Week 10 - Friday What did we talk about last time? Shortest paths - PowerPoint PPT Presentation

Week 10 - Friday

 What did we talk about last time?  Shortest paths  Matching bipartite graphs  Stable marriage

 Lab hours  Wednesdays at 5 p.m. in The Point 113  Saturdays at noon in The Point 113  CS Club  Tuesdays at 5 p.m. in The Point 113 (or next door in The Point 112)

 Used to be Königsberg, Prussia  Now called Kaliningrad, Russia  On the Pregel River, including two large islands

 In 1736, the islands were connected by seven bridges  In modern times, there are only five

 After a lazy Sunday and a bit of drinking, the citizens would challenge each other to walk around the city and try to find a path which crossed each bridge exactly once

 Can you find such a solution?  Start anywhere and find a path which crosses each bridge exactly once

 What did Euler find?  The same thing you did: nothing  But, he also proved it was impossible  Here’s how: North Shore Center East Island Island South Shore

 By simplifying the problem into a graph, the important features are clear  To arrive as many times as you leave, the degrees of each node must be even (except for the starting and ending points) North Shore Center East Island Island South Shore

 There is actually an way to find such a path on a graph where one exists:  Start with a node of odd degree if there is one  Every time we move across an edge, delete it  If you have a choice, always pick an edge whose deletion will not disconnect the graph  At the end of the algorithm, you will either have an Eulerian path or an Eulerian cycle, depending on the graph

 We can label the nodes to make a more interesting problem  Now each piece of land is associated with the Blue Prince, the Red Prince, the Bishop, or the Tavern

 The Blue Prince wants to build an 8th bridge so that he can walk starting at his castle, cross every bridge once, and end at the Tavern to brag

 Put the bridge from the Bishop’s land to the Red Prince’s land

 Furious, the Red Prince wants to build his own bridge so that only he can start at his own castle, cross all the bridges once and then end at the Tavern

 Put the bridge from the Red Prince’s land to the Blue Prince’s land

 Upset by this pettiness, the Bishop decides to build a 10th bridge which allows all citizens to cross all bridges and return to their starting point

 Put the bridge from the Center Island to the Red Prince’s land, making all pieces of land have even degree

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 A flow network is a weighted, directed graph with positive edge weights  Think of the weights as capacities , representing the maximum units that can flow across an edge  An st -flow network has a source s (where everything comes from) and a sink t (where everything goes to)

 Oil flowing from a start to a destination  Airline crews needed to man aircraft, moving from city to city  Goods being produced by factories and consumed by cities, with roads that can accommodate a certain amount of traffic

 A common flow problem is to find the maximum flow  A maximum flow is a non-negative amount of flow on each edge such that:  The maximum amount of flow gets from s to t  No edge has more flow than its capacity  The flow going into every node (except s and t ) is equal to the flow going out

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 When we were talking about matching, we mentioned augmenting paths  Augmenting paths in flows are a little different  A flow augmenting path:  Starts at s and ends at t  May cross some edges in the direction of the edge (forward edges)  May cross some edges in the opposite direction (backwards edges)  Increases the flow by the minimum of the unused capacity in the forward edges or the maximum of the flow in the backwards edges

 Ford-Fulkerson is a family of algorithms for finding the maximum flow 1. Start with zero flow on all edges 2. Find an augmenting path (increasing flow on forward edges and decreasing flow on backwards edges) 3. If you can still find an augmenting path, go back to Step 2

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 An cut in a graph partitions the graph into two disjoint sets  An st -cut is a cut such that s is in one partition and t is in the other  Think of it as a line that slices through the edges, putting s on one side and t on the other  The capacity of a cut is the sum of the capacities of the edges that the cut divides

 The smallest capacity st -cut you can make has the same capacity as the largest possible st -flow  Intuitively, it's like that cut is a set of edges that most constricts the flow from s to t

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 Our definition of Ford-Fulkerson didn't say how you pick the augmenting path  At worst, it could take O(| E | f ), where | E | is the number of edges in the graph and f is the maximum flow  That could be terrible if f has a large numerical value  Edmonds-Karp is a variation of Ford-Fulkerson that uses a breadth-first search to find a shortest augmenting path  It runs in O(| V || E | 2 )

 Binary trees are great  However, only two splits means that you have a height of log 2 n when you want to store n things  If n = 1,000,000, log 2 n = 20  What if depth was expensive? Could we have say, 10 splits?  If n = 1,000,000, log 10 n = 6

 Answer: When the tree is in secondary storage  Each read of a block from disk storage is slow  We want to get a whole node at once  Each node will give us information about lots of child nodes  We don’t have to make many decisions to get to the node we want

 A B-tree of order m has the following properties: The root has at least two subtrees unless it is a leaf 1. 2. Each nonroot and each nonleaf node holds k keys and k + 1 pointers to subtrees where m /2 ≤ k ≤ m Each leaf node holds k keys where m /2 ≤ k ≤ m 3. 4. All leaves are on the same level

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12 5 8 13 15 Insert 7 12 5 7 8 13 15

12 2 5 7 8 13 15 Insert 6 6 12 2 5 7 8 13 15

 Insert the following numbers:  86 69 81 15 100 94 8 27 56 68 92 89 38 53 88

 When the list of keys drops below half m , we have to redistribute keys  In the worst case, we have to delete a level

 Instead of requiring every non-root node to be half full, every non-root node must be at least 2/3 full  Key redistribution becomes more complex  However, the tree is fuller

 Essentially, make a B-tree such that all the leaves are tied together in a linked list  It is also necessary that all keys in a B-tree appear as leaves  Some other variations are possible, but we’ll end the list here 6 12 2 5 6 7 8 12 13 15

 Finish B-trees  Intractability and NP-completeness

 Work on Project 3  Finish Assignment 5  Due tonight!