Intersection Graphs of Maximal Convex Sub-Polygons of k ... · Willamette REU k -MSP Intersection Graphs August 4, 2017 3 / 67. Maximal Rectangles of a Polyomino: Shearer 1982, Maire

Geometric Representations of Graphs

Intersection Graphs of Maximal ConvexSub-Polygons of k -Lizards

Caroline Daugherty Josh LaisonRebecca Robinson Kyle Salois

Willamette Mathematics Consortium REU

Willamette REU k -MSP Intersection Graphs August 4, 2017 1 / 67

Outline

1 IntroductionIntersection GraphsDefinitionsUseful Lemmas

2 ResultsGraphs that are k -MSP GraphsSeparating ExamplesGeneralizations and Conjectures

3 Future Research


Intersection Graphs

A graph G is a set V (G) of points called vertices, alongside a set ofedges, which connect the vertices and form a set E(G).

An intersection graph H is a graph formed by a family of sets whereeach vertex represents a set, and two vertices have an edge betweenthem whenever the corresponding sets intersect.

A

B

C

D

E

a

b

c

d

e


Maximal Rectangles of a Polyomino: Shearer 1982, Maire 1993

A polyomino is a geometric figure formed by joining unit squares inthe plane.

We can draw an intersection graph by creating a vertex for everymaximal rectangle and connecting two vertices with an edge if theirmaximal rectangles intersect.






















Maximal Rectangles of a Polyomino

Shearer 1982, Maire 1993All intersection graphs formed by the maximal rectangles of apolyomino are perfect.

In a perfect graph, the chromatic number of every induced subgraphequals the size of the largest clique of that subgraph.An induced subgraph is a set of vertices of a graph G that areconnected by all the same edges that connect them in G.

A clique in a graph is a set of vertices where every pair has an edge.




In a perfect graph, the chromatic number of every induced subgraphequals the size of the largest clique of that subgraph.

An induced subgraph is a set of vertices of a graph G that areconnected by all the same edges that connect them in G.













Definitions

A k-lizard is a simply connected polygon such that each side hasslope in the set

{i⇡/k ,�i⇡/k , 0 | 1 i k}.

An example of a 4-lizard.


Definitions

The direction of a side on a k -lizard P is a member of the set{✓0, ✓1, ..., ✓2k�1}, where ✓i = i⇡/k , and the angle between two sides

is a measure of the interior angle of the k -lizard, and will also be amember of the set {✓0, ✓1, ..., ✓2k�1}.

θ0

θ1

θ0

θ1θ2 θ1

θ0

θ2θ3

θ2

θ0

θ1

θ3θ4

A reflex angle is an angle of measure larger than ⇡ but less than 2⇡.


Definitions



θ0

θ1

θ0

θ1θ2 θ1

θ0

θ2θ3

θ2

θ0

θ1

θ3θ4


θ1


Definitions



θ0

θ1

θ0

θ1θ2 θ1

θ0

θ2θ3

θ2

θ0

θ1

θ3θ4


θ1

θ2


Definitions



θ0

θ1

θ0

θ1θ2 θ1

θ0

θ2θ3

θ2

θ0

θ1

θ3θ4


θ1θ5

θ2


Definitions

A region Q ✓ P is a maximal convex sub-polygon, or scale, of P ifQ is a convex k -lizard that is maximal in P.


Definitions



Definitions


A graph G is a k -maximal sub-polygon graph or k -MSP graph if it isthe intersection graph of the scales of a k -lizard.


Definitions


A graph G is a k -maximal sub-polygon graph or k -MSP graph if it isthe intersection graph of the scales of a k -lizard.


Convex Subset Lemma

LemmaLet R be a convex k-lizard contained within a k-lizard L. Then R iscontained within at least one scale of L.

R

S


Convex Subset Lemma

LemmaLet R be a convex k-lizard contained within a k-lizard L. Then R iscontained within at least one scale of L.

R

S

Proof.

