# Fast Inverse Square Root

"Fast inverse square root" refers to a famous algorithm invented by William Kahan, especially as implemented in Quake III Arena.

If you prefer, there is also an excellent video covering this topic here.

Although the algorithm has been made mostly obsolete due to advances in hardware technology, it was previously used to quickly and accurately (with a maximum of one percent error) calculate the multiplicative inverse of the square root of a number $x$ ($\frac{1}{\sqrt{x}}$).

```
float y = 1 / sqrt(x);
```

The primary use of determining the inverse square root of a number is for normalizing vectors (maintaining their direction while setting their magnitude to $1$). Normalized vectors are useful in computer graphics and physics simulations because, just like the number $1$, multiplying by them doesn't change the magnitude of other vectors (while still modifying their direction).

Like numbers, vectors can be normalized by dividing them by their magnitudes. The magnitude of a vector is determined by taking the square root of the sum of the squares of its components (the two-dimensional version of this is called the Pythagorean theorem: $\sqrt{a^2+b^2}$). In other words, multiplying a vector by its inverse square root (the inverse of its magnitude) normalizes it.

The thing that makes fast inverse square root so important is, as the name implies, the computation speed of the algorithm. Division is very slow compared to the other basic arithmetic operations, so it is important to minimize its use. Even a small improvement means a lot in a function like fast inverse square root, which would be used hundreds of times per second.

## Code Analysis

The remainder of this article will be an analysis of the Quake III Arena code, which can be found here. For reference, I have copied the code below:

```
float Q_rsqrt(float number) {
long i;
float x2, y;
const float threehalfs = 1.5f;
x2 = number * 0.5f;
y = number;
i = *(long*)&y; // evil floating point bit level hacking
i = 0x5f3759DF - (i >> 1); // what the fuck?
y = *(float*)&i;
y = y * (threehalfs - (x2 * y * y)); // 1st iteration
// y = y * (threehalfs - (x2 * y * y)); // 2nd iteration
return y;
}
```

If I were to write the same function, I would modify it like this:

```
float Q_rsqrt(float y) {
float x = y * 0.5f;
long i = *(long*)&y;
i = 0x5f3759DF - (i >> 1);
y = *(float*)&i;
y *= (1.5f - (x * y * y));
return y;
}
```

### IEEE 754

In order to understand how the function works, one must first understand how floating-point variables are stored in memory. Whenever a variable is declared as a `float`

in C, it is stored according to the rules of IEEE 754. IEEE 754 was created by William Kahan, in collaboration with Intel, in the 1980s. Today, almost all computers use it to store floating-point values.

IEEE 754 is based on scientific notation, in which numbers are written as a value in the range $[1,b)$ multiplied by some power of $b$, where $b$ is the base of the number. For example, $45000_{10}=4.5*10^4$. The first number ($4.5$ in the example) is called the "*mantissa*", and the exponent of the $10$ ($4$) is just called the "*exponent*."

IEEE 754 floating-point values are stored in 32 bits. The first bit is the "*sign bit*." The sign bit is `0b1`

if the number is negative, or `0b0`

if it is positive. The sign bit comes first in order to make it easier to sort floating-point values.

The next eight bits encode the exponent. This number is right-aligned. Rather than using *two's complement* (which is a way to store signed integers where the first bit represents the negative of its typical value), the exponent is "biased" by adding $b^{n-1}-1=127$, where $b=2$ is the base and $n=8$ is the number of bits in the exponent, to its value, such that the smallest representable exponent is represented as $1$. This is done so that larger numbers visibly appear larger, and to make floating-point values easier to sort.

The final twenty-three bits encode the mantissa. These bits are left-aligned, with each bit representing $\frac{1}{2^n}$, where $n$ is the index of the bit. Since computers store values in base-two (binary) rather than base-ten (decimal), the mantissa is instead multiplied by a power of $2$, and must fall in the range $[1,2)$. Since the integer portion of the mantissa is guaranteed to be $1$ by definition, it doesn't need to be stored.

Since it would be impossible to represent $0$ with the standard described above, IEEE 754 also includes a few special rules:

- If the exponent field is all zeros, the number is "
*denormalized*," which means that the exponent becomes $1-b$, where $b$ is the bias, and the integer $1$ is not added to the mantissa. - If the exponent field is all ones and the mantissa is all zeros, the value represented is infinity.
- If the exponent field is all ones and the mantissa is not all zeros, the value represented is not a number.

For example, to store the number $3.75=1.875*2^1$ in IEEE 754:

- The number is positive, so the sign bit is
`0b0`

. - The exponent is $1$. Adding the bias of $127$, this means that the represented value must be $128$, or
`0b10000000`

. - The mantissa is $1.875$. Since the integer part doesn't need to be represented, the bits represent $0.875=\frac{7}{8}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, or
`0b11100000000000000000000`

. - All together, $3.75$ is represented as
`0b01000000011100000000000000000000`

.

