Graph Reconstruction by Permutations


 Laurel Sullivan
 3 years ago
 Views:
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 emal: Computer Scence Department St. Paul s College, Unversty of Mantoba Wnnnpeg, MB, Canada R3T 2N2 emal: 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 AxMarselle 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 NashWllams [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 parorbt 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 parorbt of q q 1, then Ω q s a parorbt 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 parorbt Ω(x, y) that s only partally contaned n G Lemma. Let Ω = Ω(x, y) be a parorbt 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 parorbt 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 nonsomorphc, 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 parorbt 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. NashWllams, The reconstructon problem, n Selected Topcs n Graph Theory, Ed. L. Benee and R. Wlson, Academc Press, 1978, pp
Affine transformations and convexity
Affne transformatons and convexty The purpose of ths document s to prove some basc propertes of affne transformatons nvolvng convex sets. Here are a few onlne references for background nformaton: http://math.ucr.edu/
More informationVolume 18 Figure 1. Notation 1. Notation 2. Observation 1. Remark 1. Remark 2. Remark 3. Remark 4. Remark 5. Remark 6. Theorem A [2]. Theorem B [2].
Bulletn of Mathematcal Scences and Applcatons Submtted: 0160407 ISSN: 789634, Vol. 18, pp 110 Revsed: 0160908 do:10.1805/www.scpress.com/bmsa.18.1 Accepted: 0161013 017 ScPress Ltd., Swtzerland
More informationMaximizing the number of nonnegative subsets
Maxmzng the number of nonnegatve subsets Noga Alon Hao Huang December 1, 213 Abstract Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what s the maxmum
More informationFoundations of Arithmetic
Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an
More informationSelfcomplementing permutations of kuniform hypergraphs
Dscrete Mathematcs Theoretcal Computer Scence DMTCS vol. 11:1, 2009, 117 124 Selfcomplementng permutatons of kunform hypergraphs Artur Szymańsk A. Paweł Wojda Faculty of Appled Mathematcs, AGH Unversty
More informationA new construction of 3separable matrices via an improved decoding of Macula s construction
Dscrete Optmzaton 5 008 700 704 Contents lsts avalable at ScenceDrect Dscrete Optmzaton journal homepage: wwwelsevercom/locate/dsopt A new constructon of 3separable matrces va an mproved decodng of Macula
More informationProblem Do any of the following determine homomorphisms from GL n (C) to GL n (C)?
Homework 8 solutons. Problem 16.1. Whch of the followng defne homomomorphsms from C\{0} to C\{0}? Answer. a) f 1 : z z Yes, f 1 s a homomorphsm. We have that z s the complex conjugate of z. If z 1,z 2
More informationAPPENDIX A Some Linear Algebra
APPENDIX A Some Lnear Algebra The collecton of m, n matrces A.1 Matrces a 1,1,..., a 1,n A = a m,1,..., a m,n wth real elements a,j s denoted by R m,n. If n = 1 then A s called a column vector. Smlarly,
More informationHMMT February 2016 February 20, 2016
HMMT February 016 February 0, 016 Combnatorcs 1. For postve ntegers n, let S n be the set of ntegers x such that n dstnct lnes, no three concurrent, can dvde a plane nto x regons (for example, S = {3,
More informationProblem Set 9 Solutions
Desgn and Analyss of Algorthms May 4, 2015 Massachusetts Insttute of Technology 6.046J/18.410J Profs. Erk Demane, Srn Devadas, and Nancy Lynch Problem Set 9 Solutons Problem Set 9 Solutons Ths problem
More informationEdge Isoperimetric Inequalities
November 7, 2005 Ross M. Rchardson Edge Isopermetrc Inequaltes 1 Four Questons Recall that n the last lecture we looked at the problem of sopermetrc nequaltes n the hypercube, Q n. Our noton of boundary
More informationThe Order Relation and Trace Inequalities for. Hermitian Operators
Internatonal Mathematcal Forum, Vol 3, 08, no, 50757 HIKARI Ltd, wwwmhkarcom https://doorg/0988/mf088055 The Order Relaton and Trace Inequaltes for Hermtan Operators Y Huang School of Informaton Scence
More informationComplete subgraphs in multipartite graphs
Complete subgraphs n multpartte graphs FLORIAN PFENDER Unverstät Rostock, Insttut für Mathematk D18057 Rostock, Germany Floran.Pfender@unrostock.de Abstract Turán s Theorem states that every graph G
More informationIntroductory Cardinality Theory Alan Kaylor Cline
Introductory Cardnalty Theory lan Kaylor Clne lthough by name the theory of set cardnalty may seem to be an offshoot of combnatorcs, the central nterest s actually nfnte sets. Combnatorcs deals wth fnte
More information2.3 Nilpotent endomorphisms
s a block dagonal matrx, wth A Mat dm U (C) In fact, we can assume that B = B 1 B k, wth B an ordered bass of U, and that A = [f U ] B, where f U : U U s the restrcton of f to U 40 23 Nlpotent endomorphsms
More informationLectures  Week 4 Matrix norms, Conditioning, Vector Spaces, Linear Independence, Spanning sets and Basis, Null space and Range of a Matrix
Lectures  Week 4 Matrx norms, Condtonng, Vector Spaces, Lnear Independence, Spannng sets and Bass, Null space and Range of a Matrx Matrx Norms Now we turn to assocatng a number to each matrx. We could
More informationWeek 2. This week, we covered operations on sets and cardinality.
