Links in edge-colored graphs

Transcription

1 Lnks n edge-coloed gaphs J.M. Becu, M. Dah, Y. Manoussaks, G. Mendy LRI, Bât. 490, Unvesté Pas-Sud 11, Osay Cedex, Fance Astact A gaph s k-lnked (k-edge-lnked), k 1, f fo each k pas of vetces x 1, y 1,, x k, y k, thee exst k pawse vetex-dsjont (espectvely edge-dsjont) paths, one pe pa x and y, = 1,,, k. Hee we deal wth the popely-edge-coloed veson of the k-lnked (kedge-lnked) polem n edge-coloed gaphs. In patcula, we gve condtons on coloed degees and/o nume of edges, suffcent fo an edge-coloed multgaph to e k-lnked (k-edge-lnked). Some of the otaned esults ae the est possle. Related conjectues ae poposed. 1 Intoducton and notaton The nvestgaton of k-lnkngs fo non coloed gaphs gave some mpotant and nteestng esults oth fom a mathematcal and algothmc pont of vew [6, 7, 8, 9, 10, 14, 15, 16, 17]. Hee we deal wth the coloed veson of the k-lnked polem n edge-coloed multgaphs. In the case of edge-coloed complete gaphs, some esults of algothmc natue fo the k-lnked polem wee aleady otaned n [11]. The study of ths type of polems has wtnessed sgnfcant development dung last decades, oth fom the pont of vew of ts theoetcal nteest and of ts domans of applcatons. In patcula, polems asng n molecula ology ae often modeled usng coloed gaphs,.e., gaphs wth coloed edges and/o vetces [13]. Gven such an edge-coloed gaph, ognal polems coespond to extact sugaphs coloed n a specfed patten. The most natual patten n such a context s that of a pope colong,.e., adjacent edges havng dffeent colos. Vaous applcatons of popely edge-coloed Hamltonan and Eulean cycles and paths ae studed n [1, 13]. Popely coloed paths and cycles have also applcatons n vaous othe felds, as n VLSI fo compactng a pogammale logcal aay [5]. Although a lage ody of wok has aleady een done [1,, 3, 4, 11], n most of that pevous wok the nume of colos was estcted to two. Fo nstance, whle t s well known that popely edge-coloed hamltonan cycles can e found effcently n -edge coloed complete gaphs, t s a long standng queston whethe thee exsts a polynomal algothm fo fndng such hamltonan cycles n edge-coloed complete gaphs wth thee colos o moe [3]. In ths pape we consde gaphs wth edges coloed wth an atay nume of colos. In patcula, we study condtons on coloed degees and/o edges suffcent fo an edge-coloed multgaph to e k-lnked (k-edge-lnked). Cuent addess: SNCF-Techncente de Neves 1, Rue Benot Fachon Vaennes-Vauzelles Cedex, Fance. Cuent addess: D.M.I., Faculty of Scence and Technology, Unvesty of Nouakchott, BP 506, Nouakchott, Mautana. Coespondng autho. E-mal addess: yanns@l.f Also Laoatoe d Infomatque et de Réseaux Télécoms (LIRT) Dépatement Géne Infomatque de L Ecole Supéeue Polytechnque de Daka B.P Fann, Daka. 1

2 Fomally, let {1,,, c} e a set of gven c colos. Thoughout the pape, G c denotes an edge-coloed multgaph so that each edge s coloed wth some colo {1,,, c} and no two paallel edges jonng the same pa of vetces have the same colo. We also suppose that G c has no solated components,.e., the undelng non-coloed gaph s connected. The vetex and edge-sets of G c ae denoted V (G c ) and E(G c ), espectvely. The ode n of G c s the nume of ts vetces. The sze m of G c s the nume of ts edges. Fo a gven colo, E (G c ) denotes the set of edges of G c on colo. When no confuson ases, we wte V, E and E nstead of V (G c ), E(G c ) and E (G c ), espectvely. When G c s not a multgaph,.e., no paallel edges etween any two vetces ae allowed, we call t gaph, as usual. If H s an nduced sugaph of G c, then NH (x) denotes the set of vetces of H, joned to x wth an edge on colo. The coloed degee of x n H, denoted y d H (x) coesponds to the cadnalty NH (x) of NH (x). Wheneve H = G c, fo smplcty, we wte N (x) (esp. d (x)) nstead of NG c(x) (esp. d Gc(x)). Fo a gven vetex x and a gven postve ntege k, the notaton d c (x) k means that fo evey {1,,, c}, d (x) k. An edge etween two vetces x and y s denoted y xy and ts colo y c(xy). Fo two gven vetces x and y and a gven colo, some tmes, to help eadng, we use the notaton x y nstead of xy E (G c ). A sugaph of G c s sad to e popely edge-coloed, f any two adjacent edges n ths sugaph dffe n colo. A popely edge-coloed path does not allow vetex epettons and any two successve edges on ths path dffe n colos. The length of a path s the nume of ts edges. A gaph s k-lnked (k-edge-lnked) wheneve fo evey k dsjont pas of vetces x 1, y 1, x, y,, x k, y k, thee exst k vetex-dsjont (edge-dsjont) popely edge-coloed paths, one pe pa x and y, = 1,,, k. The pape s oganzed as follows: In Secton, we gve condtons on coloed degees, suffcent fo the k-lnked (k-edge lnked) popety. In Secton 3, we gve condtons nvolvng oth mnmum coloed degees and the nume of edges, suffcent fo the k-lnked (k-edgelnked) popety. One of the esults of ths secton s a patal answe to an old queston y one of the authos pulshed n [10]. Though oth sectons, seveal conjectues ae poposed. Degee condtons fo k-lnked edge-coloed multgaphs Let us stat wth the followng conjectue fo k-lnked edge-coloed multgaphs nvolvng coloed degees. Conjectue.1. Let G c e a c-edge-coloed multgaph of ode n and k a non-zeo postve ntege. Thee exsts a mnmum functon f(n, k) such that f fo evey vetex x, d c (x) f(n, k), then G c s k-lnked. Poaly n the aove conjectue t suffces to set f(n, k) = n + k 1. Indeed, let A, (esp. B, C) e a complete edge-coloed multgaph of ode n k+ n k+ (espectvely, k, ). Consde the dsjont unon of A, B, C and suppose that each vetex of B s joned to each vetex of A C y c paallel edges all on dstnct colos. Although the esultng multgaph has coloed degees at least n +k, t has no k vetex-dsjont popely edge-coloed paths etween pas of vetces x and y, whee x 1 s a vetex n A, y 1 s a vetex n C and the emanng x, y vetces elong to B, k. Some suppot to the aove conjectue may e otaned fom the followng theoem.

