Continued fraction: Difference between revisions
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Line 17:
:<math>\pi = 3 + \cfrac{1}{6 + \cfrac{9}{6 + \cfrac{25}{6 + \ddots}}}</math>
;See also:
:* [[Continued fraction/Arithmetic]] for tasks that do arithmetic over continued fractions.
<br><br>
=={{header|11l}}==
<syntaxhighlight lang="11l">F calc(f_a, f_b, =n = 1000)
V r = 0.0
L n > 0
r = f_b(n) / (f_a(n) + r)
n--
R f_a(0) + r
print(calc(n -> I n > 0 {2} E 1, n -> 1))
print(calc(n -> I n > 0 {n} E 2, n -> I n > 1 {n - 1} E 1))
print(calc(n -> I n > 0 {6} E 3, n -> (2 * n - 1) ^ 2))</syntaxhighlight>
=={{header|Action!}}==
{{libheader|Action! Tool Kit}}
<syntaxhighlight lang="action!">INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit
DEFINE PTR="CARD"
DEFINE JSR="$20"
DEFINE RTS="$60"
PROC CoeffA=*(INT n REAL POINTER res)
[JSR $00 $00 ;JSR to address set by SetCoeffA
RTS]
PROC CoeffB=*(INT n REAL POINTER res)
[JSR $00 $00 ;JSR to address set by SetCoeffB
RTS]
PROC SetCoeffA(PTR p)
PTR addr
addr=CoeffA+1 ;location of address of JSR
PokeC(addr,p)
RETURN
PROC SetCoeffB(PTR p)
PTR addr
addr=CoeffB+1 ;location of address of JSR
PokeC(addr,p)
RETURN
PROC Calc(PTR funA,funB INT count REAL POINTER res)
INT i
REAL a,b,tmp
SetCoeffA(funA)
SetCoeffB(funB)
IntToReal(0,res)
i=count
WHILE i>0
DO
CoeffA(i,a)
CoeffB(i,b)
RealAdd(a,res,tmp)
RealDiv(b,tmp,res)
i==-1
OD
CoeffA(0,a)
RealAdd(a,res,tmp)
RealAssign(tmp,res)
RETURN
PROC sqrtA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(2,res)
ELSE
IntToReal(1,res)
FI
RETURN
PROC sqrtB(INT n REAL POINTER res)
IntToReal(1,res)
RETURN
PROC napierA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(n,res)
ELSE
IntToReal(2,res)
FI
RETURN
PROC napierB(INT n REAL POINTER res)
IF n>1 THEN
IntToReal(n-1,res)
ELSE
IntToReal(1,res)
FI
RETURN
PROC piA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(6,res)
ELSE
IntToReal(3,res)
FI
RETURN
PROC piB(INT n REAL POINTER res)
REAL tmp
IntToReal(2*n-1,tmp)
RealMult(tmp,tmp,res)
RETURN
PROC Main()
REAL res
Put(125) PutE() ;clear the screen
Calc(sqrtA,sqrtB,50,res)
Print(" Sqrt2=") PrintRE(res)
Calc(napierA,napierB,50,res)
Print("Napier=") PrintRE(res)
Calc(piA,piB,500,res)
Print(" Pi=") PrintRE(res)
RETURN</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Continued_fraction.png Screenshot from Atari 8-bit computer]
<pre>
Sqrt2=1.41421356
Napier=2.71828182
Pi=3.14159265
</pre>
=={{header|Ada}}==
Line 25 ⟶ 155:
Generic function for estimating continued fractions:
<
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 A (N : in Natural) return Scalar is (Scalar (Natural'(A (N))));
function B (N : in Positive) return Scalar is (Scalar (Natural'(B (N))));
Line 42 ⟶ 172:
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;
Line 78 ⟶ 208:
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.
<
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 A (N : in Natural) return Scalar is
begin
Line 105 ⟶ 235:
end loop;
return A (0) + Fraction;
end Continued_Fraction_Ada95;</
<
with Continued_Fraction_Ada95;
Line 192 ⟶ 322:
Put (Napiers_Constant.Estimate (200), Exp => 0); New_Line;
Put (Pi.Estimate (10000), Exp => 0); New_Line;
end Test_Continued_Fractions_Ada95;</
{{out}}
<pre> 1.41421356237310
Line 200 ⟶ 330:
=={{header|ALGOL 68}}==
{{works with|Algol 68 Genie|2.8}}
<syntaxhighlight lang="algol68">
PROC cf = (INT steps, PROC (INT) INT a, PROC (INT) INT b) REAL:
BEGIN
Line 226 ⟶ 356:
print (("Napier: ", cf(precision, anap, bnap), newline));
print (("Pi: ", cf(precision, api, bpi)))
</syntaxhighlight>
{{out}}
<pre>
Line 234 ⟶ 364:
Pi: +3.14159265358954e +0
</pre>
=={{header|Arturo}}==
{{trans|Nim}}
<syntaxhighlight lang="rebol">calc: function [f, n][
[a, b, temp]: 0.0
loop n..1 'i [
[a, b]: call f @[i]
temp: b // a + temp
]
[a, b]: call f @[0]
return a + temp
]
sqrt2: function [n][
(n > 0)? -> [2.0, 1.0] -> [1.0, 1.0]
]
napier: function [n][
a: (n > 0)? -> to :floating n -> 2.0
b: (n > 1)? -> to :floating n-1 -> 1.0
@[a, b]
]
Pi: function [n][
a: (n > 0)? -> 6.0 -> 3.0
b: ((2 * to :floating n)-1) ^ 2
@[a, b]
]
print calc 'sqrt2 20
print calc 'napier 15
print calc 'Pi 10000</syntaxhighlight>
{{out}}
<pre>1.414213562373095
2.718281828459046
3.141592653589544</pre>
=={{header|ATS}}==
A fairly direct translation of the [[#C|C version]] without using advanced features of the type system:
<
"share/atspre_staload.hats"
//
Line 306 ⟶ 475:
println! ("napier = ", calc(napier, 100));
println! (" pi = ", calc( pi , 100));
) (* end of [main0] *)</
=={{header|AutoHotkey}}==
<
{
return n?2.0:1.0
Line 355 ⟶ 524:
}
Msgbox, % "sqrt 2 = " . calc("sqrt2", 1000) . "`ne = " . calc("napier", 1000) . "`npi = " . calc("pi", 1000)</
Output with Autohotkey v1 (currently 1.1.16.05):
<
e = 2.718282
pi = 3.141593</
Output with Autohotkey v2 (currently alpha 56):
<
e = 2.7182818284590455
pi = 3.1415926533405418</
Note the far superiour accuracy of v2.
=={{header|Axiom}}==
Axiom provides a ContinuedFraction domain:
<
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:<
Type: Float
Line 379 ⟶ 548:
(3) 3.1415926538 39792926
Type: Float</
The value for <math>\pi</math> 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:
<
n=1 => initial
temp := 0
Line 391 ⟶ 560:
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</
=={{header|BBC BASIC}}==
{{works with|BBC BASIC for Windows}}
<
@% = &1001010
Line 410 ⟶ 579:
expr$ += STR$(EVAL(a$)) + "+" + STR$(EVAL(b$)) + "/("
UNTIL LEN(expr$) > (65500 - N)
= a0 + b1 / EVAL (expr$ + "1" + STRING$(N, ")"))</
{{out}}
<pre>
Line 420 ⟶ 589:
=={{header|C}}==
{{works with|ANSI C}}
<
#include <stdio.h>
Line 490 ⟶ 659:
return 0;
}</
{{out}}
<pre> 1.414213562
2.718281828
3.141592653</pre>
=={{header|C sharp|C#}}==
{{trans|Java}}
<syntaxhighlight lang="csharp">using System;
using System.Collections.Generic;
namespace ContinuedFraction {
class Program {
static double Calc(Func<int, int[]> f, int n) {
double temp = 0.0;
for (int ni = n; ni >= 1; ni--) {
int[] p = f(ni);
temp = p[1] / (p[0] + temp);
}
return f(0)[0] + temp;
}
static void Main(string[] args) {
List<Func<int, int[]>> fList = new List<Func<int, int[]>>();
fList.Add(n => new int[] { n > 0 ? 2 : 1, 1 });
fList.Add(n => new int[] { n > 0 ? n : 2, n > 1 ? (n - 1) : 1 });
fList.Add(n => new int[] { n > 0 ? 6 : 3, (int) Math.Pow(2 * n - 1, 2) });
foreach (var f in fList) {
Console.WriteLine(Calc(f, 200));
}
}
}
}</syntaxhighlight>
{{out}}
<pre>1.4142135623731
2.71828182845905
3.14159262280485</pre>
=={{header|C++}}==
<
#include <iostream>
#include <tuple>
Line 538 ⟶ 740:
<< calc(pi, 10000) << '\n'
<< std::setprecision(old_prec); // reset precision
}</
{{out}}
<pre>
Line 545 ⟶ 747:
3.14159265358954
</pre>
=={{header|Chapel}}==
Functions don't take other functions as arguments, so I wrapped them in a dummy record each.
