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:

F r2cf(=n1, =n2)
   [Int] r
   L n2 != 0
      (n1, V t1_n2) = (n2, divmod(n1, n2))
      n2 = t1_n2[1]
      r [+]= t1_n2[0]
   R r

print(r2cf(1, 2))
print(r2cf(3, 1))
print(r2cf(23, 8))
print(r2cf(13, 11))
print(r2cf(22, 7))
print(r2cf(14142, 10000))
print(r2cf(141421, 100000))
print(r2cf(1414214, 1000000))
print(r2cf(14142136, 10000000))