将浮点十进制值转换为分数

Question

Given a decimal floating-point value, how can you find its fractional equivalent/approximation? For example:

as_fraction(0.1) -> 1/10
as_fraction(0.333333) -> 1/3
as_fraction(514.0/37.0) -> 514/37

Is there a general algorithm that can convert a decimal number to fractional form? How can this be implemented simply and efficiently in C++?

Answer 1

First get the fractional part and then take the gcd. Use the Euclidean algorithm http://en.wikipedia.org/wiki/Euclidean_algorithm

void foo(double input)
{
    double integral = std::floor(input);
    double frac = input - integral;

    const long precision = 1000000000; // This is the accuracy.

    long gcd_ = gcd(round(frac * precision), precision);

    long denominator = precision / gcd_;
    long numerator = round(frac * precision) / gcd_;

    std::cout << integral << " + ";
    std::cout << numerator << " / " << denominator << std::endl;
}

long gcd(long a, long b)
{
    if (a == 0)
        return b;
    else if (b == 0)
        return a;

    if (a < b)
        return gcd(a, b % a);
    else
        return gcd(b, a % b);
}

Answer 2

#include <iostream>
#include <valarray> 

using namespace std;

void as_fraction(double number, int cycles = 10, double precision = 5e-4){
    int sign  = number > 0 ? 1 : -1;
    number = number * sign; //abs(number);
    double new_number,whole_part;
    double decimal_part =  number - (int)number;
    int counter = 0;
    
    valarray<double> vec_1{double((int) number), 1}, vec_2{1,0}, temporary;
    
    while(decimal_part > precision & counter < cycles){
        new_number = 1 / decimal_part;
        whole_part = (int) new_number;
        
        temporary = vec_1;
        vec_1 = whole_part * vec_1 + vec_2;
        vec_2 = temporary;
        
        decimal_part = new_number - whole_part;
        counter += 1;
    }
    cout<<"x: "<< number <<"\tFraction: " << sign * vec_1[0]<<'/'<< vec_1[1]<<endl;
}

int main()
{
    as_fraction(3.142857);
    as_fraction(0.1);
    as_fraction(0.333333);
    as_fraction(514.0/37.0);
    as_fraction(1.17171717);
    as_fraction(-1.17);
}


x: 3.14286      Fraction: 22/7                                                                                                                
x: 0.1          Fraction: 1/10                                                                                                                        
x: 0.333333     Fraction: 1/3                                                                                                                 
x: 13.8919      Fraction: 514/37                                                                                                              
x: 1.17172      Fraction: 116/99                                                                                                              
x: 1.17         Fraction: -117/100

Sometimes you would want to approximate the decimal, without needing the equivalence. Eg pi=3.14159 is approximated as 22/7 or 355/113. We could use the cycles argument to obtain these:

as_fraction(3.14159, 1);
as_fraction(3.14159, 2);
as_fraction(3.14159, 3);

x: 3.14159      Fraction: 22/7                                                                                                                
x: 3.14159      Fraction: 333/106                                                                                                             
x: 3.14159      Fraction: 355/113

Answer 3

( Too long for a comment .)

Some comments claim that this is not possible. But I am of a contrary opinion .

I am of the opinion that it is possible in the right interpretation, but it is too easy to misstate the question or misunderstand the answer.

The question posed here is to find rational approximation(s) to a given floating point value.

This is certainly possible since floating point formats used in C++ can only store rational values, most often in the form of sign/mantissa/exponent. Taking IEEE-754 single precision format as an example (to keep the numbers simpler), 0.333 is stored as 1499698695241728 * 2^(-52) . That is equivalent to the fraction 1499698695241728 / 2^52 whose convergents provide increasingly accurate approximations , all the way up to the original value: 1/3 , 333/1000 , 77590/233003 , 5586813/16777216 .

Two points of note here.

• For a variable float x = 0.333; the best rational approximation is not necessarily 333 / 1000 , since the stored value is not exactly 0.333 but rather 0.333000004291534423828125 because of the limited precision of the internal representation of floating points.

• Once assigned, a floating point value has no memory of where it came from, or whether the source code had it defined as float x = 0.333; vs. float x = 0.333000004; because both of those values have the same internal representation. This is why the (related, but different) problem of separating a string representation of a number (for example, a user-entered value) into integer and fractional parts cannot be solved by first converting to floating point then running floating point calculations on the converted value.