If R is maximal, then R is a scale.If R is not maximal, then we can extend the sides of R in at least one ofthe k directions to a side of L to create S within which R is contained.


A reflex angle in 3-MSP

Consider a ✓4 angle in the following 3-lizard. There are 3 scalestouching this reflex angle.

θ4

We define a proto-scale to be a line segment contained in the interiorof the k -lizard in a ✓i direction which touches the boundary of thek -lizard at a reflex angle.
















We define a proto-scale to be a line segment contained in the interiorof the k -lizard in a ✓i direction which touches the boundary of thek -lizard at a reflex angle.


Definitions

Let r be a reflex angle in a k -lizard L with internal angle measure ✓k+j .Then there exist j + 1 proto-scales, with each one contained in at leastone scale.

θ7

A ✓k+3 reflex angle in 4-MSP yielding 4 proto-scales


Definitions


θ7



Definitions


θ7



Definitions


θ7



Definitions


pθ7



Proto-scales Lemma

We define a g-region of a k -lizard L as a region where g scalesintersect.Lemma (Proto-scales)In a k-lizard at a reflex interior angle ✓k+j , there is a g-region, whereg � j + 1.

pθ7


Proto-scales Lemma

We define a g-region of a k -lizard L as a region where g scalesintersect.Lemma (Proto-scales)In a k-lizard at a reflex interior angle ✓k+j , there is a g-region, whereg � j + 1.

pθ7

Proof.

By definition, at a reflex angle ✓k+j , there are j + 1 distinctproto-scales that each belong to at least one scale in L.Each of these proto-scales intersect at p, so the scales formed bythem must intersect as well, so there is a g-region at p.


Ray Lemma

LemmaGiven three vertices a, b, c in a k-MSP graph with correspondingscales A,B,C such that a $ b and b $ c, but a = c, and from everypoint on the boundary of B � A there exists a ray in a ✓i direction thatintersects A, then there exists a scale D intersecting A,B, and C.

A

B

pr

C

Proof.

Since b $ c, but a = c, C must intersect boundary(B � A) at twodistinct points.Consider a point p on the boundary of both B � A and and C \ Band a ray r originating at p with ✓i direction that intersects A.


Ray Lemma

LemmaGiven three vertices a, b, c in a k-MSP graph with correspondingscales A,B,C such that a $ b and b $ c, but a = c, and from everypoint on the boundary of B � A there exists a ray in a ✓i direction thatintersects A, then there exists a scale D intersecting A,B, and C.

A

B

pr

C

Proof.

Since b $ c, but a = c, C must intersect boundary(B � A) at twodistinct points.Consider a point p on the boundary of both B � A and and C \ Band a ray r originating at p with ✓i direction that intersects A.


Ray Lemma

A

BC

pl

q

p’ε

D

Let ✏ > 0 such that B✏(p) is contained within P.Since B is convex, we can extend r by length ✏ into C � B atdirection ✓k+i to form a line segment ` with one endpoint p0.The opposite endpoint of ` should lie at a point q in ABy Convex Subset Lemma, ` is contained in at least one scaleand since it cannot be wholly contained in A, B, or C, there mustexist an additional scale D that contains `.


Parallel Sides Lemma

LemmaGiven scales A,B,C in a lizard, if a $ b, b $ c, and a and b have noshared neighbors, then the region B � A � C that borders both A andC has parallel boundary components s1 and s2 extending from thecorners of A \ B to the corners of B \ C.

A

B

C

s1

s2


Parallel Sides Lemma

Let �1 be a boundarycomponent between B � Aand B \ A, and consider theangle ↵ formed where �1reaches a side s2 of B.If ↵ ✓k�2, then we can usethe ray lemma to show thatsome scale intersects A andB.

A

B

C

s1

s2

δ1

A

B

s1

s2

δ1


End Regions

In a k -lizard L, we say that for a scales A and B, A \ B is an end

region of A if A � B is connected.

