Introduction of metric dimension of circular graphs is connected graph , The distance and diameter , Resolving sets and location number then Examples . Application in facility location problems . is has motivation (Applications in Chemistry and Networks systems). Definitions of Certain Regular Graphs. Main Results for three graphs (Prism , Antiprism and generalized Petersen graphs .

1. Metric dimension of
Circular Graphs
By:
Ruwayya Almehrezi
Amna AE Abunamous
Mouza Alnaimi
1
Amnaae6@gmail.com

2. Outline:
• Introduction, Examples & Application
• Motivation & Definitions of Certain
Regular Graphs
• Main Results
• Conclusion
2

3. • A Graph G=(V(G), E(G)) consists of a non-
empty finite set V(G) of elements called the
vertices, and a finite set E(G) of distinct
unordered pairs of distinct elements of V(G)
called the edges of the graph G.
• A graph G is connected if and only if it cannot
be expressed as the disjoint union of two
graphs in an obvious way.
3
Connected graph

4. • The distance denoted by d(u,v)
between the two vertices u and v in a
graph G is the length of a shortest u − v
path in G.
• The diameter of a connected graph G
denoted by D is the maximum distance
between any pair of distinct vertices of
G.
D= max {d(x, y): x, y є V(G)}.
4
The distance and diameter

5. • Let W={w1, w2, w3,….., wk} be an ordered set of
vertices of a connected graph G.
• The (metric) representation r(v│W) (or distance
code) of the vertex v with respect to the set W is
list of k-tuples.
r(v│W)=(d(v,w1 ) , d(v,w2),….., d(v,wk)).
5
Resolving sets and location number

6. 6
• The set W is called a Resolving Set or Locating Set is if
every vertex has distinct representation by with respect to
the set W.
• A resolving set of minimum cardinality is called a (metric)
basis for G and this cardinality is the metric dimension or
location number of G, denoted by dim(G) or loc(G).
• From above, it implies that in computing location number
of a graph, we are interested to know that to get the
distance code of each vertex distinct:
1. How many minimum number of vertices in set W we need
to choose.
2. The location for those chosen vertices is also crucial.

7. W={V1,V5 } not resolving
set.
W={V1,V2 } is a resolving
set.
7
Examples

8. A facility consists of five
rooms.
A graph model G of
five rooms with
dim(G)=2
8
Application in facility location problems

9. Motivation:
• Applications in Chemistry:
 A basic problem in chemistry is to provide
mathematical representations for a set of chemical
compounds in a way that gives distinct representations
to distinct compounds.
 The structure of a chemical compound can be
represented by a labeled graph whose vertex and edge
labels specify the atom and bond types, respectively.
9

10. 10
• Networks systems:
 Graphs are used to model different kinds of networks such
as roads, and buildings.
 The metric dimension (location number) tells us to choose
the best location and minimum nodes in the networks in
order to install the sensors/fault detecting devises so that
in case of a fault in network, one needs to check only
minimum points and not the all nodes of networks.
 It is also helpful to determine the cost of expansion when
one needs to expands the existing network.

11. Path graph (Pn)
dim (G)=1 if & only if G ≅ Pn
11

12. Cycle graph (Cn)
dim (Cn)=2 for every positive integer n ≥ 3
12

13. Complete graph (Kn)
• dim (G)=n-1 if & only if G ≅ Kn
13

14. The Petersen graph (P)
dim (P) = 3
14

15. Definition :
V(Dn)=
E(Dn)=
│V(Dn)│=2n (Order of prism)
│E(Dn)│=3n (Size of prism)
It is a 3-regular graph. For n=4, it is isomorphic to the graph of 3-cube Q3(a
member of hypercube graphs Qn).
Main results
Prism Graphs (Circular ladder Dn)
15
ui , vi: 1 ≤ i ≤ n
uiui+1,ui-2ui-1, vi vi+1,………… 1 ≤ i ≤ n

16. 16
The Prism D6 & D5
D5D6

17. Theorem :
dim ( Dn )=
2 ; n is odd
3 ; n is even
For every n ≥ 3 and n ∈ ℤ+
17

18. Proof :
The cycle u1u2………. un (Inner cycle)
v1v2…………vn (Outer cycle)
We can write n= 2k+1; The set W= {v1,vk+1} is a resolving
set.
dim (Dn) = 2
18
• When n is odd

19. Proof :
• When n is even
We can write n= 2k; The set W= {v1,v2,vk} is a
resolving set.
dim (Dn) ≤ 3
19

20. • To prove dim (Dn) ≥ 3, We use contradiction.
• Note the dim (Dn) ≠ 1 because Dn ≠ Pn
Suppose dim (Dn) = 2. We have the following possibilities.
 Both vertices are in the inner cycle. (always we get
contradiction)
20

21.  Both vertices are in outer cycle. (contradiction)
21

22.  One vertex in the outer cycle & other in inner
cycle. (Contradiction)
22

23. Definition :
V(An)=
E(An)=
We have │V(An)│=2n
│E(An)│=4n
It is a 4-regular graph.
The graphs of antiprism (An)
23
ui, vi : 1 ≤ i ≤ n
uiui+1,ui-2ui-1, vi vi+1,………… 1 ≤ i ≤ n

24. V2
U1
V1
V5
V4
V3U5
U4
U3
U2U6V6
V2
U1
V1
V5
V4
V3
U5
U4
U3
U2
24
The Antiprism A5 & A6
A5A6

25. Theorem :
dim (An)= 3
For every n ≥ 5 and n ∈ ℤ+
25

26. Proof :
• When n is even or odd; we can write n= 2k or n=2k+1
In either case, The set W= { v1,v2,vk+1} is a resolving set for An.
dim (An) = 3
26

27. Definition
The generalized Petersen graphs P(n, 2) form an important class of
3-regular graphs because of their interesting properties and being a
natural model for LAN (Local Area Network).
V(P(n, 2) ) = {ui, vi: 1 ≤ i ≤ n − 1}
E(P(n, 2) ) = {uiui+1, ui vi, vi vi+2 , ………… : 1 ≤ i ≤ n − 1}
│V(P(n,2))│=2n; │E (P(n,2))│=3n
• Note that The Petersen graph is a member of this class of graph
and using this notation, we write Petersen graphs P as P(5, 2) .
The generalized Petersen graphs P(n,2)
27

28. 28
The generalized Petersen graphs P(5,2) & P(6,2)
P(6,2) P(5,2)

29. For the generalized Petersen graphs
P(n, 2)
for every positive integer n ≥ 5 and
n ∈ ℤ+
Theorem :
dim (P(n,2))= 3
29

30. Proof :
when n = 4k
W= { u1,u2,u3}
dim (P(n,2)) = 3
30

31. 31
Proof :
when n = 4k+1
W= { u1,u2,v2k-1}
dim (P(n,2)) = 3

32. 32
Proof :
when n = 4k+2
W= { u1,u2,u3}
dim (P(n,2)) = 3

33. 33
Proof :
when n = 4k+3
W= { u1,u2,v3}
dim (P(n,2)) = 3

34. Conclusion:
• We have studied the location number of prism, antiprism and
generalized Petersen graph P(n,2) because these graphs are
most natural model for LAN (Local Area Network).
• It shows that the formulas for their location number do not
depend on order of these graphs.
• It shows that these graphs have constant location number. If
we want to expand these type of networks in future, we can do
it with minimum cost as no new fault detecting
devices/sensors will be needed to do so.
34