# The Algorithm Design Manual: Chapter 4

4-1. The Grinch is given the job of partitioning 2n players into two teams of n players each. Each player has a numerical rating that measures how good he/she is at the game. He seeks to divide the players as unfairly as possible, so as to create the biggest possible talent imbalance between team A and team B. Show how the Grinch can do the job in $O(n \log n)$ time.

Solution
First sort the players with an algorithm which runs in $O(n \log n)$ time and afterwards form two teams with the first n player in the first team and the last n players in the second team.

4-3. Take a sequence of 2n real numbers as input.Design an $O(n \log n)$ algorithm that partitions the numbers into n pairs, with the property that the partition minimizes the maximum sum of a pair. For example, say we are given the numbers (1,3,5,9). The possible partitions are ((1,3),(5,9)), ((1,5),(3,9)), and ((1,9),(3,5)). The pair sums for these partitions are (4,14), (6,12), and (10,8). Thus the third partition has 10 as its maximum sum, which is the minimum over the three partitions.

Solution We can minimize the maximum sum if we pair up the lowest and the highest element. But first we have to sort our numbers which takes $O(2n \log n)$ and create the pairs which takes $O(2n)$.

Let S be the sequence with length 2n
sort S
for i := 1 to n
pair[i] = (S[i], S[2n - i - 1])

4-6. Given two sets $S_1$ and $S_2$ (each of size n), and a number x, describe an $O(n \log n)$ algorithm for finding whether there exists a pair of elements, one from $S_1$ and one from $S_2$, that add up to x. (For partial credit, give a $\Theta (n^2)$ algorithm for this problem.)

Solution We could sort $S_2$ and then iterate through $S_1$ and calculate the second number $s_1 + s_2 = x \Leftrightarrow s_2 = x - s_1.$ Now we just have to search for $s_2$ in $S_2$ which takes $O(\log n)$ time. Therefore we take $O(2n \log 2n) + O(n \log n) = O(n \log n)$ time.

4-7. Outline a reasonable method of solving each of the following problems. Give the order of the worst-case complexity of your methods.
(a) You are given a pile of thousands of telephone bills and thousands of checks sent in to pay the bills. Find out who did not pay.
(b) You are given a list containing the title, author, call number and publisher of all the books in a school library and another list of 30 publishers. Find out how many of the books in the library were published by each company.
(c) You are given all the book checkout cards used in the campus library during the past year, each of which contains the name of the person who took out the book. Determine how many distinct people checked out at least one book.

Solution (a) We can use quick sort in this scenario which works in expected $O(n \log n)$ time and match the ith bill with the ith check which takes $O(n)$ time.

(b) We can use a dictionary quite easily which counts the books by publisher. This takes $O(k ^cdot n)$ where $k=30$ time.

(c) We can sort the list by name and then iterate through it and count each different name. This takes about $O(n \log n) + O(n) = O(n \log n)$ time.

4-9. Give an efficient algorithm to compute the union of sets A and B, where $n = max(|A|,|B|).$ The output should be an array of distinct elements that form the union of the sets, such that they appear more than once in the union.
(a) Assume that A and B are unsorted. Give an $O(n \log n)$ algorithm for the problem.
(b) Assume that A and B are sorted. Give an O(n) algorithm for the problem.

Solution
(a) We can create one array with every element of A and B. Afterwards we sort this array and iterate through it. In this iteration we just have to check if ith item differs from the (i+1)th item which takes $O(n)$ therefore we need $O(n \log n) + O(n) = O(n \log n).$

(b)

Let A got n elements and B got m elements with n >= m.

i := 1 # counter for A, A is first element
j := 1 # counter for B
while i <= n
if j >= m # no more elements in B
i := i + 1

else if A[i] == B[j] # same element
i := i + 1
j := j + 1

else if A[i] > B[j] # increase j until B[j] is big enough
j := j + 1

else if A[i] < B[j]
i := i + 1

4-12. Devise an algorithm for finding the k smallest elements of an unsorted set of n integers in $O(n + k \log n).$

Solution We can build an “unsorted” heap in $O(n)$, i.e. only bring the smallest item at the top. Now we can extract k times the smallest item which takes $O(k \log n)$ time.

