Continued fraction/Arithmetic/Construct from rational number
You are encouraged to solve this task according to the task description, using any language you may know.
To understand this task in context please see Continued fraction arithmetic
The purpose of this task is to write a function , or , which will output a continued fraction assuming:
- is the numerator
- is the denominator
The function should output its results one digit at a time each time it is called, in a manner sometimes described as lazy evaluation.
To achieve this it must determine: the integer part; and remainder part, of divided by . It then sets to and to the determined remainder part. It then outputs the determined integer part. It does this until is zero.
Demonstrate the function by outputing the continued fraction for:
- 1/2
- 3
- 23/8
- 13/11
- 22/7
- -151/77
should approach try ever closer rational approximations until boredom gets the better of you:
![{\displaystyle [1; 2, 2, 2, 2, \ldots]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68fb549d738febb493e8e3e1eac0a306663c91b2)
- 14142,10000
- 141421,100000
- 1414214,1000000
- 14142136,10000000
Try :
- 31,10
- 314,100
- 3142,1000
- 31428,10000
- 314285,100000
- 3142857,1000000
- 31428571,10000000
- 314285714,100000000
Observe how this rational number behaves differently to and convince yourself that, in the same way as may be represented as when an extra decimal place is required, may be represented as when an extra term is required.
![{\displaystyle [3;7]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b143dda8c65c1ca85c41576c4c8a3d8f5df6407e)
![{\displaystyle [3;7,\infty]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c19fef4c6c62d8e5d04683af76ddd56291d34125)
C
C does not implement Lazy evaluation and it is this particular feature which is the real challenge of this particular example. It can however be simulated. The following example uses pointers. It seems that the same data is being passed but since the function accepts pointers, the variables are being changed. One other way to simulate laziness would be to use global variables. Then although it would seem that the same values are being passed even as constants, the job is actually getting done. In my view, that would be plain cheating.
<lang C> /*Abhishek Ghosh, 8th November 2013, Rotterdam*/
- include<stdio.h>
typedef struct{ int num,den; }fraction;
fraction examples[] = {{1,2}, {3,1}, {23,8}, {13,11}, {22,7}, {-151,77}}; fraction sqrt2[] = {{14142,10000}, {141421,100000}, {1414214,1000000}, {14142136,10000000}}; fraction pi[] = {{31,10}, {314,100}, {3142,1000}, {31428,10000}, {314285,100000}, {3142857,1000000}, {31428571,10000000}, {314285714,100000000}};
int r2cf(int *numerator,int *denominator) { int quotient=0,temp;
if(denominator != 0) { quotient = *numerator / *denominator;
temp = *numerator;
*numerator = *denominator;
*denominator = temp % *denominator; }
return quotient; }
int main() { int i;
printf("Running the examples :");
for(i=0;i<sizeof(examples)/sizeof(fraction);i++) { printf("\nFor N = %d, D = %d :",examples[i].num,examples[i].den);
while(examples[i].den != 0){ printf(" %d ",r2cf(&examples[i].num,&examples[i].den)); } }
printf("\n\nRunning for %c2 :",251); /* 251 is the ASCII code for the square root symbol */
for(i=0;i<sizeof(sqrt2)/sizeof(fraction);i++) { printf("\nFor N = %d, D = %d :",sqrt2[i].num,sqrt2[i].den);
while(sqrt2[i].den != 0){ printf(" %d ",r2cf(&sqrt2[i].num,&sqrt2[i].den)); } }
printf("\n\nRunning for %c :",227); /* 227 is the ASCII code for Pi's symbol */
for(i=0;i<sizeof(pi)/sizeof(fraction);i++) { printf("\nFor N = %d, D = %d :",pi[i].num,pi[i].den);
while(pi[i].den != 0){ printf(" %d ",r2cf(&pi[i].num,&pi[i].den)); } }
return 0; }
</lang> And the run gives :
Running the examples : For N = 1, D = 2 : 0 2 For N = 3, D = 1 : 3 For N = 23, D = 8 : 2 1 7 For N = 13, D = 11 : 1 5 2 For N = 22, D = 7 : 3 7 For N = -151, D = 77 : -1 -1 -24 -1 -2 Running for √2 : For N = 14142, D = 10000 : 1 2 2 2 2 2 1 1 29 For N = 141421, D = 100000 : 1 2 2 2 2 2 2 3 1 1 3 1 7 2 For N = 1414214, D = 1000000 : 1 2 2 2 2 2 2 2 3 6 1 2 1 12 For N = 14142136, D = 10000000 : 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2 Running for π : For N = 31, D = 10 : 3 10 For N = 314, D = 100 : 3 7 7 For N = 3142, D = 1000 : 3 7 23 1 2 For N = 31428, D = 10000 : 3 7 357 For N = 314285, D = 100000 : 3 7 2857 For N = 3142857, D = 1000000 : 3 7 142857 For N = 31428571, D = 10000000 : 3 7 476190 3 For N = 314285714, D = 100000000 : 3 7 7142857
C++
<lang cpp>#include <iostream> /* Interface for all Continued Fractions
Nigel Galloway, February 9th., 2013.
- /
class ContinuedFraction { public: virtual const int nextTerm(){}; virtual const bool moreTerms(){}; }; /* Create a continued fraction from a rational number
Nigel Galloway, February 9th., 2013.
- /
class r2cf : public ContinuedFraction { private: int n1, n2; public: r2cf(const int numerator, const int denominator): n1(numerator), n2(denominator){} const int nextTerm() { const int thisTerm = n1/n2; const int t2 = n2; n2 = n1 - thisTerm * n2; n1 = t2; return thisTerm; } const bool moreTerms() {return fabs(n2) > 0;} }; /* Generate a continued fraction for sqrt of 2
Nigel Galloway, February 9th., 2013.
