Hon.Algorithms (CSCI-GA 3520), Professor Yap Fall 2012 MIDTERM EXAM. Oct 11, 2012 SOLUTION

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1 Hon.Algorithms (CSCI-GA 352), Professor Yap Fall 22 MIDTERM EXAM Oct, 22 SOLUTIONS Problem. ( Points) Consider the recurrence T(n) = 4T(n/2)+ n2 lgn. Reduce this to standard form and solve it to Θ-order. SOLUTION: (a) Let k = lgn and t(k) = T(2 k ). Then the recurrence becomes Let s(k) = t(k)/4 k. Then t(k) = 4t(k )+ 22k k. s(k) = s(k )+ k. Invoking DIC, this has the exact solution s(k) = H k usingthe Harmonic numbers with boundary condition s(k) = for k <. Plugging back, we get the exact solution T(n) = 4 lgn H lgn, which gives T(n) = Θ(n 2 (lnlgn)). Comments: Most students get perfect score for this. Problem 2. ( Points) (a) Draw the full binary tree corresponding to its compressed bit representations:,,,. (b) Is the currency system D = [,2,4,6,36] canonical? Explain. FIGURE: SOLUTION SOLUTION: (a) The tree is seen in Figure. It has 8 external nodes, so the bit string is 5 as expected. (b) NO, it is non-canonical. We know that D = [,2,4] is canonical. But D = [,2,4,6,36] is a Type B extension of D with parameters (q,q,r) = (4,2,). Thus it does not satisfy the inequalities r < q < r +q required by our theorem for canonical Type B extension. The proof of the theorem tells us how to find a counter example. A counter example is x = 48. The optimal solution is s = (,,,3,) but the greedy solution is s = (,,3,,). Comments: Many students get (b) wrong by invoking the following argument: in a currency system D = [d,...,d m ], if 2d i d i+ then the greedy algorithm is optimal. I am not sure where this idea comes from.

2 ,,,. Figure : Binary tree with compressed bit representation,,,. Problem 3. (6 Points) Recall the problem of computing the maxima of a unimodal function. In what sense can we say the stateless algorithm AL is only one step away from optimal. HINT: F k φ k / 5 = (φ ) k / 5, which monotonically decrease to very fast. SOLUTION: After k probes, AL gives a width of AL (k) = but the φ k optimal algorithm AL k gives Opt(k) = F k+ = 5. Thus Opt(k)/AL (k) = φ k+ 5/φ 2 = 5/( + φ) <. But if AL is given one extra probe, we get Opt(k)/AL (k+) = 5/φ >, i.e., AL withk+probeswouldbeatopt(k). Comments: This holds for k large enough. Actually, k does not have to be very large k at all. But we do not expect this analysis for the midterm. Now, many people repeated a mistake in my homework solution (I think). I believe I said 5/φ 2.7. Actually, it is φ 2 / 5.7. A simple reality check (like solution above) would have caught the error, since and φ 2 = +φ Sorry about that, but I did not take off points for this. Problem 4. (6 Points) Let p be a fixed odd number. Let S p denote the Standard Scheme for k-th selection problem where we divide the input set of n elements into n/p groups, each of size p, and then find the median of each group, etc. Let T p (n) denote the complexity of scheme S p. As usual, T p (n) only counts the number of comparisons. We want to understand the limit of T p (n) as p. (a) Let M(n) be the worst case number of comparisons for finding the median of n elements in the comparison tree model. Write the recurrence for T p (n) using M(n). HINT: assume p and n are so large that so that we may ignore ceiling and floors, etc. 2

3 SOLUTION: We may assume p divides n in the following. There are n/p groups, and to compute their median, we need (n/p)m(p) comparisons. We have n/p medians, so we need T p (n/p) comparisons in the recursive call. Next we to want to make sure that every element is related to the median of medians. How many unrelated elements are there? There are n/p groups, each with p/2 elements. So the number of such elements is n/2. So we need to compare each of these elements to the median. Finally, we can eliminate at least p/2 elements in n/2p groups, i.e., n/4 elements. and so recursively, we need to solve T(3n/4). The final recurrence is T(n) = M(p)(n/p)+T(n/p)+n/2+T(3n/4). () Comments: Many students ignored my HINT to assume p and n large so that you can do simple expressions like the above derivation. Otherwise, you easily get lost the details, and say some horribly wrong things. A common error is to forget the n/2 comparisons in the above recurrence. This is very instructive: if you do not make about n/2 extra comparisons, you cannot eliminate a single element for the recurrence. Figure out this remark. (b) Suppose you could (magically) prove that M(n) 2n, but you cannot exhibit any median algorithm that uses 2n comparisons. (Such a phenomenon is known to occur.) Can you exploit this information to implement the scheme S p for any given p? SOLUTION: Yes. In S p, we only need to find the median for p elements for a fixed p. Although we do not know any median algorithm that achieves 2n comparisons on n elements, for any fixed p, we can exhaustively search to find the optimal comparison tree program to find the median of p elements. This tree program uses most 2p comparisons (since we know that M(p) 2p). We can incorporate this tree program in our Scheme S p. Comments: No student understand this question. I will try to explain this in class. (c) Using the fantasy result of part (b), please solve for T p (n). In particular, find the optimal constant K p such that T p (n) K p n. What is K p as p? It is known that M(n) 2n O() and M(n) (3 ǫ)n. 3