Week 2 Ths week, we covered operatons on sets and cardnalty. Defnton 0.1 (Correspondence). A correspondence between two sets A and B s a set S contaned n A B = {(a, b) a A, b B}. A correspondence from
More informationk(k 1)(k 2)(p 2) 6(p d.
BLOCKTRANSITIVE 3DESIGNS WITH AFFINE AUTOMORPHISM GROUP Greg Gamble Let X = (Z p d where p s an odd prme and d N, and let B X, B = k. Then t was shown by Praeger that the set B = {B g g AGL d (p} s the
More informationAntivan der Waerden numbers of 3term arithmetic progressions.
Antvan der Waerden numbers of 3term arthmetc progressons. Zhanar Berkkyzy, Alex Schulte, and Mchael Young Aprl 24, 2016 Abstract The antvan der Waerden number, denoted by aw([n], k), s the smallest
More informationCOMPLEX NUMBERS AND QUADRATIC EQUATIONS
COMPLEX NUMBERS AND QUADRATIC EQUATIONS INTRODUCTION We know that x 0 for all x R e the square of a real number (whether postve, negatve or ero) s nonnegatve Hence the equatons x, x, x + 7 0 etc are not
More informationMore metrics on cartesian products
More metrcs on cartesan products If (X, d ) are metrc spaces for 1 n, then n Secton II4 of the lecture notes we defned three metrcs on X whose underlyng topologes are the product topology The purpose of
More informationSmarandacheZero Divisors in Group Rings
SmarandacheZero Dvsors n Group Rngs W.B. Vasantha and Moon K. Chetry Department of Mathematcs I.I.T Madras, Chenna The study of zerodvsors n group rngs had become nterestng problem snce 1940 wth the
More informationNUMERICAL DIFFERENTIATION
NUMERICAL DIFFERENTIATION 1 Introducton Dfferentaton s a method to compute the rate at whch a dependent output y changes wth respect to the change n the ndependent nput x. Ths rate of change s called the
More informationMATH 241B FUNCTIONAL ANALYSIS  NOTES EXAMPLES OF C ALGEBRAS
MATH 241B FUNCTIONAL ANALYSIS  NOTES EXAMPLES OF C ALGEBRAS These are nformal notes whch cover some of the materal whch s not n the course book. The man purpose s to gve a number of nontrval examples
More informationSL n (F ) Equals its Own Derived Group
Internatonal Journal of Algebra, Vol. 2, 2008, no. 12, 585594 SL n (F ) Equals ts Own Derved Group Jorge Macel BMCCThe Cty Unversty of New York, CUNY 199 Chambers street, New York, NY 10007, USA macel@cms.nyu.edu
More information12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA. 4. Tensor product
12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA Here s an outlne of what I dd: (1) categorcal defnton (2) constructon (3) lst of basc propertes (4) dstrbutve property (5) rght exactness (6) localzaton
More informationAssortment Optimization under MNL
Assortment Optmzaton under MNL Haotan Song Aprl 30, 2017 1 Introducton The assortment optmzaton problem ams to fnd the revenuemaxmzng assortment of products to offer when the prces of products are fxed.
