discrete mathematics section 2

5. There is a simple path between every pair of distinct
vertices of a connected undirected graph.
• Suppose a and b are two distinct vertices of the connected undirected
graph G =(V, E).
• we know that G is connected so by definition there is at least one path
between a and b .
• Let x0, x1, …, xn, where x0 = a and xn = b, be the vertex sequence of a path
of a least length.
• Now if this path of the least length is not simple then we have xi = xj, for
some i and j with 0 £ i < j.
• This implies that there is a path from a to b of shorter length with the
vertex sequence x0, x1, …, xi, …, xj+1, …, xn obtained by removing the
edges corresponding to the vertex sequence xi+1, …, xj.
• This shows that there is a simple path.

6. • Cut vertices (articulation points) are those vertices in the graph
whose removal along with the edges incident on them produces
subgraph with more connected components than in the original
graph.
• Cut edge (bridge) is an edge whose removal produces a graph with
more connected components than in the original graph.

7. Euler Paths and Circuits
• An Euler circuit in a graph G is a simple circuit containing every edge
of G.
• An Euler path in G is a simple path containing every edge of G.
• Example:
• Find the Euler path or circuit in the following graphs
• No Euler circuit exist,
• but the Euler path is
• 1,2,5,1,3,5,4,3,5,2,4

8. • The Euler circuit is
• 1,2,3,6,9,8,7,4,5,8,6,5,2,4,1

9. Necessary and Sufficient Conditions for Euler
Circuits and Paths
• Theorem 5:
• A connected multigraph has an Euler circuit if and only if each of its
vertices has even degree.
• Proof:
• Take a connected multigraph G =(V, E) where V and E are finite. We
can prove the theorem in two parts.

10. • Theorem 6:
• A connected multigraph has an Euler path but not Euler circuit if and
only if it has exactly two vertices of odd degree.

11. • Find Euler path or circuit?
• In the above graph, all the vertices have even degree, hence there is an
Euler circuit.
• The Euler circuit is
1,2,3,4,5,10,15,14,13,12,11,6,13,8,9,14,12,7,8,3,10,9,4,2,7,6,1

12. Hamilton paths
• A path x0, x1, …, xn-1, xn in the graph G = (V, E) is called Hamilton path if
V ={x0, x1, …, xn-1, xn} and xi≠ xj for 0 ≤ i < j ≤n.
• A circuit x0, x1, …, xn-1, xn, x0, with n > 1, in a graph G =(V, E) is a
Hamilton circuit if x0, x1, …, xn-1, xn is a Hamilton path.
• A graph with a vertex of degree one cannot have a Hamilton circuit.
• If a vertex in a graph has degree 2, then both edges incident with this
vertex must be part of Hamilton circuit.

13. • Example:
Find Hamilton circuit from the following graph if exists? What about
Hamilton path?
In graph (a) there is no Hamilton circuit since
the node c has degree 2 and both the edges
from it must be in Hamilton circuit, which is not
possible. One of the Hamilton path in the graph
(a) is a, b, c, d, e.

14. • In graph (b) we can find Hamilton circuit, the circuit can be 1,2,3,5,6,
9,8,7,4,1. Since there is circuit we can have path also.

15. • Theorem 7: Dirac’s Theorem
• If G is a simple graph with n vertices with n >= 3 such that the degree
of every vertex in G is at least n/2, then G has a Hamilton circuit.
• Theorem 8: Ore’s Theorem
• If G is a simple graph with n vertices with n >=3 such that deg(u) +
deg(v) >=n for every pair of nonadjacent vertices u and v in G, then G
has a Hamilton path.

16. Planar graphs
• A graph is called a planar if it can be drawn in the plane without any
edges crossing.
• Such drawing is called a planar representation of the graph.
• Example:
• Draw the graph below as planar representation of the graph.

17. • Theorem 9: Euler’s Formula
• Let G be a connected planar simple graph with e edges and v vertices. Let r be the
number of regions in a planar representation of G. Then r = e – v + 2.
• Example:
• Suppose that a connected planar graph has 30 edges. If a planar representation
of this graph divides the plane into 20 regions, how many vertices does this graph
have?
• Solution:
• We have, r = 20, e = 30,
• so by Euler’s formula we have
• v = e – r + 2 = 30 – 20 + 2 = 12.
• So the number of vertices is 12.

18. • Corollary 1:
• If G is a connected planar simple graph with e edges and v vertices
where v>=3, then e<=3v – 6.
• Corollary 2:
• If G is a connected planar simple graph, then G has a vertex of degree
not exceeding five.
• Corollary 3:
• If a connected planar simple graph has e edges and v vertices with
v>=3 and no circuits of length three, then e<=2v – 4.

19. • If a graph is planar, any graph obtained by removing an edge {u, v}
and adding a new vertex w together with edges {u, w} and {w, v}. Such
an operation is called an elementary subdivision.
• The graphs G1 = (V1, E1) and G2 = (V2, E2) are called homeomorphic
if they can be obtained from the same graph by a sequence of
elementary subdivisions.
• Example:
• The below example graphs are homeomorphic to the third graph.

21. • Theorem 10: Kuratowski’s Theorem
• A graph is nonplanar if and only if it contains a subgraph
homeomorphic to K3,3 or K5.
• Example:
• Determine whether the following graph is planar or not?

22. • Above we saw that the graph is homeomorphic to K5, the given graph
is not planar.

23. Graph 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.
• A chromatic number of a graph is the least number of colors needed
for a coloring of this graph.
• Theorem 11:The Four Color Theorem
• The chromatic number of a planar graph is no greater than four.
• To show the chromatic number of a graph as k we must show that the
graph can be colored using k colors and the given graph cannot be
colored using fewer than k colors

24. • Example#1
• Construct the dual graph for the map shown. Then find the number of
colors needed to color the map so that no two adjacent regions have
the same color.

25. • We can color the graph with at most 2 colors as shown in the graph.
Take R and G as red and green

26. • Example#2
• Find the chromatic number of the graph below.

27. • Lets start with vertex 1, it has adjacent vertices as 2, 3, 4, 5, 6, 9 so using
only 2 colors would suffice for the graph having the edges from 1 to its
adjacent vertices.
• However since 9 has its adjacent vertices as 1, 2, 3, 4 we cannot just color
the above graph with 2 colors because at least 1, 2 and 9 must have
different colors.
• Trying with 3 colors we found that at least 1, 2, 3, and 9 must have different
colors.
• So trying with four colors we can color the graph. Hence the chromatic
number of the above graph is 4. possible coloring is shown in the figure
below