Some algebraic approaches to graceful labellings

Andrea Vietri1

1Dipartimento di Scienze di Base e Applicate per l’IngegneriaSapienza Università di Roma

Andrea Vietri Some algebraic approaches to graceful labellings January, 9th, 2018 1 / 50

Outline

1 Graceful labelings

2 Graceful polynomials

3 Vanishing polynomials of small degree

4 Final remarks on graceful polynomials

5 Real-graceful labellings

6 Graceful hypersurfaces

Graceful labelings

Graceful labellings

Graceful labelings

A graph with e edges has a

graceful labelling

if we can assign positive integers in [0,e] to its vertices, so that thedifferences on the edges be 1,2, ...,e.

r rrr06 4

1@@��r r r r r4 0 3 1 2

r r r r r r r r r r r rr r r r r r r r r r r r17616715814913101211

519420321222123024

Graceful labelings

Graph decomposition: K13 can be decomposed into 14 copies of K4starting with a G.L. of K4 and replicating it:

r rrr06 4

1@@��

r rrr17 5

2@@�� . . .

r rrr11

12@@��

r rrr12

0@@��

counting (mod 13).

Every edge {a,b} of K13 is covered with norepetition.e.g. if b− a = ±4 (as for {1,5} or {2,11}), we choose one of the upperedges.Similarly with the differences ±1, ±2, ±3, ±5, ±6 .

Graceful labelings

Graph decomposition: K13 can be decomposed into 14 copies of K4starting with a G.L. of K4 and replicating it:

r rrr06 4

1@@��

r rrr17 5

2@@�� . . .

r rrr11

12@@��

r rrr12

0@@��

counting (mod 13). Every edge {a,b} of K13 is covered with norepetition.e.g. if b− a = ±4 (as for {1,5} or {2,11}), we choose one of the upperedges.Similarly with the differences ±1, ±2, ±3, ±5, ±6 .

Graceful labelings

A G.L. can be also regarded as a well done optimisation.

For, the best we can do in some cases is...

s ssss

@@@ ��

[0,10] \ {5} t {4}

oppures ssss

@@@ ��

[0,11] \ {6}

(cfr. F. Van Bussel, Relaxed graceful labellings of trees, Electr. J.Comb. (2002); see also the problem of Golomb rulers.)

Graceful labelings

A G.L. can be also regarded as a well done optimisation.For, the best we can do in some cases is...

s ssss

@@@ ��

[0,10] \ {5} t {4}

oppures ssss

@@@ ��

[0,11] \ {6}

Graceful labelings

A G.L. can be also regarded as a well done optimisation.For, the best we can do in some cases is...

s ssss

@@@ ��

[0,10] \ {5} t {4}

oppures ssss

@@@ ��

[0,11] \ {6}

Graceful labelings

As to K5, K6, ... gracefulness is lost (only relaxations hold).

Similarly for cycles of length 1 o 2 (mod 4) (Rosa’s Theorem: anEulerian graph having 1 or 2 edges (mod 4) is not graceful).

Rosa’s Theorem is an isolated example of happy marriage betweenalgebra and combinatorics in this area.

In order to celebrate the Golden Wedding of that marriage (2017), witha little delay we propose a generalisation of such a basic example!

Graceful labelings

Graceful polynomials

For any graph we define a family {Sn}n∈N∗ of graceful polynomials; Snhas degree n.

There is a deep connection between these polynomials and gracefullabellings.

As mentioned, the prototype S1 was employed by A. Rosa (1967) toobtain the most important class of non-graceful graphs.

A.V. (2012, 2016): generalisation to every degree (analysis of gracefultrees; construction of non-graceful graphs).

Graceful labellings

A graph G = (V ,E) is graceful if

∃ f : V → {0,1,2, ..., |E |} injective and such that

{|f (u)− f (v)| : uv ∈ E} = {1,2, ..., |E |} .

If a graph admits no such labelling, it is non-graceful.

tttt tt t

8@@@ t t

t t��@@@

ttt tt

Non-graceful

Example of graceful polynomial

How to obtain a graceful polynomial from G = rr rr rr r@@ :

Assign a variable to each vertex: rr rr rr rx4

Fix a positive integer n, say n = 3 ...

Take any set of 3 edges

u uu u ur rx1

@@ and store :

(x1 − x2)(x2 − x5)(x6 − x7) ≡ x1x2x6 + x1x2x7 + ... (mod 2)

SUM UP these polynomials over all the(8

)sets of 3 edges:

this is S3(G), the 3-rd graceful polynomial of G.

Take any set of 3 edges

u uu u ur rx1

@@ and store :

(x1 − x2)(x2 − x5)(x6 − x7) ≡ x1x2x6 + x1x2x7 + ... (mod 2)

SUM UP these polynomials over all the(8

)sets of 3 edges:

this is S3(G), the 3-rd graceful polynomial of G.