More informationREDUCTION MODULO p. We will prove the reduction modulo p theorem in the general form as given by exercise 4.12, p. 143, of [1].
REDUCTION MODULO p. IAN KIMING We wll prove the reducton modulo p theorem n the general form as gven by exercse 4.12, p. 143, of [1]. We consder an ellptc curve E defned over Q and gven by a Weerstraß
More informationDifference Equations
Dfference Equatons c Jan Vrbk 1 Bascs Suppose a sequence of numbers, say a 0,a 1,a,a 3,... s defned by a certan general relatonshp between, say, three consecutve values of the sequence, e.g. a + +3a +1
More informationFinding Dense Subgraphs in G(n, 1/2)
Fndng Dense Subgraphs n Gn, 1/ Atsh Das Sarma 1, Amt Deshpande, and Rav Kannan 1 Georga Insttute of Technology,atsh@cc.gatech.edu Mcrosoft ResearchBangalore,amtdesh,annan@mcrosoft.com Abstract. Fndng
More informationFINITELYGENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN
FINITELYGENERTED MODULES OVER PRINCIPL IDEL DOMIN EMMNUEL KOWLSKI Throughout ths note, s a prncpal deal doman. We recall the classfcaton theorem: Theorem 1. Let M be a fntelygenerated module. (1) There
More informationarxiv: v1 [math.co] 1 Mar 2014
Unonntersectng set systems Gyula O.H. Katona and Dánel T. Nagy March 4, 014 arxv:1403.0088v1 [math.co] 1 Mar 014 Abstract Three ntersecton theorems are proved. Frst, we determne the sze of the largest
More informationFACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP
C O L L O Q U I U M M A T H E M A T I C U M VOL. 80 1999 NO. 1 FACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP BY FLORIAN K A I N R A T H (GRAZ) Abstract. Let H be a Krull monod wth nfnte class
More informationLinear, affine, and convex sets and hulls In the sequel, unless otherwise specified, X will denote a real vector space.
Lnear, affne, and convex sets and hulls In the sequel, unless otherwse specfed, X wll denote a real vector space. Lnes and segments. Gven two ponts x, y X, we defne xy = {x + t(y x) : t R} = {(1 t)x +
More informationInner Product. Euclidean Space. Orthonormal Basis. Orthogonal
Inner Product Defnton 1 () A Eucldean space s a fntedmensonal vector space over the reals R, wth an nner product,. Defnton 2 (Inner Product) An nner product, on a real vector space X s a symmetrc, blnear,
More informationThe Multiple Classical Linear Regression Model (CLRM): Specification and Assumptions. 1. Introduction
ECONOMICS 5*  NOTE (Summary) ECON 5*  NOTE The Multple Classcal Lnear Regresson Model (CLRM): Specfcaton and Assumptons. Introducton CLRM stands for the Classcal Lnear Regresson Model. The CLRM s also
More informationTHE CHVÁTALERDŐS CONDITION AND 2FACTORS WITH A SPECIFIED NUMBER OF COMPONENTS
Dscussones Mathematcae Graph Theory 27 (2007) 401 407 THE CHVÁTALERDŐS CONDITION AND 2FACTORS WITH A SPECIFIED NUMBER OF COMPONENTS Guantao Chen Department of Mathematcs and Statstcs Georga State Unversty,
More informationSpectral Graph Theory and its Applications September 16, Lecture 5
Spectral Graph Theory and ts Applcatons September 16, 2004 Lecturer: Danel A. Spelman Lecture 5 5.1 Introducton In ths lecture, we wll prove the followng theorem: Theorem 5.1.1. Let G be a planar graph
More informationModule 9. Lecture 6. Duality in Assignment Problems
Module 9 1 Lecture 6 Dualty n Assgnment Problems In ths lecture we attempt to answer few other mportant questons posed n earler lecture for (AP) and see how some of them can be explaned through the concept
More informationOn intransitive graphrestrictive permutation groups
J Algebr Comb (2014) 40:179 185 DOI 101007/s1080101304825 On ntranstve graphrestrctve permutaton groups Pablo Spga Gabrel Verret Receved: 5 December 2012 / Accepted: 5 October 2013 / Publshed onlne:
More informationCollege of Computer & Information Science Fall 2009 Northeastern University 20 October 2009
College of Computer & Informaton Scence Fall 2009 Northeastern Unversty 20 October 2009 CS7880: Algorthmc Power Tools Scrbe: Jan Wen and Laura Poplawsk Lecture Outlne: Prmaldual schema Network Desgn:
More information20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The first idea is connectedness.
