Graceful polynomials of small degree

Transcription

1 Graceful polynomials of small degree Andrea Vietri 1 1 Dipartimento di Scienze di Base e Applicate per l Ingegneria Sapienza Università di Roma WAGTCN2018, Naples, September 2018 Andrea Vietri 1 / 25

2 Outline 1 Basic concepts 2 Graceful polynomials of small degree (vanishing (mod 2) ) 3 Final remarks Andrea Vietri 2 / 25

3 Basic concepts Why graceful polynomials? For any graph we define a family {S n } n N of graceful polynomials; S n has degree n. Andrea Vietri 3 / 25

4 Basic concepts Why graceful polynomials? For any graph we define a family {S n } n N of graceful polynomials; S n has degree n. There is a deep connection between these polynomials and graceful labellings. Andrea Vietri 3 / 25

5 Basic concepts Why graceful polynomials? For any graph we define a family {S n } n N of graceful polynomials; S n has degree n. There is a deep connection between these polynomials and graceful labellings. The prototype, S 1, was employed by A. Rosa (1967) to obtain the most important class of non-graceful graphs. Andrea Vietri 3 / 25

6 Basic concepts Why graceful polynomials? For any graph we define a family {S n } n N of graceful polynomials; S n has degree n. There is a deep connection between these polynomials and graceful labellings. The prototype, S 1, was employed by A. Rosa (1967) to obtain the most important class of non-graceful graphs. A.V. (2012, 2016): generalisation to every degree (analysis of graceful trees; construction of non-graceful graphs). Andrea Vietri 3 / 25

7 Basic concepts 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 Non-graceful Andrea Vietri 4 / 25

8 Basic concepts Example of graceful polynomial How to obtain a graceful polynomial from G = : Andrea Vietri 5 / 25

9 Basic concepts Example of graceful polynomial How to obtain a graceful polynomial from G = : Assign a variable to each vertex: x 1 x 4 x 2 x 3 x 5 x 6 x 7 Andrea Vietri 5 / 25

10 Basic concepts Example of graceful polynomial How to obtain a graceful polynomial from G = : x 1 Assign a variable to each vertex: x 4 Fix a positive integer n, say n = 3... x 2 x 3 x 5 x 6 x 7 Andrea Vietri 5 / 25

11 Basic concepts x 1 x 2 Take any set of 3 edges and store : x 5 x 6 x 7 (x 1 x 2 )(x 2 x 5 )(x 6 x 7 ) x 1 x 2 x 6 + x 1 x 2 x (mod 2) Andrea Vietri 6 / 25

12 Basic concepts x 1 x 2 Take any set of 3 edges and store : x 5 x 6 x 7 (x 1 x 2 )(x 2 x 5 )(x 6 x 7 ) x 1 x 2 x 6 + x 1 x 2 x (mod 2) SUM UP these polynomials over all the ( 8 3) sets of 3 edges: this is S 3 (G), the 3-rd graceful polynomial of G. Andrea Vietri 6 / 25

13 Basic concepts Formally... given a graph G... Assign variable x i to vertex v i. Andrea Vietri 7 / 25

14 Basic concepts Formally... given a graph G... Assign variable x i to vertex v i. Associate any given edge e j = v p v q to the polynomial P j = x p + x q. Andrea Vietri 7 / 25

15 Basic concepts Formally... given a graph G... Assign variable x i to vertex v i. Associate any given edge e j = v p v q to the polynomial P j = x p + x q. The n-th graceful polynomial of G is SG n (x 1, x 2,..., x V ) P j1 P j2 P jn (mod 2). 1 j 1 <j 2 <...<j n E Andrea Vietri 7 / 25

16 Basic concepts Graceful polynomials and non-graceful graphs The following Lemma provides the connection: Lemma (A.V. 2012) Let G be a graceful graph and let f i be the label of vertex v i. Then, for every positive integer n E(G), ( [ E +1 ) SG n (f 1,..., f V ) 2 ] (mod 2). n Andrea Vietri 8 / 25

17 Basic concepts Graceful polynomials and non-graceful graphs The following Lemma provides the connection: Lemma (A.V. 2012) Let G be a graceful graph and let f i be the label of vertex v i. Then, for every positive integer n E(G), ( [ E +1 ) SG n (f 1,..., f V ) 2 ] (mod 2). n Therefore, if some graceful polynomial VANISHES (mod 2)... Andrea Vietri 8 / 25

