What are the outputs of the code snippets in Part 1 and 2?
Part 1 Variable Scope
Code Output
x = 0 def foo_printx(): print(x) foo_printx() print(x)
x = 0 y = 999 def foo_printx(y):
print(y) foo_printx(x) print(x)
x = 0 def foo_printx(): x = 999 print(x)
foo_printx() print(x)
Part 2 Nested Functions
Code Output
x = 1 y = 2 def foo(y): def bar(x):
return x+y return bar(y) print(foo(x))
x = 1 y = 2 def foo(x): def bar(x):
return x+y return bar(y) print(foo(x))
Part 3 Recursion
Previously, we have shown that we can create a customized burger. Here are the ingredient prices for your convenience:
Ingredient Price
‘B’ stands for a piece of bun $0.5
‘C’ stands for cheese $0.8
‘P’ stands for patty $1.5
‘V’ stands for veggies $0.7
‘O’ stands for onions $0.4
‘M’ stands for mushroom $0.9
Your task was to write a function burgerPrice(burger) that takes in a string representing a burger and returns the price of the burger. Your task now is to write the same function burgerPrice(burger) using recursion.
Part 4 Recursion vs Iteration
B. Factorial: Given a positive number 𝑛𝑛, the value of factorial of 𝑛𝑛 (written as 𝑛𝑛!) is defined as 𝑛𝑛! = 𝑛𝑛 × (𝑛𝑛− 1)!. Additionally, the value of 0! is 1. Write one recursive and one iterative version of function fact(n) which computes the value of n!.
Part 5 Recursion vs Iteration (cont.)
A. Final Sum: Given a positive number 𝑛𝑛, the final sum is obtained by repeatedly computing the sum of all the digits of 𝑛𝑛, until the final sum is a single digit. For example, sum(52634) = 20, which is not a single digit. We then continue with sum(20) = 2 which is now a single digit. Therefore, final_sum(52634) = 2. Write one recursive and one iterative version of the function final_sum(n) which computes the final sum of n.
B. Euler Constant: The value of 𝑒𝑒𝑥𝑥 can be approximated using the formula 𝑒𝑒𝑥𝑥 𝑥𝑥0 𝑥𝑥1 𝑥𝑥2 𝑥𝑥3 . Write one recursive and one iterative version of function find_e(x, n) to find the approximation of 𝑒𝑒𝑥𝑥 up to 𝑛𝑛 + 1
𝑥𝑥𝑛𝑛 steps, i.e. the last term in the summation is .
𝑛𝑛!
