I've got a fixed point class (10.22) and I have a need of a pow, a sqrt, an exp and a log function.
Alas I have no idea where to even start on this. Can anyone provide me with some links to useful articles or, better yet, provide me with some code?
I'm assuming that once I have an exp function then it becomes relatively easy to implement pow and sqrt as they just become.
pow( x, y ) => exp( y * log( x ) )
sqrt( x ) => pow( x, 0.5 )
Its just those exp and log functions that I'm finding difficult (as though I remember a few of my log rules, I can't remember much else about them).
Presumably, there would also be a faster method for sqrt and pow so any pointers on that front would be appreciated even if its just to say use the methods i outline above.
Please note: This HAS to be cross platform and in pure C/C++ code so I cannot use any assembler optimisations.
A very simple solution is to use a decent table-driven approximation. You don't actually need a lot of data if you reduce your inputs correctly. exp(a)==exp(a/2)*exp(a/2), which means you really only need to calculate exp(x) for 1 < x < 2. Over that range, a runga-kutta approximation would give reasonable results with ~16 entries IIRC.
Similarly, sqrt(a) == 2 * sqrt(a/4) == sqrt(4*a) / 2 which means you need only table entries for 1 < a < 4. Log(a) is a bit harder: log(a) == 1 + log(a/e). This is a rather slow iteration, but log(1024) is only 6.9 so you won't have many iterations.
You'd use a similar "integer-first" algorithm for pow: pow(x,y)==pow(x, floor(y)) * pow(x, frac(y)). This works because pow(double, int) is trivial (divide and conquer).
[edit] For the integral component of log(a), it may be useful to store a table 1, e, e^2, e^3, e^4, e^5, e^6, e^7 so you can reduce log(a) == n + log(a/e^n) by a simple hardcoded binary search of a in that table. The improvement from 7 to 3 steps isn't so big, but it means you only have to divide once by e^n instead of n times by e.
[edit 2]
And for that last log(a/e^n) term, you can use log(a/e^n) = log((a/e^n)^8)/8 - each iteration produces 3 more bits by table lookup. That keeps your code and table size small. This is typically code for embedded systems, and they don't have large caches.
[edit 3]
That's still not to smart on my side. log(a) = log(2) + log(a/2). You can just store the fixed-point value log2=0.6931471805599, count the number of leading zeroes, shift a into the range used for your lookup table, and multiply that shift (integer) by the fixed-point constant log2. Can be as low as 3 instructions.
Using e for the reduction step just gives you a "nice" log(e)=1.0 constant but that's false optimization. 0.6931471805599 is just as good a constant as 1.0; both are 32 bits constants in 10.22 fixed point. Using 2 as the constant for range reduction allows you to use a bit shift for a division.
[edit 5]
And since you're storing it in Q10.22, you can better store log(65536)=11.09035488. (16 x log(2)). The "x16" means that we've got 4 more bits of precision available.
You still get the trick from edit 2, log(a/2^n) = log((a/2^n)^8)/8. Basically, this gets you a result (a + b/8 + c/64 + d/512) * 0.6931471805599 - with b,c,d in the range [0,7]. a.bcd really is an octal number. Not a surprise since we used 8 as the power. (The trick works equally well with power 2, 4 or 16.)
[edit 4]
Still had an open end. pow(x, frac(y) is just pow(sqrt(x), 2 * frac(y)) and we have a decent 1/sqrt(x). That gives us the far more efficient approach. Say frac(y)=0.101 binary, i.e. 1/2 plus 1/8. Then that means x^0.101 is (x^1/2 * x^1/8). But x^1/2 is just sqrt(x) and x^1/8 is (sqrt(sqrt(sqrt(x))). Saving one more operation, Newton-Raphson NR(x) gives us 1/sqrt(x) so we calculate 1.0/(NR(x)*NR((NR(NR(x))). We only invert the end result, don't use the sqrt function directly.
Below is an example C implementation of Clay S. Turner's fixed-point log base 2 algorithm[1]. The algorithm doesn't require any kind of look-up table. This can be useful on systems where memory constraints are tight and the processor lacks an FPU, such as is the case with many microcontrollers. Log base e and log base 10 are then also supported by using the property of logarithms that, for any base n:
logₘ(x)
logₙ(x) = ───────
logₘ(n)
where, for this algorithm, m equals 2.
A nice feature of this implementation is that it supports variable precision: the precision can be determined at runtime, at the expense of range. The way I've implemented it, the processor (or compiler) must be capable of doing 64-bit math for holding some intermediate results. It can be easily adapted to not require 64-bit support, but the range will be reduced.
When using these functions, x is expected to be a fixed-point value scaled according to the
specified precision. For instance, if precision is 16, then x should be scaled by 2^16 (65536). The result is a fixed-point value with the same scale factor as the input. A return value of INT32_MIN represents negative infinity. A return value of INT32_MAX indicates an error and errno will be set to EINVAL, indicating that the input precision was invalid.
#include <errno.h>
#include <stddef.h>
#include "log2fix.h"
#define INV_LOG2_E_Q1DOT31 UINT64_C(0x58b90bfc) // Inverse log base 2 of e
#define INV_LOG2_10_Q1DOT31 UINT64_C(0x268826a1) // Inverse log base 2 of 10
int32_t log2fix (uint32_t x, size_t precision)
{
int32_t b = 1U << (precision - 1);
int32_t y = 0;
if (precision < 1 || precision > 31) {
errno = EINVAL;
return INT32_MAX; // indicates an error
}
if (x == 0) {
return INT32_MIN; // represents negative infinity
}
while (x < 1U << precision) {
x <<= 1;
y -= 1U << precision;
}
while (x >= 2U << precision) {
x >>= 1;
y += 1U << precision;
}
uint64_t z = x;
for (size_t i = 0; i < precision; i++) {
z = z * z >> precision;
if (z >= 2U << (uint64_t)precision) {
z >>= 1;
y += b;
}
b >>= 1;
}
return y;
}
int32_t logfix (uint32_t x, size_t precision)
{
uint64_t t;
t = log2fix(x, precision) * INV_LOG2_E_Q1DOT31;
return t >> 31;
}
int32_t log10fix (uint32_t x, size_t precision)
{
uint64_t t;
t = log2fix(x, precision) * INV_LOG2_10_Q1DOT31;
return t >> 31;
}
The code for this implementation also lives at Github, along with a sample/test program that illustrates how to use this function to compute and display logarithms from numbers read from standard input.
[1] C. S. Turner, "A Fast Binary Logarithm Algorithm", IEEE Signal Processing Mag., pp. 124,140, Sep. 2010.
A good starting point is Jack Crenshaw's book, "Math Toolkit for Real-Time Programming". It has a good discussion of algorithms and implementations for various transcendental functions.
Check my fixed point sqrt implementation using only integer operations.
It was fun to invent. Quite old now.
https://groups.google.com/forum/?hl=fr%05aacf5997b615c37&fromgroups#!topic/comp.lang.c/IpwKbw0MAxw/discussion
Otherwise check the CORDIC set of algorithms. That's the way to implement all the functions you listed and the trigonometric functions.
EDIT : I published the reviewed source on GitHub here
Related
In runtime I have 2 ranges defined by their uint32_t borders a..b and c..d. The first range tends to be much greater than the second: 8 < (b - a) / (d - c) < 64.
Exact limits: a >= 0, b <= 2^31 - 1, c >= 0, d <= 2^20 - 1.
I need a routine that performs linear mapping of an integer from the first range onto the second one: f(uint32_t x) -> round_to_uint32_t((float)(x - a) / (b - a) * (d - c) + c).
