PRISM MATH 2014
Final Report
Bibi Guerrero, Lori Ashkine, Yessica Perez, Denise
Garcia, Allen Mendez
July 24, 2014
Definitions:
A graph G consists of a finite nonempty set V (G) of vertices and a set E(G) of
edges where we can view:
Vertices V (G): As points and nodes. In our project, radio stations are
represented by vertices.
Edges E(G): A set of two elements subsets of the vertices set V (G).
Path graph (Pn ): a simple graph where the vertices can be labeled v 1 , v2 , ..., vn
so that its edges are v1 v2 , v2 v3 , ..., vn−1 vn .
Figure 1: A path graph with 3 vertices, denoted by P3
Note: For figure 1, V (P3 ) = {v 1 , v2 , v3 } and E(P3 )= {a, b} ={v1 v2 , v2 v3 }
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Connected graph: A graph G is connected if there exists a path in G between
any two of its vertices.
Figure 2: A connective graph
The graph below is considered a disconneted graph because there is no path to
connect vertices B and E.
Figure 3: An example of a disconnected graph
Two vertices are adjacent if there is an edge between two vertices.
Labeled graph: A graph where each vertex of the graph are labeled.
Distance between two vertices: shortest length of any path between two vertices.
Diameter of a graph G (denoted by diam(G)): the greatest distance between
two vertices of G.
Span of f : The maximum value of f subtracting the minimum value of f .
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Distance two labeling: a labeling of a graph where interference is limited to
between vertices of distance two or less while labeling the vertices that satisfies:
2, if d(u, v) = 1;
|f (u) − f (v)| ≥
1, if d(u, v) = 2.
Figure 4: An example of a Distance-2 labeling of P5 .
A Radio labeling of G is a function f that labels all vertices a number from
{0,1,2, ...} or according to another definition all positive integers such that the
following holds true for any vertices u and v:
|f (u) − f (v)| ≥ diam(G) + 1 − d(u, v)
Figure 5: An example of a Radio labeling of a graph.
Radio number of a graph G: the smallest span of all radio labelings of G.
The optimal radio labeling of G: is a radio labeling with span equals to the
minimum span over all possible radio labelings of G.
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Figure 6: (a) The P4 above has a radio labeling with span equal to 4.(b) The
P4 below has a radio labeling with span equal to 5
Figure 6(a) would be a ”better” labeling because it has a smaller span than
figure 6(b).
Channel assignment problem: Assigning channels to transmitters to obtain
minimal span while avoiding interference.
Cartesian product of two graphs G and H: has vertex set V (G) × V (H). Means
every vertex of G × H is an ordered pair (u, v) where u is from V (G) and v
if from V (G). Two distinct vertices (u, v) and (x, y) are adjacent in G × H if
either (1) u = x and vy ∈ E(H) or (2) v = y and ux ∈ E(G).
Prism Graph: The cartesian product of P2 and Cn which is denoted by
Zn,1 = P2 × Cn
n-cycle: a graph of order n with its vertices can be labeled v 1 , v2 , ..., vn such
that its edges are v1 v2 , v2 v3 , ..., vn−1 vn , and v1 vn . It is denoted by Cn .
Trees: A connected graph that contains no cycle (no closed loops).
Bridge: An edge e of a connected graph G is a bridge of G if G-e is disconnected.
N ote: A tree is not considered a tree anymore if any edges are removed. The
edges of a tree can be viewed as a bridge. Leaf: a vertex of degree 1 in a tree.
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Forest: collection of one or more trees.
Figure 7: A tree graph
This is a tree because it does not contain a cycle.
Revision of Main Theorem in [8]:
Let n = 4k + r, where k ≥ 1, and r = 0, 1, 2, or 3. k is the quotient when n is
divided by 4, and r is our remainder.
rn(Zn,1 ) = (n − 1)φ(n, 1) + 2, where
r
φ(n, s) : r
r
r
=0
=1
=2
=3
5
s=1
k+2
k+2
k+3
k+2
Figure 8: A radio labeling of Z3,1 with span= 5
d(v1 ,v2 )=
d(v1 ,v3 )=
d(v1 ,v4 )=
d(v1 ,v5 )=
d(v1 ,v6 )=
d(v2 ,v3 )=
d(v2 ,v4 )=
d(v2 ,v5 )=
d(v2 ,v6 )=
d(v3 ,v4 )=
d(v3 ,v5 )=
d(v3 ,v6 )=
d(v4 ,v5 )=
d(v4 ,v6 )=
d(v5 ,v6 )=
6
d(va , vb )
1
1
1
2
2
1
2
1
2
2
2
1
1
1
1
diam(G)=2
|f (u) − f (v)| ≥ diam(G) + 1 − d(u, v)
Calculations of rn(Z3,1 ) based on The Main Theorem [8] :
n= 3 = 4(0) + 3, therefore k= 0 and r= 3. We have φ(3, 1) = k + 2 = (0) + 2 = 2
rn(Z3,1 ) = (3 − 1)φ(3, 1) + 2 = 2(2) + 2 = 6, but since we adapted notation from [6], rn(Z3,1 )
should be 6 − 1 = 5 as we start with label 0.
