'Converting a floating-point decimal value to a fraction
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++?
Solution 1:[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);
}
Solution 2:[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
Solution 3:[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 necessarily333 / 1000, since the stored value is not exactly0.333but rather0.333000004291534423828125because 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.
- The code to convert a
floatto 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);
}
0.333000004 = 11173626 / 16777216 * 2^(-1)
This gives the rational representation of
float val = 0.333f;as5586813/16777216.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
Solution 4:[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;
} } }
Solution 5:[5]
#include<iostream>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
/*
May be over done but seems to work as far as I have tested.
Convert a decimal to a fraction or mixed number.
write a function that derives the number of decimal places by multipling
the fractional part by powers of 10 until no fractional part is left
maybe check using a \bool is_fraction()\ function that returns true or false
and a recursive function to multiply powers of 10 , to derive the numerator by
moving the decimal place:
the denominator is the power of ten that leaves no fractional parts.
use the solutions to generate a fraction then if needed reduce it(maybe a
function
that finds the greatest common divisor).
if it reduces to a numerator over 1 print just the whole number integer.
remember to catch divide by zero, and irrational numbers should be rounded and
displayed as approx values.
*/
struct Fraction {
double value{0.0}; // The initial value entered
}; Fraction F;
class fraction {
private:
double decimal{}; // The decimal part
int whole{}; // The whole number part
int numerator{};
int denominator{};
public:
fraction() {} // default constructor
// This does the conversion, calls all the help
void get_fraction(double& n) {
F.value = n;
set_whole_part(F.value);
set_fraction_part(F.value);
set_numerator();
}
// This provides the help
double set_fraction_part(const double& );
double set_whole_part(const double& );
double get_f_part();
double get_w_part();
double set_numerator();
int set_denominator(unsigned&);
int get_denominator(){ return denominator; }
double get_numerator() { return numerator; }
int GCD(int numer, int denom);
bool is_numerator(double&);
double number{}; // bool number 0.99999999
// output operator
friend std::ostream& operator<<(std::ostream& os, fraction& f);
// input operator
friend std::istream& operator>>(std::istream& is, fraction& f);
};
double fraction::set_whole_part(const double& )
{
double w_part = F.value; // whole number part
return whole = static_cast<int>(w_part); // return an integer value
}
double fraction::set_fraction_part(const double&)
{
double f_part = F.value; // fractional number part
// returns the fractional component of float or double
return decimal = modf(F.value, &f_part);
}
double fraction::get_f_part()
{
return decimal; // decimal part
}
double fraction::get_w_part()
{
return whole;
}
// istream operator
std::istream& operator>>(std::istream& is, fraction& f) {
double n = 0.0;
is >> n;
f.get_fraction(n);
return is;
}
// ostream operator has friend status so can access private members
std::ostream& operator<<(std::ostream& os, fraction& f)
{
os << static_cast<double>(f.numerator)/f.denominator << '\n';
if (f.numerator == 0 && f.denominator == 1) {
f.numerator = (f.whole);
}
if (f.whole >= 1 && f.numerator != 0 && f.denominator != 0) {
os << f.whole<<' ' << f.numerator << '/' << f.denominator << '\n';
return os;
}
if (f.whole == 0 && f.numerator != 0 && f.denominator != 0) {
os << f.numerator << '/' << f.denominator << '\n';
return os;
}
if (f.numerator == 0 && f.denominator == 0) {
os << f.whole << '\n';
return os;
}
os << "failed to create numerator/denominator\n";
return os;
}
// returns the greatest common divisor
int fraction::GCD(int numer, int denom) {
return denom > 1 ? GCD(denom, numer % denom) : numer;
}
bool fraction::is_numerator(double& n)
{
//using the rounding information to create the break point.
//This will lose information on the original value stored.
//is the least significant digit being rounded toward zero or one?
if (n <= 0.00000001) { number = 0.00000000; return false; } // dont't round
if (n <= 0.99999999) { number = 0.99999999; return true; } // round up
return false;
}
int fraction::set_denominator(unsigned& denom)
{
return denominator = denom;
}
double fraction::set_numerator()
{
double num = get_f_part();
double dec = 0.0;
unsigned count = 0;
double f_part = 0;
while (is_numerator(num) == true)
{
num *= 10; // move one decimal place and count
f_part = num; // make a new double.decimal
double dec = modf(num, &f_part); // get the new decimal part
num = dec; // re-initialize
++count; // keeps count of shifts of fractional part
}
unsigned exp = 1; // exponenent = power of 10
for (unsigned i = 0; i < count; ++i) {
exp *= 10; // get the total shifts of decimal point
}
//promote exp to avoid arithmatic over flow
double s_dec = decimal * static_cast<double>(exp);
// cast back to an int for GCD(returns an int)
int n_dec = static_cast<int>(s_dec);
if (number == 0.999999) {
n_dec += 1; // round up
}
if ((whole != 0) && (number == 0.99999999)) {
// if rounding toward one than add one.(the bool discarded that info).
n_dec += 1;
}
unsigned reduced = GCD(n_dec, exp); // find Greatest Common Divisor
if (reduced <= 0) { // no common divisor return as is
set_denominator(exp);
return numerator = n_dec; // may produce a zero numerator
std::cerr << "could not convert fraction\n";
}
else if (reduced >= 1) { // reduce the fraction
unsigned denom = exp / reduced;
denominator=denom;
unsigned numer = n_dec / reduced;
return numerator = numer;
}
else
std::cerr << "could not convert fraction\n";
return 0;
}
int main()
{
fraction f;
for (fraction ff; std::cin >> std::setprecision(12) >> std::fixed >> f;)
std::cout << f;
return 0;
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | drescherjm |
| Solution 3 | |
| Solution 4 | |
| Solution 5 |