I am new to C++ NURBS libary. I learnt generating line (by CLine, from nurbs.h ) and save it as igs. But in case of
multiple control points, how to generate a curve ? Every other tutorial using graphics.h
(putpixel), but couldnt find anything about igs.
This should be a simple problem. But I have no idea which function can help me here.
Thanks in advance.
We have 4 control points here to begin with.
for (float t = 0.0; t <= 1.0; t += 0.2) {
double xt = 0.0, yt = 0.0;
xt = pow(1 - t, 3) * x[0] + 3 * t * pow(1 - t, 2) * x[1] + 3 * pow(t, 2) * (1 - t) * x[2]
+ pow(t, 3) * x[3];
yt = pow(1 - t, 3) * y[0] + 3 * t * pow(1 - t, 2) * y[1] + 3 * pow(t, 2) * (1 - t) * y[2]
+ pow(t, 3) * y[3];
count = count + 1;
//Math::Vector4f c(xt, yt, 0);
for (int i = 1; i < 3; i++) {
listt[i][0]= xt;
listt[i][1]= yt;
Math::Vector4f a(listt[i][0], listt[i][1],0);
myvector.push_back (&a);
}
}
......
.....
igs.Write("test.igs");
--- This is to create the points, but after that I dont know how to use the points to create a Bezier curve .
I am trying to understand this code:
void stencil(const int nx, const int ny, const int width, const int height,
double* image, double* tmp_image)
{
for (int j = 1; j < ny + 1; ++j) {
for (int i = 1; i < nx + 1; ++i) {
tmp_image[j + i * height] = image[j + i * height] * 3.0 / 5.0;
tmp_image[j + i * height] += image[j + (i - 1) * height] * 0.5 / 5.0;
tmp_image[j + i * height] += image[j + (i + 1) * height] * 0.5 / 5.0;
tmp_image[j + i * height] += image[j - 1 + i * height] * 0.5 / 5.0;
tmp_image[j + i * height] += image[j + 1 + i * height] * 0.5 / 5.0;
}
}
}
The 1-d array notation is very confusing. I am trying to convert it to a 2-d notation (which I find easier to read). Could someone point me in the right direction as to how I can accomplish this?
All this code is doing is creating a new image from an original image by taking 60% from the corresponding pixel and 10% from each neighboring pixel.
When you see tmp_image[j + i * height], read it as tmp_image[i][j].
Changing the code to literally use 2D syntax may require knowing at least one of the dimensions at compile time, whereas now it is a runtime argument. So that might be a non-starter, unless you're using C++ and want to write or use a matrix class instead of plain arrays.
I need to make a simple bandpass audio filter.
Now I've used this simple C++ class: http://www.cardinalpeak.com/blog/a-c-class-to-implement-low-pass-high-pass-and-band-pass-filters
It works well and cut off the desired bands. But when I try to change upper or lower limit with small steps, on some values of limit I hear the wrong result - attenuated or shifted in frequency (not corresponding to current limits) sound.
Function for calculating impulse response:
void Filter::designBPF()
{
int n;
float mm;
for(n = 0; n < m_num_taps; n++){
mm = n - (m_num_taps - 1.0) / 2.0;
if( mm == 0.0 ) m_taps[n] = (m_phi - m_lambda) / M_PI;
else m_taps[n] = ( sin( mm * m_phi ) -
sin( mm * m_lambda ) ) / (mm * M_PI);
}
return;
}
where
m_lambda = M_PI * Fl / (Fs/2);
m_phi = M_PI * Fu / (Fs/2);
Fs - sample rate (44.100)
Fl - lower limit
Fu - upper limit
And simple filtering function:
float Filter::do_sample(float data_sample)
{
int i;
float result;
if( m_error_flag != 0 ) return(0);
for(i = m_num_taps - 1; i >= 1; i--){
m_sr[i] = m_sr[i-1];
}
m_sr[0] = data_sample;
result = 0;
for(i = 0; i < m_num_taps; i++) result += m_sr[i] * m_taps[i];
return result;
}
Do I need to use any window function (Blackman, etc.)? If yes, how do I do this?
I have tried to multiply my impulse response to Blackman window:
m_taps[n] *= 0.42 - 0.5 * cos(2.0 * M_PI * n / double(N - 1)) +
0.08 * cos(4.0 * M_PI * n / double(N - 1));
but the result was wrong.