<syntaxhighlight lang="chapel">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));</syntaxhighlight>
=={{header|Clojure}}==
<
(defn cfrac
[a b n]
Line 556 ⟶ 797:
(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))
</syntaxhighlight>
{{out}}
<pre>
Line 568 ⟶ 809:
{{works with|GnuCOBOL}}
<
program-id. show-continued-fractions.
Line 753 ⟶ 994:
goback.
end program pi-beta.
</syntaxhighlight>
{{out}}
Line 762 ⟶ 1,003:
=={{header|CoffeeScript}}==
<
# a0 + b1 / (a1 + b2 / (a2 + b3 / ...
continuous_fraction = (f) ->
Line 796 ⟶ 1,037:
x = 2*n - 1
x * x
p Math.PI, continuous_fraction(fpi)</
{{out}}
<pre>
Line 813 ⟶ 1,054:
=={{header|Common Lisp}}==
{{trans|C++}}
<
(let ((temp 0))
(loop for n1 from n downto 1
Line 836 ⟶ 1,077:
(* (1- (* 2 n))
(1- (* 2 n))))) 10000)
'double-float))</
{{out}}
<pre>sqrt(2) = 1.4142135623730947d0
napier's = 2.7182818284590464d0
pi = 3.141592653589543d0</pre>
=={{header|D}}==
<
FP calc(FP, F)(in F fun, in int n) pure nothrow if (isCallable!F) {
Line 912 ⟶ 1,114:
calc!real(&fNapier, 200).print;
calc!real(&fPi, 200).print;
}</
{{out}}
<pre>1.4142135623730950487
2.7182818284590452354
3.1415926228048469486</pre>
=={{header|dc}}==
<syntaxhighlight lang="dc">[20k 0 200 si [li lbx r li lax + / li 1 - dsi 0<:]ds:x 0 lax +]sf
[[2q]s2[0<2 1]sa[R1]sb]sr # sqrt(2)
[[R2q]s2[d 0=2]sa[R1q]s1[d 1=1 1-]sb]se # e
[[3q]s3[0=3 6]sa[2*1-d*]sb]sp # pi
c lex lfx p
lrx lfx p
lpx lfx p</syntaxhighlight>
{{out}}
<pre>3.14159262280484694855
1.41421356237309504880
2.71828182845904523536</pre>
20 decimal places and 200 iterations.
=={{header|EasyLang}}==
<syntaxhighlight lang="easylang">
numfmt 8 0
func calc_sqrt .
n = 100
sum = n
while n >= 1
a = 1
if n > 1
a = 2
.
b = 1
sum = a + b / sum
n -= 1
.
return sum
.
func calc_e .
n = 100
sum = n
while n >= 1
a = 2
if n > 1
a = n - 1
.
b = 1
if n > 1
b = n - 1
.
sum = a + b / sum
n -= 1
.
return sum
.
func calc_pi .
n = 100
sum = n
while n >= 1
a = 3
if n > 1
a = 6
.
b = 2 * n - 1
b *= b
sum = a + b / sum
n -= 1
.
return sum
.
print calc_sqrt
print calc_e
print calc_pi
</syntaxhighlight>
=={{header|Elixir}}==
<syntaxhighlight lang="elixir">
defmodule CFrac do
def compute([a | _], []), do: a
def compute([a | as], [b | bs]), do: a + b/compute(as, bs)
def sqrt2 do
a = [1 | Stream.cycle([2]) |> Enum.take(1000)]
b = Stream.cycle([1]) |> Enum.take(1000)
IO.puts compute(a, b)
end
def exp1 do
a = [2 | Stream.iterate(1, &(&1 + 1)) |> Enum.take(1000)]
b = [1 | Stream.iterate(1, &(&1 + 1)) |> Enum.take(999)]
IO.puts compute(a, b)
end
def pi do
a = [3 | Stream.cycle([6]) |> Enum.take(1000)]
b = 1..1000 |> Enum.map(fn k -> (2*k - 1)**2 end)
IO.puts compute(a, b)
end
end
</syntaxhighlight>
{{out}}
<pre>
>elixir -e CFrac.sqrt2()
1.4142135623730951
>elixir -e CFrac.exp1()
2.7182818284590455
>elixir -e CFrac.pi()
3.141592653340542
</pre>
=={{header|Erlang}}==
<
-module(continued).
-compile([export_all]).
Line 963 ⟶ 1,275:
test_nappier (N) ->
continued_fraction(fun nappier_a/1,fun nappier_b/1,N).
</syntaxhighlight>
{{out}}
<
29> continued:test_pi(1000).
3.141592653340542
Line 973 ⟶ 1,285:
31> continued:test_nappier(1000).
2.7182818284590455
</syntaxhighlight>
=={{header|F_Sharp|F#}}==
===The Functions===
<syntaxhighlight lang="fsharp">
// I provide four functions:-
// cf2S general purpose continued fraction to sequence of float approximations
// cN2S Normal continued fractions (a-series always 1)
// cfSqRt uses cf2S to calculate sqrt of float
// π takes a sequence of b values returning the next until the list is exhausted after which it injects infinity
// Nigel Galloway: December 19th., 2018
let cf2S α β=let n0,g1,n1,g2=β(),α(),β(),β()
seq{let (Π:decimal)=g1/n1 in yield n0+Π; yield! Seq.unfold(fun(n,g,Π)->let a,b=α(),β() in let Π=Π*g/n in Some(n0+Π,(b+a/n,b+a/g,Π)))(g2+α()/n1,g2,Π)}
let cN2S = cf2S (fun()->1M)
let cfSqRt n=(cf2S (fun()->n-1M) (let mutable n=false in fun()->if n then 2M else (n<-true; 1M)))
let π n=let mutable π=n in (fun ()->match π with h::t->π<-t; h |_->9999999999999999999999999999M)
</syntaxhighlight>
===The Tasks===
<syntaxhighlight lang="fsharp">
cfSqRt 2M |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < √2 < %1.14f" (min n g) (max n g))
</syntaxhighlight>
{{out}}
<pre>
1.40000000000000 < √2 < 1.50000000000000
1.40000000000000 < √2 < 1.41666666666667
1.41379310344828 < √2 < 1.41666666666667
1.41379310344828 < √2 < 1.41428571428571
1.41420118343195 < √2 < 1.41428571428571
1.41420118343195 < √2 < 1.41421568627451
1.41421319796954 < √2 < 1.41421568627451
1.41421319796954 < √2 < 1.41421362489487
1.41421355164605 < √2 < 1.41421362489487
</pre>
<syntaxhighlight lang="fsharp">
cfSqRt 0.25M |> Seq.take 30 |> Seq.iter (printfn "%1.14f")
</syntaxhighlight>
{{out}}
<pre>
0.62500000000000
0.53846153846154
0.51250000000000
0.50413223140496
0.50137362637363
0.50045745654163
0.50015243902439
0.50005080784473
0.50001693537461
0.50000564506114
0.50000188167996
0.50000062722587
0.50000020907520
0.50000006969172
0.50000002323057
0.50000000774352
0.50000000258117
0.50000000086039
0.50000000028680
0.50000000009560
0.50000000003187
0.50000000001062
0.50000000000354
0.50000000000118
0.50000000000039
0.50000000000013
0.50000000000004
0.50000000000001
0.50000000000000
0.50000000000000
</pre>
<syntaxhighlight lang="fsharp">
let aπ()=let mutable n=0M in (fun ()->n<-n+1M;let b=n+n-1M in b*b)
let bπ()=let mutable n=true in (fun ()->match n with true->n<-false;3M |_->6M)
cf2S (aπ()) (bπ()) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < π < %1.14f" (min n g) (max n g))
</syntaxhighlight>
{{out}}
<pre>
3.13333333333333 < π < 3.16666666666667
3.13333333333333 < π < 3.14523809523810
3.13968253968254 < π < 3.14523809523810
3.13968253968254 < π < 3.14271284271284
3.14088134088134 < π < 3.14271284271284
3.14088134088134 < π < 3.14207181707182
3.14125482360776 < π < 3.14207181707182
3.14125482360776 < π < 3.14183961892940
3.14140671849650 < π < 3.14183961892940
</pre>
<syntaxhighlight lang="fsharp">
let pi = π [3M;7M;15M;1M;292M;1M;1M;1M;2M;1M;3M;1M;14M;2M;1M;1M;2M;2M;2M;2M]
cN2S pi |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < π < %1.14f" (min n g) (max n g))
</syntaxhighlight>
{{out}}
<pre>
3.14150943396226 < π < 3.14285714285714
3.14150943396226 < π < 3.14159292035398
3.14159265301190 < π < 3.14159292035398
3.14159265301190 < π < 3.14159265392142
3.14159265346744 < π < 3.14159265392142
3.14159265346744 < π < 3.14159265361894
3.14159265358108 < π < 3.14159265361894
3.14159265358108 < π < 3.14159265359140
3.14159265358939 < π < 3.14159265359140
</pre>
<syntaxhighlight lang="fsharp">
let ae()=let mutable n=0.5M in (fun ()->match n with 0.5M->n<-0M; 1M |_->n<-n+1M; n)
let be()=let mutable n=0.5M in (fun ()->match n with 0.5M->n<-0M; 2M |_->n<-n+1M; n)