- /
class SQRT2 : public ContinuedFraction { private: bool first=true; public: const int nextTerm() {if (first) {first = false; return 1;} else return 2;} const bool moreTerms() {return true;} };</lang>
Testing
1/2 3 23/8 13/11 22/7 -151/77
<lang cpp>int main() { for(r2cf n(1,2); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(3,1); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(23,8); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(13,11); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(22,7); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf cf(-151,77); cf.moreTerms(); std::cout << cf.nextTerm() << " "); std::cout << std::endl; return 0; }</lang>
- Output:
0 2 3 2 1 7 1 5 2 3 7 -1 -1 -24 -1 -2
<lang cpp>int main() { int i = 0; for(SQRT2 n; i++ < 20; std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(14142,10000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(14142136,10000000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; return 0; }</lang>
- Output:
1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 1 29 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2
Real approximations of a rational number
<lang cpp>int main() {
for(r2cf n(31,10); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(314,100); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(3142,1000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(31428,10000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(314285,100000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(3142857,1000000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(31428571,10000000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; for(r2cf n(314285714,100000000); n.moreTerms(); std::cout << n.nextTerm() << " "); std::cout << std::endl; return 0;
}</lang>
- Output:
3 10 3 7 7 3 7 23 1 2 3 7 357 3 7 2857 3 7 142857 3 7 476190 3 3 7 7142857
C#
<lang csharp>using System; using System.Collections.Generic;
class Program {
static IEnumerable<int> r2cf(int n1, int n2)
{
while (Math.Abs(n2) > 0)
{
int t1 = n1 / n2;
int t2 = n2;
n2 = n1 - t1 * n2;
n1 = t2;
yield return t1;
}
}
static void spit(IEnumerable<int> f)
{
foreach (int n in f) Console.Write(" {0}", n);
Console.WriteLine();
}
static void Main(string[] args)
{
spit(r2cf(1, 2));
spit(r2cf(3, 1));
spit(r2cf(23, 8));
spit(r2cf(13, 11));
spit(r2cf(22, 7));
spit(r2cf(-151, 77));
for (int scale = 10; scale <= 10000000; scale *= 10)
{
spit(r2cf((int)(Math.Sqrt(2) * scale), scale));
}
spit(r2cf(31, 10));
spit(r2cf(314, 100));
spit(r2cf(3142, 1000));
spit(r2cf(31428, 10000));
spit(r2cf(314285, 100000));
spit(r2cf(3142857, 1000000));
spit(r2cf(31428571, 10000000));
spit(r2cf(314285714, 100000000));
}
} </lang> Output
0 2 3 2 1 7 1 5 2 3 7 -1 -1 -24 -1 -2 1 2 2 1 2 2 3 1 1 2 1 2 2 2 2 5 3 1 2 2 2 2 2 1 1 29 1 2 2 2 2 2 2 3 1 1 3 1 7 2 1 2 2 2 2 2 2 2 1 1 4 1 1 1 1 1 2 1 6 1 2 2 2 2 2 2 2 2 2 1 594 3 10 3 7 7 3 7 23 1 2 3 7 357 3 7 2857 3 7 142857 3 7 476190 3 3 7 7142857
Common Lisp
<lang lisp>(defun r2cf (n1 n2)
(lambda ()
(unless (zerop n2)
(multiple-value-bind (t1 r)
(floor n1 n2)
(setf n1 n2 n2 r)
t1))))
- Example usage
(defun demo-generator (numbers)
(let* ((n1 (car numbers))
(n2 (cadr numbers))
(gen (r2cf n1 n2)))
(format t "~S ; ~S~%"
`(r2cf ,n1 ,n2)
(loop
:for r = (funcall gen)
:until (null r)
:collect r))))
(mapcar #'demo-generator
'((1 2)
(3 1)
(23 8)
(13 11)
(22 7)
(-151 77)
(14142 10000)
(141421 100000)
(1414214 1000000)
(14142136 10000000)
(31 10)
(314 100)
(3142 1000)
(31428 10000)
(314285 100000)
(3142857 1000000)
(31428571 10000000)
(314285714 100000000)
(3141592653589793 1000000000000000)))</lang>
Output:
(R2CF 3 1) ; (3) (R2CF 23 8) ; (2 1 7) (R2CF 13 11) ; (1 5 2) (R2CF 22 7) ; (3 7) (R2CF -151 77) ; (-2 25 1 2) (R2CF 14142 10000) ; (1 2 2 2 2 2 1 1 29) (R2CF 141421 100000) ; (1 2 2 2 2 2 2 3 1 1 3 1 7 2) (R2CF 1414214 1000000) ; (1 2 2 2 2 2 2 2 3 6 1 2 1 12) (R2CF 14142136 10000000) ; (1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2) (R2CF 31 10) ; (3 10) (R2CF 314 100) ; (3 7 7) (R2CF 3142 1000) ; (3 7 23 1 2) (R2CF 31428 10000) ; (3 7 357) (R2CF 314285 100000) ; (3 7 2857) (R2CF 3142857 1000000) ; (3 7 142857) (R2CF 31428571 10000000) ; (3 7 476190 3) (R2CF 314285714 100000000) ; (3 7 7142857) (R2CF 3141592653589793 1000000000000000) ; (3 7 15 1 292 1 1 1 2 1 3 1 14 4 2 3 1 12 5 1 5 20 1 11 1 1 1 2)
F#
<lang fsharp>let rec r2cf n d =
if d = LanguagePrimitives.GenericZero then [] else let q = n / d in q :: (r2cf d (n - q * d))
[<EntryPoint>] let main argv =
printfn "%A" (r2cf 1 2) printfn "%A" (r2cf 3 1) printfn "%A" (r2cf 23 8) printfn "%A" (r2cf 13 11) printfn "%A" (r2cf 22 7) printfn "%A" (r2cf -151 77) printfn "%A" (r2cf 141 100) printfn "%A" (r2cf 1414 1000) printfn "%A" (r2cf 14142 10000) printfn "%A" (r2cf 141421 100000) printfn "%A" (r2cf 1414214 1000000) printfn "%A" (r2cf 14142136 10000000) 0</lang>
Output
[0; 2] [3] [2; 1; 7] [1; 5; 2] [3; 7] [-1; -1; -24; -1; -2] [1; 2; 2; 3; 1; 1; 2] [1; 2; 2; 2; 2; 5; 3] [1; 2; 2; 2; 2; 2; 1; 1; 29] [1; 2; 2; 2; 2; 2; 2; 3; 1; 1; 3; 1; 7; 2] [1; 2; 2; 2; 2; 2; 2; 2; 3; 6; 1; 2; 1; 12] [1; 2; 2; 2; 2; 2; 2; 2; 2; 2; 6; 1; 2; 4; 1; 1; 2]
Haskell
This more general version generates a continued fraction from any real number (with rationals as a special case): <lang haskell>import Data.Ratio ((%))
real2cf :: (RealFrac a, Integral b) => a -> [b] real2cf x =
i : if f == 0 then [] else real2cf (1/f) where (i, f) = properFraction x
main :: IO () main = do
print $ real2cf (13 % 11) -- => [1,5,2] print $ take 20 $ real2cf (sqrt 2) -- => [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2]</lang>
J
Note that the continued fractions shown in this task differ from those in the Continued fraction task as b here is implicitly always 1.
