Continued fraction
A number may be represented as a continued fraction (see Mathworld for more information) as follows:
You are encouraged to solve this task according to the task description, using any language you may know.
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 then . is always .
For Napier's Constant, use , then . then .
For Pi, use then . .
Ada
<lang Ada>with Ada.Text_IO; use Ada.Text_IO; procedure ContFract is
type FormType is (Sqrt2, Napier, Pi); type Floaty is digits 15; package FIO is new Ada.Text_IO.Float_IO (Floaty);
procedure GetCoefs (form : FormType; n : Natural;
coefA : out Natural; coefB : out Natural)
is begin
case form is
when Sqrt2 =>
if n > 0 then coefA := 2; else coefA := 1; end if;
coefB := 1;
when Napier =>
if n > 0 then coefA := n; else coefA := 2; end if;
if n > 1 then coefB := n - 1; else coefB := 1; end if;
when Pi =>
if n > 0 then coefA := 6; else coefA := 3; end if;
coefB := (2*n - 1)**2;
end case;
end GetCoefs;
function Calc (form : FormType; n : Natural) return Floaty is
A, B : Natural;
Temp : Floaty := 0.0;
begin
for ni in reverse Natural range 1 .. n loop
GetCoefs (form, ni, A, B);
Temp := Floaty (B) / (Floaty (A) + Temp);
end loop;
GetCoefs (form, 0, A, B);
return Floaty (A) + Temp;
end Calc;
begin
FIO.Put (Calc (Sqrt2, 200), Exp => 0); New_Line; FIO.Put (Calc (Napier, 200), Exp => 0); New_Line; FIO.Put (Calc (Pi, 200), Exp => 0); New_Line;
end ContFract; </lang>
- Output:
1.41421356237310 2.71828182845905 3.14159262280485
Axiom
Axiom provides a ContinuedFraction domain: <lang Axiom>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</lang> Output:<lang Axiom> (1) 1.4142135623 730950488
Type: Float
(2) 2.7182818284 590452354
Type: Float
(3) 3.1415926538 39792926
Type: Float</lang>
The value for 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:<lang Axiom>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</lang>
CoffeeScript
<lang 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)
</lang> output
> coffee continued_fraction.coffee 1.4142135623730951 1.4142135623730951 --- 2.718281828459045 2.7182818284590455 --- 3.141592653589793 3.141592653589793 ---
D
<lang d>import std.typecons;
FP calc(FP, F)(in F fun, in int n) pure nothrow {
FP temp = 0.0;
foreach_reverse (ni; 1 .. n+1) {
immutable a_b = fun(ni);
temp = cast(FP)a_b[1] / (cast(FP)a_b[0] + temp);
}
return cast(FP)fun(0)[0] + temp;
}
// int[2] fsqrt2(in int n) pure nothrow { Tuple!(int,int) fsqrt2(in int n) pure nothrow {
return tuple(n > 0 ? 2 : 1,
1);
}
Tuple!(int,int) fnapier(in int n) pure nothrow {
return tuple(n > 0 ? n : 2,
n > 1 ? (n - 1) : 1);
}
Tuple!(int,int) fpi(in int n) pure nothrow {
return tuple(n > 0 ? 6 : 3,
(2 * n - 1) ^^ 2);
}
void main() {
import std.stdio;
writefln("%.19f", calc!real(&fsqrt2, 200));
writefln("%.19f", calc!real(&fnapier, 200));
writefln("%.19f", calc!real(&fpi, 200));
}</lang>
- Output:
1.4142135623730950487 2.7182818284590452354 3.1415926228048469486
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.
<lang factor>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</lang>
- Output:
Square root of 2: 1.414213562373095048801688724209
Napier's constant: 2.718281828459045235360287471352
Pi: 3.1415926538
Go
<lang 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())
}</lang>
- Output:
sqrt2: 1.4142135623730965 nap: 2.7182818284590455 pi: 3.141623806667839
Haskell
<lang haskell>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]</lang>
- Output:
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222665927505592755799950501152782060571 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019 3.141592653297590947683406834261190738869139611505752231394089152890909495973464508817163306557131591579057202097715021166512662872910519439747609829479577279606075707015622200744006783543589980682386
J
<lang 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 </lang>
Python
<lang python> 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
</lang>
Fast iterative version
<lang python>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)</lang>
- Output:
1.4142135623730950488016887242096980785696718753770 2.7182818284590452353602874713526624977572470937000 3.1415926839198062649342019294083175420335002640134
REXX
The $CF subroutine (for continous fractions) isn't limited to positive integers,
any form of REXX numbers (negative, exponentiated, decimal fractions) can be used.