18 Basic concepts Graceful polynomials and non-graceful graphs The following Lemma provides the connection: Lemma (A.V. 2012) Let G be a graceful graph and let f i be the label of vertex v i. Then, for every positive integer n E(G), ( [ E +1 ) SG n (f 1,..., f V ) 2 ] (mod 2). n Therefore, if some graceful polynomial VANISHES (mod 2) while E makes the binomial coefficient ODD... Andrea Vietri 8 / 25

19 Basic concepts Graceful polynomials and non-graceful graphs The following Lemma provides the connection: Lemma (A.V. 2012) Let G be a graceful graph and let f i be the label of vertex v i. Then, for every positive integer n E(G), ( [ E +1 ) SG n (f 1,..., f V ) 2 ] (mod 2). n 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. Andrea Vietri 8 / 25

20 Main problem: Choose n, and characterise all graphs for which SG n vanishes (mod 2). Andrea Vietri 9 / 25

21 Main problem: Choose n, and characterise all graphs for which SG n vanishes (mod 2). The problem seems interesting on its own right! Andrea Vietri 9 / 25

22 Main problem: Choose n, and characterise all graphs for which SG n 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...). Andrea Vietri 9 / 25

23 Main problem, degree 1 This is Rosa s polynomial. (A. Rosa, On certain valuations of the vertices of a graph, Theory of Graphs, Internat. Sympos. Rome, 1966, pp ) Andrea Vietri 10 / 25

24 Main problem, degree 1 This is Rosa s polynomial. (A. Rosa, On certain valuations of the vertices of a graph, Theory of Graphs, Internat. Sympos. Rome, 1966, pp ) Let δ p denote the degree of vertex v p. SG 1 (x 1, x 2,..., x V ) P j (x p + x q ) δ p x p 1 j E v pv q edge v p vertex Andrea Vietri 10 / 25

25 δp x p vanishes if all degrees are even (Eulerian graph). Andrea Vietri 11 / 25

26 δp x p vanishes if all degrees are even (Eulerian graph). On the other hand, ( [ E +1 ]) 2 is odd if E 1, 2 (mod 4). Therefore: 1 Andrea Vietri 11 / 25

27 δp x p vanishes if all degrees are even (Eulerian graph). On the other hand, ( [ E +1 ]) 2 is odd if E 1, 2 (mod 4). Therefore: 1 Theorem (Rosa) Eulerian graphs with E 1, 2 (mod 4) are non-graceful. Andrea Vietri 11 / 25

28 Degree 2 Denote coefficients of graceful polynomials by H. S 2 G H 2 px 2 p + H 11 pqx p x q. Andrea Vietri 12 / 25

29 Degree 2 Denote coefficients of graceful polynomials by H. S 2 G H 2 px 2 p + H 11 pqx p x q. H 2 p counts the PAIRS of edges containing v p : v p (x p + x a )(x p + x b ) v b = xp v a Andrea Vietri 12 / 25

30 Degree 2 Denote coefficients of graceful polynomials by H. S 2 G H 2 px 2 p + H 11 pqx p x q. H 2 p counts the PAIRS of edges containing v p : Therefore, H 2 p ( δ p 2 ) (mod 2) v p (x p + x a )(x p + x b ) v b = xp v a δ p 0, 1 (mod 4) v p if we ask for vanishing. Andrea Vietri 12 / 25

31 As to H 11 pq, there are two cases. G 2 1 v p v q δ p δ q choices Andrea Vietri 13 / 25

32 As to H 11 pq, there are two cases. G 2 1 G 2 v 2 p v q v p v q δ p δ q choices δ p δ q 1 choices: (x p + x q )(x p + x q ) not allowed Andrea Vietri 13 / 25

33 As to H 11 pq, there are two cases. G 2 1 G 2 v 2 p v q v p v q δ p δ q choices δ p δ q 1 choices: So we have (if we ask for vanishing): (x p + x q )(x p + x q ) not allowed δ p 0, 1 v p, (δ p, δ q ) (1, 1) G 2 1, (δ p, δ q ) (1, 1) G 2 2 all (mod 4) Andrea Vietri 13 / 25

34 From the above conditions we obtain all possible graphs: Theorem (A.V. 2012) SG 2 vanishes for complete graphs on 4d + 2 vertices, for any integer d. No other graph satisfies the requirement. Andrea Vietri 14 / 25

35 From the above conditions we obtain all possible graphs: Theorem (A.V. 2012) SG 2 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 any integer u, are non-graceful. (count the edges and reach a contradiction through the above Lemma) Andrea Vietri 14 / 25