Tacit version 1
This version is a modification of an explicit version shown in http://www.jsoftware.com/jwiki/Essays/Continued%20Fractions to comply with the task specifications. <lang j>cf=: _1 1 ,@}. (, <.)@%@-/ ::]^:a:@(, <.)@(%&x:/)</lang>
Examples
<lang j> cf each 1 2;3 1;23 8;13 11;22 7;14142136 10000000;_151 77 ┌───┬─┬─────┬─────┬───┬─────────────────────────────────┬─────────┐ │0 2│3│2 1 7│1 5 2│3 7│1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2│_2 25 1 2│ └───┴─┴─────┴─────┴───┴─────────────────────────────────┴─────────┘
cf each 14142 10000;141421 100000;1414214 1000000;14142136 10000000
┌──────────────────┬───────────────────────────┬────────────────────────────┬─────────────────────────────────┐ │1 2 2 2 2 2 1 1 29│1 2 2 2 2 2 2 3 1 1 3 1 7 2│1 2 2 2 2 2 2 2 3 6 1 2 1 12│1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2│ └──────────────────┴───────────────────────────┴────────────────────────────┴─────────────────────────────────┘
cf each 31 10;314 100;3142 1000;31428 10000;314285 100000;3142857 1000000;31428571 10000000;314285714 100000000
┌────┬─────┬──────────┬───────┬────────┬──────────┬────────────┬───────────┐ │3 10│3 7 7│3 7 23 1 2│3 7 357│3 7 2857│3 7 142857│3 7 476190 3│3 7 7142857│ └────┴─────┴──────────┴───────┴────────┴──────────┴────────────┴───────────┘</lang> This tacit version first produces the answer with a trailing ∞ (represented by _ in J) which is then removed by the last operation (_1 1 ,@}. ...). A continued fraction can be evaluated using the verb ((+%)/) and both representations produce equal results, <lang j> 3 7 =&((+ %)/) 3 7 _ 1</lang> Incidentally, J and Tcl report a different representation for -151/77 versus the representation of some other implementations; however, both representations produce equal results. <lang j> _2 25 1 2 =&((+ %)/) _1 _1 _24 _1 _2 1</lang>
Tacit version 2
Translation of python
<lang J>r2cf=:1 1{."1@}.({:,(0,{:)#:{.)^:(*@{:)^:a:</lang>
Example use:
<lang J> ((":@{.,'/',":@{:),': ',":@r2cf)@>1 2;3 1;23 8;13 11;22 7;14142136 10000000;_151 77;14142 10000;141421 100000;1414214 1000000;14142136 10000000;31 10;314 100;3142 1000;31428 10000;314285 100000;3142857 1000000;31428571 10000000;314285714 100000000 1/2: 0 2 3/1: 3 23/8: 2 1 7 13/11: 1 5 2 22/7: 3 7 14142136/10000000: 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2 _151/77: _2 25 1 2 14142/10000: 1 2 2 2 2 2 1 1 29 141421/100000: 1 2 2 2 2 2 2 3 1 1 3 1 7 2 1414214/1000000: 1 2 2 2 2 2 2 2 3 6 1 2 1 12 14142136/10000000: 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2 31/10: 3 10 314/100: 3 7 7 3142/1000: 3 7 23 1 2 31428/10000: 3 7 357 314285/100000: 3 7 2857 3142857/1000000: 3 7 142857 31428571/10000000: 3 7 476190 3 314285714/100000000: 3 7 7142857 </lang>
Explicit versions
version 1
Implemented as a class, r2cf preserves state in a separate locale. I've used some contrivances to jam the examples onto one line. <lang J> coclass'cf' create =: dyad def 'EMPTY [ N =: x , y' destroy =: codestroy r2cf =: monad define
if. 0 (= {:) N do. _ return. end.