Use the Revision of the Main Theorem [8] to Find the Radio Number
of prism graph Zn,1 :
Let n=4k+r, where r=0,1,2, or 3 . k is the quotient when n is divided by 4
and r is our remainder. We can find the minium span by following these steps
for Z3,1 :
1. Divide 3 by 4 where 3 is n and 4 is the divisor. Once solved, we obtain 0
as the quotient k and 3 as the reminder r.
2.Plug into formula: n= 4k+r −→ 3 = 4(0) + 3.
3. Use Table 1 to solve for: φ(n, s) = k + 2 −→ 0 + 2 = 2.
4. Plug into the main theorem formula:
rn(Zn,s ) = (n − 1)φ(n, s) + 2
rn(Z3,1 ) = (3 − 1)(2) + 2
= (2)(2) + 2
=6
based on [8] (based on [6], we subtract 1 which gives us a radio number of 5 ).
Table 1:
r
r
r
r
s=1
k+2
k+2
k+3
k+2
=0
=1
=2
=3
Diameter of Odds and Even Cycles:
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To find the diameter of a cycle Cn with an odd number of vertices n, use
the formula:
n−1
2
In Figure 9 for example, there are 5 vertices and therefore
diam(C5 ) =
5−1
=2
2
Figure 9: This is a cycle with 5 vertices where diam(C5 )=2. This is proven by
d(A,C)=2, d(A,D)=2, and d(u, v) ≤ 2 for all u, v ∈ V (C5 ).
To find the diameter of a cycle Cn with an even number of vertices n, use
the formula:
n
2
In Figure 10 for example, there are 6 vertices and therefore
diam(C6 ) =
6
=3
2
Figure 10: This is a cycle with 6 vertices. Due to d(A,D)=3 and d(u, v) ≤ 3 for
all u, v ∈ V (C6 ) then diam(C6 )=3 .
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Diameter for Zn,1 :
For, diam(Zn,1 ) when n is even
n+2
= diam(Cn ) + 1
2
For, diam(Zn,1 ) when n is odd
n+1
= diam(Cn ) + 1
2
Radio Labeling for Z4 k ,1 Graphs:
These are some of the characteristics for our labeling of Z4 k +1 graphs:
1. The labels of adjacent verticese alternate from evens and odds.
2. To predict our next move, we can use the following formula:
diam(Z4k,1 ) + 1
2
3. There is a change of pattern for ”every order set of 8”. Order being
0=f (x1 ) <f (x2 ) <.... <f (xn ) = span f , where f is our radio labeling. See the
table below.
x8k x8k+1
x8
x9
x16
x17
x25
x24
.
.
.
.
.
.
These characteristics can be applied to predict the optimal radio labeling of
general Z4k,1 . For example, the pattern in Z4,1 can be applied to Z8,1 .
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Figure 11: Our radio labeling of Z4,1 with a span=10.
Figure 12: Our radio labeling of Z8,1 with a span= 29.
Figure 12, has the same pattern as Figure 11 where we can begin labeling a
vertex on the outter cycle (x1 ) moving to the ”diagonal” vertex in the inner
cycle (x2 ) then back to the outter cycle (x3 ) and continue alternating. We can
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also find the position of our next vertex by following the second characteristic
mentioned under ”Radio Labeling for Z4k,1 ”.
To find the pattern of each cycle, we would use the equation:
diam(Z4k,1 ) + 1
2
In Figure 11, diam(Z4,1 )=3. Applying to the formula above, we will obtain an
answer of 2. Then rewrite it into the format (diameter, quotient), where
quotient is
diameter + 1
2
In the case of Z4,1 , it would be (3,2). This means that x1 is distances 3 away
from x2 and x2 is 2 distances away from x3 and so on. The same concept can
be applied to Figure 12 where we would obtain (5,3).
For Figure 12:
1. Alternate between the outter and inner cycles with the pattern (5,3).
2. Begin with x1 and count counterclockwise on the outter cycle with a
distance of 4 then go up the edge which will equal a total of 5 edges and leads
you to the inner cycle which would be x2 .
3. Continue counting counterclockwise from x2 in the inner cycle until you go
to the distance of 2 then go up that path onto the outter cycle which will be
x3 .
4. Continue this pattern until you get to x8 to x9 the pattern will continue the
counterclock wise direction but will stay in the inner cycle.
x8k x8k+1
x8
x9
x17
x16
x24
x25
.
.
.
.
.
.
5. After every multiples of 8 vertices, continue the steps 1 through 4.
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Radio Labeling for Z4k+1,1 :
Figure 13: Our radio labeling of Z5,1 with a span=13.
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Figure 14: Our radio labeling of Z9,1 with a span=33.
