some notes on the inverse domination conjecturesome notes on the inverse domination conjecture john...
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Outline Introduction Progress
Some Notes on the Inverse DominationConjecture
John Asplund, Joe Chaffee, James Hammer∗
Auburn University
April 20, 2013
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1 IntroductionDefinition TheoryObservationsInitial Conjecture
2 ProgressInitial ProgressStrengthening the ConjectureBaby StepsExemplary CaseSummary
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Definition Theory
Note.
Graphs in this talk are simple.
Definition
For a vertex v ∈ V (G), N(v) = {u | {u, v} ∈ E(G)} andN [v] = {v} ∪N(v). If S ⊂ V (G), N [S] = ∪v∈SN [v]. A setD is dominating if N [D] = V (G).
The domination number of G, denoted γ(G), is the smallestcardinality of a set that dominates V (G).
γ(G) = 3 in the graph below.
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Definition Theory
Note.
Graphs in this talk are simple.
Definition
For a vertex v ∈ V (G), N(v) = {u | {u, v} ∈ E(G)} andN [v] = {v} ∪N(v). If S ⊂ V (G), N [S] = ∪v∈SN [v]. A setD is dominating if N [D] = V (G).
The domination number of G, denoted γ(G), is the smallestcardinality of a set that dominates V (G).
γ(G) = 3 in the graph below.
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Definition Theory
Note.
Graphs in this talk are simple.
Definition
For a vertex v ∈ V (G), N(v) = {u | {u, v} ∈ E(G)} andN [v] = {v} ∪N(v). If S ⊂ V (G), N [S] = ∪v∈SN [v]. A setD is dominating if N [D] = V (G).
The domination number of G, denoted γ(G), is the smallestcardinality of a set that dominates V (G).
γ(G) = 3 in the graph below.
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Outline Introduction Progress
Definition Theory
Note.
Graphs in this talk are simple.
Definition
For a vertex v ∈ V (G), N(v) = {u | {u, v} ∈ E(G)} andN [v] = {v} ∪N(v). If S ⊂ V (G), N [S] = ∪v∈SN [v]. A setD is dominating if N [D] = V (G).
The domination number of G, denoted γ(G), is the smallestcardinality of a set that dominates V (G).
γ(G) = 3 in the graph below.
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Outline Introduction Progress
Definition Theory
Note.
Graphs in this talk are simple.
Definition
For a vertex v ∈ V (G), N(v) = {u | {u, v} ∈ E(G)} andN [v] = {v} ∪N(v). If S ⊂ V (G), N [S] = ∪v∈SN [v]. A setD is dominating if N [D] = V (G).
The domination number of G, denoted γ(G), is the smallestcardinality of a set that dominates V (G).
γ(G) = 3 in the graph below.
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Definition Theory
Definition
Given a minimum dominating set D of a graph G with noisolated vertices, an inverse dominating set is a set ofvertices in V (G) \D that dominates G.
The inverse domination number of G, denoted γ′(G), is thesmallest cardinality of an inverse dominating set for someminimum dominating set.
γ′(G) = 4 in the graph below.
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Definition Theory
Definition
Given a minimum dominating set D of a graph G with noisolated vertices, an inverse dominating set is a set ofvertices in V (G) \D that dominates G.
The inverse domination number of G, denoted γ′(G), is thesmallest cardinality of an inverse dominating set for someminimum dominating set.
γ′(G) = 4 in the graph below.
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Definition Theory
Definition
Given a minimum dominating set D of a graph G with noisolated vertices, an inverse dominating set is a set ofvertices in V (G) \D that dominates G.
The inverse domination number of G, denoted γ′(G), is thesmallest cardinality of an inverse dominating set for someminimum dominating set.
γ′(G) = 4 in the graph below.
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Observations
Notice.
γ(G) ≤ γ′(G)
Recall.
γ(G) ≤ α(G) (where α(G) is the maximum cardinality of a setof independent vertices)
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Observations
Notice.
γ(G) ≤ γ′(G)
Recall.
γ(G) ≤ α(G) (where α(G) is the maximum cardinality of a setof independent vertices)
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Initial Conjecture
Assertion
Kulli and Sigarkanti introduced the inverse domination numberand made the following assertion:
γ′(G) ≤ α(G)
Initial Proof
Their proof of the assertion was incorrect. This was noticed byGayla Domke, Jean Dunbar, Teresa Haynes, Steve Hedetniemi,and Lisa Markus.
Inverse Domination Conjecture
Given a graph G with no isolated vertices,
γ′(G) ≤ α(G).
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Initial Conjecture
Assertion
Kulli and Sigarkanti introduced the inverse domination numberand made the following assertion:
γ′(G) ≤ α(G)
Initial Proof
Their proof of the assertion was incorrect. This was noticed byGayla Domke, Jean Dunbar, Teresa Haynes, Steve Hedetniemi,and Lisa Markus.