Formally... given a graph G...

Assign variable xi to vertex vi .

Associate any given edge ej = vpvq to the polynomial Pj = xp + xq.

The n-th graceful polynomial of G is

SnG(x1, x2, ..., x|V |) ≡

∑1≤j1<j2<...<jn≤|E |

Pj1Pj2 · · · Pjn (mod 2).

SnG(x1, x2, ..., x|V |) ≡

∑1≤j1<j2<...<jn≤|E |

SnG(x1, x2, ..., x|V |) ≡

∑1≤j1<j2<...<jn≤|E |

Graceful polynomials and non-graceful graphs

The following Lemma provides the connection:

Lemma (A.V. 2012)Let G be a graceful graph and let fi be the label of vertex vi . Then, forevery positive integer n ≤ |E(G)|,

SnG(f1, ..., f|V |) ≡

([ |E |+1

)(mod 2).

Therefore, if some graceful polynomial VANISHES (mod 2)...... while |E | makes the binomial coefficient ODD...

... then we have a contradiction,and the graph is necessarily NON-GRACEFUL.

SnG(f1, ..., f|V |) ≡

([ |E |+1

)(mod 2).

Therefore, if some graceful polynomial VANISHES (mod 2)...

... while |E | makes the binomial coefficient ODD...... then we have a contradiction,and the graph is necessarily NON-GRACEFUL.

SnG(f1, ..., f|V |) ≡

([ |E |+1

)(mod 2).

SnG(f1, ..., f|V |) ≡

([ |E |+1

)(mod 2).

Vanishing polynomials of small degree

Vanishing polynomialsof small degree

Main problem:

Choose n, and characterise all graphs for which SnG vanishes (mod 2).

The problem seems interesting on its own right!

However, once we have these graphs, we can also find

non-graceful graphs

(choosing a suitable number of edges, if we can...).

Main problem:

non-graceful graphs

Main problem:

non-graceful graphs

Main problem, degree 1

This is Rosa’s polynomial.

(A. Rosa, On certain valuations of the vertices of a graph, Theory ofGraphs, Internat. Sympos. Rome, 1966, pp. 349-355)

Let δp denote the degree of vertex vp.

S1G(x1, x2, ..., x|V |) ≡

∑1≤j≤|E |

Pj ≡∑

vpvq edge

(xp + xq) ≡∑

vp vertex

Main problem, degree 1

This is Rosa’s polynomial.

(A. Rosa, On certain valuations of the vertices of a graph, Theory ofGraphs, Internat. Sympos. Rome, 1966, pp. 349-355)

Let δp denote the degree of vertex vp.

S1G(x1, x2, ..., x|V |) ≡

∑1≤j≤|E |

Pj ≡∑

vpvq edge

(xp + xq) ≡∑

vp vertex

∑δpxp vanishes if all degrees are even (Eulerian graph).

On the other hand,([ |E|+1

)is odd if |E | ≡ 1,2 (mod 4). Therefore:

Theorem (Rosa)

Eulerian graphs with |E | ≡ 1,2 (mod 4) are non-graceful.

Theorem (Rosa)

Degree 2

Denote coefficients of graceful polynomials byH······ .

S2G ≡

px2p +

∑H11

pqxpxq .

H2p counts the PAIRS of edges containing vp: s��AA

XXAA@@

vavbr r

vp(xp + xa)(xp + xb)

= x2p + ...

Therefore, H2p ≡

)(mod 2)

⇒ δp ≡ 0,1 (mod 4) ∀vp if we ask for vanishing.

Degree 2

S2G ≡

px2p +

∑H11

pqxpxq .

XXAA@@

vavbr r

= x2p + ...

Therefore, H2p ≡

)(mod 2)

Degree 2

S2G ≡

px2p +

∑H11

pqxpxq .

XXAA@@

vavbr r

= x2p + ...

Therefore, H2p ≡

)(mod 2)

As to H11pq , there are two cases.

r rG21

vp vq@@��

��

δpδq choices

��

r rG22

vp vq@@��

��

δpδq − 1 choices:

(xp + xq)(xp + xq) not allowed

��

So we have (if we ask for vanishing):

δp ≡ 0,1 ∀vp , (δp, δq) 6≡ (1,1)∀G21 , (δp, δq) ≡ (1,1)∀G2

all (mod 4)

r rG21

vp vq@@��

��

δpδq choices

�� r rG22

vp vq@@��

��

δp ≡ 0,1 ∀vp , (δp, δq) 6≡ (1,1)∀G21 , (δp, δq) ≡ (1,1)∀G2

all (mod 4)

r rG21

vp vq@@��

��

δpδq choices

�� r rG22

vp vq@@��

��

δp ≡ 0,1 ∀vp , (δp, δq) 6≡ (1,1)∀G21 , (δp, δq) ≡ (1,1)∀G2

all (mod 4)

From the above conditions we obtain all possible graphs:

Theorem (A.V. 2012)

S2G vanishes for complete graphs on 4d + 2 vertices, for any integer d.