And do I need to normalize taps?
I found a good free implementation of FIR filter:
http://www.iowahills.com/A7ExampleCodePage.html
...This Windowed FIR Filter C Code has two parts, the first is the
calculation of the impulse response for a rectangular window (low
pass, high pass, band pass, or notch). Then a window (Kaiser, Hanning,
etc) is applied to the impulse response. There are several windows to
choose from...
y[i] = waveform[i] × (0.42659071 – 0.49656062cos(w) + 0.07684867cos(2w))
where w = (2)i/n and n is the number of elements in the waveform
Try this I got the code from:
http://zone.ni.com/reference/en-XX/help/370592P-01/digitizers/blackman_window/
I hope this helps.
I'm trying to get fourier transforms to work, I have to do it for an assignment and I think I have it to where it should be working and i'm not sure why it's not. I think it has something to do with the complex numbers since 'i' is involved. I've looked at many references and I understand the formula but i'm having trouble programming it. this is what i have so far
void NaiveDFT::Apply( Image & img )
{
//make the fourier transform using the naive method and set that to the image.
Image dft(img);
Pixel ** dftData = dft.GetImageData();
Pixel ** imgData = img.GetImageData();
for(unsigned u = 0; u < img.GetWidth(); ++u)
{
for(unsigned v = 0; v < img.GetHeight(); ++v)
{
std::complex<double> sum = 0;
for(unsigned x = 0; x < img.GetWidth(); ++x)
{
for(unsigned y = 0; y < img.GetHeight(); ++y)
{
std::complex<double> i = sqrt(std::complex<double>(-1));
std::complex<double> theta = 2 * M_PI * (((u * x) / img.GetWidth()) + ((v * y) / img.GetHeight()));
sum += std::complex<double>(imgData[x][y]._red) * cos(theta) + (-i * sin(theta));
//sum += std::complex<double>(std::complex<double>(imgData[x][y]._red) * pow(EULER, -i * theta));
}
}
dftData[u][v] = (sum.imag() / (img.GetWidth() * img.GetHeight()));
}
}
img = dft;
}
I have a few test images i'm testing this with and i'm either getting like an all black image or like, an all gray image.
I've also tried the sum of e^(-i*2*PI*(x*u*width + y*v*height) * 1/width * height which gets the same result as expected although it's still not the desiered output.
I've also tried the sum.real() number and that doesn't look right either
if anyone has any tips or can point me in the right direction, that'd be great, at this point, i just keep trying different things and checking the output until I get what I should be getting.
thanks.
I think that there can be a problem during the multiplication with the complex term. The line:
sum += std::complex<double>(imgData[x][y]._red) * cos(theta) + (-i * sin(theta));
should be:
sum += std::complex<double>(imgData[x][y]._red) * ( cos(theta) + -i * sin(theta));
Moreover, while calculating theta you need to use double precision:
std::complex<double> theta = 2 * M_PI * ((((double)u * x) / (double)(img.GetWidth())) + (((double)v * y) / (double)(img.GetHeight())));
I am trying to implement range reduction as the first step of implementing the sine function.
I am following the method described in the paper "ARGUMENT REDUCTION FOR HUGE ARGUMENTS" by K.C. NG
I am getting error as large as 0.002339146 when using the input range of x from 0 to 20000. My error obviously shouldn't be that large, and I'm not sure how I can reduce it. I noticed that the error magnitude is associated with the input theta magnitude to cosine/sine.
I was able to obtain the nearpi.c code that the paper mentions, but I'm not sure how to utilize the code for single precision floating point. If anyone is interested, the nearpi.c file can be found at this link: nearpi.c
Here is my MATLAB code:
x = 0:0.1:20000;
% Perform range reduction
% Store constant 2/pi
twooverpi = single(2/pi);
% Compute y
y = (x.*twooverpi);
% Compute k (round to nearest integer
k = round(y);
% Solve for f
f = single(y-k);
% Solve for r
r = single(f*single(pi/2));
% Find last two bits of k
n = bitand(fi(k,1,32,0),fi(3,1,32,0));
n = single(n);
% Preallocate for speed
z(length(x)) = 0;
for i = 1:length(x)
switch(n(i))
case 0
z(i)=sin(r(i));
case 1
z(i) = single(cos(r(i)));
case 2
z(i) = -sin(r(i));
case 3
z(i) = single(-cos(r(i)));
otherwise
end
end
maxerror = max(abs(single(z - single(sin(single(x))))))
minerror = min(abs(single(z - single(sin(single(x))))))
I have edited the program nearpi.c so that it compiles. However I am not sure how to interpret the output. Also the file expects an input, which I had to input by hand, also I am not sure of the significance of the input.
