One remarkably simple formula for calculating the value of p is the so-called Madhava-Leibniz series: p4 = 1-13+15-17+19-... . Consider the recursive function below to calculate the first n terms of this formula: double computePI (int n) { if (n <= 1) { return 1.0;} int oddnum = 2 * n

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Chapter1: Introduction
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One remarkably simple formula for calculating the value of p is the so-called
Madhava-Leibniz series: p4 = 1-13+15-17+19-.... Consider the recursive function
below to calculate the first n terms of this formula:
double computePI(int n)
{
if (n <= 1) { return 1.0;}
int oddnum = 2 * n - 1;
if ((n % 2) == 0
{
}
return -1.0 oddnum + computePI(n − 1);
}
else
{
}
return 1.0 / oddnum + computePI (n - 1);
Which statements about the run-time performance of this function are true?
1.Each time this function is called it will invoke at least two more recursive calls
II.The number of recursive calls this function will make is approximately equal to the
value of the parameter variable n
III.Not counting overhead, this function will be about as efficient as an iterative
implementation of the same formula
Transcribed Image Text:One remarkably simple formula for calculating the value of p is the so-called Madhava-Leibniz series: p4 = 1-13+15-17+19-.... Consider the recursive function below to calculate the first n terms of this formula: double computePI(int n) { if (n <= 1) { return 1.0;} int oddnum = 2 * n - 1; if ((n % 2) == 0 { } return -1.0 oddnum + computePI(n − 1); } else { } return 1.0 / oddnum + computePI (n - 1); Which statements about the run-time performance of this function are true? 1.Each time this function is called it will invoke at least two more recursive calls II.The number of recursive calls this function will make is approximately equal to the value of the parameter variable n III.Not counting overhead, this function will be about as efficient as an iterative implementation of the same formula
}
In ouunu = 2*11 - 1;
if ((n % 2) == 0
{
return -1.0 / oddnum + computePI(n-1);
}
else
return 1.0 / oddnum + computePI(n - 1);
Which statements about the run-time performance of this function are true?
1.Each time this function is called it will invoke at least two more recursive calls
II.The number of recursive calls this function will make is approximately equal to the
value of the parameter variable n
III.Not counting overhead, this function will be about as efficient as an iterative
implementation of the same formula
OI, II
OI, III
© II, III
O I, II, III
Transcribed Image Text:} In ouunu = 2*11 - 1; if ((n % 2) == 0 { return -1.0 / oddnum + computePI(n-1); } else return 1.0 / oddnum + computePI(n - 1); Which statements about the run-time performance of this function are true? 1.Each time this function is called it will invoke at least two more recursive calls II.The number of recursive calls this function will make is approximately equal to the value of the parameter variable n III.Not counting overhead, this function will be about as efficient as an iterative implementation of the same formula OI, II OI, III © II, III O I, II, III
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