3 Theoem.. Let G c e a c-edge-coloed multgaph of ode n, wth n 4k, k a non-zeo postf ntege. If fo evey vetex x, d c (x) n + k 1, then Gc s k-lnked. Poof. Let G c e a c-edge-coloed multgaph of ode n, n 4k, such that fo evey vetex x, d c (x) n + k 1. We ae gong to pove y contadcton that Gc s k-lnked. Moe pecsely, we wll pove the stonge esult that, gven k pas of vetces x 1, y 1,, x k, y k of G c, each pa x and y s joned y a popely edge-coloed path of length at most 8. Assume theefoe that G c s not k-lnked. Then thee ae k dstnct vetces x 1, y 1, x, y,..., x k, y k such that thee ae no k pawse vetex-dsjont paths, one path pe pa x, y. Let us consde now a set I of nteges, such that thee ae I pawse vetex-dsjont paths of length at most 8 jonng the pas x, y, wth I. We consde I such that ts cadnalty I s the maxmum possle. Clealy I < k, fo othewse we ae fnshed. In the est of the poof we ae gong to show that we can fnd moe than I pawse vetex-dsjont paths as long as the cadnalty of I s consdeed stctly smalle than k. Ths wll contadct the maxmalty popety of I and end the poof. Clam 1. I. Poof. Assume that thee exsts at least one pa x, y, say x 1 and y 1, of vetces such that thee s no edge etween x 1 and y 1. Othewse, we ae fnshed, y consdeng the k paths defned y the k edges x y, = 1,, k. Set S = {x, y, 1 k} and let (ed) and (lue) e two fxed colos n {1,,, c}. So d G S (x 1) n + k 1 (k 1) = n k + 1. Smlaly, d G S (y 1) n k+1. Howeve N G S (x 1) NG S (y 1) n k, so NG S (x 1) NG S (y 1). Consequently, we can fnd two dstnct vetces, say u and v, n NG S (x 1) NG S (y 1). If thee s an edge etween x and y, then ths edge x y togethe wth the path x 1 u y1 pove that I. If not, then thee ae two dstnct vetces u and v n NG S (x ) NG S (y ). W.l.o.g. we may suppose u v, ut then we have found agan two paths x 1 u y1 and x v y, as desed. As I < k, n the sequel, let us suppose w.l.ofg. that 1 / I. In ode wods, we suppose that thee s no popely-edge coloed path of length at most 8 etween x 1 and y 1. Let X e the set of vetces whch ae used n ode to uld the I pawse vetex-dsjont pawse vetex-dsjont paths of length at most 8, one pe pa x and y, wth I. Clealy X 7k. Set A = N (x 1 ) (S X) and B = N (y 1 ) (S X). Then A B =, fo othewse f thee s a vetex z A B, then the path x 1 zy 1 s of length two, a contadcton to the choce of x 1 and y 1. Also A n + k 1 X S = n 8k 1. Smlaly, B n 8k 1. Set C = G (A B X S). We have A n k 1 X and B n k 1 X. Thus C = n A B S X + X, hence C 8k 1 and k 3. We dstngush now etween two Cases (I) and (II) dependng upon A and B. (I) Thee s no edge etween A and B. 3

4 Fo a colo {, } and fo each vetex x A and y B, d A C(x) = d G (B X S) (x) = d G (X S) (x) n + k 1 7k k = n 8k 1 (1) and d B C(y) = d G (A X S) (y) = d G (X S) (y) n 8k 1 () Clam. Fo evey pa u, v of dstnct vetces n A (epectvely n B), thee ae at least 3k dstnct ed-lue paths etween u and v (the ode of the colos s mpotant hee) and at least 3k dstnct lue-ed paths etween u and v. These paths do not go though a vetex of X S. Poof. We wll pove only the ed-lue case, the lue-ed case eng smla y ntechangng the ed-lue colos and applyng the same aguments. Let u and v e two dstnct vetces of A. We have d G (B X S) (u)+d G (B X S) (v) n 16k. Howeve G (B X S) < n k +1 and n 16k ( n k + 1) = n 15k 3 > 3k. Theefoe thee ae at least 3k dstnct ed-lue paths etween u and v. These paths do not go though a vetex of X S. We otan the same esults fo the ed-lue paths etween u and v n B. Clam 3. N (x 1 ) B = and N (y 1 ) A =. Poof. Assume that N (x 1 ) B s not empty. Let y e a vetex n N (x 1 ) B. As n 8k 1 > 8k, we have d B C (y) n 8k 1 C. Consequently thee s a vetex y B such that the edge yy s ed. But then we can consde the path x 1 y y y 1, a contadcton to the hypothess that thee s no path etween x 1 and y 1 of length less than 8 n G c. We otan the same esult wheneve N (y 1 ) A. Ths completes the poof of the clam. Let us set now Φ = A N (x 1 ) and Ψ = B N (y 1 ). Consde two vetces x Φ and y Ψ. Fo {, }, we have d G S (x) = d G (B S) (x) n + k 1 k + 1 = n k, snce x / N (y 1 ). Smlaly, d G S (y) = d G (A S) (x) n + k 1 k + 1 = n k, snce y / N (x 1 ). Consequently d G S (x) + d G S (y) n k. Moeove (S {x, y}) (N G S (x) N G S (y)) = and so NG S (x) N G S (y) n k. In concluson, N G S (x) N G S (y) and NG S (x) N G S (y). If N G S (x) N G S (y) s not a suset of X, then y consdeng some vetex z NG S (x) N G S (y) X we defne the path x 1 z y y 1 of length less than 8, a contadcton. So, assume that NG S (x) N G S (y) X and N G S (x) N G S (y) X Let Γ xy a suset of I such that I f and only f thee s a vetex z ( NG S (x) N G S (y)) ( N G S (x) NG S (y)) and wth the popety that any path etween x and y goes though z. Clealy, Γ xy s not empty. We defne also Γ to e a suset of I such that Γ f and only f thee ae at least two dstnct pas of vetces x, y and x, y, wth {x, x } Φ and {y, y } Ψ and satsfes Γ xy Γ x y. Let X Γ e the set of vetces whch ae used n ode to uld the Γ pawse vetex-dsjont paths of length at most 8 jonng the pas x, y, wth Γ. Thus X Γ X. Clam 4. Γ s not empty. 4

5 Poof. We have Φ n 8k 7k > 3k, snce Φ = A N (x 1 ) and NA (x 1) n + k 1 X C S = n 16k 1 3k. Also Ψ > 3k. Thus thee ae at least 3k pas of dstnct vetces n Φ Ψ. Howeve I < k. So Γ s not empty. Clam 5. Fo each Γ and any choce of two dstnct colos j and l, ethe N j (x ) Φ = and N l (y ) Ψ = o N j (x ) Ψ = and N l (y ) Φ =. Poof. Snce A B > n 16k, we may consde that thee ae at least thee vetces u, u and u of N (x ) whch elong ethe to A o to B. Assume that these thee vetces ae n A. We must show that thee s no edge etween y and A. Indeed, assume that thee s an edge vy n G c, v A. W.l.o.g. we may suppose that u v and that thee ae two vetces x Φ and y Ψ such that Γ xy and x u and x v. If c(vy ) =, then, y Clam 3, thee s a vetex w, dstnct fom u, v, x such that the path x u w v y exsts n G c. If c(vy ) =, then thee s a vetex w dstnct fom u, v, x n A (see (1)) and a vetex w dstnct fom u, v, w, x n A such that the path x u w w v y (Clam 3) exsts n G c. But then, we may otan the paths x 1 z y y1 o x 1 z y y1 (whee z s a vetex used etween x and y ) whch s n contadcton wth ou assumpton that thee s no path of length less than 8 etween x 1 and y 1. Ths completes the poof of ths clam. Clam 5 means that thee s no vetex n Φ (espectvely n Ψ) havng oth x and y as neghos, fo any Γ. As thee s no edge etween Φ and y 1 and no edge etween Ψ and x 1, then fo {, }, and fo each vetex x of Φ and each vetex y of Ψ we have, d G S(x) = d A C X(x) n + k 1 (k 1) + Γ = n k + Γ and d G S(y) = d B C X(y) n k + Γ. By summng the aove nequaltes we otan d G S (x) + d G S (y) n k + Γ. Also (S {x, y}) (NG S (x) + N G S (y)) =. Consequently N G S (x) N G S (y) n k. In concluson, we otan NG S (x) N G S (y) + Γ and N G S (x) N G S (y) + Γ. Fom now on, we ae gong to defne I + 1 dsjont paths each of length at most 8. Ths wll contadct the maxmalty popety of I and wll pemt to complete the poof of case (I). Wthout loss of genealty, let us set Γ = {, 3,..., Γ + 1}. Futhemoe, snce x and y play a symmetc ole, we may suppose that fo each Γ and fo any choce of two colos j and l, we have N j (x ) Φ = and N l (y ) Ψ =. Snce N (x ) A k and N (y ) B k, we can fnd Γ +1 dstnct vetces x, x, x 3,..., x Γ +1 n A and Γ +1 dstnct vetces y, y, y 3,..., y Γ +1 n B such that c(x x ) = c(x 1x ) = and c(y y ) = c(y 1y ) =. Recall also that fo evey pa of vetces x Φ, y Ψ, we have NG S (x) N G S (y) + Γ. In addton Φ 3k and Ψ 3k. Consequently we can fnd Γ dstnct vetces x, x 3,..., x Γ +1 n A, Γ dstnct vetces y, y 3,..., y Γ +1 n B and Γ + 1 dstnct vetces z, z, z 3,..., z Γ +1 n X Γ. Indeed, thee ae at least k pas of dstnct vetces n Φ Ψ and thee ae at most I Γ pas 5

