(* 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
let ( +$ ) = Big_int.add_big_int
let ( *$ ) = Big_int.mult_big_int
let ( =$ ) = 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 Möbius) 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 () =
  digits(int_of_string Sys.argv.(1))