Question

U.JUllal llclui SIVETULTUlas The recursive relationship for many functions follows a general pattern. T(n <= 1) = 1 T(n) = a*T(floor(n/b)) + (n^x) * ceil log2(n)"y) Since a computer cannot do fractional comparisons, floor and ceil are used in this computation Ask the user for 5 integers: • The constants a, b,xy • The input n Print out the numeric value of T(n). Here is an example of the expected output. Enter Value for a: Enter Value for b: Enter Value for x: Enter Value for y: Enter Value for n: 10 Evaluating lt (n/2) +nºi log2 (n)^0 T (10)=18

Answer 1

CODE:

import math

#inputs
a = int(input("Enter Value for a: "))
b = int(input("Enter Value for b: "))
x = int(input("Enter Value for x: "))
y = int(input("Enter Value for y: "))
n = int(input("Enter Value for n: "))

#recursive function
def T(n):
   if n <= 1:
       return 1
   else:
       return a * T(math.floor(n/b)) + (n**x) * math.ceil(math.log(n)**y)

#answer printing
print("Evaluating " + str(a)+"T(n/" + str(b) +")+n^" + str(x)+ " log2(n)^"+ str(y))
print("T(" + str(n) + ") = " + str(T(n)))

OUTPUT:

Windows PowerShell PS F:\Python\Practice> py .\HomeworkLib1.py Enter Value for a: 1 Enter Value for b: 2 Enter Value for x: 1 Enter Value for y: 0 Enter Valuefor n: 10 Evaluating 1T(n/2) +n^1 log2(n)^0 T(10) = 18 PS F:\Python\Practice>

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