Discrete Mathematics

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1 Dscrete Mathematcs 30 (00) Contents lsts avalable at ScenceDrect Dscrete Mathematcs journal homepage: The number of C 3 -free vertces on 3-partte tournaments Ana Paulna Fgueroa a, Bernardo Llano b, Rta Zuazua c a Departamento de Matemátcas Aplcadas y Sstemas, Unversdad Autónoma Metropoltana, Cuajmalpa, Artfcos 40, Colona Hdalgo, 00, Méxco, D.F., Mexco b Departamento de Matemátcas, Unversdad Autónoma Metropoltana, Iztapalapa, San Rafael Atlxco 86, Colona Vcentna, 09340, Méxco, D.F., Mexco c Facultad de Cencas, Unversdad Naconal Autónoma de Méxco, Cudad Unverstara, 0450, Méxco, D.F., Mexco a r t c l e n f o a b s t r a c t Artcle hstory: Receved July 009 Receved n revsed form March 00 Accepted June 00 Avalable onlne July 00 Keywords: Drected trangle free vertex Regular 3-partte tournament Let T be a 3-partte tournament. We say that a vertex v s C 3 -free f v does not le on any drected trangle of T. Let F 3 (T) be the set of the C 3 -free vertces n a 3-partte tournament f 3 (T) ts cardnalty. In ths paper we prove that f T s a regular 3-partte tournament, then F 3 (T) must be contaned n one of the partte sets of T. It s also shown that for every regular 3-partte tournament, f 3 (T) does not exceed n, where n s the order of T. On the 9 other h, we gve an nfnte famly of strongly connected tournaments havng n 4 C 3 - free vertces. Fnally we prove that for every c 3 there exsts an nfnte famly of strongly connected c-partte tournaments, D c (T), wth n c C 3 -free vertces. 00 Elsever B.V. All rghts reserved.. Introducton Let c be a non-negatve nteger, a c-partte or multpartte tournament s a dgraph obtaned from a complete c-partte graph by substtutng each edge wth exactly one arc. A tournament s a c-partte tournament wth exactly c vertces. The partte sets of T are the maxmal ndependent sets of T. Let D be an orented graph. The vertex set the arc set of an orented graph D are denoted by V(D) A(D), respectvely. The out-neghborhood (n-neghborhood, resp.) N + (x) (N (x), resp.) of a vertex x s the set {y V(D) xy A(D)} ({y V(D) yx A(D)}, resp.). The numbers d + (x) = N + (x) d (x) = N (x) are the out-degree the n-degree of x, respectvely. The global rregularty of an orented graph D s defned as g (D) = max x,y V(D) {max{d + (x), d (x)} mn{d + (y), d (y)}}. An orented graph D s regular (almost regular, resp.) f g (D) = 0 ( g (D), resp.). Let X, Y V(D), X domnates Y, denoted by X = Y, f Y N + (x) for every x X. An orented graph D s strongly connected f for every ordered par of vertces (x, y) there s a drected path from x to y. An m-cycle of an orented graph s a drected cycle of length m. Let T be a c-partte tournament. We say that a vertex v s C 3 -free f v does not le on any drected trangle of T. Let F 3 (T) be the set of the C 3 -free vertces n a c-partte tournament f 3 (T) ts cardnalty. For the stard termnology on dgraphs see []. The structure of cycles n multpartte tournaments has been extensvely studed, see for example [3,4,6,7] []. A very recent survey on ths topc [0] appeared wth several nterestng open problems. For nstance, the study of cycles whose length does not exceed the number of partte sets leads to varous extensons generalzatons of classc results on tournaments. Bondy [] proved that each strongly connected c-partte tournament contans an m-cycle for each m Ths work was supported by the Scholarshp Program for Posdoctoral Fellowshps sponsored by Unversdad Autónoma Metropoltana - Iztapalapa, PAPIIT IN309- Conacyt Mexco. E-mal addresses: apaulnafg@gmal.com (A.P. Fgueroa), llano@xanum.uam.mx (B. Llano), rtazuazua@gmal.com (R. Zuazua) X/$ see front matter 00 Elsever B.V. All rghts reserved. do:0.06/j.dsc

