The Reconstruction of Graphs. Dhananjay P. Mehendale Sir Parashurambhau College, Tilak Road, Pune , India. Abstract

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1 The Reconstruction of Graphs Dhananay P. Mehenale Sir Parashurambhau College, Tila Roa, Pune-4030, Inia. Abstract In this paper we iscuss reconstruction problems for graphs. We evelop some new ieas lie isomorphic extension of isomorphic graphs, partitioning of vertex sets into sets of equivalent points, subec property, etc. an evelop an approach to eal with reconstruction problem. We then iscuss complete sets of invariants for graphs an reconstruction conecture. We then begin with evelopment of few equivalent formulations of reconstruction conecture. In the last section we briefly elaborate the formulation ue to Harary its exact eman an finally procee to give a ifferent proof of reconstruction conecture using reconstructibility of graph from its spanning trees an reconstructibility of tree from its penant point elete ec of subtrees. This last proof can be use to evelop a systematic proceure to reconstruct unique graph from its ec.. Introuction: This well nown conecture, calle the reconstruction conecture, of Kelly [] an Ulam [] can be state as follows: Let G an H be two simple graphs with p points, p>, an let there exist a (-), onto map, θ : V( V(H), vi V ( ui V ( H ) such thatgi G vi Hi H ui, i, i =,, L, p, then G H. (The symbol stans for isomorphism.) The existence of such a map is calle hypomorphism. The reconstruction conecture states that hypomorphism implies isomorphism. Let v i V (. Then the inuce subgraph Gi G vi is calle the vertex elete subgraph (VDS) of G. The collection of all VDS, { Gi G vi, i =,, L, p }, is calle the ec of (vertex elete subgraphs of) G, Dec (. The graph G is reconstructible if every graph with the same ec as G is isomorphic to G. Thus, the conecture can be state as: Every graph with more than two vertices is uniquely (up to isomorphism) reconstructible from its ec.

2 In orer to reconstruct G from its ec, Dec (, it suffices to ientify which must be the neighbors (aacent vertices) of the missing vertex v i in some graph G i in the ec [3]. We procee to iscuss about how to accomplish this tas.. Preliminaries: Let G an H be two simple (p, q) graphs, i.e. having p points (vertices) an q lines (eges), an let G be isomorphic to H. Then there exists a (-), onto, aacency preserving map θ : V( V(H), vi V ( ui V ( H ) such that if a pair of vertices ( v, ) V (, the vertex set for G, are v l aacent (nonaacent) then the corresponing pair of vertices, uner the isomorphism mapθ, the pair of vertices ( u, u l ) V (H), the vertex set for H, will be aacent (nonaacent). Now, if we a a new vertex v to the vertex set of G, V (, an a new vertex u to the vertex set of H, V (H ), an also a an ege ( v, vi ) to the ege set of G, E(, an the corresponing ege ( u, ui ) e to the ege set of H, E(H), an enote the extene G an H by G an e e e H respectively, then clearly G an H will be isomorphic with the isomorphism map e e e θ : V ( G ) V ( H ) where e θ ( V ( G )) = θ ( V ( G )), an e θ ( v) = u. Few efinitions are now in orer: Definition.: Let G an H be two isomorphic graphs. The process escribe above of extening graphs G an H to their respective e e supergraphs G an H is calle the isomorphic extension of isomorphic graphs. e Definition.: Let G be an unlabelle (p, q) graph an let G be a supergraph of G obtaine by taing a (new) vertex outsie of the vertex set V( an oining it to some (unspecifie) vertex of G by an ege not e in the ege set E( is calle the extension of G to G.

3 Definition.3.: A subset V of points { u, u, L, ur } in a tree T is calle a set of equivalent points or simply an equivalent set if all the trees T + vu s, s r, obtaine from T by aing an ege vu s, obtaine by oining vertex us in set V to a new vertex v, are all isomorphic to each other. Definition.3.: A subset of points (vertices) of V( is calle a set of equivalent points or an equivalent set if the graphs obtaine by extension of G at any point among these points, achieve by oining any point among these points, one at a time, to a (new) point outsie, not in V(, leas to graphs which are all mutually isomorphic. Definition.4: A collection of subsets { V, V, L, Vm } of V(, the vertex set of G is calle partitioning of V( into equivalent sets if all subsets V i, i =,,, m are equivalent sets, m ViI V = φ, i, an V ( G ) = U Vi, where φφ is a null set. i = Theorem.: Let G be a (p, q) graph an let { V, V, L, V } be the partitioning of the vertex set of G into equivalent sets, an let { V, V, L, V } be their carinalities. The number of nonisomorphic graphs that can be achieve from a graph G by aing some r eges, r p, by extension, i.e. by oining some r points of G to an externally taen new point not in the vertex set of G, is equal to the number of (restricte ) -compositions of r, lie: r = α + α + L+ α + L+ α, where α V. 0 Proof: Any two graphs with ientical composition (though istinct ones ue to extension at istinct points from the same equivalent set) are clearly isomorphic (refer to efinition.3.). Any two graphs with istinct composition will be clearly nonisomorphic, since they are arrive at by choosing ifferent number of points for oining to the new point from the equivalent sets, iffering at least at one place, sayα. 3

4 Definition.: Let G an H be two graphs on p points. We say that Dec ( is a subec of Dec (H ), or, Dec( Dec( H ) when there exists a biection ψ : V( V( H) such that vi V ( ui V ( H ), where V (, V ( H ) are the vertex sets of G an H respectively, an Gi G vi is a subgraph of Hi H u i for all i =,,, p. Whenever Gi G vi is a proper subgraph of Hi H ui for at least one i =,,, p, we say that Dec ( is a proper subec of Dec (H ), i.e. Dec( Dec( H ). Thus, Dec ( is not a subec of Dec (H ) if there exists no map ψ (see efinition.), such that the inclusion relation, namely, Dec( Dec( H ) hols goo, i.e. this relation is violate for at least one subgraph of G an the corresponing subgraph of H. Definition.6: The subisomorphism map for two p point graphs G an H is a biection ψ : V( V( H) where vi V ( ui V ( H ) such that if a pair of vertices ( v, v l ) V ( are aacent then the corresponing pair of vertices, uner the subisomorphism map, ( u, u l ) V (H), will be aacent an when ( v, v l ) V ( are nonaacent then ( u, ) V (H) may or may not be aacent. u l Remar.: Uner the existence of subisomorphism map ψ : V ( V ( H ) G will be a spanning subgraph of H. Definition.7: Checing whether Dec( Dec( H ) hols or oesn t hol for some graphs G an H will be calle checing of subec property for graphs G an H Lemma.: Let G an G be two graphs on p points. If G is subgraph of G then Dec( G ) Dec(. Moreover, if G is a proper subgraph of G then Dec G ) Dec( ). ( G Proof: Since, G is a subgraph of G there will exist a subisomorphism map uner which every aacent vertex pair of G will be aacent in G 4