cf2S (ae()) (be()) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < e < %1.14f" (min n g) (max n g))
</syntaxhighlight>
{{out}}
<pre>
2.66666666666667 < e < 3.00000000000000
2.66666666666667 < e < 2.72727272727273
2.71698113207547 < e < 2.72727272727273
2.71698113207547 < e < 2.71844660194175
2.71826333176026 < e < 2.71844660194175
2.71826333176026 < e < 2.71828369389345
2.71828165766640 < e < 2.71828369389345
2.71828165766640 < e < 2.71828184277783
2.71828182735187 < e < 2.71828184277783
</pre>
===Apéry's constant===
See [https://tpiezas.wordpress.com/2012/05/04/continued-fractions-for-zeta2-and-zeta3/ Continued fractions for Zeta(2) and Zeta(3)] section II. Zeta(3)
<syntaxhighlight lang="fsharp">
let aπ()=let mutable n=0 in (fun ()->n<-n+1;-decimal(pown n 6))
let bπ()=let mutable n=0M in (fun ()->n<-n+1M; (2M*n-1M)*(17M*n*n-17M*n+5M))
cf2S (aπ()) (bπ()) |>Seq.map(fun n->6M/n) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.20f < p < %- 1.20f" (min n g) (max n g));;
</syntaxhighlight>
{{out}}
<pre>
1.20205479452054794521 < p < 1.20205690119184928874
1.20205690119184928874 < p < 1.20205690315781676650
1.20205690315781676650 < p < 1.20205690315959270361
1.20205690315959270361 < p < 1.20205690315959428400
1.20205690315959428400 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
</pre>
=={{header|Factor}}==
''cfrac-estimate'' uses [[Arithmetic/Rational|rational arithmetic]] and never truncates the intermediate result. When ''terms'' is large, ''cfrac-estimate'' runs slow because numerator and denominator grow big.
<
math.ranges prettyprint sequences ;
IN: rosetta.cfrac
Line 1,039 ⟶ 1,488:
PRIVATE>
MAIN: main</
{{out}}
<pre> Square root of 2: 1.414213562373095048801688724209
Line 1,046 ⟶ 1,495:
=={{header|Felix}}==
<
let a = match n with | 0 => 3.0 | _ => 6.0 endmatch in
let b = pow(2.0 * n.double - 1.0, 2.0) in
Line 1,072 ⟶ 1,521:
println$ cf_iter 200 sqrt_2; // => 1.41421
println$ cf_iter 200 napier; // => 2.71818
println$ cf_iter 1000 pi; // => 3.14159</
=={{header|Forth}}==
{{trans|D}}
<
: 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 ;
Line 1,084 ⟶ 1,533:
do i over execute frot f+ f/ -1 +loop
0 swap execute fnip f+ \ calcucate for 0
;</
{{out}}
<pre>
Line 1,096 ⟶ 1,545:
=={{header|Fortran}}==
<
implicit none
Line 1,194 ⟶ 1,643:
end function
end program examples</
{{out}}
<pre>
Line 1,201 ⟶ 1,650:
3.1415926535895435
</pre>
=={{header|FreeBASIC}}==
<syntaxhighlight lang="freebasic">#define MAX 70000
function sqrt2_a( n as uinteger ) as uinteger
return iif(n,2,1)
end function
function sqrt2_b( n as uinteger ) as uinteger
return 1
end function
function napi_a( n as uinteger ) as uinteger
return iif(n,n,2)
end function
function napi_b( n as uinteger ) as uinteger
return iif(n>1,n-1,1)
end function
function pi_a( n as uinteger ) as uinteger
return iif(n,6,3)
end function
function pi_b( n as uinteger ) as uinteger
return (2*n-1)^2
end function
function calc_contfrac( an as function (as uinteger) as uinteger, bn as function (as uinteger) as uinteger, byval iter as uinteger ) as double
dim as double r
dim as integer i
for i = iter to 1 step -1
r = bn(i)/(an(i)+r)
next i
return an(0)+r
end function
print calc_contfrac( @sqrt2_a, @sqrt2_b, MAX )
print calc_contfrac( @napi_a, @napi_b, MAX )
print calc_contfrac( @pi_a, @pi_b, MAX )</syntaxhighlight>
=={{header|Fōrmulæ}}==
{{FormulaeEntry|page=https://formulae.org/?script=examples/Continued_fraction}}
'''Solution'''
The following function definition creates a continued fraction:
[[File:Fōrmulæ - Continued fraction 01.png]]
The function accepts the following parameters:
{| class="wikitable" style="margin:auto"
|-
! Parameter !! Description
|-
| a₀ || Value for a₀
|-
| λa || Lambda expression to define aᵢ
|-
| λb || Lambda expression to define bᵢ
|-
| depth || Depth to calculate the continued fraction
|}
Since Fōrmulæ arithmetic simplifies numeric results as they are generated, the result is not a continued fraction by default.
If we want to create the structure, we can introduce the parameters as string or text expressions (or lambda expressions that produce them). Because string or text expressions are not reduced when they are operands of additions and divisions, the structure is preserved, such as follows:
[[File:Fōrmulæ - Continued fraction 02.png]]
[[File:Fōrmulæ - Continued fraction 03.png]]
'''Case 1.''' <math>\sqrt 2</math>
In this case
* a₀ is 1
* λa is n ↦ 2
* λb is n ↦ 1
Let us show the results as a table, for several levels of depth (1 to 10).
The columns are:
* The depth
* The "textual" call, in order to generate the structure
* The normal (numeric) call. Since arithmetic operations are exact by default, it is usually a rational number.
* The value of the normal (numeric) call, forced to be shown as a decimal number, by using the Math.Numeric expression (the N(x) expression)
[[File:Fōrmulæ - Continued fraction 04.png]]
[[File:Fōrmulæ - Continued fraction 05.png]]
'''Case 2.''' <math>e</math>
In this case
* a₀ is 2
* λa is n ↦ n
* λb is n ↦ 1 if n = 1, n - 1 elsewhere
[[File:Fōrmulæ - Continued fraction 06.png]]
[[File:Fōrmulæ - Continued fraction 07.png]]
'''Case 3.''' <math>\pi</math>
In this case
* a₀ is 3
* λa is n ↦ 6
* λb is n ↦ 2(n - 1)²
[[File:Fōrmulæ - Continued fraction 08.png]]
[[File:Fōrmulæ - Continued fraction 09.png]]
=={{header|Go}}==
<
import "fmt"
Line 1,255 ⟶ 1,822:
fmt.Println("nap: ", cfNap(20).real())
fmt.Println("pi: ", cfPi(20).real())
}</
{{out}}
<pre>
Line 1,262 ⟶ 1,829:
pi: 3.141623806667839
</pre>
=={{header|Groovy}}==
{{trans|Java}}
<syntaxhighlight lang="groovy">import java.util.function.Function
import static java.lang.Math.pow
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
}
static void main(String[] args) {
List<Function<Integer, Integer[]>> fList = new ArrayList<>()
fList.add({ n -> [n > 0 ? 2 : 1, 1] })
fList.add({ n -> [n > 0 ? n : 2, n > 1 ? (n - 1) : 1] })
fList.add({ n -> [n > 0 ? 6 : 3, (int) pow(2 * n - 1, 2)] })
for (Function<Integer, Integer[]> f : fList)
System.out.println(calc(f, 200))
}
}</syntaxhighlight>
{{out}}
<pre>1.4142135623730951
2.7182818284590455
3.141592622804847</pre>
=={{header|Haskell}}==
<
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 ..]
myPi = zip (3 : [6,6 ..]) ((^ 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]</syntaxhighlight>
{{out}}
<pre>
Line 1,294 ⟶ 1,901:
3.141592653297590947683406834261190738869139611505752231394089152890909495973464508817163306557131591579057202097715021166512662872910519439747609829479577279606075707015622200744006783543589980682386
</pre>
<
import Data.Bool (bool)
-- 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 ..])
