# Continued fraction

(Redirected from Continued fractions)
Continued fraction
You are encouraged to solve this task according to the task description, using any language you may know.
A number may be represented as a continued fraction (see Mathworld for more information) as follows:
${\displaystyle a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+\ddots }}}}}}}$

The task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients:

For the square root of 2, use ${\displaystyle a_{0}=1}$ then ${\displaystyle a_{N}=2}$. ${\displaystyle b_{N}}$ is always ${\displaystyle 1}$.

${\displaystyle {\sqrt {2}}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+\ddots }}}}}}}$

For Napier's Constant, use ${\displaystyle a_{0}=2}$, then ${\displaystyle a_{N}=N}$. ${\displaystyle b_{1}=1}$ then ${\displaystyle b_{N}=N-1}$.

${\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {2}{3+{\cfrac {3}{4+\ddots }}}}}}}}}$

For Pi, use ${\displaystyle a_{0}=3}$ then ${\displaystyle a_{N}=6}$. ${\displaystyle b_{N}=(2N-1)^{2}}$.

${\displaystyle \pi =3+{\cfrac {1}{6+{\cfrac {9}{6+{\cfrac {25}{6+\ddots }}}}}}}$

(The source text for these examples can also be found on Bitbucket.)

Generic function for estimating continued fractions:

generic
type Scalar is digits <>;

with function A (N : in Natural) return Natural;
with function B (N : in Positive) return Natural;
function Continued_Fraction (Steps : in Natural) return Scalar;
function Continued_Fraction (Steps : in Natural) return Scalar is
function A (N : in Natural) return Scalar is (Scalar (Natural'(A (N))));
function B (N : in Positive) return Scalar is (Scalar (Natural'(B (N))));

Fraction : Scalar := 0.0;
begin
for N in reverse Natural range 1 .. Steps loop
Fraction := B (N) / (A (N) + Fraction);
end loop;
return A (0) + Fraction;
end Continued_Fraction;

Test program using the function above to estimate the square root of 2, Napiers constant and pi:

with Continued_Fraction;

procedure Test_Continued_Fractions is
type Scalar is digits 15;

package Square_Root_Of_2 is
function A (N : in Natural) return Natural is (if N = 0 then 1 else 2);
function B (N : in Positive) return Natural is (1);

function Estimate is new Continued_Fraction (Scalar, A, B);
end Square_Root_Of_2;

package Napiers_Constant is
function A (N : in Natural) return Natural is (if N = 0 then 2 else N);
function B (N : in Positive) return Natural is (if N = 1 then 1 else N-1);

function Estimate is new Continued_Fraction (Scalar, A, B);
end Napiers_Constant;

package Pi is
function A (N : in Natural) return Natural is (if N = 0 then 3 else 6);
function B (N : in Positive) return Natural is ((2 * N - 1) ** 2);

function Estimate is new Continued_Fraction (Scalar, A, B);
end Pi;

package Scalar_Text_IO is new Ada.Text_IO.Float_IO (Scalar);
begin
Put (Square_Root_Of_2.Estimate (200), Exp => 0); New_Line;
Put (Napiers_Constant.Estimate (200), Exp => 0); New_Line;
Put (Pi.Estimate (10000), Exp => 0); New_Line;
end Test_Continued_Fractions;

### Using only Ada 95 features

This example is exactly the same as the preceding one, but implemented using only Ada 95 features.

generic
type Scalar is digits <>;

with function A (N : in Natural) return Natural;
with function B (N : in Positive) return Natural;
function Continued_Fraction_Ada95 (Steps : in Natural) return Scalar;
function Continued_Fraction_Ada95 (Steps : in Natural) return Scalar is
function A (N : in Natural) return Scalar is
begin
return Scalar (Natural'(A (N)));
end A;

function B (N : in Positive) return Scalar is
begin
return Scalar (Natural'(B (N)));
end B;

Fraction : Scalar := 0.0;
begin
for N in reverse Natural range 1 .. Steps loop
Fraction := B (N) / (A (N) + Fraction);
end loop;
return A (0) + Fraction;

type Scalar is digits 15;

package Square_Root_Of_2 is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;

function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Square_Root_Of_2;

package body Square_Root_Of_2 is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 1;
else
return 2;
end if;
end A;

function B (N : in Positive) return Natural is
begin
return 1;
end B;
end Square_Root_Of_2;

package Napiers_Constant is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;

function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Napiers_Constant;

package body Napiers_Constant is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 2;
else
return N;
end if;
end A;

function B (N : in Positive) return Natural is
begin
if N = 1 then
return 1;
else
return N - 1;
end if;
end B;
end Napiers_Constant;

package Pi is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;

function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Pi;

package body Pi is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 3;
else
return 6;
end if;
end A;

function B (N : in Positive) return Natural is
begin
return (2 * N - 1) ** 2;
end B;
end Pi;

package Scalar_Text_IO is new Ada.Text_IO.Float_IO (Scalar);
begin
Put (Square_Root_Of_2.Estimate (200), Exp => 0); New_Line;
Put (Napiers_Constant.Estimate (200), Exp => 0); New_Line;
Put (Pi.Estimate (10000), Exp => 0); New_Line;
Output:
1.41421356237310
2.71828182845905
3.14159265358954

## ALGOL 68

Works with: Algol 68 Genie version 2.8

PROC cf = (INT steps, PROC (INT) INT a, PROC (INT) INT b) REAL:
BEGIN
REAL result;
result := 0;
FOR n FROM steps BY -1 TO 1 DO
result := b(n) / (a(n) + result)
OD;
a(0) + result
END;

PROC asqr2 = (INT n) INT: (n = 0 | 1 | 2);
PROC bsqr2 = (INT n) INT: 1;

PROC anap = (INT n) INT: (n = 0 | 2 | n);
PROC bnap = (INT n) INT: (n = 1 | 1 | n - 1);

PROC api = (INT n) INT: (n = 0 | 3 | 6);
PROC bpi = (INT n) INT: (n = 1 | 1 | (2 * n - 1) ** 2);

INT precision = 10000;

print (("Precision: ", precision, newline));
print (("Sqr(2): ", cf(precision, asqr2, bsqr2), newline));
print (("Napier: ", cf(precision, anap, bnap), newline));
print (("Pi: ", cf(precision, api, bpi)))

Output:

Precision:      +10000
Sqr(2):    +1.41421356237310e  +0
Napier:    +2.71828182845905e  +0
Pi:        +3.14159265358954e  +0

## ATS

A fairly direct translation of the C version without using advanced features of the type system:

#include
//
(* ****** ****** *)
//
(*
** a coefficient function creates double values from in paramters
*)
typedef coeff_f = int -> double
//
(*
** a continued fraction is described by a record of two coefficent
** functions a and b
*)
typedef frac = @{a= coeff_f, b= coeff_f}
//
(* ****** ****** *)

fun calc
(
f: frac, n: int
) : double = let
//
(*
** recursive definition of the approximation
*)
fun loop
(
n: int, r: double
) : double =
(
if n = 0
then f.a(0) + r
else loop (n - 1, f.b(n) / (f.a(n) + r))
// end of [if]
)
//
in
loop (n, 0.0)
end // end of [calc]

(* ****** ****** *)

val sqrt2 = @{
a= lam (n: int): double => if n = 0 then 1.0 else 2.0
,
b= lam (n: int): double => 1.0
} (* end of [val] *)

val napier = @{
a= lam (n: int): double => if n = 0 then 2.0 else 1.0 * n
,
b= lam (n: int): double => if n = 1 then 1.0 else n - 1.0
} (* end of [val] *)

val pi = @{
a= lam (n: int): double => if n = 0 then 3.0 else 6.0
,
b= lam (n: int): double => let val x = 2.0 * n - 1 in x * x end
}

(* ****** ****** *)

implement
main0 () =
(
println! ("sqrt2 = ", calc(sqrt2, 100));
println! ("napier = ", calc(napier, 100));
println! (" pi = ", calc( pi , 100));
) (* end of [main0] *)

## AutoHotkey

sqrt2_a(n) ; function definition is as simple as that
{
return n?2.0:1.0
}

sqrt2_b(n)
{
return 1.0
}

napier_a(n)
{
return n?n:2.0
}

napier_b(n)
{
return n>1.0?n-1.0:1.0
}

pi_a(n)
{
return n?6.0:3.0
}

pi_b(n)
{
return (2.0*n-1.0)**2.0 ; exponentiation operator
}

calc(f,expansions)
{
r:=0,i:=expansions
; A nasty trick: the names of the two coefficient functions are generated dynamically
; a dot surrounded by spaces means string concatenation
f_a:=f . "_a",f_b:=f . "_b"

while i>0 {
; You can see two dynamic function calls here
r:=%f_b%(i)/(%f_a%(i)+r)
i--
}

return %f_a%(0)+r
}

Msgbox, % "sqrt 2 = " . calc("sqrt2", 1000) . "ne = " . calc("napier", 1000) . "npi = " . calc("pi", 1000)

Output with Autohotkey v1 (currently 1.1.16.05):

sqrt 2 = 1.414214
e = 2.718282
pi = 3.141593

Output with Autohotkey v2 (currently alpha 56):

sqrt 2 = 1.4142135623730951
e = 2.7182818284590455
pi = 3.1415926533405418

Note the far superiour accuracy of v2.