5 an Dec( G ) Dec(. When G will be a proper subgraph of G then there will exist a nonaacent vertex pair in G which will be aacent in G so when we consier a point elete subgraph of G an the corresponing point elete subgraph of G arrive at by eleting a vertex other than the one containe in the vertex pair that forms a nonaacent vertex pair in G an whose image is an aacent vertex pair in G an Dec G ) Dec( ) becomes clear. ( G Lemma.: Let Gi G vi. We have G = G i + (v, x ), where (v, x ) is the ege obtaine by oining the new point v not in V G ) to x V( G ) i. If Dec ( G ) Dec(, then subgraph of. Proof: Given Dec ( G ) Dec(. So, for x there will exist G v x r v i ( p ) G G (i.e., v, L, v G such that G x G v. We efine the biection ψ : V( G ) V( where v an v i r G then corresponing pair G v i (In fact, be aacent in while pairs G v = v. Now, if a pair of vertices v p, G G then corresponing pair v,, r, are nonaacent in x r i ( i G is, x, L, x G r x p, ( p ) G v r an x q v q will be aacent G, since = G v ). Also if pair of vertices i i, v is aacent G v v, x will v will be aacent in G, G an the corresponing pairs v i, v r may or may not be aacent in G since G x So the map is a subisomorphism map. r G v r. Remar.: The lemma will wor well also in the conition when we r have G instea of G obtaine by oining the new vertex v not in V ( Gi ) to some r points of Gi G vi, r an = egree of v i, r such that Dec ( G ) Dec(. This will lea us to the same conclusion that G. r G

6 Remar.: Let G i = G x, x V G ), an let i r r Dec ( G ) Dec(, where G is obtaine from G i by r extensions achieve by oining some r points of G i to the new point v not in V ( Gi ). There exist subgraphs G l = G v l in the Dec ( such that G i will be subgraph of Gl when v l in G will correspon to (or stan for or represents the appearance of) x in G i. There will be a point in G l an not containe in G i, say z m, of egree or ( ) epening upon whether z m is aacent or not aacent to v l. This point actually r represents the appearance of vi in G l, in such a way thatg x will be ( i subgraph ofg. Since this will happen for all l there will correspon a v l efining a x, i.e. for every r G l such that G G. So, we will thus have a subisomorphism map subgraph of l ψ : V( G r ) V( where x v l an v v i. x in x will be 3. Kelly-Ulam Conecture an a Reconstruction Algorithm : We now procee to show that hypomorphic graphs are also isomorphic, i.e. the well-nown Kelly-Ulam conecture for graphs. The proof contains an algorithm for actual reconstruction. Theorem 3.(Kelly-Ulam Conecture): Hypomorphic graphs are isomorphic. Proof : Let G an H be two simple hypomorphic graphs on p points, p>, i.e. let there exist a biection, θ : V( V(H), vi V ( ui V ( H ), where Gi G vi Hi H ui, i, i =,, L, p. Let ω i : V ( Gi ) V ( Hi ), be the corresponing isomorphism maps for i =,,, p. () Consier some subgraph Gi G vi of G an the corresponing isomorphic subgraph Hi H ui of H. () Exten isomorphically the graphs G i to G an Hi to H by aing an ege by oining some point x V( G ), the vertex set i G i 6

7 of G i, to a new point v not belonging to V ( Gi ), leaing to the formation of the graph G an by aing an ege oining the corresponing point y =ω i( x ) V( Hi ), the vertex set of H i, to a new point u, not belonging to V ( Hi ), leaing to the formation of the graph H. (Note that there are in all (p-) possibilities for aing eges, one in G i an the corresponing one uner isomorphism map in H i, since there are (p-) vertices in both G i an H i.) (3) Chec whether Dec ( G ) Dec( hols. If yes, then as per the lemma. G is a proper extension of G i towars G, org is a subgraph of G, (an automatically H will be a proper extension of H i towars H, or H will be subgraph of H). (4) Now, remove the ae ege an a a new ege by oining some new point x V( i, the vertex set of G i, to a new point v, not belonging to V ( Gi ), leaing to the formation of the graph G an by aing an ege oining the corresponing point y =ω i( x ) V ( Hi ), the vertex set of H i, to a new point u, not belonging to V ( Hi ), leaing to the formation of the graph H. Go to step (3). () Prepare the list of say, proper vertices i.e. where the extension is proper, namely { x V( G i ) s, s =,,, r}. r (6) From the pairs of proper vertices so obtaine, in number, etermine the list of proper pairs i.e. for which the extension is proper, i.e. when the extension is carrie out by oining the pair of proper vertices (in G i ) an their images uner ω i (in H i ) to the external points, respectively to v an u, leaing to the formation of the graph G an H, such that Dec ( G ) Dec(, implying (remar.) G G (an H H). r (7) Form the triplets of proper vertices, in number, an their 3 images in H such that any two of them also form a proper pair, 7