A

B

A

B

In a 3-lizard: an end region (left) and not an end region (right).

On the left, A \ B is an end region of A and B. Additionally, A \ B is a2-region since it is contained in 2 different scales. We then say A \ B isan end 2-region of A.


End Regions

In a k -lizard L, we say that for a scales A and B, A \ B is an end

region of A if A � B is connected.

A

B

A

B

In a 3-lizard: an end region (left) and not an end region (right).

On the left, A \ B is an end region of A and B. Additionally, A \ B is a2-region since it is contained in 2 different scales. We then say A \ B isan end 2-region of A.


End 2-Regions Lemma

Lemma (End 2-Regions)Any scale has at most two end 2-regions.

A

B

C

Proof.

Suppose B has three end 2-regions; B \ A, B \ C, and B \ D.The sides between B \ A and B \ C must be parallel by theParallel Sides Lemma.The sides between B \A and B \D must be parallel; thus B \D iseither not an end region, or not a 2-region.


End 2-Regions Lemma


A

B

C

Proof.

Suppose B has three end 2-regions; B \ A, B \ C, and B \ D.

The sides between B \ A and B \ C must be parallel by theParallel Sides Lemma.The sides between B \A and B \D must be parallel; thus B \D iseither not an end region, or not a 2-region.


End 2-Regions Lemma


A

B

C

Proof.

Suppose B has three end 2-regions; B \ A, B \ C, and B \ D.The sides between B \ A and B \ C must be parallel by theParallel Sides Lemma.

The sides between B \A and B \D must be parallel; thus B \D iseither not an end region, or not a 2-region.


End 2-Regions Lemma


A

B

CD

Proof.



End 2-Regions Lemma


A

B

C

D

Proof.



Summary of Lemmas

Convex Subset LemmaLet R be a convex k -lizard contained within a k -lizard L. Then R iscontained within at least one scale of L.

R

S


Summary of Lemmas

Convex Subset LemmaLet R be a convex k -lizard contained within a k -lizard L. Then R iscontained within at least one scale of L.

Proto-Scale LemmaIn a k -lizard at a reflex interior angle ✓k+j , there is a g-region, whereg � j + 1.

pθ7


Summary of Lemmas

Ray LemmaGiven scales A,B,C in a k -lizard, if a $ b, b $ c, and a = c, and fromevery point on the boundary of B � A there exists a ray in a ✓i directionthat intersects A, then there exists a scale D intersecting A,B, and C.

A

B

pr

C

Parallel Sides LemmaGiven scales A,B,C in a k -lizard, if a $ b, b $ c, and a and b haveno shared neighbors, then the region B � A � C has parallel boundarycomponents s1 and s2 from A \ B to B \ C.


Summary of Lemmas



A

B

C

s1

s2


Summary of Lemmas



End 2-Region LemmaAny scale has at most two end 2-regions.

A

B

C


Separating Examples

2-MSP 3-MSP

4-MSP


Graphs that are k -MSP Graphs

Proposition (Complete Graphs).The complete graph Kn is a k -MSP graph for all n, k 2 N, where k � 2.

K4 in 2-MSP, K5 in 3-MSP, and 4-MSP
























For k = 2, construct the "staircase" as shown previously, with n"stairs".For k � 3, construct the first scale A as a triangle with two ✓1angles and one ✓k�2 angle.Add n � 1 bumps to the longest side of A, alternating directions ofthe other two sides of A.

θ1

θk-2

θ1

θk-2

The construction of the 6-MSP representation of K4.



Proposition (Path Graphs).Pj is a k -MSP graph for all j , k 2 N, and k � 2.

The 2- and 3-MSP representation of P6.

For all k , construct the k -lizard with consecutive parallelogramsQ1, ...,Qn, intersecting such that for all 1 a n, Qa [ Qa+1 is apolygon containing one reflex angle with measure ✓k+1.



Proposition (Path Graphs).Pj is a k -MSP graph for all j , k 2 N, and k � 2.

