Graph Reconstruction by Permutations

Transcription

1 Graph Reconstructon by Permutatons Perre Ille and Wllam Kocay* Insttut de Mathémathques de Lumny CNRS UMR avenue de Lumny, Case Marselle Cedex 9, France e-mal: Computer Scence Department St. Paul s College, Unversty of Mantoba Wnnnpeg, MB, Canada R3T 2N2 e-mal: bocay@cc.umantoba.ca Ars Combnatora 86 (2008), Abstract Let G and H be graphs wth a common vertex set V, such that G = H, for all V. Let p be the permutaton of V that maps G to H, and let q denote the permutaton obtaned from p by mappng to. It s shown that certan algebrac relatons nvolvng the edges of G and the permutatons q q 1 and q q 1, where,, V are dstnct vertces, often force G and H to be somorphc. 1. Permutatons and Partal Permutatons Let G and H be graphs wth a common vertex set V, such that G = H, for all V. Ulam s conecture, also nown as the graph reconstructon conecture (see Bondy [1]), states that ths condton mples that G = H, f V 3. All graphs that we wll wor wth are smple graphs, wth the same vertex set V. We wll vew a graph G( as a set of edges, where each edge s an unordered par of dstnct vertces. We wrte V ) ( 2 for the set of all unordered pars of vertces. Thus, G V ) 2. The complement of G s G = ( V 2) G. An unordered par of vertces s denoted by [x, y]. (The square bracets are convenent when t s necessary to construct sets of unordered pars.) It wll often be convenent to wrte [x, y] G = 1 to mean that [x, y] G, and [x, y] G = 0 to mean that [x, y] G. Gven a permutaton q of V, and a vertex x V, the mage of x under q s denoted x q. Then [x, y] q = [x q, y q ]. We also wrte G q for the graph obtaned from G by permutng each edge of G by q. Let p be an somorphsm of G wth H, so that (G ) p = H. Then p s a partal permutaton of the set V. It does not map the pont. * Ths wor was partally supported by an operatng grant from the Natural Scences and Engneerng Research Councl of Canada. It was partally done whle W. Kocay was vstng the Unversty Ax-Marselle II as an nvted professor. 1

2 1.1 Defnton. A partal permutaton p of V s an nectve mappng from a proper subset X V to a proper subset Y V. A partal automorphsm of a graph G s a partal permutaton p such that for all [x, y] ( V 2) : 1) f [x, y] G and [x, y] p s defned, then [x, y] p G; and 2) f [x, y] G and [x, y] p s defned, then [x, y] p G; The nverse of a partal permutaton s also a partal permutaton. Notce that partal permutatons can be composed. If p and p are partal permutatons, so s p p 1. The result s another partal permutaton. Permutatons and partal permutatons are composed from left to rght, so that the product p p 1 means frst p, then p 1. Partal permutatons were appled to the graph reconstructon problem n Kocay [3]. We say that the mage [x, y] p p 1 exsts f p p 1 can be appled to both x and y. We are gven the graphs G and H such that (G ) p = H, for all. Notce that for all, V, p p 1 s a partal automorphsm of G, and p 1 p s a partal automorphsm of H, because p and p are somorphsms. Notce that f [x, y] p p 1 only f [x, y] p p 1 G, snce p p 1 s a partal automorphsm of G. exsts, then [x, y] G f and 1.2 Lemma. Let [x, y] be a par of vertces of V. Then the mage [x, y] p p 1 only f x, y {, p 1 }. The mage [x, y] p 1 p exsts f and only f x, y {, p }. exsts f and The partal permutaton p can be converted to a permutaton q, by extendng p to act on, defnng q =. Notce that q q 1 s n general not an automorphsm of G. 1.3 Lemma. There are two possbltes for the cycle structure of q q 1. Ether and are n the same cycle, or they are n dfferent cycles. The two cases for the cycle structure of q q 1 are llustrated n Fgure 1. We say that q q 1 s of type 1 f and are n the same cycle, and of type 2 f they are n dfferent cycles. The arrow n the dagrams ndcates the drecton of the cycle, and also ndcates those lns n the cycle whch p p 1 cannot follow. Fgure 1, The cycle structure of q q 1, types 1 and 2 2

3 We assume the basc theory of graph reconstructon. The reader s referred to the survey papers by Bondy and Hemmnger [1], Bondy [2], and Nash-Wllams [4]. In partcular, gven a fxed graph K wth fewer than V vertces, Kelly s lemma says that the number of subgraphs of G somorphc to K equals the number of subgraphs of H somorphc to K. Consequently G and H have the same number of edges. The degree of a vertex n a graph G s denoted deg(, G). By Kelly s lemma, deg(, G) = deg(, H). We wll often be concerned wth the dfference between two nearly dentcal graphs G and G. The exclusve or (also called symmetrc dfference) of graphs G and G s denoted G G = (G G ) (G G). If G and G have the same number of edges, then G G = G G, so that G G s even. 1.4 Lemma. G q H conssts of pars of the form [, x]. Proof. Ths follows because (G ) p = H. 1.5 Lemma. If deg(, G) = deg( q, G), for some and, where, then [, ] G f and only f [ q, ] H. Proof. By Kelly s lemma, deg(, G) = deg(, H) and deg( q, G) = deg( q, H). Snce deg(, G) = deg( q, G) = deg( q, H) and deg(, G ) = deg( q, H ), the concluson follows. 1.6 Corollary. If q = for some and, where, then [, ] G f and only f [, ] H. Proof. If q =, then deg(, G) = deg( q, G). The result follows from the prevous lemma. 1.7 Corollary. Gven any q, G q = H f and only f q preserves degree. Proof. If G q = H, then deg(, G) = deg( q, H) = deg( q, G), so that q preserves degree. Conversely, suppose that q preserves degree. We have deg(, G) = deg( q, G), for all. It follows that [, ] G f and only f [ q, ] H; that s, G q The automorphsm group of a graph G s denoted aut(g). 1.8 Lemma. If q q 1 aut(g) for some, then G q = G q Proof. If q q 1 aut(g), then G q q 1 = G, so that G q = G q. We show that G q Let [x, y] G. If {x, y}, then [x, y] q H. Otherwse, let [, x] G. Then snce q q 1 aut(g), we have [, x] q q 1 = [ q 1, x q q 1 ] G. If x q q 1, apply q to get [ q 1, x q q 1 ] q H, or [, x] q H. Otherwse x = q 1. We thus have, that [x, y] G f and only f [x, y] q H, for all pars [x, y], except possbly for the par [x, y] = [, q 1 ]. But G and H have the same number of edges. Therefore we also have [, q 1 ] G f and only f [, ] H, so that G q 1.9 Corollary. If q = q, for some, then G q Proof. If q = q, then q q 1 aut(g). 3