Zn,1 where n = 4k + 1
Z5,1
Z9,1
Z13,1
Z17,1
k
1
2
3
4
number of vertices to skip add by
0
3
1
4
2
5
3
6
Depending on the Prism graphs and its amount of vertices, we show how many
vertices we will skip, beginning on x1 .
For Figure 13:
1. The direction is also counterclockwise and according to the table above.
begin with the outter cycle starting with labeling 0 on v1 and will skip 0
vertex and begin adding 3 to the previous vertex label for v2 .
2. Once you have labeled every outter vertices and you are back at v1 begin to
count counterclockwise again on the outter cycle with a distance of 2 and go
on the path that leads you to the inner cycle of v8 and this is a total of a
distance of 3.
3.Now you will begin on v8 starting with radio labeling 1.
4. Continue the steps 1 through 3 but in the inner cycle now starting with v8 .
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For Figure 14:
1.Repeat steps 1, but according to the table above you will now be count by
adding 5 to the previous vertex and this time skip 1 vertex and therefore label
every other vertices by adding 4 to the previous label.
2.Once every outter vertice is labeled you will be back at v1 begin to count
counterclockwise again on the outter cycle with a distance of 4 and go up the
path that leads you to the inner cycle of v15 and this is a total of a distance of
5.
3. Now you will begin on v15 starting with radio number 1.
4. Continue the steps 1 through 3 but in the inner cycle now starting with v15 .
The pattern for 4k +1 is to follow the chart above to see how much you will
add from the previous vertex and also how many vertices you will need to skip
to label.
Radio Labeling for Z4 k +2,1 Graphs:
Figure 15: This is an example of radio labeling for Z4 k +2,1 where it equals to
Z6,1
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Figure 16: This is an example of radio labeling for Z4 k +2,1 where it equals to
Z10,1
Zn,1 where n = 4k + 1
Z6,1
Z10,1
Z14,1
Z18,1
k
1
2
3
4
# of vertices to skip add by
0
4
2
5
2
6
4
7
( Note: Number of vertices to skip is based on k.
(k − 1) if k is odd ;
)
(k) if k is even .
Depending on the Prism graphs and its k( if it is even or odd), we show how
many vertices we will skip, beginning on x1 .
For Figure 15:
1.Begin counterclockwise and according to the table above. In the graph Z6 ,
where k = 1 , k being odd you will use (k − 1). Plug in k into (k − 1) and you
will get 0. You will skip 0 vertices.
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2. Begin with the outter cycle starting with labeling 0 on v1 and will skip 0
vertex and begin adding 4 to the previous vertex label for v2 .
3. Once you have labeled every outter vertices and you are back at v1 begin to
count counterclockwise again on the outter cycle with a distance of 3 and go
on the path that leads you to the inner cycle of v11 and this is a total of a
distance of 4.
4. Now you will begin on v11 starting with radio labeling 1.
5. Continue the steps 1 through 4 but in the inner cycle now starting with v11 .
For Figure 16:
1.Begin counterclockwise and according to the table above. In the graph Z10 ,
where k = 2 , k being even you will use (k) meaning you will skip 2 vertices.
2. Begin with the outter cycle starting with labeling 0 on v1 and will skip 2
vertex and begin adding 5 to the previous vertex label for v2 .
3. Once you have labeled every outter vertices and you are back at v1 begin to
count counterclockwise again on the outter cycle with a distance of 5 and go
on the path that leads you to the inner cycle of v16 and this is a total of a
distance of 6.
4. Now you will begin on v16 starting with radio labeling 1.
5. Continue the steps 1 through 4 but in the inner cycle now starting with v16 .
The pattern for 4k +2 , 1 is to follow the chart above to see how much you
will add from the previous vertex and also how many vertices you will need to
skip to label.
Radio Labeling for Z4 k +3,1 Graphs:
Figure 17: This is an example of radio labeling for Z4 k +3,1 where it equals to
Z7,1
16
Zn,1 where n = 4k + 3
Z7,1
Z11,1
Z15,1
k
1
2
3
number of vertices to skip add by
1
3
2
4
3
5
For Figure 17:
1. The direction is counterclockwise and according to the table above, begin
with the outer cycle labeling 0 on v1 . We will skip one vertex and begin adding
3 to the previous vertex labeled v6 .
2. Once we have labeled every outer vertices and we are back at v1 begin going
clockwise moving a distance of 3 and go on the path that leads us to the inner
cycle of v12 and this will be a total of a distance of 4.
3. we will begin on v12 starting with radio-labeling1.
4. the steps 1 through 3 but now in the inner cycle.
In general,
Our Radio Labelings Pattern
Z4k+1,1
Z4k+2,1
Z4k+3,1
number of vertices to skip
add to previous label
k
−
1
k
+
2
=
diam(Z
4k+1,1 ) + 1 − (k)
k − 1 if k is odd ;
diam(Z4k+2,1 ) + 1 − (k) if k is odd ;
k+3≥
k if k is even .
diam(Z4k+2,1 ) + 1 − (k + 1) if k is even .
k
k + 2 = diam(Z4k+3,1 ) + 1 − (k + 1)
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