2 A.P. Fgueroa et al. / Dscrete Mathematcs 30 (00) {3,..., c}. In 994, Guo Volkmann [5] proved that every partte set of a strongly connected c-partte tournament T has at least one vertex that les on a cycle of length m for each m {3,..., c}. Ths result also generalzes a theorem of Gutn [8]. There are examples showng that not every vertex of a strongly connected c-partte tournament s contaned n a cycle of length m for each m {3,..., c} n general [0]. However, Zhou et al. [] proved that every vertex of a regular c-partte tournament wth at least four partte sets (c 4) s contaned n a cycle of length m for each m {3,..., c}. Volkmann [9] provded the followng nfnte famly of 4p-regular 3-partte tournaments whch shows that the prevous theorem s not vald for regular 3-partte tournaments Example (Volkmann). Let F be the famly of tournaments wth the partte sets U = U U, V = V V W = W W such that:. For every natural number p, the sets have szes W = p, W = 3p U = U = V = V = p.. The sets U V U V generate p-regular bpartte tournaments T T, respectvely. 3. V = U U = V. 4. V(T ) = W = V(T ) V(T ) = W = V(T ). Ths result leads to the natural queston: whch s the maxmum number of C 3 -free vertces on regular 3-partte tournaments? In the case of Example we have an nfnte famly of regular 3-partte tournaments such that f 3 (T) = V(T). In ths paper we prove that f T s a regular 3-partte tournament, there s at most one partte set of T contanng the vertces n F 3 (T). We also show that for every regular 3-partte tournament, f 3 (T) does not exceed V(T). On the other h, we gve an 9 nfnte famly of strongly connected tournaments havng n 4 C 3 -free vertces. Fnally we prove that for every c 3 there exsts an nfnte famly of strongly connected c-partte tournaments, D c (T), wth n c C 3 -free vertces. Ths examples show that regularty s an mportant constrant to have lots of vertces that are contaned n a drected trangle. We conclude ths secton wth the followng Remark. Let P, P, P 3 be the partte sets of a regular 3-partte tournament T. Then, r d + (x) = d (x) = r for all x V(T). = P = P = P 3. Trpartte regular tournaments Let T be a 3-partte regular tournament P be a partte set of T, u, v V(T) S V(T). For the rest of ths artcle we use the followng notaton: P + (u, v) := N + (u) N (v) P, P + (u, v) := N (u) N + (v) P, P + (S) := ( s S N+ (s)) P, P (S) := ( s S N (s)) P P (S) := P \ (P + (S) P (S)). If S = {v} for some v V(T), then we brefly wrte P + (v) P (v) to denote P + ({v}) P ({v}), respectvely. Theorem. For every regular 3-partte tournament, there s a partte set contanng F 3 (T). Proof. Let U, V, W be the partte sets of the 3-partte regular tournament T. By Remark, there exsts a postve nteger r such that r = U = V = W d + (x) = d (x) = r for every x V(T). Suppose that there exsts S = {u, v} F 3 (T) such that u U, v V uv A(T). If x W + (u, v), u les on a drected trangle (u, v, x), whch s mpossble snce S F 3 (T). Thus, W + (u, v) = W = W + (S) W (S) W + (u, v). () By defnton, W + (S), W (S) W + (u, v) are parwse dsjont sets. If W + (S) =, from Eq. () we obtan that r = d (v) W + {u} = r + > r, a contradcton. Thus, W + (S). () Analogously, W (S), snce d (u) = r < W + {v}. The sets U (v) U + (v) form a partton of U. Notce that u U (v). By Remark relaton (3), W + (S) < r, thus U + (v) because r = d + (v) = W + (S) + U + (v). By a smlar argument, Remark relaton (), V + (u) V (u) are non-empty sets. (3)