36 Degree 3 S 3 G H 3 px 3 p + H 21 p,qx 2 p x q + H 111 pqr x p x q x r (commas separate SETS of indices) Andrea Vietri 15 / 25

37 Degree 3 S 3 G H 3 px 3 p + H 21 p,qx 2 p x q + H 111 pqr x p x q x r (commas separate SETS of indices) H 3 p ( δ p 3 ) Andrea Vietri 15 / 25

38 Degree 3 S 3 G H 3 px 3 p + H 21 p,qx 2 p x q + H 111 pqr x p x q x r (commas separate SETS of indices) H 3 p ( δ p 3 ) Hp,q(G ) ( δ p ) δq, H 21 2 p,q(g2 2) ( δ p ) δq (δ 2 p 1) G 2 v a 1 G 2 v 2 p v q v p v q (x p + x q ) 2 (x p + x a ) not allowed Andrea Vietri 15 / 25

39 As to H 111 pqr, there are 4 subgraphs on 3 vertices: v p v p v p v p v q v r v q v r v q v r v q v r G 3 1 G 3 2 G 3 3 G 3 4 Andrea Vietri 16 / 25

40 As to H 111 pqr, there are 4 subgraphs on 3 vertices: v p v p v p v p v q v r v q v r v q v r v q v r G 3 1 G 3 2 G 3 3 G 3 4 For example, H 111 pqr (G 3 3 ) δ pδ q δ r δ r δ q. (forbidden repetitions of either v p v q or v p v r ) v q v p v r Andrea Vietri 16 / 25

41 System of constraints: v p δ p 3 (mod 4) G1 2 ( δp ) δq + ( δ q ) δp 0 (mod 2) 2 2 G 2 ( δp ) 2 δq + ( δ q ) δp + δ 2 2 p + δ q 0 (mod 2) (Hp,q 21 + Hq,p 21...) G1 3 δ p δ q δ r 0... G2 3 δ p (δ q δ r + 1) 0 G3 3 δ p δ q δ r + δ q + δ r 0 G4 3 δ p δ q δ r + δ p + δ q + δ r 0 Andrea Vietri 17 / 25

42 System of constraints: v p δ p 3 (mod 4) G1 2 ( δp ) δq + ( δ q ) δp 0 (mod 2) 2 2 G 2 ( δp ) 2 δq + ( δ q ) δp + δ 2 2 p + δ q 0 (mod 2) (Hp,q 21 + Hq,p 21...) G1 3 δ p δ q δ r 0... G2 3 δ p (δ q δ r + 1) 0 G3 3 δ p δ q δ r + δ q + δ r 0 G4 3 δ p δ q δ r + δ p + δ q + δ r 0 Eulerian graphs clearly satisfy this system. Andrea Vietri 17 / 25

43 If the graph has some vertices of odd degree: Andrea Vietri 18 / 25

44 If the graph has some vertices of odd degree: Theorem (A.V. 2016) S 3 G vanishes (mod 2) on G precisely in one of the following two cases. (A) G is a complete graph K 4t+2, for some positive integer t, possibly having 4u additional vertices and 4u(4t + 2) additional edges that connect these vertices to the above complete graph. (B) G is obtained by taking two complete graphs K a, K b and n additional vertices satisfying one of the following conditions: (B 1 ) : a b 2 (mod 4) and n = 0; (B 2 ) : a b n 1 (mod 4) except a = b = n = 1; (B 3 ) : a b n 3 (mod 4); every additional vertex must be fully connected to K a and K b. Andrea Vietri 18 / 25

45 Old and new non-graceful graphs, as a consequence: Theorem The following graphs are non-graceful. Type (A) in the above theorem, provided t 2 (mod 4). Type (B 1 ) with a + b 12 (mod 16); type (B 2 ) with a + b 10 (mod 16); type (B 3 ) with a + b 14 (mod 16). K 7 K 7 K 5 K 5 a = b = 7, n = 3 a = b = n = 5 Andrea Vietri 19 / 25

46 Degree 4 Not surprising: much more cases and many subgraphs to check... Andrea Vietri 20 / 25

47 Degree 4 Not surprising: much more cases and many subgraphs to check... S 4 G p H 4 px 4 p + p,q H 31 p,qx 3 p x q + pq H 22 pqx 2 p x 2 q + + p,qr Hp,qr 211 xp 2 x q x r + Hpqrs 1111 x p x q x r x s. pqrs Andrea Vietri 20 / 25

