the edge reconstruction of hamiltonian graphs

The Edge Reconstruction of Hamiltonian Graphs -

L. Pyber M A THEMA K A L /NST/ JUTE OF THE

HUNGARlAN ACADEMY OF SCIENCES. BUDAEST, HUNGARY

ABSTRACT

If a graph G on n vertices contains a Hamiltonian path, then G is reconstructible from its edge-deleted subgraphs for n sufficiently large.

0. INTRODUCTION

A graph G is said to be edge reconstructible if G is determined by its edge deleted subgraphs.

The edge reconstruction conjecture asserts that every graph with at least four edges is edge reconstructible.

There are various results stating that certain classes of graphs are edge reconstructible, e.g., the graphs with n vertices and at least n log, n edges [ 9 ] .

Another such result due to Lovasz is the following (see [4], [7]):

Theorem. path, then it is edge reconstructible for n sufficiently large.

If a graph of n vertices and rn 2 2n edges contains a Hamiltonian

Lovasz suggested [7] that perhaps the condition on the number of edges

Here we prove his conjecture. might be omitted.

Theorem I. edge reconstructible for n sufficiently large.

If a graph of n vertices contains a Hamiltonian path, then it is

We derive this theorem as a consequence of a more general result.

Definition. joint paths covering V(G).

Journal of Graph Theory, Vol. 14, No. 2, 173-179 (1990)

The Hamiltonian shortage hs(G) is the minimal number of dis-

0 1990 by John Wiley & Sons, Inc. CCC 0364-902419010201 73-07$04.00

174 JOURNAL OF GRAPH THEORY

Theorem 11. tices, and

There exists a constant (Y > 0 such that if G is a graph of n ver-

( i ) hs(G) 5 a- and d(s) 2 1 for all .v E V ( G ) . ( i i ) hs(G) 5 cY(n/log n ) and d(x ) L 2 for all .r E V(G) , then G is edge

reconstructible for n sufficiently large.

Notations and Definitions. Let G be a graph. We denote by V(G) and E(G) the set of vertices and the set of edges, respectively. e(G) = e denotes IE(G)( and n(G) = 11 denotes IV(G)I. dJu) denotes the degree of a vertex u E V(G).

Vertices of degree 1 are called endvertices. 4 ( G ) denotes the maximum degree of G and n,(G) the number of vertices of degree i . Clearly we have IE(G)I = ~ I ~ ~ ~ ' n , ( G ) . i . For e E E(G) we denote by G\e the graph obtained by deleting the edge e from G. A path P C G i s called a topological edge if the inner vertices of P are of degree 2 and the terminal vertices are of degree 2 2 . Topological edges incident to endvertices are called strings. By covering puth system we mean a set of disjoint paths 3 that cover V ( G ) (here we allow for paths of length 0).

If G and H are two graphs. then an embedding of H into G is an injection 4 from V ( H ) to V ( G ) such that for ( . v , j ) E E ( H ) we have (&), +(.v)) E E(G). By H ( G ) we denote the number of Hamiltonian cycles and by p ( G ) the number of I factors of G. By log we mean logarithm to the base 2 .

1. SOME PROPERTIES OF NON-EDGE RECONSTRUCTIBLE GRAPHS

Powerful counting methods introduced by Lovasz [6] and Muller (91 has lead to the following lemma of Nash-Williams [lo]:

Lemma 1.1. IE(G)I mod 2 we have a permutation 6 of V ( G ) such that E(G) n E(G6) = A .

If G is non-edge reconstructible then, for all A C E(G), (A( = I

The following easy but useful corollary is again due to Lovasz [ 7 ] .

Corollary 1.2. graph ( V ( G ) , T ) has at least 2"''- ' I - ' embeddings into G.

I f G is non-edge reconstructible then for all T C E(G) the I

This result enables us to consider enumeration questions in order to obtain classes of edge reconstructible graphs.

Corollary 1.3. If G is non-edge reconstructible and H ( G ) 2 1 , then

EDGE RECONSTRUCTION OF HAMILTONIAN GRAPHS 175

Proof. is fixed. I

A cycle of length n has exactly 2n embeddings into G if the image

To prove the edge reconstructibility of Hamiltonian graphs, it is sufficient to

To prove Theorems I and 11, we also need some elementary observations. show that the above inequality cannot hold for large 1 2 .

Lemma 1.4. is edge reconstructible.

The (ordered) set 3 of lengths of strings (counting multiplicities)

Proof. Let us consider the sets of lengths of strings %< in the edge-deleted subgraphs G\e, for all (x,y) = e E E(G), with dG(x) = 1 and dG(y) = 2. (The degrees of endvertices of a deleted edge e are edge reconstructible [lo].) Sup- pose we have t such edges. If t = 0, then % is of the form (1, . . . , 1) (3 has n,(G) elements).