No other graph satisfies the requirement.

By extending Rosa’s counting technique to this level we obtain:

Theorem (A.V. 2012, new proof of an old result ! )Complete graphs on either 16u + 10 or 16u + 14 vertices, with anyinteger u, are non-graceful.

(count the edges and reach a contradiction through the above Lemma)

From the above conditions we obtain all possible graphs:

Theorem (A.V. 2012)

S2G vanishes for complete graphs on 4d + 2 vertices, for any integer d.

No other graph satisfies the requirement.

By extending Rosa’s counting technique to this level we obtain:

Theorem (A.V. 2012, new proof of an old result ! )Complete graphs on either 16u + 10 or 16u + 14 vertices, with anyinteger u, are non-graceful.

(count the edges and reach a contradiction through the above Lemma)

Degree 3

S3G ≡

px3p +

∑H21

p,qx2p xq +

∑H111

pqr xpxqxr

(commas separate SETS of indices)

H3p ≡

p,q(G21) ≡

)δq , H21

p,q(G22) ≡

)δq − (δp − 1)

r rG21

vp vq@@

��

�� r rG2

vp vq@@��

��

(xp + xq)2(xp + xa) not allowed

Degree 3

S3G ≡

px3p +

∑H21

p,qx2p xq +

∑H111

pqr xpxqxr

H3p ≡

H21p,q(G2

1) ≡(δp

)δq , H21

p,q(G22) ≡

)δq − (δp − 1)

r rG21

vp vq@@

��

�� r rG2

vp vq@@��

��

Degree 3

S3G ≡

px3p +

∑H21

p,qx2p xq +

∑H111

pqr xpxqxr

H3p ≡

p,q(G21) ≡

)δq , H21

p,q(G22) ≡

)δq − (δp − 1)

r rG21

vp vq@@

��

�� r rG2

vp vq@@��

��

As to H111pqr , there are 4 subgraphs on 3 vertices:

vp r rr

For example, H111pqr (G3

3) ≡ δpδqδr − δr − δq .

(forbidden repetitions of either vpvq or vpvr ) r rr

��

As to H111pqr , there are 4 subgraphs on 3 vertices:

vp r rr

For example, H111pqr (G3

3) ≡ δpδqδr − δr − δq .

(forbidden repetitions of either vpvq or vpvr ) r rr

��

System of constraints:

∀vp δp 6≡ 3 (mod 4)

∀G21

)δq +

)δp ≡ 0 (mod 2) (H21

p,q +H21q,p ...)

∀G22

)δq +

)δp + δp + δq ≡ 0 (mod 2)

∀G31 δpδqδr ≡ 0 ...

∀G32 δp(δqδr + 1) ≡ 0

∀G33 δpδqδr + δq + δr ≡ 0

∀G34 δpδqδr + δp + δq + δr ≡ 0

Eulerian graphs clearly satisfy this system.

System of constraints:

∀vp δp 6≡ 3 (mod 4)

∀G21

)δq +

)δp ≡ 0 (mod 2) (H21

p,q +H21q,p ...)

∀G22

)δq +

)δp + δp + δq ≡ 0 (mod 2)

∀G31 δpδqδr ≡ 0 ...

∀G32 δp(δqδr + 1) ≡ 0

∀G33 δpδqδr + δq + δr ≡ 0

∀G34 δpδqδr + δp + δq + δr ≡ 0

Eulerian graphs clearly satisfy this system.

If the graph has some vertices of odd degree:

Theorem (A.V. 2016)

S3G vanishes (mod 2) on G precisely in one of the following two cases.

(A) G is a complete graph K4t+2, for some positive integer t, possiblyhaving 4u additional vertices and 4u(4t + 2) additional edges thatconnect these vertices to the above complete graph.(B) G is obtained by taking two complete graphs Ka, Kb and nadditional vertices satisfying one of the following conditions:

(B1) : a ≡ b ≡ 2 (mod 4) and n = 0;(B2) : a ≡ b ≡ n ≡ 1 (mod 4) except a = b = n = 1;(B3) : a ≡ b ≡ n ≡ 3 (mod 4);

every additional vertex must be fully connected to Ka and Kb.

If the graph has some vertices of odd degree:

Theorem (A.V. 2016)

S3G vanishes (mod 2) on G precisely in one of the following two cases.