Here is the working nearpi.c:
/*
============================================================================
Name : nearpi.c
Author :
Version :
Copyright : Your copyright notice
Description : Hello World in C, Ansi-style
============================================================================
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/*
* Global macro definitions.
*/
# define hex( double ) *(1 + ((long *) &double)), *((long *) &double)
# define sgn(a) (a >= 0 ? 1 : -1)
# define MAX_k 2500
# define D 56
# define MAX_EXP 127
# define THRESHOLD 2.22e-16
/*
* Global Variables
*/
int CFlength, /* length of CF including terminator */
binade;
double e,
f; /* [e,f] range of D-bit unsigned int of f;
form 1X...X */
// Function Prototypes
int dbleCF (double i[], double j[]);
void input (double i[]);
void nearPiOver2 (double i[]);
/*
* This is the start of the main program.
*/
int main (void)
{
int k; /* subscript variable */
double i[MAX_k],
j[MAX_k]; /* i and j are continued fractions
(coeffs) */
// fp = fopen("/src/cfpi.txt", "r");
/*
* Compute global variables e and f, where
*
* e = 2 ^ (D-1), i.e. the D bit number 10...0
* and
* f = 2 ^ D - 1, i.e. the D bit number 11...1 .
*/
e = 1;
for (k = 2; k <= D; k = k + 1)
e = 2 * e;
f = 2 * e - 1;
/*
* Compute the continued fraction for (2/e)/(pi/2) , i.e.
* q's starting value for the first binade, given the continued
* fraction for pi as input; set the global variable CFlength
* to the length of the resulting continued fraction (including
* its negative valued terminator). One should use as many
* partial coefficients of pi as necessary to resolve numbers
* of the width of the underflow plus the overflow threshold.
* A rule of thumb is 0.97 partial coefficients are generated
* for every decimal digit of pi .
*
* Note: for radix B machines, subroutine input should compute
* the continued fraction for (B/e)/(pi/2) where e = B ^ (D - 1).
*/
input (i);
/*
* Begin main loop over all binades:
* For each binade, find the nearest multiples of pi/2 in that binade.
*
* [ Note: for hexadecimal machines ( B = 16 ), the rest of the main
* program simplifies(!) to
*
* B_ade = 1;
* while (B_ade < MAX_EXP)
* {
* dbleCF (i, j);
* dbleCF (j, i);
* dbleCF (i, j);
* CFlength = dbleCF (j, i);
* B_ade = B_ade + 1;
* }
* }
*
* because the alternation of source & destination are no longer necessary. ]
*/
binade = 1;
while (binade < MAX_EXP)
{
/*
* For the current (odd) binade, find the nearest multiples of pi/2.
*/
nearPiOver2 (i);
/*
* Double the continued fraction to get to the next (even) binade.
* To save copying arrays, i and j will alternate as the source
* and destination for the continued fractions.
*/
CFlength = dbleCF (i, j);
binade = binade + 1;
/*
* Check for main loop termination again because of the
* alternation.
*/
if (binade >= MAX_EXP)
break;
/*
* For the current (even) binade, find the nearest multiples of pi/2.
*/
nearPiOver2 (j);
/*
* Double the continued fraction to get to the next (odd) binade.
*/
CFlength = dbleCF (j, i);
binade = binade + 1;
}
return 0;
} /* end of Main Program */
/*
* Subroutine DbleCF doubles a continued fraction whose partial
* coefficients are i[] into a continued fraction j[], where both
* arrays are of a type sufficient to do D-bit integer arithmetic.
*
* In my case ( D = 56 ) , I am forced to treat integers as double
* precision reals because my machine does not have integers of
* sufficient width to handle D-bit integer arithmetic.
*
* Adapted from a Basic program written by W. Kahan.
*
* Algorithm based on Hurwitz's method of doubling continued
* fractions (see Knuth Vol. 3, p.360).
*
* A negative value terminates the last partial quotient.
*
* Note: for the non-C programmers, the statement break
* exits a loop and the statement continue skips to the next
* case in the same loop.