4 Lemma 1.8 s one of the smplest possble cases of an algebrac relaton on the permutatons q mplyng the reconstructblty of G; namely G q q 1 = G mples that G s reconstructble. In ths paper, we show that reconstructblty can also be mpled by a weaer algebrac relaton. We consder the case when q q 1 s not an automorphsm of G, but where there exsts an algebrac relaton of the form G q q 1 = G q q 1 l, for sutable vertces,,, and l. Notce that f G q q 1 obtan G q q 1 = G q q 1, we can apply q to each sde of the equaton and = G, from whch Lemma 1.8 gves G Thus, the cases to consder are: 1. G q q 1 2. G q q 1 3. G q q 1 4. G q q 1 = G q q 1, where ; = G q q 1, where,, and are dstnct; = G q q 1, where,, and are dstnct; = G q q 1 l, where,,, and l are dstnct. Notce that we do not need to consder the case G q q 1 = G q q 1, because relabellng the subscpts,, as,, reduces t to case 3. In ths paper we shall consder the frst two of these possbltes. Gven any par [x, y], we can apply q q 1 to t repeatedly to obtan a cyclc sequence of pars, whch we wll denote by Ω(x, y), the par-orbt of [x, y]. It has the property = Ω(x, y). If we conugate q q 1 by q or q, we obtan the permutaton q 1 q. Therefore, f Ω s a par-orbt of q q 1, then Ω q s a par-orbt of q 1 q, and furthermore, Ω q = Ω q. If G were composed of a unon of par orbts, Ω(x, y), then snce that Ω(x, y) q q 1 Ω(x, y) q = Ω(x, y) q, we would have G q = G q, so that G = H, by Lemma 1.8. Therefore there must be at least one par-orbt Ω(x, y) that s only partally contaned n G Lemma. Let Ω = Ω(x, y) be a par-orbt of q q 1, and let C be a cycle of q q 1. Then deg(u, Ω) = deg(v, Ω) for all u, v C. Proof. There are three possbltes: x, y C; x C and y C; or x, y C. The result follows easly n each case. 2. G q q 1 = G q q 1 We assume throughout ths secton that G q q 1 (q q 1 ) 2 aut(g). 2.1 Lemma. q q 1 G q q 1 ). Proof. Snce G q q 1 G G q q 1 aut(g G q q 1 ) and q q 1 = G q q 1 and G G q q 1., we have (G G q q 1 ) q q 1 4 = G q q 1, for some. We have aut(g G q q 1 ) and q q 1 aut(g = G G q q 1. Smlarly for

5 It follows from Lemma 2.1 that f [x, y] G G q q 1 smlarly, f [x, y] G G q q 1 Ω(x, y) G G q q 1. then Ω(x, y) G G q q 1 ;, then Ω(x, y) G G q q 1 ; and f [x, y] G G q q Lemma. Let [x, y] G G q q 1. Then [x, y] p p 1 Consequently, {x, y} {, q 1 } Ø and {x, y} {, q 1 } Ø. Proof. If [x, y] G G q q 1, then [x, y] q q 1 (G G q q 1 ) q q 1 [x, y] p p 1 does not exst, snce [x, y] q q 1 1.2, we have {x, y} {, q 1 and [x, y] p p 1 G. Smlarly f [x, y] G q q 1 } Ø. Smlarly for [x, y] p p 1. then do not exst. = G q q 1 G. Hence G. By Lemma 2.3 Lemma. Let [x, y] G G q q 1, and let Ω = Ω(x, y). Then Ω (G G q q 1 ) = Ω (G q q 1 G), and Ω s even. Proof. Ths follows snce (G G q q 1 ) q q 1 Case (a). q q 1 s of type 1. = G q q 1 G. Consder a par [x, y] G G q q 1. Let C denote the cycle of q q 1 contanng and. Then q 1 and q 1 are also n C. By Lemma 2.2, we have x, y C. If C had odd length, then Ω(x, y) would be odd, a contradcton. We conclude that C has even length. Now C = 2, snce ths mples that q q 1 =, so that =, a contradcton. Therefore C has at least 4 vertces, so that t has at least 2 vertces other than {, }. There are two possbltes to consder, ether q 1 q 1 = q Theorem. If G q q 1 = G q q 1, and q q 1 and q 1 are dstnct vertces, or else s of type 1, and q 1 q 1 then G Proof. We have [x, y] G G q q 1. By Lemma 2.2, every par of Ω(x, y) must ntersect both {, q 1 C as (, q 1 } and {, q 1 }. It follows that C = 4. Refer to Fgure 2. Wrte the cycle, q 1, ). Clearly Ω(x, y) = Ω(, ). Wthout loss of generalty, we can tae Ω (G G q q 1 ) = {[, ], [ q 1, q 1 ]} and Ω (G q q 1 G) = {[, q 1 ], [, q 1 ]}. Hence, G G q q 1 = G q q 1 G = 2. We see that, q 1, q 1 and all have degree one n G G q q 1. By Lemma 1.10, they all have the same degree n G, call t α. Fgure 2, G and H when C = 4 and q 1 q 1 5

