; and he's the obvious choice since it appears in the source for one of his engines. Michael Abrash was mooted as a possible author too. Michael stands up here as x86 assembly optimiser extraordinaire, author of the legendary Zen of Assembly Language and Zen of Graphics Programming tomes, and employee of id during Quake's development where he worked alongside Carmack on optimising Quake's software renderer for the CPUs around at the time.
Asking John whether it was him or Michael returned a "not quite".
-----Original Message-----
From: John Carmack
Sent: 26 April 2004 19:51
Subject: Re: Origin of fast approximated inverse square root
At 06:38 PM 4/26/2004 +0100, you wrote:
>Hi John,
>
>There's a discussion on Beyond3D.com's forums about who the author of
>the following is:
>
>float InvSqrt (float x){
> float xhalf = 0.5f*x;
> int i = *(int*)
> i = 0x5f3759df - (i>>1);
> x = *(float*)
> x = x*(1.5f - xhalf*x*x);
> return x;
>}
>
>Is that something we can attribute to you? Analysis shows it to be
>extremely clever in its method and supposedly from the Q3 source.
>Most people say it's your work, a few say it's Michael Abrash's. Do
>you know who's responsible, possibly with a history of sorts?
Not me,
Re:What's with use of Pointers? (Score:5, Informative)
I'll take a swing at this one. It's because the author doesn't want the value of x, but the integer representation of the value at x's memory address.
If x is 3.14159, (2) will result in i==3, whereas (1) will result in whatever the 4-byte IEEE-754 representation of 3.14159 is (0x40490FD0, if Google is correct). By using (1), the author is able to use integer bitwise opeartions (>>) to perform "free" floating point operations. When i is sent back into floating point form via:
x = *(float*)
x now contains the value of the integer operation:
i = 0x5f3759df - (i >> 1);
which was presumably faster than an identical floating point operation. It's a nifty little solution, especially with regard to the selection of the magic number.
Re:x * x, right? (Score:3, Informative)
Well, you're sort of right.
f^-1(x) = x^2 is the inverse of f(x) = sqrt(x) where x > 0
Re:A famous quote (Score:4, Informative)
Re:A famous quote (Score:5, Informative)
Re:What's with use of Pointers? (Score:3, Informative)
Moreover, when compiled, the optimizer does the right thing - since access to the variable is actually a memory dereference (or a register access, but we'll ignore it for now), *(int*)& means "tell the compiler this is an int, but don't do anything in the code", whereas (int) means "create a new temporary that's the int part of the variable".
Evil math--it approximates the inverse square root (Score:2, Informative)
Seriously, they're using some evil numerical techniques that approximate the actual value of that function to withing a few percent very quickly. You could do more steps of Newton's root finding method or other complex things, but that would be slow and this is only meant to find things that are "good enough" for you to draw on someone's screen.
This function is used in graphics programming, specifically in things like the Quake 3 source code (as well as other places). First, it calculates x/2, then it gets the bits of the float x as an integer (without modifying them), then it does the "magic" step with the magic number they calculated via evil mathematics to give the best starting point they can for Newton's method using the x >> 1 to divide the float-as-integer in two (and make it positive, since your sign bit falls into the exponent of the float!
Don't mind me. I may have a degree in mathematics, and I even (somehow!) made it through at least one class on numerical analysis, but all I can tell you is that this stuff is evil
Re:What's with use of Pointers? (Score:5, Informative)
You start with some INITIAL GUESS (the real beauty of this algorithm) X(0), then apply:
X(n+1) = X(n) - f(X(n)) / f'(X(n))
where
X(n+1) is the NEXT guess after the value you 'know',
X(n) is that most recent value you know,
f(X(n)) is the function evaluated at X(n) and
f'(X(n)) is the first derivative of f(x) evaluated at X(n).
It's not foolproof and a BOTH whether it converges at al AND how FAST it converges depends on the initial guess, X(0)
The "Secant Method" is an improvement that makes it a little 'smarter,' at the expense of more computation (this is often a positive trade-off on numerical modeling codes, since the 'smarter' algorithm does tend to converge faster). There are other improvements as well, such as the Los Alamos Linear Feedback Solver (a slightly modified secant method that converges about 10-17% faster, at least for some types of problems) that I use in my own codes.
Obligatory wikipediea followup: Newton's Method [wikipedia.org]
Re:A famous quote (Score:5, Informative)
This is great for a 3D rendering application, but in a game speed is critical. This pair of calculations involves a square root and a divide. Both of thse are at least an order of magnitude slower than multiplications and additions.
So what this function does is provide a value you can multiply each component by to get a unit vector.
