How to resolve the algorithm Calkin-Wilf sequence step by step in the FreeBASIC programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Calkin-Wilf sequence step by step in the FreeBASIC programming language

Table of Contents

Problem Statement

The Calkin-Wilf sequence contains every nonnegative rational number exactly once. It can be calculated recursively as follows:

To avoid floating point error, you may want to use a rational number data type.

It is also possible, given a non-negative rational number, to determine where it appears in the sequence without calculating the sequence. The procedure is to get the continued fraction representation of the rational and use it as the run-length encoding of the binary representation of the term number, beginning from the end of the continued fraction. It only works if the number of terms in the continued fraction is odd- use either of the two equivalent representations to achieve this:

The fraction   9/4   has odd continued fraction representation     2; 3, 1,     giving a binary representation of   100011, which means   9/4   appears as the   35th   term of the sequence.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Calkin-Wilf sequence step by step in the FreeBASIC programming language

Source code in the freebasic programming language

#include "gcd.bas"

type rational
    num as integer
    den as integer
end type

dim shared as rational ONE, TWO
ONE.num = 1 : ONE.den = 1
TWO.num = 2 : TWO.den = 1

function simplify( byval a as rational ) as rational
   dim as uinteger g = gcd( a.num, a.den )
   a.num /= g : a.den /= g
   if a.den < 0 then
       a.den = -a.den
       a.num = -a.num
   end if
   return a
end function

operator + ( a as rational, b as rational ) as rational
    dim as rational ret
    ret.num = a.num * b.den + b.num*a.den
    ret.den = a.den * b.den
    return simplify(ret)
end operator

operator - ( a as rational, b as rational ) as rational
    dim as rational ret
    ret.num = a.num * b.den - b.num*a.den
    ret.den = a.den * b.den
    return simplify(ret)
end operator

operator * ( a as rational, b as rational ) as rational
    dim as rational ret
    ret.num = a.num * b.num
    ret.den = a.den * b.den
    return simplify(ret)
end operator

operator / ( a as rational, b as rational ) as rational
    dim as rational ret
    ret.num = a.num * b.den
    ret.den = a.den * b.num
    return simplify(ret)
end operator

function floor( a as rational ) as rational
    dim as rational ret
    ret.den = 1
    ret.num = a.num \ a.den
    return ret
end function

function cw_nextterm( q as rational ) as rational
    dim as rational ret = (TWO*floor(q))
    ret = ret + ONE : ret = ret - q 
    return ONE / ret
end function

function frac_to_int( byval a as rational ) as uinteger
    redim as uinteger cfrac(-1)
    dim as integer  lt = -1, ones = 1, ret = 0
    do
        lt += 1
        redim preserve as uinteger cfrac(0 to lt)
        cfrac(lt) = floor(a).num
        a = a - floor(a) : a = ONE / a
    loop until a.num = 0 or a.den = 0
    if lt mod 2 = 1 and cfrac(lt) = 1 then
        lt -= 1
        cfrac(lt)+=1
        redim preserve as uinteger cfrac(0 to lt)
    end if
    if lt mod 2 = 1 and cfrac(lt) > 1 then
        cfrac(lt) -= 1
        lt += 1
        redim preserve as uinteger cfrac(0 to lt)
        cfrac(lt) = 1
    end if
    for i as integer = lt to 0 step -1
        for j as integer = 1 to cfrac(i)
            ret *= 2
            if ones = 1 then  ret += 1
        next j
        ones = 1 - ones
    next i
    return ret
end function

function disp_rational( a as rational ) as string
    if a.den = 1 or a.num= 0 then return str(a.num)
    return str(a.num)+"/"+str(a.den)
end function

dim as rational q
q.num = 1
q.den = 1
for i as integer = 1 to 20
    print i, disp_rational(q)
    q = cw_nextterm(q)
next i

q.num = 83116
q.den = 51639
print disp_rational(q)+" is the "+str(frac_to_int(q))+"th term."

  

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