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# On the Edge-Balance Index of Flux Capacitor and L-Product of Star by Cycle Graphs - PowerPoint PPT Presentation

On the Edge-Balance Index of Flux Capacitor and L-Product of Star by Cycle Graphs. Meghan Galiardi, Daniel Perry, Hsin-hao Su Stone hill College. Labeling of the Graphs. The edges of the graph are labeled by the group Z 2 ={0, 1} The vertices are labeled according to the adjacent edges

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### On the Edge-Balance Index of Flux Capacitor and L-Product of Star by Cycle Graphs

Meghan Galiardi, Daniel Perry, Hsin-hao Su

Stone hill College

Labeling of the Graphs Star by Cycle Graphs

• The edges of the graph are labeled by the group Z2={0, 1}

• The vertices are labeled according to the adjacent edges

• Vertex labeled 0 if the number of edges adjacent labeled 0 is greater than the edges labeled 1

• Vertex labeled 1 if the number of edges adjacent labeled 1 is greater than the edges labeled 0

• Vertex unlabeled if the number of edges adjacent labeled 0 is equal to the edges labeled 1

Edge-Friendly Graphs Star by Cycle Graphs

• The graphs are said to be edge-friendly if the number of 1-edges and 0-edges differ by no more than 1.

|e(0)-e(1)|≤ 1

Example:

total edges = 12

e(0) = 6

e(1) = 6

Edge-Balance Index Set Star by Cycle Graphs

• The edge-balance index is the difference between 0-vertices and 1-vertices

EBI =|v(0) – v(1)|

• The edge-balance index set for graph G is the set of all possible edge-balance indices that G can have

• We looked for the edge-balance index sets of two types of graphs

Example:

v(0) = 3

v(1) = 2

EBI = 1

Flux Capacitor Graphs Star by Cycle Graphs

• Definition : A flux capacitor graph is composed of two different types of graphs, a star graph and a cycle graph. A star graph, St(n), consists of a center vertex and n surrounding vertices each connected to the center. A cycle graph, Cm, consists of m vertices each connected to 2 others to form a cycle where m≥3. A flux capacitor graph, FC(n, m), is a St(n) graph where on each outer vertex there is a graph Cm.

St(3)

C3

FC(3, 3)

Theorems Star by Cycle Graphs

• EBI(FC(n, m)) =

{0, 1, … , n-1} if m is odd

{0, 1, … , n} if n is odd and m is even

{0, 1, … , n-1} if n is even and m is even

How we proved it Star by Cycle Graphs

• Started with FC(n, 3) and FC(n, 4)

• First we looked for the most efficient way to label the graphs as to achieve the highest EBI

• From the highest EBI we looked at how we can rearrange the graphs to decrement the EBI by 1

• We rearranged the graphs as many times as it took to achieve EBI from the highest all the way to 0

• The results we found also generalized for any FC(n, m)

FC(n, 3) Star by Cycle GraphsEBI(FC(n, 3)) = {0, 1, … , n-1}

Most efficient way to label is to label the star with all 1-edges and then alternate the cycle with 0 and 1-edges. This creates EBI = n-1.

To decrease the EBI by one, simply switch a 0-edge and a 1-edge on one of the cycles. This changes the 0-vertex to a 1-vertex and adds an additional 0-vertex. Since v(0) was greater that v(1). This change causes the EBI to decrease by 1.

|v(0) – v(1)| = 1

|v(0) – v(1)| = 0

EBI = {0, 1}

Note: We assumed that e(0)≥e(1) and by our labeling v(0)≥v(1). The opposite can also be assumed, but the results for the EBI will still be the same so we only have to look at one case

FC(n, 4) if n is even Star by Cycle Graphs EBI(FC(n, 4)) = {0, 1, … , n-1}

When n is even the most efficient way to label the graph is shown below. This creates EBI = n-1.

• Again the edges can be rearranged to decrement the EBI by 1 each time, and all the EBI from n-1 all the way to 0.

|v(0) – v(1)| = 1

|v(0) – v(1)| = 0

EBI = {0, 1}

FC(n, 4) if n is odd Star by Cycle Graphs EBI(FC(n, 4)) = {0, 1, … , n}

The same can be done when n is odd, there is just a slightly different way of labeling the graph for the highest EBI. This creates EBI = n-1.

• Again the edges can be rearranged to decrement the EBI by 1 each time, and all the EBI from n-1 all the way to 0.

|v(0) – v(1)| = 3

|v(0) – v(1)| = 2

|v(0) – v(1)| = 1

|v(0) – v(1)| = 0

EBI = {0, 1, 2, 3}

FC(n, m) Star by Cycle Graphs

The results for EBI(FC(n, m)) generalize from FC(n, 3) and FC(n, 4)

Example: EBI(FC(4, 7)) = {0, 1, 2, 3}

|v(0) – v(1)| = 3

|v(0) – v(1)| = 2

|v(0) – v(1)| = 1

|v(0) – v(1)| =0

L-Product of Cycle by Star Star by Cycle Graphs

• Definition: An L-product of cycle by star graph is the same as a flux capacitor graph, the only difference being there is an additional cycle, Cm, on the center vertex of the star. It is represented as St(n)xLCm.

St(3)xLC3

Theorems Star by Cycle Graphs

• EBI(St(n)xLCm) =

{0, 1, … , n+1} if m is odd

{0, 1, … , n+1} if n is odd and m is even

{0, 1, … , n} if n is even and m is even

How we proved it Star by Cycle Graphs

• We started with FC(n+1, m). By removing 1 edge and merging 2 vertices we can create St(n)xLCm

• When m is odd, EBI(FC(n, m)) = {0, 1, … , n-1}

• EBI(FC(n+1, m)) = {0, 1, … , n}

• FC(n+1, m) has an even number of edges. Removing an edge will keep the graph edge friendly and cause the EBI to change at most by 1

• So EBI(St(n)xLCm) = {0, 1, … , n+1} when m is odd

How we proved it Star by Cycle Graphs

• When n+1 is even and m is even,

• When n is even EBI(FC(n, m)) = {0, 1, … , n-1}

• When n+1 is even EBI(FC(n+1, m)) = {0, 1, … , n}

• FC(n+1, m) has an even number of edges. Removing an edge will keep the graph edge friendly and cause the EBI to change at most by 1

• Starting with n+1 even and removing the edge makes n odd, while m stays even

• So EBI(St(n)xLCm) = {0, 1, … , n+1} when n is odd and m is even

How we proved it Star by Cycle Graphs

• When n+1 is odd and m is even,

• FC(n+1, m) has an odd number of edges. Removing an edge may not keep the graph edge friendly so the previous method does not work

• Results from the flux capacitor graphs could not be used so we created a most efficient way to label

• It was found EBI(St(n)xLCm) = {0, 1, … , n} when n is even and m is even

St(2)x Star by Cycle GraphsLC3

|v(0) – v(1)| = 3

|v(0) – v(1)| = 2

|v(0) – v(1)| = 1

|v(0) – v(1)| = 0

n is even, m is odd

EBI(St(n)xLCm) = {0, 1, … , n+1}

EBI = {0, 1, 2, 3}

Conclusions Star by Cycle Graphs

• EBI(FC(n, m)) =

{0, 1, … , n-1} if m is odd

{0, 1, … , n} if n is odd and m is even

{0, 1, … , n-1} if n is even and m is even

• EBI(St(n)xLCm) =

{0, 1, … , n+1} if m is odd

{0, 1, … , n+1} if n is odd and m is even

{0, 1, … , n} if n is even and m is even