Well, there's the what and why parts. As for the , I have no idea. I think it uses magic.
Re:What's with use of Pointers? (Score:3, Informative)
Newton-Raphson is a root finding method. Given a starting point it finds successively more accurate approximations to the root. The formula for it looks like: x_{n+1} = x_n - f(x_n)/f'(x_n), where f(x) is the function to find the root of, f'(x_n) is the derivative of that function and x_n are the successive approximations. Essentially, given a guess it looks at the slope of the function at that point follows a tangent line until it crosses the zero line. That point becomes the next guess.
Finding the reciprocal square root of a is equivalent to solving x^{-2} - a = 0 which means that you use f(x) = x^{-2} - a. The derivative of that is f'(x) = -2x^{-3}. Putting those into the Newton-Raphson formula gives you x_{n+1} = x_n - (x_n^{-2} - a) / -2x_n^{-3}, which simplifies down to x_{n+1} = x_n(3 - a*x_n^2)/2. This is equivalent to the last line before the return in that InvSqrt() function.
Convergence of Newton-Raphson is quadratic which means that each time you repeat that line with a previous guess you double the number of digits of accuracy (up to the limit of the hardware of course).
Re:A famous quote (Score:3, Informative)
Re:What's with use of Pointers? (Score:2, Informative)
Re:It was fast (Score:5, Informative)
rsqrtss xmm1, xmm0
about 5 cycles. And it can pipeline.
Not a fan of x86? Maybe altivec...
vrsqrtefp V2, V1
depends, but 12 cycles probably and pipelined.
On PS3's SPU it's rsqrte (6 cycles), on 3dNow it's pfrsqrt (8 cycles) both pipelined. Even PS2 had rsqrt (13 cycles). There's just no reason for software reciprocal square root. It's a cool trick, but it's not even useful anymore.
Re:A famous quote (Score:3, Informative)
Correction (Score:5, Informative)
I believe you meant to say x^(-1/2)
Too bad the people modding you up don't have math degrees. =P
Re:A famous quote (Score:3, Informative)
The inverse square root is used to calculate a normal vector [wikipedia.org]. Normal vectors, are, in turn, used in lighting calculations (to determine, for example, the light intensity for a given vertex), among other things.
AMD's 3D Now! instruction set includes a set of instructions (TFSQRTS/TFSQRTD) for approximating the inverse square root.
It may in part be related to something I did ... (Score:5, Informative)
http://web.archive.org/web/19990210111728/www.geo
And here's an updated version of the same page:
http://www.azillionmonkeys.com/qed/sqroot.html [azillionmonkeys.com]
It isn't an exact rendering of the code in question, but it explains enough for any skilled hacker to 1) understand what's going on and 2) modify the code to create the resulting code that's in the Quake 3 source. Furthermore this web page has existed since about 1997 (archive.org doesn't go back that far for some reason.)
Now *IF* in fact the code origin comes from someone who took ideas from my site, I should point out that *I* am not the originator of the idea either (though I did write the relevant code). Bruce Holloway (who I credit on the page) was the first person to point out this technique to me at around the 1997 timeframe (prior to this, I created my own method which is similar, but not really as fast). (Vesa Karvonen informed by of the technique (through a code snippet with no explanation) at roughly the same time as well.) It was a technique well known to hard core 3D accelerator and CPU implementors, and follows an intentional design idea from the IEEE-754 specification.
Prof. William Kahan, one of the key people who specified the IEEE-754 standard (the standard for floating point the many CPUs use, starting with Intel's 8087 coprocessor) apparently presented this idea, and is the source for where Bruce Holloway got the idea. The IEEE-754 standard came out around the 1982 time frame. Though, its very likely that these ideas originate from even earlier in computing history.
Re:What's with use of Pointers? (Score:3, Informative)
Subtracting the right shifted value from the magic value does NOT give the 'inverse square root.' This line of code only gives the initial guess for the Newton-Raphson algorithm.
While the initial guess is a huge part of the real beauty of the algorithm (ie, why it can converge to a sufficiently accurate value after ONLY ONE ITERATION), the 'real work' is done in the
x = x*( 1.5f - xhalf*x*x )
line. That's that application of one iteration of Newton-Raphson (rewritten a bit from what I and others posted earlier, but for this problem algebraically the same). To get more accuracy in the calculation, apply this line repeatedly.
The explanation of the why the magic value - (right shifted int represenation of the float) is a good initial guess is given in the Paper by Chris Lomont [lomont.org] linked to in article. Specifically, look on Section 4 beginning on Page 3 of Lomont's article; that's where the fun starts.