8 etermine the proper triplets for which Dec ( G 3 ) Dec( (an Dec( H ) Dec( H ) ) implying G G (an H H). (8) Continue extening isomorphically, an each time taing the proper action implie by the correctness (or incorrectness) of aing the eges as per the above given steps, till we reach the graphs G an H, where is the egrees of vertices v i, ui (the number of vertices, number of eges, an egree sequences of hypomorphic graphs are ientical). At this stage we will not only have Dec( G ) Dec( an Dec( H ) Dec( H ), moreover Dec( G ) = Dec( implying G G an Dec( H ) = Dec( H ) implying H H. An by the isomorphic extension of G i to G an Hi to H, i.e. by G H, we have G H. Proof : Since the number of vertices, number of eges, egree sequences etc. of hypomorphic graphs are ientical. We can tae the egree of vertex v i, ( vi ) an the egree of vertex u i, ( ui ) as ientical an equal to say. () Consier some subgraph Gi G vi of G an the corresponing isomorphic subgraph Hi H ui of H. () Exten isomorphically the graphs G i to G an Hi to H by aing some eges by oining some points x, x,, ( ) L x V G i, the vertex set of G i, to a new point v, not belonging to V G ), leaing to the formation of the graph ( i G an by aing some ege oining the corresponing points y, y,, ( ) L y V H i, the vertex set of H i, such that y = ω ( x ), r =,, L, to a new point u, not belonging to r i r, V ( Hi ), leaing to the formation of the graph H. Note that the total number of possibilities for aing eges at the same time by oining vertex v an some points of G i an for aing eges by oining vertex u an the corresponing points of Hi uner isomorphism map, are both equal to the binomial coefficient p. 8

9 (3) Chec each time whether Dec ( G ) = Dec( hols. If yes, then G will be a proper extension of G i to G i.e. G G (an hence H will be a proper extension of H i to H i.e. H H). If not, then the extensions are improper an the ae eges shoul be remove for trying the other choice. (4) Since the ec of G (an ec of H) is legitimate (legitimate ec: a ec which can actually be obtaine from some graph) there must exist a choice for which Dec ( G ) = Dec( implying G G an will hol. An Dec ( H ) = Dec(H ) Which will imply that H H. () We have achieve the isomorphic extension of G i to G an H to H, i.e. H, therefore, G H. i G Proof 3: Given unlabelle legitimate ec of point elete subgraphs of G we procee to show that we can reconstruct a unique graph up to isomorphism (i.e. the reconstructe graph is isomorphic to () Consier a subgraph in the ec. () A some eges by oining some points of this subgraph to a new point. (3) Chec subec property for this new graph an G. (4) If it hols then stop. Else continue the steps from step () for all possible reconstructions equal to the binomial coefficient, p. () At least one choice must exist for which the subec property will hol goo an for the graph thus prouce, say G *, we will have Dec ( G*) = Dec(. Otherwise, the ec of G will not be legitimate, a contraiction. If there exists only one choice (for aing eges), then a unique graph is thus constructe an we are through. When there is more than one choice, we have to see that all these graphs are isomorphic. But this is clear since these multiple graphs are arrive at ue to presence of more than one equivalent points in some equivalent set belonging to the equivalent partitioning of the point elete subgraph in the ec taen for extension, whose proper subset is require to be use for extension. 9

10 Remar 3.: If we carry out partitioning of V ( Gi ) into equivalent sets { V, V, L, V } an let { V, V, L, V } be their carinalities. Let s, s, L, s be the number of points in the sets V, V, L, V respectively which when oine to the external point + + prouce a graph G (a copy of such that Dec ( G ) = Dec(. It is clear that we can construct in all V i = s i i number of graphs isomorphic to + G. 4. A Counting Formula: The count of isomorphic graphs that one will construct following the proceure iscusse above in the proof 3 can be easily obtaine as follows: Theorem 4.: Let G i be a graph in the ec of G. An let { V, V, L, V } be the partitioning of the vertex set of G i into equivalent sets, an { V, V, L, V } be their carinalities. Let { r, r, L, r } be the points that are require to be oine by an ege to the new point to be taen outsie to reconstructg. Then the count of istinct isomorphic graphs that can be achieve through reconstruction is equal to the following prouct of the binomial coefficients: = V r Proof: Suppose an equivalent set V has to contribute r points to oin with the external point, then any r among the V points can be chosen (by efinition). Also such choice for each equivalent set can be mae inepenently. Hence the result.. The Reconstructibility of SQ s ( : Kelly lemma [3] shows the reconstructibility of S Q (, the number of copies of subgraph Q in G, when V ( > V ( Q). We now procee to show that the number of 0

11 copies of subgraph Q in G, SQ s (, is reconstructible, where V ( = V ( Q). Theorem.: The number SQ s ( where Q is a spanning subgraph of G, i.e. when V ( = V ( Q), is reconstructible. Proof: Consier a graph Gi G vi of G in the ec of G. Tae a new point v outsie. Since Dec ( is legitimate G is reconstructible by theorem 3. an so there shoul exist points (at least equal to egree of v i in number) among (p ) points of G i, such that any one point among which when we oin to a new point v taen outsie an form an extene graph, say G, then it will be a subgraph of G. (when no such a graph can be forme then egree of vi will be zero, G will be isconnecte, an so SQ s ( = 0 hence we are one.). Let the number of copies of spanning subgraphs Q in G be equal toα (Note that each such a copy of Q contains the newly ae ege.). Since by Kelly lemma the egree list of G is reconstructible, so let (v ) = ( v i ) = say. So there will be at least points x, x, L, x V ( G ) which when i oine, one at a time, to the point taen outsie, we will form the graphs G, G,, G which will be subgraphs of G. Let number of copies of spanning subgraphs Q in these graphs be α, α,, α. Since Dec ( is legitimate there shoul exist pairs of points (when ) among (p ) points of G i a pair among which when we oin to a new point v taen outsie an form an extene graph G then it will be a subgraph of G. Let the number of copies of spanning subgraphs Q in G equal toα (Note that here each such a copy of Q contains both the newly ae eges.). Actually there will be at least pairs ( ), mae up of points x, x,, ( ) x V G i L, such that when these pairs are oine, one at a time, to the point taen outsie we will form the graphs G, G,, G which will be subgraphs of

