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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. 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

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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)

  6. 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

  7. 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)

  8. 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

  9. 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}

  10. 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}

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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}

  18. 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

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