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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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

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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