12 G. Let the number of copies of spanning subgraphs Q in these graphs be α, α,, α. Since Dec ( is legitimate there shoul exist triplets, at least in number, of points (when 3 ) among (p ) 3 points of G i a triplet among which when we oin to a new point v taen 3 outsie an form an extene graph G then it will be a subgraph of G. Continuing on similar lines let the number of copies of spanning subgraphs Q (containing all the three newly ae eges) that we form in this case be α, α,, α. 3 Continuing on similar lines there shoul exist at least one -tuple mae up of points x, x,, ( ) L x V G i such that when all these points are oine to the point taen outsie we will form the graphg, which will be isomorphic to G itself. Let the number of copies of spanning subgraphs Q (containing all the newly ae eges) that we form in this case beα. So, let there be the points, x, x,, ( ) L x V G i such that when all these points are oine to the external point we get the extene graph G which is isomorphic to graph G, an let { α, α,, 3 α,, 3 3 α }, { α, α,, α }, { α 3, α },, { α } be the counts of graphs isomorphic to Q erive from -subsets, -subsets, 3-subsets, -subset of { x, x, L, x } respectively, such that the eges ae by oining the points of these r-subsets, r =,,, are all present in the respective graphs isomorphic to Q obtaine from each of these r-subsets, then S s r Q ( equal to sum of all theseα s, s =,,,. r Thus, SQ s ( is reconstructible.

13 6. Complexity of the Reconstruction Algorithm: The complexity of the proceure escribe above of reconstructing the graph from its ec is epenent on the number of sets in the equivalent partitioning of the vertex set of a point elete subgraph in the ec that we tae for extension. Suppose the extension at a point of the subgraph in the ec chosen for extension leas to violation of subec property then the extension of that subgraph at any other point belonging to the same equivalent set in which the first point lies will also lea to violation of subec property. We can therefore iscar that entire equivalent set for the extension purpose. Definition 6.: Let G i be a graph in the ec of G. An let { V, V, L, V } be the partitioning of the vertex set of G i into equivalent sets, an { V, V, L, V } be their carinalities. A set V i is calle vali set if for at least at one element (point) of this set the extension can be carrie out maintaining the subec property. Definition 6.: A set V i is calle invali set when it is not vali. Theorem 6.: Given the partitioning of the vertex set of some graph in the ec of G, the orer of complexity of reconstructing a graph from its ec is equal to ~ O(T ( v i ) Φ ), where T is the time require for the checing of the subec property (escribe above) an ( v i ) is egree of some vertex vi of G. An Φ is the number of invali sets in the partitioning of the vertex set of G into equivalent sets, for extension. i Proof: Tae for extension purpose some subgraph Gi G vi of G such that ( v i ) is the egree v i. () Tae some equivalent set in the partitioning of the vertex set of G i into equivalent sets an carry out extension at some element (point) of it. Chec the subec property, in time T say. () If the extension is vali, try to exten at some other elements of the same set, one by one, till one gets invaliity of subec property or till the entire set is consume. (3) If the extension is invali then iscar that entire set from further consieration for extension an repeat step () starting with some other equivalent set. The proof is now clear. G i 3

14 Example: Consier the following ec for reconstructing the graph [3]. ( Copies) ( Copies) ( Copies) Proceeing as per the algorithm (iscusse in proofs of theorem 3.), we reconstruct the following graph from the given ec: 7. The Ege-Reconstruction conecture: The most natural variation of the Kelly-Ulam conecture is an analogue for eletion of eges. Let G be a simple (n, e) graph an let the collection of all graphs {G e i }, e i E(, the ege set of G is given. The graph G is sai to be egereconstructible if it is uniquely etermine by this ec. The egereconstruction conecture ue to Harary [4] states that: Every simple (n, e) graph, e > 3, is ege-reconstructible. Greenwell [] has prove the following theorem: Theorem 7.(Greenwell, 97): A reconstructible graph with no isolate vertices is ege-reconstructible. It can be seen that: Theorem 7.: A graph with, > 0, isolate vertices is egereconstructible if an only if the graph obtaine by eleting these isolate vertices is ege-reconstructible. Proof: Ege-Kelly lemma [3] permits the ege-reconstruction of the egree sequence an thus the isolate points. Hence every egereconstruction from a ec { G e i, i =,, L, q an e i E(, the ege set of G.} of a simple (p, q) graph G, has the same number of isolate points, which are all present in each graph in the ec. 4

15 Hence, in the light of the evelopment mae so far in this paper an the theorems 7., 7., we have the following Corollary 7. (Harary): Every graph is ege-reconstructible. 8. The Direct Approach for Ege Reconstruction: By proceeing exactly on similar lines an eveloping the similar ieas for line (ege) elete ecs one can achieve the inepenent fixing of the egereconstruction conecture as follows: Ege-reconstruction conecture (Harary [4]) can be state as follows: Let G an H be two simple (p, q) graphs with p points an q lines (eges), q > 3, an let there exists a (-), onto map, calle ege-hypomorphism: θ : E( E(H), e E( f E( H ) such that G G e H H f,, =,, L, q, then G H. (The symbol stans for isomorphism.) The existence of such map is calle ege-hypomorphism an thus: the conecture states that ege-hypomorphism implies isomorphism. Let e i E( then the subgraph Gi G ei is calle the ege elete subgraph (EDS) of G. The collection of all EDS, { Gi G ei, i =,, L, q }, is calle ec of (ege elete subgraphs of) G, Dec (. The graph G is sai to be ege-reconstructible if every graph with the same ec as G is isomorphic to G. Thus, the conecture can be state as: Every graph with more than three eges is uniquely (up to isomorphism) ege-reconstructible from its ec. In orer to reconstruct G from its ec, Dec (, it suffices to ientify which must be the vertex-pair of nonaacent vertices for the missing ege e i in some graph G i in the ec. We procee to accomplish this tas. Let G an H be two simple (p, q) graphs, i.e. having p points (vertices) an q lines (eges), an let G be isomorphic to H. Then there will exist a (-), onto, aacency preserving map θ : V( V(H), vi V ( ui V ( H ) such that if pair of vertices, v, v l V (, the vertex set for G, are aacent (nonaacent) then the corresponing pair of vertices, uner the