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))
where
f ((x, y):ys)
| abs (x - y) < (1 / fromIntegral p) = x
| otherwise = 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 =
where
ratstr l a = show t ++ '.' : fracstr l n d
where
d = denominator a
fracstr 0 _ _ = []
fracstr l n d = show t ++ fracstr (l - 1) n1 d
where
(t, n1) = quotRem (10 * n) d
main
main = mapM_ putStrLn [cf2dec 200 sqrt2, cf2dec 200 napier, cf2dec 200 pie]</syntaxhighlight>
=={{header|Icon}}==
<syntaxhighlight lang="icon">
$define EVAL_DEPTH 100
# A generalized continued fraction, represented by two functions. Each
# function maps from an index to a floating-point value.
record continued_fraction (a, b)
procedure main ()
writes (" sqrt 2.0 = ")
write (evaluate_continued_fraction (continued_fraction (sqrt2_a, sqrt2_b),
EVAL_DEPTH))
writes (" e = ")
write (evaluate_continued_fraction (continued_fraction (e_a, e_b),
EVAL_DEPTH))
writes (" pi = ")
write (evaluate_continued_fraction (continued_fraction (pi_a, pi_b),
EVAL_DEPTH))
end
procedure evaluate_continued_fraction (frac, depth)
local i, retval
retval := frac.a (depth)
every i := depth to 1 by -1 do {
retval := frac.a (i - 1) + (frac.b (i) / retval)
}
return retval
end
procedure sqrt2_a (i)
return (if i = 0 then 1.0 else 2.0)
end
procedure sqrt2_b (i)
return 1.0
end
procedure e_a (i)
return (if i = 0 then 2.0 else real (i))
end
procedure e_b (i)
return (if i = 1 then 1.0 else real (i - 1))
end
procedure pi_a (i)
return (if i = 0 then 3.0 else 6.0)
end
procedure pi_b (i)
return real (((2 * i) - 1)^2)
end
</syntaxhighlight>
{{out}}
<pre>$ icon continued-fraction-task.icn
sqrt 2.0 = 1.414213562
e = 2.718281828
pi = 3.141592411
</pre>
=={{header|J}}==
<
sqrt2=: cfrac 1 1,200$2 1x
Line 1,339 ⟶ 2,014:
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 <code>(+%)/</code> instead of <code>+`%/</code>.
Line 1,346 ⟶ 2,021:
{{trans|D}}
{{works with|Java|8}}
<
import java.util.*;
import java.util.function.Function;
Line 1,370 ⟶ 2,045:
System.out.println(calc(f, 200));
}
}</
<pre>1.4142135623730951
2.7182818284590455
Line 1,393 ⟶ 2,068:
computes the continued fraction until the difference in approximations is less than or equal to delta,
which may be 0, as previously noted.
<syntaxhighlight lang="jq">
# "first" is the first triple, e.g. [1,a,b];
#
def continued_fraction( first; next; count ):
# input: [i, a, b
def cf:
if .[0] == count then 0
Line 1,419 ⟶ 2,094:
end;
[2,null] | cf;
</syntaxhighlight>
'''Examples''':
The convergence for pi is slow so we select delta = 1e-12 in that case.
<
"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)"
</syntaxhighlight>
{{Out}}
<
Value : Direct : Continued Fraction
2|sqrt : 1.4142135623730951 : 1.4142135623730951
1|exp : 2.718281828459045 : 2.7182818284590455
pi : 3.141592653589793 : 3.1415926535892935 (1e-12)</
=={{header|Julia}}==
{{works with|Julia|
High performant lazy evaluation on demand with Julias iterators.
<syntaxhighlight lang="julia">
using .Iterators: countfrom, flatten, repeated, zip
using .MathConstants: ℯ
using Printf
function
m = BigInt[a₀ 1; 1 0]
for (aᵢ, bᵢ) ∈ zip(a, b)
end
m[1]/m[2]
end
out((k, v)) = @printf "%2s: %.12f ≈ %.12f\n" k v eval(k)
foreach(out, (
:(√2) => cf(1, repeated(2)),
:ℯ => cf(2, countfrom(), flatten((1, countfrom()))),
:π => cf(3, repeated(6), (k^2 for k ∈ countfrom(1, 2)))))
</syntaxhighlight>
{{out}}
<pre>
ℯ: 2.718281828459 ≈ 2.718281828459
=={{header|Klong}}==
<syntaxhighlight lang="k">
cf::{[f g i];f::x;g::y;i::z;
f(0)+z{i::i-1;g(i+1)%f(i+1)+x}:*0}
Line 1,486 ⟶ 2,146:
cf({:[0=x;2;x]};{:[x>1;x-1;x]};1000)
cf({:[0=x;3;6]};{((2*x)-1)^2};1000)
</syntaxhighlight>
{{out}}
<pre>
Line 1,497 ⟶ 2,157:
=={{header|Kotlin}}==
{{trans|D}}
<
typealias Func = (Int) -> IntArray
Line 1,517 ⟶ 2,177:
)
for (pair in pList) println("${pair.first} = ${calc(pair.second, 200)}")
}</
{{out}}
Line 1,525 ⟶ 2,185:
pi = 3.141592622804847
</pre>
=={{header|Lambdatalk}}==
<syntaxhighlight lang="scheme">
{def gcf
{def gcf.rec
{lambda {:f :n :r}
{if {< :n 1}
then {+ {car {:f 0}} :r}
else {gcf.rec :f
{- :n 1}
{let { {:r :r}
{:ab {:f :n}}
} {/ {cdr :ab}
{+ {car :ab} :r}} }}}}}
{lambda {:f :n}
{gcf.rec :f :n 0}}}
{def phi
{lambda {:n}
{cons 1 1}}}
{gcf phi 50}
-> 1.618033988749895
{def sqrt2
{lambda {:n}
{cons {if {> :n 0} then 2 else 1} 1}}}
{gcf sqrt2 25}
-> 1.4142135623730951
{def napier
{lambda {:n}
{cons {if {> :n 0} then :n else 2} {if {> :n 1} then {- :n 1} else 1} }}}
{gcf napier 20}
-> 2.7182818284590455
{def fpi
{lambda {:n}
{cons {if {> :n 0} then 6 else 3} {pow {- {* 2 :n} 1} 2} }}}
{gcf fpi 500}
-> 3.1415926 516017554
// only 8 exact decimals for 500 iterations
// A very very slow convergence.