---

[ EDIT ] Following is the step-by-step detail of the 0.333f example.

1. The code to convert a float to an exact fraction.

#include <cfloat>
#include <cmath>
#include <limits>
#include <iostream>
#include <iomanip>

void flo2frac(float val, unsigned long long* num, unsigned long long* den, int* pwr)
{
    float mul = std::powf(FLT_RADIX, FLT_MANT_DIG);
    *den = (unsigned long long)mul;
    *num = (unsigned long long)(std::frexp(val, pwr) * mul);
    pwr -= FLT_MANT_DIG;
}

void cout_flo2frac(float val)
{
    unsigned long long num, den; int pwr;
    flo2frac(val, &num, &den, &pwr);

    std::cout.precision(std::numeric_limits<float>::max_digits10);
    std::cout << val << " = " << num << " / " << den << " * " << FLT_RADIX << "^(" << pwr << ")" << std::endl;
}

int main()
{
    cout_flo2frac(0.333f);
}

2. Output .

0.333000004 = 11173626 / 16777216 * 2^(-1)

3. This gives the rational representation of float val = 0.333f; as 5586813/16777216 .

4. What remains to be done is determine the convergents of the exact fraction, which can be done using integer calculations, only. The end result is (courtesy WA):

0, 1/3, 333/1000, 77590/233003, 5586813/16777216

Answer 4

I came up with an algorithm for this problem, but I think it is too lengthy and can be accomplished with less lines of code. Sorry about the poor indentation it is hard trying to align everything on overflow.

#include <iostream>
using namespace std;


// converts the string half of the inputed decimal number into numerical values void converting
 (string decimalNumber, float&numerator, float& denominator )

 { float number; string valueAfterPoint =decimalNumber.substr(decimalNumber.find("."    ((decimalNumber.length() -1) )); // store the value after the decimal into a valueAfterPoint 

int length = valueAfterPoint.length(); //stores the length of the value after the decimal point into length 

numerator = atof(valueAfterPoint.c_str()); // converts the string type decimal number into a float value and stores it into the numerator

// loop increases the decimal value of the numerator by multiples of ten as long as the length is above zero of the decimal

for (; length > 0; length--)  
    numerator *= 10;

do
 denominator *=10;
  while  (denominator < numerator);



// simplifies the the converted values of the numerator and denominator into simpler values for          an easier to read output 


void simplifying (float& numerator, float& denominator) { int maximumNumber = 9; //Numbers in the tenths place can only range from zero to nine so the maximum number for a position in a position for the decimal number will be nine

bool isDivisble; // is used as a checker to verify whether the value of the numerator has the       found the dividing number that will a value of zero
 // Will check to see if the numerator divided denominator is will equal to zero


   if(int(numerator) % int(denominator) == 0) {
   numerator /= denominator;
   denominator = 1;   
   return; }


  //check to see if the maximum number is greater than the denominator to simplify to lowest     form while (maximumNumber < denominator) { maximumNumber *=10;  }


 // the maximum number loops from nine to zero. This conditions stops if the function isDivisible is true 
 for(; maximumNumber > 0;maximumNumber --){

 isDivisble = ((int(numerator) % maximumNumber == 0) && int(denominator)% maximumNumber == 0);

  if(isDivisble)
 {
    numerator /= maximumNumber;  // when is divisible true numerator be devided by the max        number value for example 25/5 = numerator = 5

   denominator /= maximumNumber; //// when is divisible true denominator be devided by themax        number value for example 100/5 = denominator = 20

 }


 // stop value if numerator and denominator is lower than 17 than it is at the lowest value
 int stop = numerator + denominator;

 if (stop < 17)
 {
     return;
 } } }   

Answer 5

I agree completely with dxiv 's solution but I needed a more general function (I threw in the signed stuff for fun because my use cases only included positive values):

#include <concepts>

/**
 * \brief Multiply two numbers together checking for overflow.
 * \tparam T The unsigned integral type to check for multiplicative overflow.
 * \param a The multiplier.
 * \param b The multicland.
 * \return The result and a value indicating whether the multiplication 
 *         overflowed.
 */
template<std::unsigned_integral T>
auto mul_overflow(T a, T b) -> std::tuple<T, bool>
{
    size_t constexpr shift{ std::numeric_limits<T>::digits / 2 };
    T constexpr mask{ (T{ 1 } << shift) - T{ 1 } };
    T const a_high = a >> shift;
    T const a_low = a & mask;
    T const b_high = b >> shift;
    T const b_low = b & mask;