16 isomorphism map θ, u, u l V (H), the vertex set for H, will be aacent (nonaacent). Suppose now that the vertices, v, v l of G are nonaacent an suppose we mae them aacent an also we a an ege in the ege set of H, E(H), by oining the points u, u l V (H) such that e e u = θ ( v ), ul = θ ( vl ), an enote the extene G an H by G an H e e then clearly the supergraphs G an H will be isomorphic uner the same isomorphism map. Few efinitions are now in orer: Definition 8.: Let G an H be two isomorphic graphs. The process escribe above of extening graphs G an H to their respective e e supergraphs G an H is calle the isomorphic extension of isomorphic graphs. e Definition 8.: Let G be an unlabele (p, q) graph an let G be a supergraph of G obtaine by aing a new ege, not in the ege set E(, between some (unspecifie) nonaacent vertices of set V( is calle the e extension of G to G. Definition 8.3: The subset of nonaacent pairs of points (nonaacent vertex-pairs) of E( is calle a set of equivalent vertex-pairs if the extension of G by aing a new ege at any vertex-pair among these vertex-pairs, leas to graphs which are all isomorphic. Definition 8.4:The collection of subsets { V, V, L, Vm } of nonaacent vertex-pairs of the vertices in V(, the vertex set of G, is calle partitioning of nonaacent vertex-pairs of V( into equivalent sets if all subsets V i, i =,,, m are equivalent sets, m Vi I V = φ, i, an U Vi represents the collection of all i = the nonaacent vertex-pairs, an where φφ is a null set. Theorem 8.: Let G be a graph an let { V, V, L, V } be the partitioning of the of nonaacent vertex-pairs of the vertices in V( of G into equivalent sets, an let V, V, L, V be their carinalities. The number of nonisomorphic supergraphs that can be obtaine from a 6

17 graph G by aing a single ege is equal to, the total number of equivalent sets. Proof: Any two graphs, one obtaine by oining a nonaacent vertexpair belonging to some set, an the other obtaine by oining some other nonaacent vertex-pair belonging to the same set (though istinct ones) will be clearly isomorphic (refer to efinition.3). Any two graphs, one obtaine by oining nonaacent vertex-pair belonging some set, an the other obtaine by oining some other nonaacent vertex-pair belonging to a ifferent set will be clearly nonisomorphic. Definition 8.: Let G an H be two graphs on p points. We say that Dec ( is a subec of Dec (H ), or, Dec( Dec( H ) when there exist a biection ψ : E( E( H ) such that ei E( fi E( H ), where E(, E( H ) are the ege sets of G an H respectively, an Gi G ei is a subgraph of Hi H f i for all i =,,, p. Whenever Gi G ei is a proper subgraph of Hi H fi for at least one i {,,, p}, we say that Dec ( is a proper subec of Dec (H ), or, Dec( Dec( H ). Thus, Dec ( is not a subec of Dec(H ) if there exists no map ψ (see efinition.), such that the inclusion relation, namely, Dec( Dec( H ) hols goo, i.e. this relation is violate for at least one subgraph of G an the corresponing subgraph of H. Definition 8.6: Checing whether Dec( Dec( H ) hols (or oes not hol) for some graphs G an H is calle checing of subec property for graphs G an H. e Definition 8.7: An extension of G i to G is calle proper or vali when the extension satisfies the subec property, viz. Dec( G e ) Dec( when G i Dec (. Definition 8.8: The subisomorphism map for two p point graphs G an H is a biection ψ : V ( V ( H ) where v V ( u V ( H ) i i 7

18 such that if a pair of vertices ( v, v l ) V ( are aacent then the corresponing pair of vertices, uner the subisomorphism map θ, ( u, u l ) V (H), will be aacent an when ( v, v l ) V ( are nonaacent ( u, ) V (H) may or may not be aacent. u l Remar 8.: Uner existence of subisomorphism map G will be a spanning subgraph of H. Lemma 8.: Let G be a spanning subgraph of G, then Dec( G ) Dec(. Moreover, if G is a proper subgraph of G then Dec G ) Dec( ). ( G Proof: Since, G is a subgraph of G every aacent vertex pair of G will be aacent in G an Dec( G ) Dec(. When G will be a proper subgraph of G then the inclusion will be strict for at least one ege elete subgraph of G as a subgraph in the corresponing ege elete subgraph of G.Since, there will exist a nonaacent vertex pair in G which will be aacent in G so when we consier an ege elete subgraph of G an the corresponing ege elete subgraph of G arrive at by eleting an ege that is present in both G an G, Dec( G ) Dec( becomes clear. In short for every ege in G there is an ege in G uner the subisomorphism map. An for some nonaacent pairs of vertices in G some of the image pairs in G may be aacent. Lemma 8.: Let G an G be two graphs on p points. If Dec( G ) Dec( then G is subgraph of G. Moreover, if Dec G ) Dec( ) then G is a proper subgraph of G. ( G Proof: Suppose G is not a subgraph of G then there will not exist a subisomorphism map between their vertex sets. So, for every biection between their vertex sets there will exist a pair of vertices ( v, v l ) V ( which is aacent while the corresponing pair of vertices ( u, ul ) V ( (uner the biection) will be nonaacent. So, if we form Dec ( G ) an Dec ( G ) an chec uner the same biection whether the ege elete subgraphs of G are subgraphs of the corresponing ege elete subgraphs of G we will get violation for some subgraph obtaine by eletion of some ege other than the one corresponing to ege oining v,. v l 8

