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:

Center Island North Shore East 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)

Center Island North Shore East Island South Shore

 There is actually an way to find such a path on a graph where

• ne 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

• nly 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

8 7 9 5 6 3 4 2 1

1 5 2 4 3

 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

s t a b c d e f 4 5 3 7 4 3 7 6 4 1 5 6

 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

s t a b c d e f 4 5 3 7 4 3 7 6 4 1 5 6

 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

• ther

 Think of it as a line that slices through the edges, putting s on

• ne 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

s t a b c d e f 4 5 3 7 4 3 7 6 4 1 5 6

 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 log2

n when you want to store n things

• If n = 1,000,000, log2 n = 20

 What if depth was expensive? Could we have say, 10 splits?

• If n = 1,000,000, log10 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: 1.

The root has at least two subtrees unless it is a leaf

• 2. Each nonroot and each nonleaf node holds k keys and k + 1 pointers

to subtrees where m/2 ≤ k ≤ m

3.

Each leaf node holds k keys where m/2 ≤ k ≤ m

• 4. All leaves are on the same level

50 10 15 20 70 80 6 8 11 12 16 18 21 25 27 29 54 56 71 76 81 89

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 13 15 7 8

 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 12 13 15 6 7 8

 Finish B-trees  Intractability and NP-completeness

 Work on Project 3  Finish Assignment 5

• Due tonight!