When b - a >= d - c it is important to mantain the ratio as close to ideal as possible, otherwise in cases when element from [a; b] can be mapped on more than one integer from [c; d] it is okay to return any of these integers.
Sounds like a simple ratio problem and was already answered in many questions like
Convert a number range to another range, maintaining ratio
but here I need a really really fast solution.
This routine is a pivotal part of a specialized sorting algorithm and will be called at least once for every element of a sorted array.
SIMD solution is also acceptable if it doesn't drop overall performance.
Actual runtime division (FP and integer) is very slow so you definitely want to avoid that. The way you wrote that expression probably compiles to include a division because FP math is not associative (without -ffast-math); the compiler can't turn x / foo * bar into x * (bar/foo) for you, even though that's very good with loop-invariant bar/foo. You do need either floating point or 64-bit integers to avoid overflow in a multiply, but only FP lets you reuse a non-integer loop-invariant division result.
_mm256_fmadd_ps looks like the obvious way to go, with a pre-computed loop-invariant value for the multiplier (d - c) / (b - a). If float rounding isn't a problem for doing it strictly in order (multiply then divide), it's probably ok to do this inexact division first, outside the loop. Like
_mm256_set1_ps((d - c) / (double)(b - a)). Using double for this calculation avoids rounding error during conversion to FP of the division operands.
You're reusing the same a,b,c,d for many x, presumably coming from contiguous memory. You're using the result as part of a memory address so you do eventually need the results back from SIMD into integer registers, unfortunately. (Possibly with AVX512 scatter stores you could avoid that.)
Modern x86 CPUs have 2/clock load throughput so probably your best bet for getting 8x uint32_t back into integer registers is a vector store / integer reload, instead of spending 2 uops per element for ALU shuffle stuff. That has some latency so I'd suggest converting into a tmp buffer of maybe 16 or 32 ints (64 or 128 bytes), i.e. 2x or 4x __m256i before looping through that scalar.
Or maybe alternate converting and storing one vector then looping over the 8 elements of another one that you converted earlier. i.e. software pipelining. Out-of-order execution can hide latency but you're already going to be stretching its latency-hiding capability for cache misses for whatever you're doing with memory.
Depending on your CPU (e.g. Haswell or some Skylake), using 256-bit vector instructions might cap your max turbo slightly lower than it would otherwise. You might consider only doing vectors of 4 at once but then you're spending more uops per element.
If not SIMD, then even scalar C++ fma() is still good, for vfmadd213sd, but using intrinsics is a very convenient way to get rounding (instead of truncation) from float -> int (vcvtps2dq rather than vcvttps2dq).
Note that uint32_t <-> float conversion isn't directly available until AVX512. For scalar you can just convert to/from int64_t with truncation / zero-extension for the unsigned low half.
It's very convenient that (as discussed in comments) your inputs are range-limited so if you interpret them as signed integers they have the same value (signed non-negative). Both x and x-a (and b-a) are known to be positive and <= INT32_MAX i.e 0x7FFFFFFF. (Or at least non-negative. Zero is fine.)
Float Rounding
For SIMD, single-precision float is very good for SIMD throughput. Efficient packed-conversion to/from signed int32_t. But not every int32_t can be exactly represented as a float. Larger values get rounded to the nearest even, nearest multiple of 2^2, 2^3, or more the farther above 2^24 the value is.
Using SIMD double is possible but requires some shuffling.
I don't think float is usually a problem for the formula as-written with (float)(x-a). If the b-a input range is large, that means both ranges are large and rounding error isn't going to map all possible x values into the same output. Depending on the multiplier, the input rounding error might be worse than the output rounding error, maybe leaving some representable output floats unused for higher x-a values.
But if we want to factor out the -a * (d - c) / (b - a) part and combine it with the +c at the end, then
We potentially have precision loss from catastrophic cancellation in that value to be added.
We need to do (float)x on the raw input value. If a is huge and b-a is small, i.e. a small range near the top of the possible input range, rounding error can map all possible x values to the same float.
To make best use of FMA, we want to do the +c before converting back to integer, which again risks output rounding error if the d-c is a small output range but c is huge. In your case not a problem; with d <= 2^20 - 1 we know that float can exactly represent every output integer value in that c..d range.
If you didn't have the input range constraint, you could range-shift to/from signed before the scaling by using integer (x-a)+0x80000000U on input and ...+c+0x80000000U on output (after rounding to nearest int32_t). But that would introduce huge float rounding error for small uint32_t inputs (close to 0) which get range-shifted to close to INT_MIN.
We don't need to range-shift for the b-a or d-c because the + or - or XOR with 0x80000000U would cancel out in the subtractions.
Example:
The const vectors should be hoisted out of a loop by the compiler after this inlines,
or you can do that manually.
This requires AVX1 + FMA (e.g. AMD Piledriver or Intel Haswell or later). Untested, sorry I didn't even throw this on Godbolt to see if it compiles.
// fastest but not safe if b-a is small and a > 2^24
static inline
__m256i range_scale_fast_fma(__m256i data, uint32_t a, uint32_t b, uint32_t c, uint32_t d)
{
// avoid rounding errors when computing the scale factor, but convert double->float on the final result
double scale_scalar = (d - c) / (double)(b - a);
const __m256 scale = _mm256_set1_ps(scale_scalar);
const __m256 add = _m256_set1_ps(-a*scale_scalar + c);
// (x-a) * scale + c
// = x * scale + (-a*scale + c) but with different rounding error from doing -a*scale + c
__m256 in = _mm256_cvtepi32_ps(data);
__m256 out = _mm256_fmadd_ps(in, scale, add);
return _mm256_cvtps_epi32(out); // convert back with round to nearest-even
// _mm256_cvttps_epi32 truncates, matching C rounding; maybe good for scalar testing
}
Or a safer version, doing the input range-shift with integer: You could easily avoid FMA here if necessary for portability (just AVX1) and use an integer add for the output, too. But we know the output range is small enough that it can always exactly represent any integer
static inline
__m256i range_scale_safe_fma(__m256i data, uint32_t a, uint32_t b, uint32_t c, uint32_t d)
{
// avoid rounding errors when computing the scale factor, but convert double->float on the final result
const __m256 scale = _mm256_set1_ps((d - c) / (double)(b - a));
const __m256 cvec = _m256_set1_ps(c);
__m256i in_offset = _mm256_add_epi32(data, _mm256_set1_epi32(-a)); // add can more easily fold a load of a memory operand than sub because it's commutative. Only some compilers will do this for you.
__m256 in_fp = _mm256_cvtepi32_ps(in_offset);
__m256 out = _mm256_fmadd_ps(in_fp, scale, _mm256_set1_ps(c)); // in*scale + c
return _mm256_cvtps_epi32(out);
}
Without FMA you could still use vmulps. You might as well convert back to integer before adding c if you're doing that, although vaddps would be safe.
You might use this in a loop like
void foo(uint32_t *arr, ptrdiff_t len)
{
if (len < 24) special case;
alignas(32) uint32_t tmpbuf[16];
// peel half of first iteration for software pipelining / loop rotation
__m256i arrdata = _mm256_loadu_si256((const __m256i*)&arr[0]);
__m256i outrange = range_scale_safe_fma(arrdata);
_mm256_store_si256((__m256i*)tmpbuf, outrange);
// could have used an unsigned loop counter
// since we probably just need an if() special case handler anyway for small len which could give len-23 < 0
for (ptrdiff_t i = 0 ; i < len-(15+8) ; i+=16 ) {
// prep next 8 elements
arrdata = _mm256_loadu_si256((const __m256i*)&arr[i+8]);
outrange = range_scale_safe_fma(arrdata);
_mm256_store_si256((__m256i*)&tmpbuf[8], outrange);
// use first 8 elements
for (int j=0 ; j<8 ; j++) {
use tmpbuf[j] which corresponds to arr[i+j]
}
// prep 8 more for next iteration
arrdata = _mm256_loadu_si256((const __m256i*)&arr[i+16]);
outrange = range_scale_safe_fma(arrdata);
_mm256_store_si256((__m256i*)&tmpbuf[0], outrange);
// use 2nd 8 elements
for (int j=8 ; j<16 ; j++) {
use tmpbuf[j] which corresponds to arr[i+j]
}
}
// use tmpbuf[0..7]
// then cleanup: one vector at a time until < 8 or < 4 with 128-bit vectors, then scalar
}
These variable-names sound dumb but I couldn't think of anything better.
