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Fall 2000
Exam information
142 students took the exam. Scores ranged from 5 to 28, with a median of 17 and an average of 16.7. There were 17 scores between 23 and 30 (inclusive), 72 between 16 and 23, 50 between 8 and 15, and 3 between 1 and 7. (Three-quarters of the points on exams plus good grades on homework and projects would earn you an A–; similarly, a test grade of 16 may be projected to a B–.)
There was only a single version of the exam.
If you think we made a mistake in grading your exam, describe the mistake in writ- ing and hand the description with the exam to your discussion t.a. or to Mike Clancy. We will regrade the entire exam.
Solutions and grading standards
Problem 0 (1 point)
You could have lost the point for this problem for several reasons:
you earned some credit on a problem and did not put your name on the page; you did not indicate your discussion section and t.a.; you did not indicate your login on the EECS instructional systems; you did not identify where you were sitting.
The reason for this apparent harshness is that we need ways to recover misplaced or unstapled. We also need to know where you will expect to get your exam returned, and how to submit your grade to the grading system.
Fall 2000
Problem 1 (6 points)
In this problem, you considered the following network.
For part a, you were to display the corresponding residual graph, shown below. Also shown in bold are paths from s to t in the graph; you didn’t have to specify these, but they’re useful for the remaining parts of this problem.
For part b, you were to find an edge whose capacity could be reduced to make the given flow optimal. This edge would be found by examining the residual graph, and choosing an edge whose removal would block all paths from s to t. (Reducing the edge’s capacity to the level of its flow would saturate the edge; it thus would not appear in the residual graph.) There are two edges that qualify: the edge from s to 1 and the edge from 1 to 2. Either capacity can be reduced by 2.
Part c was to produce a corresponding cut. Again, this could be determined from the (modified) residual graph; all vertices reachable from s constitute a minimum cut. Your answer here would depend on your answer to part b. If you reduced the capacity of edge (s, 1), the cut would be {s}; if you reduced edge (1, 2), the cut would be {s, 1}. You were allowed merely to draw the cut on the diagram.
Points for this problem were allocated 3 for part a, 2 for part b, and 1 for part c. You received 2 points for an almost correct solution to part a; typically, such a solution
s
t
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Fall 2000
Each of the calls to union involve arguments with equal rank (0 in the first two calls, 1 in the next); the tie is broken by putting the characteristic element of the second argument at the root of the structure. The following call combines the two trees and also incorporates the desired path compression for D:
union (D, E or F or G);
Essentially there were five things that had to be done in this problem: four links to be defined, and one link to be changed via path compression. In general, 1 point was awarded for each of these. Illegal operations, e.g. explicit calls to find, received no credit. We attempted to determine the steps you got correct and award credit appro- priately, so that an error would not lose you all the points for remaining operations.
You were allowed to separate the path compression from the combination of {A, B, C, D} with {E, F, G, H} by including an extra call to union, for example:
union (D, A);
Other errors included operations performed out of sequence (which produced incor- rect answers because of union-by-rank) and mishandling arguments of equal rank. Not explaining the operations with words or diagrams lost you a point.
Problem 3 (6 points)
For this problem, you considered a linear programming problem whose solution, if variables were restricted to 0-1 values, would represent a vertex cover. You were first to use the possibly fractional solution values to design an approximation algorithm for finding a vertex cover, then to determine a ratio bound of the algorithm.
Here’s an algorithm. First, make a pass through the xk, changing all those with value (^1) ⁄ 2 to have value 1. Then choose for the vertex cover all vertices^ vk for which^ xk = 1.
- Can there be an edge (vj, vk) that’s not covered? No, since in order to satisfy the constraint xj + xk ≥ 1 , either one of xj and xk has value 1 or both have value 1 ⁄ 2 in the LP solution; in either case, at least one of vj and v (^) k is chosen for the cover.
- How good an approximate cover results? The value of the LP objective function x 1
- … + xn is a lower bound on the size of the optimal cover. By changing x values from 1 ⁄ 2 to 1, we at most double the value of the objective function. With all xk tak- ing on 0-1 values, the objective function represents the size of the corresponding cover. Thus the cover produced by the algorithm is at most twice the size of the optimal cover. The ratio bound is max (^1 ⁄ 2 , 2 ⁄ 1 ) = 2.
