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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 dyer@mun.ca (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.
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 2 / 25
Preliminaries Graceful labelling
An Example
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What graphs can be gracefully labelled? Lots.
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25
Preliminaries Graceful labelling
An Example
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What graphs can be gracefully labelled?
Lots.
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25
Preliminaries Graceful labelling
An Example
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What graphs can be gracefully labelled? Lots.Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 3 / 25
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 witht ≡ 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
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 witht ≡ 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
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 witht ≡ 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
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 witht ≡ 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
Preliminaries Graceful labelling
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 dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 5 / 25
Preliminaries Graceful labelling
Who cares?
Theorem
If G is graceful with m edges, then K2m+1 is G-decomposable.
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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 6 / 25
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 7 / 25
Preliminaries Skolem sequences
Skolem sequences, again.
A Skolem sequence of order 4:
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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)
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 8 / 25
Preliminaries Skolem sequences
Skolem sequences, again.
A Skolem sequence of order 4:
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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 ) (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
Preliminaries Skolem sequences
Skolem sequences, again.
A Skolem sequence of order 4:
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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 ) (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
Preliminaries Skolem sequences
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 9 / 25
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
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)
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 11 / 25
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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︷ ︸︸ ︷2
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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).
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 11 / 25
Windmills with a pendant triangle Pivots
Back to the windmills...
(0, 11, 12), (0, 7, 10), (0, 5, 9), (0, 6, 8)
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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 12 / 25
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)
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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 12 / 25
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
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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. . .
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 14 / 25
Windmills with two pendant triangles Split
Split
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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 15 / 25
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.
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 16 / 25
Windmills with two pendant triangles Results
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 `i , `j ∈ L such that `i = `j = k , and
2 if `i = `j = k with i < j , then j − i = k .
A Langford sequence of order 4 and defect 2.
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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 17 / 25
Windmills with two pendant triangles Results
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
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 18 / 25
Windmills with two pendant triangles Results
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
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 18 / 25
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.
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 19 / 25
Three pendent triangles The basics
Low hanging fruit
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 20 / 25
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 21 / 25
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 22 / 25
Three pendent triangles The hard case
One is much harder
0
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 dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 23 / 25
Three pendent triangles The hard case
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25
Three pendent triangles The hard case
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25
Three pendent triangles The hard case
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25
Three pendent triangles The hard case
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25
Three pendent triangles The hard case
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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Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 24 / 25
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!
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 25 / 25
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!
Danny Dyer dyer@mun.ca (MUN) Windmills with Pendent Triangles CanaDAM 2013 25 / 25
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