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122 lines
2.4 KiB
Plaintext
122 lines
2.4 KiB
Plaintext
/*
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* Copyright (c) 1995 David I. Bell
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* Permission is granted to use, distribute, or modify this source,
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* provided that this copyright notice remains intact.
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*
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* Calculate pi within the specified epsilon using the quartic convergence
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* iteration.
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*/
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define qpi(epsilon)
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{
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local niter, yn, ym, tm, an, am, t, tn, sqrt2, epsilon2, count, digits;
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local bits, bits2;
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if (isnull(epsilon))
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epsilon = epsilon();
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digits = digits(1/epsilon);
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if (digits <= 8) { niter = 1; epsilon = 1e-8; }
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else if (digits <= 40) { niter = 2; epsilon = 1e-40; }
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else if (digits <= 170) { niter = 3; epsilon = 1e-170; }
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else if (digits <= 693) { niter = 4; epsilon = 1e-693; }
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else {
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niter = 4;
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t = 693;
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while (t < digits) {
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++niter;
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t *= 4;
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}
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}
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epsilon2 = epsilon/(digits/10 + 1);
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digits = digits(1/epsilon2);
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sqrt2 = sqrt(2, epsilon2);
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bits = abs(ilog2(epsilon)) + 1;
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bits2 = abs(ilog2(epsilon2)) + 1;
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yn = sqrt2 - 1;
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an = 6 - 4 * sqrt2;
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tn = 2;
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for (count = 0; count < niter; count++) {
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ym = yn;
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am = an;
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tn *= 4;
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t = sqrt(sqrt(1-ym^4, epsilon2), epsilon2);
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yn = (1-t)/(1+t);
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an = (1+yn)^4*am-tn*yn*(1+yn+yn^2);
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yn = bround(yn, bits2);
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an = bround(an, bits2);
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}
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return (bround(1/an, bits));
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}
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/*
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* Print digits of PI forever, neatly formatted, using calc.
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*
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* Written by Klaus Alexander Seistrup <klaus@seistrup.dk>
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* on a dull Friday evening in November 1999.
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*
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* Inspired by an algorithm conceived by Lambert Meertens.
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*
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* See also the ABC Programmer's Handbook, by Geurts, Meertens & Pemberton,
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* published by Prentice-Hall (UK) Ltd., 1990.
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*
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*/
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define piforever()
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{
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local k = 2;
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local a = 4;
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local b = 1;
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local a1 = 12;
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local b1 = 4;
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local a2, b2, p, q, d, d1;
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local stdout = files(1);
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local first = 1, row = -1, col = 0;
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while (1) {
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/*
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* Next approximation
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*/
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p = k * k;
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q = k + k++;
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a2 = a;
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b2 = b;
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a = a1;
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a1 = p * a2 + q * a1;
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b = b1;
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b1 = p * b2 + q * b1;
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/*
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* Print common digits
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*/
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d = a // b;
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d1 = a1 // b1;
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while (d == d1) {
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if (first) {
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printf("%d.", d);
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first = 0;
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} else {
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if (!(col % 50)) {
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printf("\n");
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col = 0;
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if (!(++row % 20)) {
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printf("\n");
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row = 0;
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}
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}
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printf("%d", d);
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if (!(++col % 10))
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printf(" ");
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}
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a = 10 * (a % b);
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a1 = 10 * (a1 % b1);
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d = a // b;
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d1 = a1 // b1;
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}
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fflush(stdout);
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}
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}
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