Part a was worth 4 points, 2 for the algorithm and 2 for the argument that it pro- duces a vertex cover. The most common error was to offer no argument at all that the algorithm produces a vertex cover. Next most common was a flawed attempt at opti- mization: instead of just taking all vk for which xk > 0, some students tried to be more selective, changing some of the xk values to 0 and some to 1. An example of this approach is the following algorithm:
While there are xk’s with value 1 ⁄ 2 , pick one, change it to 1, and change all the neighboring xj with value 1 ⁄ 2 to 0.
Fall 2000
This doesn’t work for the path of length 4 shown below.
The optimal solution for the LP is x 1 = x 2 = x 3 = x 4 = 1 ⁄ 2 ; choose x 1 , set it to 1 and set x 3 to 0, then choose x 2 , set it to 1 and set x 4 to 0. The vertices thus selected are v 1 and v 2 , which are not a vertex cover. With an algorithm like this, you generally earned 1 point in part a and 0 points in part b.
Part b was worth 2 points. A correct bound alone, without some justification, earned 0 points. Many solutions provided an incorrect justification similar to the following:
Any vertex cover includes at least one endpoint from each edge. My algorithm might include both endpoints from some edges. Therefore my algorithm produces a vertex cover with at most twice as many vertices as the optimal vertex cover.
This overlooks the fact that a vertex may cover more than one edge. Consider a “star” graph with a center vertex connected to four other vertices. The center vertex alone covers all four edges, but including the other endpoint for each of the four edges results in a cover of five vertices, and of course 5 ⁄ 1 > 2.
Problem 4 (6 points)
This problem involved proving the Maximum Common Induced Subgraph problem NP-complete. (We’ll call it MCIS in the discussion below.)
You first had to provide the corresponding decision problem, which was
Given k, G 1 , and G 2 , is there a graph H with at least k vertices such that H is an induced subgraph of both G 1 and G 2?
You were allowed to say “exactly k vertices” instead of “at least k vertices”, since if there’s a common induced subgraph of size ≥ k, there is also one of size exactly k.
Next came the actual proof of NP-completeness, which has two parts: showing the problem is in NP, and reducing some known NP-complete problem to it.
Showing membership in NP involves producing, along with a “yes” answer to the decision problem, a certificate with which the answer can be checked in polynomial time. The certificate must include information that enables efficient checking for the following:
- The graph H has k vertices.
- H is an induced subgraph of G 1.
- H is an induced subgraph of G 2.
An important component of the certificate is thus the mapping from vertex numbers in H to vertex numbers in G 1 and G 2. Without this mapping, one is reduced in the worst case to a brute-force search for corresponding vertices as you did in project 1.
v 1 v 3 v4 v 2
Fall 2000
As noted above, many of you tried to reduce Subgraph Isomorphism to MCIS. You earned 1 point for this if you did it in the correct direction with the correct argu- ments but ignored the distinction between subgraph and induced subgraph. Other people tried to reduce Graph Isomorphism to MCIS. Even with a valid reduction, however, this wouldn’t show that MCIS is NP-hard since Graph Isomorphism isn’t known to be NP-complete. No points were awarded for this.
You earned 1 point for trying to reduce Clique to MCIS, 1 more point for showing that “yes” from MCIS maps to “yes” from Clique and 1 more point for showing that “no” maps to “no”.
Problem 5 (6 points)
For this problem, you were to prove that the given greedy algorithm for covering points on the real number line with unit intervals (which have length 1) produces the smallest possible set of covering intervals. (This problem, incidentally, is CLR exercise 17.2-5.)
One approach to proving greedy algorithms correct is to verify the greedy choice and optimal substructure properties. Here, the greedy choice property says that the interval chosen by the algorithm—the interval [x 1 , x 1 +1]—is part of an optimal cover. The optimal substructure property says that optimal solutions are built from optimal subsolutions. Here, it says that if [x 1 , x 1 +1] ∪ C is an optimal cover for {x 1 , x 2 , …, xn}, then C is an optimal cover for points in {x 1 , x 2 , …, xn} – [x 1 , x 1 +1].