This software pipelining is an optimization; you can just get it working / try it out with a single vector at a time used right away. (Optimize the reload of the first element from a reload to vmovd using _mm_cvtsi128_si32(_mm256_castsi256_si128(outrange)) if you want.)
Special cases
If there cases where you know (b - a) is a power of 2, you could bitscan with tzcnt or bsf, then multiply. (There are intrinsics for those, like GNU C __builtin_ctz() to count trailing zeros.)
Or can you ensure that (b - a) is always a power of 2?
Or better, if (b - a) / (d - c) is an exact power of 2 the whole thing can just be sub / right shift / add.
If you can't always ensure that you'd still need the general case sometimes, but maybe possible to do that efficiently.
My program frequently requires the following calculation to be performed:
Given:
N is a 32-bit integer
D is a 32-bit integer
abs(N) <= abs(D)
D != 0
X is a 32-bit integer of any value
Find:
X * N / D as a rounded integer that is X scaled to N/D (i.e. 10 * 2 / 3 = 7)
Obviously I could just use r=x*n/d directly but I will often get overflow from the x*n. If I instead do r=x*(n/d) then I only get 0 or x due to integer division dropping the fractional component. And then there's r=x*(float(n)/d) but I can't use floats in this case.
Accuracy would be great but isn't as critical as speed and being a deterministic function (always returning the same value given the same inputs).
N and D are currently signed but I could work around them being always unsigned if it helps.
A generic function that works with any value of X (and N and D, as long as N <= D) is ideal since this operation is used in various different ways but I also have a specific case where the value of X is a known constant power of 2 (2048, to be precise), and just getting that specific call sped up would be a big help.
Currently I am accomplishing this using 64-bit multiply and divide to avoid overflow (essentially int multByProperFraction(int x, int n, int d) { return (__int64)x * n / d; } but with some asserts and extra bit fiddling for rounding instead of truncating).
Unfortunately, my profiler is reporting the 64-bit divide function as taking up way too much CPU (this is a 32-bit application). I've tried to reduce how often I need to do this calculation but am running out of ways around it, so I'm trying to figure out a faster method, if it is even possible. In the specific case where X is a constant 2048, I use a bit shift instead of multiply but that doesn't help much.
Tolerate imprecision and use the 16 MSBits of n,d,x
Algorithm
while (|n| > 0xffff) n/2, sh++
while (|x| > 0xffff) x/2, sh++
while (|d| > 0xffff) d/2, sh--
r = n*x/d // A 16x16 to 32 multiply followed by a 32/16-bit divide.
shift r by sh.
When 64 bit divide is expensive, the pre/post processing here may be worth to do a 32-bit divide - which will certainly be the big chunk of CPU.
If the compiler cannot be coaxed into doing a 32-bit/16-bit divide, then skip the while (|d| > 0xffff) d/2, sh-- step and do a 32/32 divide.
Use unsigned math as possible.
The basic correct approach to this is simply (uint64_t)x*n/d. That's optimal assuming d is variable and unpredictable. But if d is constant or changes infrequently, you can pre-generate constants such that exact division by d can be performed as a multiplication followed by a bitshift. A good description of the algorithm, which is roughly what GCC uses internally to transform division by a constant into multiplication, is here:
http://ridiculousfish.com/blog/posts/labor-of-division-episode-iii.html
I'm not sure how easy it is to make it work for a "64/32" division (i.e. dividing the result of (uint64_t)x*n), but you should be able to just break it up into high and low parts if nothing else.
Note that these algorithms are also available as libdivide.
I've now benchmarked several possible solutions, including weird/clever ones from other sources like combining 32-bit div & mod & add or using peasant math, and here are my conclusions:
First, if you are only targeting Windows and using VSC++, just use MulDiv(). It is quite fast (faster than directly using 64-bit variables in my tests) while still being just as accurate and rounding the result for you. I could not find any superior method to do this kind of thing on Windows with VSC++, even taking into account restrictions like unsigned-only and N <= D.
However, in my case having a function with deterministic results even across platforms is even more important than speed. On another platform I was using as a test, the 64-bit divide is much, much slower than the 32-bit one when using the 32-bit libraries, and there is no MulDiv() to use. The 64-bit divide on this platform takes ~26x as long as a 32-bit divide (yet the 64-bit multiply is just as fast as the 32-bit version...).
So if you have a case like me, I will share the best results I got, which turned out to be just optimizations of chux's answer.
Both of the methods I will share below make use of the following function (though the compiler-specific intrinsics only actually helped in speed with MSVC in Windows):
inline u32 bitsRequired(u32 val)
{
#ifdef _MSC_VER
DWORD r = 0;
_BitScanReverse(&r, val | 1);
return r+1;
#elif defined(__GNUC__) || defined(__clang__)
return 32 - __builtin_clz(val | 1);
#else
int r = 1;
while (val >>= 1) ++r;
return r;
#endif
}
Now, if x is a constant that's 16-bit in size or smaller and you can pre-compute the bits required, I found the best results in speed and accuracy from this function:
u32 multConstByPropFrac(u32 x, u32 nMaxBits, u32 n, u32 d)
{
//assert(nMaxBits == 32 - bitsRequired(x));
//assert(n <= d);
const int bitShift = bitsRequired(n) - nMaxBits;
if( bitShift > 0 )
{
n >>= bitShift;
d >>= bitShift;
}
// Remove the + d/2 part if don't need rounding
return (x * n + d/2) / d;
}
On the platform with the slow 64-bit divide, the above function ran ~16.75x as fast as return ((u64)x * n + d/2) / d; and with an average 99.999981% accuracy (comparing difference in return value from expected to range of x, i.e. returning +/-1 from expected when x is 2048 would be 100 - (1/2048 * 100) = 99.95% accurate) when testing it with a million or so randomized inputs where roughly half of them would normally have been an overflow. Worst-case accuracy was 99.951172%.
For the general use case, I found the best results from the following (and without needing to restrict N <= D to boot!):
u32 scaleToFraction(u32 x, u32 n, u32 d)
{
u32 bits = bitsRequired(x);
int bitShift = bits - 16;
if( bitShift < 0 ) bitShift = 0;
int sh = bitShift;
x >>= bitShift;
bits = bitsRequired(n);
bitShift = bits - 16;
if( bitShift < 0 ) bitShift = 0;
sh += bitShift;
n >>= bitShift;
bits = bitsRequired(d);
bitShift = bits - 16;
if( bitShift < 0 ) bitShift = 0;
sh -= bitShift;
d >>= bitShift;
// Remove the + d/2 part if don't need rounding
u32 r = (x * n + d/2) / d;
if( sh < 0 )
r >>= (-sh);
else //if( sh > 0 )
r <<= sh;
return r;
}
On the platform with the slow 64-bit divide, the above function ran ~18.5x as fast as using 64-bit variables and with 99.999426% average and 99.947479% worst-case accuracy.
