65 lines
1.5 KiB
OCaml
65 lines
1.5 KiB
OCaml
(* Pi digits computed with the streaming algorithm given on pages 4, 6
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& 7 of "Unbounded Spigot Algorithms for the Digits of Pi", Jeremy
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Gibbons, August 2004. *)
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open Printf
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let zero = Z.zero
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and one = Z.one
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and three = Z.of_int 3
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and four = Z.of_int 4
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and ten = Z.of_int 10
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and neg_ten = Z.of_int (-10)
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;;
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(* Linear Fractional (aka M=F6bius) Transformations *)
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module LFT = struct
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let floor_ev (q, r, s, t) x =
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Z.((q * x + r) / (s * x + t))
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let unit = (one, zero, zero, one)
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let comp (q, r, s, t) (q', r', s', t') =
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Z.(q * q' + r * s', q * r' + r * t',
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s * q' + t * s', s * r' + t * t')
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end
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let next z = LFT.floor_ev z three
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let safe z n = (n = LFT.floor_ev z four)
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let prod z n = LFT.comp (ten, Z.(neg_ten * n), zero, one) z
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let cons z k =
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let den = 2 * k + 1 in
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LFT.comp z (Z.of_int k, Z.of_int (2 * den), zero, Z.of_int den)
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let rec digit k z n row col =
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if n > 0 then
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let y = next z in
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if safe z y then
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if col = 10 then (
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let row = row + 10 in
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printf "\t:%i\n%a" row Z.output y;
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digit k (prod z y) (n - 1) row 1
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)
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else (
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printf "%a" Z.output y;
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digit k (prod z y) (n - 1) row (col + 1)
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)
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else digit (k + 1) (cons z k) n row col
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else
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printf "%*s\t:%i\n" (10 - col) "" (row + col)
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let digits n = digit 1 LFT.unit n 0 0
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let usage () =
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prerr_endline "Usage: pi <number of digits to compute for pi>";
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exit 2
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let _ =
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let args = Sys.argv in
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if Array.length args <> 2 then usage () else
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digits (int_of_string Sys.argv.(1))
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