20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The frst dea s connectedness. Essentally, we want to say that a space cannot be decomposed
More information8.6 The Complex Number System
8.6 The Complex Number System Earler n the chapter, we mentoned that we cannot have a negatve under a square root, snce the square of any postve or negatve number s always postve. In ths secton we want
More informationSalmon: Lectures on partial differential equations. Consider the general linear, secondorder PDE in the form. ,x 2
Salmon: Lectures on partal dfferental equatons 5. Classfcaton of secondorder equatons There are general methods for classfyng hgherorder partal dfferental equatons. One s very general (applyng even to
More informationarxiv: v3 [cs.dm] 7 Jul 2012
Perfect matchng n unform hypergraphs wth large vertex degree arxv:1101.580v [cs.dm] 7 Jul 01 Imdadullah Khan Department of Computer Scence College of Computng and Informaton Systems Umm AlQura Unversty
More informationNPCompleteness : Proofs
NPCompleteness : Proofs Proof Methods A method to show a decson problem Π NPcomplete s as follows. (1) Show Π NP. (2) Choose an NPcomplete problem Π. (3) Show Π Π. A method to show an optmzaton problem
More informationCanonical transformations
Canoncal transformatons November 23, 2014 Recall that we have defned a symplectc transformaton to be any lnear transformaton M A B leavng the symplectc form nvarant, Ω AB M A CM B DΩ CD Coordnate transformatons,
More information5 The Rational Canonical Form
5 The Ratonal Canoncal Form Here p s a monc rreducble factor of the mnmum polynomal m T and s not necessarly of degree one Let F p denote the feld constructed earler n the course, consstng of all matrces
More informationRandom Walks on Digraphs
Random Walks on Dgraphs J. J. P. Veerman October 23, 27 Introducton Let V = {, n} be a vertex set and S a nonnegatve rowstochastc matrx (.e. rows sum to ). V and S defne a dgraph G = G(V, S) and a drected
More informationExercise Solutions to Real Analysis
xercse Solutons to Real Analyss Note: References refer to H. L. Royden, Real Analyss xersze 1. Gven any set A any ɛ > 0, there s an open set O such that A O m O m A + ɛ. Soluton 1. If m A =, then there
More informationOpen Systems: Chemical Potential and Partial Molar Quantities Chemical Potential
Open Systems: Chemcal Potental and Partal Molar Quanttes Chemcal Potental For closed systems, we have derved the followng relatonshps: du = TdS pdv dh = TdS + Vdp da = SdT pdv dg = VdP SdT For open systems,
More informationLecture 10: May 6, 2013
TTIC/CMSC 31150 Mathematcal Toolkt Sprng 013 Madhur Tulsan Lecture 10: May 6, 013 Scrbe: Wenje Luo In today s lecture, we manly talked about random walk on graphs and ntroduce the concept of graph expander,
More informationCaps and Colouring Steiner Triple Systems
Desgns, Codes and Cryptography, 13, 51 55 (1998) c 1998 Kluwer Academc Publshers, Boston. Manufactured n The Netherlands. Caps and Colourng Stener Trple Systems AIDEN BRUEN* Department of Mathematcs, Unversty
More informationSection 8.3 Polar Form of Complex Numbers
80 Chapter 8 Secton 8 Polar Form of Complex Numbers From prevous classes, you may have encountered magnary numbers the square roots of negatve numbers and, more generally, complex numbers whch are the
More informationErrors for Linear Systems
Errors for Lnear Systems When we solve a lnear system Ax b we often do not know A and b exactly, but have only approxmatons Â and ˆb avalable. Then the best thng we can do s to solve Âˆx ˆb exactly whch
More informationStructure and Drive Paul A. Jensen Copyright July 20, 2003
Structure and Drve Paul A. Jensen Copyrght July 20, 2003 A system s made up of several operatons wth flow passng between them. The structure of the system descrbes the flow paths from nputs to outputs.
