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pi.i is easily the most twisted and obscure INTERCAL program I've yet
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written.
works.
I say that because even I don't fully understand why it
The program writes in one number from the input, then prints
2out that many digits of pi.
There are no fancy INTERCAL tricks here;
all it does is simple integer arithmetic.
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4The algorithm used comes from an old TECO macro that was recently
discussed on alt.lang.teco.
Mark Henderson posted a translation
5into C, which at least made it easier to read, if not much easier to
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understand.
I played around with that program a bit, fixed a couple
of minor bugs which may or may not have existed in the original TECO,
7and modified it a bit to put in into a form that would be easy to
translate into INTERCAL.
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That modified program, which corresponds
closely to the algorithm in pi.i, is reproduced at the end of this
9file.
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0The INTERCAL version has only been tested up to a hundred digits,
1which took almost 30 minutes on a Sparc 1. The C version below,
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with variables declared unsigned short to correspond to the INTERCAL,
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2gives up to 535 accurate digits and then starts printing garbage,
1presumably because of an overflow. If the declarations are changed
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to short only the first 262 digits are accurate, while using int
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yields thousands of accurate digits. Though I don't understand
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what the inner loop is doing, it looks like the largest numbers
1encountered are O(n), where n is the number of digits requested.
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If, as I expect, the present INTERCAL version does fail after about
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6535 digits, it could easily be converted to a 32-bit form which
1would run as long as almost anyone would desire.
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1The only feature likely to be of utility in other INTERCAL programs
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is the new (2030) routine, which performs a 16-bit division with
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9remainder. It is faster than the (1040) routine for two reasons:
2First, it is a true 16-bit version, whereas the (1040) routine just
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shifts over its operands and then calls the 32-bit division routine
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(1550). Second, the (1550) routine generates a remainder as part of
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the computation, but then simply throws it away. I needed the
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2remainder, so I have my (2030) routine return it in .4 and the
2quotient in .3. In other respects this is just a 16-bit translation
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of the (1550) routine.
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Louis Howell
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December 30, 1991
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2#include <stdio.h>
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unsigned short line[10000];
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9main(argc,argv)
3int argc;
0char *argv[];
3{
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unsigned short n, a, i, q, v, dummy, h=0, w=0;
if (argc==2)
n=atoi(argv[1]);
else
n=20;
n+=2;
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for (i=(n*10)/3 ; i > 0 ; i--) {
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line[i-1] = 2;
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}
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for (dummy=0; dummy < n; dummy++) {
q=0;
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for (i=(n*10)/3 ; i > 0 ; i--) {
a = line[i-1] * 10 + (q*i);
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q=a/(i*2-1);
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line[i-1] = a%(i*2-1);
}
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v = w;
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w = h+(q/10);
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if (w==10) {
h = q%10;
w=0;
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v++;
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}
if (dummy!=0)
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0}
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printf("%d", v);
}
printf("\n");
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