6 of dstnct vetces of Φ Ψ whch ae joned y paths x z y and x z y of length two gong though X X Γ accodng to the defnton of Γ. Also all vetces of these paths ae dstnct fom the pevous ones. Fnally, accodng to the Clam, we can fnd Γ dstnct vetces x (3),..., x(3) Γ +1 n A and Γ dstnct vetces y(3), y(3) 3,..., y(3) Γ +1 n B, such that, x(3) 3 the paths x (3) and y y (3) fom the pevous ones. In ths way we defned the I + 1 dstnct paths and x (3) y ae n Gc and all vetces of these paths ae dstnct x 1 z y y (3) z y y1. y y As each of the aove paths has length less than 8, ths s a contadcton to the maxmalty popety of I. (II) Thee s at least one edge etween A and B. Clam 6. Thee ae two susets D and E of V (G c ) such that ) D A, E B, D 3k, E 3k and D E n + k, ) fo each vetex x D, N (x) A = and fo each vetex y E, N (y) B = and ) fo each pa of vetces x, x of D, N B (x) N B (x ) 3k and fo each pa of vetces y, y of E, N A (y) N A (y ) 3k. Poof. Let xy e an edge etween A and B, x A, y B. Assume w.l.o.g. that ts colo s lue. Then, thee s no ed edge etween y and B, fo othewse the path x 1 y z y1, z B, has length less than 8, a contadcton to ou assumptons. So, d A (y) = d A B (y) n + k 1 9k 8k + 1 = n 16k and fo each vetex x N A (y), d B (x) n 16k. Let E e a suset of B such that evey vetex u of E has at least 3 ( n 4k) neghos v n NA (y), the colo of uv s lue and suject to ths equement E s as g as possle. We must fst show that such a set E exsts and E n 60k 3k. We have B n S mn( A ) n + 6k. The wost case ases when each vetex u of N A (y) s joned wth monochomatc lue edges to each vetex of E and dstute the est of the colos on edges (f any) jonng u wth the emanng vetces of B. In fact we must show that the aveage of the lue edges etween a vetex of B E and NA (y) s at least 3 ( n 4k). In patcula, we must pove that, f E = n αk, then (d A (y) E ) d A (x) max( B ) E 3 (n 4k) (α 16)k ( n 16k) (α + 6)k 3 (n 4k) αn 10n 35k 6 In patcula, fo E = n 60k we otan α = 60 and then the pevous equaton s tue. Also fo each y E, d A (y) 3 ( n 4k) and d A (y) n 16k. Wth smla aguments we defne the 6

7 suset D of A, such that fo x NA (y), evey vetex u of D s connected to at least 3 ( n 4k) vetces v of NA (x) wth ed edges and D s as maxmum as possle. Then D n 60k 3k and fo evey x D, d B (x) 3 ( n 4k) and d B (x) n 16k. Moeove, fo each vetex x D, N (x) A = and fo each vetex y E, N (y) B =. In addton, fo each pa x, x of vetces of D, d B (x) 3 ( n 4k) and d B (x) n 16k. So d B (x)+d B (x) 5 6n 3k. Howeve B n +6k. Thus N B (x) N B (x ) n 3 38k 3k. Smlaly, fo evey pa y, y of vetces of E, we have NA (y) N A (y ) 3k. Then D E n+k snce D E n 10k n +k. Ths completes the poof of the clam. Let x, y e two vetces of G c, x D, y E. Clealy {x 1, y 1 } N (x) =, fo othewse one of the paths x 1 y1 o x 1 y y 1, y E) exsts n G c, a contadcton to the assumpton that thee s no path etween x 1 and y 1 of length less than 8. Smlaly, {x 1, y 1 } N (y) =. We have d G c (S {y}) (x) n + k 1 (k 1) 1 = n k. Analogously d G c (S {x}) (y) n k. We also have d G c (S {y}) (x)+d G c (S {x}) (y) n k and (S {x, y}) (NG c (S {y}) (x) N G c (S {x}) (y)) =. So N G c (S {y}) (x) N G c (S {x}) (y) n k. In concluson, we otan NG c (S {y}) (x) N G c (S {x}) (y). If N G c (S {y}) (x) N G c (S {x}) (y) s not a suset of X, then we can consde the path x 1 z y y1, contadctng agan ou assumptons. So we assume that NG c (S {y}) (x) N G c (S {x}) (y) X. Let Ω xy e a suset of I such that Ω xy f and only f thee s a vetex z NG c (S {y}) (x) N G c (S {x}) (y) such that the path etween x and y goes though z. The set Ω xy s not empty. We also defne a suset Ω of I such that Ω f and only f thee ae at least fou dstnct vetces x, x, y, y such that x, x D, y, y E and Ω xy Ω x y. Let X Ω e the set of vetces whch ae used n ode to uld the Ω pawse vetex-dsjont paths of length at most 8 jonng the pas x, y, wth Ω. Hence, X Ω X. Clam 7. Ω s not empty. Poof. Staghtfowad fom the fact that D k, E k and I k. Clam 8. Fo evey n Ω, ethe (N (x ) N (y )) E = and ( N (x ) N (y ) ) D = o (N (x ) N (y )) D = and ( N (x ) N (y ) ) E =. Poof. Snce D E > n +k, we can clam that thee ae at least thee vetces u, u and u of N (x ) whch elong ethe to D o to E. Assume that these thee vetces ae n A. We wll show y contadcton that thee s no ed edge etween y and E. Assume theefoe that thee s a ed edge vy n G c, wth v E. Accodng to Clam 6, thee s a lue edge vw n G c, w D. W.l.o.g. let us suppose that u w and that thee ae two vetces x D, y E such that Ω xy, x u, x w and y v. Moeove, accodng to Clam 6, thee s a vetex t of E such that t u, t y, and the path u t w exsts n G c. But then we may defne the paths x u t w v y and x 1 z y y1, whee z s a vetex used y the path etween x and y, a contadcton to ou assumptons. So (N (x ) N (y )) E = o (N (x ) N (y )) D =. Wth the same agument, we otan that ( N (x ) N (y ) ) E = o ( N (x ) N (y ) ) D =. Assume that (N (x ) N (y )) E = and ( N (x ) N (y ) ) E =. Then thee ae two vetces x D, y E such that Ω xy. Smlaly thee ae two dstnct vetces u and v of D such 7