(A) G is a complete graph K4t+2, for some positive integer t, possiblyhaving 4u additional vertices and 4u(4t + 2) additional edges thatconnect these vertices to the above complete graph.(B) G is obtained by taking two complete graphs Ka, Kb and nadditional vertices satisfying one of the following conditions:

(B1) : a ≡ b ≡ 2 (mod 4) and n = 0;(B2) : a ≡ b ≡ n ≡ 1 (mod 4) except a = b = n = 1;(B3) : a ≡ b ≡ n ≡ 3 (mod 4);

every additional vertex must be fully connected to Ka and Kb.

Old and new non-graceful graphs, as a consequence:

TheoremThe following graphs are non-graceful.Type (A) in the above theorem, provided t ≡ 2 (mod 4).Type (B1) with a + b ≡ 12 (mod 16); type (B2) with a + b ≡ 10 (mod16); type (B3) with a + b ≡ 14 (mod 16).

aaa��

��

a = b = 7 , n = 3

aaaaa��

��

a = b = n = 5

inglese italiano Traduci messaggio Disattiva per: inglese Ehrenfeucht-Fraïssé games on linear orders J K Truss (University of Leeds) This is joint work with Feresiano Mwesigye. In an n-move Ehrenfeucht-Fraïssé game on relational structures A and B, players I and II play alternately. On each move player I chooses an element of one of the structures, and player tried to ‘match it’ by choosing an element of the other structure. Player I does not have to choose from the same structure on each move. After n moves, points a1,a2, . . . ,an have been chosen in A, and points b1,b2, . . . ,bn have been chosen in B, and player II wins if the map taking ai to bi for each i is an isomorphism of induced substructures. Under these circumstances, we say that A and B are n-equivalent, written A ≡n B. The main facts about this situation are as follows. A and B are n-equivalent if and only if they satisfy the same sentences of quantifier depth at most n. Hence ≡n is an equivalence relation, and A and B are elementarily equivalent if and only if they are n-equivalent for all n. If the language is finite, then for each n there are just finitely many ≡n-classes. We consider linear orders, and also linear orders with colours, concentrating on estimates for the lengths of optimal representatives for finite coloured orders, optimal representatives for ordinals and certain special scattered liner orders.

Degree 4

Not surprising::::::much

::::::more

::::::cases and

:::::many

:::::::::::subgraphs to check...

S4G ≡

p +∑p,q

H31p,qx3

p xq +∑pq

H22pqx2

p x2q +

+∑p,qr

H211p,qr x

2p xqxr +

∑pqrs

H1111pqrs xpxqxr xs .

Strategy: (I) Find constraints for single vertices and pairs of vertices

(II) Constraints for subgraphs with 3 vertices – some ruled out by (I)

(III) Few subgraphs with 4 vertices survive. Last test: H1111pqrs ≡ 0

Degree 4

::::::more

::::::cases and

:::::many

S4G ≡

p +∑p,q

H31p,qx3

p xq +∑pq

H22pqx2

p x2q +

+∑p,qr

H211p,qr x

2p xqxr +

∑pqrs

Degree 4

::::::more

::::::cases and

:::::many

S4G ≡

p +∑p,q

H31p,qx3

p xq +∑pq

H22pqx2

p x2q +

+∑p,qr

H211p,qr x

2p xqxr +

∑pqrs

Degree 4

::::::more

::::::cases and

:::::many

S4G ≡

p +∑p,q

H31p,qx3

p xq +∑pq

H22pqx2

p x2q +

+∑p,qr

H211p,qr x

2p xqxr +

∑pqrs

Degree 4

::::::more

::::::cases and

:::::many

S4G ≡

p +∑p,q

H31p,qx3

p xq +∑pq

H22pqx2

p x2q +

+∑p,qr

H211p,qr x

2p xqxr +

∑pqrs

Some constraints for the degree 4

Constraint for single vertices: 0 ≤ δp ≤ 3 (mod 8)

Forbidden degrees of non-adjacent vertices:

(1,3), (2,2), (2,3), (3,3) (mod 4)

Forbidden degrees of adjacent vertices:

(0,0), (0,1), (3,3) (mod 4)

In particular, there is at most one vertex of degree 3 (mod 4)

etc. etc. (subgraphs with 3 vertices... nice to rephrase all this ascolour constraints)

(1,3), (2,2), (2,3), (3,3) (mod 4)

(0,0), (0,1), (3,3) (mod 4)

(1,3), (2,2), (2,3), (3,3) (mod 4)

(0,0), (0,1), (3,3) (mod 4)

(1,3), (2,2), (2,3), (3,3) (mod 4)

(0,0), (0,1), (3,3) (mod 4)

(1,3), (2,2), (2,3), (3,3) (mod 4)

(0,0), (0,1), (3,3) (mod 4)

Degree 4: in the end...

Only two very small graphs have passed ALL the exams!

TheoremLet G be a graph having at least 4 edges and some odd vertices. Thepolynomial S4

G vanishes (mod 2) on G if and only if G is a 3-cycletogether with a further edge which can be either pendent or disjointfrom the cycle.