6 We now fnd (G ) p and (G ) p n order to fnd H. We now that C q = (, q 1 q,, q ) s a cycle of q 1 q, as shown n Fgure 2. If u C, then deg(u, G) = deg(u q, H), by Lemma 1.10, snce Ω(, ) s the only par-orbt that s only partally contaned n G. We fnd that [ q 1, q 1 ] q = [, q 1 q ] H and [ q 1, q 1 ] q = [, q 1 q ] H, snce q 1 q = q 1 q. Therefore deg( q 1 q, H) = α + 1. We also fnd that [, q 1 ] q = [, q ] H and [, q 1 ] q = [, q ] H, snce q = q. It follows that deg( q, H) = α 1. But G and H must have the same degree sequences, a contradcton. We conclude that G G q q 1 = G q q 1 G = Ø, so that G = G q q 1, from whch t follows that G = H, by Lemma 1.8. The second possblty s when q 1 = q 1. Ths s llustrated n Fgure 3., and q q 1 s of type 1, and q 1 = q 1 then ether G q = H or else G q H = {[, ], [, q ]}. Furthermore, [, ] G f and only f [, ] H. If [, ] G then (G [, ]) q = H [, ]. If [, ] G then (G + [, ]) q = H + [, ]. Proof. We have [x, y] G G q q 1, where x, y C. By Lemma 2.2, every par of 2.5 Theorem. If G q q 1 = G q q 1 Ω(x, y) must ntersect both {, q 1 } and {, q 1 }. It follows that C = 4, that s C = (, q 1,, q q 1 G G q q 1 ). Notce that [, q q 1 ] p p 1. Therefore ether [x, y] = [, ] or else [x, y] = [ q 1 exsts, so that by Lemma 2.2, Ω([, q q 1 ]), q q 1 ], so that we agan have Ω(x, y) = Ω(, ). Wthout loss of generalty, we can tae Ω (G G q q 1 ) = {[, ]} and Ω (G q q 1 G) = {[ q q 1, q 1 ]}. Hence, G G q q 1 = G q q 1 G = 1. By Lemma 1.10, the vertces of C all have the same degree n G [, ], call t α. Then deg(, G) = deg(, G) = α + 1 and deg( q 1, G) = deg( q q 1, G) = α. Fgure 3, G and H when C = 4 and q 1 = q 1 We now fnd (G ) p and (G ) p n order to fnd H. Snce [ q 1, q q 1 ] G, we fnd that [ q 1, q q 1 ] q = [, q ] H and [ q 1, q q 1 ] q = [, q ] H. Therefore (Ω(, )) q H = Ø. We now that ether Ω(, q 1 ) G or Ω(, q 1 ) G. Suppose that Ω(, q 1 ) G. Then (Ω(, q 1 ) q H 4. But Ω(, q 1 ) and Ω(, ) together contan 5 edges of G. Ths s mpossble, as G and H have the same number of edges. We conclude that Ω(, q 1 ) G. We then fnd that [, q 1 ] q = [, q ] H; [, q q 1 ] q = [ q, q ] H; 6

7 and [, q 1 ] q = [, q ] H. Of the 4 edges of Ω(, q 1 ) q, at most one of them, namely [, ], can be an edge of H. Snce G and H have the same number of edges, we conclude that [, ] H. It then follows that (G [, ]) q = H [, ]. Notce that deg(, H) = deg(, H) = α + 1 and that deg( q, H) = deg( q, H) = α. If we do not have [, ] G G q q 1, then the alternatve s [ q q 1, q 1 ] G G q q 1. The analyss s very smlar. We fnd that (Ω(, )) q H and that [, q ], [ q, q ], [, q ] H but that [, ] H. It follows that (G + [, ]) q = H + [, ]. Case (b). q q 1 s of type 2. Let C denote the cycle of q q 1 cycle structure of q q 1 C 2 and C 2. By Lemma 2.2, G G q q 1 If [, q 1 ] G G q q 1 contanng, and C the cycle contanng. Refer to the llustrated n Fgure 1, for a type 2 permutaton. Suppose frst that, then by Lemma 2.1, Ω(, q 1 {[, ], [, q 1 ], [, q 1 ], [ q 1, q 1 ]}. ) G G q q 1. Snce, q 1 C, we then must have Ω(, q 1 ) = {[, q 1 ]}, whch contradcts Lemma 2.3. Smlarly f [, q 1 ] G G q q 1. It follows that G G q q 1 mples that q q 1 = q 1 q 1 q = q and C = C = 2. so that C = (, q 1 = {[, ], [ q 1, q 1 ]}. Lemma 2.1 then ). Smlarly we have C = (, q 1 ), so that 2.6 Theorem. Suppose that G q q 1 = G q q 1, that q q 1 s of type 2, and that C = C = 2. Then ether G q = H, or else G q H = {[, ], [, q ]}. Proof. We have C = (, q 1 Lemma 1.8. Otherwse G G q q 1 ) and C = (, q 1 ). If G G q q 1 = {[, ], [ q 1, q 1 ]}. Refer to Fgure 4. = Ø, then G q = H, by Fgure 4, G and H when C = C = 2 We now calculate (G ) p and (G ) p n order to fnd H. We have [ q 1 G G q q 1, q 1 ]. Therefore [ q 1, q 1 ] p = [, q ] G q H. If [, u] G q H, where u, then [, u] p 1 = [ q 1, u q 1 ] G G q q 1. It follows that u q 1 = q 1, so that u = q. Snce G q H must be even, we conclude that G q H = {[, ], [, q ]}. It follows from Theorem 2.6, that ether [, ] G and [ q 1, q 1 ] G, n whch case [, ] H and [, q ] H, or else vce versa. Notce the followng: 7

