CSCI 4260MATH 4150

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CSCI 4260-MATH 4150

• GRAPH THEORY- Lecture 22
• DISTANCE

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Distance

• Mathematical abstraction of the notion of a
  measurement (between the elements of a set S)
• Satisfies
• d(a,b) ? 0, ??a,b?S
• d(a,b) 0 iff a b
• d(a,b) d(b,a) ?a,b?S
• d(a,b) ? d(a,c) d(c,b) ?a,b, c ?S

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Metric Space

• If d(.,.) is a distance function for S, then the
  pair (S,d) is called a metric space.
• We can take measurements in this space
• The length of the shortest path function is a
  distance function over the vertices
• Today we will study this metric space
• Vertices that are far from each other, far from
  everybody else, close to each other etc
• Warning lots of definitions!

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Distance related definitions for G (V,E)

• eccentricity (of a vertex)
• e(v) max d(u,v) u ? V
• radius of G
• rad(G) min e(v) v ? V
• diameter of G
• diam(G) maxe(v) v ? V
• If e(v) rad(G), then v is a central vertex
• Cen(G) subgraph induced by central vertices

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Example

• What is the radius, diameter and center of

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Theorem

• For every nontrivial, connected graph G
• rad(G) ? diam(G) ? 2 rad(G)

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Proof

• rad(G) ? diam(G) is obvious
• diam(G) ? 2 rad(G)

u
d(u,v) diam(G)
v
A central vertex w
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Some simple properties for you to prove

• For adjacent u and v
• d(u,x) d(v,x) ? 1 for all x
• e(u) e(v) ? 1

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Theorem

• Every graph is the center of some graph

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Proof
G
What is the center of this graph? What is e(u)
for all u?
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Peripheral Vertices

• Analogous to the notion of a center
• If e(v) diam(G), then v is a peripheral vertex
• The periphery, per(G) subgraph induced by
  peripheral vertices
• What is the farthest distance from u
• If u is central, the answer is minimized
• If u is peripheral, the answer is maximized

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Question

• Is the periphery similar to the center in the
  sense that can every graph be the periphery of
  some other graph

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Theorem

• A nontrivial graph G is the periphery of some
  graph iff (?v, e(v) 1) or (?v, e(v) ? 1)

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(?v e(v) 1) or (?v e(v) ? 1) ? per(H) G for
some H

• If ?v e(v) 1 then G must be complete
• Can you find an H with per(H) G?
• If ?v e(v) ? 1, construct H by adding a new
  vertex w and connecting w to all vertices of G
• In H
• What is e(w)?
• What is the eccentrity of the remaining vertices
• What is the diameter of H
• What is the periphery of H

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per(H) G for some H ?(?v e(v) 1) or (?v e(v)
? 1)

• Let G be a graph with e(u) 1 and e(v) gt 1 for
  some u and v
• Observe that per(G) ? G
• Suppose for contradiction, ??H with per(H) G
• Clearly, G ? H
• Sort the vertices of H with respect to their ecc.
• There must be a threshold k gt 1 such that e(v)
  lt k means that v is in H - G
• Take a vertex x with e(x) 1 in G
• x is adjacent to all vertices in G
• it is peripheral it must be at least distance k
  away from some vertex w
• w can not be in G
• But then d(w,x) gt k ? e(w) gt k

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So far

• v is a peripheral vertex if e(v) diam(G)
• This means that, ?u s.t d(u,v) diam(G)
• In a sense, u and v are as far as possible
• Lets look at this property locally..i.e for a
  given vertex u, the farthest vertex v s.t. d(u,v)
  e(u) is called an eccentric vertex of u.
• A vertex is called an eccentric vertex of the
  graph if it is an eccentric vertex of some other
  vertex.

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If u is an eccentric vertex of v

• Is v an eccentric vertex of u?
• Is this true
• e(v) ? e(u)

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Set of all eccentric vertices

• The eccentric subgraph, Ecc(G), of G is the
  subgraph induced by eccentric vertices of G.
• Which graphs are eccentric subgraphs of some
  other graphs?