## Axiom

Axiom provides a ContinuedFraction domain:

get(obj) == convergents(obj).1000 -- utility to extract the 1000th value
get continuedFraction(1, repeating [1], repeating [2]) :: Float
get continuedFraction(2, cons(1,[i for i in 1..]), [i for i in 1..]) :: Float
get continuedFraction(3, [i^2 for i in 1.. by 2], repeating [6]) :: Float
Output:
(1)  1.4142135623 730950488
Type: Float

(2) 2.7182818284 590452354
Type: Float

(3) 3.1415926538 39792926
Type: Float

The value for ${\displaystyle \pi }$ has an accuracy to only 9 decimal places after 1000 iterations, with an accuracy to 12 decimal places after 10000 iterations.

We could re-implement this, with the same output:

cf(initial, a, b, n) ==
n=1 => initial
temp := 0
for i in (n-1)..1 by -1 repeat
temp := a.i/(b.i+temp)
initial+temp
cf(1, repeating [1], repeating [2], 1000) :: Float
cf(2, cons(1,[i for i in 1..]), [i for i in 1..], 1000) :: Float
cf(3, [i^2 for i in 1.. by 2], repeating [6], 1000) :: Float

## BBC BASIC

*FLOAT64
@% = &1001010

PRINT "SQR(2) = " ; FNcontfrac(1, 1, "2", "1")
PRINT " e = " ; FNcontfrac(2, 1, "N", "N")
PRINT " PI = " ; FNcontfrac(3, 1, "6", "(2*N+1)^2")
END

REM a$and b$ are functions of N
DEF FNcontfrac(a0, b1, a$, b$)
LOCAL N, expr$REPEAT N += 1 expr$ += STR$(EVAL(a$)) + "+" + STR$(EVAL(b$)) + "/("
UNTIL LEN(expr$) > (65500 - N) = a0 + b1 / EVAL (expr$ + "1" + STRING$(N, ")")) Output: SQR(2) = 1.414213562373095 e = 2.718281828459046 PI = 3.141592653588017 ## C Works with: ANSI C /* calculate approximations for continued fractions */ #include <stdio.h> /* kind of function that returns a series of coefficients */ typedef double (*coeff_func)(unsigned n); /* calculates the specified number of expansions of the continued fraction * described by the coefficient series f_a and f_b */ double calc(coeff_func f_a, coeff_func f_b, unsigned expansions) { double a, b, r; a = b = r = 0.0; unsigned i; for (i = expansions; i > 0; i--) { a = f_a(i); b = f_b(i); r = b / (a + r); } a = f_a(0); return a + r; } /* series for sqrt(2) */ double sqrt2_a(unsigned n) { return n ? 2.0 : 1.0; } double sqrt2_b(unsigned n) { return 1.0; } /* series for the napier constant */ double napier_a(unsigned n) { return n ? n : 2.0; } double napier_b(unsigned n) { return n > 1.0 ? n - 1.0 : 1.0; } /* series for pi */ double pi_a(unsigned n) { return n ? 6.0 : 3.0; } double pi_b(unsigned n) { double c = 2.0 * n - 1.0; return c * c; } int main(void) { double sqrt2, napier, pi; sqrt2 = calc(sqrt2_a, sqrt2_b, 1000); napier = calc(napier_a, napier_b, 1000); pi = calc(pi_a, pi_b, 1000); printf("%12.10g\n%12.10g\n%12.10g\n", sqrt2, napier, pi); return 0; } Output: 1.414213562 2.718281828 3.141592653 ## C++ #include <iomanip> #include <iostream> #include <tuple> typedef std::tuple<double,double> coeff_t; // coefficients type typedef coeff_t (*func_t)(int); // callback function type double calc(func_t func, int n) { double a, b, temp = 0; for (; n > 0; --n) { std::tie(a, b) = func(n); temp = b / (a + temp); } std::tie(a, b) = func(0); return a + temp; } coeff_t sqrt2(int n) { return coeff_t(n > 0 ? 2 : 1, 1); } coeff_t napier(int n) { return coeff_t(n > 0 ? n : 2, n > 1 ? n - 1 : 1); } coeff_t pi(int n) { return coeff_t(n > 0 ? 6 : 3, (2 * n - 1) * (2 * n - 1)); } int main() { std::streamsize old_prec = std::cout.precision(15); // set output digits std::cout << calc(sqrt2, 20) << '\n' << calc(napier, 15) << '\n' << calc(pi, 10000) << '\n' << std::setprecision(old_prec); // reset precision } Output: 1.41421356237309 2.71828182845905 3.14159265358954 ## Clojure (defn cfrac [a b n] (letfn [(cfrac-iter [[x k]] [(+ (a k) (/ (b (inc k)) x)) (dec k)])] (ffirst (take 1 (drop (inc n) (iterate cfrac-iter [1 n])))))) (def sq2 (cfrac #(if (zero? %) 1.0 2.0) (constantly 1.0) 100)) (def e (cfrac #(if (zero? %) 2.0 %) #(if (= 1 %) 1.0 (double (dec %))) 100)) (def pi (cfrac #(if (zero? %) 3.0 6.0) #(let [x (- (* 2.0 %) 1.0)] (* x x)) 900000)) Output: user=> sq2 e pi 1.4142135623730951 2.7182818284590455 3.141592653589793 ## CoffeeScript # Compute a continuous fraction of the form # a0 + b1 / (a1 + b2 / (a2 + b3 / ... continuous_fraction = (f) -> a = f.a b = f.b c = 1 for n in [100000..1] c = b(n) / (a(n) + c) a(0) + c # A little helper. p = (a, b) -> console.log a console.log b console.log "---" do -> fsqrt2 = a: (n) -> if n is 0 then 1 else 2 b: (n) -> 1 p Math.sqrt(2), continuous_fraction(fsqrt2) fnapier = a: (n) -> if n is 0 then 2 else n b: (n) -> if n is 1 then 1 else n - 1 p Math.E, continuous_fraction(fnapier) fpi = a: (n) -> return 3 if n is 0 6 b: (n) -> x = 2*n - 1 x * x p Math.PI, continuous_fraction(fpi) Output: > coffee continued_fraction.coffee 1.4142135623730951 1.4142135623730951 --- 2.718281828459045 2.7182818284590455 --- 3.141592653589793 3.141592653589793 --- ## Common Lisp Translation of: C++ (defun estimate-continued-fraction (generator n) (let ((temp 0)) (loop for n1 from n downto 1 do (multiple-value-bind (a b) (funcall generator n1) (setf temp (/ b (+ a temp))))) (+ (funcall generator 0) temp))) (format t "sqrt(2) = ~a~%" (coerce (estimate-continued-fraction (lambda (n) (values (if (> n 0) 2 1) 1)) 20) 'double-float)) (format t "napier's = ~a~%" (coerce (estimate-continued-fraction (lambda (n) (values (if (> n 0) n 2) (if (> n 1) (1- n) 1))) 15) 'double-float)) (format t "pi = ~a~%" (coerce (estimate-continued-fraction (lambda (n) (values (if (> n 0) 6 3) (* (1- (* 2 n)) (1- (* 2 n))))) 10000) 'double-float)) Output: sqrt(2) = 1.4142135623730947d0 napier's = 2.7182818284590464d0 pi = 3.141592653589543d0 ## Chapel Functions don't take other functions as arguments, so I wrapped them in a dummy record each. proc calc(f, n) { var r = 0.0; for k in 1..n by -1 { var v = f.pair(k); r = v(2) / (v(1) + r); } return f.pair(0)(1) + r; } record Sqrt2 { proc pair(n) { return (if n == 0 then 1 else 2, 1); } } record Napier { proc pair(n) { return (if n == 0 then 2 else n, if n == 1 then 1 else n - 1); } } record Pi { proc pair(n) { return (if n == 0 then 3 else 6, (2*n - 1)**2); } } config const n = 200; writeln(calc(new Sqrt2(), n)); writeln(calc(new Napier(), n)); writeln(calc(new Pi(), n)); ## D import std.stdio, std.functional, std.traits; FP calc(FP, F)(in F fun, in int n) pure nothrow if (isCallable!F) { FP temp = 0; foreach_reverse (immutable ni; 1 .. n + 1) { immutable p = fun(ni); temp = p[1] / (FP(p[0]) + temp); } return fun(0)[0] + temp; } int[2] fSqrt2(in int n) pure nothrow { return [n > 0 ? 2 : 1, 1]; } int[2] fNapier(in int n) pure nothrow { return [n > 0 ? n : 2, n > 1 ? (n - 1) : 1]; } int[2] fPi(in int n) pure nothrow { return [n > 0 ? 6 : 3, (2 * n - 1) ^^ 2]; } alias print = curry!(writefln, "%.19f"); void main() { calc!real(&fSqrt2, 200).print; calc!real(&fNapier, 200).print; calc!real(&fPi, 200).print; } Output: 1.4142135623730950487 2.7182818284590452354 3.1415926228048469486 ## Erlang -module(continued). -compile([export_all]). pi_a (0) -> 3; pi_a (_N) -> 6. pi_b (N) -> (2*N-1)*(2*N-1). sqrt2_a (0) -> 1; sqrt2_a (_N) -> 2. sqrt2_b (_N) -> 1. nappier_a (0) -> 2; nappier_a (N) -> N. nappier_b (1) -> 1; nappier_b (N) -> N-1. continued_fraction(FA,_FB,0) -> FA(0); continued_fraction(FA,FB,N) -> continued_fraction(FA,FB,N-1,FB(N)/FA(N)). continued_fraction(FA,_FB,0,Acc) -> FA(0) + Acc; continued_fraction(FA,FB,N,Acc) -> continued_fraction(FA,FB,N-1,FB(N)/ (FA(N) + Acc)). test_pi (N) -> continued_fraction(fun pi_a/1,fun pi_b/1,N). test_sqrt2 (N) -> continued_fraction(fun sqrt2_a/1,fun sqrt2_b/1,N). test_nappier (N) -> continued_fraction(fun nappier_a/1,fun nappier_b/1,N). Output: 29> continued:test_pi(1000). 3.141592653340542 30> continued:test_sqrt2(1000). 1.4142135623730951 31> continued:test_nappier(1000). 2.7182818284590455 ## Factor cfrac-estimate uses rational arithmetic and never truncates the intermediate result. When terms is large, cfrac-estimate runs slow because numerator and denominator grow big. USING: arrays combinators io kernel locals math math.functions math.ranges prettyprint sequences ; IN: rosetta.cfrac ! Every continued fraction must implement these two words. GENERIC: cfrac-a ( n cfrac -- a ) GENERIC: cfrac-b ( n cfrac -- b ) ! square root of 2 SINGLETON: sqrt2 M: sqrt2 cfrac-a ! If n is 1, then a_n is 1, else a_n is 2. drop { { 1 [ 1 ] } [ drop 2 ] } case ; M: sqrt2 cfrac-b ! Always b_n is 1. 2drop 1 ; ! Napier's constant SINGLETON: napier M: napier cfrac-a ! If n is 1, then a_n is 2, else a_n is n - 1. drop { { 1 [ 2 ] } [ 1 - ] } case ; M: napier cfrac-b ! If n is 1, then b_n is 1, else b_n is n - 1. drop { { 1 [ 1 ] } [ 1 - ] } case ; SINGLETON: pi M: pi cfrac-a ! If n is 1, then a_n is 3, else a_n is 6. drop { { 1 [ 3 ] } [ drop 6 ] } case ; M: pi cfrac-b ! Always b_n is (n * 2 - 1)^2. drop 2 * 1 - 2 ^ ; :: cfrac-estimate ( cfrac terms -- number ) terms cfrac cfrac-a ! top = last a_n terms 1 - 1 [a,b] [ :> n n cfrac cfrac-b swap / ! top = b_n / top n cfrac cfrac-a + ! top = top + a_n ] each ; :: decimalize ( rational prec -- string ) rational 1 /mod ! split whole, fractional parts prec 10^ * ! multiply fraction by 10 ^ prec [ >integer unparse ] bi@ ! convert digits to strings :> fraction "." ! push decimal point prec fraction length - dup 0 < [ drop 0 ] when "0" <repetition> concat ! push padding zeros fraction 4array concat ; <PRIVATE : main ( -- ) " Square root of 2: " write sqrt2 50 cfrac-estimate 30 decimalize print "Napier's constant: " write napier 50 cfrac-estimate 30 decimalize print " Pi: " write pi 950 cfrac-estimate 10 decimalize print ; PRIVATE> MAIN: main Output: Square root of 2: 1.414213562373095048801688724209 Napier's constant: 2.718281828459045235360287471352 Pi: 3.1415926538 ## Felix fun pi (n:int) : (double*double) => let a = match n with | 0 => 3.0 | _ => 6.0 endmatch in let b = pow(2.0 * n.double - 1.0, 2.0) in (a,b); fun sqrt_2 (n:int) : (double*double) => let a = match n with | 0 => 1.0 | _ => 2.0 endmatch in let b = 1.0 in (a,b); fun napier (n:int) : (double*double) => let a = match n with | 0 => 2.0 | _ => n.double endmatch in let b = match n with | 1 => 1.0 | _ => (n.double - 1.0) endmatch in (a,b); fun cf_iter (steps:int) (f:int -> double*double) = { var acc = 0.0; for var n in steps downto 0 do var a, b = f(n); acc = if (n > 0) then (b / (a + acc)) else (acc + a); done return acc; } println$ cf_iter 200 sqrt_2; // => 1.41421
println$cf_iter 200 napier; // => 2.71818 println$ cf_iter 1000 pi; // => 3.14159