8 that x u, v and u N (x ) and v N (y ). Moeove, accodng to Clam 6, thee s a vetex w of E such that w y and the path u t v exsts n G c. But then we may defne the set of paths x u w v y and x 1 z y 1, whee z s a vetex used y the path etween x and y, agan a contadcton to ou assumptons. The completes the poof of Clam 8. Now we can defne two susets Ω and Ω of Ω as follows: Ω f and only f Ω, (N (x ) N (y )) E = and ( N (x ) N (y ) ) D =. Smlaly, Ω f and only f Ω, (N (x ) N (y )) D = and ( N (x ) N (y ) ) E =. Accodng to Clam 8, we have Ω Ω = and Ω Ω = Ω. Moeove Ω, snce NG c (S {y}) (x) N G c (S {x}) (y). Wth the dfntons aove, Clam 8 means that, f Ω, then thee ae no lue edges etween D and {x, y } and no ed edges etween E and {x, y }. Smlaly, f Ω, then thee ae no ed edges etween D and {x, y } and no lue edges etween E and {x, y }. Thus, fo evey vetex x of D, d G c (S {y}) (x) n + k 1 (k 1) 1 + Ω = n k + Ω, d G c (S {y}) (x) n + k 1 k 1 + Ω = n k + Ω. Smlaly, fo evey vetex y n E, d G c (S {x}) (y) n + k 1 k 1 + Ω = n k + Ω, d G c (S {x}) (y) n + k 1 (k 1) 1 + Ω = n k + Ω. Fom the aove nequaltes we otan d G c (S {y}) (x) + d G c (S {x}) (y) n k + 4 Ω. Futhemoe (S {x, y}) (NG c (S {y}) (x) N G c (S {x}) )(y) =. So N G c (S {y}) (x) NG c (S {x}) (y) n k. In concluson we otan N G c (S {y}) (x) N G c (S {x}) (y) 4 Ω + and N G c (S {y}) (x) N G c (S {x}) (y) 4 Ω. We dstngush now etween two cases dependng upon the cadnalty of Ω. Case 1. Ω = 0. As Ω = 0, t follows that Ω = Ω. Now, we ae gong to defne Ω + 1 pawse vetex dsjont paths each of length at most 8. As Ω s a suset of I, ths wll e a contadcton wth ts maxmalty popety. Set Ω = {, 3,..., Ω +1}. Snce N (x ) D k and N (y ) E k, we can fnd Ω + 1 dstnct vetces x, x, x 3,..., x Ω +1 of D and Ω + 1 dstnct vetces y, y, y 3,..., y Ω +1 of B such that c(x x ) = c(x 1x ) = and c(y y ) = c(y 1y ) =. Recall also that fo evey two vetces x D, y E, we have NG c (S {y}) (x) N G c (S {x}) (y) 4 Ω + and D 3k and E 3k. Thus we can fnd Ω vetces x, x 3,..., x Ω +1 of D, Ω vetces y, y 3,..., y Ω +1 of E and Ω + 1 vetces z, z, z 3,..., z Ω +1 of X Ω such that the dstnct paths x z y and x z y exst n G c. All the aove mentonned sets of vetces exst n G c, snce thee ae at least k pas of dstnct vetces of D E and thee ae at most I Ω pas of dstnct vetces of D E whch ae joned y a path of length two 8

9 gong though X X Ω accodng to the defnton of Ω. Last, accodng to Clam 6, we can fnd Ω vetces x (3), x(3) 3,..., x(3) Ω +1 of D and Ω vetces y (3), y(3) 3,..., y(3) Ω +1 of E, such that the paths x (3) and y y (3) y exst n Gc. But n that way we may defne the followng Ω + 1 pawse vetex-dsjont paths and a contadcton. x (3) x 1 z y y (3) z y y1, y y Case. Ω > 0 The poof of ths second case s ased on Clams 9-1 elow. Clam 9. Thee ae at least thee dsnct pas of vetces u j and v j, j = 1,, 3, such that Ω uj v j Ω. Poof. Let X Ω e the set of vetces whch ae used n ode to defne Ω pawse vetexdsjont paths of length at most 8, one pe pa x, y, wth Ω. Then X Ω X Ω. Indeed assume thee ae at most two dstnct pas of vetces u j, v j such that Ω uj v j Ω, j = 1,. Then, y usng aguments smla to those of Case 1, we can defne Ω + 1 pawse vetex dsjont paths etween x 1 and y 1 and etween x and y of length at most 8, fo evey Ω. To defne these paths, we need to fnd Ω vetces x, x 3,..., x Ω +1 n D, Ω vetces y, y 3,..., y Ω +1 n E and Ω + 1 vetces z, z, z 3,..., z Ω +1 n X Ω such that the followng paths x z y and x z y exst n G c. Indeed all the aove-mentonned sets of vetces exst n G c ecause, as thee ae at least k pas of vetces of D E, then thee ae at most I Ω pas of dstnct vetces of D E whch ae joned y paths of length two gong though X X Ω accodng to the defnton of Ω. Also, as thee ae at most pas of dstnct vetces of D E whch ae joned y paths of length two gong though X X Ω, then thee ae at least Ω + 1 pas of dstnct vetces (u j, v j ) of D E such that Ω u j v Ω fo evey j j. Ths completes the poof of the clam. Clam 10. NG c (S {y}) (x) N G c (S {x}) (y) X. Poof. Assume that thee s a vetex z of NG c (S {y}) (x) N G c (S {x}) (y) such that z C A B. Snce Ω > 0, thee s an ntege t Ω such that we can fnd fou dstnct vetces x, u D and y, v E such that t Ω x y, u N (x t ). Also we can fnd two vetces v N (y t ) such that x, x, u, z (espectvely y, y, v, z) ae pawse dstnct. Let z t e a vetex of the path etween x t and y t such that z t NG c (S {y }) (x ) NG c (S {x }) (y ). By usng ths vetex z t and accodng to Clam 6, we can fnd two vetces u and w such that oth paths x t u u z y v v yt and x 1 z t y y 1 exst n G c. Let Θ xy a suset of I such that Θ xy f and only f thee s a vetex z N G c (S {y}) (x) N G c (S {x}) (y) such that the path etween x and y goes though z. Θ xy s not empty, snce N G c (S {y}) (x) N G c (S {x}) (y) 4 Ω and Ω > 0. We defne also a suset Θ of I 9

10 such that Θ f and only f thee ae at least fou vetces x, x Φ, y, y Ψ wth x x and y y such that Θ xy Θ x y. Let X Θ e the set of vetces whch ae used n ode to uld the Θ pawse vetex-dsjont paths each of length at most 8 jonng the pas x, y, wth Θ. Clealy X Θ X. Clam 11. Θ s not empty. Poof. As D k, E k and I k, the concluson s staghtfowad. Clam 1. Fo evey Θ, ethe (N (x ) N (y )) E = and ( N (x ) N (y ) ) D = o (N (x ) N (y )) D = and ( N (x ) N (y ) ) E =. Poof. Let Θ. If Ω the Clam s tue accodng to Clam 7. Assume theefoe that / Ω. Snce D E > n + k, thee ae at least thee vetces u, u and u of N (x ) whch elong ethe to D o to E. Assume that these thee vetces ae n D. We must show that thee s no ed edge etween y and E. Assume y contadcton that thee s a ed edge vy n G c wth v E. Accodng to Clam 6, thee s a lue edge vw n G c, w D. W.l.o.g. we may assume that u w and that thee ae two vetces x D, y E such that Θ xy, x u, x w and y v. Snce Ω > 0, y Clam 10 we can fnd an ntege t Ω and two vetces x t D and y t E such that t Ω x t y t. Futhemoe x, x t, w (espectvely y, y t, v) ae pawse dstnct. In addton, we can also fnd two dstnct vetces x t and y t such that x t N (x t ) and y t N (y t ). Now, accodng to Clam 6, thee ae two vetces w, x (3) t of E and one vetex x (3) t of D, dstnct fom the aove-mentonned ones, and such that the paths u w w, x t (3) t and y t y (3) t y exst n G c. Howeve n that way we may defne the paths x u w w v y, x 1 t zt y t 1 and x t t (3) t (3) z y y y y 1 ( z s a vetex used etween x and y and z t a vetex used etween x t and y t ) a contadcton to ou assumptons. Thus we may conclude that ethe (N (x ) N (y )) E = o (N (x ) N (y )) D =. By smla aguments we otan ( N (x ) N (y ) ) E = o ( N (x ) N (y ) ) D =. In ode to complete the pood we need to exclude the case (N (x ) N (y )) E = and ( N (x ) N (y ) ) E =. Assume theefoe that (N (x ) N (y )) E = and ( N (x ) N (y ) ) E =. Thee ae two vetces x D, y E such that Ω xy. In addton, thee ae two dstnct vetces u and v of D such that x u, v, u N (x ) and v N (y ). But, then we may defne the paths x u w v y, x 1 t zt y t 1 and x t t (3) t (3) z y y agan a contadcton to ou assumptons. Ths completes the poof of Clam 1. y y 1, Let us now set Λ = Ω Θ. We defne two new susets Λ and Λ of I as follows: We let Λ f and only f Λ, (N (x ) N (y )) E = and ( N (x ) N (y ) ) D =. Smlaly, Λ f and only f Λ, (N (x ) N (y )) D = and ( N (x ) N (y ) ) E =. Accodng to these defntons and Clam 8, we have Λ Λ = and Λ Λ = Λ. Now, n tems of Λ and Λ, Clams 8 and 13 mean that, f Λ, then thee ae no lue edges etween D and {x, y } and no ed edges etween E and {x, y }. Smlaly, f Λ, then thee ae no ed edges etween D and {x, y } and no lue edges etween E and {x, y }. We ecall that thee s no lue edge etween D and {x 1, y 1 } and no ed edge etween E and {x 1, y 1 }). 10