I was able to get more speed or more accuracy by messing with the shifting, such as trying to not shift all the way down to 16-bit if it wasn't strictly necessary, but any increase in speed came at a high cost in accuracy and vice versa.
None of the other methods I tested came even close to the same speed or accuracy, most being slower than just using the 64-bit method or having huge loss in precision, so not worth going into.
Obviously, no guarantee that anyone else will get similar results on other platforms!
EDIT: Replaced some bit-twiddling hacks with plain code that actually ran faster anyway by letting the compiler do its job.
I want to determine (in c++) if one float number is the multiplicative inverse of another float number. The problem is that i have to use a third variable to do it. For instance this code:
float x=5,y=0.2;
if(x==(1/y)) cout<<"They are the multiplicative inverse of eachother"<<endl;
else cout<<"They are NOT the multiplicative inverse of eachother"<<endl;
will output: "they are not..." which is wrong and this code:
float x=5,y=0.2,z;
z=1/y;
if(x==z) cout<<"They are the multiplicative inverse of eachother"<<endl;
else cout<<"They are NOT the multiplicative inverse of eachother"<<endl;
will output: "they are..." which is right.why is this happening?
The Float Precision Problem
You have two problems here, but both come from the same root
You can't compare floats precisely. You can't subtract or divide them precisely. You can't count anything for them precisely. Any operation with them could (and almost always does) bring some error into the result. Even a=0.2f is not a precise operation. The deeper reasons of that are very well explained by the authors of the other answers here. (My thanks and votes to them for that.)
Here comes your first and more simple error. You should never, never, never, never, NEVER use on them == or its equivalent in any language.
Instead of a==b, use Abs(a-b)<HighestPossibleError instead.
But this is not the sole problem in your task.
Abs(1/y-x)<HighestPossibleError won't work, either. At least, it won't work often enough. Why?
Let's take pair x=1000 and y=0.001. Let's take the "starting" relative error of y for 10-6.
(Relative error = error/value).
Relative errors of values are adding to at multiplication and division.
1/y is about 1000. Its relative error is the same 10-6. ("1" hasn't errors)
That makes absolute error =1000*10-6=0.001. When you subtract x later, that error will be all that remains. (Absolute errors are adding to at adding and subtracting, and the error of x is negligibly small.) Surely, you are not counting on so large errors, HighestPossibleError would be surely set lower and your program would throw off a good pair of x,y
So, the next two rule for float operations: try not to divide greater valuer by lesser one and God save you from subtracting the close values after that.
There are two simple ways to escape this problem.
By founding what of x,y has the greater abs value and divide 1 by the greater one and only later to subtract the lesser one.
If you want to compare 1/y against x, while you are working yet with letters, not values, and your operations make no errors, multiply the both sides of comparison by y
and you have 1 against x*y. (Usually you should check signs in that operation, but here we use abs values, so, it is clean.) The result comparison has no division at all.
In a shorter way:
1/y V x <=> y*(1/y) V x*y <=> 1 V x*y
We already know that such comparison as 1 against x*y should be done so:
const float HighestPossibleError=1e-10;
if(Abs(x*y-1.0)<HighestPossibleError){...
That is all.
P.S. If you really need it all on one line, use:
if(Abs(x*y-1.0)<1e-10){...
But it is bad style. I wouldn't advise it.
P.P.S. In your second example the compiler optimizes the code so, that it sets z to 5 before running any code. So, checking 5 against 5 works even for floats.
The problem is that 0.2 cannot be represented exactly in binary, because its binary expansion has an infinite number of digits:
1/5: 0.0011001100110011001100110011001100110011...
This is similar to how 1/3 cannot be represented exactly in decimal. Since x is stored in a float which has a finite number of bits, these digits will get cut off at some point, for example:
x: 0.0011001100110011001100110011001
The problem arises because CPUs often use a higher precision internally, so when you've just calculated 1/y, the result will have more digits, and when you load x to compare them, x will get extended to match the internal precision of the CPU.
1/y: 0.0011001100110011001100110011001100110011001100110011
x: 0.0011001100110011001100110011001000000000000000000000
So when you do a direct bit-by-bit comparison, they are different.
In your second example, however, storing the result into a variable means it gets truncated before doing the comparison, so comparing them at this precision, they're equal:
x: 0.0011001100110011001100110011001
z: 0.0011001100110011001100110011001
Many compilers have switches you can enable to force intermediate values to be truncated at every step for consistency, however the usual advice is to avoid doing direct comparisons between floating-point values and instead check if they differ by less than some epsilon value, which is what Gangnus is suggesting.
You will have to precisely define what it means for two approximations to be multiplicative inverses. Otherwise, you won't know what it is you're supposed to be testing.
0.2 has no exact binary representation. If you store numbers that have no exact representation with limited precision, you won't get answers that are exactly correct.
The same things happens in decimal. For example, 1/3 has no exact decimal representation. You can store it as .333333. But then you have a problem. Are 3 and .333333 multiplicative inverses? If you multiply them, you get .999999. If you want the answer to be "yes" you'll have to create a test for multiplicative inverses that isn't as simple as multiplying and testing for equality to 1.
The same thing happens with binary.
The discussions in other replies are great and so I won't repeat any of them, but there's no code. Here's a little bit of code to actually check if a pair of floats gives exactly 1.0 when multiplied.
The code makes a few assumptions/assertions (which are normally met on the x86 platform):
- float's are 32-bit binary (AKA single precision) IEEE-754
- either int's or long's are 32-bit (I decided not to rely on the availability of uint32_t)
- memcpy() copies floats to ints/longs such that 8873283.0f becomes 0x4B076543 (i.e. certain "endianness" is expected)
One extra assumption is this:
- it receives the actual floats that * would multiply (i.e. multiplication of floats wouldn't use higher precision values that the math hardware/library can use internally)
#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <assert.h>
#define C_ASSERT(expr) extern char CAssertExtern[(expr)?1:-1]
#if UINT_MAX >= 0xFFFFFFFF
typedef unsigned int uint32;
#else
typedef unsigned long uint32;
#endif
typedef unsigned long long uint64;
C_ASSERT(CHAR_BIT == 8);
C_ASSERT(sizeof(uint32) == 4);
C_ASSERT(sizeof(float) == 4);
int ProductIsOne(float f1, float f2)
{
uint32 m1, m2;
int e1, e2, s1, s2;
int e;
uint64 m;
// Make sure floats are 32-bit IEE754 and
// reinterpreted as integers as we expect
{
static const float testf = 8873283.0f;
uint32 testi;
memcpy(&testi, &testf, sizeof(testf));
assert(testi == 0x4B076543);
}
memcpy(&m1, &f1, sizeof(f1));
s1 = m1 >= 0x80000000;
m1 &= 0x7FFFFFFF;
e1 = m1 >> 23;
m1 &= 0x7FFFFF;
if (e1 > 0) m1 |= 0x800000;
memcpy(&m2, &f2, sizeof(f2));
s2 = m2 >= 0x80000000;
m2 &= 0x7FFFFFFF;
e2 = m2 >> 23;
m2 &= 0x7FFFFF;
if (e2 > 0) m2 |= 0x800000;
if (e1 == 0xFF || e2 == 0xFF || s1 != s2) // Inf, NaN, different signs
return 0;
m = (uint64)m1 * m2;
if (!m || (m & (m - 1))) // not a power of 2
return 0;
e = e1 + !e1 - 0x7F - 23 + e2 + !e2 - 0x7F - 23;
while (m > 1) m >>= 1, e++;
return e == 0;
}
const float testData[][2] =
{
{ .1f, 10.0f },
{ 0.5f, 2.0f },
{ 0.25f, 2.0f },
{ 4.0f, 0.25f },
{ 0.33333333f, 3.0f },
{ 0.00000762939453125f, 131072.0f }, // 2^-17 * 2^17
{ 1.26765060022822940E30f, 7.88860905221011805E-31f }, // 2^100 * 2^-100
{ 5.87747175411143754E-39f, 1.70141183460469232E38f }, // 2^-127 (denormalized) * 2^127
};
int main(void)
{
int i;
for (i = 0; i < sizeof(testData) / sizeof(testData[0]); i++)
printf("%g * %g %c= 1\n",
testData[i][0], testData[i][1],
"!="[ProductIsOne(testData[i][0], testData[i][1])]);
return 0;
}
Output (see at ideone.com):
0.1 * 10 != 1
0.5 * 2 == 1
0.25 * 2 != 1
4 * 0.25 == 1
0.333333 * 3 != 1
7.62939e-06 * 131072 == 1
1.26765e+30 * 7.88861e-31 == 1
5.87747e-39 * 1.70141e+38 == 1
What is striking is that whatever the rounding rule is, you expect the outcome of the two versions to be the same (either twice wrong or twice right)!