4 SOLUTION: Our recurrence, using the fact that M(p) 2p, becomes Inductively, we have T(n) = 2n+T(n/p)+n/2+T(3n/4). (2) T(n) 5n/2+K p (n/p)+k p (3n/4) ( 5 K p n + 2K p p + 3 ) 4 K p n provided K p p/(p 4). Asymptotically, K p as p. Comments: If you forgot the +n/2 term in the recurrence in part (a), then you will get K p 8. Some student get K p, which is clearly impossible. In fact, we already said that median has a 2n O() lower bound, so K p 2. You should have used this as a clue that something is wrong. Problem 5. ( Points) We consider ratio balanced trees where < ρ < is fixed. (a) Show that there are AVL trees not in RB[ρ] if ρ < 2. This means you must show that there are AVL trees whose ratio is arbitrarily small. (b) Show there are infinitely many trees in RB[ρ] that are not AVL. SOLUTION: (a) Consider the AVL tree T n where the right subtree is the complete binary tree of height h (and so has esize of 2 h+ ), and the left subtree is the minimal size AVL tree of height n (and so has esize +µ(h) 2φ h ). The ratio of T n is at most 2φ h /2 h+ and this goes to as h. (b) Let H(n) denote the maximum height of a RB[ρ] tree of size n. It is easy to see by induction that H(n) log +ρ (n+). (3) The result we want will following from: OBSERVATION: There are RB[ρ] trees with ratios arbitrarily close to lg( + ρ). Such trees cannot be AVL. Proof of observation: suppose esize(u.right) = s. Then the height of u.right is at most H(s ) log +ρ s as derived above. Also esize(u.left) > sρ. Hence the height of u.left is at least lg(sρ ). The ratio of these heights is at least lg(sρ ) log +ρ s which approaches lg(+ρ) for large s. For example, with ρ = /3, the maximum ratio is lg(+ρ) = (2 lg3).45. Comments: Most people got part (a), using basically the example shown above. Most did not get part (b) or solved it partially. A simple way to solve part(b) is to just construct infinitely many RB[ρ] trees wherethe root has height difference greater than. But our proof here gives much more information. Problem 6. (6 Points) We had looked at an algorithm for biconnected components for bigraphs. (a) Please discuss its extension to digraphs. What would the reduced digraph would look like? HINT: the structure is actually quite complex, and there seems to be no standard way to describe them. Hypergraphs seems useful here. 4

5 (b) Outline an algorithm for counting the number of biconnected components in a digraph. SOLUTION FIGURE: a 2 3 b c (i) (ii) (iii) (iv) Figure 2: Biconnectedness in digraphs SOLUTION: This was a somewhat open-ended question for you to think on your feet. You would be going down the wrong paths if you thought this was a routine question. (a) Let G = (V,E) be a digraph. We call a subset C V a biconnected set if every pair to distinct vertices u v in C are contained in a simple cycle passing through vertices in C only. We can C a biconnected component if C is a maximal set with respect to being a biconnected set. Having defined this concept, we now see that they do not lead to any simple notion of a reduced digraph. Consider Figure 2(i): C := {a,b,c} and C 2 := {,2,3} {a,b}arebiconnected components. But weseethat C C 2 = 2. But it easy to generalize this example so that C C 2 is arbitrarily large (see Figure 2(ii)-(iv)). So how do we represent these biconnected components? By a hypergraph! But these are not be arbitrary hypergraphs (what are the constraints)? Moreover, there could be exponentially many biconnected components (exercise). (b) We might try to run our biconnectivity algorithm and see what we get. But it is not clear to me how to even verify a single biconnectivity component. Comments: Basically, biconnectivity for digraphs do not seem to lead to any simplification or reduced graph. A more promising direction is to look at Eulerian graphs: these are digraphs where the indegree equals the outdegree at each node. We can think of them as a generalization of bigraphs. 5