11 Thus fo each vetex x of D we have, d G c (S {y}) (x) n + k 1 (k 1) 1 + Λ = n k + Λ, d G c (S {y}) (x) n + k 1 k 1 + Λ = n k + Λ. Smlaly, fo each vetex y of E, d G c (S {x}) (y) n + k 1 k 1 + Λ = n k + Λ, d G c (S {x}) (y) n + k 1 (k 1) 1 + Λ = n k + Λ. By summng the aove nequaltes we otan, d G c (S {y}) (x) + d G c (S {x}) (y) n k + 4 Λ. Also, (S {x, y}) (NG c (S {y}) (x) N G c (S {x}) )(y) =. So N G c (S {y}) (x) N G c (S {x}) (y) n k. We conclude that N G c (S {y}) (x) N G c (S {x}) (y) 4 Λ +. Smlaly, NG c (S {y}) (x) N G c (S {x}) (y) 4 Λ. Let X Λ e the set of vetces whch ae used n ode to defne the Λ pawse vetexdsjont paths of length at most 8, jonng the pas x, y, fo each Λ. We have X Λ X. In the sequel, we shall defne Λ + 1 dstnct paths each of length at most 8. Ths wll contadct the maxmalty popety of Λ, and wll pemt to complete the poof of the theoem. Set Λ = {, 3,..., Λ + 1}. Recall that fo each Λ, N (x ) D k and N (y ) E k. Respectvely, fo each Λ, N (x ) D k and N (y ) E k. Thus, we can fnd Λ + 1 dstnct vetces x, x, x 3,..., x Ω +1 n D and Λ + 1 vetces y, y, y 3,..., y Ω +1 n B. Futhemoe, fo evey Λ, c(x 1 x ) =, c(y 1 y ) =, c(x x ) =. Also fo evey Λ, c(x x ) =. In addton, fo evey Λ, c(y y ) =. Fnally, fo evey Λ, c(y y ) =. Moeove, we can fnd Λ vetces x, x 3,..., x Λ +1 of D, Λ vetces y, y 3,..., y Λ +1 of E and Λ + 1 vetces z, z, z 3,..., z Λ +1 of X Λ such that the paths x z y f Λ o x z y f Λ and x z y exst n G c. Indeed, thee ae at least k pas of dstnct vetces of D E and thee ae at most I Λ pas of dstnct vetces of D E whch ae joned y a path of length two gong though X X Λ accodng to the defnton of Λ. Assume fst Λ Λ. Fst, we consde the vetces x, y, z wth Λ (ecall that fo x D, y E, we have NG c (S {y}) (x) N G c (S {x}) (y) 4 Λ Λ ). Next we consde the vetces x, y, z wth Λ (agan fo x D, y E, t holds NG c (S {y}) (x) N G c (S {x}) (y) 4 Λ + Λ + Λ ). Assume next Λ < Λ. In that case, we consde fst the vetces x, y, z wth Λ (ecall agan that fo x D, y E, NG c (S {y}) (x) N G c (S {x}) (y) 4 Λ + Λ ). Next we consde the vetces x, y, z wth Λ (as fo x D, y E, t holds NG c (S {y}) (x) N G c (S {x}) (y) 4 Λ Λ + Λ ). Fnally, accodng to Clam 6, we can fnd Λ vetces x (3) Λ vetces y (3), y(3) 3,..., y(3) Λ +1 follows: If Λ we defne the paths x Futehmoe, f Λ, we defne y y (3), x(3) 3,..., x(3) Λ +1 of D and of E. By usng these vetces we may defne a set of paths as (3) y o f Λ we defne y o f Λ, then x y (3) (3) y.. 11

12 In that way we may defne Λ + 1 dstnct paths as follows: Fo evey Λ, x (3) z y y (3) Fo evey Λ, Fnally, fo = 1 x (3) x 1 z y y (3) z y y1. y y y. y. Ths contadcts the maxmalty popety of Λ and completes the poof of the theoem. Theoem.3. Let G c e a c-edge-coloed multgaph of ode n and k a non-zeo postve ntege. If fo evey vetex x, d c (x) n, then Gc s k-edge-lnked. Poof. Let x, y, 1 k, e k dstnct vetces of G c. We shall pove a stonge esult, namely, that we can fnd k pawse edge-dsjont paths of length at most two, one pe pa x and y. Assume fst that fo some, the edge x y exsts n G c. Then ths edge defnes a path etween x and y. Ths choce wll not affect the est of the poof, as any path etween anothe pa of vetces x j, y j, 1 j k gong though the edge x y has length at least thee. In the sequel, we can theefoe assume that thee ae no edges x y n E(G c ), fo each = 1,,, k. Let us choose two colos, say (ed) and (lue). As d (x ) n and d (y ) n, we otan d (x ) + d (y ) n, fo each = 1,..., k. As thee s no edge x y, we can fnd two dstnct vetces, say a and, n G c such that a N (x ) N (y ) and N (x ) N (y ). If a / {x j, y j, 1 j k}, then we consde the path x a y. These two edges x a and a y ae not used y paths jonng othe pas of vetces x j, y j, j, snce we clam that the length of these othe paths s at most two. Afte the choce of such paths, t suffces to constuct pawse edge-dsjont paths wth the emanng pas of vetces x, y such that {a,, 1 k} {x, y, 1 k}. We shall complete the poof y showng that, n the wost case (whch s {a,, 1 k} {x, y, 1 k}), we can constuct k pawse edge-dsjont paths of length at most two, one pe pa x, y, fo each = 1,..., k. Assume theefoe that fo any and j, 1 j k, we have x j N (x ) N (y ) o y j N (x ) N (y ). Now, let us choose and goup togethe the popely edge-coloed paths of the fom x x +1 y whch ae pawse edge-dsjont ones. We change the ode of the pas x q, y q and we swap, f necessay, x q and y q and a q wth q n ode to maxmze the cadnalty of each goup. Let d e the cadnalty of a maxmal goup. W.l.o.g., ths goup can e consdeed as the one defned y x 1 x y 1,..., x d x d+1 y d (d s consdeed modulo k). If d = k, then the poof has done snce thee ae k pawse edge-dsjont paths of length at most two, one pe pa x, y, = 1,..., k, as clamed. Othewse, we use the same pocess n ode to fnd the next maxmal goup of pawse edge-dsjont paths fo the emanng pas of vetces x, y, = d + 1,..., k. Ths s possle, snce f a = x j, wth > d and j d, then we can consde the path x j y, whch uses new edges not aleady used e pevously defned goups of paths. Ths pocess s fnte, snce at each step the nume of the emanng pas not lnked yet deceases stctly. Hence, at the end of the pocess, we have 1