In a nutshell, from Lomont, the initial guess is computed "by multiplying the exponent by -1/2, and then picking bits to minimize error." I have not worked through all the details on why this is a good initial guess, but will reemphasize that getting a good result from Newton-Raphson DEPENDS on the initial guess. So, it is not surprising that someone hunted for a general algorithm to find a good initial guess.
Re:A famous quote (Score:5, Informative)
Slow:
const float length = sqrt( v.x*v.x + v.y*v.y + v.z*v.z );
v.x
v.y
v.z
Fast:
const float recip_length = InvSqrt( v.x*v.x + v.y*v.y + v.z*v.z );
v.x *= recip_length;
v.y *= recip_length;
v.z *= recip_length;
The 2nd version has no divides, and no call to sqrt, which makes it *loads* faster.
Re:It was fast (Score:2, Informative)
Re:A famous quote (Score:3, Informative)
You are correct, the inverse to x^n is x^(1/n), from which it follows that the inverse to a square root is the inverse inverse square, or just square.
What TFA calls "inverse square root" is really "inverse of the square root", a small but significant difference.
Not that it matters much, cause those who use the function tend to know what is meant.
Regards,
--
*Art
Re: A better question: (Score:5, Informative)
It starts by taking a guess at the right answer, and then improving the guess until it's accurate enough to use.
The first step depends heavily on the fact that a floating point number on a computer is represented as a significand (aka mantissa) and an exponent (a power of two). For the moment, consider taking just the square root of X instead of its inverse. You could separate out the exponent part of the floating point number, divide it by two, and then put the result back together with the original significand, and have a reasonable starting point.
From there, you could improve your guesses to get a better approximation. The simplest version of that would be like a high-low game -- you split the difference between the current guess and the previous guess, and then add or subtract that depending on whether your previous guess was high or low. Eventually, you'll get arbitrarily close to the correct answer.
This can take quite a few iterations to get to the right answer though. To improve that, Newton-Raphson looks at the curve of the function you're working with, and projects a line tangent to the curve at the point of the current guess. Where that line crosses the origin gives you the next guess. That's probably a lot easier to understand from picture [sosmath.com].
In this case, we're looking for the inverse square root, which changes the curve, but not the basic idea. As a general rule, the closer your first guess, the fewer iterations you need to get some particular level of accuracy. That's the point of the:
While the originator of this constant is unknown, and some of it is rather obscure, the basic idea of most of it is fairly simple: we start by shifting the original number right a bit. This divides both the mantissa and the exponent part by two, with the possibility that IF the exponent was odd, it shifts a bit from the exponent into the mantissa. The subtraction from the magic number then does a couple of things. For one thing, if a bit from the exponent was shifted into the mantissa, it removes it. The rest of the subtraction is trickier. If memory serves, it's based on the harmonic mean of the difference between sqrt(x) and (x/2) for every possible floating point number of the size you're using.
This is where the fact that it's 1/sqrt(x) instead of sqrt(x) means a lot: 1/sqrt(x) is a curve, but it's a fairly flat curve -- much flatter than sqrt(x). The result is that we can approximate a point on the curve fairly accurately with a line. In this case, it's really two lines, which gets it a bit closer still.
From there, the number has had a bit of extra tweaking done -- it doesn't actually give the most accurate first guess, but its errors are often enough in the opposite direction from those you get in the Newton-Raphson iteration steps that it gives slightly more accurate final results.
Re:It was fast (Score:3, Informative)
Re:What's with use of Pointers? (Score:5, Informative)
The trick of this function is to take the 32 bits of data that are really a float, but process it as if it's an integer. So you take that cumbersome number 21 as a float, then BAM! presto, turn it directly to an integer not through type conversion, but by simply treating those same 32 bits as if they were representing an integer all along.
Let's use the number 21 as an example in the function call.
The binary representation of 21 as a float is 01000001 10101000 00000000 00000000 (broken into 8-bit words for clarity). The function then goes to create a new integer i, whose value is also 01000001 10101000 00000000 00000000 (which happens to be 1101529088 in decimal). The magical line of the code, i = 0x5f3759df - (i>>1), takes that integer i, shifts its bits one to the right (turning our 01000001 10101000 00000000 00000000 into 00100000 11010100 00000000 00000000, or 550764544 in decimal), then subtracts it (still doing integer math here) from 0x5f3759df (which is 01011111 00110111 01011001 11011111 or 1597463007 in decimal), and winds up with 00111110 01100011 01011001 11011111 (or 1046698463 in decimal).