A proof of the greedy choice property can construct a suitable cover as follows. First, we assume that some optimal cover C does not contain [x 1 , x 1 +1]. One of its intervals must contain x 1 , so assume that interval is [y, y+1] where y < x 1 ≤ y+1. Since y < x 1 , y+1 < x 1 +1, so the interval [x 1 , x 1 +1] contains at least as many of the xk’s as the interval [x 1 , y+1]. Moreover, since x 1 is the leftmost point of those to be covered, there are no points in the interval [y, x 1 ). Thus [x 1 , x 1 +1] contains at least as many of the xk’s as [y, y+1], and substituting [x 1 , x 1 +1] for [y, y+1] in C results in a cover of all the xk’s with exactly as many intervals.
A variation on this proof would assume that no optimal cover contains [x 1 , x 1 +1] and then go on to derive a contradiction using the above approach.
We prove optimal substructure—if [x 1 , x 1 +1] ∪ C is an optimal cover for {x 1 , x 2 , …, xn}, then C is an optimal cover for points in {x 1 , x 2 , …, x (^) n} – [x 1 , x 1 +1]—by induction on the number of greedy choices. First, the base case: If one interval covers all the points, C and the resulting set of points are both empty, so we have a trivial “optimal substructure”. Now suppose that two or more greedy choices are made, that [x 1 , x 1 +1] ∪ C is an optimal cover for {x 1 , x 2 , …, xn}, and that C is not an optimal cover for points in {x 1 , x 2 , …, xn} – [x 1 , x 1 +1]. Thus there is another cover C' with one fewer interval. But C', together with the interval [x 1 , x 1 +1], covers all the points in {x 1 , x 2 , …, xn} and has one fewer interval than C. This contradicts the assumption that C is an optimal cover of {x 1 , x 2 , …, xn}.
A more concise proof that the algorithm works, provided by Francis Hsu, is the fol- lowing. Consider the left endpoints of the intervals chosen by the greedy algorithm.
Fall 2000
All these xk’s must be covered; moreover, no unit interval can cover more than one of them since by construction the difference between one of them and the next is more than 1. Thus any legal cover must contain at least as many intervals as the cover produced by the greedy algorithm.
For solutions that split the proof into greedy choice and optimal substructure sec- tions, we awarded up to 3 points per section. Of the 3 greedy choice points, you earned 1 by noting that [x 1 , x 1 +1] included the most points among all intervals con- taining x 1. You earned the other 2 by showing the existence of an optimal cover con- taining [x 1 , x 1 +1]; your proof had to include some mention of substituting [x 1 , x 1 +1] for some other interval in an optimal cover to get either of these points. Of the 3 opti- mal substructure points, you earned 1 for noting some kind of equivalent substruc- ture. The other 2 points were for removing [x 1 , x 1 +1], invoking the induction hypothesis, and then putting [x 1 , x 1 +1] back into the cover thus derived.
Many solutions attempted a proof by induction on the number of points, claiming that if the greedy algorithm works on all sets of n points, it will also work on all sets of n+1 points. This approach won’t work; it essentially goes backward from the induction of the optimal substructure property. Still, it could earn as many as 2 points: 1 for the base case and 1 for analyzing cases on the last interval. Your case analysis had to be along the following lines to get this point: either the last inter- val—not necessarily an interval beginning at xn!—contains xn+1, or it doesn’t, and in the latter case, either xn+1 – xn > 1 or xn+1 – xn ≤ 1. A few other solutions tried a proof by contradiction, and may have earned 1 point for limiting consideration to the smallest set of n points that the algorithm doesn’t work for.
A few other common errors were the following. In the greedy choice proof, some stu- dents wanted to show that [x 1 , x 1 +1] must be in an optimal cover. Some students tried to vary the sequence in which intervals are chosen, noting, say, that the choice of the interval [x 2 , x 2 +1] before [x 1 , x 1 +1] would not lead to a better cover. This, how- ever, doesn’t say anything about covers produced in other ways. Finally, some stu- dents concluded incorrectly in a proof by contradiction that if an interval [xk, xk+1] was not in an optimal cover, then xk was not included in any of the intervals of the cover.
