How to resolve the algorithm Continued fraction/Arithmetic/Construct from rational number step by step in the Sidef programming language
How to resolve the algorithm Continued fraction/Arithmetic/Construct from rational number step by step in the Sidef 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 Sidef programming language
Source code in the sidef programming language
func r2cf(num, den) {
func() {
den || return nil
var q = 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.nude)) }
print "\n"
}
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