Most probably, in the first case a promotion to higher accuracy in the FPU registers takes place when evaluating x==1/y, whereas z= 1/y really stores the single-precision result.
Other contributors have explaine why 5==1/0.2 can fail, I needn't repeat that.
I was always wondering how I can make a function which calculates the power (e.g. 23) myself. In most languages these are included in the standard library, mostly as pow(double x, double y), but how can I write it myself?
I was thinking about for loops, but it think my brain got in a loop (when I wanted to do a power with a non-integer exponent, like 54.5 or negatives 2-21) and I went crazy ;)
So, how can I write a function which calculates the power of a real number? Thanks
Oh, maybe important to note: I cannot use functions which use powers (e.g. exp), which would make this ultimately useless.
Negative powers are not a problem, they're just the inverse (1/x) of the positive power.
Floating point powers are just a little bit more complicated; as you know a fractional power is equivalent to a root (e.g. x^(1/2) == sqrt(x)) and you also know that multiplying powers with the same base is equivalent to add their exponents.
With all the above, you can:
Decompose the exponent in a integer part and a rational part.
Calculate the integer power with a loop (you can optimise it decomposing in factors and reusing partial calculations).
Calculate the root with any algorithm you like (any iterative approximation like bisection or Newton method could work).
Multiply the result.
If the exponent was negative, apply the inverse.
Example:
2^(-3.5) = (2^3 * 2^(1/2)))^-1 = 1 / (2*2*2 * sqrt(2))
AB = Log-1(Log(A)*B)
Edit: yes, this definition really does provide something useful. For example, on an x86, it translates almost directly to FYL2X (Y * Log2(X)) and F2XM1 (2x-1):
fyl2x
fld st(0)
frndint
fsubr st(1),st
fxch st(1)
fchs
f2xmi
fld1
faddp st(1),st
fscale
fstp st(1)
The code ends up a little longer than you might expect, primarily because F2XM1 only works with numbers in the range -1.0..1.0. The fld st(0)/frndint/fsubr st(1),st piece subtracts off the integer part, so we're left with only the fraction. We apply F2XM1 to that, add the 1 back on, then use FSCALE to handle the integer part of the exponentiation.
Typically the implementation of the pow(double, double) function in math libraries is based on the identity:
pow(x,y) = pow(a, y * log_a(x))
Using this identity, you only need to know how to raise a single number a to an arbitrary exponent, and how to take a logarithm base a. You have effectively turned a complicated multi-variable function into a two functions of a single variable, and a multiplication, which is pretty easy to implement. The most commonly chosen values of a are e or 2 -- e because the e^x and log_e(1+x) have some very nice mathematical properties, and 2 because it has some nice properties for implementation in floating-point arithmetic.
The catch of doing it this way is that (if you want to get full accuracy) you need to compute the log_a(x) term (and its product with y) to higher accuracy than the floating-point representation of x and y. For example, if x and y are doubles, and you want to get a high accuracy result, you'll need to come up with some way to store intermediate results (and do arithmetic) in a higher-precision format. The Intel x87 format is a common choice, as are 64-bit integers (though if you really want a top-quality implementation, you'll need to do a couple of 96-bit integer computations, which are a little bit painful in some languages). It's much easier to deal with this if you implement powf(float,float), because then you can just use double for intermediate computations. I would recommend starting with that if you want to use this approach.
The algorithm that I outlined is not the only possible way to compute pow. It is merely the most suitable for delivering a high-speed result that satisfies a fixed a priori accuracy bound. It is less suitable in some other contexts, and is certainly much harder to implement than the repeated-square[root]-ing algorithm that some others have suggested.
If you want to try the repeated square[root] algorithm, begin by writing an unsigned integer power function that uses repeated squaring only. Once you have a good grasp on the algorithm for that reduced case, you will find it fairly straightforward to extend it to handle fractional exponents.
There are two distinct cases to deal with: Integer exponents and fractional exponents.
For integer exponents, you can use exponentiation by squaring.
def pow(base, exponent):
if exponent == 0:
return 1
elif exponent < 0:
return 1 / pow(base, -exponent)
elif exponent % 2 == 0:
half_pow = pow(base, exponent // 2)
return half_pow * half_pow
else:
return base * pow(base, exponent - 1)
The second "elif" is what distinguishes this from the naïve pow function. It allows the function to make O(log n) recursive calls instead of O(n).
For fractional exponents, you can use the identity a^b = C^(b*log_C(a)). It's convenient to take C=2, so a^b = 2^(b * log2(a)). This reduces the problem to writing functions for 2^x and log2(x).
The reason it's convenient to take C=2 is that floating-point numbers are stored in base-2 floating point. log2(a * 2^b) = log2(a) + b. This makes it easier to write your log2 function: You don't need to have it be accurate for every positive number, just on the interval [1, 2). Similarly, to calculate 2^x, you can multiply 2^(integer part of x) * 2^(fractional part of x). The first part is trivial to store in a floating point number, for the second part, you just need a 2^x function over the interval [0, 1).
The hard part is finding a good approximation of 2^x and log2(x). A simple approach is to use Taylor series.
Per definition:
a^b = exp(b ln(a))
where exp(x) = 1 + x + x^2/2 + x^3/3! + x^4/4! + x^5/5! + ...
where n! = 1 * 2 * ... * n.
In practice, you could store an array of the first 10 values of 1/n!, and then approximate
exp(x) = 1 + x + x^2/2 + x^3/3! + ... + x^10/10!
because 10! is a huge number, so 1/10! is very small (2.7557319224⋅10^-7).
Wolfram functions gives a wide variety of formulae for calculating powers. Some of them would be very straightforward to implement.
For positive integer powers, look at exponentiation by squaring and addition-chain exponentiation.
Using three self implemented functions iPow(x, n), Ln(x) and Exp(x), I'm able to compute fPow(x, a), x and a being doubles. Neither of the functions below use library functions, but just iteration.
Some explanation about functions implemented:
(1) iPow(x, n): x is double, n is int. This is a simple iteration, as n is an integer.
(2) Ln(x): This function uses the Taylor Series iteration. The series used in iteration is Σ (from int i = 0 to n) {(1 / (2 * i + 1)) * ((x - 1) / (x + 1)) ^ (2 * n + 1)}. The symbol ^ denotes the power function Pow(x, n) implemented in the 1st function, which uses simple iteration.
(3) Exp(x): This function, again, uses the Taylor Series iteration. The series used in iteration is Σ (from int i = 0 to n) {x^i / i!}. Here, the ^ denotes the power function, but it is not computed by calling the 1st Pow(x, n) function; instead it is implemented within the 3rd function, concurrently with the factorial, using d *= x / i. I felt I had to use this trick, because in this function, iteration takes some more steps relative to the other functions and the factorial (i!) overflows most of the time. In order to make sure the iteration does not overflow, the power function in this part is iterated concurrently with the factorial. This way, I overcame the overflow.
