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courses:cs211:winter2011:journals:charles:chapter6 [2011/03/30 03:22] – [6.6 Sequence Alignment] gouldc | courses:cs211:winter2011:journals:charles:chapter6 [2011/03/30 04:29] – [6.8 Shortest Paths in a Graph] gouldc | ||
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This is because there are three possibilities for the mth letter in string 1 and the nth letter in string 2: they form a pair in the optimal alignment, the mth letter in string 1 is not in any pair in the optimal alignment, or the nth letter in string 2 is not in any pair in the optimal alignment. The dynamically programmed algorithm for this problem runs in O(mn) time because every pair (i,j) must be considered from 1...m and 1...n, and each of these takes at most constant time. It also takes O(mn) space. | This is because there are three possibilities for the mth letter in string 1 and the nth letter in string 2: they form a pair in the optimal alignment, the mth letter in string 1 is not in any pair in the optimal alignment, or the nth letter in string 2 is not in any pair in the optimal alignment. The dynamically programmed algorithm for this problem runs in O(mn) time because every pair (i,j) must be considered from 1...m and 1...n, and each of these takes at most constant time. It also takes O(mn) space. | ||
===== 6.7 Sequence Alignment in Linear Space via Divide and Conquer ===== | ===== 6.7 Sequence Alignment in Linear Space via Divide and Conquer ===== | ||
+ | This section shows us an improved algorithm for sequence alignment. It still runs in O(mn) time but it decreases the space requirement from O(mn) to O(m+n). We realize that computing a single column in the memoized array in the last problem only requires us to know about the current column and the previous column. Therefore it is possible to arrive at the solution while only maintaining a m-by-2 array instead of a m-by-n array. This idea will work nicely for building up to the solution //value// (i.e., the total cost of all gaps and mismatches)... but we also want to know the optimal // | ||
+ | We do this by incorporating divide-and-conquer principles into our algorithm. I'm slightly confused by the divide-and-conquer component to this problem. In class we did it with the graph in Figure 6.19. We started in the top left and found the shortest path to some midpoint, then we did the same thing from the bottom right. I will need to look in more depth at the pseudo-code on page 288 to get a better grasp of the concept. | ||
===== 6.8 Shortest Paths in a Graph ===== | ===== 6.8 Shortest Paths in a Graph ===== | ||
+ | This section is about finding the shortest path between two nodes in a directed graph. Each edge has a weight. If there are no negative cycles (cycles that decrease the total cost of the path), then we want to find the path from s -> t that minimizes total edge costs... the shortest path. | ||
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+ | starting node s and destination node t with minimum total cost. Each edge in the graph has an associated weight. | ||
+ | This section is about using a memoized array to keep track of the shortest paths in a graph. The new method is contrasted with Dijkstra' |