Inverse Domination Conjecture
Given a graph G with no isolated vertices,
γ′(G) ≤ α(G).
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Initial Conjecture
Assertion
Kulli and Sigarkanti introduced the inverse domination numberand made the following assertion:
γ′(G) ≤ α(G)
Initial Proof
Their proof of the assertion was incorrect. This was noticed byGayla Domke, Jean Dunbar, Teresa Haynes, Steve Hedetniemi,and Lisa Markus.
Inverse Domination Conjecture
Given a graph G with no isolated vertices,
γ′(G) ≤ α(G).
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Initial Progress
True Fact (Johnson, Prier, and Walsh)
If G has no isolated vertices and γ(G) ≤ 4 then γ′(G) ≤ α(G).
The Problem
However, their method is difficult to extend to γ(G) ≥ 5.
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Initial Progress
True Fact (Johnson, Prier, and Walsh)
If G has no isolated vertices and γ(G) ≤ 4 then γ′(G) ≤ α(G).
The Problem
However, their method is difficult to extend to γ(G) ≥ 5.
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Strengthening the Conjecture
Definition
A fixed inverse domination number of G, denoted Γ′D(G), is thesmallest cardinality of an inverse dominating set for a fixedminimum dominating set D.
Observation
Clearly γ(G) ≤ γ′(G) ≤ Γ′D(G).
Conjecture
If G has no isolated vertices then Γ′D(G) ≤ α(G) for anyminimum dominating set D.
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Strengthening the Conjecture
Definition
A fixed inverse domination number of G, denoted Γ′D(G), is thesmallest cardinality of an inverse dominating set for a fixedminimum dominating set D.
Observation
Clearly γ(G) ≤ γ′(G) ≤ Γ′D(G).
Conjecture
If G has no isolated vertices then Γ′D(G) ≤ α(G) for anyminimum dominating set D.
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Strengthening the Conjecture
Definition
A fixed inverse domination number of G, denoted Γ′D(G), is thesmallest cardinality of an inverse dominating set for a fixedminimum dominating set D.
Observation
Clearly γ(G) ≤ γ′(G) ≤ Γ′D(G).
Conjecture
If G has no isolated vertices then Γ′D(G) ≤ α(G) for anyminimum dominating set D.
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Notes
Based on the observation that γ′(G) ≤ Γ′D(G), ourconjecture is certainly a strengthening of the inversedomination conjecture.
Also, note that for G = K2,m, γ′(G) = 2, but there exists a
minimum dominating set D such that Γ′D(G) = m.
However, in this extremal case and a few others we’vestudied, Γ′D(G) ≤ α(G)
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Notes
Based on the observation that γ′(G) ≤ Γ′D(G), ourconjecture is certainly a strengthening of the inversedomination conjecture.
Also, note that for G = K2,m, γ′(G) = 2, but there exists a
minimum dominating set D such that Γ′D(G) = m.
However, in this extremal case and a few others we’vestudied, Γ′D(G) ≤ α(G)
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Notes
Based on the observation that γ′(G) ≤ Γ′D(G), ourconjecture is certainly a strengthening of the inversedomination conjecture.
Also, note that for G = K2,m, γ′(G) = 2, but there exists a
minimum dominating set D such that Γ′D(G) = m.
However, in this extremal case and a few others we’vestudied, Γ′D(G) ≤ α(G)
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Baby Steps
Sub-Conjecture
If G has no isolated vertices and γ(G) = 5 then Γ′D(G) ≤ α(G)for any minimum dominating set D.
Let D be a minimum dominating set of G and let I be amaximum independent set where I ∩D is as small as possible.For illustration, suppose |I ∩D| = 3
Goal: Form an inverse dominating set using I \D or reachsome contradiction.
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Baby Steps
Sub-Conjecture
If G has no isolated vertices and γ(G) = 5 then Γ′D(G) ≤ α(G)for any minimum dominating set D.
Let D be a minimum dominating set of G and let I be amaximum independent set where I ∩D is as small as possible.For illustration, suppose |I ∩D| = 3
Goal: Form an inverse dominating set using I \D or reachsome contradiction.
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Baby Steps
Sub-Conjecture
If G has no isolated vertices and γ(G) = 5 then Γ′D(G) ≤ α(G)for any minimum dominating set D.
Let D be a minimum dominating set of G and let I be amaximum independent set where I ∩D is as small as possible.For illustration, suppose |I ∩D| = 3
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Goal: Form an inverse dominating set using I \D or reachsome contradiction.
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Exemplary Case
Suppose both d0 and d1 are adjacent to I \D.
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Exemplary Case
What is dominating the neighborhood of I ∩D inV (G) \ (I ∩D)?
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Exemplary Case
What is dominating the neighborhood of I ∩D inV (G) \ (I ∩D)?Answer: I \D
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Exemplary Case
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A lot of case analysis is used, but the idea is that if this were notthe case, either |I| would be bigger, |I ∩D| would be smaller, orwe could form an inverse dominating set of the proper size.