3 484 A.P. Fgueroa et al. / Dscrete Mathematcs 30 (00) Let x U (v) w W + (S). Snce v F 3 (T), we have that the drected path (x, v, w) s not contaned n a drected trangle of T. Ths mples that xw A(T), thus U (v) = W + (S). Even more, for every y V (u) w W + (S), the drected path (y, u, w) s not a drected trangle of T, snce u F 3 (T), thus V (u) = W + (S). From relatons (4) (5), for every w W + (S) we have that U (v) V (u) {v} N (w) U (v) + V (u) + r. On the other h, for every x U + (v) w W (S), we have that the drected path (w, v, x) s not n a drected trangle of T, then W (S) = U + (v). Moreover, for every y V + (u) w W (S), we have that the drected path (w, u, y) s not n a drected trangle of T, thus W (S) = V + (u). Thus, from relatons (7) (8), for every w W (S) we have that U + (v) V + (s) {u} N (w ) U + (v) + V + (u) + r. Snce T s regular U + (v) + U (v) = r V + (u) + V (u) = r. If we sum Eqs. (6) (9) we get a contradcton. Thus, u v must be elements of the same partton set of T. Remark 3. Let C be a 4-cycle that contans vertces from all the partte sets of a regular 3-partte tournament. Then, the number of C 3 -free vertces belongng to C s at most one. Proof. Snce C s not an nduced cycle, we have that at least three of ts vertces les n a drected trangle, thus at most one of them can be C 3 -free. Theorem 4. Let T be a 3-partte regular tournament wth partte sets V 0, V V. If u, v F 3 (T) V 0, then there exsts {, } such that V + (u, v) = V + Proof. Let V 0, V V be the partte sets of a r-regular 3-partte tournament S = {u, v} F 3 (T) V 0. It s enough to prove that f V + (u, v), then V + (u, v) = V + Let x V + (u, v). If there exsts x V + (u, v), then (u, x, v, x ) s a drected 4-cycle of T contanng two vertces n F 3 (T), whch by Remark 3 s mpossble. Thus, V + If y V + (u, v) y V + (u, v), then (v, y, u, y ) s a drected 4-cycle of T contanng two vertces of F 3 (T). Then, V + By Remark, r = N + (u) = V + r = N (v) = V by relatons (0) (), r = V + (S) + V (S ) =, =, whch s a contradcton because + (u, v) + V + + (u, v), () + (u, v) + V + (u, v), (3) r = V + + (u, v) + V (S) for = {, } (4) (5) (6) (7) (8) (9) (0) ()

4 A.P. Fgueroa et al. / Dscrete Mathematcs 30 (00) V + (u, v) for = {, }. Therefore, V + (u, v) s empty. Thus, we have proved that V + (u, v) = V + Colorally. Let T be a 3-partte regular tournament such that F 3 (T). Then, there exsts at least one partte set P such that P = P + (F 3 (T)) P (F 3 (T)). Proof. Let P P be partte sets of T not contanng F 3 (T). Suppose that P P + (F 3(T)) P (F 3(T)) P P + (F 3(T)) P (F 3(T)). Then, there exst u, v F 3 (T) such that P + (u, v). By Theorem 4, P = P + ({u, v}) P ({u, v}). Snce P P + (F 3(T)) P (F 3(T)), there exsts w F 3 (T) such that P P + ({u, w}) P ({u, w}). Then, by relaton (4), P P + ({v, w}) P ({v, w}). Applyng Theorem 4, we have that P = P + ({u, w}) P ({u, w}) P = P + ({v, w}) P ({v, w}). Thus, P + (u, v) P + ({u, w}) P ({v, w}) N+ (w) N (w) =, whch s a contradcton because P + (u, v). Theorem 5. Let T be a r-regular 3-partte tournament. Then, f 3 (T) < r 3. Proof. Let T be a 3-partte tournament, by Theorem there exst partte sets P P of T such that P F 3 (T) = for =,. Denote by P + P the sets P + (F 3 (T)) P (F 3 (T)), respectvely for {, }. By Corollary, we can assume that P = P + P. By defnton P = P \ (P + P ), then P = P + P P. We may assume f 3(T) > 0 for the rest of the proof. Let v P w P+ P. There exsts x F 3(T) such that vx A(T) xw A(T). Snce the drected path (v, x, w) s not n a drected trangle of T, then vw A(T). Thus, P = P+ (5) P = P. Let v P + w P P. There exsts x F 3(T) such that xv A(T) wx A(T). Snce the drected path (w, x, v) s not n a drected trangle of T, then wv A(T). Thus, P = P+ (7) (4) (6) P = P+. (8) From relatons (5) (6), we have that P + P are not empty sets, otherwse, d (v) > r for every vertex v P whch s mpossble. Analogously, from relatons (7) (8), we have that P P+ are not empty sets, otherwse, d (v) > r for every vertex v P. Let s = {uv A(T) u P +, v P+ }, s = {uv A(T) u P +, v P+ }, t = {uv A(T) u P, v P } t = {uv A(T) u P, v P }. If s = 0, then d (x) > P + + P + P + f 3(T) > r, whch s mpossble. Thus s 0. Analogously, s, t, t 0. Notce that s + s = P + P+ t + t = P P. We consder the followng cases. Case. P + r Subcase.. s P+ P+ From the regularty of T, the relatons (5), (6) the defntons of P + r P + = {uv A(T) v P+ } s + f 3 (T) P + + P P+. By the hypothess, r P + P+ P+ + f 3 (T) P + + P P+. s, we have that

