UNIVERSITY of OSLO. Faculty of Mathematics and Natural Sciences. INF 4130/9135: Algorithms: Design and efficiency Date of exam: 8th December 2016

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1 UNIVERIY of OLO Faculty of Mathematics and Natural ciences Exam in: INF 4130/9135: Algorithms: Design and efficiency Date of exam: 8th December 2016 Exam hours: 09:00 13:00 (4 hours) Exam paper consists of: 6 pages (including attachment) Attachment: 1 page Permitted materials: All written and printed Make sure that your copy of this examination paper is complete before answering. You can give your answers in English or in Norwegian, as you like. Read the text carefully, and good luck! Including answers Assignment 1 (16%) Note: his exercise can be answered without any knowledge of bio-informatics, DNA or amino acids, we are really just looking at strings. Question 1.a We are given a string = AAA. Your tasks: (1) Draw an uncompressed suffix tree for string. (2) Use your drawing to explain what the specific running time for checking whether the pattern P = AAA is a substring of is. Let processing of one letter in P represent one step, and use that as the unit in your answer. Do not include the running time for building the suffix tree. (3) Give a general expression for the running time for a pattern of n letters. Answer 1.a he suffixes of = AAA are as follows: A AA AAA 1

2 A A A earching for the pattern P = AAA will have a running time of 6 (steps) (indicated in white). he general expression for a pattern of length n is O(n). Question 1.b is part of a slightly longer string ꞌ = AAGGAAAA. his longer string happens to be a DNA sequence encoding the amino acid sequence a = KDK. he encoding is simply a translation where triples of letters in the DNA sequence are mapped into letters of a different alphabet. he alphabet for DNA sequences consists of four letters; so there are 4 3 = 64 possible triplets that map to an alphabet of about 20 letters representing the amino acids. In our example we have the following five mappings: AAG = K, GA = D, AAA = K, =, =. Your tasks: (1) Draw an uncompressed suffix tree for string a. (2) Use your drawing to explain what the specific running time for checking whether the pattern P a = K is a substring of a is. Let processing of one letter in P a represent one step, and use that as the unit in your answer. Do not include the running time for building the suffix tree. 2

3 (3) ompare this running time with the one from Question 1.a. In reality we searched for the same string. What part does the encoding (the length of the input) play here? Answer 1.b he suffixes of = AAA are as follows: K DK KDK D K K D K earching for the pattern P a = K will have a running time of 2 (steps) (indicated in white). When we represent three letters from 1.a as one letter here in 1.b, comparing two letters will not be directly comparable operations. In 1.a we can assume that one letter comparison will consist of 2 bit comparisons (assuming four letters in the alphabet). In 1.b on letter comparison will consist of 2*3 bit comparisons. his makes up for the difference in running time when we only count steps: 6 steps (6*2 = 12 bit comparisons ) versus 2 steps (2*2*3 = 12 bit comparisons ). Question 1.c he Horspool variant if the Boyer-Moore algorithm is also an option for checking whether the pattern P = AAA is part of the string = AAA. ompute the shift values as used by the Horspool algorithmm, the alphabet is A = {A,,G, (the four letters used to make up DNA sequences). Answer 1.c hift (A) = 3, hift () = 1, hift (G) = 6, hift() = 2. 3

4 Assignment 2 (20%) We are given a boolean matrix B with values either (true) or F (false), and with indices i from 0 to m-1 and j from 0 to n-1. he aim is to find the size of the largest (connected) square within the matrix containing only values (and by size of a square we mean the length of the edges). For a given matrix B we shall denote this maximal size max(b), and we will find its value using dynamic programming. By definition we shall say that, if B contains only F-values, then max(b)=0. A sub-square of the matrix with only -values is called a -square. In the figure below a matrix B (with m=6 and n=7) is given, and blank entries have value F. We see that in this case max(b) = 3. We will use an integer matrix of size m x n as a table for our computation. We give two alternatives for what the element [i, j] ( i=0 to n-1, j=0 to n-1 ) should contain, but it turns out that only one of them will work properly for our task. In 2.b you will have to decide which one this is. 1. [i, j] contains the size of the largest -square within the rectangular area between B[0,0] and B[i, j]. 2. [i, j] contains the size of the largest -square that includes B[i, j] in the rectangular area between B[0,0] and B[i, j]. Question 2.a he initialization of the first row and the first column (for i = 0 or j = 0) of the table will depend on whether you use rule 1 or 2 above. Describe for each of these cases how you would initialize the matrix. You can describe it either in English/Norwegian, or by sketching a few program lines for each of the cases. Answer 2.a he answers for Rule 1 and for Rule 2 are: // Rule 1 (Value 1 if a occurres earlier or now in the row/column, otherwise 0): boolean haveeen; int i, j; haveeen = false; for (i = 0; i < m; i++) { if (B[i,0] == true) { haveeen = true; // or only: if (B[i,0]) { haveeen = true; if (haveeen == true) { [i, 0] = 1; else { [i, 0] = 0; haveeen = false; for (j = 0; j < n; j++) { 4