8 [, ] G f and only f [, ] H [, q 1 [, q 1 [ q 1 [, q 1 ] G f and only f [, q 1 ] p = [, q ] H ] G f and only f [, q 1 ] p = [, q ] H, q 1 ] G f and only f [, q ] H and [, q ] H ] G and [, q 1 ] G f and only f [ q, q ] H We can construct graphs G and H satsfyng Theorem 2.6, as shown n Fgure 5. Snce G and H have the same number of edges, there are only two possbltes for the four pars connectng C and C. Ether only [, ] G and [, ] H, or else [, q 1 ], [, q 1 ], [ q 1, q 1 ] G, and [, q ], [, q ], [ q, q ] H. In Fgure 5, the cy- are represented by ellpses of vertces that are alternately whte and blac. cles of q q 1 There may be other edges between the cycles, and there may be addtonal cycles, as well. It s evdent from the constructon that (G ) q = H and that (G ) q = H, and that C = C = 2. In ths example, G and H are somorphc. However, ths s not necessarly so, as we can add addtonal edges between the two cycles, as well as addtonal cycles. The resultng graphs would then not be hypomorphc, although we do not now how to prove ths n general. Fgure 5, Possble graphs G and H when C = C = 2 Consder now the case when C = 1 and C 2. Snce q =, by Corollary 1.6, we have [, ] G f and only f [, ] H. Now [, ] G f and only f [, ] q = [, ] G q. Therefore [, ] G q H. Suppose that [, u] G q H. Snce u, we have [, u] p 1 G G q q 1. By Lemma 2.2, t follows that { q 1, u q 1 } {, q 1 } = Ø and { q 1, u q 1 } {, } Ø. The second condton mples that u = q. Hence G q H {[, q ]}. But snce G q H s even, t follows that G q We summarse the prevous results n the followng theorem. 2.7 Theorem. Suppose that G q q 1 = G q q 1, and that q q 1 s of type 2. If C C or f C = C 3 then G If C = C = 2, then ether G q = H or else G q H = {[, ], [, q ]}. 8

9 There remans the stuaton when C = C = 1. We then have q = and q =. Lemma 2.2 now only says that [x, y] G G q q 1 satsfes {x, y} {, } = Ø. Snce q q 1 fxes both and, t acts as an automorphsm of G {, }, and so (G {, }) q = (G {, }) q = H {, }. The vertces of G are organzed nto cycles of q q 1. Snce (q q 1 ) 2 aut(g), we now that f or s oned to a vertex on a cycle of q q 1, then t s also oned to every second vertex on the cycle. We can construct graphs G and H satsfyng these requrements, as shown n Fgure 6. In the dagram on the left, the cycles of q q 1 are represented by ellpses of vertces that are alternately whte and blac. It s evdent from the constructon that (G ) q = H and that (G ) q = H, and that C = C = 1. In ths example, G and H are not hypomorphc. However, we do not now how to prove ths n general. It may even be that (q q 1 ) 2 s the dentty, so that the cycles of q q 1 have length at most two. Ths case s llustrated n the dagram on the rght. It s very easy to see that the graphs of ths example are not hypomorphc, but more complcated examples could be constructed. Fgure 6, Possble graphs G and H when C = C = 1 3. G q q 1 = G q q 1, where,, and are dstnct. We assume throughout ths secton that G q q 1 Notce that (G ) q q 1 = G q q Lemma. Let G q q 1 q q 1 = G q q 1 = G q q 1 = G. Ths gves:. Let G = G q q 1 = G q q 1. = G q q 1, where,, and are dstnct. Then q q 1 aut(g ). Snce p p 1 s a partal automorphsm of G, t follows that G and G are nearly dentcal. We are concerned wth G G, the dfference between G and G. Suppose that [x, y] G G. If [x, y] p p 1 partal automorphsm of G, and snce q q 1 exsted, then we could apply p p 1 [x, y] G = [x, y] G, a contradcton. It follows that [x, y] p p 1 Lemma 1.2, {x, y} {, q 1 that {x, y} {, q 1 to t. Snce p p 1 s a maps edges of G to edges of G, we obtan } = Ø. Snce G can also be wrtten as G q q 1 } Ø s also a requrement. Ths gves: 9 does not exst, so that by, we conclude

10 3.2 Lemma. If {, q 1 } and {, q 1 } are dsont, then G and G can dffer n at most 4 pars, [, ], [, q 1 ], [, q 1 ], and [ q 1, q 1 ]. We wll prove that when {, q 1 } and {, q 1 } are dsont, G and H are always somorphc; that s, the algebrac condton G q q 1 = G q q 1 mples the reconstructblty of G, when {,, q 1, q 1 } = 4. The proof conssts of a number of cases, determned by the cycle structure of the permutaton q q 1. We wll prove that n each case, H s equal to one of G q, G q, or G q. Notce that the graphs G, G, H, G q, G q, and G q all have the same number of edges. Therefore any symmetrc dfference of the form G G, G H, G q H, etc., all have an even number of pars, and that G G = G G, etc. We can conclude that f G G has four pars, that G G = G G = 2. The technque used n each case s based on the followng dea: Gven G G, we apply the mappngs p and p to ts pars. Applyng p and p to pars of G wll result n pars of H. Snce G = G q q 1 = G q q 1, applyng p or p to pars of G wll result n pars of G q. Therefore, applyng p and p to pars of G G wll gve pars of H G q and applyng p and p to pars of G G wll gve pars of G q H. So gven G G, we can often obtan G q H. We loo at the cycle structure of q q 1. We assume that G and H are non-somorphc, hypomorphc graphs, and obtan a contradcton n each case. 3.3 Lemma. G q H conssts of all pars [x, y] p and [x, y] p whch exst, such that [x, y] G G. Proof. Recall that q and q both map G to G q, and that p and p both map pars of G to H. Therefore, gven [x, y] G G, f [x, y] p exsts, then t s a par of G q H, and smlarly for [x, y] p. Conversely, f [, w] G q H, then at least one of [, w] p 1 and [, w] p 1 exsts, snce, gvng a par [x, y] G G wth the requred propertes. 3.4 Lemma. If () or () s a cycle of q q 1, then G G 4. Proof. Suppose that () s a cycle of q q 1. We have q q 1 =, so that q =. By Lemma 1.6, [, ] G f and only f [, ] H. We also now that [, ] G f and only f [, ] q = [, ] (G ) q = G q. It then follows that [, ] G G f and only f [, ] G q H, a contradcton to Lemma 1.4. Therefore [, ] G G, so that G G 4, by Lemma 3.2. The proof s smlar f () s a cycle. 3.5 Theorem. G G = 4 s mpossble. Proof. Suppose that G G = 4. By the precedng lemmas, we have G G = {[, ], [, q 1 Therefore p p 1 ], [, q 1 ], [ q 1 exsts, whether q q 1 Wthout loss of generalty, we can assume that [, q 1 of G, f necessary). Apply p p 1 ] G G (by consderng G nstead ] G G, snce q q 1 s an automorphsm of G, and p p 1, q 1 ]}. We also now that () and () are not cycles of q q 1. s of type 1 or 2. Consder the par [, q 1 ] G G. to get [ q q 1, q 1 s a partal automorphsm of G. Comparng [ q q 1 four pars of G G, we conclude that ether q q 1 10 = or q 1., q 1 ] wth the

