Problem 1
(a) ['levels', 'of']
(b) ['This', ['is', 'a'], 'n', 'e', 's']
(c) ['nested', 'nesting']
(d) 21
(e) [['n', 'st', 'd'], ['n', 'sting']]
(f) 0
(g) 'nestingnested'
(h) [1, 5]
(i) 'if'
(j) ['This', 'list', 'several', 'levels']
 
Problem 2(a)
-----------
(i) ['help', 'hlep', 'hple', 'help', 'hepl', 'help']
(ii) False
     True
     False
     False

Problem 2(b)
-----------
(i)
0 4 9
5 7 9
5 5 6
6 6 6
7 6

(ii)
0 5 11
0 2 4
3 3 4

(iii)
0 5 11
0 2 4
3 3 4
4 4 4
5 4

Problem 3
---------
(a)
f = open("presidents.txt", "r")
line = f.readline().rstrip()
presidentList = []
while line:
    [firstName, lastName, lifeSpan] = line.split()
    [birthYear, deathYear] = lifeSpan.split("-")
    presidentList.append([int(deathYear) - int(birthYear), firstName + " " + lastName])
    line = f.readline()

presidentList.sort(reverse = True)

for president in presidentList:
    print president[1]

f.close()

(b)
f = open("matrix.txt", "r")

mat = []
zeroes = 0
for line in f:
   mat.append([0]*zeroes + map(int, line.split()))
   zeroes = zeroes + 1

print mat

for i in range(len(mat)):
    for j in range(i+1, len(mat)):
        mat[j][i] = mat[i][j]

print mat

Problem 4
---------
def findBitonicMax(L):

    left = 0
    right = len(L)-1
    
    while left <= right:
        # Special Case 1: problem is of size 1
        if left == right:
            return left
    
        # Special Case 2: problem is of size 2
        if left == right-1:
            if L[left] > L[right]:
                return left
            else:
                return right
    
        # Typical case, when the list to be searched L[left..right]
        # consists of 3 or more elements. Note that mid-1 and mid+1
        # are valid indices in this case.
        mid = (left + right)/2
        if L[mid] > L[mid-1] and L[mid] > L[mid+1]: #mid is max
            return mid
        elif L[mid] < L[mid+1]: # max is to the right of mid
            left = mid + 1
        elif L[mid] < L[mid-1]: # max is to the left of mid
            right = mid - 1