5 if (B[0,j] == true) { haveeen = true; if (haveeen == true) { [0, j] = 1; else { [0, j] = 0; // Rule 2 (Value 1 if the corresponding entry in B is, otherwise 0): int i, j; for (i = 0; i < m; i++) { if (B[i,0] == true) { [i, 0] = 1; else { [i, 0] = 0; for (j = 0; j < n; j++) { if (B[0,j] == true) { [0, j] = 1; else { [0, j] = 0; Question 2.b You shall here find a recurrence formula that can solve the problem above for i > 0 and j > 0. For this you should choose the most suited of the rules 1 and 2 above, and write the recurrence formula you obtain with this rule. Explain why your formula will work, and why it won t miss any of the possible -squares. Answer 2.b It seems difficult to solve this problem with rule 1 above, but with rule 2 we can try with the following recurrence formula : for 0 < i < m and 0 < j < m: if (B[i, j] == false) { // Or if (! B[i, j] ) [i, j] = 0; else { [i, j] = min([i-1, j], [i, j-1], [i-1, j-1]) + 1; Here the first case is easy to verify: When B[i, j] == false we can obviously not have any - square that includes B[i, j] at all, and then [i, j] = 0 is correct. However, if B[i, j] == true, then [i, j] should be at least 1, and from a simple drawing one will see that [i, j] cannot be larger than any of the three values: (1): [i, j-1] +1 (2): [i-1, j] +1 (3): [i-1, j-1] +1 F somewhere here ase (1), and by symmetry case (2), above: j j B: : i i 3? B[i, j-1] [i, j-1] annot be larger than 4 5

6 F somewhere here j B: : i ase (3): i 3? j B[i-1, j-1] [i-1, j-1] annot be larger than 4 Also if: m = min( [i-1, j], [i, j-1], [i-1, j-1] ) + 1 then we will know that there will be a -square (with B[i, j] as its lower right corner) of size m as indicated by the following figure: All here, because of (3) B: i j All here, because of (2) All here, because of (1) here, because of the if-test hus, as all possible -squares must have a lower right corner, and we have been through all possibilities for such a corner, we must find the largest of these by the above procedure. Question 2.c Write a procedure/method/function max( ) that receives the matrix B and its size as parameter(s), generates and fills the matrix, and delivers the size of the largest -square of B. Also include the initialization. Note that we are only interested in the size of the largest - square, not where it is positioned. You may generate the matrix by the following statement (or something corresponding to that): = new integer array[m, n] 6

7 Answer 2.c In a suitably language, it can be written as follows: integer function max(boolean array B[m, n], integer m, integer n) { integer i, j, maxize; for (i = 0; i < m; i++) { if (B[i,0] == true) { [i, 0] = 1; else { [i, 0] = 0; for (j = 1; j < n; j++) { // We don t look at B[0, 0] twice if (B[0,j] == true) { [0, j] = 1; else { [0, j] = 0; maxize = 0; for ( i =1, i < m, i++) { for (j = 1; j < n; j++) { if (B[i, j] == false) { [i, j] = 0; else { [i, j] = 1 + min([i-1, j], [i, j-1], [i-1, j-1]); if ([i, j] > maxize) maxize = [i, j]; // an be moved to the else-case above return maxize; Question 2.d Evaluate briefly whether or not a top-down (recursive) dynamic programming solution would be preferable for the above problem. Answer 2.d his was probably not easy, so we considered it a reasonable answer to say something like: here seem to be no obvious way to avoid making recursive calls for all the n x m entries of B. Notice e.g. that if B contain only F-values except for one -value, this -value can occur anywhere in B (e.g. in the corners), so we cannot stop the recursion without looking into all corners. It is not a good answer if you indicate that you can avoid making recurive calls if B[i, j] = false. his will not work! hus, with the well-known overhead connected to calling recursive functions, a top-down method will probably use more time than a bottom up version. If we are very eager to find ways to avoid looking through all entries in B, one way we could consider is to try to detect situations satisfying the following conditions: (1) you have already found a -square of size s, (2) you know that no -square in the explored area can be extended (e.g. because of a full column/row of F-values) and (3) the length of the smallest dimension of the unexplored territory do not exceed s. If all these conditions are satisfied, 7