19 If we combine lemma. an lemma. we have Theorem 8.: Let G an G be two graphs on p points. Then G be a subgraph of G if an only if Dec( G ) Dec(. Moreover, G is a proper subgraph of G if an only if Dec G ) Dec( ). ( G 9. Ege-Reconstruction Conecture: We now procee to show that ege-hypomorphic graphs are also isomorphic, i.e. the egereconstruction conecture is true for all graphs. Theorem 9. (Ege-Reconstruction Conecture): Ege-Hypomorphic graphs are isomorphic. Proof : Let G an H be two simple ege-hypomorphic graphs on q lines, q > 3, i.e. let there exists a biection, θ : E( E(H), ei E( fi E( H ) such that Gi G ei Hi H fi, i, i =,, L, q. Letω i : V ( Gi ) V ( Hi ), be the corresponing isomorphism maps for i =,,, q. () Consier some subgraph Gi G ei of G an the corresponing isomorphic subgraph Hi H fi of H. () Exten isomorphically the graphs G i to G an Hi to H by aing an ege by oining some nonaacent vertex-pair ( x, x ) E( Gi ), the ege set of G i, leaing to the formation of the graph G an by aing an ege oining the corresponing nonaacent vertex-pair y, y ) E( H ), y = ω ( x ), y = ω ( x ), leaing to the ( i i i formation of the graph H. (Note that there are in all Φ number of possibilities for aing eges, one in G i an the corresponing one uner isomorphism map in H i, where Φ is the count of nonaacent vertex-pairs in both G i an H i.) (3) Chec whether Dec ( G ) Dec( hols. If yes, then G will be a proper extension of G i towars G (an automatically H will be 9

20 a proper extension of H i towars H). If not, then the extensions are improper an the ae eges shoul be remove. (3.) When the extension is proper, we are one, an we stop. It is clear that Dec ( G ) = Dec ( when the extension is proper, since G i is short of only one ege than G. (3.) When the extension is improper then remove the ae eges, both in G i an Hi an again go to step (). Again, exten isomorphically the graphs G i to G an Hi to H by aing an ege (other than the one previously ae an foun improper among the (Φ -) eges), leaing to the formation of the graph G an the corresponing graph H. Go to step (3) an chec subec property for graphs G an G. (4) Continue extening isomorphically, an each time taing the proper action implie by the correctness (or incorrectness) of aing the eges as per the above given steps, till we reach the stop action in step (3.). At this stage (completing step (3.)) we will not only have Dec( G ) Dec( an Dec( H ) Dec( H ) but we will achieve more. We will actually have Dec ( G ) = Dec( an Dec ( H ) = Dec( H ). Thus, we have achieve the isomorphic extension of G i to G an Hi to H, i.e. G H. () Note that if the isomorphic extension of G i fails at each aition, i.e. if we cannot a an ege ing i then clearly Dec ( will be illegitimate. Proof : Given unlabelle legitimate ec of ege elete subgraphs of G we procee to show that we can ege-reconstruct a unique graph up to isomorphism (i.e. the ege-reconstructe graph will be isomorphic to ()Tae a subgraph in the ec. ()A some an ege by aing an ege between some nonaacent vertex-pair of this subgraph. (3)Chec subec property for this new graph an G. (4) If it hols then stop. Else continue the steps from step () for all possible reconstructions, in all equal to Φ, the total number of nonaacent vertex-pairs in the subgraph uner consieration. () At least one choice must exist for which the subec property will hol goo an for the graph thus prouce, say G *, we will have Dec ( G*) = Dec(. Otherwise, the ec of G will not be 0

21 legitimate, a contraiction. If there exists only one choice, then a unique graph is thus constructe an we are through. When there is more than one choice, we have to see that all these graphs are isomorphic. But this is clear since these multiple graphs are arrive at having the same ec ue to presence of more than one equivalent vertex-pairs in equivalent set to which the vertex-pair, causing vali extension, belongs. 0. A Simple Count: The count of isomorphic graphs that one will construct from a given legitimate ec of ege elete subgraphs following the proceure iscusse above in the proof can be easily obtaine as follows: Theorem 0.: Let G i be a graph in the Dec (. The count of istinct isomorphic graphs that can be achieve through ege-reconstruction is equal to the carinality of the equivalent sets of nonaacent vertex pairs in the partitioning of nonaacent vertex-pairs of the vertices in V( G i ) into equivalent sets for extension. Proof: Every equivalent set give rise to only one graph up to isomorphism by aing an ege.. Complexity of the Ege-Reconstruction Algorithm: The complexity of the proceure escribe above of ege-reconstructing the graph from its ec is epenent on the number of sets in the equivalent partitioning of the of nonaacent vertex-pairs of an ege elete subgraph in the ec that we tae for extension. Suppose the extension by oining a nonaacent vertex-pair of the subgraph in the ec chosen for extension leas to violation of subec property then the extension at any other nonaacent vertex-pair of that subgraph belonging to the same equivalent set in which the first nonaacent vertex-pair lies will also lea to violation of subec property. We can therefore iscar that entire equivalent set for the extension purpose. Definition.: Let G i be a graph in the ec ofg. An let { V, V, L, V } be the partitioning of the nonaacent vertex-pairs of G i into equivalent sets, an { V, V, L, V } be their carinalities. A set V i is calle vali set if the extension, by maing a nonaacent vertexpair aacent, maintains the subec property.

22 Definition.: A set V i is calle invali set when it is not vali. Theorem.: The orer of complexity of ege-reconstructing a graph from its ec is equal to ~ O(T Φ ), where T is the time require for the checing of the subec property an Φ is the number of sets in the partitioning of the vertex set of G into equivalent sets, for extension. i Proof: Tae for extension purpose some subgraph Gi G ei of G. () Tae some equivalent set an carry out extension at some element (a nonaacent vertex-pair) of it. Chec the subec property, in time T say. () If the extension is vali, stop. (3) If the extension is invali then iscar that entire set from further consieration for extension an repeat step () starting with some other equivalent set. The proof is now clear.. Complete Set of Invariants an Reconstruction: One more approach to reconstruction problem is to evise a complete set of invariants that etermines a graph uniquely up to isomorphism an use that complete set for reconstruction. We now procee to iscuss it in this section. An invariant of a graph G is a number associate with G which has same value for every graph isomorphic to G. For example, the number of points an the number of lines of a graph are clearly invariant numbers. A complete set of invariants is that set of numbers which completely etermine a graph up to isomorphism. No ecent complete set of invariants for a graph is nown [6]. We now propose to associate a vector having certain numbers as components with a graph an show that these components of the vector together form a complete set of invariants for that graph. Using this complete set one can see that Kelly-Ulam conecture for graphs follows. 3. Preliminaries: Let G be a (p, q) labele connecte graph, i.e. a graph containing p points an q lines. We enote by C r ( the number of labele (r, r ) trees in G, r p, an let us enote by C ( the vector of tree counts in G, thus:

23 C ( = ( C ( ), C ( ),, C r (,, C p ( ). G G We thus have C ( G ) = p an C ( = q, the invariant numbers mentione above. Theorem 3.: Let G an H be two labele connecte graphs. G H if an only if C ( = C (H ). Proof: When G H there exists an aacency preserving biection between the vertex sets of G an H, so essentially we can eep G on H (or H on such that one graph completely hies the other graph an so C ( = C (H ) becomes clear. Conversely, let C ( = C (H ). We procee to show that for every (, ) tree in G there exists a (, ) tree in H isomorphic to it for all p which leas to the outcome that for every spanning tree in G there exists a spanning tree in H isomorphic to it ( = p case). Using the fact that a graph is reconstructible from all its spanning trees we settle the claim. Since there is only one (, 0), (, ) an (3, ) tree up to isomorphism therefore for every (, 0), (, ) an (3, ) tree in G there is a (, 0), (, ) an (3, ) tree in H isomorphic to it. Note that there are 4 4 two (4, 3) trees T an T up to isomorphism as shown below an let 4 they be in quantity αi an β i 4, i =, in G an H respectively: : 4 T 4 T There are three types of (, 4) treest, T, isomorphism as shown below an let they be in quantity to 3 in G an H respectively: T 3 up to α i an β i, i = T T T 3 3

24 Let T an T be two (nonisomorphic) trees on points. Let ( T ) an ( T ) be their egree sequences written in nonincreasing orer, namely, ( T ) = ( T ) ( T ) 3 ( T ) ( T ) ( T ) = ( T ) ( T ) 3 ( T ) ( T ) Definition 3.: ( T ) = ( T ) if ( T ) = ( T ) for all = to an ( T ) > ( T ) if ( T ) = ( T ) for = to s, s <, then ( T ) > ( T ) for = (s+). Definition 3.: T > T if ( T ) > ( T ) or when ( T ) = ( T ) ( ) then the count of trees of type T u is more in T than in T while the ( ) ( ) count of trees of type T is less in T than T when T > ( ) v T. v Thus, the above efinition is recursive an starts with T > T. Now, from equality C ( = C (H ) we have α + α = β + β (.) so, we have ( α β ) + ( α β ) = 0 (.) Now, clearly T > T. If α > β then for equation (.) to hol we must have α < β. So, let ( α β ) = u then ( α β ) = u. 4 4 Since a T tree contains three (3, ) trees while a T tree contains two (3, ) trees, therefore G will contain u (3u u = u) number of (3, ) trees more than H, a contraiction. Thus, for every (4, 3) tree in G there is a (4, 3) tree in H isomorphic to it. Similarly, we clearly have T > T > T 3. If α > 4 u 4 β then either α < β or α 3 < β 3 or both. It is easy to chec from this that G will 4 4 contain more trees of type T an less trees T than H, a contraiction. We have thus shown that the equality of C ( an C (H ), = to implies that for every (, ) tree in G there is a (, ) tree in H isomorphic to it. 4

25 Now we procee by inuction an assume that the equality of C ( an C (H ), = to, implies that for every (, ) tree in G there is a (, ) tree in H isomorphic to it an procee to show the same for = ( + ). Let T ( +) ( +) to isomorphism an let α an, = to l, be the types of ( +, ) trees up ( +) β be their count in the labele graphs G an H respectively. Also, let T, = to m, be the types of (, ) trees up to isomorphism an suppose by inuction that their count in the labele graphs G an H is equal i.e. α = β for all = to m. Also, we assume without loss of generality that ( + ) ( + ) ( + ) ( +) T > T > T > > T an T > T > 3 T 3 > > an our aim is to show that T m ( +) α = l ( +) β for all = to l. For this ( +) we procee by inuction on the number of places r where α an ( +) β iffer in their count. Now, if they iffer in only one place i.e. ( +) ( +) if r = then without loss of generality let α = β for all except = s. But this is impossible because this will violate the equality of ( +) ( +) C( + ) ( an C( + ) ( H ). So, let r =, i.e. let α s > β s an ( +) ( +) α t < β t, for some s < t. But this will imply that there will exist trees T u an T v, T u > T v, u < v, such that the count of trees T u as ( +) subtrees in T will be more in G than H, while the count of trees s ( +) as subtrees in T t will be less in G than H. This in turn implies that α u > β u while α v < β v, a contraiction. Assuming that any r = w terms ifferent is impossible we can see that an aitional iffering term, to mae r = w +, at upper or lower en can t avoi existence of two or more terms lie α u > β u an α v < β v which is a contraiction. ( +) ( +) Hence, α = β for all = to l. Thus, C ( = C (H ) implies that for every labele spanning tree of G there will exist a labele spanning tree in H T v

26 isomorphic to it. Now, it is a well-nown fact that a graph is uniquely reconstructible from all its spanning trees, therefore G H. We thus have the following Theorem 3.: The components of C ( forms a complete set of invariants for G. 4. Kelly-Ulam Conecture for (Connecte) Graphs: Let G an H be two simple connecte (since the case for isconnecte graphs is alreay settle [3]) graphs with p points, p>, an let there exist a (-), onto map, θ : V( V(H), vi V ( ui V ( H ) such thatgi G vi Hi H ui, i, i =,, L, p, then G H. (The symbol stans for isomorphism.) The existence of such a map is calle hypomorphism an thus: the conecture states that hypomorphism implies isomorphism. Let v i V (. Then the inuce subgraph Gi G vi is calle the vertex elete subgraph (VDS) of G. The collection of all VDS, { Gi G vi, i =,, L, p }, is calle the ec of (vertex elete subgraphs of) G, Dec (. The graph G is reconstructible if every graph with the same ec as G is isomorphic to G. Thus, the conecture can be state as: Every graph with more than two vertices is uniquely (up to isomorphism) reconstructible from its ec. We procee to show that in the light of theorems 3., 3. it suffices to have the nowlege of the egree of the missing vertex vi for establishing the reconstructibility of G which we have since the egree sequence of a graph is reconstructible [3]. Theorem.(Kelly-Ulam Conecture): Hypomorphic graphs are isomorphic. Proof: Let G an H be two simple connecte hypomorphic graphs on p points, p>, i.e. let there exist a biection, θ : V( V(H), vi V ( ui V ( H ), where G G v H H u, i, i =,, L p. i i i i, 6

