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**Graphs, Planar graphs Graph coloring**

Discrete Math Graphs, Planar graphs Graph coloring 1

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Planar graph A graph is called planar if it can be drawn in the plane without any edges crossing. Such a drawing of a graph, with no crossings, is called a planar representation of the graph A crossing is the intersection of the lines or arcs representing the edges at a point other than a common endpoint

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**Is the graph planar? Can we draw this graph in a plane without**

any crossings? If so how? a1 e7 e1 a2 e6 a3 e2 e5 e3 e4 a4 a5

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Is the graph planar? First move a5 to the left to remove the crossings, This removes crossings e5-e3, e5-e6, e4-e1, e4-e6, e4-e2 a1 e5 a5 e1 e7 e4 e6 a2 a3 e2 e3 a4

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Is the graph planar? Then move a4 to the right to remove the remaining crossing e2-e6 a4 e2 a1 e5 a5 e3 e1 e7 e4 e6 a2 a3

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Your turn: Is this a planar graph D G C F E B A

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**Your turn: solution 1 Is this a planar graph A D G C F D G C F E B E B**

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**Your turn: solution 2 Is this a planar graph A D G D G C F E E B B C A**

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Euler’s formula Theorem: Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar represetnation of G. Then r = e – v + 2 Proof by inductive construction of a sequence of subgraphs. Begin with a single edge, add one more edge each step

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**Euler’s formula: proof (1)**

Pick an arbitrary edge of G and the vertices at each end of that arbitrary edge to be our first subgraph G1 Obtain graph Gn from graph Gn-1 by arbitrarily picking an edge incident with a vertex that is part of Gn-1, but is not itself part of Gn-1 and adding it to Gn-1 Adding the vertex at the other end of the edge we just added to Gn-1 to Gn-1

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**Euler’s formula: proof (2)**

Now prove by induction BASE CASE: For an arbitrary edge and the vertices it is incident upon e=1 v=2 so r1=1-2+2=1. Cleary the plane is a single region surrounding this graph INDUCTIVE HYPOTHESIS: Euler’s formula rn=en-vn+2 holds for n=p INDUCTIVE STEP: show Euler’s formula holds for n=p+1 , beginning with the inductive hypothesis

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**Euler’s formula: proof (3)**

If the selected edge is incident upon a second node that is already a part of Gn. Then the new edge must be inside one of the existing regions The number of edges increases by 1 The number of nodes does not change. The new edge splits one region of the plane into two regions (otherwise lines representing edges would cross. So the number of regions increases by one. en+1-vn+2=en+1-vn+1+2=rn+1

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**Euler’s formula: proof (3)**

If the selected edge is incident upon a second node that will be added to Gn. The number of edges increases by 1 The number of nodes increases by 1 No new regions are created en+1-(vn+1)+2=en+1-vn+1+2=rn+1 Using the principle of mathematical induction we have established that Euler’s formula holds

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Example Suppose that a connected simple planar graph has 25 vertices and 12 edges of degree 3 and 13 edges of degree 4. How many regions does this graph divide the plane into? The graph has 0.5(4*13+3*12 )=(52+36)/2=88/2=44 edges So =21 regions are created

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Corollary If G is a connected planar simple graph with e edges and v vertices, where v≥3, then e ≤3v-6

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Corollary If G is a connected planar simple graph, the G has a vertex of degree not exceeding five Proof: If there is one vertex or two vertices. A simple graph will be a vertex or an edge containing two vertices. In both of these cases the degree of the vertices cannot exceed 5. If there are at least 3 vertices, lets prove by contradiction

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**Corollary (proof continued)**

If G is a connected planar simple graph, the G has a vertex of degree not exceeding five If there are at least 3 vertices, lets prove by contradiction we know e≤3v-6 from previous corallary. So 2e≤6v-12. Assume the degree of every vertex is at least 6, then by the handshake theorem 2e= deg(v)≥6v . But this contradicts 2e≤6v-12 so there must be a vertex with degree <6

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Another corallary In a connected planar simple graph has e edges and v vertices with, v≥3, and no circuits of length 3, then e≤2v-4

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Is a graph planar? The corollaries on the previous slide can be used to help prove graphs are not planar. If you have a planar graph and any one of the sets of conditions specified in the corollaries do not hold, then your graph cannot be planar.

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**Elementary subdivision**

If a graph is planar, then any subgraph made by replacing an edge (u,v) with a vertex w and two edges (u,w) and (w,v) will also be planar: This operation is called an elementary subdivision IF a graph is planar, then any subgraph made by removing one edge is also planar.

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Homeomorphic Given two planar graphs, G and H. H is and G are homeomorphic if both H and G by produced by applying a two series of elementary subdivisions to the same original graph Thus by definition a planar graph is homeomorphic to itself because if you apply and empty sequence of elementary subdivisions to the graph you end up with the same graph

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**Example Are these graphs homeomorphic G1 = Original graph**

Homeomorphic to itself G2 Homeomorphic to G1?

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**Example G1 = Original graph Homeomorphic to itself G2**

Homeomorphic to G1? YES

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Kuratowski’s Theorem A graph is non planar if and only if it contains a subgraph homeomorphic to K3,3 or K5 Remember K3,3 is the complete bipartite graph with 3 vertices in each set Remember K5 is the complete graph with 5 vertices

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Maps If we think of a map we think of a series of areas on a page that represent countries, provinces, etc We can think of a map as a graph.The edges of the graph are the boundaries between the countries. The nodes of the graph are either on bondaries between contries or at points where more than 2 countries touch.

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**Dual graph We can also represent a map by a dual graph**

Each region of the map is represented by a vertex When two regions share a boundary of more than one point, the vertices representing those regions are connected by edges in the dual graph The dual graph of a map must be a planar graph

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**Maps and coloring A planar graph divides the plane into regions**

Those regions can be though of as “countries” on a map. Customarily when we draw a map we want to color adjacent countries different colors so it is each to see where the borders (edges) are What is the minimum number of colors we need to color the map?

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Example A B A B E C E D C D

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Your turn: solution Construct the dual graph B D C E A F G

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Your turn: solution G D G A B A B E D E C F C F

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Coloring A coloring of a simple graph is the assignment of a color to each vertex of the graph so that no two adjacent vertices are assigned the same color. Clearly each graph can be colored by assigning a different color to each vertex For many graphs the graph can be colored using fewer colors than the number of vertices

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Chromatic number The chromatic number of a graph is the least number of colors needed for a coloring of that graph. The chromatic number of a graph G is denoted χ(G).

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The four color theorem The chromatic number of a planar graph is no greater than four First proved 1879, Kempe Proof shown invalid 1890, Heawood Proved again (similar argument) Appell and Haken 1976 Used a computer to exhaustively prove for about 2000 cases

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**What are the chromatic numbers**

K3,3 K5

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**What are the chromatic numbers**

K3,3 K5 35

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