8 you can obviously stop the search for a larger -square. However, to detect such a situation in an efficient way seems difficult. In addition, it will probably not occur very often. Assignment 3 (16%) Which of the following languages are undecidable? Prove your answers. Question 3.a {M uring machine M never writes anything on its tape when started on a blank input Question 3.b {M uring machine M never writes an A on its tape when started on a blank input Question 3.c {M Whenever uring machine M writes an A on its tape, it writes a B immediately afterwards, when started on a blank input Answer 3.a, 3.b and 3.c a. Decidable without righting anything on its tape, after a constant number of steps (equal to the number of states of M), M must repeat the same state and reveal that it is in an infinite loop. o a decision procedure needs to simulate M only a constant number of steps to find the answer. b. Undecidable proof by a modification of the standard reduction from the Halting problem, where M is imulate M on x (the M and the x that are the instance of Halting) and writes an A. It must also be secured that an A cannot be written otherwise (for ex. By replacing A in the alphabet by another symbol). wap Yes and No to get the answer to Halting. c. Undecidable the proof is the same as in (b). Assignment 4 (12%) When NP-complete problems occur in practice there are often special conditions that might make the problem simpler. Are the following simplified versions of the corresponding known NP-complete problems still NP-complete? Prove your answers. Question 4.a While studying DM (three-dimensional matching), we used an example with triples as compatibility constraints, each consisting of a boy, a girl and a car. However, in a real case all that matters is that the boy and the girl like the same car model of which there are of course many instances available. Model this problem as a graph problem. Is the problem still NP-complete? 8

9 Question 4.b Regarding the clique problem (to determine if input graph G has a complete graph of K nodes), one might say that the cliques are a subject of concern only when they become too large. Is the clique problem still NP-complete if the input instance is only a graph G, and the associated question is whether G has a clique that includes at least one half of its vertices? Answer 4.a and 4.b a. he problem is no longer NP-complete. It becomes the matching problem when a compatible pair (a boy and a girl) have no car model that they both like, their compatibility (edge) is removed; then the problem is solved by a bipartite matching algorithm. b. he problem remains NP-complete. he proof is an easy reduction from the clique, where we create an instance G of the new problem by adding m vertices to the input graph G. m is chosen so that m + k = (m + n) / 2. Each of the new m vertices is made adjacent to every vertex in G, and to all other new vertices. hen the new graph G has a clique that includes half of its vertices ((m + n) / 2) if and only if G has a clique of size k. Assignment 5 (16%) We are given the following set P of points in the plane: p 1 p 2 p 3 p 4 p 5 p 6 p 7 Figure 1. Point set P Question 5.a he convex hull of point set P is uniquely defined. Describe the convex hull of P by listing its corners in a well-ordered manner. Answer 5.a he corners must be given either clockwise or counter clockwise order, from an arbitrarily chosen staring point when we describe a polygon. ounter clockwise is the most common way. o, for instance: 9

10 H = p6, p7, p3, p1, p2. (H = p1, p3, p7, p6, p2 is also OK) (BU H = p1, p2, p3, p6, p7 is NO correct) Question 5.b here are many triangulations of the point set P. Draw the Delaunay triangulation of P. Justify/prove your answer. Answer 5.b ee figure. Quadrangles p6, p5, p4, p2 and p5, p7, p3, p4 may have confused the students. It is hard to see by looking at the drawing whether these four points lie on the same circle or not. Assuming they do, both possible triangulations of each of the quadrangles are legal Delaunay triangulations. Assuming they don t, one must argue that p5 lies outside the circumcircle of triangles p6, p4, p2 and p7, p4, p3. An argument based on circumcircles is what is required, both assumptions for quadrangles p6, p5, p4, p2 and p5, p7, p3, p4 (all four points on the same circle, or p5 outside) should be considered correct. Question 5.c Draw another triangulation of P and explain why it is not a Delaunay triangulation. Answer 5.c A triangulation containing p6, p1, p2 as a triangle will for instance be particularly bad. You may use the copy of Figure 1 included twice in the appendix to give your answers for Question 5.b and 5.c above. ear out the page and hand it in with the white version of your answers. 10