4-13. You wish to store a set of n numbers in either a max-heap or a sorted array. For each application below, state which data structure is better, or if it does not matter. Explain your answers.
(a) Want to find the maximum element quickly.
(b) Want to be able to delete an element quickly.
(c) Want to be able to form the structure quickly.
(d) Want to find the minimum element quickly.

Solution (a) They are both equally fast. In the max-heap it’s the first element in the sorted array the last one.
(b) A max-heap is here better because it takes only $O(\log n)$ instead of $O(n)$ for a sorted array.

(c) Both the heap and the sorted array take $O(n \log n)$ time to be formed.

(d) The minimum element in a sorted array is the first. In a max-heap every leaf could be the minimum element, therefore it takes $O(n).$

4-14. Give an $O(n \log k)$-time algorithm that merges k sorted lists with a total of n elements into one sorted list. (Hint: use a heap to speed up the elementary $O(kn)$– time algorithm).

Solution We can basically do an heap sort on these lists. We iterate from i := 1 to n and select the smallest item from the head of each list which takes $O(1) + O(\log n)$ time. Therefore the algorithm takes $O(n \log n)$ time.

4-15. (a) Give an efficient algorithm to find the second-largest key among n keys. You can do better than $2n - 3$ comparisons.
(b) Then, give an efficient algorithm to find the third-largest key among n keys. How many key comparisons does your algorithm do in the worst case? Must your algorithm determine which key is largest and second-largest in the process?

Solution (a) There’s a faster method for construction heaps which runs in $O(n)$. Afterwards we just have to call find-min two times which takes $O(\log n).$

(b) We can use the same method here as well. The largest second-largest key is implicitly found by constructing the heap.

4-16. Use the partitioning idea of quicksort to give an algorithm that finds the median element of an array of n integers in expected $O(n)$ time.

Solution The partition part of quicksort basically can help us. It determines partitions which are bigger respectively smaller than the pivot element. We just have to find n/2 elements which are smaller or equal to our element m which is then the median. This takes expected $O(n)$ time.

4-19. An inversion of a permutation is a pair of elements that are out of order.
(a) Show that a permutation of n items has at most $n(n-1) / 2$ inversions. Which permutation(s) have exactly $n(n-1) / 2$ inversions?
(b) Let P be a permutation and Pr be the reversal of this permutation. Show that $P$ and $P^r$ have a total of exactly $n(n-1) / 2$ inversions.
(c) Use the previous result to argue that the expected number of inversions in a random permutation is $n(n-1) / 4.$

Solution (a) Let’s take for example the items 1..5. We can get the maximum inversions if we reverse this list, i.e. [5, 4, 3, 2, 1] or more general [n, n-1, …, 1]. Now we can start at the right and count the inversions. Sublist              Inversions                  #Inversions
1                                                     0
2, 1                 (2,1)                            1
3, 2, 1              (3, 2), (3, 1)                   2
4, 3, 2, 1           (4, 3), (4, 2), (4, 1)           3
5, 4, 3, 2, 1        (5, 4), (5, 3), (5, 2), (5,1)    4

Therefore we got $sum_{i=0}^{n} (i-1) = \frac{n(n-1)}{2}$ inversions at most.

(b) Let’s prove this relation. Assume we got a permutation with n items. Let’s add an other item. The reversal for the old permutation got $\frac{n(n-1)}{2}$ inversions. The (n+1)th item adds n inversions which are: $(n+1, n), (n+1, n-1), .... (n+1, 1).$ Therefore we get $\frac{n(n-1)}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n(n+1)}{2}$ and we’re done.

(c) A rough approach is to take the smallest and the highest value and assume that inversions are uniformly distribute. Therefore we get $\frac{0 + \frac{n(n-1)}{2}}{2} = \frac{n(n+1)}{4}$

4-24. Let A[1..n] be an array such that the first $n - sqrt{n}$ elements are already sorted (though we know nothing about the remaining elements). Give an algorithm that sorts A in substantially better than $n \log n$ steps.

Solution Merge sort can used quite nicely. We need to sort the remaining $sqrt{n}$ which takes $sqrt{n} \log sqrt{n}$ time. Afterwards we have to use merge on our old and new sorted lists which takes $O(n)$ time.

