On the edge balance index of flux capacitor and l product of star by cycle graphs
This presentation is the property of its rightful owner.
Sponsored Links
1 / 18

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


  • 55 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

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

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


On the edge balance index of flux capacitor and l product of star by cycle graphs

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

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

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

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

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

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

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

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

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

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

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

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


Theorems1

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 it1

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


On the edge balance index of flux capacitor and l product of star by cycle graphs

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


On the edge balance index of flux capacitor and l product of star by cycle graphs

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 x l c 3

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

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


  • Login