13 found k pawse edge-dsjont paths of length at most two, one pe pa x, y, = 1,..., k. Ths completes the poof of the theoem. 3 Mnmum coloed degees and nume of edges suffcent fo the k-lnked popety n edge-coloed multgaphs Let us stat wth the followng theoem nvolvng mnmum nume of edges suffcent fo the k-lnked popety. Theoem 3.1. Let G c e a c-edge-coloed multgaph of ode n and k a non-zeo postve ntege, n k. If m c n(n 1) c(n k + 1) + 1, then G c s k-lnked. Poof. By nducton on n. Fo n = k the statement s tue. Indeed, n ths patcula case, t s an easy execce to see that all edges x y ae pesent n G c, = 1,,, k. The theoem s also tue fo small values of n and k. Let us fx k dstnct vetces x 1, y 1,, x k, y k n G c. Assume next that fo some, say = 1, thee exsts a popely edge-coloed path of length two etween x 1 and y 1 n G c. Let z denote the ndemedate vetex of ths path. In ths case consde the [ gaph G = G c \ {x 1, z, y 1 } on n 3 vetces. It has at least n ] 9n m [3c(n 3) + c] = c + k edges, thus t s (k 1)-lnked y nducton. The k 1 paths of G togethe wth the path etween x 1 and y 1 though z defne agan the k pawse-vetex-dsjont paths n G c, as desed. Assume fst that fo some, say = 1, thee exsts a path of length one etween x 1 and y 1 n G c. Consde ( the gaph G = G c \ {x 1, y 1 } on n vetces. It has at least n ) 7n m (nc 3c) = c + k edges. Hence G s (k 1)-lnked y nducton. Theefoe we can fnd k 1 pawse vetex-dsjont paths etween each pa x, y n G, =, 3,, k. These k 1 paths of G togethe wth the edge x 1 y 1 defne the k pawse vetex-dsjont paths n G c, as desed. Assume fnally that fo each = 1,,, k, thee exst no path etween x and y of length at most two n G c. Thus c edges ae mssng n G c etween x and y, fo othewse a path of length one could e defned etween x and y n G c. Futhemoe, fo each vetex z / {x 1, x,, x k }, at least c edges ae mssng etween z and {x, y } n G c, = 1,,, k. By summng the mssng edges fo a gven pa x and y we otan c(n k) + c. Theefoe fo the k pas, we conclude that at least kc(n k + 1) edges ae mssng n G c. But then, the nume of edges of G c s at most c n(n 1) kc(n k + 1). Fo k 1, we otan c n(n 1) kc(n k + 1) < c n(n 1) c(n k + 1) + 1, a contadcton. Ths completes the poof of the theoem. Aove theoem s the est possle. Indeed, let us consde a c-edge-coloed multgaph on n 13

14 vetces, n k, otaned as follows: Consde the dsjont unon of an solated vetex x 1 and a c-edge-coloed complete multgaph on n 1 vetces. Then add all possle edges on all possle colos etween x 1 and k fxed vetces, say x, y,, x k, y k, of the complete gaph. The esultng gaph, although t has c n(n 1) c(n k + 1) edges, t s not k-lnked. In fact, let y 1 e a vetex of the complete gaph, othe than x, y,, x k, y k. Then thee ae no k pawse vetex-dsjont paths, one pe pa x and y, 1 k, as any path etween x 1 and y 1 go though the est of vetces x, y,, x k, y k. In the est of the secton we deal wth condtons nvolvng oth mnmum coloed degees and nume of acs, suffcent fo the k-lnked (k-edge lnked) popety. Moe pecsely, let, k, e two fxed postve non-zeo nteges. We ae lookng fo functons f(n,, k) and g(n,, k) such that f a coloed multgaph G c on n vetces has coloed degees at least and f the nume of ts edges s at least f(n,, k) (espectvely, at least g(n,, k)), then t s k-lnked (espectvely k-edge-lnked). In ode to state ou conjectues late, let us fst defne the extemal gaph H c (t 1, t, t 3, t 4, t 5 ) (o shotly H c ) as follows: Let t 1, t, t 3, t 4, t 5 e gven non-zeo postve nteges. Let now A 1, A, A 3, A 4, A 5 e fve c-edge-coloed multgaphs on t 1, t, t 3, t 4 and t 5 vetces, espectvely. We defne A 1 (espectvely A, A 5 ) to e a complete edge-coloed multgaph, so that etween each two vetces of A 1 (espectvely of A, A 5 ) thee ae all possle multcoloed edges, one edge pe each avalale colo. The gaph A 3 (espectvely A 4 ) s complete and monochomatc on one fxed colo, say ed, (espectvely, say lue). We defne now H c to e the dsjont unon of A 1, A, A 3, A 4, A 5 y addng all edges on all possle colos etween A 5 and A 1 A A 3 A 4, all lue and ed edges etween A 3 and A 4, all ed edges etween A 1 A and A 3 and all lue edges etween A 1 A and A 4. The gaph H c has the nteestng popety that any path etween a vetex x of A 1 and a vetex y of A go though the set A 5. The extemal gaph H c helps to state the conjectue elow fo k-lnked multgaphs. Conjectue 3.. Let G c e a c-edge-coloed multgaph of ode n and k, e two non-zeo postve nteges. Assume that fo evey vetex x, d c (x), k 1 n + k. ) f c =, n 6 10k + 14, and m f 1 (n,, k) = n n( 4k + 7) + ( k + )(3 k + 3) + ( k + 3) + 1, ) f c =, n 6 10k + 14, and m f (n,, k) = 3n 4 + n(k 5 ) k(k 3) + 11, ) f c 3 and m f 3 (n,, k, c) = c [ n n( 4k + 7) + ( k + 3)( + 1) ] + 1, then G c s k-lnked. If tue, Conjectue 3. s the est possle. Indeed, let us consde the followng extemal gaphs: Fo Case (), we consde the gaph H c (1, n + k 3, k +, k +, k ) wth f 1 (n,, k) 1 edges. Choose now k pas of vetces, x 1 A 1, y 1 A and x, y A 5, k. Then thee ae no k pawse vetex-dsjont paths one pe pa x, y snce any path etween x 1 and y 1 goes though vetces of A 5. Howeve all vetces of A 5 ae aleady used y the paths jonng the othe pas of vetces x, y, =,, k. Fo Case () we consde the gaph H c (1, 1, n k, n k, k ). It has f (n,, k) 1 edges. Howeve, as n the pevous case, thee ae no k pawse vetex-dsjont popely edge-coloed paths, fo x 1 A 1, y 1 A and x, y A 5, wth =,, k. Fnally, fo Case () we consde the gaph H c ( k+3, n 1, 0, 0, k ) wth f 3 (n,, k, c) 1 edges. Agan, thee ae no k pawse vetex-dsjont popely edge-coloed paths x, y fo 14

