Re: Vieta Accelerated Message #2 Posted by hugh steers on 1 Dec 2008, 3:51 p.m., in response to message #1 by Egan Ford
Hello Egan,
Thanks for posting your interesting article on Viète acceleration. Your Pi shootout is excellent and the HPGCC tutorial is well worth reading for anyone interested in native code development for the 50g. Incidentally, do you know if there is now a way to perform the “S>EXE” transform on the PC rather than the calculator? This was always a problem for me since larger lumps of code would run out of memory for this transformation.
Also, I claim you only need k+1 numbers to calculate V(k,n). What you do, is fill out a k+1 array with V(0,n) .. V(0,n+k) and reduce this “row” k times with the recurrence relation until finally the first slot of the array is the value of V(k,n).
Take a look at this code which illustrates what im talking about. Im only using native doubles here in place of bignums, but the calculation should be the same.
double kreminski(int k, int n)
{
// create array of k+1 values
double* kc = new double[k+1];
int i;
int m;
double r = 0;
double t = 1;
// fill out V(0,n) for n to n+k
m = n + k;
for (i = 1; i <= m; ++i)
{
r = sqrt(r + 2);
t *= 2/r;
if (i >= n)
kc[in] = t;
}
// reduce table row by row until top
int j;
double k4 = 1;
m = k;
for (j = 0; j < k; ++j)
{
k4 *= 4;
for (i = 0; i < m; ++i)
kc[i] = (k4*kc[i+1]  kc[i])/(k41);
m; // one shorter each time
}
t = kc[0]; // top left is answer!
delete [] kc;
return t;
}
is this a winner?
cheers,
 hugh.