If t 2 1 then let e be the edge for which Lfe is minimal with respect to the lexicographic order. Adding 1 to the ( n , ( G ) - t)th element of Ye , we obtain 3. I

Lemma 1.5. edges of length >2n,(G) + 2.

If G is non-edge reconstructible, then G has no topological

Proof. Suppose G has a string S of length s 2 2. Then s - Is%, for other- wise if e is the final edge of S then G\e and % uniquely determine one endvertex of e in G\e and the o ther i s an isolated ver tex , i . e . , G i s edge reconstructible. Therefore n,(G) 2 s.

Suppose there is a topological edge P of length >2n1(G) + 2). Let e be the middle edge of this (path) P. Given G\e, we know that e connects two endvertices of G\e; moreover, as there are two strings of length ?n,(G) + 1 in G\e, e has to connect the endvertices incident to these two strings (paths). G is edge reconstructible, a contradiction. I

II. HAMILTONIAN CIRCUITS

Corollary 1.3 shows that we have to find an estimate for H(G) that is "some- what better" than 2P'G)-"(C' . Let us note that there is a trivial upper bound H(G) 5

. It follows from the fact that the cycles (subgraphs with all degrees even) of G form a vector space over the 2 element field and the dimension of this vector space is exactly e(G) - h(G) + 1 (see [ I ] ) .

Lovasz suggested the investigation of H(G) for 3 regular graphs. This case turns out to be the core of our proof.

First we need some results on p ( G ) , the number of 1 factors. Using the theory of matrix functions (e.g. , permanents and hafnians), the following is proved in the book of Minc 181:

2efC)-nlGl+ I


Lemma 2.1. For a graph G with degrees d, , . . . , d,, we have

Lemma 2.2. Suppose G is a graph with A(G) 5 3. Then H(G) 5 6" 7' 'c- '"

Proof. We might suppose that n,,(G) = 0 and r i l = (G) = 0. Let us consider the subgraph D G induced by vertices of degree 3 in G. The comple- ment of a Hamiltonian cycle of G is a 1 factor of D: therefore H(G) 5 p ( D ) .

We have n(D) = 2(e - n ) and by Lemma 2.1 our statement follows. I

Next we prove a trivial reduction lemma.

Lemma 2.3. incident to u with

Suppose u E V(G) and dG(u) = d . There exists an edge e of G

P m f . Let e l , . . . , e, denote the edges incident to d. Each Hamiltonian cycle I of G avoids exactly d - 2 of them. Therefore CP;, H(G\e,) = (d - 2)H(G).

Lemma 2.4. Suppose G is a graph with A(G) 5 4. Then H(G) 5 c'-" where

Proof. 2e = 2n,(G) + 3n,(G) + 4n,(G). Therefore e - n = in,(G) + n,(G). It follows that G has at least e - n vertices of degree 23 .

Suppose there is a vertex u E V(G) such that u is of degree 4 and its neighbors are of degree 4 and 2. By Lemma 2.3 we might delete an edge e incident to u such that 2H(G\e) 3 H ( G ) . Let us observe that n,(G) + n,(G) = n,(G\e) + n,(G\e) by the choice of u. We delete edges in the same way until we arrive at a graph GI where each vertex of degree 4 has at least one neighbor of degree 3. If we deleted r , edges then, by Lemma 2.3, 2'IH(G,) 2 H(G). By the choice of G we have n 3 ( G , ) + n,(G,) = nJG) + n,(G) and 3n,(G,) 2

n,(G,). Therefore n,(G,) 2 ace - n ) . Now we again start an edge-deleting process. If u E GI is a vertex of degree 4 we delete an edge e incident to u such that

2H(G,k) 2 H ( G , ) . Clearly we have n,(G,k) 2 n,(G,). We delete edges in the same way until we amve at a graph G, with A(G2) 5 3 .

Now we have n,(G,) 2 a(e - n) and if we delete t , edges, then 2'W(G,) 2

H(G,). We have e - n - r l - tz = e(G2) - n(G,) = r, = p,(G2) 2 i ( e - n ) . I


By Lemma 2.2 we have H(G,) 5 G(1'3)r3. Therefore using 2'lH(G1) 2 H(G)) we have

(because 61'3 < 2,tl + t, + t3 = e - n and f3 2 $(e - n)). 1

Lemma 2.5. For an arbitrary graph, G we have

H(G) 5 ce- .

Proof. If G has a vertex u with d(u) 2 5 then by Lemma 2.3 we might delete an edge e incident to u such that fH(G\e) 2 H(G). Deleting edges in the same way, we obtain a graph GI with A(G,) I 4. If we deleted r edges then H(G) I ($Y(G,). As e - n = t + @(GI) - n(G,)) and c > i, our statement follows from Lemma 2.4. I

111. COVERING PATH SYSTEMS

Lemma 3.1. If G is a non-edge reconstructible graph with n,(G) = 0 and hs(G) = p , then e - n 1 f((n - P)/(2nl(G) + 2) - P - n,(G)).

