CMPSCI 611: Theory of Algorithms

David Mix Barrington

Spring, 2006

Midterm Exam Solutions

There are seven questions for 100 total points. The scale is A=90, B=60. The first four problems are "true/false with explanation".

• Question 1 (10): (true/false with explanation) Let G be a connected undirected graph with positive edge weights. Let e be a particular edge of G such that for any other edge e', the weight w(e) is strictly less than w(e'). Then if T is any minimum spanning tree of G, then e is one of the edges of T.

  TRUE. Let T be a spanning tree that does not contain e. Add e to T, creating a graph U that has a cycle. Find an edge on that cycle other than e and call it e'. Removing e' from U cannot cause U to become disconnected because the endpoints of e' are still connected through the cycle. So U - {e'} is a spanning tree, because it is a connected graph with the right number of edges. The weight of this tree is strictly less than the weight of T, because we added e to T and removed e' and w(e) < w(e'). So T cannot have been an MST, since there is another tree of strictly smaller weight.

• Question 2 (10): (true/false with explanation) Let G be a weighted directed graph with exactly one source s and exactly one sink t. Let (A,B) be a maximum cut in G, that is, A and B are disjoint sets whose union is V, s ∈ A, t ∈ B, and the sum of the weights of all edges from A to B is the maximum for any two such sets. Now let H be the weighted directed graph obtained by adding 1 to the weight of each edge in G. Then (A,B) must still be a maximum cut in H.

  FALSE. First note that the min-cut-max-flow theorem is not relevant here, because we are talking about maximum cuts. (I actually meant to phrase the question in terms of minimum cuts, and the answer would have been similar.) There is no reason why adding 1 to all edges might not promote one cut ahead of another, and all we need is an example. Mine has edges (s,4,a), (a,1,b), (a,1,c), (a,1,d), (b,1,t), (c,1,t), and (d,1,t). The maximum cut in this graph has just s on one side and has size 4. But in H, all the weight-1 edges become weight 2 and there are two cuts of weight 6 while the previous cut goes from weight 4 to 5. A single counterexample is enough to refute the given statement.

• Question 3 (10): (true/false with explanation) Suppose that given two n-bit numbers X and Y, each divided into three pieces of n/3 bits each, I can determine the product XY using six multiplications of one n/3-bit piece by another, plus some fixed number of additions. Then I can use this method to devise a method to multiply any two n-bit numbers in O(n^1+log₃2) time.

  TRUE. (Though no such method is known, and it may well have been proved that none can exist.) We would have a recurrence for the time T(n) to multiply two n-bit numbers, with T(n) = 6T(n/3) + O(n). (The O(n) is because we can add two n-bit numbers in O(n) time and we are told that this method uses only O(1) additions.) This solves by the Master Theorem to T(n) = n^β, where β = log₃6. (Here α = 1, so we are in the α<β case of the Master Theorem.) All we need to check is that log₃6 equals the given exponent 1 + log₃2, which is true because log₃6 = log₃3 + log₃2.

• Question 4 (10): (true/false with explanation) Let G be a complete n-vertex undirected graph with positive integer edge weights, and let A be the matrix such that A_ij is the weight on the edge from i to j. (For each i, A_ii = 0.) Then for each i and j, A^n-1_ij gives the shortest-path distance from i to j, where the matrix multiplication is carried out using ordinary integer addition and multiplication.

  FALSE. This all looks good until the end where it says "ordinary integer addition and multiplication" instead of min-plus. But although this statement does not describe the min-plus matrix method from lecture, we still have to present an argument that this alleged method does not also work. A single counterexample suffices to refute the claim. Let the graph be a triangle with all weights equal to 1:

  
      0 1 1                    0 1 1                   2 1 1 
  A = 1 0 1   A^2 (min-plus) = 1 0 1  A^2 (ordinary) = 1 2 1
      1 1 0                    1 1 0                   1 1 2
  

  The statement would have it that the shortest-path distance from vertex 1 to itself is 2, whereas we know that it is zero.

• Question 5 (15): Let E be a finite set of intervals on the line, so that each interval e_i has a start point s_i and a finish point f_i. Let I be the set of all sets of intervals in E such that no two intervals in X overlap. That is, there do not exist e_i and e_j in X such that s_i < s_j < f_i. Prove that I is a subset system (which means that it is closed downward under inclusion). Prove that it need not be a matroid.