## Forth

Translation of: D
: fsqrt2 1 s>f 0> if 2 s>f else fdup then ;
: fnapier dup dup 1 > if 1- else drop 1 then s>f dup 1 < if drop 2 then s>f ;
: fpi dup 2* 1- dup * s>f 0> if 6 else 3 then s>f ;
( n -- f1 f2)
: cont.fraction ( xt n -- f)
1 swap 1+ 0 s>f \ calculate for 1 .. n
do i over execute frot f+ f/ -1 +loop
0 swap execute fnip f+ \ calcucate for 0
;
Output:
' fsqrt2  200 cont.fraction f. cr 1.4142135623731
ok
' fnapier 200 cont.fraction f. cr 2.71828182845905
ok
' fpi     200 cont.fraction f. cr 3.14159268391981
ok

## Fortran

module continued_fractions
implicit none

integer, parameter :: long = selected_real_kind(7,99)

type continued_fraction
integer :: a0, b1
procedure(series), pointer, nopass :: a, b
end type

interface
pure function series (n)
integer, intent(in) :: n
integer :: series
end function
end interface

contains

pure function define_cf (a0,a,b1,b) result(x)
integer, intent(in) :: a0
procedure(series) :: a
integer, intent(in), optional :: b1
procedure(series), optional :: b
type(continued_fraction) :: x
x%a0 = a0
x%a => a
if ( present(b1) ) then
x%b1 = b1
else
x%b1 = 1
end if
if ( present(b) ) then
x%b => b
else
x%b => const_1
end if
end function define_cf

pure integer function const_1(n)
integer,intent(in) :: n
const_1 = 1
end function

pure real(kind=long) function output(x,iterations)
type(continued_fraction), intent(in) :: x
integer, intent(in) :: iterations
integer :: i
output = x%a(iterations)
do i = iterations-1,1,-1
output = x%a(i) + (x%b(i+1) / output)
end do
output = x%a0 + (x%b1 / output)
end function output

end module continued_fractions

program examples
use continued_fractions

type(continued_fraction) :: sqr2,napier,pi

sqr2 = define_cf(1,a_sqr2)
napier = define_cf(2,a_napier,1,b_napier)
pi = define_cf(3,a=a_pi,b=b_pi)

write (*,*) output(sqr2,10000)
write (*,*) output(napier,10000)
write (*,*) output(pi,10000)

contains

pure integer function a_sqr2(n)
integer,intent(in) :: n
a_sqr2 = 2
end function

pure integer function a_napier(n)
integer,intent(in) :: n
a_napier = n
end function

pure integer function b_napier(n)
integer,intent(in) :: n
b_napier = n-1
end function

pure integer function a_pi(n)
integer,intent(in) :: n
a_pi = 6
end function

pure integer function b_pi(n)
integer,intent(in) :: n
b_pi = (2*n-1)*(2*n-1)
end function

end program examples
Output:
1.4142135623730951
2.7182818284590455
3.1415926535895435

## Go

package main

import "fmt"

type cfTerm struct {
a, b int
}

// follows subscript convention of mathworld and WP where there is no b(0).
// cf[0].b is unused in this representation.
type cf []cfTerm

func cfSqrt2(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
f[n] = cfTerm{2, 1}
}
f[0].a = 1
return f
}

func cfNap(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
f[n] = cfTerm{n, n - 1}
}
f[0].a = 2
f[1].b = 1
return f
}

func cfPi(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
g := 2*n - 1
f[n] = cfTerm{6, g * g}
}
f[0].a = 3
return f
}

func (f cf) real() (r float64) {
for n := len(f) - 1; n > 0; n-- {
r = float64(f[n].b) / (float64(f[n].a) + r)
}
return r + float64(f[0].a)
}

func main() {
fmt.Println("sqrt2:", cfSqrt2(20).real())
fmt.Println("nap: ", cfNap(20).real())
fmt.Println("pi: ", cfPi(20).real())
}
Output:
sqrt2: 1.4142135623730965
nap:   2.7182818284590455
pi:    3.141623806667839

import Data.List (unfoldr)
import Data.Char (intToDigit)

-- continued fraction represented as a (possibly infinite) list of pairs
sqrt2, napier, myPi :: [(Integer, Integer)]
sqrt2 = zip (1 : [2,2..]) [1,1..]
napier = zip (2 : [1..]) (1 : [1..])
myPi = zip (3 : [6,6..]) (map (^2) [1,3..])