The 2- and 3-MSP representation of P6.

For all k , construct the k -lizard with consecutive parallelogramsQ1, ...,Qn, intersecting such that for all 1 a n, Qa [ Qa+1 is apolygon containing one reflex angle with measure ✓k+1.


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn


Cycles are not k -MSP Graphs

TheoremCj is not a k-MSP graph when j � 4 and k � 2.




Proof.

Assume there exists a k -MSPrepresentation of Cj .j scales must intersect two"neighbor" scales. Connect one pointin each intersection with linesegments.Note that each line segment must becontained in one of the j scales.The j points and j line segmentsmust form a closed path P, whichmust be contained in the k -lizard.

PA

B

C




Proof cont.

Construct a convex k -lizard R suchthat R ✓ P is maximal inside P.The vertices of R must intersect atleast three distinct sides of P, thus Rmust contain points from at leastthree distinct scales.

PR




Proof cont.

R must be contained in a scale S ofL by the Convex Subset Lemma.S corresponds to a vertex withdegree at least three, but all verticesin Cj must have degree two. So Cj isnot a k -MSP graph.

R

S


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

Cn(n>3)


Induced Cycles in k -MSP Graphs

TheoremC5 is an induced subgraph for 3 and 4-MSP but not for 2-MSP.

Proof (C5 is not an induced subgraph for 2-MSP).

Recall from Shearer 1982 and Maire 1993, all 2-MSP graphs areperfect.C5 is not perfect, so it cannot be a subgraph of a 2-MSP graph.

Maire also proved that C4 is the largest induced cycle that can bemade in 2-MSP.




Proof (C5 is not an induced subgraph for 2-MSP).

Recall from Shearer 1982 and Maire 1993, all 2-MSP graphs areperfect.C5 is not perfect, so it cannot be a subgraph of a 2-MSP graph.

Maire also proved that C4 is the largest induced cycle that can bemade in 2-MSP.




Proof (C5 is an induced subgraph for 3 and 4-MSP).

See the constructions for an induced C5 with 3 and 4-MSP.


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

Cn(n>3)

inducedC5



PropositionCn is an induced subgraph of a k -MSP graph when n � 3 and k � 5.

Along one "direction," alternate two types of triangles: one withangle measures (✓1, ✓1, ✓k�2), and another with angle measures(✓1, ✓2, ✓k�3)

Construct n � 3 of these triangles, then complete the cycle with 3paralellograms

θ1

θk-3

θ2

θ1

θ1

θk-2





ConjectureIn 3-MSP and 4-MSP, there is an upper bound on the size of thelargest possible induced cycle.

Alternating triangle construction fails because there is only onetriangle possible in each case (60, 60, 60 in 3-MSP or 45, 45, 90 in4-MSP)

k -MSP Maximum Induced Cycle2-MSP 4 (Maire)3-MSP 124-MSP 16

(5+)-MSP unbounded


Tree Graphs

TheoremAll tree graphs are 2-MSP graphs.

a

b c

d e

A

B C

D E


Tree Graphs


a

b c

d e f

A

BC

D E

F


Tree Graphs


a

b c

d e f

g A

BC

D E

F

G


Tree Graphs

TheoremTree graphs are k-MSP graphs for all k � 3 if and only if they arecaterpillars.

Proof (()

Recall the construction for Pn for any k .We can extend any scale corresponding to any vertex of P.On any of the scales, we can add triangles with interior angles equal to✓k+1.

A 3-lizard whose graph is a caterpillar


Tree Graphs

TheoremTree graphs are k-MSP graphs for all k � 3 if and only if they arecaterpillars.

Proof (()

Recall the construction for Pn for any k .We can extend any scale corresponding to any vertex of P.On any of the scales, we can add triangles with interior angles equal to✓k+1.