5 486 A.P. Fgueroa et al. / Dscrete Mathematcs 30 (00) Snce P + > 0, the last equaton s equvalent to r P+ + f 3(T) + P = P+ + f 3(T) + P P +. Snce T s regular, P + + P = r. Thus, r f 3 (T) + r P+, whch mples that P + f 3 (T). Fnally by the hypothess of Case, f 3 (T) r 4. Subcase.. s P+ P+ From relatons (7) (8) the defntons of P + s, t follows that r P + = {uv A(T) v P+ } s + f 3 (T) P + + P P+ + P P+. If we proceed as we dd n the prevous subcase, by the hypothess s P+ P+ the fact that P + + P + P = r, we obtan that f 3 (T) P+. If f 3 (T) r 3, from the last relaton we obtan that P+ 3 r. On the other h, from relatons (5) (6) the defntons of t P, t follows that r P = {uv A(T) u P } t + f 3 (T) P + P+ P + P P. Snce P > 0, P 0, t 0, P + r f 3 3(T) r we have that 3 r P + + P + f 3(T) + t P > P+ + f 3(T) r. whch s mpossble, thus f 3 (T) < r 3. Case. P r Subcase.. t P P From the regularty of T, relaton (7) the defntons of P r P = {uv A(T) u P } t + f 3 (T) P + P+ P. t we have that By the hypothess, snce P > 0 P+ + P = r, the last equaton s equvalent to P f 3 (T). Fnally, by the hypothess of Case, f 3 (T) r 4. Subcase.. t P P From relatons (5) (6) the defntons of P t, t follows that r P = {uv A(T) u P } t + f 3 (T) P + P+ P + P P. By the hypothess, t P P f 3 (T) P. the fact that P + + P + P = r, we have that

6 A.P. Fgueroa et al. / Dscrete Mathematcs 30 (00) Fg.. T 0 (n). If f 3 (T) r 3, from the last relaton we obtan that P 3 r. On the other h, from relatons (7) (8) the defntons of s P +, t follows that r P + = {uv A(T) v P+ } s + f 3 (T) P + + P P+ + P P+. Snce P + > 0, P 0, s 0, P 3 r f 3(T) r 3, then r P + P + f 3(T) + s P + > P + f 3(T) r whch s mpossble. Thus f 3 (T) < r Strong c-partte tournaments wth lots of C 3 -free vertces Frst we gve an nfnte famly of strongly connected 3-partte tournaments havng n 4 C 3 -free vertces (see Fg. ). Example. Let T 0 (n) be the 3-partte tournament wth n vertces partte sets X u, Y w Z v, v such that: A(T 0 (n)) = {(x, v) x X, v Y Z {v, w}} {(y, v) y Y, v {v, u}} {(z, v) z Z, v Y {u, w}} {(v, v) v X Y {u}} {(v, u), (u, w), (w, v ), (w, v )}. Note that T 0 (n) s a strongly connected: for every x X, y Y z Z there exsts the drected cycle (v, u, w, v, x, z, y) n T 0 (n). Observe that F 3 (T 0 (n)) = X Y Z. Thus, for every n there s a strongly connected 3-partte tournament wth n 4 C 3 -free vertces. The regularty s also mportant to prove that every vertex s n a drected trangle n the case of c-partte tournaments, c 4. In the followng example we gve an nfnte famly of c-partte tournaments on n vertces havng n c C 3 -free vertces (see Fg. ). Example 3. Let T 0 (n c + 3) be the trpartte tournament on n c + 3 vertces of Example T be a tournament wth order c 3. Let D c (T)be the c-partte tournament wth vertex set V(T 0 (n c + 3)) V(T) such that A(D c (T)) = A(T 0 (n c + 3)) A(T) {xu x V(T)} {yx y V(T 0 (n c + e)) \ {u} x V(T)}. It s clear that D c (T) s strongly connected because T 0 (n c + 3) s strongly connected for every vertex x of T, xu wx are arcs of A(D c (T)). If C 3 s a drected trangle wth at least one vertex n X Y Z, then t cannot be contaned n T 0 (n c + 3). If C 3 has only one vertex a X Y Z then a s a source of C 3. If C 3 has only one vertex of b V(T) then b s a snk of C 3. Thus, f 3 (D c (T)) = X Y Z = n c. So we have proved the followng Theorem 6. For every c 3 there exsts an nfnte famly of strongly connected c-partte tournaments, D c (T) wth n c C 3 - free vertces. Fnally we conjecture that for every tournament T, the maxmum number of C 3 -free vertces of T s V(T) as suggested by Example.

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