    T const low_low{ a_low * b_low };
    if (!(a_high || b_high))
    {
        return { low_low, false };
    }

    bool overflowed = a_high && b_high;
    T const low_high{ a_low * b_high };
    T const high_low{ a_high * b_low };

    T const ret{ low_low + ((low_high + high_low) << shift) };
    return
    {
        ret,
        overflowed
        || ret < low_low
        || (low_high >> shift) != 0
        || (high_low >> shift) != 0
    };
}

/**
 * \brief Converts a floating point value to a numerator and
 * denominator pair.
 *
 * If the floating point value is larger than the maximum that the Tout
 * type can hold, the results are silly.
 *
 * \tparam Tout The integral output type.
 * \tparam Tin The floating point input type.
 * \param f The value to convert to a numerator and denominator.
 * \return The numerator and denominator.
 */
template <std::integral Tout, std::floating_point Tin>
auto to_fraction(Tin f) -> std::tuple<Tout, Tout>
{
    const Tin multiplier
    {
        std::pow(std::numeric_limits<Tin>::radix, 
                 std::numeric_limits<Tin>::digits)
    };
    uint64_t denominator{ static_cast<uint64_t>(multiplier) };
    int power;
    Tout num_fix{ 1 };
    if constexpr (std::is_signed_v<Tout>)
    {
        num_fix = f < static_cast<Tin>(0) ? -1 : 1;
        f = std::abs(f);
    }

    uint64_t numerator
    {
        static_cast<uint64_t>(std::frexp(f, &power) * multiplier)
    };
    uint64_t const factor
    {
        static_cast<uint64_t>(std::pow(
            std::numeric_limits<Tin>::radix, std::abs(power)))
    };
    if (power > 0)
    {
        while(true)
        {
            auto const [res, overflow]{ mul_overflow(numerator, factor) };
            if (!overflow)
            {
                numerator = res;
                break;                    
            }
            numerator >>= 1;
            denominator >>= 1;
        }
    }
    else
    {
        while (true)
        {
            auto const [res, overflow]{ mul_overflow(denominator, factor) };
            if (!overflow)
            {
                denominator = res;
                break;
            }
            numerator >>= 1;
            denominator >>= 1;
        }
    }

    // get the results into the requested sized integrals.
    while ((numerator > std::numeric_limits<Tout>::max()
            || denominator > std::numeric_limits<Tout>::max())
           && denominator > 1)
    {
        numerator >>= 1;
        denominator >>= 1;
    }

    return 
    {
        num_fix * static_cast<Tout>(numerator),
        static_cast<Tout>(denominator)
    };
}

You can call this like:

auto [n, d] { to_fraction<int8_t>(-124.777f) };

And you get n=-124 , d=1 ;

auto [n, d] { to_fraction<uint64_t>(.33333333333333) };

gives n=6004799503160601 , d=18014398509481984

Answer 6

#include<iostream>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<string>
#include<vector>
#include <exception>
#include <sstream>

// note using std = c++11
// header section

#ifndef rational_H
#define rational_H

struct invalid : std::exception {
    const char* what() const noexcept { return "not a number\n"; }};

struct Fraction {
public:
    long long value{0};
    long long numerator{0};
    long long denominator{0};
}; Fraction F;

class fraction : public Fraction{

public:
    fraction() {}
    void ctf(double &);
    void get_fraction(std::string& w, std::string& d, long double& n) {
        F.value = (long long )n;
        set_whole_part(w);
        set_fraction_part(d);
        make_fraction();
    }
    long long set_whole_part(std::string& w) {
        return whole = std::stoll(w);
    }
    long long set_fraction_part(std::string& d) {
         return decimal = std::stoll(d);
    }
    void make_fraction();
    bool cmpf(long long&, long long&, const long double& epsilon);
    int Euclids_method(long long&, long long&);

    long long get_f_part() { return decimal; };
    void convert(std::vector<long long>&);
    bool  is_negative{ false };

    friend std::ostream& operator<<(std::ostream& os, fraction& ff);
    struct get_sub_length;

private:
    long long whole{ 0 };
    long long decimal{ 0 };
};
#endif // rational_H

