Prof. Swarat Chaudhuri COMP 482: Design and Analysis of Algorithms Spring 2012 Lecture 17

Q1: Longest palindromic subsequence Give an algorithm to find the longest subsequence of a given string A that is a palindrome. “amantwocamelsacrazyplanacanalpanama” 2

Q1-a: Palindromes (contd.) Every string can be decomposed into a sequence of palindromes. Give an efficient algorithm to compute the smallest number of palindromes that makes up a given string. 3

6.5 RNA Secondary Structure

5 RNA Secondary Structure RNA. String B = b 1 b 2  b n over alphabet { A, C, G, U }. Secondary structure. RNA is single-stranded so it tends to loop back and form base pairs with itself. This structure is essential for understanding behavior of molecule. G U C A GA A G CG A U G A U U A G A CA A C U G A G U C A U C G G G C C G Ex: GUCGAUUGAGCGAAUGUAACAACGUGGCUACGGCGAGA complementary base pairs: A-U, C-G

6 RNA Secondary Structure Secondary structure. A set of pairs S = { (b i, b j ) } that satisfy: n [Watson-Crick.] S is a matching and each pair in S is a Watson- Crick complement: A-U, U-A, C-G, or G-C. n [No sharp turns.] The ends of each pair are separated by at least 4 intervening bases. If (b i, b j )  S, then i < j - 4. n [Non-crossing.] If (b i, b j ) and (b k, b l ) are two pairs in S, then we cannot have i < k < j < l. Free energy. Usual hypothesis is that an RNA molecule will form the secondary structure with the optimum total free energy. Goal. Given an RNA molecule B = b 1 b 2  b n, find a secondary structure S that maximizes the number of base pairs. approximate by number of base pairs

7 RNA Secondary Structure: Examples Examples. C GG C A G U U UA A UGUGGCCAU GG C A G U UA A UGGGCAU C GG C A U G U UA A GUUGGCCAU sharp turncrossingok G G 44 base pair

8 RNA Secondary Structure: Subproblems First attempt. OPT(j) = maximum number of base pairs in a secondary structure of the substring b 1 b 2  b j. Difficulty. Results in two sub-problems. n Finding secondary structure in: b 1 b 2  b t-1. n Finding secondary structure in: b t+1 b t+2  b n-1. 1 tn match b t and b n OPT(t-1) need more sub-problems

9 Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1  b j. n Case 1. If i  j - 4. – OPT(i, j) = 0 by no-sharp turns condition. n Case 2. Base b j is not involved in a pair. – OPT(i, j) = OPT(i, j-1) n Case 3. Base b j pairs with b t for some i  t < j - 4. – non-crossing constraint decouples resulting sub-problems – OPT(i, j) = 1 + max t { OPT(i, t-1) + OPT(t+1, j-1) } Remark. Same core idea in CKY algorithm to parse context-free grammars. take max over t such that i  t < j-4 and b t and b j are Watson-Crick complements

10 Bottom Up Dynamic Programming Over Intervals Q. What order to solve the sub-problems? A. Do shortest intervals first. Running time. O(n 3 ). RNA(b 1,…,b n ) { for k = 5, 6, …, n-1 for i = 1, 2, …, n-k j = i + k Compute M[i, j] return M[1, n] } using recurrence i 6789 j

6.8 Shortest Paths

12 Shortest Paths Shortest path problem. Given a directed graph G = (V, E), with edge weights c vw, find shortest path from node s to node t. Ex. Nodes represent agents in a financial setting and c vw is cost of transaction in which we buy from agent v and sell immediately to w. s 3 t allow negative weights

13 Shortest Paths: Failed Attempts Dijkstra. Can fail if negative edge costs. Re-weighting. Adding a constant to every edge weight can fail. u t sv st

14 Shortest Paths: Negative Cost Cycles Negative cost cycle. Observation. If some path from s to t contains a negative cost cycle, there does not exist a shortest s-t path; otherwise, there exists one that is simple. st W c(W) <

15 Shortest Paths: Dynamic Programming Def. OPT(i, v) = length of shortest v-t path P using at most i edges. n Case 1: P uses at most i-1 edges. – OPT(i, v) = OPT(i-1, v) n Case 2: P uses exactly i edges. – if (v, w) is first edge, then OPT uses (v, w), and then selects best w-t path using at most i-1 edges Remark. By previous observation, if no negative cycles, then OPT(n-1, v) = length of shortest v-t path.