(4) fPow(x, a): x and a are both doubles. This function does nothing but just call the other three functions implemented above. The main idea in this function depends on some calculus: fPow(x, a) = Exp(a * Ln(x)). And now, I have all the functions iPow, Ln and Exp with iteration already.
n.b. I used a constant MAX_DELTA_DOUBLE in order to decide in which step to stop the iteration. I've set it to 1.0E-15, which seems reasonable for doubles. So, the iteration stops if (delta < MAX_DELTA_DOUBLE) If you need some more precision, you can use long double and decrease the constant value for MAX_DELTA_DOUBLE, to 1.0E-18 for example (1.0E-18 would be the minimum).
Here is the code, which works for me.
#define MAX_DELTA_DOUBLE 1.0E-15
#define EULERS_NUMBER 2.718281828459045
double MathAbs_Double (double x) {
return ((x >= 0) ? x : -x);
}
int MathAbs_Int (int x) {
return ((x >= 0) ? x : -x);
}
double MathPow_Double_Int(double x, int n) {
double ret;
if ((x == 1.0) || (n == 1)) {
ret = x;
} else if (n < 0) {
ret = 1.0 / MathPow_Double_Int(x, -n);
} else {
ret = 1.0;
while (n--) {
ret *= x;
}
}
return (ret);
}
double MathLn_Double(double x) {
double ret = 0.0, d;
if (x > 0) {
int n = 0;
do {
int a = 2 * n + 1;
d = (1.0 / a) * MathPow_Double_Int((x - 1) / (x + 1), a);
ret += d;
n++;
} while (MathAbs_Double(d) > MAX_DELTA_DOUBLE);
} else {
printf("\nerror: x < 0 in ln(x)\n");
exit(-1);
}
return (ret * 2);
}
double MathExp_Double(double x) {
double ret;
if (x == 1.0) {
ret = EULERS_NUMBER;
} else if (x < 0) {
ret = 1.0 / MathExp_Double(-x);
} else {
int n = 2;
double d;
ret = 1.0 + x;
do {
d = x;
for (int i = 2; i <= n; i++) {
d *= x / i;
}
ret += d;
n++;
} while (d > MAX_DELTA_DOUBLE);
}
return (ret);
}
double MathPow_Double_Double(double x, double a) {
double ret;
if ((x == 1.0) || (a == 1.0)) {
ret = x;
} else if (a < 0) {
ret = 1.0 / MathPow_Double_Double(x, -a);
} else {
ret = MathExp_Double(a * MathLn_Double(x));
}
return (ret);
}
It's an interesting exercise. Here's some suggestions, which you should try in this order:
Use a loop.
Use recursion (not better, but interesting none the less)
Optimize your recursion vastly by using divide-and-conquer
techniques
Use logarithms
You can found the pow function like this:
static double pows (double p_nombre, double p_puissance)
{
double nombre = p_nombre;
double i=0;
for(i=0; i < (p_puissance-1);i++){
nombre = nombre * p_nombre;
}
return (nombre);
}
You can found the floor function like this:
static double floors(double p_nomber)
{
double x = p_nomber;
long partent = (long) x;
if (x<0)
{
return (partent-1);
}
else
{
return (partent);
}
}
Best regards
A better algorithm to efficiently calculate positive integer powers is repeatedly square the base, while keeping track of the extra remainder multiplicands. Here is a sample solution in Python that should be relatively easy to understand and translate into your preferred language:
def power(base, exponent):
remaining_multiplicand = 1
result = base
while exponent > 1:
remainder = exponent % 2
if remainder > 0:
remaining_multiplicand = remaining_multiplicand * result
exponent = (exponent - remainder) / 2
result = result * result
return result * remaining_multiplicand
To make it handle negative exponents, all you have to do is calculate the positive version and divide 1 by the result, so that should be a simple modification to the above code. Fractional exponents are considerably more difficult, since it means essentially calculating an nth-root of the base, where n = 1/abs(exponent % 1) and multiplying the result by the result of the integer portion power calculation:
power(base, exponent - (exponent % 1))
You can calculate roots to a desired level of accuracy using Newton's method. Check out wikipedia article on the algorithm.
I am using fixed point long arithmetics and my pow is log2/exp2 based. Numbers consist of:
int sig = { -1; +1 } signum
DWORD a[A+B] number
A is number of DWORDs for integer part of number
B is number of DWORDs for fractional part
My simplified solution is this:
//---------------------------------------------------------------------------
longnum exp2 (const longnum &x)
{
int i,j;
longnum c,d;
c.one();
if (x.iszero()) return c;
i=x.bits()-1;
for(d=2,j=_longnum_bits_b;j<=i;j++,d*=d)
if (x.bitget(j))
c*=d;
for(i=0,j=_longnum_bits_b-1;i<_longnum_bits_b;j--,i++)
if (x.bitget(j))
c*=_longnum_log2[i];
if (x.sig<0) {d.one(); c=d/c;}
return c;
}
//---------------------------------------------------------------------------
longnum log2 (const longnum &x)
{
int i,j;
longnum c,d,dd,e,xx;
c.zero(); d.one(); e.zero(); xx=x;
if (xx.iszero()) return c; //**** error: log2(0) = infinity
if (xx.sig<0) return c; //**** error: log2(negative x) ... no result possible
if (d.geq(x,d)==0) {xx=d/xx; xx.sig=-1;}
i=xx.bits()-1;
e.bitset(i); i-=_longnum_bits_b;
for (;i>0;i--,e>>=1) // integer part
{
dd=d*e;
j=dd.geq(dd,xx);
if (j==1) continue; // dd> xx
c+=i; d=dd;
if (j==2) break; // dd==xx
}
for (i=0;i<_longnum_bits_b;i++) // fractional part
{
dd=d*_longnum_log2[i];
j=dd.geq(dd,xx);
if (j==1) continue; // dd> xx
c.bitset(_longnum_bits_b-i-1); d=dd;
if (j==2) break; // dd==xx
}
c.sig=xx.sig;
c.iszero();
return c;
}
//---------------------------------------------------------------------------
longnum pow (const longnum &x,const longnum &y)
{
//x^y = exp2(y*log2(x))
int ssig=+1; longnum c; c=x;
if (y.iszero()) {c.one(); return c;} // ?^0=1
if (c.iszero()) return c; // 0^?=0
if (c.sig<0)
{
c.overflow(); c.sig=+1;
if (y.isreal()) {c.zero(); return c;} //**** error: negative x ^ noninteger y
if (y.bitget(_longnum_bits_b)) ssig=-1;
}
c=exp2(log2(c)*y); c.sig=ssig; c.iszero();
return c;
}
//---------------------------------------------------------------------------
where:
_longnum_bits_a = A*32
_longnum_bits_b = B*32
_longnum_log2[i] = 2 ^ (1/(2^i)) ... precomputed sqrt table
_longnum_log2[0]=sqrt(2)
_longnum_log2[1]=sqrt[tab[0])
_longnum_log2[i]=sqrt(tab[i-1])
longnum::zero() sets *this=0
longnum::one() sets *this=+1
bool longnum::iszero() returns (*this==0)
bool longnum::isnonzero() returns (*this!=0)
bool longnum::isreal() returns (true if fractional part !=0)
bool longnum::isinteger() returns (true if fractional part ==0)
int longnum::bits() return num of used bits in number counted from LSB
longnum::bitget()/bitset()/bitres()/bitxor() are bit access
longnum.overflow() rounds number if there was a overflow X.FFFFFFFFFF...FFFFFFFFF??h -> (X+1).0000000000000...000000000h
int longnum::geq(x,y) is comparition |x|,|y| returns 0,1,2 for (<,>,==)
All you need to understand this code is that numbers in binary form consists of sum of powers of 2, when you need to compute 2^num then it can be rewritten as this
2^(b(-n)*2^(-n) + ... + b(+m)*2^(+m))
where n are fractional bits and m are integer bits. multiplication/division by 2 in binary form is simple bit shifting so if you put it all together you get code for exp2 similar to my. log2 is based on binaru search...changing the result bits from MSB to LSB until it matches searched value (very similar algorithm as for fast sqrt computation). hope this helps clarify things...