A 3-lizard whose graph is a caterpillar


Tree Graphs

Proof ()), by contradiction.

a b c d e

f

g

Recall that NC7 (the smallest non-caterpillar with 7 vertices) is asubgraph of every tree that is not a caterpillar, so it suffices toshow that NC7 is not a k -MSP graph for all k � 3.By the End 2-Regions Lemma, the corresponding scale C has atmost two end 2-regions and has parallel sides.


Tree Graphs

Proof ()) cont.

By the Proto-scales Lemma, the only angle that creates a 2-regionis ✓k+1.

θk+1 θk+1C

F

Note that from any point on boundary(F � C), there exists a ray ina ✓i direction that intersects C.By the Ray Lemma, we cannot form another scale intersecting Fwithout creating a scale that intersects both C and F .

So a k -MSP tree graph G must be a caterpillar.


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

NC7

Cn(n>3)

inducedC5


Seagull Graphs

We define the family of seagull graphs, Seagulln, to be a clique with nvertices joined to the third vertex of P5, with a pendant vertex adjacentto a different vertex of the clique.

Seagull2(NC7) (left) and Seagull3 (right)


Seagull Graphs

TheoremSeagullk�1 is not a k-MSP graph, but is a (k � 1)-MSP graph.

Proof (Seagullk�1 is not a k -MSP graph).

Since C has two end 2-regions, it must have parallel sidesbetween them (by Parallel Sides Lemma), so C � G cannot be anend region.Therefore, we just need to consider the reflex angles at C [ G.


Seagull Graphs

TheoremSeagullk�1 is not a k-MSP graph, but is a (k � 1)-MSP graph.

Proof (Seagullk�1 is not a k -MSP graph).

C

G

{ Kk-1

Since C has two end 2-regions, it must have parallel sidesbetween them (by Parallel Sides Lemma), so C � G cannot be anend region.Therefore, we just need to consider the reflex angles at C [ G.


Seagull Graphs

Case I: C [ G has one reflex angle.

G must intersect C at at least two points, one of which is the reflexangle.To avoid second reflex angle, G must intersect at a corner of C.However, this causes G to intersect an end 2-region of C, makingit no longer a 2-region.


Seagull Graphs

Case I: C [ G has one reflex angle.G must intersect C at at least two points, one of which is the reflexangle.To avoid second reflex angle, G must intersect at a corner of C.However, this causes G to intersect an end 2-region of C, makingit no longer a 2-region.

θk+1C

D

G


Seagull Graphs

Case II: C [ G has two reflex angles with measures ✓k+i and ✓k+j .

We do not want to create any more than k � 2 scales (notincluding C) for the k � 1 clique of the seagull, so each anglecannot create more than k � 2 proto-scales.This implies the maximum value for either i or j is k � 3.The minimum value for i or j is 1, else the angles would not bereflex.We then split Case II into three subcases.


Seagull Graphs

Case II: C [ G has two reflex angles with measures ✓k+i and ✓k+j .

We do not want to create any more than k � 2 scales (notincluding C) for the k � 1 clique of the seagull, so each anglecannot create more than k � 2 proto-scales.This implies the maximum value for either i or j is k � 3.The minimum value for i or j is 1, else the angles would not bereflex.We then split Case II into three subcases.


Seagull Graphs

Case II.a: i + j < k

If i + j < k , the sides of G are not parallel and point towards eachother.This means G � C forms a shape such that from every point onthe boundary of G � C, we can construct a ray that intersects C.By the Ray Lemma, this means any scale intersecting G wouldalso intersect C.


Seagull Graphs

Case II.a: i + j < k

θk+i θk+jC

G

If i + j < k , the sides of G are not parallel and point towards eachother.This means G � C forms a shape such that from every point onthe boundary of G � C, we can construct a ray that intersects C.By the Ray Lemma, this means any scale intersecting G wouldalso intersect C.


Seagull Graphs

Case II.b: i + j = kIf i + j = k , the sides of G are parallel.We create ij scales with construction of G.