// definitions/source

struct get_sub_length {
    size_t sub_len{};
    size_t set_decimal_length(size_t& n) {
        sub_len = n;
        return sub_len;
    }
    size_t get_decimal_length() { return sub_len; }
}; get_sub_length slen;

struct coefficient {
    std::vector<long long>coef;
}; coefficient C;

//compare's the value returned by convert with the original 
// decimal value entered.
//if its within the tolarence of the epsilon consider it the best
//approximation you can get.
//feel free to experiment with the epsilon.
//for better results.
 
bool fraction::cmpf(long long& n1, long long& d1, const long double& epsilon = 0.0000005) 
{
long double ex = pow(10, slen.get_decimal_length());
long long  d = get_f_part();       // the original fractional part to use for comparison.
long double  a = (long double)d / ex;
long double b = ((long double)d1 / (long double)n1);
if ((fabs(a - b) <= epsilon)) { return true; }
return false;
}

//Euclids algorithm returns the cofficients of a continued fraction through recursive division,
//for example: 0.375 -> 1/(375/1000) (note: for the fractional portion only).
// 1000/375 -> Remainder of 2.6666.... and  1000 -(2*375)=250,using only the integer value
// 375/250 -> Remainder of 1.5  and   375-(1*250)=125,
// 250/125 -> Remainder of 2.0  and   250-(2*125)=2
//the coefficients of the continued fraction are the integer values 2,1,2
// These are generally written [0;2,1,2] or [0;2,1,1,1] were 0 is the whole number value.

int fraction::Euclids_method(long long& n_dec, long long& exp) 
{

    long long quotient = 0;

    if ((exp >= 1) && (n_dec != 0)) {
        quotient = exp / n_dec;

        C.coef.push_back(quotient);

        long long divisor = n_dec;
        long long dividend = exp - (quotient * n_dec);

        Euclids_method(dividend, divisor); // recursive division 
    }
    return 0;
}

 // Convert is adding the elements stored in coef as a simple continued fraction
// which should result in a good approximation of the original decimal number.

void fraction::convert(std::vector<long long>& coef) 
{
    std::vector<long long>::iterator pos;
    pos = C.coef.begin(), C.coef.end();
    long long n1 = 0;
    long long n2 = 1;
    long long d1 = 1;
    long long d2 = 0;

    for_each(C.coef.begin(), C.coef.end(), [&](size_t pos) {

        if (cmpf(n1, d1) == false) {

            F.numerator = (n1 * pos) + n2;
            n2 = n1;
            n1 = F.numerator;

            F.denominator = (d1 * pos) + d2;
            d2 = d1;
            d1 = F.denominator;
        }
    });

    //flip the fraction back over to format the correct output.
    F.numerator = d1;
    F.denominator = n1;
}

// creates a fraction from the decimal component
// insures its in its abs form to ease calculations.

void fraction::make_fraction() {

    size_t count = slen.get_decimal_length();
    long long n_dec = decimal;
    long long exp = (long long)pow(10, count);

    Euclids_method(n_dec, exp);
    convert(C.coef);
}

std::string get_w(const std::string& s) 
{
    std::string st = "0";
    std::string::size_type pos;
    pos = s.find(".");
        if (pos - 1 == std::string::npos) {
            st = "0";
            return st;
        }
        else { st = s.substr(0, pos);
        return st;
        }

    if (!(s.find("."))){
            st = "0";
        return st;
    }
    return st;
 }

std::string get_d(const std::string& s)
{ 
    std::string st = "0";
    std::string::size_type pos;
        pos = s.find(".");
        if (pos == std::string::npos) {
            st = "0";
            return st;
        }
        std::string sub = s.substr(pos + 1);
            st = sub;
                size_t sub_len = sub.length(); 
                    slen.set_decimal_length(sub_len);
        return st;
}

void fraction::ctf(double& nn)
{
        //using stringstream for conversion to string
        std::istringstream is;
        is >> nn;
        std::ostringstream os;
        os << std::fixed << std::setprecision(14) << nn;

        std::string s = os.str();

        is_negative = false; //reset for loops
        C.coef.erase(C.coef.begin(), C.coef.end()); //reset for loops

        long double n = 0.0;
        int m = 0;

        //The whole number part will be seperated from the decimal part leaving a pure fraction.
        //In such cases using Euclids agorithm would take the reciprocal 1/(n/exp) or exp/n.
        //for pure continued fractions the cf must start with 0 + 1/(n+1/(n+...etc
        //So the vector is initilized with zero as its first element.