27 Let ω i : V ( Gi ) V ( Hi ), be the corresponing isomorphism maps for i =,,, p. Consier some subgraph Gi G vi of G an the corresponing isomorphic subgraph Hi H ui of H. If we will exten isomorphically the graphs G i to G an Hi to H by aing an ege by oining some point x V( G ) i, the vertex set of G i, to a new point v not belonging to V ( Gi ), leaing to the formation of the graph G an by aing an ege oining the corresponing point y =ω i( x ) V( Hi ), the vertex set of H i, to a new point u, not belonging to V ( Hi ), leaing to the formation of the graph H, then we have essentially forme two isomorphic spanning graphs. Since Gi Hi we have for every subgraph of G i there is the corresponing isomorphic subgraph of H i. Let ( v i ) = ( u i ) =, the egrees of vertices v i an u i respectively. Now, we now that there are some neighbors to v i an corresponing image points uner isomorphism as neighbors to u i in G i an H i respectively (which if we can specify then we achieve the reconstruction of G an H). But whatever may be these points, once we fix their choice, among ( p ) possibilities, then it is easy to see that: () Every pair of spanning trees, one in G i an the other corresponing isomorphic one in H i, will give rise to exactly trees by isomorphic extension of these isomorphic trees counte in respectively ( G ) C( p ) i an C ( H ). Therefore, we have C p ( ) = ( C ( G )) an H ( p ) i G G H ( p ) i C p ( ) = ( C( p ) ( Hi ) ), where C p ( ) an C p ( ) enotes the count of the spanning trees in G an H respectively such that the egree of v i an u i equal to one. It is important to note at this uncture that the spanning trees thus obtaine may not be the spanning trees of G an H, but are efinitely equal in count. () Let us enote by C p ( an C p ( H ) the count of the spanning trees that exist in respectively G an H such that the egrees of vertices v i an u i equal to two. We construct trees that are equal in count to trees in G an H as follows: we carry out all possible isoint partitioning of the sets of (p-) points of G an H into two sets V ( ) an V ( ) such i i G i G i 7

28 that V ( Gi ) = V ( G i ) U V ( G i ) an V ( Hi ) = V ( H i ) U V ( H i ) such that if x V ( G ) then y = ω i ( x ) V ( H ) an if i i x V ( G ) i then y = ω ( ) ( ) i x V H i. Let an be the points among the points chosen in G i an the images of these points uner isomorphism that one gets in H i, such that the points belong to V ( ) an their images to V ( ) while points belong to G i V ( G i ) an their images to V ( H i ) an an a up to. It is easy to see that every subtree that spans the vertex set V ( G i ) an its isomorphic image that spans the vertex set V ( H i ) can be extene in ways by oining its point among the points to vertex vi an by oining the image point among the image points to vertex u i.similarly, every subtree that spans the vertex set V ( G i ) an its isomorphic image that spans the vertex set V ( H i ) can be extene in ways by oining its point among the points to vertex vi an by oining the image point among the image points to vertex u i. Also, note that the choice of a point for extension among the points an can be mae inepenently, so there will be in all spanning trees that can be obtaine from each pair of isomorphic trees, one that spans the vertex set V ( G i ) an its isomorphic image that spans the vertex set V ( H i ) while the other that spans the vertex set V ( G i ) an its isomorphic image that spans the vertex set V ( H i ), by oining its point among the points to vertex vi an by oining the corresponing image point among the image points to vertex u i an by oining its point among the points to vertex vi an by oining the image point among the image points to vertex u i. Thus, by consiering: (a) All the possible isoint partitioning of vertex set V ( Gi ) an the corresponing automatic partitioning that follows by consiering the images uner isomorphism map in V ( Hi ) (b) All the possible values of an offere by these partitioning of sets (c) All the possible subtrees that span the partitione sets () All the possible extensions offere by each subtree an its isomorphic image H i 8

29 (e) All the possible spanning trees with egree of v i an u i equal to two an collecting them together we can have the count C p ( = C p ( H ). Continuing with isoint partitioning of V ( Gi ) an the corresponing automatic partitioning that follows from isomorphism of V ( Hi ) into three subsets, four subsets,..., subsets, etc. an forming all possible spanning trees with three, four,,, extension one from each subtree that spans the partitione subset an from the point that is among the chosen points etc. we can have the count of C 3 p (, 4 G 3 H 4 H C p ( ),, C p ( equal to C p ( ), C p ( ),, C p (H ) respectively. Aing the respective counts, we thus have, C p ( = C p (H ). Note that as far as the count is concerne the same outcome will result for any choice of points. Now, by Kelly lemma respectively S Q (, the count of the copies of subgraph Q in G is reconstructible when we are given a legitimate ec of graph G, Dec (, an the carinality of vertex sets V (Q) is less than the carinality of vertex set V ( [3]. Thus, from hypomorphism between G an H we are assure of the existence of an isomorphic subtree in H corresponing to every subtree of G when the carinality of vertex set of that tree is less than the carinality of vertex set V (. Thus, we have C ( = C (H ), = to (p ). Hence C ( = C (H ). Thus, we have the unique components C (, = to (p ) by reconstructibility of S Q ( an these unique components leas to a unique C p ( so using theorem. at this stage we have a unique G (up to isomorphism) from the given ec. 6. Another Complete Set of Invariants: We now see that if we efine a vector as follows by moifying the above consiere invariant vector by replacing the components representing the count of the labele trees of a particular size by count of the labele trees with multiplicity i.e. a particular labele (r, r ) tree is counte times if it occurs as a subtree of number of istinct (r+, r) labele trees that exist in G, then the components of this new vector also forms a complete set of invariants for G an by proceeing on similar lines we get the same evelopment as follows: Let G be a (p, q) labele connecte graph, i.e. a graph containing p points an q lines. Let us enote by C *( the vector of tree counts in G, thus: 9