### The Magic Constant

Since the floating-point values that fast inverse square root is designed to work with are always positive (the square root of a negative number is imaginary), the sign bit is always `0b0`

, so it can be disregarded. Since the eight-bit exponent is followed by the twenty-three-bit mantissa, if given a binary exponent $e$ and a binary mantissa $m$, the bit representation of a floating-point number can be calculated as $2^{23}e+m$, since multiplying $e$ by $2^{23}$ shifts it twenty-three digits to the left. Reverse-engineering this formula yields the following method to get the value of the mantissa $m$ and the exponent $e$ from a given set of bits:

The $127$ comes from the bias in the exponent bits in IEEE 754.

Taking the logarithm of the above equation and simplifying yields the following:

$\log_2(1+\frac{m}{2^{23}})+e-127$For small values of $x$, $\log_2(1+x)\approx x$. A corrective term, $\mu$, is added to this, yielding $\log_2(1+x)\approx x+\mu$. $\mu\approx0.043$ gives the smallest average error for numbers in the range $[0,1]$. Since $\frac{m}{2^{23}}$ is in the range $[0,1]$, the formula can be simplified further to the following:

$\frac{m}{2^{23}}+\mu+e-127$The above equation can be rearranged to:

$\frac{2^{23}e+m}{2^{23}}+\mu-127$The appearance of $2^{23}e+m$ here is important because, as explained previously, that is the bit representation. In other words, the bit representation of a number is its own logarithm, albeit scaled and shifted by some constants.

### Evil Floating-Point Bit-Level Hacking

The first notable line in the fast inverse square root implementation in Quake III Arena is similar to the following:

```
long i = *(long*)&y;
```

Usually, when casting a value from one type to another in C, the bits that represent that value must be changed so that the value remains consistent. For example, `1`

as a `float`

is represented as `0b00111111100000000000000000000000`

, while `1`

as a `char`

is represented as `0b00000001`

. This line circumvents this automatic process by telling C that the pointer to the `float`

`y`

, `&y`

, is a pointer to a `long`

(`long*`

), and then dereferencing it (`*`

). This allows `i`

, which is a `long`

, to have the exact same bit representation as `y`

, which is a `float`

. This is useful because it is possible to use bit manipulation techniques on `long`

s but not `float`

s.

One such bit manipulation technique is that shifting a binary (base-two) number to the right one place (`i >> 1`

) halves it, and shifting it to the left one place (`i << 1`

) doubles it, both of which are very fast operations. This is similar to the way that shifting a decimal (base-ten) number to the right one place (such as $100$ to $10$) divides it by $10$.

Note that, after the next step, the same operation is applied in reverse:

```
y = *(float*)&i;
```

### Applying the Magic Constant

Note that halving the exponent of a number yields the square root of that number ($x^{\frac{1}{2}}=\sqrt{x}$), and that negating the exponent yields its inverse ($x^{-1}=\frac{1}{x}$). Therefore, $x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$.

As explained previously, `i`

now contains `log(y)`

. Using this fact, it is possible to simplify the algorithm by finding the logarithm of the inverse square root of `y`

, rather than finding the inverse square root directly.

The second notable line of code is similar to the following:

```
i = 0x5f3759DF - (i >> 1);
```

The `- (i >> 1)`

part of the line is effectively halving the logarithm of `y`

, while the constant `0x5f3759DF`

comes from the scaling and shifting applied to the logarithm previously. The constant can be reverse-engineered as the function $\Gamma$ as follows:

Find the logarithm of both sides:

$\log_2(\Gamma)=\log_2(\frac{1}{\sqrt{y}})=-\frac{\log_2(y)}{2}$Replace both sides with the bit representation:

$\frac{2^{23}e_\Gamma+m_\Gamma}{2^{23}}+\mu-127=-\frac{\frac{2^{23}e_y+m_y}{2^{23}}+\mu-127}{2}$Solve for the bits of $\Gamma$:

$2^{23}e_\Gamma+m_\Gamma=\frac{3(2^{23})(127-\mu)-(2^{23}e_y+m_y)}{2}$$-\frac{2^{23}e_y+m_y}{2}$ is equivalent to `- (i >> 1)`

in the code, which leaves $\frac{3(2^{23})(127-\mu)}{2}$, which makes up the constant `0x5f3759DF`

.

### Newton's Method

Newton's method is an iterative algorithm which produces successively better approximations to the roots of a real-valued function. In other words, every time we run our approximation through Newton's method, it will become closer to the real value.

The approximation that we've computed thus far is already so close to the real value that just one iteration of Newton's method will bring the error within $1\%$.

Newton's method is accomplished in the original code similar to this:

```
y *= (1.5f - (x * y * y));
```

$f(y)=\frac{1}{y^2}-x\implies{y=\frac{1}{\sqrt{x}}}$