48 Degree 4 Not surprising: much more cases and many subgraphs to check... S 4 G p H 4 px 4 p + p,q H 31 p,qx 3 p x q + pq H 22 pqx 2 p x 2 q + + p,qr Hp,qr 211 xp 2 x q x r + Hpqrs 1111 x p x q x r x s. pqrs Strategy: (I) Find constraints for single vertices and pairs of vertices Andrea Vietri 20 / 25

49 Degree 4 Not surprising: much more cases and many subgraphs to check... S 4 G p H 4 px 4 p + p,q H 31 p,qx 3 p x q + pq H 22 pqx 2 p x 2 q + + p,qr Hp,qr 211 xp 2 x q x r + Hpqrs 1111 x p x q x r x s. pqrs Strategy: (I) Find constraints for single vertices and pairs of vertices (II) Constraints for subgraphs with 3 vertices some ruled out by (I) Andrea Vietri 20 / 25

50 Degree 4 Not surprising: much more cases and many subgraphs to check... S 4 G p H 4 px 4 p + p,q H 31 p,qx 3 p x q + pq H 22 pqx 2 p x 2 q + + p,qr Hp,qr 211 xp 2 x q x r + Hpqrs 1111 x p x q x r x s. pqrs 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: H 1111 pqrs 0 Andrea Vietri 20 / 25

51 Some constraints for the degree 4 Constraint for single vertices: 0 δ p 3 (mod 8) Andrea Vietri 21 / 25

52 Some constraints for the degree 4 Constraint for single vertices: 0 δ p 3 (mod 8) Forbidden degrees of non-adjacent vertices: (1, 3), (2, 2), (2, 3), (3, 3) (mod 4) Andrea Vietri 21 / 25

53 Some constraints for the degree 4 Constraint for single vertices: 0 δ p 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) Andrea Vietri 21 / 25

54 Some constraints for the degree 4 Constraint for single vertices: 0 δ p 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) Andrea Vietri 21 / 25

55 Some constraints for the degree 4 Constraint for single vertices: 0 δ p 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 as colour constraints) Andrea Vietri 21 / 25

56 Degree 4: in the end... Only two very small graphs have passed ALL the exams! Andrea Vietri 22 / 25

57 Degree 4: in the end... Theorem Only two very small graphs have passed ALL the exams! Let G be a graph having at least 4 edges and some odd vertices. The polynomial SG 4 vanishes (mod 2) on G if and only if G is a 3-cycle together with a further edge which can be either pendent or disjoint from the cycle. Andrea Vietri 22 / 25

58 Degree 4: in the end... Theorem Only two very small graphs have passed ALL the exams! Let G be a graph having at least 4 edges and some odd vertices. The polynomial SG 4 vanishes (mod 2) on G if and only if G is a 3-cycle together with a further edge which can be either pendent or disjoint from the cycle. Here the advantages for gracefulness are NONE. Andrea Vietri 22 / 25

59 Degree 4: in the end... Theorem Only two very small graphs have passed ALL the exams! Let G be a graph having at least 4 edges and some odd vertices. The polynomial SG 4 vanishes (mod 2) on G if and only if G is a 3-cycle together with a further edge which can be either pendent or disjoint from the cycle. Here the advantages for gracefulness are NONE. One graph is graceful, the other is not exercise before sleeping. Andrea Vietri 22 / 25

60 Final remarks Final remarks Degree 5 is being analysed... Andrea Vietri 23 / 25

61 Final remarks Final remarks Degree 5 is being analysed... Are there efficient techniques for higher degrees? Andrea Vietri 23 / 25

62 Final remarks Final remarks Degree 5 is being analysed... Are there efficient techniques for higher degrees? Why only vanishing (mod 2)? Andrea Vietri 23 / 25

63 Final remarks 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! Andrea Vietri 23 / 25

64 Final remarks 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 the chromatic polynomial...) Andrea Vietri 23 / 25

65 For Further Reading Final remarks J.A. Gallian A Dynamic Survey of Graph Labeling Electr. J. Comb. 16, DS6 (2015). A. Rosa On certain valuations of the vertices of a graph Theory of Graphs (Internat. Sympos. Rome, 1966) pp , Gordon and Breach, New York; Dunod, Paris, A. Vietri Necessary conditions on graceful labels: a study case on trees and other examples Util. Math. 89 (2012), pp A. Vietri On the Graceful Polynomials of a graph submitted... Andrea Vietri 24 / 25

66 Final remarks THANK YOU. Andrea Vietri 25 / 25