15 x 1 A 1, y 1 A and x, y A 5, =,, k. In the sequel, we shall pove Conjectue 3. fo k = 1 and, c non fxed. But fo convenent easons we wll pove the cases c =, c = 3 and c 4 sepaately, n Theoems 3.3, 3.4, 3.5, espectvely. Theoem 3.3. Let G c e a -edge-coloed multgaph of ode n and a non-zeo postve ntege. Assume that fo evey vetex x, d c (x), n 1. ) f n and m n n( + 3) , ) f n and m 3n 4 3n + 1, then G c s lnked. Poof. The poof s y contadcton. Assume that, although condtons of theoem ae fullflled, thee s no path etween two gven vetces x and y of G c. Let R e a functon denotng the nume of edges of the complement of G c. In othe wods, R denotes the nume of edges to e added to G c n ode to ecome a complete -edge coloed multgaph of ode n. Clealy a -edge coloed multgaph on n vetces has n(n 1) edges. Unde the hypothess that thee s no path etween x and y, t wll e enough to show that f n (espectvely n 6 + 4), then R s at least n(n 1) [n n( +3) ] = n( +) 3 5 (espectvely n(n 1) [ 3n 4 3n ] = n 4 + n ). Ths wll e a contadcton wth the nume of edges of Gc. Let A, A, C, D e fou susets of V (G c ) such that: fo each z A, thee s a path fom x to z endng y a ed edge and thee s no path fom x to z endng y a lue edge n G c. fo each z A, thee s a path fom x to z endng y a lue edge and thee s no path fom x to z endng y a ed edge n G c. fo evey z C, thee ae at least two (not necessaly dsjont) paths fom x to z n G c, the fst path endng y a ed edge and the second one y a lue edge. D = V (G c ) ( A A C {x} ) Accodng to pevous defntons, the followng two Clams 1 and ae ovous. Clam 1. Thee s no lue edge etween x and A and no ed edge etween x and A. Clam. Thee s no edge etween D and C {x}, no lue edge etween D and A and no ed edge etween D and A. Clam 3. Thee s no lue edge n A and y symmety thee s no ed edge n A. Poof. Assume that thee s a lue edge, say uv, n A. Let P denote a popely edge-coloed path fom x to u such that the last edge of ths path s ed. We may suppose that v s not an ntenal vetex of P, fo othewse we can consde v nstead of u and then consde the segment of P etween x and v nstead of P. Then the path P uv jons x to v n G c and ts last edge uv s lue. Thus we conclude that v C, a contadcton snce A and C ae vetex-dsjont y defnton. 15

16 Clam 4. Fo evey z C, thee s at most one lue edge, say zu, etween z and A and f ths unque edge zu exsts n G c, then thee s no ed edge zx n G c. Poof. Assume y contadcton that thee ae at least two lue edges, say zu and zv n G c u, v A. Consde a path P fom x to z whose last edge s ed. Clealy such a path exsts n G c, y the defnton of C. If u s not on ths path, then u C, snce P zu defnes a path fom x to u whose last edge s lue, a contadcton to the defntons of A and C. Smla aguments hold f we consde v nstead of u. Consequently, we conclude that oth u and v elong to P. Let u (espectvely u + ) denote the pedecesso (espectvely the sucesso) of u, when we go fom x to z along P. Analogously we defne v and v + and z. As u and v ae oth vetces of A, the edges u u and v v ae oth ed. Futehmoe, as P s a popely edge-coloed path, t follows that oth edges uu + and vv + ae lue. Now y consdeng the path x u uzz v v etween x and v, we conclude that v C, a contadcton to the defntons of A and C. Ths poves that thee exsts at most one edge etween each vetex z C and A. Remans to pove that f fo some vetex z C, ths unque edge, say zu, u A exsts n G c, then the edge zx (f any) s not a ed one. Assume theefoe that a ed edge xz exsts n G c. But then the path x z u exsts well n G c and ts last edge s a lue one. Thus u C, agan a contadcton, snce A and C ae vetex dsjont y defnton. Ths completes the poof of the clam. Clam 5. Fo evey z C, thee s at most one ed edge etween z and A and f ths edge exsts n G c, then the edge zx (f any), s not a lue one. Poof. Smla to that of pevous clam. Now we ae eady to detemne R. Set A = a, A = a, C = c and D = d. Clealy a + a + c + d = n 1. Then, R = a + a + d(c + 1) + d(a + a ) + a(a 1) + a (a 1) + c(a + a ) = d(c + 1) + (d + c + 1)(a + a ) + a +a a+a = d(c + 1) + (d + c + 1 )(a + a ) + (a+a ) a a. We need to mnmze R. Let us fst fx a + a. Set a + a = a and consde a = a = a. Then, R = d(c + 1) + a(d + c + 1 ) + a 4 As a s fxed, then d + c s also fxed, snce a + c + d + 1 = n. We dstngush now etween two cases dependng upon a and. Fst case a Fo c = 0 and d = n 1 a, we otan R = (n 1 a) + a(n a 1 ) + a 4 = 3a 4 + a(n 5 ) + (n 1) If we consde R as a functon of a, then the mnmum values of R ae otaned fo a = o fo a = n. In patcula, R() = n( + 1) 3 5 and R(n ) = n 4 + n. Now y compang R() and R(n ) we may see that fo n 6 + 4, R() R(n ). Othewse, f n 6 + 4, then R(n ) R(). Ths s n contadcton wth the nume of edges of G c 16

17 and completes the poof of that case. Second case a < In ths second case we consde c = a and d = n a 1. Fo these patcula values of c and d we otan, R = (n a 1)( a + 1) + a(n a 1 ) + a 4 = a 4 a + n( + 1) ( + 1) If we consde R as a functon of a, we can see that the mnmum values of R ae otaned fo a = 1 o a =. Futhemoe R( 1) = n( + 1) ( + 1) and R() = n( + 1) As each of these two values s geate than n( + ) 3 5, ths contadcts the hypothess on the nume of edges of G c and completes the poof of the case and of the theoem. Theoem 3.4. Let G c e a 3-edge-coloed multgaph of ode n and a non-zeo postve ntege. Assume that fo evey vetex x, d c (x), n 1. If m 3 [n n( +3)+( +)+]+1, then G c s lnked. Poof. The poof s y contadcton. Let x and y e two vetces of G c. Assume that thee s no popely edge-coloed path etween x and y. As n pevous theoem, let R e a functon denotng the nume of edges of the complement of G c. In othe wods, R denotes the nume of edges to e added to G c n ode to ecome a complete 3-edge coloed multgaph of ode n. Clealy the nume of edges of a complete 3-edge coloed multgaph of ode n s 3 n(n 1). Set λ = 3 n(n 1) 3 [n n( + 3) + ( + ) + ] = 3n( + 1) 3( + ) 3. In ode to otan a contadcton, unde the hypothess that thee s no path etween x and y, t wll e enough to show that R λ. Let A, D, E e fve susets of V (G c ), 1 3, such that : Fo evey z A, 1 3, thee s a path fom x to z endng y an edge on colo and thee s no path fom x to z endng y an edge on a colo dffeent than. Fo evey z D, thee ae at least two (not necessaly dsjont) paths fom x to z, the fst one endng y an edge on colo and the second on endng y an edge on colo j, 1 j 3. E = V (G c ) (( 1 3 A ) D {x} ) Accodng to pevous defntons, the thee Clams 1,, and 3 elow ae ovous. Clam 1. Fo each = 1,, 3, thee s no edge on colo j etween x and A, 1 j 3. Clam. Thee s no edge etween E and D {x}. Clam 3. Fo each = 1,, 3, thee s no edge of colo j etween E and A, 1 j 3. Clam 4. Fo each = 1,, 3, thee s no edge on colo j n A, 1 j 3. 17

