def mystery(L, x): if L==[]: return False if L[0] == x: return True L.pop(0) return mystery(L,...

def mystery(L, x):

    if L==[]:

        return False

    if L[0] == x:

        return True

    L.pop(0)

    return mystery(L, x)

What is the input or length size of the function mystery?

What is the final output of mystery([1,3,5,7], 0)?

Explain in one sentence what mystery does?

What is the smallest input that mystery can have? Does the recursive call have smaller inputs? Why?

Assuming the recursive call in mystery is correct, use this assumption to explain in a few sentences why mystery is correct?

Expert Solution

What is the input or length size of the function mystery?
Input is a list L and an element x

What is the final output of mystery([1,3,5,7], 0)?
final output of mystery([1,3,5,7], 0) is False.
Because value 0 not found in the list [1,3,5,7]

Explain in one sentence what mystery does?
mystery checks for the value of x in the list L.
if x found in L then returns True, otherwise returns False

What is the smallest input that mystery can have?
smallest input that mystery can have is []

Does the recursive call have smaller inputs? Why?
Yes.
For example if mystery([1],2) is getting called then the first recursive call is mystery([], 2)
mystery([], 2) is the smaller input

Assuming the recursive call in mystery is correct, use this assumption to explain in a few sentences why mystery is correct?
For example
mystery([1,3,5,7], 0)
This compares the value of 1 with 0. and both are not equals and then it makes a recursive call mystery([3,5,7], 0) 
This compares the value of 3 with 0. and both are not equals and then it makes a recursive call mystery([5,7], 0)
This compares the value of 5 with 0. and both are not equals and then it makes a recursive call mystery([7], 0)
This compares the value of 7 with 0. and both are not equals and then it makes a recursive call mystery([], 0)
mystery([], 0) return False

venereology answered 2 hours ago

import Stack def stack_mystery(myStack): mystery=0 while not myStack.isEmpty(): value=myStack.pop() if v

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