// Here is a quicker version without any obvious pattern
{def pi
{lambda {:n}
{cons {A.get :n {A.new 3 7 15 1 292 1 1 1 2 1 3 1 14 2 1 1}} 1}}}
{gcf pi 15}
-> 3.1415926 53589793
// Much quicker, 15 exact decimals after 15 iterations
</syntaxhighlight>
=={{header|Lua}}==
{{trans|C}}
<syntaxhighlight lang="lua">function calc(fa, fb, expansions)
local a = 0.0
local b = 0.0
local r = 0.0
local i = expansions
while i > 0 do
a = fa(i)
b = fb(i)
r = b / (a + r)
i = i - 1
end
a = fa(0)
return a + r
end
function sqrt2a(n)
if n ~= 0 then
return 2.0
else
return 1.0
end
end
function sqrt2b(n)
return 1.0
end
function napiera(n)
if n ~= 0 then
return n
else
return 2.0
end
end
function napierb(n)
if n > 1.0 then
return n - 1.0
else
return 1.0
end
end
function pia(n)
if n ~= 0 then
return 6.0
else
return 3.0
end
end
function pib(n)
local c = 2.0 * n - 1.0
return c * c
end
function main()
local sqrt2 = calc(sqrt2a, sqrt2b, 1000)
local napier = calc(napiera, napierb, 1000)
local pi = calc(pia, pib, 1000)
print(sqrt2)
print(napier)
print(pi)
end
main()</syntaxhighlight>
{{out}}
<pre>1.4142135623731
2.718281828459
3.1415926533405</pre>
=={{header|Maple}}==
<syntaxhighlight lang="maple">
contfrac:=n->evalf(Value(NumberTheory:-ContinuedFraction(n)));
contfrac(2^(0.5));
contfrac(Pi);
contfrac(exp(1));
</syntaxhighlight>
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<
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@@First@Fold[{{#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</
{{out}}
<pre>
Line 1,549 ⟶ 2,342:
=={{header|Maxima}}==
<
for i from n step -1 thru 2 do z: b[i]/(a[i] + z), a[1] + z)$
Line 1,571 ⟶ 2,364:
fpprec: 20$
x: cfeval(cf_pi(10000))$
bfloat(x - %pi); /* 2.4999999900104930006b-13 */</
=={{header|NetRexx}}==
<
* Derived from REXX ... Derived from PL/I with a little "massage"
* SQRT2= 1.41421356237309505 <- PL/I Result
Line 1,626 ⟶ 2,419:
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:-)
Line 1,637 ⟶ 2,430:
See [[#REXX|Rexx]] for a better computation
=={{header|Nim}}==
<syntaxhighlight lang="nim">proc calc(f: proc(n: int): tuple[a, b: float], n: int): float =
var a, b, temp = 0.0
for i in countdown(n, 1):
(a, b) = f(i)
temp = b / (a + temp)
(a, b) = f(0)
a + temp
proc sqrt2(n: int): tuple[a, b: float] =
if n > 0:
(2.0, 1.0)
else:
(1.0, 1.0)
proc napier(n: int): tuple[a, b: float] =
let a = if n > 0: float(n) else: 2.0
let b = if n > 1: float(n - 1) else: 1.0
(a, b)
proc pi(n: int): tuple[a, b: float] =
let a = if n > 0: 6.0 else: 3.0
let b = (2 * float(n) - 1) * (2 * float(n) - 1)
(a, b)
echo calc(sqrt2, 20)
echo calc(napier, 15)
echo calc(pi, 10000)</syntaxhighlight>
{{out}}
<pre>1.414213562373095
2.718281828459046
3.141592653589544</pre>
=={{header|OCaml}}==
<
and nap = 2, fun n -> (max 1 (n-1), n)
and root2 = 1, fun n -> (1, 2) in
Line 1,652 ⟶ 2,478:
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):
<pre>sqrt(2) = 1.414213562373095
Line 1,660 ⟶ 2,486:
=={{header|PARI/GP}}==
Partial solution for simple continued fractions.
<
back(vector(100,i,2-(i==1)))</
Output:
<pre>%1 = 1.4142135623730950488016887242096980786</pre>
=={{header|Pascal}}==
This console application is written in Delphi, which allows the results to be displayed to 17 correct decimal places (Free Pascal seems to allow only 16). As in the jq solution, we aim to work forwards and stop as soon the desired precision has been reached, rather than guess a suitable number of terms and work backwards. In this program, the continued fraction is converted to an infinite sum, each term after the first being the difference between consecutive convergents. The convergence for pi is very slow (as others have noted) so as well as the c.f. in the task description an alternative is given from the Wikipedia article "Continued fraction".
<syntaxhighlight lang="pascal">
program ContFrac_console;
{$APPTYPE CONSOLE}
uses
SysUtils;
type TCoeffFunction = function( n : integer) : extended;
// Calculate continued fraction as a sum, working forwards.
// Stop on reaching a term with absolute value less than epsilon,
// or on reaching the maximum number of terms.
procedure CalcContFrac( a, b : TCoeffFunction;
epsilon : extended;
maxNrTerms : integer = 1000); // optional, with default
var
n : integer;
sum, term, u, v : extended;
whyStopped : string;
begin
sum := a(0);
term := b(1)/a(1);
v := a(1);
n := 1;
repeat
sum := sum + term;
inc(n);
u := v;
v := a(n) + b(n)/u;
term := -term * b(n)/(u*v);
until (Abs(term) < epsilon) or (n >= maxNrTerms);
if n >= maxNrTerms then whyStopped := 'too many terms'
else whyStopped := 'converged';
WriteLn( SysUtils.Format( '%21.17f after %d terms (%s)',
[sum, n, whyStopped]));
end;
//---------------- a and b for sqrt(2) ----------------
function a_sqrt2( n : integer) : extended;
begin
if n = 0 then result := 1
else result := 2;
end;
function b_sqrt2( n : integer) : extended;
begin
result := 1;
end;
//---------------- a snd b for e ----------------
function a_e( n : integer) : extended;
begin
if n = 0 then result := 2
else result := n;
end;
function b_e( n : integer) : extended;
begin
if n = 1 then result := 1
else result := n - 1;
end;
//-------- Rosetta Code a and b for pi --------
function a_pi( n : integer) : extended;
begin
if n = 0 then result := 3
else result := 6;
end;
function b_pi( n : integer) : extended;
var
temp : extended;
begin
temp := 2*n - 1;
result := temp*temp;
end;
//-------- More efficient a and b for pi --------
function a_pi_alt( n : integer) : extended;
begin
if n = 0 then result := 0
else result := 2*n - 1;
end;
function b_pi_alt( n : integer) : extended;
var
temp : extended;
begin
if n = 1 then
result := 4
else begin
temp := n - 1;
result := temp*temp;
end;
end;
//---------------- Main routine ----------------
// Unlike Free Pascal, Delphi does not require
// an @ sign before the function names.
begin
WriteLn( 'sqrt(2)');
CalcContFrac( a_sqrt2, b_sqrt2, 1E-20);
WriteLn( 'e');
CalcContFrac( a_e, b_e, 1E-20);
WriteLn( 'pi');
CalcContFrac( a_pi, b_pi, 1E-20);
WriteLn( 'pi (alternative formula)');
CalcContFrac( a_pi_alt, b_pi_alt, 1E-20);
end.
</syntaxhighlight>
{{out}}
<pre>
sqrt(2)
1.41421356237309505 after 27 terms (converged)
e
2.71828182845904524 after 20 terms (converged)
pi
3.14159265383979293 after 1000 terms (too many terms)
pi (alternative formula)
3.14159265358979324 after 29 terms (converged)
</pre>
=={{header|Perl}}==
<syntaxhighlight lang="perl">use strict;
use warnings;
no warnings 'recursion';
use experimental 'signatures';
sub continued_fraction ($a, $b, $n = 100) {
$a->() + ($n and $b->() / continued_fraction($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++
printf "π ≈ %.9f\n", continued_fraction do { my $n; sub { $n++ ? 6 : 3 } }, do { my $n; sub { (2*$n++ + 1)**2 } },
printf "π/2 ≈ %.9f\n", continued_fraction do { my $n; sub { 1/($n++
{{out}}
<pre>√2 ≈ 1.414213562
Line 1,685 ⟶ 2,635:
π/2 ≈ 1.570717797</pre>
=={{header|
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">precision</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">10000</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">continued_fraction</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">f</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">steps</span><span style="color: #0000FF;">=</span><span style="color: #000000;">precision</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">atom</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">steps</span> <span style="color: #008080;">to</span> <span style="color: #000000;">1</span> <span style="color: #008080;">by</span> <span style="color: #0000FF;">-</span><span style="color: #000000;">1</span> <span style="color: #008080;">do</span>
<span style="color: #0000FF;">{</span><span style="color: #000000;">a</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">}</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">f</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">res</span> <span style="color: #0000FF;">:=</span> <span style="color: #000000;">b</span> <span style="color: #0000FF;">/</span> <span style="color: #0000FF;">(</span><span style="color: #000000;">a</span> <span style="color: #0000FF;">+</span> <span style="color: #000000;">res</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #0000FF;">{</span><span style="color: #000000;">a</span><span style="color: #0000FF;">}</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">f</span><span style="color: #0000FF;">(</span><span style="color: #000000;">0</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">a</span> <span style="color: #0000FF;">+</span> <span style="color: #000000;">res</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">sqr2</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">return</span> <span style="color: #0000FF;">{</span><span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">0</span><span style="color: #0000FF;">?</span><span style="color: #000000;">1</span><span style="color: #0000FF;">:</span><span style="color: #000000;">2</span><span style="color: #0000FF;">),</span><span style="color: #000000;">1</span><span style="color: #0000FF;">}</span> <span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">nap</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">return</span> <span style="color: #0000FF;">{</span><span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">0</span><span style="color: #0000FF;">?</span><span style="color: #000000;">2</span><span style="color: #0000FF;">:</span><span style="color: #000000;">n</span><span style="color: #0000FF;">),</span><span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span><span style="color: #0000FF;">?</span><span style="color: #000000;">1</span><span style="color: #0000FF;">:</span><span style="color: #000000;">n</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)}</span> <span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">pi</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">return</span> <span style="color: #0000FF;">{</span><span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">0</span><span style="color: #0000FF;">?</span><span style="color: #000000;">3</span><span style="color: #0000FF;">:</span><span style="color: #000000;">6</span><span style="color: #0000FF;">),</span><span style="color: #7060A8;">power</span><span style="color: #0000FF;">(</span><span style="color: #000000;">2</span><span style="color: #0000FF;">*</span><span style="color: #000000;">n</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">)}</span> <span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Precision: %d\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">precision</span><span style="color: #0000FF;">})</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Sqr(2): %.10g\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">continued_fraction</span><span style="color: #0000FF;">(</span><span style="color: #000000;">sqr2</span><span style="color: #0000FF;">)})</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Napier: %.10g\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">continued_fraction</span><span style="color: #0000FF;">(</span><span style="color: #000000;">nap</span><span style="color: #0000FF;">)})</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Pi: %.10g\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">continued_fraction</span><span style="color: #0000FF;">(</span><span style="color: #000000;">pi</span><span style="color: #0000FF;">)})</span>
<!--</syntaxhighlight>-->
{{Out}}
<pre>
Precision: 10000
Sqr(2): 1.414213562
Napier: 2.718281828
Pi: 3.141592654
</pre>
=={{header|Picat}}==
For Pi a test is added with a higher precision (200 -> 2000) to get a better result.
