Graceful Labellings of Triangular Cacti Danny Dyer 1 , Ian Payne 2 , - - PowerPoint PPT Presentation

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graceful labellings of triangular cacti

Graceful Labellings of Triangular Cacti Danny Dyer 1 , Ian Payne 2 , - - PowerPoint PPT Presentation

May 17, 2023 375 likes •788 views

Graceful Labellings of Triangular Cacti Danny Dyer 1 , Ian Payne 2 , Nabil Shalaby 1 , Brenda Wicks 1 Department of Mathematics and Statistics Memorial University of Newfoundland 2 Department of Pure Mathematics University of Waterloo CanaDAM

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SLIDE 1

Graceful Labellings of Triangular Cacti

Danny Dyer1, Ian Payne2, Nabil Shalaby1, Brenda Wicks

1Department of Mathematics and Statistics

Memorial University of Newfoundland

2Department of Pure Mathematics

University of Waterloo

CanaDAM 2013

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 1 / 25

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SLIDE 2

Preliminaries Graceful labelling

Graceful labelling

For a graph G = (V , E) on m edges, an injection f : V → {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 injection f : V → {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 of G. A graph is (near) graceful if it admits a (near) graceful labelling.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 2 / 25

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SLIDE 3

Preliminaries Graceful labelling

An Example

1 2 3 7 8 9 10 12 15 9 10 12 15 7 8 11 14 6 13 4 5 1 2 3

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25

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SLIDE 4

Preliminaries Graceful labelling

An Example

1 2 3 7 8 9 10 12 15 9 10 12 15 7 8 11 14 6 13 4 5 1 2 3 What graphs can be gracefully labelled?

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25

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SLIDE 5

Preliminaries Graceful labelling

An Example

1 2 3 7 8 9 10 12 15 9 10 12 15 7 8 11 14 6 13 4 5 1 2 3 What graphs can be gracefully labelled? Lots.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25

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SLIDE 6

Preliminaries Graceful labelling

Some conjectures

Ringel-Kotzig Conjecture All trees are graceful.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 4 / 25

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SLIDE 7

Preliminaries Graceful labelling

Some conjectures

Ringel-Kotzig Conjecture All trees are graceful. Ringel famously called efforts to prove this “a disease.”

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 4 / 25

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SLIDE 8

Preliminaries Graceful labelling

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 with t ≡ 2, 3 mod 4 are near graceful.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 4 / 25

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SLIDE 9

Preliminaries Graceful labelling

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 with t ≡ 2, 3 mod 4 are near graceful. Gallian suggests this is “hopelessly difficult.”

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 4 / 25

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SLIDE 10

Preliminaries Graceful labelling

Windmills

A regular Dutch windmill is a triangular cactus in which all blocks have a common vertex that we will call the central vertex. The blocks will be called vanes. 11 12 5 9 7 10 6 8 11 12 5 9 7 10 6 8 1 4 3 2 Bermond (1979) showed that all regular Dutch windmills are graceful or near graceful.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 5 / 25

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SLIDE 11

Preliminaries Graceful labelling

Who cares?

Theorem If G is graceful with m edges, then K2m+1 is G-decomposable. 5 6 7 8 9 10 11 12

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 6 / 25

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SLIDE 12

Preliminaries Skolem sequences

Skolem sequences

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

1 for every k ∈ {1, 2, . . . , n} there exist exactly two elements si, sj ∈ S

such that si = sj = k, and

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

A Skolem sequence of order 4: 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 7 / 25

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SLIDE 13

Preliminaries Skolem sequences

Skolem sequences, again.

A Skolem sequence of order 4: 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 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)

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 8 / 25

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SLIDE 14

Preliminaries Skolem sequences

Skolem sequences, again.

A Skolem sequence of order 4: 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 Sometimes we just write the pairs of indices. (1, 5), (2, 4), (3, 6), (7, 8) These pairs are useful. (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)

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 8 / 25

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SLIDE 15

Preliminaries Skolem sequences

Skolem sequences, again.