4-29.
Mr. B. C. Dull claims to have developed a new data structure for priority queues that supports the operations Insert, Maximum, and Extract-Max—all in $O(1)$ worst-case time. Prove that he is mistaken. (Hint: the argument does not involve a lot of gory details—just think about what this would imply about the $Omega(n \log n)$ lower bound for sorting.)

Solution If insert, maximum and extract-max would be possible in $O(1)$ we could use the following algorithm to sort data.

A[1...n]
for i := 1 to n
insert(A[i])

for i := 1 to n
A[i] = maximum()
Extract-max()

This would sort data in $O(2n) = O(n)$ which is smaller than lower bound $Theta(n \log n).$

4-30. A company database consists of 10,000 sorted names, 40% of whom are known as good customers and who together account for 60% of the accesses to the database. There are two data structure options to consider for representing the database:
• Put all the names in a single array and use binary search.
• Put the good customers in one array and the rest of them in a second array. Only if we do not find the query name on a binary search of the first array do we do a binary search of the second array.
Demonstrate which option gives better expected performance. Does this change if linear search on an unsorted array is used instead of binary search for both options?

Solution Single array, binary search: $\log 10000 = 4$
Good and bad, binary search: $0.6 cdot \log 4000 + 0.4 cdot (\log 4000 + \log 6000) approx 5.11$
The first variant is a bit faster here.

Single array, unsorted: $10000$
Good and bad, unsorted: $0.6 cdot 4000 + 0.4 cdot (4000 + 6000) = 6400$
However in this case the second variant is far superior.

4-32. Consider the numerical 20 Questions game. In this game, Player 1 thinks of a number in the range 1 to n. Player 2 has to figure out this number by asking the fewest number of true/false questions. Assume that nobody cheats.
(a) What is an optimal strategy if n in known?
(b) What is a good strategy is n is not known?

Solution (a) Binary search.
(b) We can start asking if the number is between 1 and two. If not we can double our range from two to four, four and eight, etc.

4-35. Let M be an $n x m$ integer matrix in which the entries of each row are sorted in increasing order (from left to right) and the entries in each column are in increasing order (from top to bottom). Give an efficient algorithm to find the position of an integer x in M, or to determine that x is not there. How many comparisons of x with matrix entries does your algorithm use in worst case?

Solution

for j:= 1 to m
if m[j] <= x <= m[n][j]
do binary search on this row
if found return value

return Not found

This algorithm runs in $O(m\log n)$ time.

4-39. Design and implement a parallel sorting algorithm that distributes data across several processors. An appropriate variation of mergesort is a likely candidate. Mea- sure the speedup of this algorithm as the number of processors increases. Later, compare the execution time to that of a purely sequential mergesort implementation. What are your experiences?

from random import randint

def mergesort(lst):
n = len(lst)
if n < 2:
return lst
else:
middle = n / 2

left = mergesort(lst[:middle])
right = mergesort(lst[middle:])

return merge(left, right)

def merge(left, right):
result = []
i = 0
j = 0

while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
j += 1

result += left[i:] # append the rest
result += right[j:]
return result

unsorted = [randint(0, 150000) for i in xrange(0, 10 ** 6)]
mergesort(unsorted)


from random import randint
from multiprocessing import Process, Queue

def mergesort(lst):
n = len(lst)
if n < 2:
return lst
else:
middle = n / 2

left = mergesort(lst[:middle])
right = mergesort(lst[middle:])

return merge(left, right)

def merge(left, right):
result = []
i = 0
j = 0

while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
j += 1

result += left[i:] # append the rest
result += right[j:]
return result

def callMergesort(lst, q):
q.put(mergesort(lst))

def main():
unsorted = [randint(0, 150000) for i in xrange(0, 10 ** 6)]
middle = 5 * 10 ** 5

q = Queue()

p1 = Process(target = callMergesort, args = (unsorted[:middle], q))
p2 = Process(target = callMergesort, args = (unsorted[middle:], q))

p1.start()
p2.start()

merge(q.get(), q.get())

if __name__ == '__main__':
main()

For small to medium sizes the threaded version is a bit slower than the non-threaded one. However, for very big sizes the threaded version works faster.

