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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

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

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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

• 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

• 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

• 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

• 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

• 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

• 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)EBI(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 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 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)

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

• 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

• 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

• 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

• 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

• 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)xLC3

|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

• 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