Here the advantages for gracefulness are NONE.

One graph is graceful, the other is not – exercise before sleeping.

Final remarks on graceful polynomials

Final remarkson graceful polynomials

Final remarks

Degree 5 is being analysed...

Are there efficient techniques for higher degrees?

Why only vanishing (mod 2)?

We have a FAMILY OF POLYNOMIALS, let’s study it !

(modest aim: surpassing the fascinating popularity of thechromatic polynomial...)

Final remarks

For Further Reading

J.A. GallianA Dynamic Survey of Graph LabelingElectr. J. Comb. 16, DS6 (2015).

A. RosaOn certain valuations of the vertices of a graphTheory of Graphs (Internat. Sympos. Rome, 1966) pp. 349-355,Gordon and Breach, New York; Dunod, Paris, 1967.

A. VietriNecessary conditions on graceful labels: a study case on treesand other examplesUtil. Math. 89 (2012), pp. 275-287.

A. VietriGraceful polynomials of graphs, and their vanishing (mod 2)March 2016, submitted.

Real-graceful labellings

Fundamental ingredient in Rosa’s Theorem: double counting of labels(mod 2):

Is (mod 2) absolutely necessary?Counting (mod 2) drastically reduces the information, but...... it is the typical remedy agains the unpredictability of sign.

For, we can sum up (mod 2) the differences on the edges – and in themeantime sum up the labels on vertices – regardless of the SIGN.

Actually, there is at least an alternative way of managing theoscillations of sign!

Is (mod 2) absolutely necessary?Counting (mod 2) drastically reduces the information, but...

... it is the typical remedy agains the unpredictability of sign.

We can look at the SUM OF POWERS OF 2.

(however, any other base would work)

As an example, let us go back to the graceful graph K4:r rrr06 4

1@@��

Let us sum the terms 2a−b + 2b−a where {a,b} are the labels on eachedge – therefore, orientation does not count, similarly as counting(mod 2).

(20−4 + 24−0) + (24−1 + 21−4) + (26−1 + 21−6) + (20−6 + 26−0) +(20−1 + 21−0) + (24−6 + 26−4)...

We can look at the SUM OF POWERS OF 2.

(however, any other base would work)

As an example, let us go back to the graceful graph K4:r rrr06 4

1@@��

Let us sum the terms 2a−b + 2b−a where {a,b} are the labels on eachedge – therefore, orientation does not count, similarly as counting(mod 2).

(20−4 + 24−0) + (24−1 + 21−4) + (26−1 + 21−6) + (20−6 + 26−0) +(20−1 + 21−0) + (24−6 + 26−4)...

The contribution of all edges (because of gracefulness) is 2t + 2−t , forall t in 1,2, ...,e. (here e = 6).

TOTAL : 21 + 22 + · · ·+ 2e + 2−1 + 2−2 + · · ·+ 2−e =

(2e+1 − 1

2− 1− 20

((2−1)e+1 − 1

2−1 − 1− (2−1)0

= 2e+1 − 2−e − 1 .

In our example we obtain 27 − 2−6 − 1.

The contribution of all edges (because of gracefulness) is 2t + 2−t , forall t in 1,2, ...,e. (here e = 6).

TOTAL : 21 + 22 + · · ·+ 2e + 2−1 + 2−2 + · · ·+ 2−e =

(2e+1 − 1

2− 1− 20

((2−1)e+1 − 1

2−1 − 1− (2−1)0

= 2e+1 − 2−e − 1 .

In our example we obtain 27 − 2−6 − 1.

Theorem: (A.V. , 2011)Let G be a graph with e edges. If an injective labelling of the vertices ofG, with natural numbers not exceeding e, gives 2e+1 − 2−e − 1 as thetotal sum, then such a labelling is graceful.

Hint of proof via the Lemma:

Let n be a positive integer. Among all sums ofthe form∑

i ai2i equal ton and such that ai ∈ N for all i , the sum corresponding to the bynaryrepresentation of n minimises the quantity

∑i ai , and it is the only one

to have this property (induction on n etc.).Example:25 = 3 · 1 + 1 · 2 + 3 · 4 + 1 · 8 = 1 · 1 + 3 · 8 = 1 · 1 + 1 · 8 + 1 · 16 = ...

Theorem: (A.V. , 2011)Let G be a graph with e edges. If an injective labelling of the vertices ofG, with natural numbers not exceeding e, gives 2e+1 − 2−e − 1 as thetotal sum, then such a labelling is graceful.

Hint of proof via the Lemma:

Let n be a positive integer. Among all sums ofthe form∑

i ai2i equal ton and such that ai ∈ N for all i , the sum corresponding to the bynaryrepresentation of n minimises the quantity

∑i ai , and it is the only one

to have this property (induction on n etc.).Example:25 = 3 · 1 + 1 · 2 + 3 · 4 + 1 · 8 = 1 · 1 + 3 · 8 = 1 · 1 + 1 · 8 + 1 · 16 = ...

