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Path Elongation and rReduced Cutting
Numbers of Cycles
Brad Bailey
Dianna Spence
North Georgia College & State University
Joint Mathematics Meetings 2010
Agenda


Background
New Terms
-What is a “reduced cutting number”?

Min/Max Problems
-Minimum or maximum edges for given cutting #

Path Elongation
-Initial results
Notation and Assumptions

All graphs considered are simple and
connected unless otherwise stated.

V(G) = vertex set of G

E(G) = edge set of G

dist(u, v, G) is length of shortest path from
u to v in graph G

k(G) is number of components in G
Background

Imagine: Find a parade route through a city


Starts and ends at same place
Does not “disconnect” city when closed to traffic
Edges = Streets
Vertices = Intersections
Definitions
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
For a cycle C contained within a simple
connected graph G, the cutting number of
cycle C, denoted C#(C,G), is the number of
components in G – E(C).
For a simple connected graph G, the cutting
number of graph G is
C#(G) = max{C#(C,G) for all cycles C in G}
If G is acyclic, C#(G) = 0.
Example



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Cycle with cutting
number 1
Cycle with cutting
number 2
Cycle with cutting
number 3
Therefore, graph
has cutting
number 3
More Definitions


kr(G) denotes the number of components of
G with order at least r.
The graph G(C,r) is the graph that results
from graph G by removing the edges of C
and then deleting any components of order
less than r.
C
G
G(C,3)
Extension of Definition

For a cycle C contained within a simple
connected graph G, the r-reduced cutting
number of cycle C, denoted C#r(C,G), is the
number of components in G – E(C) with order
at least r.
C#2(C,G) = 3
C
C#3(C,G) = 1
C#4(C,G) = 1
C#5(C,G) = 0
G
Notes on r-Reduced Cutting #

C#r(C,G) = kr(G – E(C)) = k(G(C,r))

The cutting number as originally defined is
simply the 1-reduced cutting number.
Min/Max Problems
Definitions

mr(k,n) is the minimum number of edges in a
simple connected graph on n vertices with
r-reduced cutting number k

Mr (k,n) is the maximum number of edges in a
simple connected graph on n vertices with
r-reduced cutting number k
Results for m1(k,n) – Minimum
m1(2,n) = n+2 for n  4
...
m1 (k,n) = n for 3 ≤ k ≤ n
Results for mr(k,n) – Minimum
 n  1
For 1  r  
, n  4, mr(2,n) = n

 2 
n
n
For 1  r    and 3  k  , mr (k,n) = n for
r
2
3≤k≤n
Mr(k,n) – Maximum
 n  1
For n  5, M1 (2,n) = 
2
 2 
For n  5,
 r   n  r (k  1) 
  min{ n, 2r (k  1)}
Mr (k,n) = (k  1)   
2
2
 


Path Elongation
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Motivation:
 What is the impact on the length of a shortest path
between two vertices when a cycle is removed?
Definition:
 Choose vertices u and v in graph G.
Then for some cycle C:

If u and v are in the same component of G-E(C), then
pe(u, v, C,G) = dist(u,v,G - E(C)) – dist(u,v,G).

If u and v are not in the same component of G-E(C) then
we consider the path elongation to be infinite.
Example
v
Cycle C1
pe(u,v,C1,G) = 5 – 3 = 2
Cycle C2
pe(u,v,C2,G) = infinite
u
Ready observations

For complete graphs, path elongation is
always less than or equal to 1.

The only graphs for which path elongation is
finite for all cycles and pairs of vertices are
the graphs with cutting number 1.
Path Elongation for a pair of
vertices

pe1(u, v, G) is the maximum of pe(u, v, C, G)
over all cycles C.
v
The pe1(u,v,C1,G) = 0
Cycle C1
Cycle C2
The pe1(u,v,C2,G) = 2
u
Therefore, pe1(u,v,G) = 2
Path Elongation of a cycle

pe2(C, G) is the maximum of pe(u,v,C,G)
over all pairs of vertices u and v in G.

Theorem:
pe2(C, G) =max{pe(u,v,C,G): u and v in C}.
v'
v
u
u'
Path Elongations of a graph
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
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pe1(G) = max{pe1(u, v, G) : u and v in G}
pe2(G) = max{pe2(C, G) : C in G}
Theorem:
For a connected graph G, pe1(G) = pe2(G).
Questions