*
* The call modf ( l / 2, &l0 ) assigns the integer portion of
* half of L to L0.
*/
int dbleCF (double i[], double j[])
{
double k,
l,
l0,
j0;
int n,
m;
n = 1;
m = 0;
j0 = i[0] + i[0];
l = i[n];
while (1)
{
if (l < 0)
{
j[m] = j0;
break;
};
modf (l / 2, &l0);
l = l - l0 - l0;
k = i[n + 1];
if (l0 > 0)
{
j[m] = j0;
j[m + 1] = l0;
j0 = 0;
m = m + 2;
};
if (l == 0) {
/*
* Even case.
*/
if (k < 0)
{
m = m - 1;
break;
}
else
{
j0 = j0 + k + k;
n = n + 2;
l = i[n];
continue;
};
}
/*
* Odd case.
*/
if (k < 0)
{
j[m] = j0 + 2;
break;
};
if (k == 0)
{
n = n + 2;
l = l + i[n];
continue;
};
j[m] = j0 + 1;
m = m + 1;
j0 = 1;
l = k - 1;
n = n + 1;
continue;
};
m = m + 1;
j[m] = -99999;
return (m);
}
/*
* Subroutine input computes the continued fraction for
* (2/e) / (pi/2) , where e = 2 ^ (D-1) , given pi 's
* continued fraction as input. That is, double the continued
* fraction of pi D-3 times and place a zero at the front.
*
* One should use as many partial coefficients of pi as
* necessary to resolve numbers of the width of the underflow
* plus the overflow threshold. A rule of thumb is 0.97
* partial coefficients are generated for every decimal digit
* of pi . The last coefficient of pi is terminated by a
* negative number.
*
* I'll be happy to supply anyone with the partial coefficients
* of pi . My ARPA address is mcdonald#ucbdali.BERKELEY.ARPA .
*
* I computed the partial coefficients of pi using a method of
* Bill Gosper's. I need only compute with integers, albeit
* large ones. After writing the program in bc and Vaxima ,
* Prof. Fateman suggested FranzLisp . To my surprise, FranzLisp
* ran the fastest! the reason? FranzLisp's Bignum package is
* hand coded in assembler. Also, FranzLisp can be compiled.
*
*
* Note: for radix B machines, subroutine input should compute
* the continued fraction for (B/e)/(pi/2) where e = B ^ (D - 1).
* In the case of hexadecimal ( B = 16 ), this is done by repeated
* doubling the appropriate number of times.
*/
void input (double i[])
{
int k;
double j[MAX_k];
/*
* Read in the partial coefficients of pi from a precalculated file
* until a negative value is encountered.
*/
k = -1;
do
{
k = k + 1;
scanf ("%lE", &i[k]);
printf("hello\n");
printf("%d", k);
} while (i[k] >= 0);
/*
* Double the continued fraction for pi D-3 times using
* i and j alternately as source and destination. On my
* machine D = 56 so D-3 is odd; hence the following code:
*
* Double twice (D-3)/2 times,
*/
for (k = 1; k <= (D - 3) / 2; k = k + 1)
{
dbleCF (i, j);
dbleCF (j, i);
};
/*
* then double once more.
*/
dbleCF (i, j);
/*
* Now append a zero on the front (reciprocate the continued
* fraction) and the return the coefficients in i .
*/
i[0] = 0;
k = -1;
do
{
k = k + 1;
i[k + 1] = j[k];
} while (j[k] >= 0);
/*
* Return the length of the continued fraction, including its
* terminator and initial zero, in the global variable CFlength.
*/
CFlength = k;
}
/*
* Given a continued fraction's coefficients in an array i ,
* subroutine nearPiOver2 finds all machine representable
* values near a integer multiple of pi/2 in the current binade.
*/
void nearPiOver2 (double i[])
{
int k, /* subscript for recurrences (see
handout) */
K; /* like k , but used during cancel. elim.