-- approximate a continued fraction after certain number of iterations
approxCF :: (Integral a, Fractional b) => Int -> [(a, a)] -> b
approxCF t =
foldr (\(a,b) z -> fromIntegral a + fromIntegral b / z) 1 . take t

-- infinite decimal representation of a real number
decString :: RealFrac a => a -> String
decString frac = show i ++ '.' : decString' f where
(i,f) = properFraction frac
decString'
= map intToDigit . unfoldr (Just . properFraction . (10*))

main :: IO ()
main = mapM_ (putStrLn . take 200 . decString .
(approxCF 950 :: [(Integer, Integer)] -> Rational))
[sqrt2, napier, myPi]
Output:
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222665927505592755799950501152782060571
2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019
3.141592653297590947683406834261190738869139611505752231394089152890909495973464508817163306557131591579057202097715021166512662872910519439747609829479577279606075707015622200744006783543589980682386
import Data.Ratio

-- ignoring the task-given pi sequence: sucky convergence
-- pie = zip (3:repeat 6) (map (^2) [1,3..])

pie = zip (0:[1,3..]) (4:map (^2) [1..])
sqrt2 = zip (1:repeat 2) (repeat 1)
napier = zip (2:[1..]) (1:[1..])

-- truncate after n terms
cf2rat n = foldr (\(a,b) f -> (a%1) + ((b%1) / f)) (1%1) . take n

-- truncate after error is at most 1/p
cf2rat_p p s = f $map (\i -> (cf2rat i s, cf2rat (1+i) s))$ map (2^) [0..]
where f ((x,y):ys) = if abs (x-y) < (1/fromIntegral p) then x else f ys

-- returns a decimal string of n digits after the dot; all digits should
-- be correct (doesn't mean it's the best approximation! the decimal
-- string is simply truncated to given digits: pi=3.141 instead of 3.142)
cf2dec n = (ratstr n) . cf2rat_p (10^n) where
ratstr l a = (show t) ++ '.':fracstr l n d where
d = denominator a
(t, n) = quotRem (numerator a) d
fracstr 0 _ _ = []
fracstr l n d = (show t)++ fracstr (l-1) n1 d where (t,n1) = quotRem (10 * n) d

main = do
putStrLn $cf2dec 200 sqrt2 putStrLn$ cf2dec 200 napier
putStrLn $cf2dec 200 pie ## J cfrac=: +% / NB. Evaluate a list as a continued fraction sqrt2=: cfrac 1 1,200$2 1x
pi=:cfrac 3, , ,&6"0 *:<:+:>:i.100x
e=: cfrac 2 1, , ,~"0 >:i.100x

NB. translate from fraction to decimal string
NB. translated from factor
dec =: (-@:[ (}.,'.',{.) ":@:<.@:(* 10x&^)~)"0

100 10 100 dec sqrt2, pi, e
1.4142135623730950488016887242096980785696718753769480731766797379907324784621205551109457595775322165
3.1415924109
2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274

Note that there are two kinds of continued fractions. The kind here where we alternate between a and b values, but in some other tasks b is always 1 (and not included in the list we use to represent the continued fraction). The other kind is evaluated in J using (+%)/ instead of +%/.

## Java

Translation of: D
Works with: Java version 8
import static java.lang.Math.pow;
import java.util.*;
import java.util.function.Function;

public class Test {
static double calc(Function<Integer, Integer[]> f, int n) {
double temp = 0;

for (int ni = n; ni >= 1; ni--) {
Integer[] p = f.apply(ni);
temp = p[1] / (double) (p[0] + temp);
}
return f.apply(0)[0] + temp;
}

public static void main(String[] args) {
List<Function<Integer, Integer[]>> fList = new ArrayList<>();
fList.add(n -> new Integer[]{n > 0 ? 2 : 1, 1});
fList.add(n -> new Integer[]{n > 0 ? n : 2, n > 1 ? (n - 1) : 1});
fList.add(n -> new Integer[]{n > 0 ? 6 : 3, (int) pow(2 * n - 1, 2)});

for (Function<Integer, Integer[]> f : fList)
System.out.println(calc(f, 200));
}
}
1.4142135623730951
2.7182818284590455
3.141592622804847

## jq

Works with: jq version 1.4

We take one of the points of interest here to be the task of representing the infinite series a0, a1, .... and b0, b1, .... compactly, preferably functionally. For the type of series typically encountered in continued fractions, this is most readily accomplished in jq 1.4 using a filter (a function), here called "next", which, given the triple [i, [a[i], b[i]], will produce the next triple [i+1, a[i+1], b[i+1]].

Another point of interest is avoiding having to specify the number of iterations. The approach adopted here allows one to specify the desired accuracy; in some cases, it is feasible to specify that the computation should continue until the accuracy permitted by the underlying floating point representation is achieved. This is done by specifying delta as 0, as shown in the examples.

We therefore proceed in two steps: continued_fraction( first; next; count ) computes an approximation based on the first "count" terms; and then continued_fraction_delta(first; next; delta) computes the continued fraction until the difference in approximations is less than or equal to delta, which may be 0, as previously noted.

# "first" is the first triple,
# e.g. [1,a,b]; count specifies the number of terms to use.
def continued_fraction( first; next; count ):
# input: [i, a, b]]
def cf:
if .[0] == count then 0
else next as $ab | .[1] + (.[2] / ($ab | cf))
end ;
first | cf;

# "first" and "next" are as above;
# if delta is 0 then continue until there is no detectable change.
def continued_fraction_delta(first; next; delta):
def abs: if . < 0 then -. else . end;
def cf:
# state: [n, prev]
.[0] as $n | .[1] as$prev
| continued_fraction(first; next; $n+1) as$this
| if $prev == null then [$n+1, $this] | cf elif delta <= 0 and ($prev == $this) then$this
elif (($prev -$this)|abs) <= delta then $this else [$n+1, $this] | cf end; [2,null] | cf; Examples: The convergence for pi is slow so we select delta = 1e-12 in that case. "Value : Direct : Continued Fraction", "2|sqrt : \(2|sqrt) : \(continued_fraction_delta( [1,1,1]; [.[0]+1, 2, 1]; 0))", "1|exp : \(1|exp) : \(2 + (1 / (continued_fraction_delta( [1,1,1]; [.[0]+1, .[1]+1, .[2]+1]; 0))))", "pi : \(1|atan * 4) : \(continued_fraction_delta( [1,3,1]; .[0]+1 | [., 6, ((2*. - 1) | (.*.))]; 1e-12)) (1e-12)" Output:$ jq -M -n -r -f Continued_fraction.jq
Value  : Direct  : Continued Fraction
2|sqrt : 1.4142135623730951 : 1.4142135623730951
1|exp  : 2.718281828459045  : 2.7182818284590455
pi  : 3.141592653589793  : 3.1415926535892935 (1e-12)

## Julia

function sqrt2(a,n)
if a
return n>0?2.0:1.0
else
return 1.0
end
end

function napier(a,n)
if a
return n>0?Float64(n):2.0
else
return n>1?n-1.0:1.0
end
end

function _pi(a,n)
if a
return n>0?6.0:3.0
else
return (2.0*n-1.0)^2.0 # exponentiation operator
end
end

function calc(f,expansions)
a=true;b=false
r=0.0
for i=expansions:-1:1
r=f(b,i)/(f(a,i)+r)
end
return f(a,0)+r
end

print("""
sqrt 2 = $(calc(sqrt2, 1000)) e =$(calc(napier, 1000))
pi = $(calc(_pi, 1000)) """) Output: sqrt 2 = 1.4142135623730951 e = 2.7182818284590455 pi = 3.141592653340542 ## Mathematica / Wolfram Language sqrt2=Function[n,{1,Transpose@{Array[2&,n],Array[1&,n]}}]; napier=Function[n,{2,Transpose@{Range[n],Prepend[Range[n-1],1]}}]; pi=Function[n,{3,Transpose@{Array[6&,n],Array[(2#-1)^2&,n]}}]; approx=Function[l, N[Divide@@[email protected][{{#2.#[[;;,1]],#2.#[[;;,2]]},#[[1]]}&,{{l[[2,1,1]]l[[1]]+l[[2,1,2]],l[[2,1,1]]},{l[[1]],1}},l[[2,2;;]]],10]]; r2=approx/@{sqrt2@#,napier@#,pi@#}&@10000;r2//TableForm Output: 1.414213562 2.718281828 3.141592654 ## Maxima cfeval(x) := block([a, b, n, z], a: x[1], b: x[2], n: length(a), z: 0, for i from n step -1 thru 2 do z: b[i]/(a[i] + z), a[1] + z)$

cf_sqrt2(n) := [cons(1, makelist(2, i, 2, n)), cons(0, makelist(1, i, 2, n))]$cf_e(n) := [cons(2, makelist(i, i, 1, n - 1)), append([0, 1], makelist(i, i, 1, n - 2))]$

