Dscrete Mathematcs Theoretcal Computer Scence DMTCS vol. 11:1, 2009, 117 124 Self-complementng permutatons of k-unform hypergraphs Artur Szymańsk A. Paweł Wojda Faculty of Appled Mathematcs, AGH Unversty of Scence Technology, Al. Mckewcza 30, 30-053 Kraków, Pol receved January 20, 2008, revsed January 16, 2009, accepted January 23, 2009. A k-unform hypergraph H = (V ; E) s sad to be self-complementary whenever t s somorphc wth ts complement H = (V ; `V k E). Every permutaton σ of the set V such that σ(e) s an edge of H f only f e E s called selfcomplementng. 2-self-comlementary hypergraphs are exactly self complementary graphs ntroduced ndependently by Rngel (1963) Sachs (1962). For any postve nteger n we denote by λ(n) the unque nteger such that n = 2 λ(n) c, where c s odd. In the paper we prove that a permutaton σ of [1, n] wth orbts O 1,..., O m s a self-complementng permutaton of a k-unform hypergraph of order n f only f there s an nteger l 0 such that k = a2 l + s, a s odd, 0 s < 2 l the followng two condtons hold: () n = b2 l+1 + r, r {0,..., 2 l 1 + s}, () P :λ( O ) l O r. For k = 2 ths result s the very well known characterzaton of self-complementng permutaton of graphs gven by Rngel Sachs. Keywords: Self-complementng permutatons, k-unform hypergraphs 1 Introducton Let V be a set of n elements. The set of all k-subsets of V s denoted by ( V. A k-unform hypergraph H conssts of a vertex-set V (H) an edge-set E(H) ( ) V (H) k. Two k-unform hypergraphs G H are somorphc, f there s a bjecton σ : V (G) V (H) such that e E(G) f only f {σ(x) x e} E(H). The complement of a k- unform hypergraph H s the hypergraph H such that V (H) = V (H) the edge set of whch conssts of all k-subsets of V (H) not n E(H) (n other words E(H) = ( ) V (H) k E). A k-unform hypergraph H s called self-complementary (s-c for short) f t s somorphc wth ts complement H. Isomorphsm of a k-unform self-complementary hypergraph onto ts complement s called self-complementng permutaton (or s-c permutaton). 1365 8050 c 2009 Dscrete Mathematcs Theoretcal Computer Scence (DMTCS), Nancy, France
118 Artur Szymańsk A. Paweł Wojda The 2-unform self-complementary hypergraphs are exactly self-complementary graphs. Ths class of graphs has been ndependently dscovered by Rngel Sachs who proved the followng. Theorem 1 (Rngel (Rn63) Sachs (Sac62)) Let n be a postve nteger. A permutaton σ of [1, n] s a self-complementng permutaton of a self-complementary graph of order n f only f all the orbts of σ have ther cardnaltes congruent to 0 (mod 4) except, possbly, one orbt of cardnalty 1. Observe that by Theorem 1 an s-c graph of order n exsts f only f n 0 or n 1 (mod 4) or, equvalently, whenever ( n 2) s even. In (SW) we prove that a smlar result s true for k-unform hypergraphs. Theorem 2 ((SW)) Let k n be postve ntegers, k n. A k-unform self-complementary hypergraph of order n exsts f only f ( n s even. A smple crteron for evenness of ( n has been gven n (Gla99) ( then redscovered n (KHRM58)). Theorem 3 ((Gla99; KHRM58)) Let k n be postve ntegers, k = + =0 c 2 n = + =0 d 2, where c, d {0, 1} for every. ( n s even f only f there s 0 such that c 0 = 1 d 0 = 0. Theorem 3 asserts that ( n s even f only f k has 1 n a certan bnary place whle n has 0 n the correspondng bnary place. For example, ( 27 13) s even snce 13 = 1 2 3 + 1 2 2 + 0 2 1 + 1 2 0 27 = 1 2 4 + 1 2 3 + 0 2 2 + 1 2 1 + 1 2 0 (so we have c 2 = 1 d 2 = 0). Except for Theorem 1 whch s a characterzaton of the self-complementng permutatons for graphs, there are already two publshed results characterzng the permutatons of k-unform s-c hypergraphs for k > 2. Namely, Kocay n (Koc92) (see also (Pal73)) Szymańsk n (Szy05) have characterzed the s-c permutatons of s-c k-unform hypergraphs