Question

Please answer ALL parts completely-- answer with clear explanation will get a thumbs up!

Problem 2: Programming [2 points] Consider the following pseudo code where a function has been defined for a non negative integer x. Note that the function calls itself (within the while loop) making it a recursive function. def func (x) print x while i<x: i = i+1 func (x-1) (a) How many times does the program print if we call func (0) and func (1), respectively? (b) Suppose T(1) represents the number of times the function prints when called with an argument 1, T(2) represents the number of times the function prints when called with an argument 2, ..., T(n - 1) represents the number of times the function prints when called with an argument n -1, and T(n) represents the number of times the function prints when called with an argument n We can now make the following argument: func (n) will print once (it prints the number n) and will follow this up by calling func (n-1), n times. According to our notation, each such call prints T(n-1) times. Therefore T(n)=1+nT(n-1). Based on this information, does the program run in polynomial time? Provide an explanation for your answer.

Answer 1

(a) The code snippet is :

def func(x):

i = 0

print x

while i < x:

i = i + 1

func(x - 1)

If we call this function by 'func(0)', then the program will print only once as the value of 'x' in the function will be 0. So, x = 0 will be printed once and when the while loop condition will be checked, it will be false as 'i' is also 0, or 0 < 0 which is a false condition. Hence the function will print 'x' just once or 1 time.

If we call this function by 'func(1)', then the program will print twice as the value of 'x' in the function will be 1. So, x = 1 will be printed first and then the while loop condition will be checked, which is true as 'i' which is 0 is smaller than 'x', or 0 < 1 or the program will enter the while loop where the value of 'i' will be incremented by 1 to be i = 0 + 1 = 1. Now the function will be called again by passing x - 1 as the argument or 1 - 1 = 0. So, now i = 1 and value of 'x' on recursive call is 0, which will be printed. So again the while loop won't be executed as 'i' is not lesser than 'x', 1 < 0. Hence the function will print 'x' twice or 2 times.

(b) In this function, we can see that the recursive call is made inside a while loop. According to the notation of representing the number of times 'n' the function prints to be denoted as T(n), the relation obtained is T(n) = 1 + nT(n - 1). Assuming n > 1 for calculating the time complexity of the function. Also, T(0) = 1 since the function prints once when 0 is passed to it as argument. Now, we can say that T(0) = O(1) since the function has to just print once when x = 0. Then,

T(n) = 1 + n((1 + nT(n - 2)), since T(n - 1) = 1 + nT(n - 2)

Expnading this equation, we get a polynomial time running relation like

or, T(n) = (1 + n) + n²T(n - 2)

or, T(n) = (1 + n + n²) + n³T(n - 3)

or, T(n) = (1 + n + n² + n³) + n⁴T(n - 4) ... and so on. This implies

T(n) = n * (n - 1) * (n - 2) * (n - 3) ... T(0) + n for n number of prints

or, T(n) = n! + n which means the polynomial time run complexity of this algorithm is O(n!)

or, T(n) = O(2ⁿ) as n! is almost equal to 2ⁿ in terms of polynomial runtime complexities.