cf_pi(n) := [cons(3, makelist(6, i, 2, n)), cons(0, makelist((2*i - 1)^2, i, 1, n - 1))]$cfeval(cf_sqrt2(20)), numer; /* 1.414213562373097 */ % - sqrt(2), numer; /* 1.3322676295501878*10^-15 */ cfeval(cf_e(20)), numer; /* 2.718281828459046 */ % - %e, numer; /* 4.4408920985006262*10^-16 */ cfeval(cf_pi(20)), numer; /* 3.141623806667839 */ % - %pi, numer; /* 3.115307804568701*10^-5 */ /* convergence is much slower for pi */ fpprec: 20$
x: cfeval(cf_pi(10000))$bfloat(x - %pi); /* 2.4999999900104930006b-13 */ ## NetRexx /* REXX *************************************************************** * Derived from REXX ... Derived from PL/I with a little "massage" * SQRT2= 1.41421356237309505 <- PL/I Result * 1.41421356237309504880168872421 <- NetRexx Result 30 digits * NAPIER= 2.71828182845904524 * 2.71828182845904523536028747135 * PI= 3.14159262280484695 * 3.14159262280484694855146925223 * 07.09.2012 Walter Pachl * 08.09.2012 Walter Pachl simplified (with the help of a friend) **********************************************************************/ options replace format comments java crossref savelog symbols class CFB public properties static Numeric Digits 30 Sqrt2 =1 napier=2 pi =3 a =0 b =0 method main(args = String[]) public static Say 'SQRT2='.left(7) calc(sqrt2, 200) Say 'NAPIER='.left(7) calc(napier, 200) Say 'PI='.left(7) calc(pi, 200) Return method get_Coeffs(form,n) public static select when form=Sqrt2 Then do if n > 0 then a = 2; else a = 1 b = 1 end when form=Napier Then do if n > 0 then a = n; else a = 2 if n > 1 then b = n - 1; else b = 1 end when form=pi Then do if n > 0 then a = 6; else a = 3 b = (2*n - 1)**2 end end Return method calc(form,n) public static temp=0 loop ni = n to 1 by -1 Get_Coeffs(form,ni) temp = b/(a + temp) end Get_Coeffs(form,0) return (a + temp) Who could help me make a,b,sqrt2,napier,pi global (public) variables? This would simplify the solution:-) I got this help and simplified the program. However, I am told that 'my' value of pi is incorrect. I will investigate! Apparently the coefficients given in the task description are only good for an approximation. One should, therefore, not SHOW more that 15 digits. See http://de.wikipedia.org/wiki/Kreiszahl See Rexx for a better computation ## OCaml let pi = 3, fun n -> ((2*n-1)*(2*n-1), 6) and nap = 2, fun n -> (max 1 (n-1), n) and root2 = 1, fun n -> (1, 2) in let eval (i,f) k = let rec frac n = let a, b = f n in float a /. (float b +. if n >= k then 0.0 else frac (n+1)) in float i +. frac 1 in Printf.printf "sqrt(2)\t= %.15f\n" (eval root2 1000); Printf.printf "e\t= %.15f\n" (eval nap 1000); Printf.printf "pi\t= %.15f\n" (eval pi 1000); Output (inaccurate due to too few terms): sqrt(2) = 1.414213562373095 e = 2.718281828459046 pi = 3.141592653340542 ## PARI/GP Partial solution for simple continued fractions. back(v)=my(t=contfracpnqn(v));t[1,1]/t[2,1]*1. back(vector(100,i,2-(i==1))) Output: %1 = 1.4142135623730950488016887242096980786 ## Perl We'll use closures to implement the infinite lists of coeffficients. sub continued_fraction { my ($a, $b,$n) = (@_[0,1], $_[2] // 100);$a->() + ($n &&$b->() / continued_fraction($a,$b, $n-1)); } printf "√2 ≈ %.9f\n", continued_fraction do { my$n; sub { $n++ ? 2 : 1 } }, sub { 1 }; printf "e ≈ %.9f\n", continued_fraction do { my$n; sub { $n++ || 2 } }, do { my$n; sub { $n++ || 1 } }; printf "π ≈ %.9f\n", continued_fraction do { my$n; sub { $n++ ? 6 : 3 } }, do { my$n; sub { (2*$n++ + 1)**2 } }, 1_000; printf "π/2 ≈ %.9f\n", continued_fraction do { my$n; sub { 1/($n++ || 1) } }, sub { 1 }, 1_000; Output: √2 ≈ 1.414213562 e ≈ 2.718281828 π ≈ 3.141592653 π/2 ≈ 1.570717797 ## Perl 6 Works with: rakudo version 2015-10-31 sub continued-fraction(:@a, :@b, Int :$n = 100)
{
my $x = @a[$n - 1];
$x = @a[$_ - 1] + @b[$_] /$x for reverse 1 ..^ $n;$x;
}

printf "√2 ≈%.9f\n", continued-fraction(:a(1, |(2 xx *)), :b(Nil, |(1 xx *)));
printf "e ≈%.9f\n", continued-fraction(:a(2, |(1 .. *)), :b(Nil, 1, |(1 .. *)));
printf "π ≈%.9f\n", continued-fraction(:a(3, |(6 xx *)), :b(Nil, |((1, 3, 5 ... *) X** 2)));
Output:
√2 ≈ 1.414213562
e  ≈ 2.718281828
π  ≈ 3.141592654

A more original and a bit more abstract method would consist in viewing a continued fraction on rank n as a function of a variable x:

${\displaystyle \mathrm {CF} _{3}(x)=a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+x}}}}}}}$

Or, more consistently:

${\displaystyle \mathrm {CF} _{3}(x)=a_{0}+{\cfrac {b_{0}}{a_{1}+{\cfrac {b_{1}}{a_{2}+{\cfrac {b_{2}}{a_{3}+{\cfrac {b_{3}}{x}}}}}}}}}$

Viewed as such, ${\displaystyle \mathrm {CF} _{n}(x)}$ could be written recursively:

${\displaystyle \mathrm {CF} _{n}(x)=\mathrm {CF} _{n-1}(a_{n}+{\frac {b_{n}}{x}})}$

Or in other words:

${\displaystyle \mathrm {CF} _{n}=\mathrm {CF} _{n-1}\circ f_{n}=\mathrm {CF} _{n-2}\circ f_{n-1}\circ f_{n}=\ldots =f_{0}\circ f_{1}\ldots \circ f_{n}}$

where ${\displaystyle f_{n}(x)=a_{n}+{\frac {b_{n}}{x}}}$

Perl6 has a builtin composition operator. We can use it with the triangular reduction metaoperator, and evaluate each resulting function at infinity (any value would do actually, but infinite makes it consistent with this particular task).

sub continued-fraction(@a, @b) {
map { .(Inf) }, [\o] map { @a[$_] + @b[$_] / * }, ^Inf
}

printf "√2 ≈ %.9f\n", continued-fraction((1, |(2 xx *)), (1 xx *))[10];
printf "e ≈ %.9f\n", continued-fraction((2, |(1 .. *)), (1, |(1 .. *)))[10];
printf "π ≈ %.9f\n", continued-fraction((3, |(6 xx *)), ((1, 3, 5 ... *) X** 2))[100];
Output:
√2 ≈ 1.414213552
e  ≈ 2.718281827
π  ≈ 3.141592411

## PL/I

/* Version for SQRT(2) */
test: proc options (main);
declare n fixed;

denom: procedure (n) recursive returns (float (18));
declare n fixed;
n = n + 1;
if n > 100 then return (2);
return (2 + 1/denom(n));
end denom;

put (1 + 1/denom(2));

end test;
Output:
1.41421356237309505E+0000

Version for NAPIER:

test: proc options (main);
declare n fixed;

denom: procedure (n) recursive returns (float (18));
declare n fixed;
n = n + 1;
if n > 100 then return (n);
return (n + n/denom(n));
end denom;

put (2 + 1/denom(0));

end test;
2.71828182845904524E+0000

Version for SQRT2, NAPIER, PI

/* Derived from continued fraction in Wiki Ada program */

continued_fractions: /* 6 Sept. 2012 */
procedure options (main);
declare (Sqrt2 initial (1), napier initial (2), pi initial (3)) fixed (1);

Get_Coeffs: procedure (form, n, coefA, coefB);
declare form fixed (1), n fixed, (coefA, coefB) float (18);

select (form);
when (Sqrt2) do;
if n > 0 then coefA = 2; else coefA = 1;
coefB = 1;
end;
when (Napier) do;
if n > 0 then coefA = n; else coefA = 2;
if n > 1 then coefB = n - 1; else coefB = 1;
end;
when (Pi) do;
if n > 0 then coefA = 6; else coefA = 3;
coefB = (2*n - 1)**2;
end;
end;
end Get_Coeffs;

Calc: procedure (form, n) returns (float (18));
declare form fixed (1), n fixed;
declare (A, B) float (18);
declare Temp float (18) initial (0);
declare ni fixed;

do ni = n to 1 by -1;
call Get_Coeffs (form, ni, A, B);
Temp = B/(A + Temp);
end;
call Get_Coeffs (form, 0, A, B);
return (A + Temp);
end Calc;

put edit ('SQRT2=', calc(sqrt2, 200)) (a(10), f(20,17));
put skip edit ('NAPIER=', calc(napier, 200)) (a(10), f(20,17));
put skip edit ('PI=', calc(pi, 99999)) (a(10), f(20,17));

end continued_fractions;
Output:
SQRT2=     1.41421356237309505
NAPIER=    2.71828182845904524
PI=        3.14159265358979349

## Prolog

continued_fraction :-
% square root 2
continued_fraction(200, sqrt_2_ab, V1),
format('sqrt(2) = ~w~n', [V1]),

% napier
continued_fraction(200, napier_ab, V2),
format('e = ~w~n', [V2]),

% pi
continued_fraction(200, pi_ab, V3),
format('pi = ~w~n', [V3]).