The Lemma has a Corollary which goes in our direction:

Corollary:Let P be a positive integer. If a non-trivial identity

∑1≤i≤P

2i +∑

1≤i≤P

1≤j≤J

aj2j +∑

1≤k≤K

holds for some positive integers J, K and with every aj , a′k in N, then∑j

aj +∑

a′k 6= 2P .

Now the proof of the theorem is closer...

The Lemma has a Corollary which goes in our direction:Corollary:Let P be a positive integer. If a non-trivial identity

∑1≤i≤P

2i +∑

1≤i≤P

1≤j≤J

aj2j +∑

1≤k≤K

holds for some positive integers J, K and with every aj , a′k in N, then∑j

aj +∑

a′k 6= 2P .

Now the proof of the theorem is closer...

Note 1: Almost an excercise... although it seems original!

Nota 2: Changing the base, we obtain other characterisations.

Let us now us particularly strange exponents: ...Definition: (MAIN)Let G be a graph with e edges. An injective mapping γ , from thevertices of G to the real interval [0,e] is a real-graceful labelling if∑

{u,v} spigolo di G

(2γ(u)−γ(v) + 2γ(v)−γ(u)

)= 2e+1 − 2−e − 1 .

Note 1: Almost an excercise... although it seems original!Nota 2: Changing the base, we obtain other characterisations.

{u,v} spigolo di G

(2γ(u)−γ(v) + 2γ(v)−γ(u)

)= 2e+1 − 2−e − 1 .

Let us now us particularly strange exponents: ...

Definition: (MAIN)Let G be a graph with e edges. An injective mapping γ , from thevertices of G to the real interval [0,e] is a real-graceful labelling if∑

{u,v} spigolo di G

(2γ(u)−γ(v) + 2γ(v)−γ(u)

)= 2e+1 − 2−e − 1 .

{u,v} spigolo di G

(2γ(u)−γ(v) + 2γ(v)−γ(u)

)= 2e+1 − 2−e − 1 .

Example: we know that C14 is not graceful;

let us nevertheless construct the oscillating classical labelling (alsowith the typical jump) but with an unknown:(14,0,13,1,12,2,11,3,10,5,9,6,8,x) .Now we calculate x by imposing that

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 26 + 2−6 ;

(we require that the differences arising from x contribute to the totalsum as the classical missing differences, ±1 e ±6)

(2x )2(1 + 26)− (2x ) · 4(

218 +52

212 + 26)

+ 16 · 218(26 + 1) = 0 ⇒

appr. x1 = 14.01 (NO because > 14), x2 = 7.99 (this is admissible!).

Example: we know that C14 is not graceful;let us nevertheless construct the oscillating classical labelling (alsowith the typical jump) but with an unknown:(14,0,13,1,12,2,11,3,10,5,9,6,8,x) .

Now we calculate x by imposing that

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 26 + 2−6 ;

(2x )2(1 + 26)− (2x ) · 4(

218 +52

212 + 26)

+ 16 · 218(26 + 1) = 0 ⇒

Example: we know that C14 is not graceful;let us nevertheless construct the oscillating classical labelling (alsowith the typical jump) but with an unknown:(14,0,13,1,12,2,11,3,10,5,9,6,8,x) .Now we calculate x by imposing that

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 26 + 2−6 ;

(2x )2(1 + 26)− (2x ) · 4(

218 +52

212 + 26)

+ 16 · 218(26 + 1) = 0 ⇒

Example: we know that C14 is not graceful;let us nevertheless construct the oscillating classical labelling (alsowith the typical jump) but with an unknown:(14,0,13,1,12,2,11,3,10,5,9,6,8,x) .Now we calculate x by imposing that

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 26 + 2−6 ;

(2x )2(1 + 26)− (2x ) · 4(

218 +52

212 + 26)

+ 16 · 218(26 + 1) = 0 ⇒

Note: When the cycle length increases, x tends to an adjacent label:the cycle behaves not that ORIGINALLY...But without the little jump, the equation would be not resolvable!

(14,0,13,1,12,2,11,3,10,4,9,5,8, x)

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 22 + 2−2 ⇒

(2x )2 · (1 + 26)− (2x ) · 27 · 212 + 16 · 224 + 16 · 218 = 0 ⇒

IMPOSSIBLE (∆ < 0) .Therefore, our expoinential equation makes some sense...

(14,0,13,1,12,2,11,3,10,4,9,5,8, x)

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 22 + 2−2 ⇒

(2x )2 · (1 + 26)− (2x ) · 27 · 212 + 16 · 224 + 16 · 218 = 0 ⇒

IMPOSSIBLE (∆ < 0) .

Therefore, our expoinential equation makes some sense...