C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs


C θk+2 θk+3


Seagull Graphs

Recall that we cannot create any more than k � 2 scales, so weneed ij k � 2.Since i + j = k , without a loss of generality, we can make thesubstitution i = k � j into the inequality to yield j � k � 2.However, we know that j must be smaller than k � 3. )(


Seagull Graphs

Case II.c: i + j > k , in other words i + j � k + 1

If i + j � k + 1, the sides of G are not parallel and point indirections away from each other.WLOG, the minimum number for i is 4, since if i were any smaller,i + j 6� k + 1.If i = 4, then 4 + j � k + 1, which implies j � k � 3.However, we know j k � 3, so we just consider the case wherej = k � 3.After substitution, we determine that ij > k � 1. )(


Seagull Graphs


C

G

θk+i θk+j

If i + j � k + 1, the sides of G are not parallel and point indirections away from each other.WLOG, the minimum number for i is 4, since if i were any smaller,i + j 6� k + 1.

If i = 4, then 4 + j � k + 1, which implies j � k � 3.However, we know j k � 3, so we just consider the case wherej = k � 3.After substitution, we determine that ij > k � 1. )(


Seagull Graphs


C

G

θk+i θk+j

If i + j � k + 1, the sides of G are not parallel and point indirections away from each other.WLOG, the minimum number for i is 4, since if i were any smaller,i + j 6� k + 1.If i = 4, then 4 + j � k + 1, which implies j � k � 3.However, we know j k � 3, so we just consider the case wherej = k � 3.After substitution, we determine that ij > k � 1. )(


Seagull Graphs

Case III: There are three or more reflex angles at C [ G.

We know that no matter the angle measurements, any two reflexangles create too many scales.Adding an additional reflex angle will never remove scales.Thus, it directly follows that three or more reflex angles would alsocreate too many scales.

Therefore, Seagullk�1 is not a k -MSP graph.


Seagull Graphs

Case III: There are three or more reflex angles at C [ G.We know that no matter the angle measurements, any two reflexangles create too many scales.Adding an additional reflex angle will never remove scales.Thus, it directly follows that three or more reflex angles would alsocreate too many scales.

Therefore, Seagullk�1 is not a k -MSP graph.


Seagull Graphs

Proof (Seagullk�1 is a (k � 1)-MSP graph.)We propose the following construction:

Let m = k � 1. Construct the representation for P5 in m-MSP.


Seagull Graphs

Construct G such that the two reflex angles at the intersection are✓2m�1 and ✓m+1.

θ2m-1θm+1

We have formed m = k � 1 total scales.


Seagull Graphs


θ2m-1θm+1


Seagull Graphs


θ2m-1θm+1


Seagull Graphs


θ2m-1θm+1


Seagull Graphs


θ2m-1θm+1

We have formed m = k � 1 total scales.


Seagull Graphs

We know that we can create the pendant vertex since G has parallelsides.

θ2m-1θm+1


Seagull Graphs

Example.Seagull3 is a 2- and 3-MSP graph but not a 4-MSP graph.

a b c d e

f g

hA

B

C

D

E

F

G

H

A

B

C

D

E

F

G

H

ConjectureSeagullk is a j-MSP graph for all j k .


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

NC7 Seagull3

Cn(n>3)

inducedC5


Turtle Graphs

We define the family of turtle graphs, Turtlen, p, to be a clique with nvertices, to which p pendant vertices are attached, each to distinctvertices of the clique.

Turtle6,4


Turtle Graphs

TheoremTurtlen,k is a k -MSP graph for all k and some n.

Turtle9,4 in 4-MSP


Turtle Graphs


Turtle9,4 in 4-MSP


Turtle Graphs


Turtle9,4 in 4-MSP


Turtle Graphs


Turtle9,4 in 4-MSP


Turtle Graphs


Turtle9,4 in 4-MSP


Turtle Graphs

TheoremTurtlen,k+1 is not a k -MSP graph for any n 2 N

Proof.

