Graceful Labellings of Triangular Cacti · Graceful Labellings of Triangular Cacti Danny Dyer1, Ian...

Graceful Labellings of Triangular Cacti

Danny Dyer1, Ian Payne2, Nabil Shalaby1, Brenda Wicks

1Department of Mathematics and StatisticsMemorial University of Newfoundland

2Department of Pure MathematicsUniversity of Waterloo

CanaDAM 2013

Danny Dyer [email protected] (MUN) Windmills with Pendent Triangles CanaDAM 2013 1 / 25

Preliminaries Graceful labelling

Graceful labelling

For a graph G = (V ,E ) on m edges, an injection f : V 7→ {0, 1, 2, . . . ,m}with the property that for all edges uv ∈ E ,{|f (u)− f (v)|} = {1, 2, 3, . . . ,m} is a graceful labelling of G .

(Coined as a β-valuation by Rosa in 1967; graceful by Golomb in 1972.)

For a graph G = (V ,E ) on m edges, an injectionf : V 7→ {0, 1, 2, . . . ,m + 1} with the property that for all edges uv ∈ E ,{|f (u)− f (v)|} = {1, 2, 3, . . . ,m− 1,m + 1} is a near graceful labelling ofG .

A graph is (near) graceful if it admits a (near) graceful labelling.



An Example

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What graphs can be gracefully labelled? Lots.



An Example

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What graphs can be gracefully labelled?

Lots.



An Example

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What graphs can be gracefully labelled? Lots.Danny Dyer [email protected] (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25


Some conjectures

Ringel-Kotzig Conjecture

All trees are graceful.

Ringel famously called efforts to prove this “a disease.”

A triangular cactus is a connected graph all of whose blocks are triangles.

Rosa’s Conjecture

All triangular cacti with t ≡ 0, 1 mod 4 blocks are graceful, and those witht ≡ 2, 3 mod 4 are near graceful.

Gallian suggests this is “hopelessly difficult.”



Windmills

A regular Dutch windmill is a triangular cactus in which all blocks have acommon vertex that we will call the central vertex. The blocks will becalled vanes.

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Bermond (1979) showed that all regular Dutch windmills are graceful ornear graceful.Danny Dyer [email protected] (MUN) Windmills with Pendent Triangles CanaDAM 2013 5 / 25


Who cares?

Theorem

If G is graceful with m edges, then K2m+1 is G-decomposable.

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Preliminaries Skolem sequences

Skolem sequences

A Skolem sequence of order n is a sequence S = (s1, s2, . . . , s2n) of 2nintegers satisfying the conditions

1 for every k ∈ {1, 2, . . . , n} there exist exactly two elements si , sj ∈ Ssuch that si = sj = k , and

2 if si = sj = k with i < j , then j − i = k .

A Skolem sequence of order 4:

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Skolem sequences, again.


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Sometimes we just write the pairs of indices.

(1, 5), (2, 4), (3, 6), (7, 8)

These pairs are useful.

(ai , bi )(7, 8)(2, 4) →(3, 6)(1, 5)





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(1, 5), (2, 4), (3, 6), (7, 8)


(ai , bi ) (i , ai + n, bi + n)(7, 8) (1, 11, 12)(2, 4) → (2, 6, 8)(3, 6) (3, 7, 10)(1, 5) (4, 5, 9)





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(1, 5), (2, 4), (3, 6), (7, 8)


(ai , bi ) (i , ai + n, bi + n) (0, ai + n, bi + n)(7, 8) (1, 11, 12) (0, 11, 12)(2, 4) → (2, 6, 8) → (0, 6, 8)(3, 6) (3, 7, 10) (0, 7, 10)(1, 5) (4, 5, 9) (0, 5, 9)



Graceful labelling from Skolem sequences

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(0, 11, 12), (0, 6, 8), (0, 7, 10), (0, 5, 9)

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Windmills with a pendant triangle Something harder

Windmills with a pendant triangle

Now something a little harder...


Windmills with a pendant triangle Pivots

Back to Skolem sequences

A number i (1 ≤ i ≤ n) is a pivot of a Skolem sequence if bi + i ≤ 2n.

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We see 2 is a pivot.

(ai , bi ) (0, ai + n, bi + n)(7, 8) (0, 11, 12)(2, 4) → (0, 6, 8)(3, 6) (0, 7, 10)(1, 5) (0, 5, 9)



Back to Skolem sequences

A number i (1 ≤ i ≤ n) is a pivot of a Skolem sequence if bi + i ≤ 2n.

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We see 2 is a pivot.

(ai , bi ) (0, ai + n, bi + n)(7, 8) (0, 11, 12)(2, 4) → ��(0, 6, 8)→ (2, 8, 10)(3, 6) (0, 7, 10)(1, 5) (0, 5, 9)

Replace the pivot’s triple (0, aj + n, bj + n) with (j , aj + j + n, bj + j + n).