===Recursion===
{{trans|Prolog}}
<syntaxhighlight lang="picat">go =>
% square root 2
continued_fraction(200, sqrt_2_ab, V1),
printf("sqrt(2) = %w (diff: %0.15f)\n", V1, V1-sqrt(2)),
% napier
continued_fraction(200, napier_ab, V2),
printf("e = %w (diff: %0.15f)\n", V2, V2-math.e),
% pi
continued_fraction(200, pi_ab, V3),
printf("pi = %w (diff: %0.15f)\n", V3, V3-math.pi),
% get a better precision
continued_fraction(20000, pi_ab, V3b),
printf("pi = %w (diff: %0.15f)\n", V3b, V3b-math.pi),
nl.
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 = A + Temp.
continued_fraction(N, Compute_ab, Tmp, V) =>
call(Compute_ab, N, A, B),
Tmp1 = B / (A + Tmp),
N1 = N - 1,
continued_fraction(N1, Compute_ab, Tmp1, V).
% 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).</syntaxhighlight>
{{out}}
<pre>
sqrt(2) = 1.414213562373095 (diff: 0.000000000000000)
e = 2.718281828459046 (diff: 0.000000000000000)
pi = 3.141592622804847 (diff: -0.000000030784946)
pi = 3.141592653589762 (diff: -0.000000000000031)
</pre>
===Iterative===
{{trans|Python}}
(from Python's Fast Iterative version)
<syntaxhighlight lang="picat">continued_fraction_it(Fun, N) = Ret =>
Temp = 0.0,
foreach(I in N..-1..1)
[A,B] = apply(Fun,I),
Temp := B / (A + Temp)
end,
F = apply(Fun,0),
Ret = F[1] + Temp.
fsqrt2(N) = [cond(N > 0, 2, 1),1].
fnapier(N) = [cond(N > 0, N,2), cond(N>1,N-1,1)].
fpi(N) = [cond(N>0,6,3), (2*N-1) ** 2].</syntaxhighlight>
Which has exactly the same output as the recursive version.
=={{header|PicoLisp}}==
<syntaxhighlight lang="picolisp">(scl 49)
(de fsqrt2 (N A)
(default A 1)
(cond
((> A (inc N)) 2)
(T
(+
(if (=1 A) 1.0 2.0)
(*/ `(* 1.0 1.0) (fsqrt2 N (inc A))) ) ) ) )
(de pi (N A)
(default A 1)
(cond
((> A (inc N)) 6.0)
(T
(+
(if (=1 A) 3.0 6.0)
(*/
(* (** (dec (* 2 A)) 2) 1.0)
1.0
(pi N (inc A)) ) ) ) ) )
(de napier (N A)
(default A 0)
(cond
((> A N) (* A 1.0))
(T
(+
(if (=0 A) 2.0 (* A 1.0))
(*/
(if (> 1 A) 1.0 (* A 1.0))
1.0
(napier N (inc A)) ) ) ) ) )
(prinl (format (fsqrt2 200) *Scl))
(prinl (format (napier 200) *Scl))
(prinl (format (pi 200) *Scl))</syntaxhighlight>
{{out}}
<pre>
1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134
</pre>
=={{header|PL/I}}==
<
test: proc options (main);
declare n fixed;
Line 1,772 ⟶ 2,802:
put (1 + 1/denom(2));
end test;</
{{out}}
<pre> 1.41421356237309505E+0000 </pre>
Version for NAPIER:
<
declare n fixed;
Line 1,788 ⟶ 2,818:
put (2 + 1/denom(0));
end test;</
<pre> 2.71828182845904524E+0000 </pre>
Version for SQRT2, NAPIER, PI
<
continued_fractions: /* 6 Sept. 2012 */
Line 1,834 ⟶ 2,864:
put skip edit ('PI=', calc(pi, 99999)) (a(10), f(20,17));
end continued_fractions;</
{{out}}
<pre>
Line 1,843 ⟶ 2,873:
=={{header|Prolog}}==
<
% square root 2
continued_fraction(200, sqrt_2_ab, V1),
Line 1,885 ⟶ 2,915:
pi_ab(0, 3, _).
pi_ab(N, 6, V) :-
V is (2 * N - 1)*(2 * N - 1).</
{{out}}
<pre> ?- continued_fraction.
Line 1,896 ⟶ 2,926:
=={{header|Python}}==
{{works with|Python|2.6+ and 3.x}}
<
import itertools
try: zip = itertools.izip
Line 1,962 ⟶ 2,992:
cf = CF(Pi_a(), Pi_b(), 950)
print(pRes(cf, 10))
#3.1415926532</
===Fast iterative version===
{{trans|D}}
<
def calc(fun, n):
Line 1,988 ⟶ 3,018:
print calc(fsqrt2, 200)
print calc(fnapier, 200)
print calc(fpi, 200)</
{{out}}
<pre>1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134</pre>
=={{header|Quackery}}==
<syntaxhighlight lang="quackery"> [ $ "bigrat.qky" loadfile ] now!