Assume by way of contradiction thatthere exists a k -lizard which givesthe k -MSP graph Turtlen,k+1.Let p1, ..., pk+1 be the pendantvertices in Turtlen,k+1, andq1, ..., qk+1 the vertices in the graphadjacent to each pendant vertex.

pa qaqbpb


Turtle Graphs


Proof cont.

By the parallel sides lemma, sincepa $ qa, and qa has anotherneighbor qb, Qa �Pa �Qb contains apair of parallel sides.For each qa, 1 a k + 1, thelemma holds, so we have k + 1 pairsof parallel sides. Since we have kdirections for the sides, two pairs ofparallel sides share a direction.

pa qaqbpb

s1 s2 s3 s4

Qa Qb


Turtle Graphs


Proof cont.

Case 1: If the sides are arrangeds1, s2, s3, s4, then Qa and Qb do notintersect.

pa qaqbpb

s1 s2 s3 s4

Qa Qb


Turtle Graphs


Proof cont.

Case 1: If the sides are arrangeds1, s2, s3, s4, then Qa and Qb do notintersect.Case 2: If the sides are arrangeds1, s3, s2, s4, then there is a scale Qcbetween s3 and s2. By the raylemma, a scale attached to Qa alsointersects Qc , so Qa cannot have apendant vertex.

pa qaqbpb

s1 s2

s3 s4

Qa

Qb Pb

Pa

Qc


Turtle Graphs


Proof cont

Case 3: If the sides are arrangeds1, s3, s4, s2, then we can draw a rayfrom any point along s1 or s2 thatintersects Qa and Qb. By the raylemma, a scale Pa intersecting Qacreates a scale that intersects Pa,Qa, and Qc , so Qa cannot have apendant vertex.

pa qaqbpb

s1 s2s3 s4

QbQa Pa


Turtle Graphs


Proof cont

Case 4: If two sides intersect,we have two subcases:

If in s1, s2, s3, s4, if s2 and s3intersect, then Qa and Qbdon’t intersect.If in s1, s3, s4, s2, if s1 and s3intersect, then again by theray lemma, a scale Paintersecting Qa creates ascale that intersects Pa, Qa,and Qb, so Qa cannot havea pendant vertex.

s1 s2

s3 s4Qa

Qb

s1s2

s3 s4

QbQa

Pa


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

NC7 Seagull3

Cn(n>3)

Turtlen,4

Turtlen,3

inducedC5


Separating Examples

2-MSP 3-MSP

4-MSP

Kn Pn

NC7 Seagull3

Cn(n>3)

???

???

Turtlen,4

Turtlen,3

inducedC5


Generalized Separating Examples

(k-1)-MSP k-MSP

(k+1)-MSP

Kn Pn

Seagullk-1 Seagullk

Cn(n>3)

Turtlen, kØ

TurtleSeagulln, k

Turtlen, k+1


Seagull-Turtle Graph

ConjectureThe Seagull-Turtle Graph, SeaTurtk ,n,k , is a k -MSP graph but notj-MSP for j 6= k .


Seagull-Turtle Graph

ConjectureThe Seagull-Turtle Graph, SeaTurtk ,n,k , is a k -MSP graph but notj-MSP for j 6= k .


Future Research

There are several other things we would like to consider:Considering k -snakes, which require integer side lengths (Kaplan,2017)Requiring maximal sub-polygons also have integer side lengths

Does 4-MSP reduce to 2-MSP?

Non-simply connected k -lizards


Acknowledgements

We gratefully acknowledge:Josh Laison for his invaluable contributions to the projectRichard Moy for his ideas, uplifting spirit and encouragement tohave funErin McNicholas and Josh Laison for organizing the WillametteMathematics Consortium REUThe support of NSF grant DMS 1460982


Documents

Intersection Graphs of Maximal Convex Sub-Polygons of k ... · Willamette REU k -MSP Intersection Graphs August 4, 2017 3 / 67. Maximal Rectangles of a Polyomino: Shearer 1982, Maire