% code for continued fractions
continued_fraction(N, Compute_ab, V) :-
continued_fraction(N, Compute_ab, 0, V).

continued_fraction(0, Compute_ab, Temp, V) :-
call(Compute_ab, 0, A, _),
V is A + Temp.

continued_fraction(N, Compute_ab, Tmp, V) :-
call(Compute_ab, N, A, B),
Tmp1 is B / (A + Tmp),
N1 is N - 1,
continued_fraction(N1, Compute_ab, Tmp1, V).

% specific codes for examples
% definitions for square root of 2
sqrt_2_ab(0, 1, 1).
sqrt_2_ab(_, 2, 1).

% definitions for napier
napier_ab(0, 2, _).
napier_ab(1, 1, 1).
napier_ab(N, N, V) :-
V is N - 1.

% definitions for pi
pi_ab(0, 3, _).
pi_ab(N, 6, V) :-
V is (2 * N - 1)*(2 * N - 1).
Output:
?- continued_fraction.
sqrt(2) = 1.4142135623730951
e       = 2.7182818284590455
pi      = 3.141592622804847
true .

## Python

Works with: Python version 2.6+ and 3.x
from fractions import Fraction
import itertools
try: zip = itertools.izip
except: pass

# The Continued Fraction
def CF(a, b, t):
terms = list(itertools.islice(zip(a, b), t))
z = Fraction(1,1)
for a, b in reversed(terms):
z = a + b / z
return z

# Approximates a fraction to a string
def pRes(x, d):
q, x = divmod(x, 1)
res = str(q)
res += "."
for i in range(d):
x *= 10
q, x = divmod(x, 1)
res += str(q)
return res

# Test the Continued Fraction for sqrt2
def sqrt2_a():
yield 1
for x in itertools.repeat(2):
yield x

def sqrt2_b():
for x in itertools.repeat(1):
yield x

cf = CF(sqrt2_a(), sqrt2_b(), 950)
print(pRes(cf, 200))
#1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147

# Test the Continued Fraction for Napier's Constant
def Napier_a():
yield 2
for x in itertools.count(1):
yield x

def Napier_b():
yield 1
for x in itertools.count(1):
yield x

cf = CF(Napier_a(), Napier_b(), 950)
print(pRes(cf, 200))
#2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901

# Test the Continued Fraction for Pi
def Pi_a():
yield 3
for x in itertools.repeat(6):
yield x

def Pi_b():
for x in itertools.count(1,2):
yield x*x

cf = CF(Pi_a(), Pi_b(), 950)
print(pRes(cf, 10))
#3.1415926532

### Fast iterative version

Translation of: D
from decimal import Decimal, getcontext

def calc(fun, n):
temp = Decimal("0.0")

for ni in xrange(n+1, 0, -1):
(a, b) = fun(ni)
temp = Decimal(b) / (a + temp)

return fun(0)[0] + temp

def fsqrt2(n):
return (2 if n > 0 else 1, 1)

def fnapier(n):
return (n if n > 0 else 2, (n - 1) if n > 1 else 1)

def fpi(n):
return (6 if n > 0 else 3, (2 * n - 1) ** 2)

getcontext().prec = 50
print calc(fsqrt2, 200)
print calc(fnapier, 200)
print calc(fpi, 200)
Output:
1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134

## Racket

### Using Doubles

This version uses standard double precision floating point numbers:

#lang racket
(define (calc cf n)
(match/values (cf 0)
[(a0 b0)
(+ a0
(for/fold ([t 0.0]) ([i (in-range (+ n 1) 0 -1)])
(match/values (cf i)
[(a b) (/ b (+ a t))])))]))

(define (cf-sqrt i) (values (if (> i 0) 2 1) 1))
(define (cf-napier i) (values (if (> i 0) i 2) (if (> i 1) (- i 1) 1)))
(define (cf-pi i) (values (if (> i 0) 6 3) (sqr (- (* 2 i) 1))))

(calc cf-sqrt 200)
(calc cf-napier 200)
(calc cf-pi 200)

Output:

1.4142135623730951
2.7182818284590455
3.1415926839198063

### Version - Using Doubles

This versions uses big floats (arbitrary precision floating point):

#lang racket
(require math)
(bf-precision 2048) ; in bits

(define (calc cf n)
(match/values (cf 0)
[(a0 b0)
(bf+ (bf a0)
(for/fold ([t (bf 0)]) ([i (in-range (+ n 1) 0 -1)])
(match/values (cf i)
[(a b) (bf/ (bf b) (bf+ (bf a) t))])))]))

(define (cf-sqrt i) (values (if (> i 0) 2 1) 1))
(define (cf-napier i) (values (if (> i 0) i 2) (if (> i 1) (- i 1) 1)))
(define (cf-pi i) (values (if (> i 0) 6 3) (sqr (- (* 2 i) 1))))

(calc cf-sqrt 200)
(calc cf-napier 200)
(calc cf-pi 200)

Output:

(bf #e1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727350138462309122970249248360558507372126441214970999358960036439214262599769155193770031712304888324413327207659690547583107739957489062466508437105234564161085482146113860092820802430986649987683947729823677905101453725898480737256099166805538057375451207262441039818826744940289448489312217214883459060818483750848688583833366310472320771259749181255428309841375829513581694269249380272698662595131575038315461736928338289219865139248048189188905788104310928762952913687232022557677738108337499350045588767581063729)
(bf #e2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901157383418793070215408914993488416750924476146066808226480016847741185374234544243710753907774499206955170276183860626133138458300075204493382656029760673711320070932870912744374704723624212700454495421842219077173525899689811474120614457405772696521446961165559468253835854362096088934714907384964847142748311021268578658461064714894910680584249490719358138073078291397044213736982988247857479512745588762993966446075)
(bf #e3.14159268391980626493420192940831754203350026401337226640663040854412059241988978103217808449508253393479795573626200366332733859609651462659489470805432281782785922056335606047700127154963266242144951481397480765182268219697420028007903565511884267297358842935537138583640066772149177226656227031792115896439889412205871076985598822285367358003457939603015797225018209619662200081521930463480571130673429337524564941105654923909951299948539893933654293161126559643573974163405197696633200469475250152247413175932572922175467223988860975105100904322239324381097207835036465269418118204894206705789759765527734394105147)

## REXX

### version 1

The   cf   subroutine   (for Continued Fractions)   isn't limited to positive integers.
Any form of REXX numbers (negative, exponentiated, decimal fractions) can be used.
Note the use of negative fractions for the   ß   terms when computing   ½  .

There isn't any practical limit for the decimal digits that can be used, although 100k digits would be a bit unwieldy to display.

A generalized       function was added to calculate a few low integers   (and also   1/2).

More code is used for nicely formatting the output than the continued fraction calculation.

/*REXX program  calculates and displays  values of  various  continued fractions.       */
parse arg terms digs .
if terms=='' | terms=="," then terms=500
if digs=='' | digs=="," then digs=100
numeric digits digs /*use 100 decimal digits for display.*/
b.=1 /*omitted ß terms are assumed to be 1.*/
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=2; call tell '√2', cf(1)
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=1; do N=2 by 2 to terms; a.N=2; end; call tell '√3', cf(1) /*also: 2∙sin(π/3) */
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=2 /* ___ */
do N=2 to 17 /*generalized √ N */
b.=N-1; NN=right(N, 2); call tell 'gen √'NN, cf(1)
end /*N*/
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=2; b.=-1/2; call tell 'gen √ ½', cf(1)
/*══════════════════════════════════════════════════════════════════════════════════════*/
do j=1 for terms; a.j=j; if j>1 then b.j=a.p; p=j; end; call tell 'e', cf(2)
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=1; call tell 'φ, phi', cf(1)
/*══════════════════════════════════════════════════════════════════════════════════════*/
a.=1; do j=1 for terms; if j//2 then a.j=j; end; call tell 'tan(1)', cf(1)
/*══════════════════════════════════════════════════════════════════════════════════════*/
do j=1 for terms; a.j=2*j+1; end; call tell 'coth(1)', cf(1)
/*══════════════════════════════════════════════════════════════════════════════════════*/
do j=1 for terms; a.j=4*j+2; end; call tell 'coth(½)', cf(2) /*also: [e+1]÷[e-1] */
/*══════════════════════════════════════════════════════════════════════════════════════*/
terms=100000
a.=6; do j=1 for terms; b.j=(2*j-1)**2; end; call tell 'π, pi', cf(3)
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
cf: procedure expose a. b. terms; parse arg C;  !=0; numeric digits 9+digits()
do k=terms by -1 for terms; d=a.k+!;  !=b.k/d
end /*k*/
return !+C
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell: parse arg ?,v; $=left(format(v)/1,1+digits()); w=50 /*50 bytes of terms*/ aT=; do k=1; _=space(aT a.k); if length(_)>w then leave; aT=_; end /*k*/ bT=; do k=1; _=space(bT b.k); if length(_)>w then leave; bT=_; end /*k*/ say right(?,8) "="$ ' α terms='aT ...
if b.1\==1 then say right("",12+digits()) ' ß terms='bT ...
a=; b.=1; return /*only 50 bytes of α & ß terms ↑ are displayed. */

