(* Pi digits computed with the sreaming algorithm given on pages 4, 6
   & 7 of "Unbounded Spigot Algorithms for the Digits of Pi", Jeremy
   Gibbons, August 2004. *)

open Printf;;
open Big_int;;

let ( !$ ) = Big_int.big_int_of_int
and ( +$ ) = Big_int.add_big_int
and ( *$ ) = Big_int.mult_big_int
and ( =$ ) = Big_int.eq_big_int
;;

let zero = Big_int.zero_big_int
and one = Big_int.unit_big_int
and three = !$ 3
and four = !$ 4
and ten = !$ 10
and neg_ten = !$(-10)
;;

(* Linear Fractional (aka M=F6bius) Transformations *)
module LFT = struct

  let floor_ev (q, r, s, t) x = div_big_int (q *$ x +$ r) (s *$ x +$ t);;

  let unit = (one, zero, zero, one);;

  let comp (q, r, s, t) (q', r', s', t') =
    (q *$ q' +$ r *$ s', q *$ r' +$ r *$ t',
     s *$ q' +$ t *$ s', s *$ r' +$ t *$ t')
;;

end
;;

let next z = LFT.floor_ev z three
and safe z n = (n =$ LFT.floor_ev z four)
and prod z n = LFT.comp (ten, neg_ten *$ n, zero, one) z
and cons z k =
  let den = 2 * k + 1 in
  LFT.comp z (!$ k, !$(2 * den), zero, !$ den)
;;

let rec digit k z n row col =
  if n > 0 then
    let y = next z in
    if safe z y then
      if col = 10 then (
        let row = row + 10 in
        printf "\t:%i\n%s" row (string_of_big_int y);
        digit k (prod z y) (n - 1) row 1
      )
      else (
        print_string(string_of_big_int y);
        digit k (prod z y) (n - 1) row (col + 1)
      )
    else digit (k + 1) (cons z k) n row col
  else
    printf "%*s\t:%i\n" (10 - col) "" (row + col)
;;

let digits n = digit 1 LFT.unit n 0 0
;;

let usage () =
  prerr_endline "Usage: pi_big_int <number of digits to compute for pi>";
  exit 2
;;

let main () =
  let args = Sys.argv in
  if Array.length args <> 2 then usage () else
  digits (int_of_string Sys.argv.(1))
;;

main ()
;;