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Exemplary Case
Let X be a set of no more than |I ∩D| vertices dominating eachvertex in I ∩D
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So (I ∪X) \D is an inverse dominating set whereΓ′D(G) ≤ α(G).
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Exemplary Case
Let X be a set of no more than |I ∩D| vertices dominating eachvertex in I ∩D
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So (I ∪X) \D is an inverse dominating set whereΓ′D(G) ≤ α(G).
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Exemplary Case
Suppose both d0 and d1 are not adjacent to I \D.
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Then I \D is not dominated by D . . . a contradiction!
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Exemplary Case
Suppose both d0 and d1 are not adjacent to I \D.
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Then I \D is not dominated by D . . . a contradiction!
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Exemplary Case
Suppose only d1 is adjacent to I \D. (We say D′ is the set ofvertices not adjacent to I \D; so, |D′| = 1.)
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Exemplary Case
Let Y = V (G) \DLet Z be a maximal independent set in Y that contains atleast one vertex that is adjacent to a vertex in I ∩D andthen is adjacent to as many vertices in D \ I as possible.
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Exemplary Case
Let Y = V (G) \DLet Z be a maximal independent set in Y that contains atleast one vertex that is adjacent to a vertex in I ∩D andthen is adjacent to as many vertices in D \ I as possible.
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Exemplary Case
Let U be the set of at most |D \ (D′ ∪ (I ∩D))| vertices inV (G) \D adjacent to all of the vertices inD \ (D′ ∪ (I ∩D)).Let X be the set of at most |(I ∩D) \D0| vertices inV (G) \D adjacent to every vertex in (I ∩D) \D0.
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Exemplary Case
Let U be the set of at most |D \ (D′ ∪ (I ∩D))| vertices inV (G) \D adjacent to all of the vertices inD \ (D′ ∪ (I ∩D)).Let X be the set of at most |(I ∩D) \D0| vertices inV (G) \D adjacent to every vertex in (I ∩D) \D0.
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Exemplary Case
If |D′| = 0 or if |D′| = 2then Γ′D(G) ≤ α(G) asbefore.
If |Z| ≤ |(I \D) ∪D0| − 1then Z ∪X ∪ U is aninverse dominating set ofan appropriate size.
If |Z| > |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is alarger independent set thanI.
If |Z| = |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is anindependent set withsmaller intersection to D.
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Exemplary Case
If |D′| = 0 or if |D′| = 2then Γ′D(G) ≤ α(G) asbefore.
If |Z| ≤ |(I \D) ∪D0| − 1then Z ∪X ∪ U is aninverse dominating set ofan appropriate size.
If |Z| > |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is alarger independent set thanI.
If |Z| = |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is anindependent set withsmaller intersection to D.
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Exemplary Case
If |D′| = 0 or if |D′| = 2then Γ′D(G) ≤ α(G) asbefore.
If |Z| ≤ |(I \D) ∪D0| − 1then Z ∪X ∪ U is aninverse dominating set ofan appropriate size.
If |Z| > |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is alarger independent set thanI.
If |Z| = |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is anindependent set withsmaller intersection to D.
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Exemplary Case
If |D′| = 0 or if |D′| = 2then Γ′D(G) ≤ α(G) asbefore.
If |Z| ≤ |(I \D) ∪D0| − 1then Z ∪X ∪ U is aninverse dominating set ofan appropriate size.
If |Z| > |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is alarger independent set thanI.
If |Z| = |(I \D) ∪D0| thenZ ∪ ((I ∩D) \D0) is anindependent set withsmaller intersection to D.
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Summary
In Conclusion
There are similar arguments for the cases when|I ∩D| = {1, 2}; however, some oddities occur.
The benefit of this approach is that if sub-conjecture is nottrue, it would point to a possible counterexample.
This approach is also a plausible approach in that Drs.Peter Johnson and Jessica McDonald (Auburn University)have used this technique to re-prove the inverse dominationconjecture for claw-free graphs.
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Summary
In Conclusion
There are similar arguments for the cases when|I ∩D| = {1, 2}; however, some oddities occur.
The benefit of this approach is that if sub-conjecture is nottrue, it would point to a possible counterexample.
This approach is also a plausible approach in that Drs.Peter Johnson and Jessica McDonald (Auburn University)have used this technique to re-prove the inverse dominationconjecture for claw-free graphs.
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Summary
In Conclusion
There are similar arguments for the cases when|I ∩D| = {1, 2}; however, some oddities occur.
The benefit of this approach is that if sub-conjecture is nottrue, it would point to a possible counterexample.
This approach is also a plausible approach in that Drs.Peter Johnson and Jessica McDonald (Auburn University)have used this technique to re-prove the inverse dominationconjecture for claw-free graphs.
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War Eagle!
Thank You For YourKind Attention !
Figure: War Eagle !
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