RV =. <.@:(%/) N
N =: ({. , |/)@:|. N
RV
)
cocurrent'base' CF =: conew'cf'
Until =: conjunction def 'u^:(-.@:v)^:_'
(,. }.@:}:@:((,r2cf__CF)Until(_-:{:))@:(8[create__CF/)&.>)1 2;3 1;23 8;13 11;22 7;14142136 10000000;_151 77 Note 'Output' ┌─────────────────┬─────────────────────────────────┐ │1 2 │0 2 │ ├─────────────────┼─────────────────────────────────┤ │3 1 │3 │ ├─────────────────┼─────────────────────────────────┤ │23 8 │2 1 7 │ ├─────────────────┼─────────────────────────────────┤ │13 11 │1 5 2 │ ├─────────────────┼─────────────────────────────────┤ │22 7 │3 7 │ ├─────────────────┼─────────────────────────────────┤ │14142136 10000000│1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2│ ├─────────────────┼─────────────────────────────────┤ │_151 77 │_2 25 1 2 │ └─────────────────┴─────────────────────────────────┘ )</lang>
version 2
<lang J> f =: 3 : 0
a =. {.y
b =. {:y
out=. <. a%b
while. b > 1 do.
'a b' =. b; b|a
out=. out , <. a%b
end.
)
f each 1 2;3 1;23 8;13 11;22 7;14142136 10000000;_151 77
┌───┬─┬─────┬─────┬───┬───────────────────────────────────┬─────────┐ │0 2│3│2 1 7│1 5 2│3 7│1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2 _│_2 25 1 2│ └───┴─┴─────┴─────┴───┴───────────────────────────────────┴─────────┘</lang>
version 3
translation of python:
<lang J>r2cf=:3 :0
'n1 n2'=. y r=. while.n2 do. 'n1 t1 n2'=. n2,(0,n2)#:n1 r=.r,t1 end.
)</lang>
Example:
<lang J> r2cf each 1 2;3 1;23 8;13 11;22 7;14142136 10000000;_151 77 ┌───┬─┬─────┬─────┬───┬─────────────────────────────────┬─────────┐ │0 2│3│2 1 7│1 5 2│3 7│1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2│_2 25 1 2│ └───┴─┴─────┴─────┴───┴─────────────────────────────────┴─────────┘</lang>
Julia
<lang Julia># It's most appropriate to define a Julia iterable object for this task
- Julia doesn't have Python's yield, the closest to it is produce/consume calls with Julia tasks
- but for various reasons they don't work out for this task
- This solution works with two integers, a Julia rational or a real
type R2cf n1::Union{Int,Float64} # numerator or real n2::Int # denominator or 1 if real t1::Int # generated coefficient f::Float64 # aux. field for working with real inputs end
- constructors for all possible input types
R2cf(n::Union{Int,Float64})=R2cf(n,1,0,0.1) R2cf(n1::Int,n2::Int)=R2cf(n1,n2,0,0.1) R2cf(r::Rational{Int})=R2cf(num(r),den(r),0,0.1)
- methods to make our object iterable
Base.start(::R2cf)=nothing
- generates the next coefficient
function Base.next(cf::R2cf,s) if typeof(cf.n1)==Int cf.n1, (cf.t1, cf.n2) = cf.n2, divrem(cf.n1, cf.n2) else cf.t1,cf.f=divrem(cf.n1,1) if cf.f!=0.0 cf.n1=1/cf.f end end return (cf.t1,nothing) end
- returns true if we've prepared the continued fraction
function Base.done(cf::R2cf,s) if typeof(cf.n1)==Int return cf.n2==0 else return cf.f==0.0 end end
- tell Julia that this object always returns ints (all coeffs are integers)
Base.eltype(::Type{R2cf})=Int
- overload the default collect function so that we can collect the first maxiter coeffs of infinite continued fractions
- array slicing doesn't work as Julia crashes before the slicing due to our infinitely long array
function Base.collect(itr::R2cf,maxiter::Int) r=Array{Int,1}() # all results are ints i=1 for v=itr push!(r,v) i+=1 if i==maxiter break end end return r end
- test cases according to task description with outputs in comments
println(collect(R2cf(1,2))) # => [0, 2] println(collect(R2cf(3,1))) # => [3] println(collect(R2cf(23,8))) # => [2, 1, 7] println(collect(R2cf(13,11))) # => [1, 5, 2] println(collect(R2cf(22,7))) # => [3, 7] println(collect(R2cf(14142,10000))) # => [1, 2, 2, 2, 2, 2, 1, 1, 29] println(collect(R2cf(141421,100000))) # => [1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 3, 1, 7, 2] println(collect(R2cf(1414214,1000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] println(collect(R2cf(14142136,10000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2]
println(collect(R2cf(13//11))) # => [1, 5, 2] println(collect(R2cf(2 ^ 0.5), 20)) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]</lang>
Mathematica / Wolfram Language
Mathematica has a build-in function ContinuedFraction. <lang mathematica>ContinuedFraction[1/2] ContinuedFraction[3] ContinuedFraction[23/8] ContinuedFraction[13/11] ContinuedFraction[22/7] ContinuedFraction[-151/77] ContinuedFraction[14142/10000] ContinuedFraction[141421/100000] ContinuedFraction[1414214/1000000] ContinuedFraction[14142136/10000000]</lang>
- Output:
{0, 2}
{3}
{2, 1, 7}
{1, 5, 2}
{3, 7}
{-1, -1, -24, -1, -2}
{1, 2, 2, 2, 2, 2, 1, 1, 29}
{1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 3, 1, 7, 2}
{1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12}