  I is a subset system because if we take any set of intervals X in I and remove one or more of the intervals, the remaining set Y has no overlaps. This is because if two intervals in Y overlapped, they would also overlap in X and X would not be a valid set in I.

  We only need one counterexample to the Exchange Property to show that I need not be a matroid. Let E consist of the intervals (1,3), (2,5), and (4,6). The sets A = {(1,3), (4,6)} and B = {(2,5)} are both in I. If I satisfied the Exchange Property, we would be able to take one of the two intervals in A and add it to B with the resulting set being in I. But both these intervals overlap with (2,5). A and B are also maximal sets of different size, and thus provide a counterexample to the Cardinality Theorem.

• Question 6 (25): Let E be a finite set of intervals as in Question 5, with the additional property that each interval e_i has a positive weight w_i. Consider the optimization problem of finding a non-overlapping set of intervals that maximizes the total weight of the intervals in the set. You may assume that all the start and finish points are distinct (and that no start point equals any finish point).
  • (a,10) Give an example showing that the simplest greedy strategy (always choosing the heaviest remaining interval that does not overlap with those already chosen) can fail to find the optimal set of intervals.

    Let E be the set of intervals above and assign weight 2 to each interval in A and weight 3 to the interval in B. Then the greedy strategy results in B, with total weight 3, whereas the optimal solution is A, with total weight 4.

  • (b,15) Explain a dynamic programming algorithm that finds the optimal set in time that is polynomial in the number of intervals. (Hint: First sort the intervals by finish time, than for each i from 1 to n, compute the best set of intervals among the first i to finish.)

    Sort and renumber the intervals in order of finish time as suggested. For each i from 1 to n, we calculate the subset S_i of the first i intervals that is in I and has the largest total weight. Clearly S₁ = {1} as no overlap is possible. Once we have S_i, we calculate S_i+1 as follows. We consider two candidate sets. The first is just S_i. The second is S_j &union; {i+1}, where j is the index of the last interval to finish before interval i+1 starts. This second set contains no overlap, and because S_j was the best choice among the first j intervals, it is the best choice among the first i+1 intervals if we restrict to sets containing interval i+1. To get S_i+1, we just take the higher-scoring of these two candidate sets. Once we have S_n, this is the final answer. The running time is O(n log n) to sort and O(n log n) to compute the n sets. (Computing the second candidate set requires finding the particular value of j, which we can do in O(log n) time by binary search. There are other ways to do the job that are still polynomial time.)

• Question 7 (20): Let E be a set of elements and let C be a set of attributes, such that each element may have zero, one, or more than one attribute. Recall from HW#2 that a traversal is a set T ⊆ E such that there exists a subset A of C, with |A| = |T|, and a one-to-one function f from T to A such that each element a in T has attribute f(a) (perhaps along with other attributes). That is, it is possible to pick an attribute held by each element in T such that no attribute is used twice.

  Given the information as to which element has which attribute, describe how to define a flow network such that a maximum flow through the network can be used to find a traversal of maximum possible size. Explain carefully why the maximum flow has to be of the form necessary to define the traversal.

  We make a vertex for each element of E and a vertex for each element of C, and add the following edges, all of weight 1:

  • An edge from s to each element of E
  • An edge from e to c whenever element e has attribute c
  • An edge from each element of c to t

  Any traversal of size k corresponds to a flow of size k through this graph, where we take the edges from s to each element of T, from each a in T to its corresponding f(a), and from each element of A to t. Furthermore, any integer flow in this directed graph corresponds to a traversal, because we may take T to be the set of endpoints of edges from S and A to be the startpoints of edges to T, and the intermediate edges give us the function f.

  However, for full credit on the problem you must argue that the maximum flow in this graph must be an integer flow, rather than for example having fractional flows along edges that would not necessarily correspond to a traversal. (This is what the last sentence of the question meant.) The reason we know that the maximum flow is an integer flow is that the Ford-Fulkerson algorithm gets a maximum flow, and it only deals with integer flows if all the capacities in the flow diagram are integers.

Last modified 4 April 2006