A Skolem sequence of order 4: 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 Sometimes we just write the pairs of indices. (1, 5), (2, 4), (3, 6), (7, 8) These pairs are useful. (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)

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 8 / 25

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SLIDE 16

Preliminaries Skolem sequences

Graceful labelling from Skolem sequences

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 (0, 11, 12), (0, 6, 8), (0, 7, 10), (0, 5, 9) 11 12 5 9 7 10 6 8 11 12 5 9 7 10 6 8 1 4 3 2

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 9 / 25

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SLIDE 17

Windmills with a pendant triangle Something harder

Windmills with a pendant triangle

Now something a little harder...

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 10 / 25

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SLIDE 18

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. 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 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)

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 11 / 25

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SLIDE 19

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. 4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 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).

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 11 / 25

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SLIDE 20

Windmills with a pendant triangle Pivots

Back to the windmills...

(0, 11, 12), (0, 7, 10), (0, 5, 9), (0, 6, 8) 11 12 5 9 7 10 11 12 5 9 7 10 1 4 3

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 12 / 25

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SLIDE 21

Windmills with a pendant triangle Pivots

Back to the windmills...

(0, 11, 12), (0, 7, 10), (0, 5, 9), ✘✘✘

✘

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

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 12 / 25

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SLIDE 22

Windmills with two pendant triangles Base cases

Windmills with two pendant triangles

independent stacked split double

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 13 / 25

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SLIDE 23

Windmills with two pendant triangles Independent

Independent

4 1 8 2 5 3 7 4 4 5 1 6 1 7 5 8 6 9 8 10 7 11 2 12 3 13 2 14 6 15 3 16

(6, 7)

→ (0, 14, 15) → (1, 15, 16) (12, 14) → (0, 20, 22) → (2, 22, 24) 11 16 24 21 1 15 2 22 . . .

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 14 / 25

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SLIDE 24

Windmills with two pendant triangles Split

Split

4 1 2 2 7 3

2

4 4 5 3 6 8 7 6 8 3 9

7

10 5 11 1 12 1 13 6 14 8 15 5 16 (1, 5) → (0, 9, 13) → (4, 13, 17) (2, 4) → (0, 10, 12) → (2, 12, 14) 14 17 4 13 2 12

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 15 / 25

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SLIDE 25

Windmills with two pendant triangles Results

Some results

Theorem A Dutch windmill with one pendant triangle is either graceful or near graceful. Theorem A Dutch windmill with two pendant triangles is either graceful or near graceful.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 16 / 25

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SLIDE 26

Windmills with two pendant triangles Results

A note on method

A Langford sequence of order n and defect d is a sequence L = (ℓ1, ℓ2, . . . , ℓ2n) of 2n integers satisfying the conditions

1 for every k ∈ {d, d + 1, . . . , d + n − 1} there exist exactly two

elements ℓi, ℓj ∈ L such that ℓi = ℓj = k, and

2 if ℓi = ℓj = k with i < j, then j − i = k.

A Langford sequence of order 4 and defect 2. 5 1 2 2 4 3 2 4 3 5 5 6 4 7 3 8

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 17 / 25

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SLIDE 27

Windmills with two pendant triangles Results

A way forward

Let Sn be a Skolem sequence, and Ln+1,m−n be a Langford sequence of

rder m − n with defect n + 1.

Sn Ln+1,m−n

Sm

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 18 / 25

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SLIDE 28

Windmills with two pendant triangles Results

A way forward

Let Sn be a Skolem sequence, and Ln+1,m−n be a Langford sequence of

rder m − n with defect n + 1.

Sn Ln+1,m−n

Sm
Weird structure

More vanes

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 18 / 25

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SLIDE 29

Three pendent triangles The basics

Three pendent triangles

We can push this problem a littler farther. To do this, we use a combination of our previous methods. Unfortunately, there are 11 cases to consider. Some follow directly from the previous constructions, or from new Skolem sequences. One is a big problem.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 19 / 25

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SLIDE 30

Three pendent triangles The basics

Low hanging fruit

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 20 / 25

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SLIDE 31

Three pendent triangles A little harder...

A little harder...

