02393 C++ Programming Exercises  Assignment 6 -Solved

A class for fractions of integers 
The goal of the exercise of this week is to 
implement a class of fractions of integers supporting some basic operations like 
addition and multiplication. You can use the following interface for the class as 
a starting point and modify it at will: 
class fraction { 
private: 
// Internal representation of a fraction as two integers 
int numerator; 
int denominator; 
public: 
// Class constructor 
fraction(int n, int d); 
// Methods to update the fraction 
void add(fraction f); 
void mult(fraction f); 
void div(fraction f); 
// Display method 
void display(void); 
}; 
The main idea above is that a fraction a
b is represented by two integers 
(the numerator a and the denominator b) and several methods are provided to 
support arithmetic operations on fractions. For example: 
fraction(int n, int m) constructs the fraction 
n
m 
; 
void add(fraction f) updates a fraction by adding fraction f to it. 
void mult(fraction f) updates a fraction by multiplying it by fraction f. 
void div(fraction f) updates a fraction by dividing it by fraction f. 
Your program should read sequences of simple expressions from cin of the 
form: a / b + c / d, a / b * c / d or a / b div c / d and provide the 
result as a simplifified fraction. For example, if the input is 
1 / 4 + 1 / 2 
your program should provide the following output to cout 
3 / 4 
which results from 1·2+1
·4 
4·2 
= 6
8 which after simplifification is 
3
4 
.Challenge 
The suggested internal representation poses some limits on the ac
tual domain of integral numbers that the fraction class can cover. For example, 
the largest number would be INT 1 MAX and the smallest positive fraction would be 
1 
INT MAX . A possible remedy to mitigate this would be to exploit the fact that every 
integer can be represented by a product of prime numbers (see e.g. https:// 
en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic). This was 
indeed what we exploited in assignment 2. For example 
137200 = 2 · 2 · 2 · 2 · 5 · 5 · 7 · 7 · 7 
To avoid repeating the same prime number many times, we could equivalently 
represent 137200 with exponentiation of prime numbers: 
137200 = 24 · 52 · 73 
We could thus represent all the exponents in an array of exponents, where 
the i-th element of the array represents how many times the i-th prime number 
occurs.1 So the representation of 137200 would be 
array index 0 1 2 3 4 5 
. . . 
array 
4 
0 
2 3 0 
0 
. . . 
corresponding prime number 2 3 5 7 11 13 
. . . 
Arithmetic operations can also exploit such representation. For instance, the 
multiplication of two numbers in such representation can be obtained by adding 
the corresponding exponents. For instance, multiplying the factorizations of 12 
and 10 would be done as follows 
primes 2 3 5 
12 
2 
1 0 
10 
1 0 
1 
Adding exponents gives: 
120 
3 
1 
1 
Division can be obtained in a similar manner, by substracting the exponents. 
Addition may be more involved. 
The challenge is to implement the class of fractions with this representation 
technique. 
1Formally, the factorization of a natural number n > 0 consists of the (uniquely determined) 
exponents k1, k2, . . . such that n = pk
1
1 
· pk22 · . . . where p1, p2, . . . are the prime numbers. 
2