Induction

Induction

Great Theoretical Ideas In Computer Science Victor Adamchik CS 15-251 Lecture 8 Graphs - I Carnegie Mellon University Plan Graph Representations Counting Trees Cayleys Formula Prfer Sequence Minimum Spanning Trees Planar Graphs

Eulers Polyhedra Theorem Definition A graph G is a pair (V,E) where V is a set of vertices (or nodes) E is a set of edges connecting the vertices A A self-loop is an edge that connects to the same vertex twice A multi-edge is a set of two or B more edges that have the same two vertices A graph is simple if it has no multi-edges or self-loops.

C D E More terms Directed: an edge is an ordered pair of vertices Undirected: edge is unordered pair of vertices Weighted: (a cost associated with an edge) Path (is a sequence of vertices) Cycle (the start and end vertices are the same) Acyclic Connected or Disconnected The degree of a vertex (in an undirected graph is the number of edges associated with it.) The handshaking theorem Let G=(V,E) be an undirected graph

with V vertices and E edges. Then 2 E deg(x) xV In a directed graph: E indeg(x) outdeg(x) xV xV Exercise Given a graph with 7 vertices; 3 of them of degree two and 4 of degree one. Is this graph is connected? 2 E deg(x)

xV No, the graph has only 5 edges. Representing Graphs Adjacency List or Adjacency Matrix Vertex X is adjacent to vertex Y if and only if there is an edge (X, Y) between them. Adjacency List Representation C D E A

B A B E B E D C A D

D null E C Adjacency Matrix Representation 2 3 4 0 1

0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0

1 1 0 0 1 1 0 0 0 Representing Graphs Adjacency List Representation is used for representation of the sparse graphs. Adjacency Matrix Representation is used for representation of the dense graphs. Trees A tree is a connected simple

graph without cycles. Not Tree Not Tree Theorem: Let G be a graph with V nodes and E edges The following are equivalent (TFAE) : 1. G is a tree (connected, acyclic) 2. Every two nodes of G are joined by a unique path 3. G is connected and V = E + 1 4. G is acyclic and V = E + 1 5. G is acyclic and if any two nonadjacent nodes are joined by an edge, the resulting To prove this, it suffices to show

123451 Well just show 1234 and leave the rest to the reader 1 2 1. G is a tree (connected, acyclic) 2. Every two nodes of G are joined by a unique path Proof: (by contradiction) Assume G is a tree that has two nodes connected by two different paths: Then there exists a cycle! 23 2. Every two nodes of G are

joined by a unique path 3. G is connected and V = E + 1 Proof: (by strong induction) Assume true for every graph with < V vertices Let G have V nodes and let x and y be adjacent G1 x y G2 Then V = V1 + V2 = E1 + E2 + 2 = E + 1

3 4 3. G is connected and V = E + 1 4. G is acyclic and V = E + 1 Proof: by contradiction Assume, G has a cycle with k vertices. Start adding nodes and edges until you cover the whole graph. Number of edges in the graph will be at least V, since the cycle has k vertices and k edges. Corollary: Every nontrivial tree has at least two vertices of degree 1. Proof (by contradiction): Assume all but one of the vertices in the tree have degree at least 2. Can we? In any graph, sum of the degrees = 2E Under our assumption 2E = degi 2(V1)+1 Then the total number of edges in the

tree is at least E (2V-1)/2 = V - 1/2 > V-1 Contradiction, since in a tree E = V - 1 Cayleys Formula Using the above property, we can now begin to discuss Cayleys formula that tells us how many different trees we can construct on n vertices. How many labeled trees are there with three nodes? Two labeled trees with the same set of labels are isomorphic iff they have the same adjacency matrix. 2

1 3 3 1 2 1 2 3 3

2 1 1 3 2 2 3 1 How many labeled trees are there with four nodes? 1

4 2 3 These are called spanning trees, for a complete graph of 4 vertices. How many labeled trees are there with five nodes?