11 Suppose frst that q q 1 = q 1. We have two pars, [, q 1 The remanng two pars of G G are [, ], [, q 1 [, q 1 of q q 1 ] to obtan [ p p 1, q 1. Therefore t s a thrd par of G G, whch s mpossble. Consequently, we must have q q 1 two pars, [, q 1 ], [, q 1 and p p 1 to [ q 1, q 1 whether these pars exst or not. q 1 Suppose frst that [ q 1 = or p 1 ], [ q 1, q 1 ] G G. to ] G G. We now apply p p 1 ]. Now ths par must exst, snce () and () are not cycles =. It follows that q q 1 ] G G. Therefore [, ], [ q 1 ], to get [ q 1, q 1 and q 1 = q 1 and has exactly 4 vertces. p p 1, q 1 p p 1 ] and [ q 1 ] and [ q 1, q 1, q 1, q 1 s of type 1. We have ] G G. Apply p p 1 p p 1 ]. We must consder p p 1 ] do not exst. Then by Lemma1.2,, q 1 } = 4, we must have q 1 = q 1 and q 1 = or p 1. As {,, q 1. Ths s llustrated n Fgure 7. We fnd that the cycle of q q 1 contanng Fgure 7, G G = 4, [ q 1, q 1 p p 1 ] does not exst We now calculate G q H, usng Lemma 3.3, by applyng p and p to pars of G G. The result s G q H = {[, q ], [, ], [, ]}, snce q = q. Ths s mpossble, as G q H must be even. There remans the stuaton when [ q 1, q 1 p p 1 ] or [ q 1, q 1 p p 1 ] exsts. Suppose that [ q 1 p p 1, q 1 p p 1 [ q 1, q 1 ] = [ q 1, q 1 Ths stuaton s llustrated n Fgure 8. ] exsts. As ths par must be n G G, we conclude that ]. It follows that ( q 1 ) q q 1 = q 1, or equvalently, (q 1 q ) 2 =. Fgure 8, G G = 4, [ q 1, q 1 p p 1 ] = [ q 1, q 1 ] Let C denote the cycle of q q 1 contanng and. Snce q q 1 aut(g ), the vertces of C all have the same degree n G, call t α. The cycle of q q 1 contanng 11

12 q 1 {[, q 1 s ( q 1, q 1 ). Let β denote the degree of q 1 ], [, q 1 the edges [, q 1 ]} and G G = {[, ], [ q 1 ], [, q 1 ] and addng the edges [, ], [ q 1 and q 1 n G. Now G G =, q 1 ]}, e, G s formed from G by removng, q 1 ]. These pars are ndcated n the dagram. Consequently, all vertces of C also have degree α n G; and q 1, q 1 both have degree β n G. It follows that deg(v, G) = deg(v, G ), for all v V. But G = G q q 1, so that deg(v, G) = deg(v q q 1, G q q 1 have the same degree n G, for all v V. We show that α = β. ) = deg(v q q 1 Consder the par-orbt Ω( q 1, G ) = deg(v q q 1, G), e, v and v q q 1, q 1 ) wth respect to q q 1. It conssts of a cyclc sequence of pars (P 0, P 1,..., P m ), where P 0 = [ q 1, q 1 ], and P l+1 = P q q 1 l, where l = 0, 1, 2,..., m, and addton of subscrpts s reduced modulo m + 1. As G = G q q 1, we conclude that P l G f and only f P l+1 G. Now P 0 G G. Therefore P m G. Hence, there s some l {0, 1,..., m 1}, such that P l G but P l+1 G. Then P l+1 G G. It follows that P l+1 s ether [, q 1 ] or [, q 1 ]. Now deg( q 1, G) = deg( q 1, G) = β. As deg(v, G) = deg(v q q 1, G), we conclude that α = deg(, G) = deg(, G) = deg( q 1, G) = deg( q 1, G) = β. We now fnd H G q = {[, q 1 q ]} and G q H = {[, q ]}, usng Lemma 3.3. Therefore, f v { q 1 q, q }, then deg(v, G q ) = deg(v, H). In G q, the vertces of C q all have degree α; furthermore and q 1 q have degree β, whch equals α. Now H s formed from G q by removng the edge [, q ] and addng the edge [, q 1 q ]. Hence, n H, the vertces of C q all have degree α, except for q, whch has degree α 1. Also, deg(, H) = α, but deg( q 1 q, H) = α + 1. But G and H must have the same degree sequences, a contradcton. Ths completes the proof that G G = 4 s mpossble. We now turn to the stuaton when G G = 2. There are 6 ways of choosng two edges out of {[, ], [, q 1 ], [, q 1 ], [ q 1, q 1 ]}. However, due to the symmetry between and, some of these wll be equvalent. We loo at each case n turn. 3.6 Lemma. If G G = 2, then [, ] G G. Proof. Suppose that [, ] G G and that some par [x, y] G G. Then snce [, ] p and [, ] p do not exst, we fnd that H G q = Ø and that G q H Ø, whch s mpossble. A smlar result holds f [, ] G G and [x, y] G G. 3.7 Lemma. If G G = {[, q 1 Proof. Suppose that [, q 1 ], [ q 1, q 1 ]}, then G q ] G G and that [ q 1, q 1 ] G G. Apply p and p to obtan [, q ] H G q and [, q 1 q ], [, q 1 q ] G q H. Snce H G q = G q H = 1, we conclude that q 1 q = q 1 q. It then follows that H G q = (G G ) q. But ths equals G q G q. Cancellng G q leaves G q Interchangng and n the above lemma gves: 12

