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CS 336 March 19, 2012 Tandy Warnow

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**Basic Graph Terminology**

Nodes, vertices, edges, degrees, paths, cycles, connected components, adjacency, isolated vertices, trees, forests Directed graphs: indegree, outdegree, trees

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**Advanced terminology Cliques Independent sets**

Chromatic number and vertex colorings Eulerian cycles and Eulerian paths Hamiltonian paths Matchings Dominating Set Vertex Cover

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**Paths, Connected Components, etc.**

A path is a sequence of vertices v1, v2, …, vn so that vi is adjacent to vi+1 for i=1,2,…,n-1. A simple path is one that does not have repeated vertices. A graph is connected if every pair of vertices in the graph is connected by some path. A connected component is a maximal subset of the vertices that is connected.

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Cycles A cycle in a graph is a path that starts and ends at the same vertex. A simple cycle is a cycle that does not have any repeated vertices (other than the start and end vertex). A graph is acylic if it has no simple cycles.

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**Trees Two types: rooted and unrooted**

Unrooted (simplest): acylic connected graph Rooted: take an unrooted tree, pick one node to be the root, and direct all edges away from the root. Voila!

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Theorems about trees Let T be a connected acyclic graph (i.e., a tree) with n vertices (n>0). Then: T has at least one leaf (node with degree 0 or 1). T has n-1 edges. Every edge in T is a cut-edge. Every tree can be 2-colored.

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**Theorem: Every tree has at least one leaf (node of degree 1)**

Theorem: For any tree T with at least one vertex, T has at least one leaf (node with degree 0 or 1). Proof: If n=1, then T is a single vertex which is a leaf. Else, n>1. Let P be a longest simple path in T, so P=v1,v2,…,vk. If vk has degree 1, we are done. Otherwise, vk has at least two neighbors, and so some neighbor w other than vk-1. If w is in P, then we have a simple cycle in T, contradicting that T is a tree. If w is not in P, then we can extend P and get a longer path, contradicting that P is a longest simple path in T. Hence, vk has degree 1, and we are done.

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**Theorem: Any tree with n>0 nodes has n-1 edges**

Proof: by induction on n. Base case: n=1 (trivial) Inductive hypothesis: for some positive n, any tree on n nodes has exactly n-1 edges. Let T be a tree on n+1 nodes. We want to show T has exactly n edges.

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**Proof (cont’d) Let v be a node in T with degree 1.**

Remove v from T. The result is a tree T’ with n nodes, and hence n-1 edges (by the inductive hypothesis) T’ contains one fewer edge and one fewer vertex (node) than T, and so T has n edges.

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**Theorem: every edge in a tree is a cut-edge**

Proof (by contradiction). Suppose T is a tree, e=(v,w) is an edge in T that is not a cut-edge. Then G=T-{e} (but keeping v and w) is connected. Hence there is a simple path P from v to w in G. Since e is not in G, P does not include edge e. Therefore, we can form a simple cycle C by adding edge e to P. Since every edge in C is in T, this means that T is not acyclic, contradicting the assumption that T is a tree (connected acyclic graph).

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Vertex Coloring A (proper) vertex coloring of a graph is a function c: V -> {1,2,…,k}, s.t. no two adjacent vertices are mapped to the same color. The chromatic number of a graph is the minimum number of colors needed to properly color the graph. How many colors does a tree need?

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2-coloring a tree Theorem: every connected acyclic graph (i.e., tree) can be 2-colored. Proof: by induction on the number of vertices.

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**Proof that every tree can be 2-colored**

Let G be a tree on n vertices. The base case is n=1. Clearly every tree on 1 vertex can be 2-colored. The Inductive Hypothesis is that for some positive integer n, any tree on n vertices can be 2-colored. Let G be a tree with n+1 vertices. We want to show that G can be 2-colored.

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Proof (cont’d) Let v be a node in G that has degree 1, and let w be its unique neighbor in G. Consider the graph G’ formed by deleting v (and its incident edge but not w) from G. G’ is also acyclic (why?) and has n-1 vertices. Therefore, by the inductive hypothesis, G’ can be 2-colored. We extend the coloring from G’ to G, by letting c(v) be 1 if c(w)=2, and c(v)=2 if c(w)=1. Note that this coloring is proper for G. Hence G can be 2-colored.

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**Structural Induction This was a proof by structural induction.**

Proofs by structural induction can be applied more generally!

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**Theorem about rooted trees**

A rooted tree in which every node has 0 or 2 children is called a “binary tree” Theorem: every binary tree with n nodes has (n-1)/2 internal nodes (defined to be nodes with more than 0 children). Proof: by strong induction on n. Base case: n=1. Such a tree has no internal nodes, so it is true.