Back to the windmills...

(0, 11, 12), (0, 7, 10), (0, 5, 9), (0, 6, 8)

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Back to the windmills...

(0, 11, 12), (0, 7, 10), (0, 5, 9), ��(0, 6, 8)→ (2, 8, 10)

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Windmills with two pendant triangles Base cases

Windmills with two pendant triangles

independent stacked

split double


Windmills with two pendant triangles Independent

Independent

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16︸︷︷︸(6, 7) → (0, 14, 15) → (1, 15, 16)

(12, 14) → (0, 20, 22) → (2, 22, 24)

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. . .


Windmills with two pendant triangles Split

Split

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(1, 5) → (0, 9, 13) → (4, 13, 17)(2, 4) → (0, 10, 12) → (2, 12, 14)

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Windmills with two pendant triangles Results

Some results

Theorem

A Dutch windmill with one pendant triangle is either graceful or neargraceful.

Theorem

A Dutch windmill with two pendant triangles is either graceful or neargraceful.



A note on method

A Langford sequence of order n and defect d is a sequenceL = (`1, `2, . . . , `2n) of 2n integers satisfying the conditions

1 for every k ∈ {d , d + 1, . . . , d + n − 1} there exist exactly twoelements ì , `j ∈ L such that ì = `j = k , and

2 if ì = `j = k with i < j , then j − i = k .

A Langford sequence of order 4 and defect 2.

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A way forward

Let Sn be a Skolem sequence, and Ln+1,m−n be a Langford sequence oforder m − n with defect n + 1.

Sn Ln+1,m−n

︷︸︸︷Sm



A way forward

Let Sn be a Skolem sequence, and Ln+1,m−n be a Langford sequence oforder m − n with defect n + 1.

Sn Ln+1,m−n

︷︸︸︷Sm

︸︷︷︸︸︷︷︸Weird structure More vanes


Three pendent triangles The basics

Three pendent triangles

We can push this problem a littler farther. To do this, we use acombination of our previous methods.

Unfortunately, there are 11 cases to consider.

Some follow directly from the previous constructions, or from new Skolemsequences.

One is a big problem.


Three pendent triangles The basics

Low hanging fruit


Three pendent triangles A little harder...

A little harder...

But there are seven more cases! Some are similar.

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(4, 5) → (0, 12, 13) → (1, 13, 14)(8, 11) → (0, 16, 19) → (3, 19, 22)(1, 6) → (0, 9, 14) → (5, 14, 19)

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Three pendent triangles A little harder...

A new trick4

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(0, ai + n, bi + n) (0, i , bi + n)(0, 11, 12) (0, 1, 12)

(0, 6, 8) (0, 2, 8)(0, 7, 10) (0, 3, 10)(0, 5, 9) (0, 4, 9)

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Three pendent triangles The hard case

One is much harder

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A bad pivoting structure. We use our trick, and some extra “shifting”,beyond pivoting. Instead of shifting (0, ai + n, bi + n) to(i , ai + i + n, bi + i + n), we could shift it to (k , ai + k + n, bi + k + n), forany k.Danny Dyer [email protected] (MUN) Windmills with Pendent Triangles CanaDAM 2013 23 / 25


One is much harder

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(0, 23, 24)(0, 18, 20)(0, 19, 22)(0, 25, 35)(0, 27, 42)

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One is much harder

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(0, 23, 24)

��(0, 18, 20) → (2, 20, 22)

��(0, 19, 22) → (3, 22, 25)(0, 25, 35)(0, 27, 42)

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One is much harder

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��(0, 23, 24) → (0, 1, 24)

��(0, 18, 20) → (2, 20, 22)

��(0, 19, 22) → (3, 22, 25)(0, 25, 35)(0, 27, 42)

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One is much harder

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��(0, 23, 24) → ��(0, 1, 24) → (3, 4, 27)

��(0, 18, 20) → (2, 20, 22)

��(0, 19, 22) → (3, 22, 25)(0, 25, 35)(0, 27, 42)

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One is much harder

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��(0, 23, 24) → ��(0, 1, 24) → (3, 4, 27)

��(0, 18, 20) → (2, 20, 22)

��(0, 19, 22) → (3, 22, 25)(0, 25, 35)

��(0, 27, 42) → (0, 15, 42)

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Conclusions and Open Questions

Conclusions and Open Questions

Theorem

A Dutch windmill with three pendant triangles is either graceful or neargraceful.

1 Define S(n) to be the number of triangular cacti of order n which canbe labelled using Skolem sequences and their pivots. Define T (n) to

be the number of triangular cacti of order n. What is limn→∞

S(n)

T (n)?

2 Can we use the Langford sequence construction to gracefully labelother graphs?

Thank you!


Graceful Labellings of Triangular Cacti · Graceful Labellings of Triangular Cacti Danny Dyer1, Ian...

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