CS431 homework 2. 8 June Question 1 (page 54, problem 2.3). Is lg n = O(n)? Is lg n = Ω(n)? Is lg n = Θ(n)?

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1 CS1 homework June 011 Question 1 (page, problem.). Is lg n = O(n)? Is lg n = Ω(n)? Is lg n = Θ(n)? Answer. Recall the definition of big-o: for all functions f and g, f(n) = O(g(n)) if there exist constants c and N such that for all n > N, f(n) c g(n). For big-omega, replace the with in the above definition. A function f(n) is big-theta of g(n) if f(n) = O(g(n)) and f(n) = Ω(g(n)). If we choose c = 1 and N = 0, then we immediately see that for all n > 0, lg n = 1 n. Therefore lg n = O(n). Since n < n for all n 1, it follows that lg n < lg n = n for all n 1. Even with a constant multiplicative factor on the left-hand side of the inequality, there is a great enough n such that the inequality will be true. This means that lg n Ω(n). Since lg n is not Ω(n), it is not Θ(n). Question. Assume you have a DNA base sequence P of n bases and are given a sequence S of m bases. The problem is to find the place in S where P first occurs, if there is one. 1. Describe a simple exhaustive search algorithms to solve this problem.. Assume and P = TCGATG S = TCGACTCGAATTGCCTCGATGGATCCATATCG. How many comparisons does your algorithm take before it finds where P occurs?. Assume P does not occur in S. How many comparisons does your algorithm take before it discovers this? Answer. 1. On input P = p 1 p p m and S = s 1 s s n the naïve exhaustive search algorithm is as follows: for i = 1 up to n m + 1 1

2 for j = 1 up to m if s i+j p j, break out of this inner loop, and continue to the next iteration of the outer loop if the inner loop terminates, output i if the outer loop terminates, then P does not exist in S. Each number in the following sum is the total number of comparisons when starting at position 1, position, and so on, in the string S: This sum equals.. In the worst case this algorithm performs m comparisons at n m + 1 positions in the string S. Therefore the running time of this algorithm if the string P does not exist in the string S is O(m(n m + 1)). Question (page 1, problem.10). A rook stands on the upper left square of a chessboard. Two players take turns moving the rook either horizontally to the right or vertically downward (as many squares as they want). The player who can place the rook on the lower right square of the chessboard wins. Who will win? Describe the winning strategy. Answer. Recall the analysis of the game we saw in class in which two players took turns moving a king from the upper left corner of the chess board to the bottom right corner. We will start analyzing from the bottom right corner and note that if player one were to start at any position on the bottom row or the right-most column, he or she could certainly win by simply moving the rook to the bottom right corner. See Figure 1a. We also notice that from position g, player one must lose, since he or she will move either down one or right one, allowing player two to move to the final position. See Figure 1b. Now from any position in column g above g, and from any position in row to the right of g, player one can force player two to lose by moving directly to position g. See Figure 1c. At position f, player one must lose, because moving either right or down by one or two allows player two to win. See Figure 1d. This pattern continues all the way back up towards the top left position. See Figure 1e. This means that if both players play optimally, player one will lose if he or she starts with the rook initially in the top left position. Therefore player two will win. The winning strategy for player two follows. On the first turn, player one must move the rook either right along the top row or down along the left-most column. Regardless of where player one moves the rook, player two should move the rook to the diagonal (where all the s are). This forces player one to a losing position for the next turn. Repeat this strategy after each move by player one, and player two will always win.

3 Figure 1: Analyzing the rook problem described in Question. (a) Player one can win from the bottom row or the right-most column. (b) Player one must lose from position g. W W W W W W W W W W W W W W (c) Player one can force player two to lose by moving directly to g. (d) Player one loses at f because any move allows player two to win. WW WW WW WWWWWW W WW WW WW WW WWWWWW W (e) The complete analysis of losing and winning starting positions for player one. W WWWWWW WW WWWWW WWW WWWW WWWW WWW WWWWW WW WWWWWW W