13 3.8 Lemma. If G G = {[, q 1 ], [ q 1, q 1 ]}, then G q There remans one case. 3.9 Theorem. G G = {[, q 1 ], [, q 1 ]} s mpossble. Proof. Wthout loss of generalty, suppose that [, q 1 If [, q 1 ] p p 1 ] G G and that [, q 1 ] G G. were to exst, t would also be a par of G G, as p p 1 automorphsm of G and an automorphsm of G ; that s, we would have [ q 1 s a partal, q q 1 ] G G. Ths would mae G G > 1, whch s mpossble. We conclude that [, q 1 does not exst. Ths mples that p p 1 does not exst; hence p =. Smlarly, we fnd ] p p 1 p =. Ths s llustrated n Fgure 9, whch shows the cycles () and () of q q 1, and the cycle C contanng q 1 and q 1. It follows from Lemma 3.3 that H G q = {[, ]} and G q H = {[, ]}. Fgure 9, G G = {[, q 1 ], [, q 1 ]} We show that G and H have dfferent degree sequences, by observng that G and G q have the same degree sequences, but that G q and H have dfferent degree sequences. We have G G = {[, q 1 ], [, q 1 ]}, and G q H = {[, ], [, ]}. If v {,, q 1, q 1 }, then deg(v, G) = deg(v, G ) so that deg(v q, G q ) = deg(v q, (G ) q ) = deg(v q, G q ). Snce v q {,, }, we have deg(v, G) = deg(v q, G q ) = deg(v q, H). Snce q = and q =, we also have deg(v, G) = deg(v q, H) when v {, }. We must stll compare deg(v, G) and deg(v q, H) when v { q 1, q 1 }. Now G G = {[, q 1 ]}. Snce q q 1 aut(g ), t follows that n G, s adacent to every vertex of C. In G, all vertces of C have a common degree, call t α. Then deg( q 1, G ) = deg( q 1, G ) = α. Snce G G = {[, q 1 ]}, we conclude that deg( q 1, G) = deg( q 1, G ) 1 = α 1, and that deg( q 1, G) = deg( q 1, G ) + 1 = α + 1. Then deg( q 1, G) = α 1 = deg(, G q ) 1. Snce G q H = {[, ], [, ]}, we have deg(, H) = deg(, G q ) = α. Smlarly, deg( q 1, G) = α = deg( q 1 q, G q ) = q, H). Hence deg( q 1, G) = deg( q 1 deg( q 1 dfferent degree sequences, a contradcton. 13 q, H). We conclude that G and H have

14 We summarse the precedng lemmas and theorems as the followng: 3.10 Theorem. Let G and H be hypomorphc graphs. If G q q 1 = G q q 1, where,, and are dstnct, and {,, q 1, q 1 } = 4, then H s equal to one of G q, G q, or G q. We thus have an algebrac condton forcng the reconstructblty of G. We must stll deal wth the case when {,, q 1, q 1 } {,, q 1, q 1 } 3 The possble cases are: 1. = q 1 2. q 1 = q 1 3. = q 1 Case 1. = q 1 (or symmetrcally, = q 1 ) and = q 1 In ths secton we assume that {,, q 1, q 1 } = 3 and that = q Lemma. G G conssts of pars of the form [, x] or [, q 1 ]. Proof. As n Lemma 3.2, we fnd that f [x, y] G G, then [x, y] p p 1 not exst. Hence {x, y} {, q 1 } Ø and {x, y} {, q 1 reduces to ether {x, y}, or [x, y] = [, q 1 ]. and [x, y] p p 1 do } Ø. Snce = q 1, ths Notce that q q 1 maps q 1 to q 1 = so that q q 1 does not fx the pont. 4.2 Lemma. Gven [, x] G G, for some x. If [, x] p p 1 exsts, then q and G G = {[, ], [, q 1 Proof. We have q q 1 ], [, q 1 ], [, q q 1 ]}. = q 1. Snce p p 1 automorphsm of G, we have [, x] p p 1 that ether x q q 1 = [ q 1 = or. Now we can not have x q q 1 s both an automorphsm of G and a partal, x q q 1 ] G G. By Lemma 4.1, we fnd =, because then x = q 1, but then x p p 1 does not exst, contrary to assumpton. It follows that x = q q 1 and that q, snce x p p 1 exsts. We have found that [, q q 1 ], [, q 1 ] G G, and that these pars are ether both n G G, or both n G G. There must be at least two more pars to balance these, and they must be pars [, x], such that x p p 1 does not exst. The only possbltes are [, ] and [, q 1 ]. Ths completes the proof. Notce that t follows from Lemma 4.2 that G G and G G are {[, q q 1 and {[, ], [, q 1 ]}. Let C denote the cycle of q q 1 contanng. 14 ], [, q 1 ]} contanng, and C the cycle