for, respectvely, k = 3 k = 4. Ths work s a contnuaton of the work of (SW) (Woj06). We generalze all the results mentoned above by gvng a characterzaton of the s-c permutatons of k-unform hypergraphs for any ntegers k n. 2 Result Any postve nteger n may be wrten n the form n = 2 l c, where c s an odd nteger. Moreover, l c are unquely determned. We wrte then λ(n) = l. Note that n the bnary expanson of n, λ(n) s the ndex of the frst 1 bt. For any set A we shall wrte λ(a) n place of λ( A ), for short. In the proof of our man result we shall need the followng lemma proved n (Woj06). Lemma 1 Let k, m n be postve ntegers, let σ : V V be a permutaton of a set V, V = n, wth orbts O 1,..., O m. σ s a self-complementng permutaton of a self-complementary k-unform hypergraph, f only f, for every p {1,..., k} for every decomposton k = k 1 +... + k p of k (k j > 0 for j = 1,..., p), for every subsequence of orbts O 1,..., O p such that k j O j for j = 1,..., p, there s a subscrpt j 0 {1,..., p} such that λ(k j0 ) < λ(o j0 )
Self-complementng permutatons of k-unform hypergraphs 119 Gven any nteger l 0. If the bnary expanson of k s 1 bt n poston l, then k can be wrtten n the form k = a l 2 l + s l, where a l s odd 0 s l < 2 l. Theorem 4 Let k n be ntegers, k n. A permutaton σ of [1, n] wth orbts O 1,..., O m s a selfcomplementng permutaton of a k-unform hypergraph of order n f only f there s a nonnegatve nteger l such that k = a l 2 l + s l, where a l s odd 0 s l < 2 l, the followng two condtons hold: () n = b l 2 l+1 + r l, r l {0,..., 2 l 1 + s l }, () :λ(o ) l O r l. Proof: Suffcency. By contradcton. Let n, k, l, a l, b l, s l r l be ntegers verfyng the condtons of the theorem, let σ be a permutaton of [1, n] wth orbts O 1,..., O m verfyng (), let us suppose that σ s not a s-c permutaton of any k-unform s-c hypergraph of order n. Then, by Lemma 1, there s a decomposton of k = k 1 + + k t a subsequence of orbts O 1,..., O t such that 0 < k j O j (1) λ(k j ) λ(o j ) (2) for j = 1,..., t. Snce a l s odd, we have k 2 l + s l (mod 2 l+1 ). By (2), j: λ(o j )>l k j 0 (mod 2 l+1 ). Therefore k = t k j = j=1 j: λ(o j )>l k j + j: λ(o j ) l k j j: λ(o j ) l k j (mod 2 l+1 ) Hence, by (1), () () we have j:λ(o j ) l k j j:λ(o j ) l O j < 2 l+1, therefore a contradcton. 2 l + s l = j: λ(o j ) l k j j: λ(o j ) l O j r l < 2 l + s l Necessty. Let 1 k n let σ be a permutaton of the set [1, n] wth orbts O 1,..., O m. Let us suppose that for every nteger l such that k = a l 2 l + s l, where a l s odd postve nteger, 0 s l < 2 l, n = b l 2 l+1 + r l, 0 r l < 2 l+1 we have ether or r l {2 l + s l,..., 2 l+1 1} r l {0,..., 2 l 1 + s l } : λ(o ) l O > r l We shall prove that σ s not a s-c permutaton of any s-c k-unform hypergraph of order n. For ths purpose we shall gve two clams.
120 Artur Szymańsk A. Paweł Wojda Clam 1 For every nonnegatve nteger l such that k = a l 2 l + s l, where a l s odd 0 s l < 2 l, we have O 2 l + s l Proof of Clam 1. : λ(o ) l Let us wrte :λ(o ) l O :λ(o )>l O n ther bnary forms: :λ(o ) l :λ(o )>l O = O = e j 2 j j=0 f j 2 j where e j, f j {0, 1} for every j. Observe that f j = 0 for j = 0,..., l therefore We shall consder two cases. j=0 l e j 2 j = r l (3) Case 1. r l {0,..., 2 l + s l 1} :λ(o O ) l > r l. We have n 2 l+1 (otherwse r l = n = :λ(o O ) l ). Snce j=0 e j2 j > r l, by (3), we obtan j=0 e j2 j 2 l+1 > 2 l + s l. j=0 Case 2. r l {2 l + s l,..., 2 l+1 1}. We have :λ(o ) l O = j=0 e j2 j l j=0 e j2 j = r l 2 l + s l, the clam s proved. Clam 2 Let α 1,..., α q λ 1,..., λ q be ntegers such that 0 < α, 0 λ λ(α ) λ l for = 1,..., q q =1 α 2 l. Then there are β 1,..., β q such that for every = 1,..., q 0 β α (4) ether β = 0 or λ(β ) λ (5) q β = 2 l (6) =1 Proof of Clam 2. The