4 Question (page 1, problem.1). Two players play the following game with two sequences of length n and m nucleotides. At every turn a player must delete two nucleotides from one sequence (either the first or the second) and one nucleotide from the other. The player who cannot move loses. Who will win? Describe the winning strategy for each n and m. Answer. et the string of length m be called u and the string of length n be called v. Notice that either m or n (or both) may be zero, representing an empty string. We can view this game as a two player game on a chessboard of m + 1 rows and n + 1 columns in which there is a knight in the top left corner, similar to the game described in Question. The only valid moves for the knight in this game are two moves right and one move down, or two moves down and one move right. The two players take turns moving the knight, and the one who cannot make a move loses. Why is this the correct notion for the original game described in this question? Every move to the right represents a deletion of a letter from v, and every move down represents a deletion of a letter from u. Notice that placing the knight in the right-most column means that n characters have been deleted from v, so it is the empty string at this point. Similarly, placing the knight in the bottom column means that m characters have been deleted from u, so it is the empty string at this point. The board is of size (m + 1) (n + 1) to accomodate the possibility that either u or v is the empty string. As before, we start by examining the bottom right region of the board, where any position along the bottom row or the right-most column is a losing position. In addition, no move can be made from g, so it is also a losing position. See Figure a (which shows the game on an board, so m = n = ). Player one can force player two into a losing position from most of the positions outside these extreme positions, though. See Figure b. The analysis continues until we get the pattern seen in Figure c. The pattern would continue outward for a board of arbitrary size. Now we just need to determine under what conditions (that is, board sizes) player one will win or lose if the knight starts in the top right corner. It will be a little bit simpler to determine under what conditions player one will lose than to determine under what conditions player one will win. If m + 1 represents the number of rows and n + 1 represents the number of columns, then the following positions contain s: m = 0 and n 0, m = and n, m = and n,... n = 0 and m 0, n = and m, n = and m,... Expressed more concisely, m = n = 1, m = n =, m = n =,... m = k and n k for some k N (1) n = k and m k for some k N () m = n = k + 1 for some k N ()

5 Figure : Analyzing the knight problem described in Question for an board. (a) The bottom row and the right-most column are losing positions in the knight game. (b) Player one can force player two into a losing position from the next layer out. WW WW WW WWWWWW (c) The complete analysis of winning and losing starting positions for player one. WW WW WW WW WWWW WW WWW WW WW WWWWWW

6 (Here, N is the set of natural numbers including 0, {0, 1,,...}.) Therefore, if a board with m+1 rows and n +1 columns meets any of these three conditions, player one will lose. Under the complementary set of conditions, player one will win. Translating this back to the original problem, if the strings are of length m and n respectively, then player one will lose if m and n meet any of these conditions. The winning strategy is to move the knight onto one of these strips of losing positions, in order to force the next player to be in a losing position. Question. Describe how RNA works to carry information DNA and helps in making cell proteins. In particular, 1. What are the different forms of RNA and how is their structure different?. What are the different functions of the different forms of RNA?. What do ribosomes do and where are they located? Note: Don t write a book on this, just a page or so will do. This process is discussed on pages of the text but you may need a bit more research than that here. Answer. There are several different forms of ribonucleic acid (RNA). The main three significant types of RNA in humans (to know about for a bioinformatics class) are messenger RNA (mrna), ribosomal RNA (rrna), and transfer RNA (trna). Messenger RNA is a transcription of a portion of DNA which codes for a protein. In the nucleus, a molecule called RNA polymerase forms strands of mrna which are complementary to corresponding portions of DNA. The mrna is a molecular sequence of nucleotides, and each triple of nucleotides (called a codon ) codes for an amino acid, according to a fixed translation table. The translation table has size =, since there are four possible nucleotide bases (A, G, C, and T). Ribosomal RNA makes up the ribosome, an organelle which exists in the cytoplasm outside of the nucleus of the cell, which synthesizes proteins coded by mrna. The ribosome is more of a static collection of rrna and proteins which can be likened to a factory. The mrna is the template along which the ribosome moves, and as the ribosome reads each codon, it appends the amino acid delivered by the appropriate transfer RNA to a growing chain of amino acids called a polypeptide chain which will eventually become part of a protein. Each transfer RNA has an anticodon component, which matches a codon, and a site which binds to the amino acid to which the matching codon corresponds. There is at least one trna in the cell for each codon. When the ribosome reads a codon, it also allows a corresponding trna molecule to bind, and subsequently helps move the amino acid from the trna to the growing polypeptide chain created using previous trna molecules corresponding to prior codons read from mrna.

7 Table 1: One possible dynamic programming table for computing the longest common subsequence of GAGTACA and GCTAGGA. G A G T A C A G C T A 1 G 1 G 1 A 1 Question. Use the CS method in section. of the textbook to find the longest common subsequence of the two sequences v = GAGTACA and w = GCTAGGA. Answer. See Table 1 for one possible dynamic programming table, filled from the top left corner outward using the algorithm shown in the textbook, given the two strings GAGTACA and GCTAGGA. Once the table is filled in, we follow the path from the bottom right corner back to the top left corner (shown in bold) to get the alignment of the two strings. This results in the alignment G--AGTA-A GCTAG--CA giving the longest common subsequence GAGA of length four.

Fun and Games on a Chess Board

Fun and Games on a Chess Board Olga Radko November 19, 2017 I Names of squares on the chess board Color the following squares on the chessboard below: c3, c4, c5, c6, d5, e4, f3, f4, f5, f6 What letter