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