Problem 3: Deep (50pts)

When nested mutable objects appear, the behavior of your program might be confusing. Consider the example below

>>> inner1 = [1, 2, 3]
>>> inner2 = [4, 5, 6]
>>> a = [inner1, inner2]
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b1 = a
>>> b2 = a[:]
>>> b3 = [a[0][:], a[1][:]]
>>> b1 == b2 and b2 == b3 and b1 == b3
True
>>> inner1[0] = 222
>>> b1
[[222, 2, 3], [4, 5, 6]]
>>> b2
[[222, 2, 3], [4, 5, 6]]
>>> b3
[[1, 2, 3], [4, 5, 6]]

1. We call b1 an alias of a, since they are just different names for the same list.
2. We call b2 a shallow-copy of a, which means b2 is not a, but the contents of b2 are the same objects as the contents of a.
3. We call b3 a deep-copy of a, as all the contents of b3 (i.e., lists of integers) are created via slicing and are not the same objects as the contents of a.

Now you need to implement a function deep to create a deep-copy of a given non-empty list, whose contents are non-empty lists of integers. In the real-world, objects may be arbitrarily nested (e.g., lists of lists of lists of ...), but in this problem, you only need to handle a simple situation, where the contents of the given list are only lists of integers.

Hint: try to use list comprehension to make your answer concise

def deep(l):
    """Returns a deep-copy of the given list of lists.
    >>> a = [[2, 2, 2], [5, 0, 2], [5, 0, 3]]
    >>> b = deep(a)
    >>> b
    [[2, 2, 2], [5, 0, 2], [5, 0, 3]]
    >>> a is b
    False
    >>> a[0] is b[0]
    False
    >>> a[1] is b[1]
    False
    >>> a[2] is b[2]
    False
    >>> len(a) == len(b)
    True
    """
    "*** YOUR CODE HERE ***"