output

√2 = 1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
√3 = 1.732050807568877293527446341505872366942805253810380628055806979451933016908800037081146186757248576   α terms=1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 ...
gen √ 2 = 1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
gen √ 3 = 1.732050807568877293527446341505872366942805253810380628055806979451933016908800037081146186757248576   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
gen √ 4 = 2                                                                                                       α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 ...
gen √ 5 = 2.236067977499789696409173668731276235440618359611525724270897245410520925637804899414414408378782275   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 ...
gen √ 6 = 2.449489742783178098197284074705891391965947480656670128432692567250960377457315026539859433104640235   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ...
gen √ 7 = 2.645751311064590590501615753639260425710259183082450180368334459201068823230283627760392886474543611   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 ...
gen √ 8 = 2.828427124746190097603377448419396157139343750753896146353359475981464956924214077700775068655283145   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 ...
gen √ 9 = 3                                                                                                       α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 ...
gen √10 = 3.162277660168379331998893544432718533719555139325216826857504852792594438639238221344248108379300295   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 ...
gen √11 = 3.316624790355399849114932736670686683927088545589353597058682146116484642609043846708843399128290651   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 ...
gen √12 = 3.464101615137754587054892683011744733885610507620761256111613958903866033817600074162292373514497151   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 ...
gen √13 = 3.605551275463989293119221267470495946251296573845246212710453056227166948293010445204619082018490718   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 ...
gen √14 = 3.741657386773941385583748732316549301756019807778726946303745467320035156306939027976809895194379572   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 ...
gen √15 = 3.872983346207416885179265399782399610832921705291590826587573766113483091936979033519287376858673518   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ...
gen √16 = 4                                                                                                       α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 ...
gen √17 = 4.123105625617660549821409855974077025147199225373620434398633573094954346337621593587863650810684297   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 ...
gen √ ½ = 0.707106781186547524400844362104849039284835937688474036588339868995366239231053519425193767163820786   α terms=2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...
ß terms=-0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 ...
e = 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427   α terms=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
φ, phi = 1.618033988749894848204586834365638117720309179805762862135448622705260462818902449707207204189391137   α terms=1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ...
tan(1) = 1.557407724654902230506974807458360173087250772381520038383946605698861397151727289555099965202242984   α terms=1 1 3 1 5 1 7 1 9 1 11 1 13 1 15 1 17 1 19 1 21 1 ...
coth(1) = 1.313035285499331303636161246930847832912013941240452655543152967567084270461874382674679241480856303   α terms=3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 ...
coth(½) = 2.163953413738652848770004010218023117093738602150792272533574119296087634783339486574409418809750115   α terms=6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 ...
π, pi = 3.141592653589792988470143264530440384041017830472772036746332303472711537960073664096818977224037083   α terms=6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 ...

Note:   even with 200 digit accuracy and 100,000 terms, the last calculation of pi is only accurate to 15 digits.

### version 2 derived from PL/I

/* REXX **************************************************************
* Derived from PL/I with a little "massage"
* SQRT2= 1.41421356237309505 <- PL/I Result
* 1.41421356237309504880168872421 <- REXX Result 30 digits
* NAPIER= 2.71828182845904524
* 2.71828182845904523536028747135
* PI= 3.14159262280484695
* 3.14159262280484694855146925223
* 06.09.2012 Walter Pachl
**********************************************************************/

Numeric Digits 30
Parse Value '1 2 3 0 0' with Sqrt2 napier pi a b
Say left('SQRT2=' ,10) calc(sqrt2, 200)
Say left('NAPIER=',10) calc(napier, 200)
Say left('PI=' ,10) calc(pi, 200)
Exit

Get_Coeffs: procedure Expose a b Sqrt2 napier pi
Parse Arg form, n
select
when form=Sqrt2 Then do
if n > 0 then a = 2; else a = 1
b = 1
end
when form=Napier Then do
if n > 0 then a = n; else a = 2
if n > 1 then b = n - 1; else b = 1
end
when form=pi Then do
if n > 0 then a = 6; else a = 3
b = (2*n - 1)**2
end
end
Return

Calc: procedure Expose a b Sqrt2 napier pi
Parse Arg form,n
Temp=0
do ni = n to 1 by -1
Call Get_Coeffs form, ni
Temp = B/(A + Temp)
end
call Get_Coeffs form, 0
return (A + Temp)

### version 3 better approximation

/* REXX *************************************************************
* The task description specifies a continued fraction for pi
* that gives a reasonable approximation.
* Literature shows a better CF that yields pi with a precision of
* 200 digits.
* http://de.wikipedia.org/wiki/Kreiszahl
* 1
* pi = 3 + ------------------------
* 1
* 7 + --------------------
* 1
* 15 + ---------------
* 1
* 1 + -----------
*
* 292 + ...
*
* This program uses that CF and shows the first 50 digits
* PI =3.1415926535897932384626433832795028841971693993751...
* PIX=3.1415926535897932384626433832795028841971693993751...
* 201 correct digits
* 18.09.2012 Walter Pachl
**********************************************************************/

pi='3.1415926535897932384626433832795028841971'||,
'693993751058209749445923078164062862089986280348'||,
'253421170679821480865132823066470938446095505822'||,
'317253594081284811174502841027019385211055596446'||,
'229489549303819644288109756659334461284756482337'||,
'867831652712019091456485669234603486104543266482'||,
'133936072602491412737245870066063155881748815209'||,
'209628292540917153643678925903600113305305488204'||,
'665213841469519415116094330572703657595919530921'||,
'861173819326117931051185480744623799627495673518'||,
'857527248912279381830119491298336733624'
Numeric Digits 1000
al='7 15 1 292 1 1 1 2 1 3 1 14 2 1 1 2 2 2 2 1 84 2',
'1 1 15 3 13 1 4 2 6 6 99 1 2 2 6 3 5 1 1 6 8 1 7 1 2',
'3 7 1 2 1 1 12 1 1 1 3 1 1 8 1 1 2 1 6 1 1 5 2 2 3 1',
'2 4 4 16 1 161 45 1 22 1 2 2 1 4 1 2 24 1 2 1 3 1 2',
'1 1 10 2 5 4 1 2 2 8 1 5 2 2 26 1 4 1 1 8 2 42 2 1 7',
'3 3 1 1 7 2 4 9 7 2 3 1 57 1 18 1 9 19 1 2 18 1 3 7',
'30 1 1 1 3 3 3 1 2 8 1 1 2 1 15 1 2 13 1 2 1 4 1 12',
'1 1 3 3 28 1 10 3 2 20 1 1 1 1 4 1 1 1 5 3 2 1 6 1 4'
a.=3
Do i=1 By 1 while al<>''
Parse Var al a.i al
End
pix=calc(194)
Do e=1 To length(pi)
If substr(pix,e,1)<>substr(pi,e,1) Then Leave
End
Numeric Digits 50
Say 'PI ='||(pi+0)||'...'
Say 'PIX='||(pix+0)||'...'
Say (e-1) 'correct digits'
Exit

Get_Coeffs: procedure Expose a b a.
Parse Arg n
a=a.n
b=1
Return

Calc: procedure Expose a b a.
Parse Arg n
Temp=0
do ni = n to 1 by -1
Call Get_Coeffs ni
Temp = B/(A + Temp)
end
call Get_Coeffs 0
return (A + Temp)

## Ruby

require 'bigdecimal'

# square root of 2
sqrt2 = Object.new
def sqrt2.a(n); n == 1 ? 1 : 2; end
def sqrt2.b(n); 1; end

# Napier's constant
napier = Object.new
def napier.a(n); n == 1 ? 2 : n - 1; end
def napier.b(n); n == 1 ? 1 : n - 1; end

pi = Object.new
def pi.a(n); n == 1 ? 3 : 6; end
def pi.b(n); (2*n - 1)**2; end

# Estimates the value of a continued fraction _cfrac_, to _prec_
# decimal digits of precision. Returns a BigDecimal. _cfrac_ must
# respond to _cfrac.a(n)_ and _cfrac.b(n)_ for integer _n_ >= 1.
def estimate(cfrac, prec)
last_result = nil
terms = prec

loop do
# Estimate continued fraction for _n_ from 1 to _terms_.
result = cfrac.a(terms)
(terms - 1).downto(1) do |n|
a = BigDecimal cfrac.a(n)
b = BigDecimal cfrac.b(n)
digits = [b.div(result, 1).exponent + prec, 1].max
result = a + b.div(result, digits)
end
result = result.round(prec)

if result == last_result
return result
else
# Double _terms_ and try again.
last_result = result
terms *= 2
end
end
end

