How to resolve the algorithm Continued fraction/Arithmetic/Construct from rational number step by step in the Nim programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Continued fraction/Arithmetic/Construct from rational number step by step in the Nim programming language

Table of Contents

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:

Step by Step solution about How to resolve the algorithm Continued fraction/Arithmetic/Construct from rational number step by step in the Nim programming language

Source code in the nim programming language

iterator r2cf*(n1, n2: int): int =
  var (n1, n2) = (n1, n2)
  while n2 != 0:
    yield n1 div n2
    n1 = n1 mod n2
    swap n1, n2

#———————————————————————————————————————————————————————————————————————————————————————————————————

when isMainModule:

  from sequtils import toSeq

  for pair in [(1, 2), (3, 1), (23, 8), (13, 11), (22, 7), (-151, 77)]:
    echo pair, " -> ", toSeq(r2cf(pair[0], pair[1]))

  echo ""
  for pair in [(14142, 10000), (141421, 100000), (1414214, 1000000), (14142136, 10000000)]:
    echo pair, " -> ", toSeq(r2cf(pair[0], pair[1]))

  echo ""
  for pair in [(31,10), (314,100), (3142,1000), (31428,10000), (314285,100000),
              (3142857,1000000), (31428571,10000000), (314285714,100000000)]:
    echo pair, " -> ", toSeq(r2cf(pair[0], pair[1]))


  

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