Proof. By Lemma 1.5 any path of length f in G contains at least ( f / ( 2 n l ( G ) + 2 ) ) - I vertices of degree 2 3 . For a covering path system 8 = {PI, . . . , Pp} we have at least

n - p 2n,(G) + 2 - P

such vertices. Therefore

Lemma 3.2. If G is a graph with hs(G) = p and 9 = {PI , . . . , P,} is a minimal covering path system, then the graph (8, V(G)) has less than 2p * ce-" . nZp embeddings into G.

Proof. With a fixed embedding q of (9, V(G)) into G we might associate a Hamiltonian cycle of a graph Go 3 G in the following way: As the image q(9) C G is again a minimal covering path system, there are no edges of G between the endvertices of different paths of q(8). Therefore we might connect the last endvertex of q(PJ to the first endvertex of cp(P,+,) (with Pp+, = PI). This way we obtain a graph Go with e(G,) = e + p where the new edges and q(9)


form a Hamiltonian cycle. Given such a graph G,,, a Hamiltonian cycle contain- ing the p new edges. the "first" new edge (i.e., the one that connects cp(P,) to cp(P2)) uniquely determines ( ~ ( 3 ) . (The first edge is an ordered one.)

The number of such graphs Go is at most (:Y. H(G,,) 5 c"-"+" for all such graphs by Lemma 2.5. and there are 2 p possible f in t edges. Therefore the number of embeddings of (9. V(G)) into G is at most

3.3 Main Lemma. Suppose G is a graph with hs(G) = p 5 ~ ( e - n)/log n where E > 0 is defined by 2" = ( 2 / c ) > 1 . Then G is edge reconstructible.

Proof. By Lemma 3.2 the number of embeddings of a fixed minimal cov-

contradicts Corollary 1.2 - ering path system (V(G) , 3) is less than 2p . n2p . ct'-'' 5 2 p * 22e'c-n'c'r-n' - <

2'-"',>-l(2p . , - l ~ l~ - ! , l + /~ - l l < = 2 ~ - n + p + I c i t i ~ - p i P I I = 2 (we used E ( e - n ) 2 p ) . I

3.4. The Proof of Theorem 11. For a graph G with hs(G) = p we have nl(G) 5 2p. By Lemma 3.1 we have e((e-n)/log 17) 2 -(s - 3 p ) = x, and for a suitable cr if n is sufficiently large p 5 a- leads to x > p . (This inequation is quadratic in p ) . By Lemma 3.3 the proof of I l ( i ) is complete.

For a graph G with n,(G) = 0 we have

and for a suitable a! if 11 is sufficiently large p 5 a(n/log n ) leads to J > p . By Lemma 3.3 the proof of II(ii) is complete. I

IV. ADDITIONAL REMARKS

Our results are valid only for very large 1 1 . The strongest possible condition in our setting is Hamiltonicity. Using more involved estimates it can be shown that this condition implies the edge reconstructibility of graphs with n 2 240 vertices. However. we are still far from the optimal result:

Problem. Prove that all Hamiltonian graphs are edge reconstructible. The sufficient conditions for edge reconstructibility in Theorem I1 seem to be rather general ones. However, we were unable to find natural classes or graphs % with hs(G) = o(n(G)) for G E 93, which are not necessarily Hamiltonian.

On the contrary, by a result of Brown [3] there exist 3 connected planar graphs P with hs(P) 2 [ ( n - 10)/3]n arbitrarily large.

Another negative example is the class <?Fr of r connected r regular graphs.


Example. many graphs G E 9, with hs(G) 2 a,n(G).

For all r 2 3 there exists a constant 0 < a, < 1 and infinitely

Proof. It is known [5] that there exist graphs on n vertices ( n arbitrarily large) in Sr where the longest cycle is of length sns(r), E(r) < 1 . The same can be proved for the longest paths (e.g., using a result of [2]). Therefore there exists a graph G E 9, with hs(G) > i r + 1. Now if H E 9, then we might construct a new graph I? E 3, in the usual way: We replace each edge of H by a path of length 3. We omit the original vertices, and for a u E V(G) take n ( H ) copies of G\u. fi is constructed by identifying the r vertices of degree r - 1 of the ith copy with the r neighbors of the ith original vertex of H for all i .

Let 9 be a covering path system of I?. As hs(G) > i r + 1 it follows that each copy of G contains at least one endvertex of some path in 9. Therefore hs(I?) 2 (n(fi))/2n(G), which proves our statement. I

ACKNOWLEDGMENT

The author wishes to thank L. Lovasz for his continued help.

References

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the edge reconstruction of hamiltonian graphs

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