(14,0,13,1,12,2,11,3,10,4,9,5,8, x)

2x−8 + 28−x + 2x−14 + 214−x = 21 + 2−1 + 22 + 2−2 ⇒

(2x )2 · (1 + 26)− (2x ) · 27 · 212 + 16 · 224 + 16 · 218 = 0 ⇒

IMPOSSIBLE (∆ < 0) .Therefore, our expoinential equation makes some sense...

Let us see what happens with complete graphs. Es. K5 .Let us leave only one label free: s s

@@@ ��

��

missing differences: ±3, ±4, ±5, ±6. Therefore:

2x−10 + 210−x + 2x−8 + 28−x + 2x−1 + 21−x + 2x + 2−x =∑

3≤i≤6

(2i + 2−i) .

Putting 2x = η , we have: 1541η2 − 123120η + 1313792 = 0. Appr.solutions for x : 3.66 e 6.07, both admissible.

@@@ ��

��

2x−10 + 210−x + 2x−8 + 28−x + 2x−1 + 21−x + 2x + 2−x =∑

3≤i≤6

(2i + 2−i) .

@@@ ��

��

2x−10 + 210−x + 2x−8 + 28−x + 2x−1 + 21−x + 2x + 2−x =∑

3≤i≤6

(2i + 2−i) .

In order to label K6 we start for example with labels 15, 0, 1, 13, 4. Itremains to cover T = {5,6,7,8,10}. Let {pi} be the labels 15, 0, ... :

η2∑

1≤i≤5

215−pi − 215−10η(

210∑t∈T

2t +∑t∈T

210−t)

1≤i≤5

215+pi = 0 .

Appr. x = 5.81 e 10.87 (both admissible because smaller than 15).

For K7 we use 21, 0, 1, 19, 4, 14 and find x = 5.19 e 15.50.

Questions:1) Do there exist labels in some sense “ optimal ”?2) What is the largest number of integer labels one can ask for, whenthe order increases? (for K3, see later on)

η2∑

1≤i≤5

215−pi − 215−10η(

210∑t∈T

2t +∑t∈T

210−t)

1≤i≤5

215+pi = 0 .

η2∑

1≤i≤5

215−pi − 215−10η(

210∑t∈T

2t +∑t∈T

210−t)

1≤i≤5

215+pi = 0 .

η2∑

1≤i≤5

215−pi − 215−10η(

210∑t∈T

2t +∑t∈T

210−t)

1≤i≤5

215+pi = 0 .

Questions:1) Do there exist labels in some sense “ optimal ”?

2) What is the largest number of integer labels one can ask for, whenthe order increases? (for K3, see later on)

η2∑

1≤i≤5

215−pi − 215−10η(

210∑t∈T

2t +∑t∈T

210−t)

1≤i≤5

215+pi = 0 .

Graceful hypersurfaces

Let us come back to the main definition.

Given a graph G with v vertices, all its real-graceful labellings (inparticular the CLASSICAL ones, if any) are, actually, points of ahypersurface of Rv depending on the graph.

Let us write 2xi−xj + 2xj−xi as

Xi(Xs = 2xs ) .

Real-graceful labellings are the points X in the closed space [1,2e]v ,with pairwise distinct coordinates, belonging to the hypersurface

IG :∑

{ui ,uj} edge

)= 2e+1 − 2−e − 1 .

Given a graph G with v vertices, all its real-graceful labellings (inparticular the CLASSICAL ones, if any) are, actually, points of ahypersurface of Rv depending on the graph.Let us write 2xi−xj + 2xj−xi as

Xi(Xs = 2xs ) .

IG :∑

{ui ,uj} edge

)= 2e+1 − 2−e − 1 .

Given a graph G with v vertices, all its real-graceful labellings (inparticular the CLASSICAL ones, if any) are, actually, points of ahypersurface of Rv depending on the graph.Let us write 2xi−xj + 2xj−xi as

Xi(Xs = 2xs ) .

IG :∑

{ui ,uj} edge

)= 2e+1 − 2−e − 1 .

Classical graceful labellings correspond to the points of IG ∩ [1,2e]v

whose coordinates are distinct and in in {1,2,4,8, ...,2e}.

Problem A:Studying this particular class of hypersurfaces.Problem B:Are there non-graceful graphs as a consequence of the absence ofsuch points on IG? (very strong requirement...).Problem C:Can we reach a classical point along a curve which starts from anon-classical point, using a suitable recursion?Example: K4 . Start with 0, 0, 0, x . Find x (if any):

whose coordinates are distinct and in in {1,2,4,8, ...,2e}.Problem A:Studying this particular class of hypersurfaces.