puts estimate(sqrt2, 50).to_s('F')
puts estimate(napier, 50).to_s('F')
puts estimate(pi, 10).to_s('F')
Output:
$ruby cfrac.rb 1.41421356237309504880168872420969807856967187537695 2.71828182845904523536028747135266249775724709369996 3.1415926536 ## Rust use std::iter; // Calculating a continued fraction is quite easy with iterators, however // writing a proper iterator adapter is less so. We settle for a macro which // for most purposes works well enough. // // One limitation with this iterator based approach is that we cannot reverse // input iterators since they are not usually DoubleEnded. To circumvent this // we can collect the elements and then reverse them, however this isn't ideal // as we now have to store elements equal to the number of iterations. // // Another is that iterators cannot be resused once consumed, so it is often // required to make many clones of iterators. macro_rules! continued_fraction { ($a:expr, $b:expr ;$iterations:expr) => (
($a).zip($b)
.take($iterations) .collect::<Vec<_>>().iter() .rev() .fold(0 as f64, |acc: f64, &(x, y)| { x as f64 + (y as f64 / acc) }) ); ($a:expr, $b:expr) => (continued_fraction!($a, $b ; 1000)); } fn main() { // Sqrt(2) let sqrt2a = (1..2).chain(iter::repeat(2)); let sqrt2b = iter::repeat(1); println!("{}", continued_fraction!(sqrt2a, sqrt2b)); // Napier's Constant let napiera = (2..3).chain(1..); let napierb = (1..2).chain(1..); println!("{}", continued_fraction!(napiera, napierb)); // Pi let pia = (3..4).chain(iter::repeat(6)); let pib = (1i64..).map(|x| (2 * x - 1).pow(2)); println!("{}", continued_fraction!(pia, pib)); } Output: 1.4142135623730951 2.7182818284590455 3.141592653339042 ## Scala Works with: Scala version 2.9.1 Note that Scala-BigDecimal provides a precision of 34 digits. Therefore we take a limitation of 32 digits to avoiding rounding problems. object CF extends App { import Stream._ val sqrt2 = 1 #:: from(2,0) zip from(1,0) val napier = 2 #:: from(1) zip (1 #:: from(1)) val pi = 3 #:: from(6,0) zip (from(1,2) map {x=>x*x}) // reference values, source: wikipedia val refPi = "3.14159265358979323846264338327950288419716939937510" val refNapier = "2.71828182845904523536028747135266249775724709369995" val refSQRT2 = "1.41421356237309504880168872420969807856967187537694" def calc(cf: Stream[(Int, Int)], numberOfIters: Int=200): BigDecimal = { (cf take numberOfIters toList).foldRight[BigDecimal](1)((a, z) => a._1+a._2/z) } def approx(cfV: BigDecimal, cfRefV: String): String = { val p: Pair[Char,Char] => Boolean = pair =>(pair._1==pair._2) ((cfV.toString+" "*34).substring(0,34) zip cfRefV.toString.substring(0,34)) .takeWhile(p).foldRight[String]("")((a:Pair[Char,Char],z)=>a._1+z) } List(("sqrt2",sqrt2,50,refSQRT2),("napier",napier,50,refNapier),("pi",pi,3000,refPi)) foreach {t=> val (name,cf,iters,refV) = t val cfV = calc(cf,iters) println(name+":") println("ref value: "+refV.substring(0,34)) println("cf value: "+(cfV.toString+" "*34).substring(0,34)) println("precision: "+approx(cfV,refV)) println() } } Output: sqrt2: ref value: 1.41421356237309504880168872420969 cf value: 1.41421356237309504880168872420969 precision: 1.41421356237309504880168872420969 napier: ref value: 2.71828182845904523536028747135266 cf value: 2.71828182845904523536028747135266 precision: 2.71828182845904523536028747135266 pi: ref value: 3.14159265358979323846264338327950 cf value: 3.14159265358052780404906362935452 precision: 3.14159265358 For higher accuracy of pi we have to take more iterations. Unfortunately the foldRight function in calc isn't tail recursiv - therefore a stack overflow exception will be thrown for higher numbers of iteration, thus we have to implement an iterative way for calculation: object CFI extends App { import Stream._ val sqrt2 = 1 #:: from(2,0) zip from(1,0) val napier = 2 #:: from(1) zip (1 #:: from(1)) val pi = 3 #:: from(6,0) zip (from(1,2) map {x=>x*x}) // reference values, source: wikipedia val refPi = "3.14159265358979323846264338327950288419716939937510" val refNapier = "2.71828182845904523536028747135266249775724709369995" val refSQRT2 = "1.41421356237309504880168872420969807856967187537694" def calc_i(cf: Stream[(Int, Int)], numberOfIters: Int=50): BigDecimal = { val cfl = cf take numberOfIters toList var z: BigDecimal = 1.0 for (i <- 0 to cfl.size-1 reverse) z=cfl(i)._1+cfl(i)._2/z z } def approx(cfV: BigDecimal, cfRefV: String): String = { val p: Pair[Char,Char] => Boolean = pair =>(pair._1==pair._2) ((cfV.toString+" "*34).substring(0,34) zip cfRefV.toString.substring(0,34)) .takeWhile(p).foldRight[String]("")((a:Pair[Char,Char],z)=>a._1+z) } List(("sqrt2",sqrt2,50,refSQRT2),("napier",napier,50,refNapier),("pi",pi,50000,refPi)) foreach {t=> val (name,cf,iters,refV) = t val cfV = calc_i(cf,iters) println(name+":") println("ref value: "+refV.substring(0,34)) println("cf value: "+(cfV.toString+" "*34).substring(0,34)) println("precision: "+approx(cfV,refV)) println() } } Output: sqrt2: ref value: 1.41421356237309504880168872420969 cf value: 1.41421356237309504880168872420969 precision: 1.41421356237309504880168872420969 napier: ref value: 2.71828182845904523536028747135266 cf value: 2.71828182845904523536028747135266 precision: 2.71828182845904523536028747135266 pi: ref value: 3.14159265358979323846264338327950 cf value: 3.14159265358983426214354599901745 precision: 3.141592653589 ## Sidef Translation of: Perl func continued_fraction(a, _, (0)) { a() } func continued_fraction(a, b, n=100) { a() + (b() / continued_fraction(a, b, n-1)); } var f = Hash.new( "√2" => [ do { var n = 0; { n++ ? 2 : 1 } }, { 1 }, ], "e" => [ do { var n = 0; { n++ || 2 } }, do { var n = 0; { n++ || 1 } }, ], "π" => [ do { var n = 0; { n++ ? 6 : 3 } }, do { var n = 0; { (2*(n++) + 1)**2 } }, 1_000, ], "π/2" => [ do { var n = 0; { 1/(n++ || 1) } }, { 1 }, 1_000, ] ); f.each { |k,v| printf("%3s ≈ %.9f\n", k, continued_fraction(v...)); } Output: e ≈ 2.718281828 π ≈ 3.141592653 √2 ≈ 1.414213562 π/2 ≈ 1.570011909 ## Tcl Works with: tclsh version 8.6 Translation of: Python Note that Tcl does not provide arbitrary precision floating point numbers by default, so all result computations are done with IEEE doubles. package require Tcl 8.6 # Term generators; yield list of pairs proc r2 {} { yield {1 1} while 1 {yield {2 1}} } proc e {} { yield {2 1} while 1 {yield [list [incr n]$n]}
}
proc pi {} {
set n 0; set a 3
while 1 {
yield [list $a [expr {(2*[incr n]-1)**2}]] set a 6 } } # Continued fraction calculator proc cf {generator {termCount 50}} { # Get the chunk of terms we want to work with set terms [list [coroutine cf.c$generator]]
while {[llength $terms] <$termCount} {
lappend terms [cf.c]
}
rename cf.c {}

# Merge the terms to compute the result
set val 0.0
foreach pair [lreverse $terms] { lassign$pair a b
set val [expr {$a +$b/$val}] } return$val
}

# Demonstration
puts [cf r2]
puts [cf e]
puts [cf pi 250]; # Converges more slowly
Output:
1.4142135623730951
2.7182818284590455
3.1415926373965735

## XPL0

The number of iterations (N) needed to get the 13 digits of accuracy was determined by experiment.

include c:\cxpl\codes;
int N;
real A, B, F;
[Format(1, 15);
A:= 2.0; B:= 1.0; N:= 16;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=1 do [F:= B/(A+F); N:= N-1];
RlOut(0, 1.0+F); CrLf(0);
RlOut(0, sqrt(2.0)); CrLf(0);

N:= 13;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=2 do [F:= float(N-1)/(float(N)+F); N:= N-1];
RlOut(0, 2.0 + 1.0/(1.0+F)); CrLf(0);
RlOut(0, Exp(1.0)); CrLf(0);

N:= 10000;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=1 do [F:= float(sq(2*N-1))/(6.0+F); N:= N-1];
RlOut(0, 3.0+F); CrLf(0);
RlOut(0, ACos(-1.0)); CrLf(0);
]
Output:
16
1.414213562372820
1.414213562373100
13
2.718281828459380
2.718281828459050
10000
3.141592653589540
3.141592653589790

## zkl

fcn cf(fa,fb,a0){fcn(fa,fb,a0,n){
a0 + [n..1,-1].reduce(
'wrap(p,n){ fb(n)/(fa(n)+p) },0.0) }.fp(fa,fb,a0)
}

cf creates a function that calculates the continued fraction from the bottom up. The new function takes a single parameter, n, which is used to calculate the nth term.

sqrt2:=cf((2.0).noop,(1.0).noop,1.0);
sqrt2(200) : "%.20e".fmt(_).println();
nap:=cf((0.0).create,fcn(n){ (n==1) and 1.0 or (n-1).toFloat() },2.0);
println(nap(15) - (1.0).e);
pi:=cf((6.0).noop,fcn(n){ n=2*n-1; (n*n).toFloat() },3.0);
println(pi(1000) - (1.0).pi);

(1.0).create(n) --> n, (1.0).noop(n) --> 1.0

Output:
1.41421356237309514547e+00
1.33227e-15
-2.49251e-10