4-45. Given a search string of three words, find the smallest snippet of the document that contains all three of the search words—i.e. , the snippet with smallest number of words in it. You are given the index positions where these words in occur search strings, such as word1: (1, 4, 5), word2: (4, 9, 10), and word3: (5, 6, 15). Each of the lists are in sorted order, as above.

Solution We can sort these positions with its identifier. Afterwards we iterative through the list and put identifiers on a stack. If we got all of them, we found the snippet. We replace each identifier if we find a nearer one. At the end, we just have to search for the smallest snippet in all snippets.

# The Algorithm Design Manual: Chapter 1

Yeah, new book series! I had laying this book around for about two and an half years and only read about a quarter of it but never worked through it. So, I decided to work through it and post interesting problems and solutions online.
The book consists of two parts. The first part treats the algorithmic basics. The second part is just a reference of different algorithmic problems and ways to solve it. I won’t cover the second part, mainly because there are no exercises.

1-1. Show that a + b can be less than min(a, b).
Solution: 1-1. For any $a, b < 0: a + b < \text{min}(a,b)$. For example: $a = -5, b = -3. a + b = -8 < \text{min}(-5, -3) = -5.$

1-2. Show that a × b can be less than min(a, b).
Solution For example for $a = -5$ and $b = 3$ the result is $-5 * 3 = -15 < \text{min}(-5, 3) = -5.$

1-5. The knapsack problem is as follows: given a set of integers $S = {s_1, s_2, . . . , s_n}$, and a target number T, find a subset of S which adds up exactly to T. For example, there exists a subset within $S = {1,2,5,9,10}$ that adds up to $T = 22$ but not $T = 23$.
Find counterexamples to each of the following algorithms for the knapsack problem. That is, giving an S and T such that the subset is selected using the algorithm does not leave the knapsack completely full, even though such a solution exists.
(a) Put the elements of S in the knapsack in left to right order if they fit, i.e. the first-fit algorithm.
(b) Put the elements of S in the knapsack from smallest to largest, i.e. the best-fit algorithm.
(c) Put the elements of S in the knapsack from largest to smallest.
Solution:
(a) $S = {1, 2}, T = 2$
(b) $S = {1, 2}, T = 2$
(c) $S = {2, 3, 4}, T = 5$

1-16. Prove by induction that n3 + 2n is divisible by 3 for all n ≥ 0.
Solution: The base case is $n = 0, 0^3 + 2*0 = 0 \text{ mod } 3 = 0$ which is true.
We can assume that this holds up to n. For n + 1 we get: $(n+1)^3 + 2(n+1) = n^3 + 3n^2 + 3n + 1 + 2n + 1$ $= n^3 + 3n^2 + 5n + 2 = (n^3 + 2n) + (3 (n^2 + n)).$
The first term is parenthesis is our assumption and the second term is obviously divisible by 3, therefore we showed that our assumption is true.

1-26. Implement the two TSP heuristics of Section 1.1 (page 5). Which of them gives better-quality solutions in practice? Can you devise a heuristic that works better than both of them?
Solution:

from random import randint

class Point:
def __init__(self, ID, x, y):
self.ID = ID
self.x = x
self.y = y

def distance(self, other):
return ( (self.x - other.x) ** 2 + (self.y - other.y) ** 2 ) ** 0.5

def closestPoint(P, visited, p):
shortestDistance = float("Inf")
shortestPoint = None

for pi in P:
if visited[pi.ID] == False:
if p.distance(pi) < shortestDistance:
shortestDistance = p.distance(pi)
shortestPoint = pi

return shortestPoint

def nearestNeighbour(P):
visited = [False] * ( len(P) + 1) # ID starts with 1
p = P

visited = True # not existing
visited = True # first node

i = 0

finalPath = [p]

while False in visited:
i += 1

pi = closestPoint(P, visited, p)

if pi == None: # last point
break

visited[pi.ID] = True

p = pi
finalPath.append(p)

finalPath.append(P)

return finalPath

def closestPair(P):
n = len(P)
finalPath = []

vertexChain = [[p] for p in P]

