#include using namespace std; /* INSERT YOUR RAT CLASS UP HERE */ double continu

Question

#include

using namespace std;

/*
   INSERT YOUR RAT CLASS UP HERE
*/

double continued_frac_1(int a[], int i) {

}

int* continued_frac_2(int a[], int i) {

}

Rat continued_frac_3(int a[], int i) {

}

int main() {
   int array[] = {3, 7, 16, -1};
   cout << "Continued fractions function 1: " <<
       continued_frac_1(array, 0) << endl;
  
   int *a = continued_frac_2(array, 0);
   cout << "Continued fractions function 2: " <<
       a[0] << '/' << a[1] << endl;
   delete []a;
  
   cout << "Continued fractions function 3: " <<
       continued_frac_3(array, 0) << endl;;
  
   return 0;
}

Expected Output:

Continued fractions function 1: 3.14159
Continued fractions function 2: 355/113
Continued fractions function 3: 3 16/113

-----------------------------------------------------------------------------------------------------------

1. Idea for solving problem 1:

double f(a[], i) {
if a[i+1] = -1: #base case
    _____________
return a[i] + 1 / ___________ //recursive call
}

2. Pseudo-code:

int* f(a[], i) {
if (a[i+1] == -1)
    return array with the right numerator and denominator
r = f(a, ______) //recursive call
new_frac = add a[i] and fraction (1 / r) //see trick
reduce and return new_frac
}

Trick: a + (1 / (b/c)) = a + (c / b) = (a*b + c) / b

3. You should use your Rat class to solve this one.

Rat f(a[], i) {
if (a[i+1] == -1)
    return Rat object with the right numerator and denominator
Rat r = f(a, ______) //recursive call
return Rat object that with sum of a[i] + (1 / r) //see hint
}

Your Rat object should be "smart" enough to reduce itself to lowest terms

Hint: r is some fraction, say b/c. (1 / r) = (1 / (b/c)) = c / b

NOTE: Your actual code will not have a one-to-one correspondence
with the pseudo-code above.

Continued Fraction The assignment involves computing continued fractions, and contains three parts. The general form of a continued fraction is: A find continued fraction is an expression of the from where alpha 0 is an integer, all other alpha i are positive integers, and n is a non-negative integer. We can specify a continued fraction by an array of integers containing the values a0 ... a0 Write a function that takes an array of integers as specified above (and ending with a -1) and returns the value of the fraction as a double. Write a function that takes an array of integers as specified above (and ending with a -1). Your function will represent the value of the continued fraction as a "regular" fraction, p / q. in lowest terms. We are looking for the result of doing the fractional arithmetic and keeping everything in integers as we work our way through. The function returns a 2 element integer array, v. with v[0] = p and v[l] = q. Write recursive functions that, given the above representation of a continued fraction, returns p and q such that p / q represent the value of the continued fraction as a "regular" fraction in lowest terms, as in question 2 above.

Explanation / Answer

double continued_frac_1(int a[], int i) {
if a[i+1] == -1: #base case
return a[i] + 1 /continued_frac_1(a,i+1); //recursive call
}
int* continued_frac_2(int a[], int i) {
int final[2];
if (a[i+1] == -1){
return final;
}
final[0]=a[i];
r = continued_frac_2(a, i+1) //recursive call
new_frac = a[i]+(1 / r);
final[1]=new_frac;
return final;
}
Rat continued_frac_3(int a[], int i) {
if (a[i+1] == -1)
return Rat object with the right numerator and denominator
Rat r = f(a, continued_frac_3) //recursive call
return a[i]+1/r;
}

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