5 labelings 5x4x3 labelings 5!/2 labelings 125 labeled trees Cayleys Formula (1889) The number of labeled trees on n nodes is nn-2 Put another way, it counts the number of spanning trees

of a complete graph Prfer Encoding (1918) We are going to find a bijection between the set of sequences and the set of labeled trees. A Prfer sequence is a sequence of n2 numbers, each being one of the numbers 1 through n. We should initially note that indeed there are nn2 Prfer sequences for any given n. bijection: T(n) -> P(n-2) Encoding a tree into a Prfer sequence Take a tree and label vertices from 1 to n in any manner. 4 4

2 1 3 1 5 6 3 5 6 Take the vertex with the smallest label

whose degree is equal to 1, delete it from the tree and write down the value of its only neighbor. Repeat this process until only two vertices remain. So now we have a sequence of n 2 elements encoded from our tree. Encoding a tree into a Prfer sequence 4 2 1 3 4

5 1 6 Sequence: 3 5 6 Sequence: 5 Encoding a tree into a Prfer sequence 4

4 1 3 5 1 6 Sequence: 5, 1 5 6 Sequence: 5,1,1

Encoding a tree into a Prfer sequence 1 5 5 6 Sequence: 5, 1, 1, 5 6 Sequence: 5, 1, 1, 5 Encoding a tree into a Prfer sequence

4 2 1 3 5 P = 5, 1, 1, 5 6 Notice that all of the vertices of degree 1 do not occur in P. No leaves in P. Every vertex in P has degree equal to 1 + r, where r is the number of times that vertex appears in our sequence P.

Exercise: write the Prfer sequence 1 2 6 4 3 5 7 8 Sequence: 1, 2, 1, 3, 3, 5 Reconstructing a tree Given P = {a1,,an-2} and the list L = {1,,

n} Let k be the smallest number in L that is not in P. Let aj be the fist number in the Prfer sequence P. Connect k and aj with an edge. Remove k from L and aj from P. Repeat this process until all elements of P have been exhausting (n-2 times) Connect the last two vertices in L with an edge. Reconstructing a tree L=1,2,3,4,5,6 L=1,3,4,5,6 2

3 2 5 P = 5, 1, 1, 5 5 1 P = 1, 1, 5 Reconstructing a tree L=1,4,5,6 L=1,5,6 2

3 2 5 3 1 1 5 4 4 P = 1, 5 P=5

L=5,6 2 3 1 5 4 6 Exercise Given P = {1, 2, 1, 3, 3, 5}. Reconstruct a tree. Exercise L = {1,2,3,4,5,6,7,8} P = {1, 2, 1, 3, 3, 5} 8

5 3 4 1 6 2 7 Bijection between Prfer Sequences and Labeled Trees Let T be a set of labeled tree of n

vertices Let P be a set of Prfer sequences of length n-2 A map f: T-> P is a bijection. A map f: T-> P is injective. We need to show that two different trees T1 ,T2 generate different Prfer sequences. By Induction on the number of vertices. Base case: n = 2, two vertices joined by an edge. Assume its true for n, prove it for n+1. Take the lowest-labeled leaf in T1 and in T2. Case 1: Those two leaves are different Case 2: Same, but neighbors not Case 3: Leaves and neighbors are the same A map f: T-> P is surjective. We need to show that any sequence P={a1,

,an-2} generates at least one tree on L={1, , n} By Induction on the number of vertices. Base case: n = 2, P = {}. Assume its true for n, prove it for n+1. Take the lowest vkL s.t. vkP Consider P=P\a1 and L= L\vk. By IH there is T. Form T from T by adding vk joined with a1. Since a1 is internal, T is a tree. The Minimum Spanning Tree ORD 10 1 DEN

Minimum spanning tree (MST) is a spanning tree of 4 a weighted graph with minimum total edge weight PIT 6 7 9 STL 8 DFW The weight of a spanning tree is the

sum of the weights on all the edges which comprise the spanning tree. 3 DCA 5 2 ATL The MST Fred Hackers algorithm: Find ALL spanning trees and then pick one with the minimum cost. Whats wrong with this idea?

The number of spanning trees in Kn is nn-2 The Minimum Spanning Tree Boruvkas Algorithm (1926) Kruskals Algorithm (1956) Prim's Algorithm (1957) Property of the MST Lemma: Let X be any subset of the vertices of G, and let edge e be the smallest edge connecting X to G-X. Then e is part of the minimum spanning tree. Property of the MST - proof Let T be the MST && e not in T G-X e is the

smallest edge X e u v Property of the MST - proof There exists a unique path in T from u to v. G-X e is the smallest edge X e

u v e is not in T Property of the MST - proof Let T be the MST && e not in T G-X e is the smallest edge f X e u

v f>e Since T1 =T f + e < T thus T is not the MST Prim's Algorithm algorithm builds a tree one VERTEX at a time. - Start with an arbitrary vertex as component C - Expand C by adding a vertex having the minimum weight edge of the graph having exactly one end point in C. - Continue to grow the tree until C gets all vertices. Prim's Algorithm