15 4.3 Theorem. If G G = {[, ], [, q 1 q q 1 must be of type 2, and G q H = {[, ], [, q ]}. Proof. Wthout loss of generalty, let G G ], [, q 1 ], [, q q 1 ]} and q, then C = 2, = {[, q 1 ], [, q q 1 ]} and G G = {[, ], [, q 1 ]}. Apply p and p to obtan H G q = {[, q ]} and G q H = {[, ]}. The case when q q 1 s of type 1 s llustrated n Fgure 10; type 2 s llustrated n Fgure 11. A dotted lne ndcates a par that s nown not to be an edge of G. Fgure 10, G G = {[, ], [, q 1 Notce that q 1 two were equal, then q q 1 ], [, q 1 ], [, q q 1 ]}, q q 1 of type 1, by assumpton. We also now that q 1 would map [, q q 1 have the same degree n G, call t α. Smlarly,, q q 1 ] G G to [, q 1 q q 1, for f these ] G G, whch s mpossble, as q q 1 aut(g ). We have, q 1, q 1 C ; therefore these vertces all C, so that these vertces also have the same degree n G, call t β. If q q 1 s of type 1, then C = C and α = β. Fgure 11, G G = {[, ], [, q 1 ], [, q 1 ], [, q q 1 ]}, q q 1 of type 2 We show that f C 3, then G and H have dfferent degree sequences. Ths s done by observng that G and G q have the same degree sequence, and showng that G q and H have dfferent degree sequences. Suppose that C 3, so that, q 1, and q 1 are dstnct vertces. Snce G G = {[, q 1 ], [, q q 1 15 ]} and G G = {[, ], [, q 1 ]}, we

16 have deg(, G) = deg( q 1, G) = α 1 and deg( q 1, G) = α + 1. Any other vertces of C have degree α n G. Smlarly, deg( q q 1, G) = β + 1, whereas all other vertces of C have degree β n G. If v {, q 1, q 1, q q 1 }, then deg(v, G) = deg(v, G ) Hence deg(v q, G q ) = deg(v q, (G ) q ) = deg(v q, G q ). Snce G q H = {[, ], [, q ]}, we now that v q {,, q }, so that deg(v, G) = deg(v q, G q ) = deg(v q, H). The vertces of {, q 1, q 1, q q 1 } have degrees {α 1, α 1, α, β + 1} n G. Snce G q H = {[, ]} and H G q = {[, q ]}, we have deg( q, H) = deg(, H) = α; deg( q 1 q, H) = α; deg( q 1 q, H) = deg(, H) = α 1; and deg( q q 1 q, H) = deg( q, H) = β + 1. Thus, the degrees of v q, when v {, q 1, q 1, q q 1 }, are {α, α, α 1, β + 1}. If C 3, ths s not possble. We conclude that C = 2, so that q q 1 s of type 2. that q 1 We have been unable to prove that C = 2 s not possble n Theorem 4.3. We fnd = q 1, so that the degrees n G of v {, q 1, q q 1 } are the same as the degrees n H of v q. Lemmas 4.2 and Theorem 4.3 handle the stuaton when there s a par [, x] G G such that [, x] p p 1 every [, x] G G. Snce p p 1 follows that ether x = or x = q 1 exsts. We now consder the cases when [, x] p p 1 [, q 1 ]. G G contans exactly two of these pars. 4.4 Lemma. If [, x] p p 1 then q =. Proof. If [, q 1 ] p p 1 does not exst, for = q 1, we conclude that x p p 1 does not exst. It. The possble pars for G G are [, ], [, q 1 ], and does not exst, for any [, x] G G, and [, q 1 ] G G, were to exst, t would equal a par of G G, snce p p 1 automorphsm of both G and G. It would follow that [, q 1 ether [, ] or [, q 1 ]. Ths mples that q =. ] p p 1 = [, q q 1 s a partal ] equals 4.5 Lemma. G G = {[, ], [, q 1 ]} s mpossble. Proof. Calculatng G q H = {[, q ]} gves a contradcton, snce G q H must be even. 4.6 Lemma. G G = {[, ], [, q 1 ]} s mpossble. Proof. Calculatng G q H = {[, ]}, usng Lemma 3.3, gves a contradcton, snce G q H must be even. 4.7 Lemma. If G G = {[, q 1 ], [, q 1 ]}, then G q Proof. Wthout loss of generalty, assume that G G = {[, q 1 ]} and that G G = {[, q 1 ]}. By Lemma 4.4, q =. Calculate H G q = {[, ]}, usng Lemma 3.3, and G q H = {[, q ]} = {[, ]}, snce q =. Let C be the cycle of q q 1 contanng, as llustrated n Fgure 12. In G, all vertces of C have the same degree, call t α. Let β = deg(, G ). Suppose frst that C 3, so that q 1 16 q 1. Snce G G = {[, q 1 ]}

17 and G G = {[, q 1 ]}, we have deg(, G) = deg( q 1, G) = α 1; deg( q 1, G) = α + 1; and deg(, G) = β + 1. If v {,, q 1, q 1 }, then deg(v, G) = deg(v, G ) so that deg(v q, G q ) = deg(v q, (G ) q ) = deg(v q, G q ). Snce v q {,,, q 1 q }, and snce G q H = {[, ], [, ]}, we conclude that deg(v q, G q ) = deg(v q, H). If v =, we also have deg(, G) = β + 1 = deg( q, H). The degrees of {, q 1, q 1 } n G are {α 1, α 1, α}. The degrees of v q n H are {α 1, α, α}, so that G and H have dfferent degree sequences, a contradcton. We conclude that C = 2, so that q 1 G G = {[, q 1 ]}. We then fnd that q maps [, q 1 = q 1, and ] G G to [, ] H G q, and [, q 1 ] G G s mapped to [, ] G q H. Therefore (G G ) q = H G q. Snce (G ) q = G q, t follows that G q = H, as requred. Fgure 12, G G = {[, q 1 ], [, q 1 ]} Case 2. q 1 = q 1 q 1 In ths secton we assume that {,, q 1 s a fxed pont of q q 1., q 1 } = 3 and that q 1 = q 1. Notce that 4.8 Lemma. G G conssts of pars of the forms [ q 1, x] or [, ]. Proof. As n Lemma 3.3, we fnd that f [x, y] G G, then [x, y] p p 1 not exst. Hence {x, y} {, q 1 } = Ø and {x, y} {, q 1 reduces to ether q 1 {x, y}, or [x, y] = [, ]. 4.9 Theorem. If q 1 } = Ø. Snce q 1 = q 1, and q q 1 s of type 2, then G q Proof. Let X be the set of vertces x such that [ q 1 set of all pars [ q 1 x X, f x p p 1 G. If x p p 1 tmes to obtan x p p 1, x] G G. Let [ q 1, x] such that x X. By Lemma 4.8, (G G ) [, ] = [ q 1 exsts, then x p p 1 X, snce p p 1 does not exst, then snce q q 1 X. Thus X q q 1 and [x, y] p p 1 do = q 1, ths, X] denote the, X]. Gven s a partal automorphsm of G and a number of s of type 2, we can apply p p 1 = X. By Lemma 3.3, H G q = [, X q ], snce 17