A lot of approaches are given in other answers. Here is something that I thought may be useful in case of integral powers.
In the case of integer power x of nx, the straightforward approach would take x-1 multiplications. In order to optimize this, we can use dynamic programming and reuse an earlier multiplication result to avoid all x multiplications. For example, in 59, we can, say, make batches of 3, i.e. calculate 53 once, get 125 and then cube 125 using the same logic, taking only 4 multiplcations in the process, instead of 8 multiplications with the straightforward way.
The question is what is the ideal size of the batch b so that the number of multiplications is minimum. So let's write the equation for this. If f(x,b) is the function representing the number of multiplications entailed in calculating nx using the above method, then
Explanation: A product of batch of p numbers will take p-1 multiplications. If we divide x multiplications into b batches, there would be (x/b)-1 multiplications required inside each batch, and b-1 multiplications required for all b batches.
Now we can calculate the first derivative of this function with respect to b and equate it to 0 to get the b for the least number of multiplications.
Now put back this value of b into the function f(x,b) to get the least number of multiplications:
For all positive x, this value is lesser than the multiplications by the straightforward way.
maybe you can use taylor series expansion. the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor's series are named after Brook Taylor who introduced them in 1715.
The standard representation of constant e as the sum of the infinite series is very inefficient for computation, because of many division operations. So are there any alternative ways to compute the constant efficiently?
Since it's not possible to calculate every digit of 'e', you're going to have to pick a stopping point.
double precision: 16 decimal digits
For practical applications, "the 64-bit double precision floating point value that is as close as possible to the true value of 'e' -- approximately 16 decimal digits" is more than adequate.
As KennyTM said, that value has already been pre-calculated for you in the math library.
If you want to calculate it yourself, as Hans Passant pointed out, factorial already grows very fast.
The first 22 terms in the series is already overkill for calculating to that precision -- adding further terms from the series won't change the result if it's stored in a 64 bit double-precision floating point variable.
I think it will take you longer to blink than for your computer to do 22 divides. So I don't see any reason to optimize this further.
thousands, millions, or billions of decimal digits
As Matthieu M. pointed out, this value has already been calculated, and you can download it from Yee's web site.
If you want to calculate it yourself, that many digits won't fit in a standard double-precision floating-point number.
You need a "bignum" library.
As always, you can either use one of the many free bignum libraries already available, or re-invent the wheel by building your own yet another bignum library with its own special quirks.
The result -- a long file of digits -- is not terribly useful, but programs to calculate it are sometimes used as benchmarks to test the performance and accuracy of "bignum" library software, and as stress tests to check the stability and cooling capacity of new machine hardware.
One page very briefly describes the algorithms Yee uses to calculate mathematical constants.
The Wikipedia "binary splitting" article goes into much more detail.
I think the part you are looking for is the number representation:
instead of internally storing all numbers as a long series of digits before and after the decimal point (or a binary point),
Yee stores each term and each partial sum as a rational number -- as two integers, each of which is a long series of digits.
For example, say one of the worker CPUs was assigned the partial sum,
... 1/4! + 1/5! + 1/6! + ... .
Instead of doing the division first for each term, and then adding, and then returning a single million-digit fixed-point result to the manager CPU:
// extended to a million digits
1/24 + 1/120 + 1/720 => 0.0416666 + 0.0083333 + 0.00138888
that CPU can add all the terms in the series together first with rational arithmetic, and return the rational result to the manager CPU: two integers of perhaps a few hundred digits each:
// faster
1/24 + 1/120 + 1/720 => 1/24 + 840/86400 => 106560/2073600
After thousands of terms have been added together in this way, the manager CPU does the one and only division at the very end to get the decimal digits after the decimal point.
Remember to avoid PrematureOptimization, and
always ProfileBeforeOptimizing.
If you're using double or float, there is an M_E constant in math.h already.
#define M_E 2.71828182845904523536028747135266250 /* e */
There are other representions of e in http://en.wikipedia.org/wiki/Representations_of_e#As_an_infinite_series; all the them will involve division.
I'm not aware of any "faster" computation than the Taylor expansion of the series, i.e.:
e = 1/0! + 1/1! + 1/2! + ...
or
1/e = 1/0! - 1/1! + 1/2! - 1/3! + ...
Considering that these were used by A. Yee, who calculated the first 500 billion digits of e, I guess that there's not much optimising to do (or better, it could be optimised, but nobody yet found a way, AFAIK)
EDIT
A very rough implementation
#include <iostream>
#include <iomanip>
using namespace std;
double gete(int nsteps)
{
// Let's skip the first two terms
double res = 2.0;
double fact = 1;
for (int i=2; i<nsteps; i++)
{
fact *= i;
res += 1/fact;
}
return res;
}
int main()
{
cout << setprecision(50) << gete(10) << endl;
cout << setprecision(50) << gete(50) << endl;
}
Outputs
2.71828152557319224769116772222332656383514404296875
2.71828182845904553488480814849026501178741455078125
This page has a nice rundown of different calculation methods.
This is a tiny C program from Xavier Gourdon to compute 9000 decimal digits of e on your computer. A program of the same kind exists for π and for some other constants defined by mean of hypergeometric series.
[degolfed version from https://codereview.stackexchange.com/a/33019 ]
#include <stdio.h>
int main() {
int N = 9009, a[9009], x;
for (int n = N - 1; n > 0; --n) {
a[n] = 1;
}
a[1] = 2;
while (N > 9) {
int n = N--;
while (--n) {
a[n] = x % n;
x = 10 * a[n-1] + x/n;
}
printf("%d", x);
}
return 0;
}
This program [when code-golfed] has 117 characters. It can be changed to compute more digits (change the value 9009 to more) and to be faster (change the constant 10 to another power of 10 and the printf command). A not so obvious question is to find the algorithm used.
I gave this answer at CodeReviews on the question regarding computing e by its definition via Taylor series (so, other methods were not an option). The cross-post here was suggested in the comments. I've removed my remarks relevant to that other topic; Those interested in further explanations migth want to check the original post.
The solution in C (should be easy enough to adapt to adapt to C++):
#include <stdio.h>
#include <math.h>
int main ()
{
long double n = 0, f = 1;
int i;
for (i = 28; i >= 1; i--) {
f *= i; // f = 28*27*...*i = 28! / (i-1)!
n += f; // n = 28 + 28*27 + ... + 28! / (i-1)!
} // n = 28! * (1/0! + 1/1! + ... + 1/28!), f = 28!
n /= f;
printf("%.64llf\n", n);
printf("%.64llf\n", expl(1));
printf("%llg\n", n - expl(1));
printf("%d\n", n == expl(1));
}
Output:
2.7182818284590452354281681079939403389289509505033493041992187500
2.7182818284590452354281681079939403389289509505033493041992187500
0
1
There are two important points:
This code doesn't compute 1, 1*2, 1*2*3,... which is O(n^2), but computes 1*2*3*... in one pass (which is O(n)).