        C.coef.push_back(m);
        std::cout << '\n';
    
        if (s == "q") { // for loop structures
            exit(0);
        }

        if (s.front() == '-') { // flag negative values. 
            is_negative = true; // represent nagative in output
            s.erase(remove(s.begin(), s.end(), '-'), s.end()); // using abs
        }

        // w, d, seperate the string components
        std::string w = get_w(s); 
        std::string d = get_d(s);

        try
        {
            if (!(n = std::stold(s))) {throw invalid(); } // string_to_double()
            get_fraction(w, d, n);
        } 
        catch (std::exception& e) {
            std::cout << e.what();
            std::cout <<'\n'<< std::endl;
        }
}

// The ostream formats and displays the various outputs

std::ostream& operator<<(std::ostream& os, fraction& f) 
{
    std::cout << '\n';
    if (f.is_negative == true) {
        os << "The coefficients are [" << '-' << f.whole << ";";
            for (size_t i = 1; i < C.coef.size(); ++i) {
                os << C.coef[i] << ',';
            }
            std::cout << "]" << '\n';
        os << "The cf is: " << '-' << f.whole;
            for (size_t i = 1; i < C.coef.size(); ++i) {
                os << "+1/(" << C.coef[i];
            }
            for (size_t i = 1; i < C.coef.size(); ++i) {
                os << ')';
            }
            std::cout << '\n';

        if (F.value >= 1 && F.numerator == 0 && F.denominator == 1) {
            F.numerator = abs(f.whole);
                os << '-' << F.numerator << '/' << F.denominator << '\n';
                return os;
        }
        else if (F.value == 0 && F.numerator == 0 && F.denominator == 1) {
                os << F.numerator << '/' << F.denominator << '\n';
                return os;
        }
        else if (F.value == 0 && F.numerator != 0 && F.denominator != 0) {
                os << '-' << abs(F.numerator) << '/' << abs(F.denominator) << '\n';
                return os;
        }
        else if (F.numerator == 0 && F.denominator == 0) {
                os << '-' << f.whole << '\n';
                return os;
        }
        else
                os << '-' << (abs(f.whole) * abs(F.denominator) + abs(F.numerator)) << '/' << abs(F.denominator) << '\n';
    }

    if (f.is_negative == false) {

        os << "The coefficients are [" << f.whole << ";";
             for (size_t i = 1; i < C.coef.size(); ++i) {
                os << C.coef[i] << ',';
            }
            std::cout << "]" << '\n';
        os << "The cf is: " << f.whole;
            for (size_t i = 1; i < C.coef.size(); ++i) {
                os << "+1/(" << C.coef[i];
            }
            for (size_t i = 1; i < C.coef.size(); ++i) {
                os << ')';
            }
            std::cout << '\n';

        if (F.value >= 1 && F.numerator == 0 && F.denominator == 1) {
            F.numerator = abs(f.whole);
                os << F.numerator << '/' << F.denominator << '\n';
                return os;
        }
        else if (F.value == 0 && F.numerator != 0 && F.denominator != 0) {
                os << abs(F.numerator) << '/' << abs(F.denominator) << '\n';
                return os;
        }
        else if (F.numerator == 0 && F.denominator == 0) {
            os << f.whole << '\n';
            return os;
        }
        else
            os << (abs(f.whole) * abs(F.denominator) + abs(F.numerator)) << '/' << abs(F.denominator) << '\n';

            os << f.whole << ' ' << F.numerator << '/' << F.denominator << '\n';
    }
    return os;
}

int main() 
{
    fraction f;
    double s = 0;
    std::cout << "Enter a number to convert to a fraction\n";
    std::cout << "Enter a \"q\" to quit\n";
    // uncomment for a loop

    while (std::cin >> s) {
        f.ctf(s);
        std::cout << f << std::endl;
    }

    // comment out these lines if you want the loop

    //std::cin >> s; 
    //f.ctf(s);
    //std::cout << f << std::endl;
 }