[ 1 min
[ table
[ 1 1 ]
[ 2 1 ] ] do ] is sqrt2 ( n --> n/d )
[ dup 2 min
[ table
[ drop 2 1 ]
[ 1 ]
[ dup 1 - ] ] do ] is napier ( n --> n/d )
[ dup 1 min
[ table
[ drop 3 1 ]
[ 2 * 1 - dup *
6 swap ] ] do ] is pi ( n --> n/d )
[ ]'[ temp put
0 1
rot times
[ i 1+
temp share do
v+ 1/v ]
0 temp take do v+ ] is cf ( n --> n/d )
1000 cf sqrt2 10 point$ echo$ cr
1000 cf napier 10 point$ echo$ cr
1000 cf pi 10 point$ echo$ cr</syntaxhighlight>
{{out}}
<pre>1.4142135624
2.7136688544
3.1413776152</pre>
=={{header|Racket}}==
===Using Doubles===
This version uses standard double precision floating point numbers:
<
#lang racket
(define (calc cf n)
Line 2,014 ⟶ 3,083:
(calc cf-napier 200)
(calc cf-pi 200)
</syntaxhighlight>
Output:
<
1.4142135623730951
2.7182818284590455
3.1415926839198063
</syntaxhighlight>
===Version - Using Doubles===
This versions uses big floats (arbitrary precision floating point):
<
#lang racket
(require math)
Line 2,044 ⟶ 3,113:
(calc cf-napier 200)
(calc cf-pi 200)
</syntaxhighlight>
Output:
<
(bf #e1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727350138462309122970249248360558507372126441214970999358960036439214262599769155193770031712304888324413327207659690547583107739957489062466508437105234564161085482146113860092820802430986649987683947729823677905101453725898480737256099166805538057375451207262441039818826744940289448489312217214883459060818483750848688583833366310472320771259749181255428309841375829513581694269249380272698662595131575038315461736928338289219865139248048189188905788104310928762952913687232022557677738108337499350045588767581063729)
(bf #e2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901157383418793070215408914993488416750924476146066808226480016847741185374234544243710753907774499206955170276183860626133138458300075204493382656029760673711320070932870912744374704723624212700454495421842219077173525899689811474120614457405772696521446961165559468253835854362096088934714907384964847142748311021268578658461064714894910680584249490719358138073078291397044213736982988247857479512745588762993966446075)
(bf #e3.14159268391980626493420192940831754203350026401337226640663040854412059241988978103217808449508253393479795573626200366332733859609651462659489470805432281782785922056335606047700127154963266242144951481397480765182268219697420028007903565511884267297358842935537138583640066772149177226656227031792115896439889412205871076985598822285367358003457939603015797225018209619662200081521930463480571130673429337524564941105654923909951299948539893933654293161126559643573974163405197696633200469475250152247413175932572922175467223988860975105100904322239324381097207835036465269418118204894206705789759765527734394105147)
</syntaxhighlight>
=={{header|Raku}}==
(formerly Perl 6)
{{Works with|rakudo|2015-10-31}}
<syntaxhighlight lang="raku" line>sub continued-fraction(:@a, :@b, Int :$n = 100)
{
my $x = @a[$n - 1];
$x = @a[$_ - 1] + @b[$_] / $x for reverse 1 ..^ $n;
$x;
}
printf "√2 ≈%.9f\n", continued-fraction(:a(1, |(2 xx *)), :b(Nil, |(1 xx *)));
printf "e ≈%.9f\n", continued-fraction(:a(2, |(1 .. *)), :b(Nil, 1, |(1 .. *)));
printf "π ≈%.9f\n", continued-fraction(:a(3, |(6 xx *)), :b(Nil, |((1, 3, 5 ... *) X** 2)));</syntaxhighlight>
{{out}}
<pre>√2 ≈ 1.414213562
e ≈ 2.718281828
π ≈ 3.141592654</pre>
A more original and a bit more abstract method would consist in viewing a continued fraction on rank n as a function of a variable x:
:<math>\mathrm{CF}_3(x) = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + x}}}</math>
Or, more consistently:
:<math>\mathrm{CF}_3(x) = a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{a_3 + \cfrac{b_3}{x}}}}</math>
Viewed as such, <math>\mathrm{CF}_n(x)</math> could be written recursively:
:<math>\mathrm{CF}_n(x) = \mathrm{CF}_{n-1}(a_n + \frac{b_n}{x})</math>
Or in other words:
:<math>\mathrm{CF}_n= \mathrm{CF}_{n-1}\circ f_n = \mathrm{CF}_{n-2}\circ f_{n-1}\circ f_n=\ldots=f_0\circ f_1 \ldots \circ f_n</math>
where <math>f_n(x) = a_n + \frac{b_n}{x}</math>
Raku has a builtin composition operator. We can use it with the triangular reduction metaoperator, and evaluate each resulting function at infinity (any value would do actually, but infinite makes it consistent with this particular task).
<syntaxhighlight lang="raku" line>sub continued-fraction(@a, @b) {
map { .(Inf) }, [\o] map { @a[$_] + @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];</syntaxhighlight>
{{out}}
<pre>√2 ≈ 1.414213552
e ≈ 2.718281827
π ≈ 3.141592411</pre>
=={{header|REXX}}==
Line 2,067 ⟶ 3,177:
More code is used for nicely formatting the output than the continued fraction calculation.
<
parse arg terms digs .
if terms=='' | terms=="," then terms=500
Line 2,109 ⟶ 3,219:
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'''
<pre>
Line 2,157 ⟶ 3,267:
===version 2 derived from [[#PL/I|PL/I]]===
<
* Derived from PL/I with a little "massage"
* SQRT2= 1.41421356237309505 <- PL/I Result
Line 2,200 ⟶ 3,310:
end
call Get_Coeffs form, 0
return (A + Temp)</
===version 3 better approximation===
<
* The task description specifies a continued fraction for pi
* that gives a reasonable approximation.
Line 2,274 ⟶ 3,384:
end
call Get_Coeffs 0
return (A + Temp)</
=={{header|Ring}}==
<
# Project : Continued fraction
Line 2,296 ⟶ 3,406:
eval("temp3=" + expr + "1" + str)
return a0 + b1 / temp3
</syntaxhighlight>
Output:
<pre>
Line 2,302 ⟶ 3,412:
e = 2.718281828459046
PI = 3.141592653588017
</pre>
=={{header|RPL}}==
This task demonstrates how both global and local variables, arithmetic expressions and stack can be used together to build a compact and versatile piece of code.
{{works with|Halcyon Calc|4.2.7}}
{| class="wikitable"
! RPL code
! Comment
|-
|
≪
4 ROLL ROT → an bn
≪ 'N' STO 0 '''WHILE''' N 2 ≥ '''REPEAT'''
an + INV bn * EVAL
'N' 1 STO- '''END'''
an + / + EVAL
'N' PURGE
≫ ≫ ‘'''→CFRAC'''’ STO
|
'''→CFRAC''' ''( a0 an b1 bn N -- x ) ''
Unstack an and bn
Loop from N to 2
Calculate Nth fraction
Decrement counter
Calculate last fraction with b1 in stack, then add a0
Discard N variable - not mandatory but hygienic
|}
{{in}}
<pre>
1 2 1 1 100 →CFRAC
2 'N' 1 'N-1' 100 →CFRAC
3 6 1 '(2*N-1)^2' 1000 →CFRAC
1 1 1 1 100 →CFRAC
1 '2-MOD(N,2)' 1 1 100 →CFRAC
</pre>
{{out}}
<pre>
5: 1.41421356237
4: 2.71828182846
3: 3.14159265334
2: 1.61803398875
1: 1.73205080757
</pre>
=={{header|Ruby}}==
<
# square root of 2
Line 2,351 ⟶ 3,504:
puts estimate(sqrt2, 50).to_s('F')
puts estimate(napier, 50).to_s('F')
puts estimate(pi, 10).to_s('F')</
{{out}}
<pre>$ ruby cfrac.rb
Line 2,359 ⟶ 3,512:
=={{header|Rust}}==
<
use std::iter;
Line 2,405 ⟶ 3,558:
println!("{}", continued_fraction!(pia, pib));
}
</syntaxhighlight>
{{out}}
Line 2,417 ⟶ 3,570:
{{works with|Scala|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.
<
import Stream._
val sqrt2 = 1 #:: from(2,0) zip from(1,0)
Line 2,447 ⟶ 3,600:
println()
}
}</
{{out}}
<pre>sqrt2:
Line 2,464 ⟶ 3,617:
precision: 3.14159265358</pre>
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:
<
import Stream._
val sqrt2 = 1 #:: from(2,0) zip from(1,0)
Line 2,498 ⟶ 3,651:
println()
}
}</
{{out}}
<pre>sqrt2:
Line 2,518 ⟶ 3,671:
The following code relies on a library implementing SRFI 41 (lazy streams). Most Scheme interpreters include an implementation.