*/
double p[MAX_k], /* product of the q's (see
handout) */
q[MAX_k], /* successive tail evals of CF (see
handout) */
j[MAX_k], /* like convergent numerators (see
handout) */
tmp, /* temporary used during cancellation
elim. */
mk0, /* m[k - 1] (see
handout) */
mk, /* m[k] is one of the few ints (see
handout) */
mkAbs, /* absolute value of m sub k
*/
mK0, /* like mk0 , but used during cancel.
elim. */
mK, /* like mk , but used during cancel.
elim. */
z, /* the object of our quest (the argument)
*/
m0, /* the mantissa of z as a D-bit integer
*/
x, /* the reduced argument (see
handout) */
ldexp (), /* sys routine to multiply by a power of
two */
fabs (), /* sys routine to compute FP absolute
value */
floor (), /* sys routine to compute greatest int <=
value */
ceil (); /* sys routine to compute least int >=
value */
/*
* Compute the q's by evaluating the continued fraction from
* bottom up.
*
* Start evaluation with a big number in the terminator position.
*/
q[CFlength] = 1.0 + 30;
for (k = CFlength - 1; k >= 0; k = k - 1)
q[k] = i[k] + 1 / q[k + 1];
/*
* Let THRESHOLD be the biggest | x | that we are interesed in
* seeing.
*
* Compute the p's and j's by the recurrences from the top down.
*
* Stop when
*
* 1 1
* ----- >= THRESHOLD > ------ .
* 2 |j | 2 |j |
* k k+1
*/
p[0] = 1;
j[0] = 0;
j[1] = 1;
k = 0;
do
{
p[k + 1] = -q[k + 1] * p[k];
if (k > 0)
j[1 + k] = j[k - 1] - i[k] * j[k];
k = k + 1;
} while (1 / (2 * fabs (j[k])) >= THRESHOLD);
/*
* Then mk runs through the integers between
*
* k + k +
* (-1) e / p - 1/2 & (-1) f / p - 1/2 .
* k k
*/
for (mkAbs = floor (e / fabs (p[k]));
mkAbs <= ceil (f / fabs (p[k])); mkAbs = mkAbs + 1)
{
mk = mkAbs * sgn (p[k]);
/*
* For each mk , mk0 runs through integers between
*
* +
* m q - p THRESHOLD .
* k k k
*/
for (mk0 = floor (mk * q[k] - fabs (p[k]) * THRESHOLD);
mk0 <= ceil (mk * q[k] + fabs (p[k]) * THRESHOLD);
mk0 = mk0 + 1)
{
/*
* For each pair { mk , mk0 } , check that
*
* k
* m = (-1) ( j m - j m )
* 0 k-1 k k k-1
*/
m0 = (k & 1 ? -1 : 1) * (j[k - 1] * mk - j[k] * mk0);
/*
* lies between e and f .
*/
if (e <= fabs (m0) && fabs (m0) <= f)
{
/*
* If so, then we have found an
*
* k
* x = ((-1) m / p - m ) / j
* 0 k k k
*
* = ( m q - m ) / p .
* k k k-1 k
*
* But this later formula can suffer cancellation. Therefore,
* run the recurrence for the mk 's to get mK with minimal
* | mK | + | mK0 | in the hope mK is 0 .
*/
K = k;
mK = mk;
mK0 = mk0;
while (fabs (mK) > 0)
{
p[K + 1] = -q[K + 1] * p[K];
tmp = mK0 - i[K] * mK;
if (fabs (tmp) > fabs (mK0))
break;
mK0 = mK;
mK = tmp;
K = K + 1;
};
/*
* Then
* x = ( m q - m ) / p
* K K K-1 K
*
* as accurately as one could hope.
*/
x = (mK * q[K] - mK0) / p[K];
/*
* To return z and m0 as positive numbers,
* x must take the sign of m0 .
*/
x = x * sgn (m0);
m0 = fabs (m0);
/*d
* Set z = m0 * 2 ^ (binade+1-D) .
*/
z = ldexp (m0, binade + 1 - D);
/*
* Print z (hex), z (dec), m0 (dec), binade+1-D, x (hex), x (dec).
*/
printf ("%08lx %08lx Z=%22.16E M=%17.17G L+1-%d=%3d %08lx %08lx x=%23.16E\n", hex (z), z, m0, D, binade + 1 - D, hex (x), x);
}
}
}
}
Theory
First let's note the difference using single-precision arithmetic makes.
[Equation 8] The minimal value of f can be larger. As double-precision numbers are a super-set of the single-precision numbers, the closest single to a multiple of 2/pi can only be farther away then ~2.98e-19, therefore the number of leading zeros in fixed-arithmetic representation of f must be at most 61 leading zeros (but will probably be less). Denote this quantity fdigits.