More information3.1 Expectation of Functions of Several Random Variables. )' be a kdimensional discrete or continuous random vector, with joint PMF p (, E X E X1 E X
Statstcs 1: Probablty Theory II 37 3 EPECTATION OF SEVERAL RANDOM VARIABLES As n Probablty Theory I, the nterest n most stuatons les not on the actual dstrbuton of a random vector, but rather on a number
More informationPolynomials. 1 More properties of polynomials
Polynomals 1 More propertes of polynomals Recall that, for R a commutatve rng wth unty (as wth all rngs n ths course unless otherwse noted), we defne R[x] to be the set of expressons n =0 a x, where a
More informationLecture 12: Discrete Laplacian
Lecture 12: Dscrete Laplacan Scrbe: Tanye Lu Our goal s to come up wth a dscrete verson of Laplacan operator for trangulated surfaces, so that we can use t n practce to solve related problems We are mostly
More informationn ). This is tight for all admissible values of t, k and n. k t + + n t
MAXIMIZING THE NUMBER OF NONNEGATIVE SUBSETS NOGA ALON, HAROUT AYDINIAN, AND HAO HUANG Abstract. Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what
More informationCHAPTER 14 GENERAL PERTURBATION THEORY
CHAPTER 4 GENERAL PERTURBATION THEORY 4 Introducton A partcle n orbt around a pont mass or a sphercally symmetrc mass dstrbuton s movng n a gravtatonal potental of the form GM / r In ths potental t moves
More informationProblem Solving in Math (Math 43900) Fall 2013
Problem Solvng n Math (Math 43900) Fall 2013 Week four (September 17) solutons Instructor: Davd Galvn 1. Let a and b be two nteger for whch a b s dvsble by 3. Prove that a 3 b 3 s dvsble by 9. Soluton:
More informationTHE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens
THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of
More informationDIFFERENTIAL FORMS BRIAN OSSERMAN
DIFFERENTIAL FORMS BRIAN OSSERMAN Dfferentals are an mportant topc n algebrac geometry, allowng the use of some classcal geometrc arguments n the context of varetes over any feld. We wll use them to defne
More information= s j Ui U j. i, j, then s F(U) with s Ui F(U) G(U) F(V ) G(V )
1 Lecture 2 Recap Last tme we talked about presheaves and sheaves. Preshea: F on a topologcal space X, wth groups (resp. rngs, sets, etc.) F(U) or each open set U X, wth restrcton homs ρ UV : F(U) F(V
More informationCharacter Degrees of Extensions of PSL 2 (q) and SL 2 (q)
Character Degrees of Extensons of PSL (q) and SL (q) Donald L. Whte Department of Mathematcal Scences Kent State Unversty, Kent, Oho 444 Emal: whte@math.kent.edu July 7, 01 Abstract Denote by S the projectve
More informationPerron Vectors of an Irreducible Nonnegative Interval Matrix
Perron Vectors of an Irreducble Nonnegatve Interval Matrx Jr Rohn August 4 2005 Abstract As s well known an rreducble nonnegatve matrx possesses a unquely determned Perron vector. As the man result of
More informationA CHARACTERIZATION OF ADDITIVE DERIVATIONS ON VON NEUMANN ALGEBRAS
Journal of Mathematcal Scences: Advances and Applcatons Volume 25, 2014, Pages 112 A CHARACTERIZATION OF ADDITIVE DERIVATIONS ON VON NEUMANN ALGEBRAS JIA JI, WEN ZHANG and XIAOFEI QI Department of Mathematcs
More informationA padic PERRONFROBENIUS THEOREM
A padic PERRONFROBENIUS THEOREM ROBERT COSTA AND PATRICK DYNES Advsor: Clayton Petsche Oregon State Unversty Abstract We prove a result for square matrces over the padc numbers akn to the PerronFrobenus
More informationEconomics 101. Lecture 4  Equilibrium and Efficiency
Economcs 0 Lecture 4  Equlbrum and Effcency Intro As dscussed n the prevous lecture, we wll now move from an envronment where we looed at consumers mang decsons n solaton to analyzng economes full of
More informationOn cyclic of Steiner system (v); V=2,3,5,7,11,13
On cyclc of Stener system (v); V=,3,5,7,,3 Prof. Dr. Adl M. Ahmed Rana A. Ibraham Abstract: A stener system can be defned by the trple S(t,k,v), where every block B, (=,,,b) contans exactly Kelementes
More informationIntroduction to Vapor/Liquid Equilibrium, part 2. Raoult s Law:
CE304, Sprng 2004 Lecture 4 Introducton to Vapor/Lqud Equlbrum, part 2 Raoult s Law: The smplest model that allows us do VLE calculatons s obtaned when we assume that the vapor phase s an deal gas, and
More informationLecture 4: November 17, Part 1 Single Buffer Management
Lecturer: Ad Rosén Algorthms for the anagement of Networs Fall 20032004 Lecture 4: November 7, 2003 Scrbe: Guy Grebla Part Sngle Buffer anagement In the prevous lecture we taled about the Combned Input
More informationChapter 5. Solution of System of Linear Equations. Module No. 6. Solution of Inconsistent and Ill Conditioned Systems
Numercal Analyss by Dr. Anta Pal Assstant Professor Department of Mathematcs Natonal Insttute of Technology Durgapur Durgapur713209 emal: anta.bue@gmal.com 1 . Chapter 5 Soluton of System of Lnear Equatons
More informationTHERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q.
THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. IAN KIMING We shall prove the followng result from [2]: Theorem 1. (BllngMahler, 1940, cf. [2]) An ellptc curve defned over Q does not have a
More informationCALCULUS CLASSROOM CAPSULES
CALCULUS CLASSROOM CAPSULES SESSION S86 Dr. Sham Alfred Rartan Valley Communty College salfred@rartanval.edu 38th AMATYC Annual Conference Jacksonvlle, Florda November 8, 202 2 Calculus Classroom Capsules
More informationRandić Energy and Randić Estrada Index of a Graph
EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No., 202, 8896 ISSN 3075543 www.ejpam.com SPECIAL ISSUE FOR THE INTERNATIONAL CONFERENCE ON APPLIED ANALYSIS AND ALGEBRA 29 JUNE 02JULY 20, ISTANBUL
More informationThe L(2, 1)Labeling on Product of Graphs
Annals of Pure and Appled Mathematcs Vol 0, No, 05, 939 ISSN: 79087X (P, 790888(onlne Publshed on 7 Aprl 05 wwwresearchmathscorg Annals of The L(, Labelng on Product of Graphs P Pradhan and Kamesh
More informationFormulas for the Determinant
page 224 224 CHAPTER 3 Determnants e t te t e 2t 38 A = e t 2te t e 2t e t te t 2e 2t 39 If 123 A = 345, 456 compute the matrx product A adj(a) What can you conclude about det(a)? For Problems 40 43, use
More informationModule 2. Random Processes. Version 2 ECE IIT, Kharagpur
Module Random Processes Lesson 6 Functons of Random Varables After readng ths lesson, ou wll learn about cdf of functon of a random varable. Formula for determnng the pdf of a random varable. Let, X be
More information1 GSW Iterative Techniques for y = Ax
1 for y = A I m gong to cheat here. here are a lot of teratve technques that can be used to solve the general case of a set of smultaneous equatons (wrtten n the matr form as y = A), but ths chapter sn
More informationDistribution of subgraphs of random regular graphs
Dstrbuton of subgraphs of random regular graphs Zhcheng Gao Faculty of Busness Admnstraton Unversty of Macau Macau Chna zcgao@umac.mo N. C. Wormald Department of Combnatorcs and Optmzaton Unversty of Waterloo
More informationFirst day August 1, Problems and Solutions
FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 August 4, 997, Plovdv, BULGARIA Frst day August, 997 Problems and Solutons Problem. Let {ε n } n= be a sequence of postve
More informationSpectral graph theory: Applications of CourantFischer
Spectral graph theory: Applcatons of CourantFscher Steve Butler September 2006 Abstract In ths second talk we wll ntroduce the Raylegh quotent and the Courant Fscher Theorem and gve some applcatons for
More informationThe Second Eigenvalue of Planar Graphs
Spectral Graph Theory Lecture 20 The Second Egenvalue of Planar Graphs Danel A. Spelman November 11, 2015 Dsclamer These notes are not necessarly an accurate representaton of what happened n class. The
More informationTransfer Functions. Convenient representation of a linear, dynamic model. A transfer function (TF) relates one input and one output: ( ) system
Transfer Functons Convenent representaton of a lnear, dynamc model. A transfer functon (TF) relates one nput and one output: x t X s y t system Y s The followng termnology s used: x y nput output forcng
More informationLinear Approximation with Regularization and Moving Least Squares
Lnear Approxmaton wth Regularzaton and Movng Least Squares Igor Grešovn May 007 Revson 4.6 (Revson : March 004). 5 4 3 0.5 3 3.5 4 Contents: Lnear Fttng...4. Weghted Least Squares n Functon Approxmaton...