11 Assignment 6 (20%) Below you find questions from different areas discussed in the course. In your answers you can use facts included in the curriculum, even if no proof is given there. Question 6.a Alice claims that, in a general graph with a given matching M, there will always be an augmenting path (relative to M) if there exists a matching with more edges than in M. Bob says that Alice s claim is wrong If you think Alice is right give an argument showing that she is, otherwise give an example showing that Bob is right. Answer 6.a Here, Alice is right. he way one can prove this from facts discussed in the course is to look at the algorithm (with trees and blossoms etc.) for finding a matching with as many edges as possible in a general graph. It is stated there (without any proof) that if we have a matching M and there exists a larger matching, then this algorithm will find an augementing path. hus, it follows that Alice is right. Question 6.b In a (general) graph G, we shall say that a subset of nodes such that all edges have at least one end in the set is an edge covering set. We can easily prove that the size of is at least as large as (the number of edges in) any matching M in G (and you need not prove that). We look at the following claims: Alice: In a (general) graph, there will always be a matching M and an edge covering set so that the sizes of M and are the same. Bob says that Alice s claim is wrong If you think Alice is right give an argument showing that she is, otherwise give an example showing that Bob is right. Answer 6.b Here Bob is right. he simple example is a triangle graph where a matching can have at most one edge, but two nodes are needed to cover all the edges. However, for bipartite graphs Alice s claim is correct. Question 6.c We consider the N-puzzle from the mandatory exercise, and the following two claims about heuristics for the A* algorithm: Alice: If we allow diagonal moves, and the cost of each move is 1, then the sum of the Manhattan distances between the current position and the correct position for all tiles is a non-monotone heuristic, but simply counting the number of misplaced tiles will be a monotone heuristic. he empty position does not count as a tile. Bob says that Alice s claim is wrong. If you think Alice is right, give an argument showing that she is, otherwise give an example showing that Bob is right. 11

12 Answer 6.c Here Alice is right. For the first part of her claim a simple drawing of a 2x2 board should suffice as an example: Here only one move is needed, but its Manhattan heuristic will be 2. his is against the requirements for monotone heuristics, which says that that the heuristics for a situation should never exceed the fewest number of steps to the final position. For the second part, we first state that the heuristic of the correct (final) situation is 0, as required. We then observe that in each move, at most one tile can move from an incorrect to a correct position, so that the heuristic for the whole board can diminish at most 1 in one move. hus, if we do one move from situation s1 to situations s2, we will know that h(s1) 1+h(s2), and the second requirement for monotonicity is therby satisfried. Question 6.d Alice claims that no NP-hard problems have a polynomial-time ɛ-approximation algorithm for ɛ = 0.1 Bob says that Alice s claim is wrong If you think Alice is right give an argument showing that she is, otherwise give an example showing that Bob is right. Answer 6.d Bob is right. It was shown in class that the 0-1 Knapsack and some other problems have Polynomial ime Approximation chemes algorithms that take /epsilon as part of the input instance, and produce a solution to that instance that is within the /epsilon interval from the optimum, in polynomial time (where the shape of the polynomial depends on /epsilon). Hence this is also true for the specific /epsilon, 0.1 Question 6.e Alice claims that all NP-hard problems have a polynomial-time ɛ-approximation algorithm for ɛ = 0.1 Bob says that Alice s claim is wrong If you think Alice is right give an argument showing that she is, otherwise give an example showing that Bob is right. Answer 6.e Bob is right. It was shown in class that the raveling alesperson s Problem has no /epsilon approximation for any constant /epsilon, unless P=NP. [END] 12

13 Attachment for answering Assignment 5 ide nr.: Emne : INF 4130 Kandidatnr. : Dato : 8. des, 2016 p 1 p 2 p 3 p 4 p 5 p 6 p 7 p 1 p 2 p 3 p 4 p 5 p 6 p 7 6