C={a} 4 A 1 1 B 2 C 2 8 D 5

E heap d-1 c-1 b-4 e-oo foo 7 3 F Prim's Algorithm C={a,d} A 4 1

1 B 2 C 2 8 D 5 E heap c-1 b-4 e-oo f-oo

7 3 F Prim's Algorithm C={a,d} A 4 1 1 B 2

C 2 E heap 8 D 5 7 3 F c-1 b-4 e-5 f-7

decreaseKe Prim's Algorithm C={a,d,c} a 4 1 1 b c 2 8

d 5 2 e heap b-4 e-5 f-7 7 3 f Prim's Algorithm C={a,d,c}

a 4 1 1 b c 2 8 d 5

2 e heap b-2 e-2 f-7 7 3 f Prim's Algorithm C={a,d,c,b} a 4 1

1 b c 2 e e-2 f-7 d 5 2 heap

8 7 3 f Prim's Algorithm C={a,d,c,b,e, 4 f} a 1 1 b

c 2 8 d 5 2 e 7 3 f Weight = 1+1+2+2+3 = 9

What is the worst-case runtime complexity of Prim's Algorithm? We maintain a PQ of vertices: We run deleteMin V times We update the queue E times O(V*log V + E*log V) Kruskal's Algorithm algorithm builds a tree one EDGE at a time. - Start with all vertices as a forest - Choose the cheapest edge and joint correspondent vertices (subject to cycles) - Continue to grow the forest Kruskal's Algorithm Start with minimal weight edges.

There are three edges of weight 1 4 1 2 8 1 3 2 6 11 6

4 8 9 2 2 5 5 1 3

4 7 1 8 5 7 6 8 3 9 Kruskal's Algorithm

There are three edges of weight 2 4 1 2 8 1 3 2 6 11 6

4 8 9 2 2 5 5 1 3

4 7 1 8 5 7 6 8 3 9 Kruskal's Algorithm

1 2 2 2 2 1 3 1 6 5 4

7 4 1 8 3 9 Planar Graphs A graph is planar if it can be drawn in the plane without crossing edges =

Planar Graphs A graph is planar if it can be drawn in the plane without crossing edges A graph is planar if and only if it can be embedded in a sphere. This is useful because often a sphere is more convenient to work with. A sphere can be 1-1 mapped (except 1 point) to the plane and vice-versa. E.g. the stereographic projection: Faces A planar graph when drawn in the plane, splits the plane into disjoint faces.

4 faces Eulers Formula If G is a connected planar graph with V vertices, E edges and F faces, then VE+F=2 Generalized for any polyhedron, For a cube: v=8 e=12 f=6 Proof of Eulers Formula For connected arbitrary planar graphs V E + F = 2 The proof is by induction on edges. Start with a single edge and 2 vertices:

V=2, E=1, F=1. Check. Add the edges in an order so that what weve added so far is connected. There are two cases to consider. (1)The edge connects two vertices already there. (2)The edge connects the current graph to a new vertex In case (1) we add a new edge (E++) and we split one face in two (F++). So V-E+F is preserved. In case (2) we add a new vertex (V++) and a new edge (E++). So again V-E+F is Theorem: In any connected planar graph with at least 3 vertices: E = O(V) E 3V - 6

By means of this theorem we can prove, for example, that a complete graph K5 is not planar K5 has 5 vertices and 10 edges, thus E = 10 3x5 6 = 9 which is clearly false Theorem: In any connected planar graph with at least 3 vertices: E 3V - 6 Proof. 1. If the graph has no cycles, E = V-1 V V +(2V-6) = 3V 6, since V 3, and therefore 2V-6 0, Theorem: In any connected planar graph with at least 3 vertices:

E 3V - 6 Proof (cont.) 2. If the graph has a cycle. We will count the number of pairs (edge, face). Each face is bounded by at least 3 edges: (edge, face) 3 F Each edge is associated with at most 2 faces: (edge, face) 2 E It follows, 3F2E Theorem: In any connected planar graph with at least 3 vertices: E 3V - 6 Proof (cont.) We found, 3 F 2 E By Eulers theorem : 2=VE+F

6 = 3V 3E + 3F 3V 3E + 2E = 3V - E QED Planar Graphs Theorem: In any connected planar graph with V vertices, E edges and F faces, then VE+F=2 Theorem: In any connected planar graph with at least 3 vertices: E3V-6 Lemma: In any connected planar graph with at least 3 vertices: 3F2E K5 can be embedded on the torus Embedding a graph onto a surface means drawing the graph on the surface such that no edges cross.

Always there is a surface so any graph can be embedded to. More embeddings Blanua graph on a trefoil knot Cayleys Formula Prfer Encoding Minimum Spanning Trees Planar Graphs Eulers Polyhedra Theorem Heres What You Need to Know

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