It starts from smaller numbers. If we tried to compute
1/1 + 1/2 + 1/6 + ... + 1/20!
and tried to add it 1/21!, we'd be adding
1/21! = 1/51090942171709440000 = 2E-20,
to 2.something, which has no effect on the result (double holds about 16 significant digits). This effect is called underflow.
However, when we start with these numbers, i.e., if we compute 1/32!+1/31!+... they all have some impact.
This solution seems in accordance to what C computes with its expl function, on my 64bit machine, compiled with gcc 4.7.2 20120921.
You may be able to gain some efficiency. Since each term involves the next factorial, some efficiency may be obtained by remembering the last value of the factorial.
e = 1 + 1/1! + 1/2! + 1/3! ...
Expanding the equation:
e = 1 + 1/(1 * 1) + 1/(1 * 1 * 2) + 1/(1 * 2 * 3) ...
Instead of computing each factorial, the denominator is multiplied by the next increment. So keeping the denominator as a variable and multiplying it will produce some optimization.
If you're ok with an approximation up to seven digits, use
3-sqrt(5/63)
2.7182819
If you want the exact value:
e = (-1)^(1/(j*pi))
where j is the imaginary unit and pi the well-known mathematical constant (Euler's Identity)
There are several "spigot" algorithms which compute digits sequentially in an unbounded manner. This is useful because you can simply calculate the "next" digit through a constant number of basic arithmetic operations, without defining beforehand how many digits you wish to produce.
These apply a series of successive transformations such that the next digit comes to the 1's place, so that they are not affected by float rounding errors. The efficiency is high because these transformations can be formulated as matrix multiplications, which reduce to integer addition and multiplication.
In short, the taylor series expansion
e = 1/0! + 1/1! + 1/2! + 1/3! ... + 1/n!
Can be rewritten by factoring out fractional parts of the factorials (note that to make the series regular we've moved 1 to the left side):
(e - 1) = 1 + (1/2)*(1 + (1/3)*(1 + (1/4)...))
We can define a series of functions f1(x) ... fn(x) thus:
f1(x) = 1 + (1/2)x
f2(x) = 1 + (1/3)x
f3(x) = 1 + (1/4)x
...
The value of e is found from the composition of all of these functions:
(e-1) = f1(f2(f3(...fn(x))))
We can observe that the value of x in each function is determined by the next function, and that each of these values is bounded on the range [1,2] - that is, for any of these functions, the value of x will be 1 <= x <= 2
Since this is the case, we can set a lower and upper bound for e by using the values 1 and 2 for x respectively:
lower(e-1) = f1(1) = 1 + (1/2)*1 = 3/2 = 1.5
upper(e-1) = f1(2) = 1 + (1/2)*2 = 2
We can increase precision by composing the functions defined above, and when a digit matches in the lower and upper bound, we know that our computed value of e is precise to that digit:
lower(e-1) = f1(f2(f3(1))) = 1 + (1/2)*(1 + (1/3)*(1 + (1/4)*1)) = 41/24 = 1.708333
upper(e-1) = f1(f2(f3(2))) = 1 + (1/2)*(1 + (1/3)*(1 + (1/4)*2)) = 7/4 = 1.75
Since the 1s and 10ths digits match, we can say that an approximation of (e-1) with precision of 10ths is 1.7. When the first digit matches between the upper and lower bounds, we subtract it off and then multiply by 10 - this way the digit in question is always in the 1's place where floating-point precision is high.
The real optimization comes from the technique in linear algebra of describing a linear function as a transformation matrix. Composing functions maps to matrix multiplication, so all of those nested functions can be reduced to simple integer multiplication and addition. The procedure of subtracting the digit and multiplying by 10 also constitutes a linear transformation, and therefore can also be accomplished by matrix multiplication.
Another explanation of the method:
http://www.hulver.com/scoop/story/2004/7/22/153549/352
The paper that describes the algorithm:
http://www.cs.ox.ac.uk/people/jeremy.gibbons/publications/spigot.pdf
A quick intro to performing linear transformations via matrix arithmetic:
https://people.math.gatech.edu/~cain/notes/cal6.pdf
NB this algorithm makes use of Mobius Transformations which are a type of linear transformation described briefly in the Gibbons paper.
From my point of view, the most efficient way to compute e up to a desired precision is to use the following representation:
e := lim (n -> inf): (1 + (1/n))^n
Especially if you choose n = 2^x, you can compute the potency with just x multiplications, since:
a^n = (a^2)^(n/2), if n % 2 = 0
The binary splitting method lends itself nicely to a template metaprogram which produces a type which represents a rational corresponding to an approximation of e. 13 iterations seems to be the maximum - any higher will produce a "integral constant overflow" error.
#include <iostream>
#include <iomanip>
template<int NUMER = 0, int DENOM = 1>
struct Rational
{
enum {NUMERATOR = NUMER};
enum {DENOMINATOR = DENOM};
static double value;
};
template<int NUMER, int DENOM>
double Rational<NUMER, DENOM>::value = static_cast<double> (NUMER) / DENOM;
template<int ITERS, class APPROX = Rational<2, 1>, int I = 2>
struct CalcE
{
typedef Rational<APPROX::NUMERATOR * I + 1, APPROX::DENOMINATOR * I> NewApprox;
typedef typename CalcE<ITERS, NewApprox, I + 1>::Result Result;
};
template<int ITERS, class APPROX>
struct CalcE<ITERS, APPROX, ITERS>
{
typedef APPROX Result;
};
int test (int argc, char* argv[])
{
std::cout << std::setprecision (9);
// ExpType is the type containing our approximation to e.
typedef CalcE<13>::Result ExpType;
// Call result() to produce the double value.
std::cout << "e ~ " << ExpType::value << std::endl;
return 0;
}
Another (non-metaprogram) templated variation will, at compile-time, calculate a double approximating e. This one doesn't have the limit on the number of iterations.
#include <iostream>
#include <iomanip>
template<int ITERS, long long NUMERATOR = 2, long long DENOMINATOR = 1, int I = 2>
struct CalcE
{
static double result ()
{
return CalcE<ITERS, NUMERATOR * I + 1, DENOMINATOR * I, I + 1>::result ();
}
};
template<int ITERS, long long NUMERATOR, long long DENOMINATOR>
struct CalcE<ITERS, NUMERATOR, DENOMINATOR, ITERS>
{
static double result ()
{
return (double)NUMERATOR / DENOMINATOR;
}
};
int main (int argc, char* argv[])
{
std::cout << std::setprecision (16);
std::cout << "e ~ " << CalcE<16>::result () << std::endl;
return 0;
}
In an optimised build the expression CalcE<16>::result () will be replaced by the actual double value.
Both are arguably quite efficient since they calculate e at compile time :-)
#nico Re:
..."faster" computation than the Taylor expansion of the series, i.e.:
e = 1/0! + 1/1! + 1/2! + ...
or
1/e = 1/0! - 1/1! + 1/2! - 1/3! + ...
Here are ways to algebraically improve the convergence of Newton’s method:
https://www.researchgate.net/publication/52005980_Improving_the_Convergence_of_Newton's_Series_Approximation_for_e
It appears to be an open question as to whether they can be used in conjunction with binary splitting to computationally speed things up. Nonetheless, here is an example from Damian Conway using Perl that illustrates the improvement in direct computational efficiency for this new approach. It’s in the section titled “𝑒 is for estimation”:
http://blogs.perl.org/users/damian_conway/2019/09/to-compute-a-constant-of-calculusa-treatise-on-multiple-ways.html
(This comment is too long to post as a reply for answer on Jun 12 '10 at 10:28)
From wikipedia replace x with 1