More informationPhysics 5153 Classical Mechanics. Principle of Virtual Work1
P. Guterrez 1 Introducton Physcs 5153 Classcal Mechancs Prncple of Vrtual Work The frst varatonal prncple we encounter n mechancs s the prncple of vrtual work. It establshes the equlbrum condton of a mechancal
More informationAGC Introduction
. Introducton AGC 3 The prmary controller response to a load/generaton mbalance results n generaton adjustment so as to mantan load/generaton balance. However, due to droop, t also results n a nonzero
More informationDiscrete Mathematics
Dscrete Mathematcs 3 (0) 6 40 Contents lsts avalable at ScVerse ScenceDrect Dscrete Mathematcs journal homepage: www.elsever.com/locate/dsc Hamltonan cycles wth all small even chords Guantao Chen a, Katsuhro
More informationEvery planar graph is 4colourable a proof without computer
Peter Dörre Department of Informatcs and Natural Scences Fachhochschule Südwestfalen (Unversty of Appled Scences) Frauenstuhlweg 31, D58644 Iserlohn, Germany Emal: doerre(at)fhswf.de Mathematcs Subject
More informationA 2D Bounded Linear Program (H,c) 2D Linear Programming
A 2D Bounded Lnear Program (H,c) h 3 v h 8 h 5 c h 4 h h 6 h 7 h 2 2D Lnear Programmng C s a polygonal regon, the ntersecton of n halfplanes. (H, c) s nfeasble, as C s empty. Feasble regon C s unbounded
More informationFinding Primitive Roots PseudoDeterministically
Electronc Colloquum on Computatonal Complexty, Report No 207 (205) Fndng Prmtve Roots PseudoDetermnstcally Ofer Grossman December 22, 205 Abstract Pseudodetermnstc algorthms are randomzed search algorthms
More informationNumerical Heat and Mass Transfer
Master degree n Mechancal Engneerng Numercal Heat and Mass Transfer 06FnteDfference Method (Onedmensonal, steady state heat conducton) Fausto Arpno f.arpno@uncas.t Introducton Why we use models and
More informationOn the Multicriteria Integer Network Flow Problem
BULGARIAN ACADEMY OF SCIENCES CYBERNETICS AND INFORMATION TECHNOLOGIES Volume 5, No 2 Sofa 2005 On the Multcrtera Integer Network Flow Problem Vassl Vasslev, Marana Nkolova, Maryana Vassleva Insttute of
More informationMath 261 Exercise sheet 2
Math 261 Exercse sheet 2 http://staff.aub.edu.lb/~nm116/teachng/2017/math261/ndex.html Verson: September 25, 2017 Answers are due for Monday 25 September, 11AM. The use of calculators s allowed. Exercse
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 12 10/21/2013. Martingale Concentration Inequalities and Applications
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.65/15.070J Fall 013 Lecture 1 10/1/013 Martngale Concentraton Inequaltes and Applcatons Content. 1. Exponental concentraton for martngales wth bounded ncrements.
More informationSPECIAL SUBSETS OF DIFFERENCE SETS WITH PARTICULAR EMPHASIS ON SKEW HADAMARD DIFFERENCE SETS
SPECIAL SUBSETS OF DIFFERENCE SETS WITH PARTICULAR EMPHASIS ON SKEW HADAMARD DIFFERENCE SETS ROBERT S. COULTER AND TODD GUTEKUNST Abstract. Ths artcle ntroduces a new approach to studyng dfference sets
More informationThe Geometry of Logit and Probit
The Geometry of Logt and Probt Ths short note s meant as a supplement to Chapters and 3 of Spatal Models of Parlamentary Votng and the notaton and reference to fgures n the text below s to those two chapters.
More informationSolutions for Tutorial 1
Toc 1: Semdrect roducts Solutons for Tutoral 1 1. Show that the tetrahedral grou s somorhc to the semdrect roduct of the Klen four grou and a cyclc grou of order three: T = K 4 (Z/3Z). 2. Show further
More information