[Equation Before 9] Consequently, instead of 121 bits, y must be accurate to fdigits + 24 (non-zero significant bits in single-precision) + 7 (extra guard bits) = fdigits + 31, and at most 92.
[Equation 9] "Therefore, together with the width of x's exponent, 2/pi must contain 127 (maximal exponent of single) + 31 + fdigits, or 158 + fdigits and at most 219 bits.
[Subsection 2.5] The size of A is determined by the number of zeros in x before the binary point (and is unaffected by the move to single), while the size of C is determined by Equation Before 9.
For large x (x>=2^24), x looks like this: [24 bits, M zeros]. Multiplying it by A, whose size is the first M bits of 2/pi, will result in an integer (the zeros of x will just shift everything into the integers).
Choosing C to be starting from the M+d bit of 2/pi will result in the product x*C being of size at most d-24. In double precision, d is chosen to be 174 (and instead of 24, we have 53) so that the product will be of size at most 121. In single, it is enough to choose d such that d-24 <= 92, or more precisely, d-24 <= fdigits+31. That is, d can be chosen as fdigits+55, or at most 116.
As a result, B should be of size at most 116 bits.
We are therefore left with two problems :
Computing fdigits. This involves reading ref 6 from the linked paper and understanding it. Might not be that easy. :) As far as I can see, that's the only place where nearpi.c is used.
Computing B, the relevant bits of 2/pi. Since M is bounded below by 127, we can just compute the first 127+116 bits of 2/pi offline and store them in an array. See Wikipedia.
Computing y=x*B. This involves multipliying x by a 116-bits number. This is where Section 3 is used. The size of the blocks is chosen to be 24 because 2*24 + 2 (multiplying two 24-bits numbers, and adding 3 such numbers) is smaller than the precision of double, 53 (and because 24 divides 96). We can use blocks of size 11 bits for single arithmetic for similar reasons.
Note - the trick with B only applies to numbers whose exponents are positive (x>=2^24).
To summarize - first, you have to solve the problem with double precision. Your Matlab code doesn't work in double precision too (try removing single and computing sin(2^53), because your twooverpi only has 53 significant bits, not 175 (and anyway, you can't directly multiply such precise numbers in Matlab). Second, the scheme should be adapted to work with single, and again, the key problem is representing 2/pi precisely enough, and supporting multiplication of highly-precise numbers. Last, when everything works, you can try and figure out a better fdigits to reduce the number of bits you have to store and multiply.
Hopefully I'm not completely off - comments and contradictions are welcome.
Example
As an example, let us compute sin(x) where x = single(2^24-1), which has no zeros after the significant bits (M = 0). This simplifies finding B, as B consists of the first 116 bits of 2/pi. Since x has precision of 24 bits and B of 116 bits, the product
y = x * B
will have 92 bits of precision, as required.
Section 3 in the linked paper describes how to perform this product with enough precision; the same algorithm can be used with blocks of size 11 to compute y in our case. Being drudgery, I hope I'm excused for not doing this explicitly, instead relying on Matlab's symbolic math toolbox. This toolbox provides us with the vpa function, which allows us to specify the precision of a number in decimal digits. So,
vpa('2/pi', ceil(116*log10(2)))
will produce an approximation of 2/pi of at least 116 bits of precision. Because vpa accepts only integers for its precision argument, we usually can't specify the binary precision of a number exactly, so we use the next-best.
The following code computes sin(x) according to the paper, in single precision :
x = single(2^24-1);
y = x * vpa('2/pi', ceil(116*log10(2))); % Precision = 103.075
k = round(y);
f = single(y - k);
r = f * single(pi) / 2;
switch mod(k, 4)
case 0
s = sin(r);
case 1
s = cos(r);
case 2
s = -sin(r);
case 3
s = -cos(r);
end
sin(x) - s % Expected value: exactly zero.
(The precision of y is obtained using Mathematica, which turned out to be a much better numerical tool than Matlab :) )
In libm
The other answer to this question (which has been deleted since) lead me to an implementation in libm, which although works on double-precision numbers, follows the linked paper very thoroughly.
See file s_sin.c for the wrapper (Table 2 from the linked paper appears as a switch statement at the end of the file), and e_rem_pio2.c for the argument reduction code (of particular interest is an array containing the first 396 hex-digits of 2/pi, starting at line 69).