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Theorem

• A nontrivial graph G is the eccentric subgraph of
  some graph iff (?v, e(v) 1) or (?v, e(v) ? 1)
• Proof pretty much the same as the previous
  theorem

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Lets further analyze local distances

• If v is an eccentric vertex of u, then v is
  farther from u then any other vertex.
• Lets change this requirement to then v is
  farther from u then any other neighbor of v.
• Such a v is called a boundary vertex of u
• That is if d(u,v) ? d(u,w) for all w ? N(v)
• A vertex is a boundary vertex of the graph if
  its boundary vertex of some other vertex

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Lets look at an example
z not peripheral but an eccentric vertex of w
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s
2
4
4
3
3
2
w boundary vertex of s but not an eccentric
vertex
Peripheral vertex
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Intuition for boundary vertices

• If v is a boundary vertex of u
• Then, if you reach v from u via a geodesic, you
  can not proceed farther
• In the sense that there is no neighbor of v with
  a longer geodesic (from u)
• Some vertices can not be boundary vertices.

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Theorem

• No cut-vertex of a connected graph is a boundary
  vertex.

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Proof

• Suppose not. Let v be a cut vertex and a boundary
  vertex of u.

v
u
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Comment

• Therefore no cut vertex is an eccentric vertex or
  a peripheral vertex either.

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Simplicial vertices

• A vertex v is simplicial (or complete or extreme)
  if the subgraph induced by N(v) is complete.
• Suppose v is simplicial
• Take u ? N(v)
• d(u,v) 1
• d(u,w) 1 for all w in N(w)
• So v is a boundary vertex of u. In fact

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Theorem

• Let G be a connected graph.
• A vertex v is a boundary vertex of every other
  vertex (other than v)
• iff
• v is a complete (simplicial) vertex

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Proof simplicial ? boundary of any vertex

• Let v be a complete vertex. Pick u ? v
• Fix a u-v geodesic and consider w in N(v)

x
v
u
w
What can you say about d(u,w)?
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Proof boundary of any vertex ? simplicial

• contrapositive
• v is not simplicial ? not boundary vertex of
  some vertex u
• Let u and w be two neighbors of v that are not
  adjacent
• d(u,w) gt d(u,v)

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Recap

• We proved that if v is simplicial then it is a
  boundary vertex of everybody else.
• How about vertices such that everybody else is
  their boundary vertices?

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Theorem

• Let G be a connected, non-trivial graph.
• Let u be a vertex.
• Every v ? u is a boundary vertex of uiff
• e(u) 1

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Proof e(u) 1 ? every v is a boundary vertex of
u

• Pick any v
• Take w a neighbor of v
• Since e(u) 1, d(u,v) ? 1
• Since e(u) 1, d(u,w) ? 1

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every v is a boundary vertex of u ? e(u) 1

• Suppose not.
• Since e(u) gt 1, there is an x with d(u,x) 2
• Consider the path x-y-u
• Vertex y is not a boundary vertex of u
• d(u,y) 1 and d(u,x) 2
• Contradicts with the assumption that every vertex
  is a boundary vertex of u

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Finally

• Vertices that are not boundary vertices?
• To understand such vertices we need the notion of
  betweenness.
• Vertex y lies between x and z ifd(x,z) d(x,y)
  d(y,z)
• A vertex v is an interior vertex of a graph if
  for every u ? v, there is some w such that v lies
  between u and w.

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Theorem

• A vertex v is a boundary vertex of G iff v is not
  an interior vertex of G.

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Proof boundary ? not interior

• Suppose not. Let v be a boundary vertex (of, say
  u). Suppose v is also an interior vertex.
• Then, there must be some w such that v lies
  between u and w.
• Can v be a boundary vertex of u?

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Proof not interior ? boundary

• Suppose v is not an interior vertex.
• This means that there is some u and
• for every w (other than u and v) v does not lie
  between u and w.
• Let x be any neighbor of v
• d(u,x) lt d(u,v) d(v,x) d(u,v) 1
• Sod(u,x) ? d(u,v) and v is a boundary vertex of
  u.

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What have we learned

• The vertices of a graph can be partitioned into
• Interior and boundary vertices
• Some boundary vertices are eccentric (as far as
  possible from some other vertex)
• Some eccentric vertices are peripheral (furthest
  possible overall)

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Administrative business

• Next lecture presentations
• Also course evaluation
• Pick up graded hw3