Differences

This shows you the differences between two versions of the page.

--- courses:cs211:winter2011:journals:charles:chapter6 [2011/03/30 03:30] – [6.7 Sequence Alignment in Linear Space via Divide and Conquer] gouldc
+++ courses:cs211:winter2011:journals:charles:chapter6 [2011/04/05 05:29] – [6.8 Shortest Paths in a Graph] gouldc
@@ Line 48: / Line 48: @@
 ===== 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 //sequence// itself.
+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 =====
+Problem outline: We want to find the shortest path between two nodes in a directed graph. We cannot use Dijkstra's Algorithm because there can be negative weights. (We won't know whether we have grown the blob in the right way because there could be a negative edge in the future that provides a shorter path to some node already in the blob.) Therefore we want to build up partial solutions. The first partial solution is the cost of s->t that uses at most one edge; if s doesn't have an edge to t, then the value we put in the memoized array is infinity. (6.22) says that the longest possible optimal path has n-1 edges. So at most we would do n-1 iterations of this.
+Here we assume no negative cycles because otherwise the optimal path would follow that cycle infinitely many times and so reduce the cost of the path to negative infinity. But that would be stupid. We saw from class that like the problem above we can fill any column by referring only to the column itself and the column that came before it.
+(6.25) says that the shortest path algorithm can run in O(mn) time.
+===== 6.9 Shortest Paths and Distance Vector Protocols =====