Problem B:Are there non-graceful graphs as a consequence of the absence ofsuch points on IG? (very strong requirement...).Problem C:Can we reach a classical point along a curve which starts from anon-classical point, using a suitable recursion?Example: K4 . Start with 0, 0, 0, x . Find x (if any):

whose coordinates are distinct and in in {1,2,4,8, ...,2e}.Problem A:Studying this particular class of hypersurfaces.Problem B:Are there non-graceful graphs as a consequence of the absence ofsuch points on IG? (very strong requirement...).

Problem C:Can we reach a classical point along a curve which starts from anon-classical point, using a suitable recursion?Example: K4 . Start with 0, 0, 0, x . Find x (if any):

whose coordinates are distinct and in in {1,2,4,8, ...,2e}.Problem A:Studying this particular class of hypersurfaces.Problem B:Are there non-graceful graphs as a consequence of the absence ofsuch points on IG? (very strong requirement...).Problem C:Can we reach a classical point along a curve which starts from anon-classical point, using a suitable recursion?

Example: K4 . Start with 0, 0, 0, x . Find x (if any):

whose coordinates are distinct and in in {1,2,4,8, ...,2e}.Problem A:Studying this particular class of hypersurfaces.Problem B:Are there non-graceful graphs as a consequence of the absence ofsuch points on IG? (very strong requirement...).Problem C:Can we reach a classical point along a curve which starts from anon-classical point, using a suitable recursion?Example: K4 . Start with 0, 0, 0, x . Find x (if any):

Intersection between IG and the line η1 = η2 = η3 = 1, in R4 ; putη = 2x ;

3(1 + 1)2

1 · 1+ 3

(1 + η)2

1 · η= 27 − 2−6 − 1 + 2 · 6 ⇒

(A := 127− 164

... ) 3η2 + (6− A)η + 3 = 0

Appr. η = 40.30, η = 0.02. Only the first solutions falls in [1,64].Appr. x = 5.33.

r rrr00 0

5.33...@@�� ?

r rrr01 4

6@@��

3(1 + 1)2

1 · 1+ 3

(1 + η)2

1 · η= 27 − 2−6 − 1 + 2 · 6 ⇒

(A := 127− 164

... ) 3η2 + (6− A)η + 3 = 0

r rrr00 0

5.33...@@�� ?

r rrr01 4

6@@��

3(1 + 1)2

1 · 1+ 3

(1 + η)2

1 · η= 27 − 2−6 − 1 + 2 · 6 ⇒

(A := 127− 164

... ) 3η2 + (6− A)η + 3 = 0

r rrr00 0

5.33...@@�� ?

r rrr01 4

6@@��

Note: Starting with 4 equal coordinates leads to no solution.......................................................

the case of K3 (triangle, with labels x , y , z) is very intuitive; let us studythe corresponding surface:Put λ = 2x , µ = 2y , ν = 2z .

(λ+ µ)2

(λ+ ν)2

(µ+ ν)2

Homogeneous coordinates [λ, µ, ν] −→ projective curve in P2 .

Now ν 6= 0 ⇒ 8(X 2Y + XY 2 + X 2 + Y 2 + X + Y )− 119XY = 0

(X = λ/ν, Y = µ/ν). We obtain a real cubic Γ.

Note: Starting with 4 equal coordinates leads to no solution.......................................................the case of K3 (triangle, with labels x , y , z) is very intuitive; let us studythe corresponding surface:

Put λ = 2x , µ = 2y , ν = 2z .

(λ+ µ)2

(λ+ ν)2

(µ+ ν)2

Now ν 6= 0 ⇒ 8(X 2Y + XY 2 + X 2 + Y 2 + X + Y )− 119XY = 0

Note: Starting with 4 equal coordinates leads to no solution.......................................................the case of K3 (triangle, with labels x , y , z) is very intuitive; let us studythe corresponding surface:Put λ = 2x , µ = 2y , ν = 2z .

(λ+ µ)2

(λ+ ν)2

(µ+ ν)2

Now ν 6= 0 ⇒ 8(X 2Y + XY 2 + X 2 + Y 2 + X + Y )− 119XY = 0

(λ+ µ)2

(λ+ ν)2

(µ+ ν)2

Now ν 6= 0 ⇒ 8(X 2Y + XY 2 + X 2 + Y 2 + X + Y )− 119XY = 0

(λ+ µ)2

(λ+ ν)2

(µ+ ν)2

Now ν 6= 0 ⇒ 8(X 2Y + XY 2 + X 2 + Y 2 + X + Y )− 119XY = 0

Es. (4,8) ∈ Γ. In fact log2 4, log2 8, log2 1 are graceful labels (note:ν = 1, so we have set a label equal to 0).Another “classical graceful” point: (2,8) (labels: 1,3,0 instead of 2,3,0).And so forth.

For Further Reading

A. VietriReal-graceful labellings: a generalisation of graceful labellingsArs Comb. 102, 2011, pp. 359-364.

Some algebraic approaches to graceful labellings

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