d = float("INF")
for i in xrange(1, n-1):
d = float("INF")
for i1, v1 in enumerate(vertexChain):
for i2, v2 in enumerate(vertexChain):
if i1 != i2:
s = v1[-1]
t = v2[-1]

if s.distance(t) <= d:
sm = i1
tm = i2
d = s.distance(t)

vertexChain[sm] += vertexChain[tm]
vertexChain.pop(tm)

vertexChain += vertexChain.pop(1)

return vertexChain
return finalPath

def getPathSum(P):
distance = 0
for i in xrange(1, len(P)):
distance += ((P[i].x - P[i-1].x) ** 2 + (P[i].y - P[i-1].y) ** 2)

return distance ** 0.5

P_rect = [Point(1, 0, 0), Point(2, 0, 5), Point(3, 5, 0), Point(4, 5, 5)]
P_line = [Point(1, 0, 5), Point(2, 0, 0), Point(3, 0, 10), Point(3, 0, 15)]
P_rand = [Point(i, randint(0, 15), randint(0,15)) for i in xrange(0, 10)]

print "**Rectangle**"
print "> NearestNeighbour: %i" % getPathSum(nearestNeighbour(P_rect))
print "> ClosestPair: %i" % getPathSum(closestPair(P_rect))

print "**Line**"
print "> NearestNeighbour: %i" % getPathSum(nearestNeighbour(P_line))
print "> ClosestPair: %i" % getPathSum(closestPair(P_line))

print "**Rand**"
print "> NearestNeighbour: %i" % getPathSum(nearestNeighbour(P_rand))
print "> ClosestPair: %i" % getPathSum(closestPair(P_rand))


NearestNeighbour createst shorter paths for small input graphs. For midsize input paths and big input graphs closestPair creates smaller paths. However nearestNeighbour is much faster. In conclusion, it depends on the application which heuristic is more suitable.

1-28. Write a function to perform integer division without using either the / or * operators. Find a fast way to do it.
Solution: A simple way to perform integer division is substracting the divisor and counting each substraction. This only works for positive numbers. (more in the comments)

def divideBySub(a, b):
count = 0
while a >= 0:
a -= b
count += 1

return count

If we can find a bigger divisor we could speed it up. An easy way is to multipy the divisor by itself as long as it divides the denominator with a rest.

def divideBySubFaster(a, b):
countB = 0
lastB = b

while a % b == 0:
lastB = b
b += b
countB += 1

b -= lastB # revert last change
countB -= 1 # revert last change

result = divideBySub(a, b)
for i in xrange(0, countB):
result += result

return result

This works quite nice and can speed up the process substantially.

# SPOJ: 1728. Common Permutation

Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of b.

Solution: That’s short and nice problem. You have to find all letters which are in both strings. I’d actually like to see some other implementations of this. I bet there are some languages which handle this in a smart way.

def intersection(lst1, lst2):
same = []
if len(lst1) > len(lst2):
for k in lst1:
try: # checks if element k is in lst2
lst2.remove(k)
same.append(k)
except:
pass
else:
for k in lst2:
try:
lst1.remove(k)
same.append(k)
except:
pass
same.sort()
return same

while True:
try:
inp1 = raw_input()
inp2 = raw_input()

common = intersection(list(inp1), list(inp2))
print "".join(common)
except:
break


# SPOJ: 3374. Scavenger Hunt

He does a poor job, though, and wants to learn from Bill’s routes. Unfortunately Bill has left only a few notes for his successor. Bill never wrote his routes completely, he only left lots of little sheets on which he had written two consecutive steps of the routes. […]
This made much sense, since one step always required something from the previous step. George however would like to have a route written down as one long sequence of all the steps in the correct order.

Solution:

scenarios = int(raw_input())

for i in xrange(0, scenarios):
S = int(raw_input())

myDict = dict()
myRevDict = dict()

for j in xrange(0, S - 1):
fromLocation, toLocation = raw_input().split()
myDict[fromLocation] = toLocation
myRevDict[toLocation] = fromLocation

# find start
start = ""
for key in myDict.keys():
if key not in myRevDict:
start = key
break

# print message
print "Scenario #%i:" % (i + 1)

# print path
currentKey = start
for l in xrange(0, S - 1):
print currentKey
currentKey = myDict[currentKey]

print currentKey
print ""