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Week 11 - Monday
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What did we talk about last time? Binomial theorem and Pascal's triangle Conditional probability Bayes’ theorem
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As I was going to St. Ives I crossed the path of seven wives Every wife had seven sacks Every sack had seven cats Every cat had seven kittens Kittens, cats, sacks, wives How many were going to St. Ives?
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Let sample space S be a union of mutually disjoint events B 1, B 2, B 3, … B n Let A be an event in S Let A and B 1 through B n have non-zero probabilities For B k where 1 ≤ k ≤ n
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Bayes' theorem is often used to evaluate tests that can have false positives and false negatives Consider a test for a disease that 1 in 5000 people have The false positive rate is 3% The false negative rate is 1% What's the probability that a person who tests positive for the disease has the disease? Let A be the event that the person tests positively for the disease Let B 1 be the event that the person actually has the disease Let B 2 be the event that the person does not have the disease Apply Bayes' theorem
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If events A and B are events in a sample space S, then these events are independent if and only if P(A B) = P(A)∙P(B) This should be clear from conditional probability If A and B are independent, then P(B|A) = P(B)
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Student Lecture
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A graph G is made up of two finite sets Vertices: V(G) Edges: E(G) Each edge is connected to either one or two vertices called its endpoints An edge with a single endpoint is called a loop Two edges with the same sets of endpoints are called parallel Edges are said to connect their endpoints Two vertices that share an edge are said to be adjacent A graph with no edges is called empty
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Graphs can be used to represent connections between arbitrary things Streets connecting towns Links connecting computers in a network Friendships between people Enmities between people Almost anything…
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We can represent graphs in many ways One is simply by listing all the vertices, all the edges, and all the vertices connected by each edge Let V(G) = {v 1, v 2, v 3, v 4, v 5, v 6 } Let E(G) = {e 1, e 2, e 3, e 4, e 5, e 6, e 7 } Edges connect the following vertices: Draw the graph with the given connections EdgeVertices e1e1 {v 1, v 2 } e2e2 {v 1, v 3 } e3e3 e4e4 {v 2, v 3 } e5e5 {v 5, v 6 } e6e6 {v5}{v5} e7e7 {v6}{v6}
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Graphs can (generally) be drawn in many different ways We can label graphs to show that they are the same Label these two graphs to show they are the same:
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A simple graph does not have any loops or parallel edges Let n be a positive integer A complete graph on n vertices, written K n, is a simple graph with n vertices such that every pair of vertices is connected by an edge Draw K 1, K 2, K 3, K 4, K 5 A complete bipartite graph on (m, n) vertices, written K m,n is a simple graph with a set of m vertices and a disjoint set of n vertices such that: There is an edge from each of the m vertices to each of the n vertices There are no edges among the set of m vertices There are no edges among the set of n vertices Draw K 3,2 and K 3,3 A subgraph is a graph whose vertices and edges are a subset of another graph
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The degree of a vertex is the number of edges that are incident on the vertex The total degree of a graph G is the sum of the degrees of all of its vertices What's the relationship between the degree of a graph and the number of edges it has? What's the degree of a complete graph with n vertices? Note that the number of vertices with odd degree must be even… why?