<
(import (rnrs base (6))
(srfi :41 streams))
Line 2,567 ⟶ 3,720:
(stream-cons 3
(stream-cycle (build-stream (lambda (n) (expt (- (* 2 (+ n 1)) 1) 2)))
(stream-constant 6))))</
Test:
<
1.4142135623730951
> (cf->real napier)
2.7182818284590455
> (cf->real pi)
3.141592653589794</
=={{header|Sidef}}==
<syntaxhighlight lang="ruby">func continued_fraction(a, b, f, n = 1000, r = 1) {
f(func (r) {
r < n ? (a(r) / (b(r) + __FUNC__(r+1))) : 0
}(r))
}
var
"
"√2" => [ {
"e" => [ { _ }, { _ }, { 1 + 1/_ } ],
"π" => [ { (2*_ - 1)**2 }, { 6 }, { 3 + _ } ],
"τ" => [ { _**2 }, { 2*_ + 1 }, { 8 / (1 + _) } ],
)
for k in (params.keys.sort) {
printf("%
}</
{{out}}
<pre>
e ≈ 2.7182818284590452353602874713526624977572470937
π ≈ 3.14159265383979292596359650286939597045138933078
τ ≈ 6.28318530717958647692528676655900576839433879875
φ ≈ 1.61803398874989484820458683436563811772030917981
√2 ≈ 1.41421356237309504880168872420969807856967187538
</pre>
=={{header|Swift}}==
{{trans|Rust}}
<syntaxhighlight lang="swift">extension BinaryInteger {
@inlinable
public func power(_ n: Self) -> Self {
return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *)
}
}
public struct CycledSequence<WrappedSequence: Sequence> {
private var seq: WrappedSequence
private var iter: WrappedSequence.Iterator
init(seq: WrappedSequence) {
self.seq = seq
self.iter = seq.makeIterator()
}
}
extension CycledSequence: Sequence, IteratorProtocol {
public mutating func next() -> WrappedSequence.Element? {
if let ele = iter.next() {
return ele
} else {
iter = seq.makeIterator()
return iter.next()
}
}
}
extension Sequence {
public func cycled() -> CycledSequence<Self> {
return CycledSequence(seq: self)
}
}
public struct ChainedSequence<Element> {
private var sequences: [AnySequence<Element>]
private var iter: AnyIterator<Element>
private var curSeq = 0
init(chain: ChainedSequence) {
self.sequences = chain.sequences
self.iter = chain.iter
self.curSeq = chain.curSeq
}
init<Seq: Sequence>(_ seq: Seq) where Seq.Element == Element {
sequences = [AnySequence(seq)]
iter = sequences[curSeq].makeIterator()
}
func chained<Seq: Sequence>(with seq: Seq) -> ChainedSequence where Seq.Element == Element {
var res = ChainedSequence(chain: self)
res.sequences.append(AnySequence(seq))
return res
}
}
extension ChainedSequence: Sequence, IteratorProtocol {
public mutating func next() -> Element? {
if let el = iter.next() {
return el
}
curSeq += 1
guard curSeq != sequences.endIndex else {
return nil
}
iter = sequences[curSeq].makeIterator()
return iter.next()
}
}
extension Sequence {
public func chained<Seq: Sequence>(with other: Seq) -> ChainedSequence<Element> where Seq.Element == Element {
return ChainedSequence(self).chained(with: other)
}
}
func continuedFraction<T: Sequence, V: Sequence>(
_ seq1: T,
_ seq2: V,
iterations: Int = 1000
) -> Double where T.Element: BinaryInteger, T.Element == V.Element {
return zip(seq1, seq2).prefix(iterations).reversed().reduce(0.0, { Double($1.0) + (Double($1.1) / $0) })
}
let sqrtA = [1].chained(with: [2].cycled())
let sqrtB = [1].cycled()
print("√2 ≈ \(continuedFraction(sqrtA, sqrtB))")
let napierA = [2].chained(with: 1...)
let napierB = [1].chained(with: 1...)
print("e ≈ \(continuedFraction(napierA, napierB))")
let piA = [3].chained(with: [6].cycled())
let piB = (1...).lazy.map({ (2 * $0 - 1).power(2) })
print("π ≈ \(continuedFraction(piA, piB))")
</syntaxhighlight>
{{out}}
<pre>√2 ≈ 1.4142135623730951
e ≈ 2.7182818284590455
π ≈ 3.141592653339042</pre>
{{trans|Java}}
<syntaxhighlight lang="swift">
import Foundation
func calculate(n: Int, operation: (Int) -> [Int])-> Double {
var tmp: Double = 0
for ni in stride(from: n, to:0, by: -1) {
var p = operation(ni)
tmp = Double(p[1])/(Double(p[0]) + tmp);
}
return Double(operation(0)[0]) + tmp;
}
func sqrt (n: Int) -> [Int] {
return [n > 0 ? 2 : 1, 1]
}
func napier (n: Int) -> [Int] {
var res = [n > 0 ? n : 2, n > 1 ? (n - 1) : 1]
return res
}
func pi(n: Int) -> [Int] {
var res = [n > 0 ? 6 : 3, Int(pow(Double(2 * n - 1), 2))]
return res
}
print (calculate(n: 200, operation: sqrt));
print (calculate(n: 200, operation: napier));
print (calculate(n: 200, operation: pi));
</syntaxhighlight>
=={{header|Tcl}}==
Line 2,621 ⟶ 3,912:
{{trans|Python}}
Note that Tcl does not provide arbitrary precision floating point numbers by default, so all result computations are done with IEEE <code>double</code>s.
<
# Term generators; yield list of pairs
Line 2,661 ⟶ 3,952:
puts [cf r2]
puts [cf e]
puts [cf pi 250]; # Converges more slowly</
{{out}}
<pre>1.4142135623730951
2.7182818284590455
3.1415926373965735</pre>
=={{header|VBA}}==
{{trans|Phix}}<syntaxhighlight lang="vb">Public Const precision = 10000
Private Function continued_fraction(steps As Integer, rid_a As String, rid_b As String) As Double
Dim res As Double
res = 0
For n = steps To 1 Step -1
res = Application.Run(rid_b, n) / (Application.Run(rid_a, n) + res)
Next n
continued_fraction = Application.Run(rid_a, 0) + res
End Function
Function sqr2_a(n As Integer) As Integer
sqr2_a = IIf(n = 0, 1, 2)
End Function
Function sqr2_b(n As Integer) As Integer
sqr2_b = 1
End Function
Function nap_a(n As Integer) As Integer
nap_a = IIf(n = 0, 2, n)
End Function
Function nap_b(n As Integer) As Integer
nap_b = IIf(n = 1, 1, n - 1)
End Function
Function pi_a(n As Integer) As Integer
pi_a = IIf(n = 0, 3, 6)
End Function
Function pi_b(n As Integer) As Long
pi_b = IIf(n = 1, 1, (2 * n - 1) ^ 2)
End Function
Public Sub main()
Debug.Print "Precision:", precision
Debug.Print "Sqr(2):", continued_fraction(precision, "sqr2_a", "sqr2_b")
Debug.Print "Napier:", continued_fraction(precision, "nap_a", "nap_b")
Debug.Print "Pi:", continued_fraction(precision, "pi_a", "pi_b")
End Sub</syntaxhighlight>{{out}}
<pre>Precision: 10000
Sqr(2): 1,4142135623731
Napier: 2,71828182845905
Pi: 3,14159265358954 </pre>
=={{header|Visual Basic .NET}}==
{{trans|C#}}
<syntaxhighlight lang="vbnet">Module Module1
Function Calc(f As Func(Of Integer, Integer()), n As Integer) As Double
Dim temp = 0.0
For ni = n To 1 Step -1
Dim p = f(ni)
temp = p(1) / (p(0) + temp)
Next
Return f(0)(0) + temp
End Function
Sub Main()
Dim fList = {
Function(n As Integer) New Integer() {If(n > 0, 2, 1), 1},
Function(n As Integer) New Integer() {If(n > 0, n, 2), If(n > 1, n - 1, 1)},
Function(n As Integer) New Integer() {If(n > 0, 6, 3), Math.Pow(2 * n - 1, 2)}
}
For Each f In fList
Console.WriteLine(Calc(f, 200))
Next
End Sub
End Module</syntaxhighlight>
{{out}}
<pre>1.4142135623731
2.71828182845905
3.14159262280485</pre>
=={{header|Wren}}==
{{trans|D}}
<syntaxhighlight lang="wren">var calc = Fn.new { |f, n|
var t = 0
for (i in n..1) {
var p = f.call(i)
t = p[1] / (p[0] + t)
}
return f.call(0)[0] + t
}
var pList = [
["sqrt(2)", Fn.new { |n| [(n > 0) ? 2 : 1, 1] }],
["e ", Fn.new { |n| [(n > 0) ? n : 2, (n > 1) ? n - 1 : 1] }],
["pi ", Fn.new { |n| [(n > 0) ? 6 : 3, (2*n - 1) * (2*n - 1)] }]
]
for (p in pList) System.print("%(p[0]) = %(calc.call(p[1], 200))")</syntaxhighlight>
{{out}}
<pre>
sqrt(2) = 1.4142135623731
e = 2.718281828459
pi = 3.1415926228048
</pre>
=={{header|XPL0}}==
The number of iterations (N) needed to get the 13 digits of accuracy was determined by experiment.
<
int N;
real A, B, F;
Line 2,693 ⟶ 4,085:
RlOut(0, 3.0+F); CrLf(0);
RlOut(0, ACos(-1.0)); CrLf(0);
]</
{{out}}
<pre>
Line 2,708 ⟶ 4,100:
=={{header|zkl}}==
<
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(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
{{out}}
Line 2,729 ⟶ 4,121:
=={{header|ZX Spectrum Basic}}==
{{trans|BBC_BASIC}}
<
20 LET a0=2: LET b1=1: LET a$="N": LET b$="N": PRINT "e = ";: GO SUB 1000
30 LET a0=3: LET b1=1: LET a$="6": LET b$="(2*N+1)^2": PRINT "PI = ";: GO SUB 1000
Line 2,739 ⟶ 4,131:
1035 FOR i=1 TO n: LET p$=p$+")": NEXT i
1040 PRINT a0+b1/VAL (e$+"1"+p$)
1050 RETURN</
|