Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Discrete Math Graphs, Planar graphs Graph coloring.

Similar presentations


Presentation on theme: "1 Discrete Math Graphs, Planar graphs Graph coloring."— Presentation transcript:

1 1 Discrete Math Graphs, Planar graphs Graph coloring

2 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 2

3 Is the graph planar?  Can we draw this graph in a plane without any crossings? If so how? 3 a1 a5a4 a2a3 e4 e7 e6 e5e2 e3 e1

4  First move a5 to the left to remove the crossings, This removes crossings e5-e3, e5-e6, e4-e1, e4-e6, e4-e2 Is the graph planar? 4 a1 a5 a4 a2a3 e5 e4 e7 e6 e3 e1 e2

5  Then move a4 to the right to remove the remaining crossing e2-e6 Is the graph planar? 5 a1 a5 a4 a2a3 e5 e4 e7 e6 e3 e1 e2

6 Your turn: 6 B D C E A G F  Is this a planar graph

7 Your turn: solution 1 7 B D C E A G F  Is this a planar graph B D C E A G F

8 Your turn: solution 2 8 B D C E A G F  Is this a planar graph B D C E A G F

9 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 9

10 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 G 1  Obtain graph G n from graph G n-1 by  arbitrarily picking an edge incident with a vertex that is part of G n-1, but is not itself part of G n-1 and adding it to G n-1  Adding the vertex at the other end of the edge we just added to G n-1 to G n-1 10

11 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 r n =e n -v n +2 holds for n=p  INDUCTIVE STEP: show Euler’s formula holds for n=p+1, beginning with the inductive hypothesis 11

12 Euler’s formula: proof (3)  If the selected edge is incident upon a second node that is already a part of G n. Then the new edge must be inside one of the existing regions 12  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.  e n +1-v n +2=e n+1 -v n+1 +2=r n+1

13 Euler’s formula: proof (3)  If the selected edge is incident upon a second node that will be added to G n. 13  The number of edges increases by 1  The number of nodes increases by 1  No new regions are created  e n +1-(v n +1)+2=e n+1 -v n+1 +2=r n+1  Using the principle of mathematical induction we have established that Euler’s formula holds

14 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 14

15 Corollary  If G is a connected planar simple graph with e edges and v vertices, where v≥3, then e ≤3v- 6 15

16 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 16

17 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 17

18 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 18

19 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. 19

20 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. 20

21 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 21

22 Example  Are these graphs homeomorphic 22 G1 = Original graph Homeomorphic to itself G2 Homeomorphic to G1?

23 Example 23 G1 = Original graph Homeomorphic to itself G2 Homeomorphic to G1? YES

24 Kuratowski’s Theorem  A graph is non planar if and only if it contains a subgraph homeomorphic to K 3,3 or K 5  Remember K 3,3 is the complete bipartite graph with 3 vertices in each set  Remember K 5 is the complete graph with 5 vertices 24

25 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. 25

26 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 26

27 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? 27

28 Example 28 B D C E A B D C E A

29 Your turn: solution  Construct the dual graph 29 B D C E A F G

30 Your turn: solution 30 B D C E A B D C E A F G G F

31 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 31

32 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). 32

33 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 33

34 What are the chromatic numbers K 3,3 K 5 34

35 What are the chromatic numbers K 3,3 K 5 35


Download ppt "1 Discrete Math Graphs, Planar graphs Graph coloring."

Similar presentations


Ads by Google