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Used to be Königsberg, Prussia Now called Kaliningrad, Russia On the Pregel River, including two large islands
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In 1736, the islands were connected by seven bridges In modern times, there are only five
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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
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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
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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
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A walk from v to w is a finite alternating sequence of adjacent vertices and edges of G, starting at vertex v and ending at vertex w A walk must begin and end at a vertex A path from v to w is a walk that does not contain a repeated edge A simple path from v to w is a path that does contain a repeated vertex A closed walk is a walk that starts and ends at the same vertex A circuit is a closed walk that does not contain a repeated edge A simple circuit is a circuit that does not have a repeated vertex other than the first and last
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We can always pin down a walk unambiguously if we list each vertex and each edge traversed How would we notate a walk that starts at v 1 and ends at v 2 and visits every edge exactly once in the following graph? However, if a graph has no parallel edges, then a sequence of vertices uniquely determines the walk v1v1 v2v2 v3v3 e1e1 e2e2 e3e3 e4e4
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Vertices v and w of G are connected iff there is a walk from v to w Graph G is connected iff all pairs of vertices v and w are connected to each other A graph H is a connected component of a graph G iff H is a subgraph of G H is connected No connected subgraph of G has H as a subgraph and contains vertices or edges that are not in H A connected component is essentially a connected subgraph that cannot be any larger Every (non-empty) graph can be partitioned into one or more connected components
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What if you want to find an Euler circuit of your own? If a graph is connected, non-empty, and every node in the graph has even degree, the graph has an Euler circuit Algorithm to find one: 1. Pick an arbitrary starting vertex 2. Move to an adjacent vertex and remove the edge you cross from the graph ▪ Whenever you choose such a vertex, pick an edge that will not disconnected the graph 3. If there are still uncrossed edges, go back to Step 2
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An Euler circuit has to visit every edge of a graph exactly once A Hamiltonian circuit must visit every vertex of a graph exactly once (except for the first and the last) If a graph G has a Hamiltonian circuit, then G has a subgraph H with the following properties: H contains every vertex of G H is connected H has the same number of edges as vertices Every vertex of H has degree 2 In some cases, you can use these properties to show that a graph does not have a Hamiltonian circuit In general, showing that a graph has or does not have a Hamiltonian circuit is NP-complete (widely believed to take exponential time) Does the following graph have a Hamiltonian circuit? e b ac d
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As you presumably know, a matrix is a rectangular array of elements An m x n matrix has m rows and n columns
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There are many, many different ways to represent a graph If you get tired of drawing pictures or listing ordered pairs, a matrix is not a bad way To represent a graph as an adjacency matrix, make an n x n matrix, where n is the number of vertices Let the nonnegative integer at a ij give the number of edges from vertex i to vertex j A simple graph will always have either 1 or 0 for every location
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What is the adjacency matrix for the following graph? What about for this one? v1v1 v1v1 v3v3 v3v3 v2v2 v2v2 v3v3 v3v3 v2v2 v2v2 v1v1 v1v1
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Draw a graph corresponding to this matrix
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What's the adjacency matrix of this graph? Note that the matrix is symmetric In a symmetric matrix, a ij = a ji for all 1 ≤ i ≤ n and 1 ≤ j ≤ n All undirected graphs have a symmetric matrix representation v1v1 v1v1 v4v4 v4v4 v2v2 v2v2 v3v3 v3v3
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To multiply matrices A and B, it must be the case that A is an m x k matrix and that B is a k x n matrix Then, the i th row, j th column of the result is the dot product of the i th row of A with the j th column of B In other words, we could compute element c ij in the result matrix C as follows:
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Multiply matrices A and B
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Matrix multiplication is associative That is, A(BC) = (AB)C Matrix multiplication is not commutative AB is not always equal to BA (for one thing, BA might not even be legal if AB is) There is an n x n identity matrix I such that, for any m x n matrix A, AI = A I is all zeroes, except for the diagonal (where row = column) which is all ones We can raise square matrices to powers using the following recursive definition A 0 = I, where I is the n x n identity matrix A k = AA k-1, for all integers k ≥ 1
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Is A symmetric? Compute A 0 Compute A 1 Compute A 2 Compute A 3
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We can find the number of walks of length k that connect two vertices in a graph by raising the adjacency matrix of the graph to the k th power Raising a matrix to the zero th power means you can only get from a vertex to itself (identity matrix) Raising a matrix to the first power means that the number of paths of length one from one vertex to another is exactly the number of edges between them The result holds for all k, but we aren't going to prove it
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Graph isomorphism Trees
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Keep reading Chapter 10
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