Problem Statement

The purpose of this task is to write a function

r 2 c f

(

i n t

{\displaystyle {\mathit {r2cf}}(\mathrm {int} }

N

1

,

i n t

{\displaystyle N_{1},\mathrm {int} }

N

2

)

{\displaystyle N_{2})}

, or

r 2 c f

(

F r a c t i o n

{\displaystyle {\mathit {r2cf}}(\mathrm {Fraction} }

N )

{\displaystyle N)}

, which will output a continued fraction assuming: 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

N

1

{\displaystyle N_{1}}

divided by

N

2

{\displaystyle N_{2}}

. It then sets

N

1

{\displaystyle N_{1}}

to

N

2

{\displaystyle N_{2}}

and

N

2

{\displaystyle N_{2}}

to the determined remainder part. It then outputs the determined integer part. It does this until

a b s

(

N

2

)

{\displaystyle \mathrm {abs} (N_{2})}

is zero. Demonstrate the function by outputing the continued fraction for:

2

{\displaystyle {\sqrt {2}}}

should approach

[ 1 ; 2 , 2 , 2 , 2 , … ]

{\displaystyle [1;2,2,2,2,\ldots ]}

try ever closer rational approximations until boredom gets the better of you: Try : Observe how this rational number behaves differently to

2

{\displaystyle {\sqrt {2}}}

and convince yourself that, in the same way as

3.7

{\displaystyle 3.7}

may be represented as

3.70

{\displaystyle 3.70}

when an extra decimal place is required,

[ 3 ; 7 ]

{\displaystyle [3;7]}

may be represented as

[ 3 ; 7 , ∞ ]

{\displaystyle [3;7,\infty ]}

when an extra term is required.

Let's start with the solution:

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)
}

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)
}

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
}