def recPower(a, n): # raises a to the n-th power print "n =", n if n == 0: return 1 else: factor = recPower(a, n/2) if n%2 == 0: # n is even return factor*factor else: # n is odd return factor*factor*a if __name__ == "__main__": print recPower(2,25)

def count_nesting(L): nesting = 0 for item in L: if isinstance(item, list): nesting = max(nesting, 1 + count_nesting(item)) return nesting if __name__ == '__main__': x = range(10) print x, count_nesting(x) x = [1, [4, 5], 3, 6, 8, [3, [2]]] print x, count_nesting(x)

Write a recursive function *count_a(L)* which takes a list L and returns the number of times character 'a' occurs in L. The recursion should operate on the following principle:

- If the list is empty (it has length zero), return 0.
- If the list contains one or more element, consider the first element of L.
- If the first element is not a list, check whether the first element is equal to 'a' and return the value 0 or 1, depending on the outcome of the comparison
- If the first element is a list, make a recursive call using the first element as the argument. The call will return the number of 'a's occurring in this list.

- After handling the first element of the list, make a recursive call on the list without the first element. Add the value returned from this recursive call to the value generated in step 2.
- Return the value representing the solution for list L.

The Python function `isinstance(test, list)`

returns True when variable `test`

represents a list and False otherwise.

Here are some example lists that you can use to test your function:

L1 = [1, 2, 3, 4] L2 = [ ['a', 4, [4]], 'a', [2, [4], [3, ['a']]], 'ab'] L3 = [[[['a']]], ['a', 'a', 'a'], []] print L1 print count_a(L1) print L2 print count_a(L2) print L3 print count_a(L3)

def count_a(L): if not len(L): return 0 count_zero = 0 if isinstance(L[0], list): count_zero = count_a(L[0]) elif L[0] == 'a': count_zero = 1 return count_zero + count_a(L[1:]) if __name__ == '__main__': L1 = [1, 2, 3, 4] L2 = [['a', 4, [4]], 'a', [2, [4], [3, ['a']]], 'ab'] L3 = [[[['a']]], ['a', 'a', 'a'], []] for L in [L1, L2, L3]: print L print count_a(L)