{1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2}
PARI/GP
<lang parigp>apply(contfrac,[1/2,3,23/8,13/11,22/7,-151/77])</lang>
- Output:
[[0, 2], [3], [2, 1, 7], [1, 5, 2], [3, 7], [-2, 25, 1, 2]]
Perl
To do output one digit at a time, we first turn off buffering to be pedantic, then use a closure that yields one term per call. <lang perl>$|=1;
sub rc2f {
my($num, $den) = @_;
return sub { return unless $den;
my $q = int($num/$den);
($num, $den) = ($den, $num - $q*$den);
$q; };
}
sub rcshow {
my $sub = shift; print "["; my $n = $sub->(); print "$n" if defined $n; print "; $n" while defined($n = $sub->()); print "]\n";
}
rcshow(rc2f(@$_))
for ([1,2],[3,1],[23,8],[13,11],[22,7],[-151,77]);
print "\n"; rcshow(rc2f(@$_))
for ([14142,10000],[141421,100000],[1414214,1000000],[14142136,10000000]);
print "\n"; rcshow(rc2f(314285714,100000000));</lang>
- Output:
[0; 2] [3] [2; 1; 7] [1; 5; 2] [3; 7] [-1; -1; -24; -1; -2] [1; 2; 2; 2; 2; 2; 1; 1; 29] [1; 2; 2; 2; 2; 2; 2; 3; 1; 1; 3; 1; 7; 2] [1; 2; 2; 2; 2; 2; 2; 2; 3; 6; 1; 2; 1; 12] [1; 2; 2; 2; 2; 2; 2; 2; 2; 2; 6; 1; 2; 4; 1; 1; 2] [3; 7; 7142857]
Perl 6
Straightforward implementation: <lang perl6>sub r2cf(Rat $x is copy) {
gather loop {
$x -= take $x.floor; last unless $x > 0; $x = 1 / $x;
}
}
say r2cf(.Rat) for <1/2 3 23/8 13/11 22/7 1.41 1.4142136>;</lang>
- Output:
(0 2) (3) (2 1 7) (1 5 2) (3 7) (1 2 2 3 1 1 2) (1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2)
As a silly one-liner: <lang perl6>sub r2cf(Rat $x is copy) { gather $x [R/]= 1 while ($x -= take $x.floor) > 0 }</lang>
Python
<lang python>def r2cf(n1,n2):
while n2: n1, (t1, n2) = n2, divmod(n1, n2) yield t1
print(list(r2cf(1,2))) # => [0, 2] print(list(r2cf(3,1))) # => [3] print(list(r2cf(23,8))) # => [2, 1, 7] print(list(r2cf(13,11))) # => [1, 5, 2] print(list(r2cf(22,7))) # => [3, 7] print(list(r2cf(14142,10000))) # => [1, 2, 2, 2, 2, 2, 1, 1, 29] print(list(r2cf(141421,100000))) # => [1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 3, 1, 7, 2] print(list(r2cf(1414214,1000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] print(list(r2cf(14142136,10000000))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2]</lang> This version generates it from any real number (with rationals as a special case): <lang python>def real2cf(x):
while True:
t1, f = divmod(x, 1)
yield int(t1)
if not f:
break
x = 1/f
from fractions import Fraction from itertools import islice
print(list(real2cf(Fraction(13, 11)))) # => [1, 5, 2] print(list(islice(real2cf(2 ** 0.5), 20))) # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]</lang>
Racket
<lang racket>
- lang racket
(define ((r2cf n d))
(or (zero? d)
(let-values ([(q r) (quotient/remainder n d)])
(set! n d)
(set! d r)
q)))
(define (r->cf n d)
(for/list ([i (in-producer (r2cf n d) #t)]) i))
(define (real->cf x places)
(define d (expt 10 places)) (define n (exact-floor (* x d))) (r->cf n d))
(map r->cf
'(1 3 23 13 22 -151)
'(2 1 8 11 7 77))
(real->cf (sqrt 2) 10) (real->cf pi 10) </lang>
- Output:
'((0 2) (3) (2 1 7) (1 5 2) (3 7) (-1 -1 -24 -1 -2)) '(1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 1 3 8 9 1 20 1 2) '(3 7 15 1 292 1 1 6 2 13 3 1 12 3)
REXX
Programming notes:
- Increase numeric digits to a higher value to generate more terms.
- Two subroutines, sqrt and pi, were included here to demonstrate terms for √ 2 and pi.
- The subroutine $maxfact was included and is only needed if the number used for r2cf is a decimal fraction.
- Checks were included to verify that the arguments being passed to r2cf' are indeed numeric and also not zero.
- This version also handles negative numbers.
<lang rexx>/*REXX program converts a decimal or rational fraction to a continued fraction. */ numeric digits 230 /*determines how many terms to be gened*/ say ' 1/2 ──► CF: ' r2cf( '1/2' ) say ' 3 ──► CF: ' r2cf( 3 ) say ' 23/8 ──► CF: ' r2cf( '23/8' ) say ' 13/11 ──► CF: ' r2cf( '13/11' ) say ' 22/7 ──► CF: ' r2cf( '22/7 ' ) say ' ___' say '───────── attempts at √ 2.' say '14142/1e4 ──► CF: ' r2cf( '14142/1e4 ' ) say '141421/1e5 ──► CF: ' r2cf( '141421/1e5 ' ) say '1414214/1e6 ──► CF: ' r2cf( '1414214/1e6 ' ) say '14142136/1e7 ──► CF: ' r2cf( '14142136/1e7 ' ) say '141421356/1e8 ──► CF: ' r2cf( '141421356/1e8 ' ) say '1414213562/1e9 ──► CF: ' r2cf( '1414213562/1e9 ' ) say '14142135624/1e10 ──► CF: ' r2cf( '14142135624/1e10 ' ) say '141421356237/1e11 ──► CF: ' r2cf( '141421356237/1e11 ' ) say '1414213562373/1e12 ──► CF: ' r2cf( '1414213562373/1e12 ' ) say '√2 ──► CF: ' r2cf( sqrt(2) ) say say '───────── an attempt at pi' say 'pi ──► CF: ' r2cf( pi() ) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ r2cf: procedure; parse arg g 1 s 2; $=; if s=='-' then g = substr(g,2)