18 X q = X q. Snce G G and H G q are even we conclude that G G = [ q 1, X], so that (G G ) q = G q G q = [, X q ] = G q H so that G q 4.10 Theorem. If q 1 = q 1, and q q 1 G q H = {[, ], [, q ]} Proof. Let C denote the cycle of q q 1 Theorem 4.9, let X denote the set of vertces x such that [ q 1 s of type 1, then ether G = H or else contanng and. Refer to Fgure 13. As n (G G ) [, ] = [ q 1, X]. Let X = X C. Then (X ) q q 1 partal automorphsm of both G and G, and C s a cycle of q q 1 agan have G q, x] G G. By Lemma 4.8, = X, snce p p 1 s a. If Xq q 1 = X, we = H, as n Theorem 4.9. We dvde the cycle C nto two subsets: P 1, the vertces on the path from to ; and P 2, the vertces on the path from q 1 P 1 P 2 = C. If x X P l, where l = 1 or 2, then we can apply ether p p 1 to q 1. Then or ts nverse to x as many tmes as needed to obtan P l X. Consequently we can assume that X C s ether P 1 or P 2. Suppose frst that X C = P 1. By Lemma 3.3, G q H = [, (X ) p ] [, (X ) p ]. Let m = P 1. Then G q H = X + m 1, whereas [ q 1, X] = X + m. Snce G q H and G G are both even, we conclude that [, ] G G. We then have (G G ) q = G q G q = [ q 1, X] q {[, ] q } = [, X q ] {[, q ]}. Comparng ths wth the expresson for G q H gves G q H = {[, ], [, q ]}. Fgure 13, q 1 = q 1 Suppose now that X C = P 2. Let l = P 2. We fnd that G q H = [, X p ] [, X p ], snce, X. Therefore G q H = X + l + 1, whereas [ q 1, X] = X + l. It follows that [, ] G G. As above, we obtan G q H = {[, ], [, q ]}. We have been unable to prove that G q = H n ths case. If we attempt to use the degree sequence as n Theorem 3.5, we can proceed as follows. Consder the case when X C = P 1, and assume that [ q 1, P 1 ] G G. Then [ q 1, ] G G so that [, ] G q. Snce G q H = {[, ], [, q ]}, we have [, ] G q H and [, q ] H G q. Therefore 18

19 [, ] G G. Let m 1 = (G G) [ q 1, X ] and m 2 = (G G ) [ q 1, X ]. Then G G = m and G G = m + m 2, so that m = m + m 2. In G, the vertces of C all have the same degree, call t α. Let β = deg( q 1, G ). Then deg( q 1, G) = β +m+m 2 m 1 = β +1. Also, deg(, G) = deg(, G) = α. The other vertces of P 1 have degree α + 1 n G. We use G q H = {[, ]} and H G q = {[, q ]} to obtan the degrees n H. We have β + 1 = deg( q 1, G) = deg(, G q ) = deg(, H) + 1, so that deg(, H) = β. Also, α = deg(, G) = deg( q, G q ) = deg( q, H) 1, so that deg( q, H) = α + 1. If v {, q }, then deg(v, G q ) = deg(v, H). The only concluson drawn s that α = β. Case 3. = q 1 and = q 1 In ths secton we assume that {,, q 1, q 1 } = 2 and that = q 1 and = q 1. These condtons weaen the mplcatons of the relaton G q q 1 = G q q 1, and do not seem suffcent to settle ths case. We have only the followng smple lemma and observatons Lemma. G G conssts of pars of the forms [, x] or [, x]. Proof. As n Lemma 3.3, we fnd that f [x, y] G G, then [x, y] p p 1 do not exst. Hence {x, y} {, q 1 q 1 =, ths reduces to ether {x, y}, or {x, y}. Notce that q q 1 = ( q 1 ) q q 1 } = Ø and {x, y} {, q 1 and [x, y] p p 1 } = Ø. Snce q 1 = and = q 1 =. Therefore q q 1 s of type 1. Suppose exsts, then [, x q q 1 ] s also n G G, s a partal automorphsm of G, and an automorphsm of G. If t does not that [, x] G G, for some x. If [, x] p p 1 snce p p 1 exst, then x s ether or q 1 [, q 1. Thus, the possble dfference pars are [, q 1 ] (by symmetry); and pars of the form [, x], [, x q q 1 ]. 5. Concluson ], [, ], and The algebrac relatons G q q 1 = G q q 1 or G q q 1 = G q q 1 are suffcent to force the reconstructblty of G n many cases. The second relaton mples that one of G q, G q, or G q always equals H when {,, q 1, q 1 } = 4. When {,, q 1, q 1 } = 3, the results are less conclusve; G q and H can sometmes dffer n up to two edges, that s G q H 2. When {,, q 1, q 1 } = 2, the condtons are not strong enough to force G Other algebrac relatons for whch smlar results lely hold are G q q 1 = G q q 1 and G q q 1 = G q q 1 l, where,, and l are dstnct. References 1. J.A. Bondy and R.H. Hemmnger, Graph reconstructon a survey, J. Graph Th. 1, (1977), pp J.A. Bondy, A graph reconstructor s manual, n Surveys n Combnatorcs, Ed. A.D. Keedwell, Cambrdge Unversty Press, pp

20 3. W. Kocay, Hypomorphsms, orbts, and reconstructon, J. Comb. Th. (B) 44 (1988), pp C.St.J. Nash-Wllams, The reconstructon problem, n Selected Topcs n Graph Theory, Ed. L. Benee and R. Wlson, Academc Press, 1978, pp