There isn't any realistic limit on the precision that can be used, although 100,000 digits would be a bit unwieldly to display.
<lang rexx>
/*calculate some continous fractions. */
numeric digits 60 /*let's use sixty digits for show&tell*/ terms=100 /* ... and an even one hundred terms. */ /*──────────────────────────────────────────────────────────────────────*/ a=copies(' 2',terms); call tell 'sqrt(2)',$cf(1 a) /*──────────────────────────────────────────────────────────────────────*/
do j=1 for terms a=a j end; call tell 'e',$cf(2 a,1 a)
/*──────────────────────────────────────────────────────────────────────*/ a=copies(' 6',terms)
do j=1 for terms b=b (2*j-1)**2 end; call tell 'pi',$cf(3 a,b)
/*──────────────────────────────────────────────────────────────────────*/ a=copies(' 1',terms); call tell 'phi',$cf(1 a) /*──────────────────────────────────────────────────────────────────────*/ a=copies(' 1 2',terms%2); call tell 'sqrt(3)',$cf(1 a)
/*also: 2∙sin(π/3) */
/*──────────────────────────────────────────────────────────────────────*/
do j=1 for terms%2 by 2 a=a j 1 end; call tell 'tan(1)',$cf(1 a)
/*──────────────────────────────────────────────────────────────────────*/
do j=1 for terms a=a 2*j+1 end; call tell 'coth(1)',$cf(1 a)
/*──────────────────────────────────────────────────────────────────────*/
do j=1 for terms
a=a 4*j+2
end; call tell 'coth(½)',$cf(2 a)
/*also: [e+1] / [e-1] */
exit
/*────────────────────────────────$CF subroutine────────────────────────*/ $cf: procedure; parse arg c x,y; !=0; numeric digits digits()*2
do j=words(x) to 1 by -1 a=word(x,j); b=word(y,j); if b== then b=1 d=a+!; if d=0 then call divzero /*just in case divisor is bogus.*/ !=b/d end
return !+c
divzero: say; say '***error!***'; say 'division by zero.'; say; exit 13 tell:parse arg _,v;say right(_,8) '=' format(v) 'a terms='left(a,40);a=;b=;return </lang> Output:
sqrt(2) = 1.41421356237309504880168872420969807856967187537694807317668 a terms= 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
e = 2.71828182845904523536028747135266249775724709369995957496697 a terms= 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
pi = 3.14159241097198067426258886021672643729365090674537669567319 a terms= 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
phi = 1.61803398874989484820458683436563811772030781841853559947667 a terms= 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
sqrt(3) = 1.73205080756887729352744634150587236694280525381038062805521 a terms= 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
tan(1) = 1.55740772465490223050697480745836017308725077238152003838395 a terms= 1 1 3 1 5 1 7 1 9 1 11 1 13 1 15 1 17 1
coth(1) = 1.31303528549933130363616124693084783291201394124045265554315 a terms= 3 5 7 9 11 13 15 17 19 21 23 25 27 29 3
coth(½) = 2.16395341373865284877000401021802311709373860215079227253357 a terms= 6 10 14 18 22 26 30 34 38 42 46 50 54 5
Ruby
<lang 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')</lang>
- Output:
$ ruby cfrac.rb 1.41421356237309504880168872420969807856967187537695 2.71828182845904523536028747135266249775724709369996 3.1415926536
Tcl
Note that Tcl does not provide arbitrary precision floating point numbers by default, so all result computations are done with IEEE doubles.
<lang tcl>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</lang>
- 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. <lang XPL0>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); ]</lang>
Output:
16 1.414213562372820 1.414213562373100 13 2.718281828459380 2.718281828459050 10000 3.141592653589540 3.141592653589790