else s =
if pos('.',g)\==0 then do
if \datatype(g,'N') then call serr 'not numeric:' g
g = $maxfact(g)
end
if pos('/',g)==0 then g = g"/"1
parse var g n '/' d
if \datatype(n,'W') then call serr "a numerator isn't an integer:" n
if \datatype(d,'W') then call serr "a denominator isn't an integer:" d
n = abs(n) /*ensure numerator is positive. */
if d=0 then call serr 'a denominator is zero'
do while d\==0 /*where the rubber meets the road*/
$ = $ s || (n%d) /*append another number to list. */
_ = d
d = n // d /* % is int div, // is modulus.*/
n = _
end /*while*/
return strip($)
/*──────────────────────────────────────────────────────────────────────────────────────*/ pi: return 3.1415926535897932384626433832795028841971693993751058209749445923078164062862,
|| 089986280348253421170679821480865132823066470938446095505822317253594081284,
|| 811174502841027019385211055596446229489549303819644288109756659334461284756,
|| 48233786783165271 /* ··· should ≥ NUMERIC DIGITS */
/*──────────────────────────────────────────────────────────────────────────────────────*/ serr: say; say '***error!***'; say; say arg(1); say; exit 13 /*──────────────────────────────────────────────────────────────────────────────────────*/ sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); h=d+6; numeric form
m.=9; numeric digits; parse value format(x,2,1,,0) 'E0' with g 'E' _ .; g=g*.5'e'_%2
do j=0 while h>9; m.j=h; h=h%2+1; end /*j*/
do k=j+5 to 0 by -1; numeric digits m.k; g=(g+x/g)*.5; end /*k*/
numeric digits d; return g/1
/*──────────────────────────────────────────────────────────────────────────────────────*/ $maxFact: procedure; parse arg x 1 _x,y; y=10**(digits()-1); b=0; h=1; a=1; g=0
do while a<=y & g<=y; n=trunc(_x); _=a; a=n*a+b; b=_; _=g; g=n*g+h; h=_
if n=_x | a/g=x then do; if a>y|g>y then iterate; b=a; h=g; leave; end
_x=1/(_x-n); end; return b'/'h</lang>
output
1/2 ──► CF: 0 2
3 ──► CF: 3
23/8 ──► CF: 2 1 7
13/11 ──► CF: 1 5 2
22/7 ──► CF: 3 7
___
───────── attempts at √ 2.
14142/1e4 ──► CF: 1 2 2 2 2 2 1 1 29
141421/1e5 ──► CF: 1 2 2 2 2 2 2 3 1 1 3 1 7 2
1414214/1e6 ──► CF: 1 2 2 2 2 2 2 2 3 6 1 2 1 12
14142136/1e7 ──► CF: 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2
141421356/1e8 ──► CF: 1 2 2 2 2 2 2 2 2 2 2 3 4 1 1 2 6 8
1414213562/1e9 ──► CF: 1 2 2 2 2 2 2 2 2 2 2 2 1 1 14 1 238 1 3
14142135624/1e10 ──► CF: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 5 4 1 8 4 2 1 4
141421356237/1e11 ──► CF: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 1 4 1 2 1 63 2 1 1 1 4 2
1414213562373/1e12 ──► CF: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 1 11 2 3 2 1 1 1 25 1 2 3
√2 ──► CF: 1 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 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 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 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 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 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 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 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 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 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 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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3
───────── an attempt at pi
pi ──► CF: 3 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 1 120 2 1 1 3 1 23 1 15 1 3 7 1 16 1 2 1 21 2 1 1 2 9 1 6 4
Ruby
<lang ruby># Generate a continued fraction from a rational number
def r2cf(n1,n2)
while n2 > 0 n1, (t1, n2) = n2, n1.divmod(n2) yield t1 end
end</lang>
Testing
Test 1: <lang ruby>[[1,2], [3,1], [23,8], [13,11], [22,7], [-151,77]].each do |n1,n2|
print "%10s : " % "#{n1} / #{n2}"
r2cf(n1,n2) {|n| print "#{n} "}
puts
end</lang>
- Output:
1 / 2 : 0 2
3 / 1 : 3
23 / 8 : 2 1 7
13 / 11 : 1 5 2
22 / 7 : 3 7
-151 / 77 : -2 25 1 2
Test 2: <lang ruby>(5..8).each do |digit|
n2 = 10 ** (digit-1)
n1 = (Math.sqrt(2) * n2).round
print "%-8s / %-8s : " % [n1, n2]
r2cf(n1,n2) {|n| print "#{n} "}
puts
end</lang>
- Output:
14142 / 10000 : 1 2 2 2 2 2 1 1 29 141421 / 100000 : 1 2 2 2 2 2 2 3 1 1 3 1 7 2 1414214 / 1000000 : 1 2 2 2 2 2 2 2 3 6 1 2 1 12 14142136 / 10000000 : 1 2 2 2 2 2 2 2 2 2 6 1 2 4 1 1 2
Test 3: <lang ruby>a =[ [31,10],
[314,100],
[3142,1000],
[31428,10000],
[314285,100000],
[3142857,1000000],
[31428571,10000000],
[314285714,100000000]
]
a.each do |n1,n2|
print "%-9s / %-9s : " % [n1, n2]
r2cf(n1,n2) {|n| print "#{n} "}
puts
end</lang>
- Output:
31 / 10 : 3 10 314 / 100 : 3 7 7 3142 / 1000 : 3 7 23 1 2 31428 / 10000 : 3 7 357 314285 / 100000 : 3 7 2857 3142857 / 1000000 : 3 7 142857 31428571 / 10000000 : 3 7 476190 3 314285714 / 100000000 : 3 7 7142857
Rust
<lang rust> struct R2cf {
n1: i64, n2: i64
}
// This iterator generates the continued fraction representation from the // specified rational number. impl Iterator for R2cf {
type Item = i64;
fn next(&mut self) -> Option<i64> {
if self.n2 == 0 {
None
}
else {
let t1 = self.n1 / self.n2;
let t2 = self.n2;
self.n2 = self.n1 - t1 * t2;
self.n1 = t2;
Some(t1)
}
}
}
fn r2cf(n1: i64, n2: i64) -> R2cf {
R2cf { n1: n1, n2: n2 }
}
macro_rules! printcf {
($x:expr, $y:expr) => (println!("{:?}", r2cf($x, $y).collect::<Vec<_>>()));
}
fn main() {
printcf!(1, 2); printcf!(3, 1); printcf!(23, 8); printcf!(13, 11); printcf!(22, 7); printcf!(-152, 77);
printcf!(14_142, 10_000); printcf!(141_421, 100_000); printcf!(1_414_214, 1_000_000); printcf!(14_142_136, 10_000_000);
printcf!(31, 10); printcf!(314, 100); printcf!(3142, 1000); printcf!(31_428, 10_000); printcf!(314_285, 100_000); printcf!(3_142_857, 1_000_000); printcf!(31_428_571, 10_000_000); printcf!(314_285_714, 100_000_000);
} </lang>
- Output:
[0, 2] [3] [2, 1, 7] [1, 5, 2] [3, 7] [-1, -1, -37, -2] [1, 2, 2, 2, 2, 2, 1, 1, 29] [1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 3, 1, 7, 2] [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12] [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2] [3, 10] [3, 7, 7] [3, 7, 23, 1, 2] [3, 7, 357] [3, 7, 2857] [3, 7, 142857] [3, 7, 476190, 3] [3, 7, 7142857]
Sidef
<lang ruby>func r2cf(num, den) {
func() {
den.is_zero && return()
var q = int(num/den)
(num, den) = (den, num - q*den)
return q
}
}
func showcf(f) {
print "["
var n = f()
print "#{n}" if defined(n)
print "; #{n}" while defined(n = f())
print "]\n"
}
[
[1/2, 3/1, 23/8, 13/11, 22/7, -151/77], [14142/10000, 141421/100000, 1414214/1000000, 14142136/10000000], [314285714/100000000],
].each { |seq|
seq.each { |r| showcf(r2cf(r.parts)) }
print "\n"
}</lang>
- Output:
[0; 2] [3] [2; 1; 7] [1; 5; 2] [3; 7] [-1; -1; -24; -1; -2] [1; 2; 2; 2; 2; 2; 1; 1; 29] [1; 2; 2; 2; 2; 2; 2; 3; 1; 1; 3; 1; 7; 2] [1; 2; 2; 2; 2; 2; 2; 2; 3; 6; 1; 2; 1; 12] [1; 2; 2; 2; 2; 2; 2; 2; 2; 2; 6; 1; 2; 4; 1; 1; 2] [3; 7; 7142857]
Tcl
Direct translation
<lang tcl>package require Tcl 8.6
proc r2cf {n1 {n2 1}} {
# Convert a decimal fraction (e.g., 1.23) into a form we can handle
if {$n1 != int($n1) && [regexp {\.(\d+)} $n1 -> suffix]} {
set pow [string length $suffix] set n1 [expr {int($n1 * 10**$pow)}] set n2 [expr {$n2 * 10**$pow}]
}
# Construct the continued fraction as a coroutine that yields the digits in sequence
coroutine cf\#[incr ::cfcounter] apply {{n1 n2} {
yield [info coroutine] while {$n2 > 0} { yield [expr {$n1 / $n2}] set n2 [expr {$n1 % [set n1 $n2]}] } return -code break
}} $n1 $n2
}</lang> Demonstrating: <lang tcl>proc printcf {name cf} {
puts -nonewline "$name -> "
while 1 {
puts -nonewline "[$cf],"
} puts "\b "
}
foreach {n1 n2} {
1 2 3 1 23 8 13 11 22 7 -151 77 14142 10000 141421 100000 1414214 1000000 14142136 10000000 31 10 314 100 3142 1000 31428 10000 314285 100000 3142857 1000000 31428571 10000000 314285714 100000000 3141592653589793 1000000000000000
} {
printcf "\[$n1;$n2\]" [r2cf $n1 $n2]
}</lang>
- Output:
[1;2] -> 0,2 [3;1] -> 3 [23;8] -> 2,1,7 [13;11] -> 1,5,2 [22;7] -> 3,7 [-151;77] -> -2,25,1,2 [14142;10000] -> 1,2,2,2,2,2,1,1,29 [141421;100000] -> 1,2,2,2,2,2,2,3,1,1,3,1,7,2 [1414214;1000000] -> 1,2,2,2,2,2,2,2,3,6,1,2,1,12 [14142136;10000000] -> 1,2,2,2,2,2,2,2,2,2,6,1,2,4,1,1,2 [31;10] -> 3,10 [314;100] -> 3,7,7 [3142;1000] -> 3,7,23,1,2 [31428;10000] -> 3,7,357 [314285;100000] -> 3,7,2857 [3142857;1000000] -> 3,7,142857 [31428571;10000000] -> 3,7,476190,3 [314285714;100000000] -> 3,7,7142857 [3141592653589793;1000000000000000] -> 3,7,15,1,292,1,1,1,2,1,3,1,14,4,2,3,1,12,5,1,5,20,1,11,1,1,1,2
Objectified version
<lang tcl>package require Tcl 8.6
- General generator class based on coroutines
oo::class create Generator {
constructor {} {
coroutine [namespace current]::coro my Apply
}
destructor {
catch {rename [namespace current]::coro {}}
}
method Apply {} {
yield
# Call the method (defined in subclasses) that actually produces values
my Produce my destroy return -code break
}
forward generate coro
method unknown args {
if {![llength $args]} { tailcall coro } next {*}$args
}
# Various ways to get the sequence from the generator
method collect {} {
set result {} while 1 { lappend result [my generate] } return $result
}
method take {n {suffix ""}} {
set result {} for {set i 0} {$i < $n} {incr i} { lappend result [my generate] } while {$suffix ne ""} { my generate lappend result $suffix break } return $result