But there are seven more cases! Some are similar. 5 1 8 2 6 3 1 4 1 5 5 6 7 7 3 8 6 9 8 10 3 11 4 12 2 13 7 14 2 15 4 16 (4, 5) → (0, 12, 13) → (1, 13, 14) (8, 11) → (0, 16, 19) → (3, 19, 22) (1, 6) → (0, 9, 14) → (5, 14, 19) 22 15 5 14 3 19 1 13

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 21 / 25

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SLIDE 32

Three pendent triangles A little harder...

A new trick

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 (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) 11 12 5 9 7 10 6 8 11 12 5 9 7 10 6 8 1 4 3 2 1 12 4 9 3 10 2 8 1 12 4 9 3 10 2 8 11 5 7 6

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 22 / 25

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SLIDE 33

Three pendent triangles The hard case

One is much harder

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), for any k.

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 23 / 25

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SLIDE 34

Three pendent triangles The hard case

One is much harder

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 10 9 8 10 15 11 16 12 14 13 11 14 6 15 7 16 13 17 8 18 10 19 12 20 6 21 9 22 7 23 5 24 11 25 15 26 14 27 16 28 5 29 13 30 9 31 12 32

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

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25

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SLIDE 35

Three pendent triangles The hard case

One is much harder

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 10 9 8 10 15 11 16 12 14 13 11 14 6 15 7 16 13 17 8 18 10 19 12 20 6 21 9 22 7 23 5 24 11 25 15 26 14 27 16 28 5 29 13 30 9 31 12 32

(0, 23, 24)

✘✘✘✘ ✘

(0, 18, 20) → (2, 20, 22)

✘✘✘✘ ✘

(0, 19, 22) → (3, 22, 25) (0, 25, 35) (0, 27, 42) 25 35 3 22 2 20

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25

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SLIDE 36

Three pendent triangles The hard case

One is much harder

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 10 9 8 10 15 11 16 12 14 13 11 14 6 15 7 16 13 17 8 18 10 19 12 20 6 21 9 22 7 23 5 24 11 25 15 26 14 27 16 28 5 29 13 30 9 31 12 32 ✘✘✘✘ ✘

(0, 23, 24) → (0, 1, 24)

✘✘✘✘ ✘

(0, 18, 20) → (2, 20, 22)

✘✘✘✘ ✘

(0, 19, 22) → (3, 22, 25) (0, 25, 35) (0, 27, 42) 25 35 3 22 2 20

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25

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SLIDE 37

Three pendent triangles The hard case

One is much harder

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 10 9 8 10 15 11 16 12 14 13 11 14 6 15 7 16 13 17 8 18 10 19 12 20 6 21 9 22 7 23 5 24 11 25 15 26 14 27 16 28 5 29 13 30 9 31 12 32 ✘✘✘✘ ✘

(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) 25 35 3 22 4 27 2 20

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25

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SLIDE 38

Three pendent triangles The hard case

One is much harder

4 1 2 2 3 3 2 4 4 5 3 6 1 7 1 8 10 9 8 10 15 11 16 12 14 13 11 14 6 15 7 16 13 17 8 18 10 19 12 20 6 21 9 22 7 23 5 24 11 25 15 26 14 27 16 28 5 29 13 30 9 31 12 32 ✘✘✘✘ ✘

(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) 25 35 3 22 4 27 2 20

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25

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SLIDE 39

Conclusions and Open Questions

Conclusions and Open Questions

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

1 Define S(n) to be the number of triangular cacti of order n which can

be labelled using Skolem sequences and their pivots. Define T(n) to be the number of triangular cacti of order n. What is lim

n→∞

S(n) T(n)?

2 Can we use the Langford sequence construction to gracefully label

ther graphs?

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 25 / 25

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SLIDE 40

Conclusions and Open Questions

Conclusions and Open Questions

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

1 Define S(n) to be the number of triangular cacti of order n which can

be labelled using Skolem sequences and their pivots. Define T(n) to be the number of triangular cacti of order n. What is lim

n→∞

S(n) T(n)?

2 Can we use the Langford sequence construction to gracefully label

ther graphs?

Thank you!

Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 25 / 25