CMPSCI 311: Theory of Algorithms

David Mix Barrington

Fall, 2003

Practice Exam #2 Solutions

Exam posted Monday 3 November 2003

Solutions posted Thursday 6 October 2003

Actual (in-class) exam is Monday 10 November 2003

Questions in black, solutions in blue.

Directions: Each of the first three questions is a statement that may be true or false. Please state whether it is true or false -- you will get five points for a correct boolean answer and there is no penalty for a wrong guess. Then justify your answer -- the remaining points will depend on the quality and validity of your justification.

• Question 1 (10): (true/false with justification) If G is an undirected graph, any depth-first search of G starting from vertex v will contain the longest simple path starting at v. (A simple path is defined to be a path that never revisits a vertex.)

  FALSE. A DFS tree can have back edges that are part of a simple path but are not part of the forest. (The question is a little vague, it should have said "the DFS forest contains the path".) For example, suppose G has nodes {A,B,C,D} and edges (A,B), (A,C), (B,C), and (B,D). A DFS starting at A and looking at edges in alphabetical order will take (A,B), (B,C), and (B,D) as the tree edges, giving the DFS tree depth two. But there is a simple path of length three from A, going A to C to B to D.

• Question 2 (10): (true/false with justification) Let u be a string of length &Theta(1) and v a string of length n. Then the Boyer-Moore method to determine whether u is a substring of v is asymptotically faster than the brute force method.

  FALSE. If m is the length of u, brute force takes Θ(mn) = Θ(n). Boyer-Moore takes Ω(n) in the worst case, which asymptotically is no better though it is better by a constant factor.

• Question 3 (10): (true/false with justification) If A is a 3-by-3 matrix of integers, the matrix Aⁿ can be computed in O(log n) time, assuming that individual integers can be added or multiplied in O(1) time.

  TRUE. By the repeated squaring method we can compute Aⁿ using O(log n) matrix multiplications. And multiplying two 3 by 3 matrices by the brute-force method takes O(3³) = O(1) integer operations. So the total time is O(log n) times O(1) times O(1), or O(log n).

• Question 4 (20): Consider a max-heap implemented using pointers rather than an array, so that the root has pointers to two smaller heaps, all of whose elements are smaller than the root's element. Give an algorithm to find the smallest element in the heap, and argue that your algorithm always runs in O(n) time on a heap with n elements.

  Recursively find the minimum element in each of the two subheaps (whose roots are the children of the original heap's root. Compare these two elements and return the smaller. (There are some base cases -- if the heap is size one return the root's value, if it is size two return the value or the single leaf, which is the left child of the root.) This is clearly correct because the minimum value is in one of these two subheaps unless both are empty.

  To find the time we use a recurrence, where T(n) is the time of this algorithm on a heap with n nodes. We have T(1) = O(1) and T(2) = O(1) as base cases. Then if m is the number of nodes in the left subheap we have T(n) = T(a) + T(n-a-1) + O(1), which solves to T(n) = O(n). (For the inductive step of the proof, if T(a) ≤ ca, T(n-a-1) ≤ c(n-a-1), and the O(1) term is ≤ c, then the total is ≤ cn as desired.)

• Question 5 (25): The triangle problem is to input an undirected graph and determine whether it contains three distinct nodes a, b, and c such that all three edges (a,b), (a,c), and (b,c) are in the graph. Let n be the number of nodes in G and e the number of edges.

  • (a,5) Describe a brute-force algorithm to solve the triangle problem in O(n³) time.

    For each triple of distinct nodes a, b, and c, check whether the three given edges exist and return true if they all do. If you complete the loop without returning, return false. This is O(n³) because we can have three nested loops ranging over all a, all b, and all c respectively, and testing for distinctness and for the existence of the three nodes is O(1).

  • (b,10) Explain how the triangle problem can be solved in o(n³ time using the Solovay-Strassen algorithm for matrix multiplication. (You may quote the performance of Solovay-Strassen.)

    If A is the adjacency matrix of G, we know that for any i and j, A³_i,j is the number of paths of length exactly three from i to j. Thus the graph has a triangle iff for some i, A³_i,i is positive. We can calculate A³ using two matrix multiplications on n by n matrices, which can each be done in o(n³) time by Solovay-Strassen. Then checking for a positive entry on the diagonal takes only O(n) time, which is also o(n³).

  • (c,10) Explain how the triangle problem can be solved in O(n+e) time. (Hint: Use depth-first search.)

    In O(n+e) time we can create a DFS forest for G, recording the depth of each node. Then in O(n+e) time we can check each edge to see whether it goes from a node to its grandparent.

    We must prove that there is an edge to a grandparent in the DFS forest iff G contains a triangle. The forward direction is obvious, as the node, parent, and grandparent fulfill the condition to be a triangle. For the other direction, suppose that there is a triangle in G. Let x by the node in the triangle that is reached first in the DFS. Because all neighbors of x are searched before leaving x, one of the other two triangle nodes will be found first -- call this node y. The third node z must be found while y is being searched, because z is a neighbor of y and all neighbors of y are searched before leaving y. So x is the parent of y and the grandparent of z, and the edge (x,z) connects z to its grandparent.

• Question 6 (25): Let A be an array containing n elements, all distinct. The successor of an element x in A is the element y such that x is smaller than y but no element z is strictly between x and y. Every element except the maximum has a successor.

  • (a,5) Explain how to find the successor of x in O(n) time whether A is a sorted array or not.

    Make one pass through the elements, keeping a variable "candidate". The first time you see an element larger than x, set "candidate" to that element. (If you never see such an element, return that x has no successor.) Later if you see an element smaller than "candidate" but bigger than x, set "candidate" to that element. At the end, return "candidate". You have taken O(n) time because you spent O(1) time with each element. And you have returned the correct successor because when the correct successor is found "candidate" will be set to it, and "candidate" will not be changed after that happens.

  • (b,10) Explain how to find the successor of x in O(log n) time if A is sorted. Would you recommend presorting A to make finding the successor easier?

    If we knew where x was in the array, we would need only O(1) time because we just look at the element following x. Finding x in the array takes only O(log n) time by binary search.

    Presorting the array would take O(n log n) time. For one successor query this is clearly slower than doing the search. If we were doing a larger number of successor queries on the same array, such as Ω(n) of them, presorting might be worthwhile.

  • (c,10) Suppose you have a table allowing you to find the successor of any element in O(1) time. Explain how to sort A in O(n) time.

    Find the smallest element in O(n) time and make it element 0 of the sorted array. Then for each i from 1 through n-1, find the successor of element i-1 of the sorted array and make it element i of the sorted array. The n successor queries take O(1) time each by assumption, so we have O(n) + O(n) = O(n) total time.

  Last modified 6 November 2003