}
}
oo::class create R2CF {
superclass Generator
variable a b
# The constructor converts other kinds of fraction (e.g., 1.23, 22/7) into a
# form we can handle.
constructor {n1 {n2 1}} {
next; # Delegate to superclass for coroutine management if {[regexp {(.*)/(.*)} $n1 -> a b]} { # Nothing more to do; assume we can ignore second argument here } elseif {$n1 != int($n1) && [regexp {\.(\d+)} $n1 -> suffix]} { set pow [string length $suffix] set a [expr {int($n1 * 10**$pow)}] set b [expr {$n2 * 10**$pow}] } else { set a $n1 set b $n2 }
}
# How to actually produce the values of the sequence
method Produce {} {
while {$b > 0} { yield [expr {$a / $b}] set b [expr {$a % [set a $b]}] }
}
}
proc printcf {name cf {take ""}} {
if {$take ne ""} {
set terms [$cf take $take \u2026]
} else {
set terms [$cf collect]
} puts [format "%-15s-> %s" $name [join $terms ,]]
}
foreach {n1 n2} {
1 2 3 1 23 8 13 11 22 7 -151 77 14142 10000 141421 100000 1414214 1000000 14142136 10000000 31 10 314 100 3142 1000 31428 10000 314285 100000 3142857 1000000 31428571 10000000 314285714 100000000 3141592653589793 1000000000000000
} {
printcf "\[$n1;$n2\]" [R2CF new $n1 $n2]
}
- Demonstrate parsing of input in forms other than a direct pair of decimals
printcf "1.5" [R2CF new 1.5] printcf "23/7" [R2CF new 23/7]</lang>
- Output:
[1;2] -> 0,2 [3;1] -> 3 [23;8] -> 2,1,7 [13;11] -> 1,5,2 [22;7] -> 3,7 [-151;77] -> -2,25,1,2 [14142;10000] -> 1,2,2,2,2,2,1,1,29 [141421;100000]-> 1,2,2,2,2,2,2,3,1,1,3,1,7,2 [1414214;1000000]-> 1,2,2,2,2,2,2,2,3,6,1,2,1,12 [14142136;10000000]-> 1,2,2,2,2,2,2,2,2,2,6,1,2,4,1,1,2 [31;10] -> 3,10 [314;100] -> 3,7,7 [3142;1000] -> 3,7,23,1,2 [31428;10000] -> 3,7,357 [314285;100000]-> 3,7,2857 [3142857;1000000]-> 3,7,142857 [31428571;10000000]-> 3,7,476190,3 [314285714;100000000]-> 3,7,7142857 [3141592653589793;1000000000000000]-> 3,7,15,1,292,1,1,1,2,1,3,1,14,4,2,3,1,12,5,1,5,20,1,11,1,1,1,2 1.5 -> 1,2 23/7 -> 3,3,2
XPL0
<lang XPL0>include c:\cxpl\codes; real Val;
proc R2CF(N1, N2, Lev); \Output continued fraction for N1/N2 int N1, N2, Lev; int Quot, Rem; [if Lev=0 then Val:= 0.0; Quot:= N1/N2; Rem:= rem(0); IntOut(0, Quot); if Rem then [ChOut(0, if Lev then ^, else ^;); R2CF(N2, Rem, Lev+1)]; Val:= Val + float(Quot); \generate value from continued fraction if Lev then Val:= 1.0/Val; ];
int I, Data; [Data:= [1,2, 3,1, 23,8, 13,11, 22,7, 0]; Format(0, 15); I:= 0; while Data(I) do
[IntOut(0, Data(I)); ChOut(0, ^/); IntOut(0, Data(I+1)); ChOut(0, 9\tab\); ChOut(0, ^[); R2CF(Data(I), Data(I+1), 0); ChOut(0, ^]); ChOut(0, 9\tab\); RlOut(0, Val); CrLf(0); I:= I+2];
]</lang>
- Output:
1/2 [0;2] 5.000000000000000E-001 3/1 [3] 3.000000000000000E+000 23/8 [2;1,7] 2.875000000000000E+000 13/11 [1;5,2] 1.181818181818180E+000 22/7 [3;7] 3.142857142857140E+000
zkl
Two iterators; one light weight, one heavy weight.
Light weight, explicit state: <lang zkl>fcn r2cf(nom,dnom){ // -->Walker (iterator)
Walker.tweak(fcn(state){
nom,dnom:=state;
if(dnom==0) return(Void.Stop);
n,d:=nom.divr(dnom);
state.clear(dnom,d);
n
}.fp(List(nom,dnom))) // partial application (light weight closure)
}</lang> Heavy weight, implicit state: <lang zkl>fcn r2cf2(nom,dnom){ // -->Generator (heavy weight Walker)
Utils.Generator(fcn(nom,dnom){
while(dnom){
n,d:=nom.divr(dnom); nom,dnom=dnom,d; vm.yield(n);
}
Void.Stop;
},nom,dnom)
}</lang> Both of the above return an iterator so they function the same: <lang zkl>foreach nom,dnom in (T(T(1,2), T(3,1), T(23,8), T(13,11), T(22,7), T(14142,10000), T(141421,100000), T(1414214,1000000), T(14142136,10000000))){
r2cf(nom,dnom).walk(25).println(); // print up to 25 numbers
}</lang>
- Output:
L(0,2) L(3) L(2,1,7) L(1,5,2) L(3,7) L(1,2,2,2,2,2,1,1,29) L(1,2,2,2,2,2,2,3,1,1,3,1,7,2) L(1,2,2,2,2,2,2,2,3,6,1,2,1,12) L(1,2,2,2,2,2,2,2,2,2,6,1,2,4,1,1,2)