18 Poof. Assume that thee s an edge uv n A on some colo j, 1 j 3. Let P denote a popely edge-coloed path fom x to u n G c. We may suppose that v does not elong to P, fo othewse, we may exchange u and v and, nstead of P, consde the segment of P fom x to v. Moeove the colo of the last edge of P s. But then we may conclude that the path P uv exsts n G c and ts last edge s on colo j. Thus we otan that v C, a contadcton to the defntons of A and C. Clam 5. Fo each and j, 1 j 3, thee s no edge on colo l etween A and A j, l and l j. Poof. Assume that thee s an edge, say uv, on colo l etween A and A j, u A and v A j, l, j. Let P denote a popely edge-coloed path etween x and u. W.l.o.g. we may suppose that v does not elong to P, fo othewse, as n pevous clam, we can exchange u and v. Moeove the colo of the last edge of P s. In that case, y consdeng the path P uv, we conclude that v C. Ths s a contadcton wth the defntons of A j and C. Clam 6. Fo evey z D and fo each = 1,, 3, thee s at most one edge on colo, etween z and j A j. Futhemoe f ths unque edge etween z and j A j exsts, then thee exst no edge zx on colo j n G c, j. Poof. Assume y contadcton that thee ae two dstnct edges, say zu and zv, oth on colo etween z and j A j, u, v j A j. By the defnton of D, thee s a path, say P, fom x to z whose last edge s on colo l, l. If the vetex u s not on P, then u D, snce P zu s a path of G c jonng x to u. Smla aguments hold fo v. Consequently, n what follows we may suppose that oth u and v elong to P. Suppose w.l.ofg. that u s efoe v when we walk fom x to z along P. Let u (espectvely u + ) denote the pedecesso (espectvely the sucesso) of u. Analogously, we defne v and v + and z. Let q (espectvely q ) e the colo of edge u u (espectvely v v) As u and v ae oth vetces of j A j, q and q. Futhemoe, as P s popely edge-coloed, the colo of edge vv + s dffeent fom q. Now y consdeng the path x u uzz v v etween x and v, we conclude that v D, a contadcton to the defntons of A j, j and D. Ths poves that thee exsts at most one edge etween each vetex z D and j A j. Remans to pove that f fo some vetex z D, ths unque edge, say zu, u j A j exsts n G c, then the edge zx (f any) s not on colo j, j. Assume theefoe that an edge xz on colo j exsts n G c. But then the path x j z u exsts well n G c and ts last edge s on colo. Thus u D, agan a contadcton, snce j A j and D ae vetex dsjont y defnton. Ths completes the poof of the clam. Now we ae eady to detemne R. Set A = a, D = d and E = e. Clealy 3 =1 a + d + e + 1 = n. Then, that s, R = 3 a + 3e(d + 1) + e =1 3 a + =1 3 =1 a (a 1) R = 3e(d + 1) + (e + d ) a + =1 3 =1 + d a a + 1 a a j, =1 a a j. j j 18

19 Set 3 =1 a = a. To mnmze R, wth a fxed, we must consde a = a 3, fo each = 1,, 3. Fo these patcula values of a we otan, R = 3e(d + 1) + a(e + d + 1 ) + 3 a As a s fxed, we may suppose that d+e s also fxed, snce a+d+e+1 = n. Now we dstngush etween thee cases dependng upon n, and a. Fst case n 3 + and a 3. By takng d = 0 and e = n 1 a, we otan, R = 3(n a 1) + a(n a 1 ) + 3 a = a a((n 1 ) 3) + 3(n 1) If we consde R as a functon of a, we can see that the mnmum values of R ae otaned fo a = 3 o a = n. In patcula, R(n ) = 3 n n 1 3 and R(3) = ( (n 1 ) 3) +3(n 1) = 3n(+1) 1 ( + 1) 3. It suffces to show that R(n ) λ 0 and R(3) λ 0. Howeve, R(n ) λ = 3 n ( )n + 3( + ) We see easly that R(n ) λ 0, f n 3 +. Smlaly, fo n 3 + R(a = 3) λ = 3n Ths completes the poof of ths case. Second case n 3 + and a < 3 Fo d = a 3 and e = n a 3 1, we otan R = 3(n a 3 1)( a 3 +1)+a(n a 1 )+ 3 a = 3 a +a(n )+3n(+1) 3(+1) We can see that R s mnmum fo a = 0 o a = 3. Futhemoe R(0) = λ = 3(n 1)(+1) = 3n( + 1) 3( + ) 3 Now we can vefy that R(3) R(0). Indeed, R(3) R(0) = (n )( n 3 3 ) But n 3 +, so R(3) R(0) 0. Ths completes the poof of ths second case. Thd case n < 3 +. By the hypothess of the theoem, n +. Set n = 3 + ɛ whee ɛ s an ntege, 0 < ɛ. Clealy a + d + e + 1 = 3 + ɛ = n. To maxmze a, we take a = 3( ɛ), d = ɛ, e = ɛ + 1. Howeve fo d = a 3, e = n a 3 1 and fo any a < 3( ɛ), we have R = 3( ɛ a 3 + 1)( a 3 + 1) + a(3 ɛ a + 3 ) + 3 a R = 3 a + ( ɛ)a + 3( + 3 ɛ ɛ + 1) f(ɛ, ) + g(ɛ, ) The mnmum values of R ae otaned fo a = 0 o a = 3( ɛ). In patcula, R(0) = 3( ɛ + 1)( + 1) = 3 [ + (3 ɛ) + 1 ɛ ] and R(3 3ɛ) = 3( + 3 ɛ + 1 ɛ ) But R(3 3ɛ) R(0) = 3ɛ( ɛ) 0. Ths completes the poof of ths last case and of the theoem. The pevous esults deal wthn at most 3 colos. followng theoem. Fo moe than 3 colos, we have the 19

20 Theoem 3.5. Let G c e a c-edge-coloed multgaph of ode n, c > 3 and an ntege. Assume that fo evey vetex x, d c (x), 1 n 1. If m c [ n n( + 3) + ( + ) + ] + 1, then G c s lnked. Poof. By contadcton. Let x and y two vetces of G c. Assume that thee s no path etween x and y n G c. Let R e a functon countng the nume of edges n the complement of G c. The man pupose s to show that R c(n( + 1) ( + ) 1) whch wll e n contadcton wth m. Snce thee s no path etween x and y n G c, we may suppose that thee s no path ethe etween x and y n the sugaph G c of Gc contanng the edges of colo, +1 and + (modulo c) of G c, fo evey fxed colo = 1,,, c. As fo R, n a smla way let us defne R fo each such sugaph G c. Now R 3 (n(+1) (+) 1). Then R = c 3 R c(n(+1) (+) 1), snce evey colo s used thee tmes. Ths completes the poof of the theoem. Let us tun now ou attenton to suffcent condtons nvolvng mnmum degees and nume of acs guaantyng the k-edge-lnked popety. Moe pecsely, let us fomulate the followng conjectue. Conjectue 3.6. Let G c e a c-edge-coloed multgaph of ode n and k, e two non-zeo postve nteges. Assume that fo evey vetex x, d c (x), n 1. ) f c =, n and m g 1 (n, k, ) = n n( + 3) , ) If c =, n and m g (n, k, ) = 3n 4 3n + 1, ) f c 3 and m g 3 (n, k,, c) = c [ n n( + 3) + ( + ) + ] + mn(k 1, ) + 1, then G c s k-edge-lnked. If tue, ths conjectue should e the est possle. Indeed : Fo (), we consde the -edge-coloed multgaph H c (1, n 1,,, 0). Although t has g 1 (n, k, ) 1 edges, t s no 1-edge lnked. In patcula, thee s no popely edge-coloed path etween x 1 and y 1, fo any choce of vetces x 1 A 1 and y 1 A. Fo (), we consde the the -edge-coloed multgaph H c (1, 1, n 1, n 1, 0) havng g (n, k, ) 1 edges. As n pevous Case () thee s no popely edge-coloed path etween any pa of vetces x 1 A 1, y 1 A. Fnally fo (), we consde the c-edge-coloed multgaph H c (+1, n 1, 0, 0, 0). If +1 k then we add k 1 edges etween A 1 and A. Now f we consde x A 1 and y A, 1 k, then we can not fnd k pawse edge-dsjont paths one pe pa x, y, snce thee ae at most k 1 edges etween A 1 and A. Thus H c s not k-edge-lnked. Othewse, f k then we add edges etween A 1 and A. If we select x A 1 and y A, 1 + 1, then, agan, we can not fnd + 1 pawse edge-dsjont paths, one pe pa x, y, snce thee ae at most edges etween A 1 and A. Thus, although H c has c[n n(+3)+(+)+] + mn(k 1, ) edges, t s not ( + 1)-edge-lnked fo + 1 < k. By Theoem 3.5, Conjectue 3.6 aove s tue fo k = 1 and, c non fxed. Also n Theoem 3.8 stated late we pove that ths conjectue emans tue fo = 1, c = and k non fxed. In vew of Theoem 3.8, let us pove the followng lemma. Lemma 3.7. Let G c e a -edge-coloed multgaph of ode n 5. Assume that fo evey vetex x, d c (x) 1. ) If n 10, and m n 5n + 11, 0