Now, for its next trick, it takes that wonky integer 1046698463, and turns it back into a floating point number, by the same trick used above, i.e. simply by looking at those same 32 bits, and pretending they're a float, not an int. The binary representation of 1046698463, 00111110 01100011 01011001 11011111, is the same as 0.22202251851558685 in float.
From here on out, it's all floating math. Apply the Newton-Rhapson method (thats the next line), we get x = 0.22202251851558685 * (1.5 - ( (21*.5) * 0.22202251851558685^2 )) = 0.218117811. We return this value at the closing of the function. As it turns out, the inverse square root of 21 is 0.21821789... (thanks Google calc). So, I have no idea WHY the Float to Int to Float trick works, but it works very well.
Whew!
Re:It might be damn smart.. (Score:3, Informative)
Clever trick! (Score:5, Informative)
The algorithm is simple Newton-Raphson -- make a good initial guess, then iterate making the guess better. I think Newton-Raphson on 1/sqrt picks up 5-6 bits each try in the line "x = x*(1.5f - xhalf*x*x)". It can be repeated to get a more accurate result each time it's repeated.
The problem with Newton-Raphson is making a good first guess--otherwise, you need an extra iteration or two. And that's what the magic number is doing, making a good first guess.
So let's work out what a good first guess would look like for 1/sqrt(f), to see where this code came from.
Floating Point numbers are stored with a mantissa and an exponent: f = mantissa * (2 ^ exponent), where exponent is 8-bits wide and the mantissa is 23-bits wide.
Let's take an example: 1/sqrt(16) would have f = 1.0 * (2 ^ 4). We want the result 0.25 which is f = 1.0 * ( 2 ^ -2).
So our first guess should take our exponent, negate it, and cut it in half. (Try more examples to see that this works--it's basically the definition of 1/sqrt(f)). We'll ignore the mantissa--if we can just get within a factor of 2 of the answer in one step, we're doing pretty well.
Unfortunately, the exponent is stored in FP numbers in an offset format. In memory, The mantissa is in the low 23 bits, and the most-significant bit is the sign (which will be 0 if we're taking roots). For now, let's just assume we have our exponents as 8-bit values, to work out what we need to do with the +127 offset.
We want new_actual_exp = -(actual_exp)/2. But in memory, exp = (actual_exp + 127). Or, actual_exp = exp - 127.
Substituting gives (new_exp - 127) = -(exp - 127)/2. Simplify this to: new_exp = 127 - (exp - 127)/2 => new_exp = 3*127/2 - (exp / 2).
Now the exponent is shifted 23 places in memory, so let's write out our code (and ignore the mantissa completely for now...): rewriting as hex: Well, first, it's arguable whether it should be 0x5f000000 or 0x5f400000 (The "4" is actually in the mantissa). I'm guessing resolving that dilemma led to the original author discovering that choosing a particular pattern of bits in the mantissa can help make the initial guess even more accurate, leading to the 0x5f3759df constant.
I haven't worked it out, but Chris Lomont http://www.lomont.org/Math/Papers/2003/InvSqrt.pd
Re:It might be damn smart.. (Score:2, Informative)
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * )
i = 0x5f3759df - ( i >> 1 );
y = * ( float * )
y = y * ( threehalfs - ( x2 * y * y ) );
Re:It might be damn smart.. (Score:2, Informative)
The only non-portable assumptions made by the code are: int is 32 bits, and float is stored in IEEE-754 format.
The first assumption is still true in 64 bit gcc, and even if it isn't you can just write int32_t instead.
The f after 0.5 says 0.5 is a float not a double (Score:2, Informative)
The f after 0.5 and the 1.5 simply means "use the floating point version of this number". Multiplying two numbers which have different types (e.g. multiplying an int and a float) or doing assignment between different types means that one of the numbers will have to be automatically converted to the other's type [wikipedia.org] and which conversion happens depends on the types of both numbers. By default, I think type of the literal 0.5 (without any letters after it) is a double. By adding the f (assuming x is a float) the types of the operands are the same so there is no implicit type conversion thus eliminating the chance of the wrong conversion happening automatically. You care about what type of multiply is done because certain types can lead to unwanted loss of precision.
As for 1)
Well if you iterate by one step you don't need a loop. Presumably most looping iteration happens because your first guess wasn't accurate enough. If your first guess WAS accurate enough why would you needlessly calculate more accuracy? I'm only speculating though - I don't know if this is really the case.
Re:Bad programmer, no cookie (Score:3, Informative)
Re:Lookup Table? (Score:3, Informative)
Performance-wise, you are much better doing several calculations than a single lookup.
Re:A famous quote (Score:2, Informative)