Pi by Rabinowitz Spigot Formula

[bencooper]

I have here translated a PASCAL version of Rabinowitz Spigot Formula for calculating Pi to any number of digits.
It is quite fast but unfortunately is accurate to only 6 digits.
(The code is based on  a string of 2's each with a different radix which is converted to decimal from the "back-end")

I'm not sure if the code is translated correctly (can't be far wrong ?) or if I'm struggling with rounding-off errors (32-bit integers required). The attached has the LB code and the original PASCAL code 

Look forward to advice, please !

Attachments:

PiRabin.bas (2.84 KB)

[bencooper]

'Program Pi_Spigot PASCAL
'const n =1000;
'len =10*n div 3;
'var i, j, k, q, x, nines, predigit : integer;
'a : array[1..len] of longint;
'begin
'for j := 1 to len do a[j] := 2; {Start with 2s}
'nines := 0; predigit := 0 {First predigit is a 0}
'for j := 1 to n do
'begin q := 0;
'for i := len downto 1 do {Work backwards}
'begin
'x := 10*a+q*i;
'a := x mod (2*i-1);
'q := x div (2*i-1);
'end;
'a[1] := q mod 10; q := q div 10;
'if q = 9 then nines := nines + 1
'else if q=10 then
'begin write(predigit+1);
'for k := 1 to nines do write(O); {zeros}
'predigit := O; nines := O
'end
'else begin
'write(predigit), predigit := q;
'if nines <> O then
'begin
'for k := 1 to nines do write(9);
'nines := 0
'end
'end
'end,
'writeln(predigit);
'end.

[Rod]

If you search Rosetta Code site there are various solutions for various BASIC languages. Might give you a pointer or two.

rosettacode.org/wiki/Category:Liberty_BASIC

[tsh73]

Found better formatted Pascal code
With a C code alongside
craftofcoding.wordpress.com/2018/04/26/the-problem-with-pi-ii/

So I did tweak it until it run as seemingly should

I compared 100 characters it outputs with first part from the page.
Got 100% match

n =100                       ' number of digits of pi
L =INT(10*n/3)
'var i, j, k, q, x, nines, predigit : integer;
DIM a(L)                      ' of longint;

FOR j = 1 TO L
  a(j) = 2                   '{Start with 2s}
NEXT j
  nines= 0
  predigit = 0               '{First predigit is a 0}
  FOR j = 1 TO n
      q = 0
      FOR i = L TO 1 STEP -1 '{Work backwards}
        x= 10*a(i)+q*i
        a(i) = x MOD (2*i-1)
        q = INT(x/(2*i-1))
      NEXT i
      a(1) = q MOD 10
      q =INT(q/10)
      IF q = 9 THEN
        nines = nines + 1
        ELSE
          IF q=10 THEN
            PRINT (predigit+1);
            FOR k = 1 to nines
              PRINT "0";             '{zeros}
            NEXT k
              predigit = 0
              nines = 0
           'END IF
          ELSE
            PRINT predigit;
            predigit =q
            IF nines <> 0 THEN
             FOR k = 1 TO nines
              PRINT "9";
             NEXT k
              nines = 0
            END IF
          END IF
      END IF
  NEXT j
   PRINT predigit ;
PRINT
'FOR x=1 TO L
'  PRINT a(x);
'NEXT x
PRINT "Program Terminated"
END


[bencooper]

Thanks to Rod and Anatoly for helpful responses !
I checked but couldn't immediately find what Anatoly "tweaked" but it worked a dream !
1000 digits in 33 seconds, with no demand on memory AFAIK !!

[tsh73]

I checked but couldn't immediately find what Anatoly "tweaked"

Basically, moving last "PRINT predigit ;" after last "next j"
Uncommenting all prints ('PRINT "0";, 'PRINT predigit)
And making sure there is a ( ; ) after every print.
Also I moved "Next K" up but it does not matter.