16 Shortest Paths: Implementation Analysis.  (mn) time,  (n 2 ) space. Finding the shortest paths. Maintain a "successor" for each table entry. Shortest-Path(G, t) { foreach node v  V M[0, v]   M[0, t]  0 for i = 1 to n-1 foreach node v  V M[i, v]  M[i-1, v] foreach edge (v, w)  E M[i, v]  min { M[i, v], M[i-1, w] + c vw } }

17 Shortest Paths: Practical Improvements Practical improvements. n Maintain only one array M[v] = shortest v-t path that we have found so far. n No need to check edges of the form (v, w) unless M[w] changed in previous iteration. Theorem. Throughout the algorithm, M[v] is length of some v-t path, and after i rounds of updates, the value M[v] is no larger than the length of shortest v-t path using  i edges. Overall impact. n Memory: O(m + n). n Running time: O(mn) worst case, but substantially faster in practice.

18 Bellman-Ford: Efficient Implementation Push-Based-Shortest-Path(G, s, t) { foreach node v  V { M[v]   successor[v]   } M[t] = 0 for i = 1 to n-1 { foreach node w  V { if (M[w] has been updated in previous iteration) { foreach node v such that (v, w)  E { if (M[v] > M[w] + c vw ) { M[v]  M[w] + c vw successor[v]  w } If no M[w] value changed in iteration i, stop. }

19 Dynamic Programming Summary Recipe. n Characterize structure of problem. n Recursively define value of optimal solution. n Compute value of optimal solution. n Construct optimal solution from computed information. Dynamic programming techniques. n Binary choice: weighted interval scheduling. n Multi-way choice: segmented least squares. n Adding a new variable: knapsack. n Dynamic programming over intervals: RNA secondary structure. Top-down vs. bottom-up: different people have different intuitions. Viterbi algorithm for HMM also uses DP to optimize a maximum likelihood tradeoff between parsimony and accuracy CKY parsing algorithm for context-free grammar has similar structure

6.10 Negative Cycles in a Graph

21 Detecting Negative Cycles Lemma. If OPT(n,v) = OPT(n-1,v) for all v, then no negative cycles. Pf. Bellman-Ford algorithm. Lemma. If OPT(n,v) < OPT(n-1,v) for some node v, then (any) shortest path from v to t contains a cycle W. Moreover W has negative cost. Pf. (by contradiction) n Since OPT(n,v) < OPT(n-1,v), we know P has exactly n edges. n By pigeonhole principle, P must contain a directed cycle W. n Deleting W yields a v-t path with < n edges  W has negative cost. v t W c(W) < 0

22 Detecting Negative Cycles Theorem. Can detect negative cost cycle in O(mn) time. n Add new node t and connect all nodes to t with 0-cost edge. n Check if OPT(n, v) = OPT(n-1, v) for all nodes v. – if yes, then no negative cycles – if no, then extract cycle from shortest path from v to t v t

23 Detecting Negative Cycles: Summary Bellman-Ford. O(mn) time, O(m + n) space. n Run Bellman-Ford for n iterations (instead of n-1). n Upon termination, Bellman-Ford successor variables trace a negative cycle if one exists. n See p. 288 for improved version and early termination rule.

Q2. Arbitrage Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currenct into more than one unit of the same currency. For example, suppose that 1 US dollar buys 0.7 British pound, 1 British pound buys 9.5 French francs, and 1 French franc buys 0.16 US dollar. Then, by converting currencies, a trader can start with a US dollar and buy 0.7 x 9.5 x 0.16 = 1.064$ US dollars, thus turning a profit of 6.4 percent. Suppose that we are given n currencies c 1,…, c n and an n x n table R of exchange rates, such that one unit of currency c i buys R[i,j] units of currency c j. Give an efficient algorithm to determine whether or not there exists a sequence of currencies (c i1, …, c ik ) such that R[i 1, i 2 ] x R[i 2, i 3 ] x… x R[i k-1,i k ] x R[i k,i 1 ] > 1. Give an efficient algorithm to print out such a sequence if one exists. Analyze the running time of your algorithm. 24

Q3. Number of shortest paths Suppose we have a directed graph with costs on the edges. The costs may be positive or negative, but every cycle in the graph has a strictly positive cost. We are also given two nodes v, w. Give an efficient algorithm that computes the number of shortest v-w paths in G. 25