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Proof, cont’d. Strong Inductive hypothesis: for some n>0, and for all positive integers k up to n, all rooted binary trees with k nodes have (k-1)/2 internal nodes. Let T have n+1 nodes, and let the children of the root be A and B. (We know the root has two children, since if it had no children, T would have 1 node, contradicting our hypothesis.) We want to show Int(T) = n/2

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**We want to show Int(T) = n/2**

TA, the subtree of T rooted at A, is a binary tree; let nA be the number of nodes in TA TB, the subtree of T rooted at B, is a binary tree; let nB be the number of nodes in TB Let Int(T) be the number of internal nodes of T, and Int(TA) and Int(TB) be similarly defined.

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**We want to show Int(T) = n/2**

Then nA and nB are both at most n, and by the inductive hypothesis Int(TA) = (nA-1)/2 Int(TB ) = (nB-1)/2 Therefore Int(T) = (nA-1)/2 + (nB-1)/2 + 1

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**We want to show Int(T) = n/2**

We have established that Int(T) = (nA-1)/2 + (nB-1)/2 + 1 Simplifying this, we get Int(T) = (nA-1 + nB )/2 = (nA + nB)/2 Note nT = nA + nB + 1 Therefore, Int(T) = (nT - 1)/2 Recall that nT = n+1. Therefore, Int(T) = n/2 Q.E.D.

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Genome Assembly Given a DNA sequence, technology can allow you to get a collection of k-mers (substrings of length k) that come from analyses of the sequence. From these k-mers, your objective is to come up with the sequence.

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**Genome Assembly Let X be a very long DNA sequence**

Consider all k-mers in X, with k big enough so that no k-mer appears two or more times Goal: reconstruct X from its set of k-mers

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**Genome Assembly, attempt #1**

Approach 1: Make a node for each k-mer, and put a directed edge from v to w if the k-1 suffix of v is the k-1 prefix of w. Create the graph for the following string, using k=5 ACATAGGATTCAC

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**Genome Assembly, attempt #1**

Approach 1: Make a node for each k-mer, and put a directed edge from v to w if the k-1 suffix of v is the k-1 prefix of w. Every such graph has a Hamiltonian Path, as long as no k-mer appears more than once!

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Hamiltonian Path A Hamiltonian Path in a graph visits every node exactly once

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**Genome Assembly Attempt #1**

Create the graph for the following string, using k=5 ACATAGGATTCAC Does the graph have a Hamiltonian Path? Is it unique? Can you reconstruct the sequence from the path?

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Hamiltonian Path A Hamiltonian Path in a graph visits every node exactly once Determining if a graph has a Hamiltonian Path is NP-Complete So this approach to Genome Assembly is computationally intensive (infeasible)

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Eulerian Cycles An Eulerian cycle is one that goes through every edge exactly once It is easy to see that if a graph has an Eulerian cycle, then every node has even degree. The converse is also true, but a bit harder to prove. For directed graphs, the cycle will need to follow the direction of the edges (also called “arcs”). In this case, a graph has an Eulerian cycle if and only if the indegree is equal to the outdegree for every node.

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Eulerian Paths An Eulerian path is one that goes through every edge exactly once It is easy to see that if a graph has an Eulerian path, then all but 2 nodes have even degree. The converse is also true, but a bit harder to prove. For directed graphs, the cycle will need to follow the direction of the edges (also called “arcs”). In this case, a graph has an Eulerian path if and only if the indegree(v)=outdegree(v) for all but 2 nodes (x and y), where indegree(x)=outdegree(x)+1, and indegree(y)=outdegree(y)-1.

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**de Bruijn Graph Input: the set of k-mers for the DNA sequence**

Output: the de Bruijn Graph Vertices: the (k-1)-mers Directed edges: from v->w if the (k-2)-suffix of v is the (k-2)-prefix of w, and the k-mer formed by starting with v and ending with w is one of the k-mers in the input

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de Bruijn Graph If the k-mer set comes from a sequence and no k-mer appears more than once in the sequence, then the de Bruijn graph has an Eulerian path!

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**Using de Bruijn Graphs Given: set of k-mers from a DNA sequence**

Algorithm: Construct the de Bruijn graph Find an Eulerian path in the graph The path defines a sequence with the same set of k-mers as the original

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de Bruijn Graph Create the de Bruijn graph for the following string, using k=5 ACATAGGATTCAC Find the Eulerian path Is the Eulerian path unique? Reconstruct the sequence from this path

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