How to resolve the algorithm Greedy algorithm for Egyptian fractions step by step in the FreeBASIC programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Greedy algorithm for Egyptian fractions step by step in the FreeBASIC programming language

Table of Contents

Problem Statement

An   Egyptian fraction   is the sum of distinct unit fractions such as: Each fraction in the expression has a numerator equal to   1   (unity)   and a denominator that is a positive integer,   and all the denominators are distinct   (i.e., no repetitions).
Fibonacci's   Greedy algorithm for Egyptian fractions   expands the fraction

x y

{\displaystyle {\tfrac {x}{y}}}

to be represented by repeatedly performing the replacement

(simplifying the 2nd term in this replacement as necessary, and where

⌈ x ⌉

{\displaystyle \lceil x\rceil }

is the   ceiling   function).

For this task,   Proper and improper fractions   must be able to be expressed.

Proper  fractions   are of the form

a b

{\displaystyle {\tfrac {a}{b}}}

where

a

{\displaystyle a}

and

b

{\displaystyle b}

are positive integers, such that

a < b

{\displaystyle a<b}

,     and improper fractions are of the form

a b

{\displaystyle {\tfrac {a}{b}}}

where

a

{\displaystyle a}

and

b

{\displaystyle b}

are positive integers, such that   a ≥ b.

(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.) For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Greedy algorithm for Egyptian fractions step by step in the FreeBASIC programming language

Source code in the freebasic programming language

' version 16-01-2017
' compile with: fbc -s console

#Define max 30

#Include Once "gmp.bi"

Dim Shared As Mpz_ptr num(max), den(max)

Function Egyptian_fraction(fraction As String, ByRef whole As Integer, range As Integer = 0) As Integer

    If InStr(fraction,"/") = 0 Then
        Print "Not a fraction, program will end"
        Sleep 5000, 1
        End
    End If

    Dim As Integer i, count

    Dim As Mpz_ptr tmp_num, tmp_den, x, y, q
    tmp_num = Allocate(Len(__Mpz_struct)) : Mpz_init(tmp_num)
    tmp_den = Allocate(Len(__Mpz_struct)) : Mpz_init(tmp_den)
    x = Allocate(Len(__Mpz_struct)) : Mpz_init(x)
    y = Allocate(Len(__Mpz_struct)) : Mpz_init(y)
    q = Allocate(Len(__Mpz_struct)) : Mpz_init(q)

    For i = 1 To max ' clear the list
        Mpz_set_ui(num(i), 0)
        Mpz_set_ui(den(i), 0)
    Next

    i = InStr(fraction,"/")
    Mpz_set_str(x, Left(fraction, i -1), 10)
    Mpz_set_str(y, Right(fraction, Len(fraction) - i), 10)

    ' if it's a improper fraction make it proper fraction
    If Mpz_cmp(x , y) > 0  Then
        Mpz_fdiv_q(q, x, y)
        whole = Mpz_get_ui(q)
        Mpz_fdiv_r(x, x, q)
    Else
        whole = 0
    End If

    Mpz_gcd(q, x, y) ' check if reduction is possible
    If Mpz_cmp_ui(q, 1) > 0 Then
        If range <> 0 Then ' return if we do a range test
            Return -1
        Else
            Mpz_fdiv_q(x, x, q)
            Mpz_fdiv_q(y, y, q)
        End If
    End If

    Mpz_set(num(count), x)
    Mpz_set(den(count), y)
    ' Fibonacci's Greedy algorithm for Egyptian fractions
    Do
        If Mpz_cmp_ui(num(count), 1) = 0 Then Exit Do
        Mpz_set(x, num(count))
        Mpz_set(y, den(count))
        Mpz_cdiv_q(q, y, x)
        Mpz_set_ui(num(count), 1)
        Mpz_set(den(count), q)
        Mpz_mul(tmp_den, y, q)
        Mpz_neg(y, y)
        Mpz_mod(tmp_num, y, x)
        count += 1
        Mpz_gcd(q, tmp_num, tmp_den) ' check if reduction is possible
        If Mpz_cmp_ui(q, 1) > 0 Then
            Mpz_fdiv_q(tmp_num, tmp_num, q)
            Mpz_fdiv_q(tmp_den, tmp_den, q)
        End If
        Mpz_set(num(count), tmp_num)
        Mpz_set(den(count), tmp_den)
    Loop

    Mpz_clear(tmp_num) : Mpz_clear(tmp_den)
    Mpz_clear(x) : Mpz_clear(y) :Mpz_clear(q)

    Return count

End Function

Sub prt_solution(fraction As String, whole As Integer, count As Integer)

    Print fraction; " = ";

    If whole <> 0 Then
        Print "["; Str(whole); "] + ";
    End If

    For i As Integer = 0 To count
        Gmp_printf("%Zd/%Zd ", num(i), den(i))
        If i <> count Then Print "+ ";
    Next
    Print

End Sub

' ------=< MAIN >=------

Dim As Integer n, d, number, improper, max_term,  max_size
Dim As String str_in, max_term_str, max_size_str, m_str
Dim As ZString Ptr gmp_str : gmp_str = Allocate(1000000)

For n = 0 To max
    num(n) = Allocate(Len(__Mpz_struct)) : Mpz_init(num(n))
    den(n) = Allocate(Len(__Mpz_struct)) : Mpz_init(den(n))
Next

Data "43/48", "5/121", "2014/59"
' 4/121 = 12/363 = 11/363 + 1/363 = 1/33 + 1/363
' 5/121 = 4/121 + 1/121 = 1/33 + 1/121 + 1/363
' 2014/59 = 34 + 8/59
' 8/59 = 1/8 + 5/472 = 1/8 + 4/472 + 1/472 = 1/8 + 1/118 + 1/472

For n = 1 To 3
    Read str_in
    number = Egyptian_fraction(str_in, improper)
    prt_solution(str_in, improper, number)
    Print
Next

Dim As Integer a = 1 , b = 99

Do
    For d = a To b
        For n = 1 To d -1
            str_in = Str(n) + "/" + Str(d)
            number = Egyptian_fraction(str_in, improper,1)
            If number = -1 Then Continue For ' skip
            If number > max_term Then
                max_term = number
                max_term_str = str_in
            ElseIf number = max_term Then
                max_term_str += ", " & str_in
            End If
            Mpz_get_str(gmp_str, 10, den(number))
            If Len(*gmp_str) > max_size Then
                max_size = Len(*gmp_str)
                max_size_str = str_in
                m_str = *gmp_str
            ElseIf max_size = Len(*gmp_str) Then
                max_size_str += ", " & str_in
            End If
        Next
    Next
    Print
    Print "for 1 to"; Len(Str(b)); " digits"
    Print "Largest number of terms is"; max_term +1; " for "; max_term_str
    Print "Largest size for denominator is"; max_size; " for "; max_size_str

    If b = 999 Then Exit Do
    a = b +1 : b = b * 10 +9
Loop

For n = 0 To max
    Mpz_clear(num(n))
    Mpz_clear(den(n))
Next

DeAllocate(gmp_str)

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End

  

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