1.
- False. Although this is a common base case, other bases cases may make sense, like a list with one item in it, for example rec_max.
- True. When the mystery function is called 10 gets associated with the parameter x, but the original x variable also exists and has it's own value.  Updating the parameter to 2 in the function does not change the external x variable.
- False. It depends on which version of DFS as well as how the next states are returned.

2.
a b c output
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1 (>= threshold is 1)
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1

3.
def count_even(list):
    if list == []:
        return 0
    else:
        count = count_even(list[1:])
        
        if list[0]%2 == 0:
            return 1 + count
        else:
            return count

4.
a. No problem.
b. Aliasing problem.
c. No problem.
d. Aliasing problem.
e. No problem.

5. 
a ---->10
       ^ ^
c -----| |
d -------|

b ----> [1, 2, 3, 4, 5]

6. DFS: No.  It will depend on how the state space is searched.  If we're not careful, DFS could get stuck searching s -> a -> s -> a ...

BFS: Yes.  Even though BFS will also explores s -> a -> s -> a ... it will also explore other paths at the same time and will eventually find the solution.

7. Memory.  At any given point, BFS has to keep track of an entire row/level of the state space.  If the branching factor (number of next states for a given state) is high and/or the goal state is far from the starting state, the row/level can be huge.

8. T/F
- False.  This is only true if there is a single unique goal state.  Otherwise, since they search the state space in different order, they can find different goal states.
- False.  m[1] represents the second row.  m[0] would represent the first row.
- False.  BFS/DFS could get lucky or, if the heuristic function isn't good, then the informed search algorithm could take longer.

9. 
def matrix_transpose(m):
    t = []
    
    for c in range(len(m)):
        new_row = []
        
        for r in range(len(m)):
            new_row.append(m[r][c])
            
        t.append(new_row)
        
    return t

Note if you didn't want to assume that the matrix was square, something like the following would work:

def matrix_transpose(m):
    t = []
    rows = len(m)
    cols = len(m[0])
    
    for c in range(cols):
        new_row = []
        
        for r in range(rows):
            new_row.append(m[r][c])
            
        t.append(new_row)
        
    return t

10. Set the weights on in1 and in2 to be -1 and the threshold to be -1.  If in1 and in2 are both 1, then the weighted sum will be -2 and below the threshold and will output 0.  If either in1 or in2 are 0 (or both are) then the weighted sum will be greater than or equal to -1, so it will output 1.

11.
i. print(dog)
Charlie of type dog and is not hungry

ii. print(cat)
Lucy of type cat and is hungry

iii. print(cat.same_type(cat2))
True