: C Square Root Functions: Richard Marks May 06, 2010, 08:55:02 PM
I do not take credit for this code snippet. I'm posting it here for my own reference, and others.
The code was made available in André LaMothe's book: Tricks of the 3D Game Programming Gurus (ISBN 0-672-31835-0): float Sqrt(float value) It is at least 3x faster than the standard sqrt() function.{ float result; _asm { mov eax, value sub eax, 0x3F800000 sar eax, 1 add eax, 0x3F800000 mov result, eax } return result; } The code depends on the format of the float data type. I assume this code will only work on Intel/AMD PCs. I've tried to understand HOW it works to no avail. I just will use it as is. The following implementation of a square root function is my own design. : unsigned int Sqrt(unsigned int value) This code executes about 2x faster than the standard sqrt() function!{ unsigned int result = 0; for (unsigned int i = 128; i != 0; i >>= 1) { result += i; if (result * result > value) { result -= i; } } return result; } The following implementation is by an unnamed member in the NDS homebrew development scene. : unsigned int Sqrt(unsigned int n) I was disappointed to find that their code executes 4x SLOWER than the standard sqrt() function!{ unsigned int a; for (a = 0; n >= (2 * a) + 1; n -= (2 * a++) + 1); return a; } No soup for you! If you've got an implementation of calculating the square root of a number, then post it here. I tested 1,000,000 calculations using each function and timed the calculations using a high precision timer. : Re: C Square Root Functions: kingthomas May 06, 2010, 09:48:41 PM
This one comes from the Taylor Series for √(1 + x):
: // calculates the square root based on the Taylor series // higher precision is slower, but more accurate // default is somewhat arbitrary float sqrt(float operand, int precision = 5) { // can't square root a negative (though this series might give you an interesting answer) if (operand < 0) return operand; // x = operand - 1, since the sum is for x + 1 int x = operand - 1; // the first is always 1 float result = 1; for (int n = 1; n < precision; n++) { result += (pow(-1, n) * factorial(2 * n)) / ((1 - 2 * n) * pow(factorial(n), 2) * pow(4, n)) * pow(x, n); } return result; } By the way, it only works for 0 - 2. HA HA HA! |

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