exstence of β 1,..., β q verfyng (4)-(5) q =1 β 2 l s very easy. Indeed, t s mmedate that β 1 = 2 λ1, β 2 =... β q = 0 s a sequence wth the desred propertes. So let us suppose that β 1,..., β q s a sequence verfyng (4)-(5) q =1 β 2 l such that q =1 β s maxmal. If q =1 β = 2 l then the proof s complete. So let us suppose that q =1 β < 2 l. Then there s 0 {1,..., q} such that β 0 < α 0. Observe that β 0 + 2 λ 0 α 0. The sequence β 1,..., β q defned by β 0 = β 0 + 2 λ 0 β = β for 0 also verfes (4)-(5) q =1 β 2 l, whch contradcts the maxmalty of the sum q =1 β, the clam s proved. We shall use our clams to construct a decomposton of k n the form k = k 1 +... + k m such that
Self-complementng permutatons of k-unform hypergraphs 121 (1) k 1,..., k m are nonnegatve ntegers, (2) k O for = 1,..., m, (3) λ(k ) λ(o ) whenever k > 0 By Lemma 1, ths wll mply that σ s not a s-c permutaton of any k-unform s-c hypergraph. Let us wrte k n ts bnary form: k = 2 lt + 2 lt 1 +... + 2 l1 + 2 l0 where l 0 < l 1 <... < l t. By Clam 1, :λ(o ) l 0 O 2 l0. Hence, by Clam 2, there are nonnegatve ntegers k (0) 1, k(0) 2,..., k(0) m such that k (0) = 0 for such that λ(o ) > l 0 k (0) O for = 1,..., m λ(k (0) ) λ(o ) whenever k (0) > 0 m =1 k (0) = 2 l0 Note that, for = 1,..., m, we have λ( O k (0) ) λ(o ). Let us suppose that we have already constructed k (j) 1,..., k(j) m, (j t), such that k (j) = 0 for such that λ(0 ) > l O for = 1,..., m k (j) λ(k (j) m =0 ) λ(o ) whenever k (j) > 0 k (j) = 2 lj + 2 lj 1 + + 2 l0 λ( O k (j) ) λ(o ) If j = t, then we have already found a desred decomposton of k. If j < t, then, by Clam 1, we have :λ(o ) l j+1 ( O k (j) ) 2 lj+1. λ( O k (j) ) λ(o ) for every {1,..., m} such that O k (j) > 0. Hence, by Clam 2, there are β 1,..., β m such that β = 0 for such that λ(o ) > l j+1 0 β O k (j) for = 1,..., m λ(o ) λ(β ) for = 1,..., m whenever β 0 m β = 2 lj+1 =1
122 Artur Szymańsk A. Paweł Wojda Thus we may defne for every = 1,..., m k (j+1) = k (j) + β to obtan the sequence (k (j+1) 1,..., k (j+1) m ) verfyng for every {1,..., m} k (j+1) = 0 for such that λ(o ) > l j+1 k (j+1) O λ(k (j+1) ) λ(o ) whenever k (j+1) > 0 m =1 k (j+1) = 2 lj+1 + 2 lj + + 2 l0 It s clear that k = m =1 k(t) the proof of Theorem 4 s complete. Theorem 4 mples very easly the followng theorem frst proved by Kocay. Corollary 1 (Kocay (Koc92)) σ s a self-complementng permutaton of a self-complementary 3-unform hypergraph f only f ether all the orbts of σ have even cardnaltes, or else, t has 1 or 2 fxed ponts the all remanng orbts of σ have ther cardnaltes beng multples of 4. For k = 2 l Theorem 4 may be wrtten as follows. Corollary 2 Let l n be nonnegatve ntegers, 2 l < n, let 0 r < 2 l+1 be such that n r (mod 2 l+1 ). A permutaton σ of [1, n] wth orbts O 1,... O m s a self-complementng permutaton of a 2 l -unform self-complementary hypergraph f only f () r {0,..., 2 l 1} () :λ(o ) l O r. Theorem 2 for l = 1 (.e. for graphs) s exactly Theorem 1, for l = 2 the followng theorem proved by Szymańsk n (Szy05). Corollary 3 A permutaton σ s self-complementng permutaton of a 4-unform hypergraph of order n f only f n r( mod 8) wth r = 0, 1, 2 or 3, the sum of the cardnaltes of orbts whch are not multples of 8 s at most 3. Acknowledgements The research was partally supported by AGH local grant No 11 420 04.
Self-complementng permutatons of k-unform hypergraphs 123 References [Gla99] J.W.L. Glasher. On the resdue of a bnomal coeffcent wth respect to a prme modulus. Quarterly Journal of Mathematcs, 30:150 156, 1899. [KHRM58] S.H. Kmball, T.R. Hatcher, J.A. Rley, L. Moser. Soluton to problem e1288: Odd bnomal coeffcents. Amer. Math. Monthly, 65:368 369, 1958. [Koc92] W. Kocay. Reconstructng graphs as subsumed graphs of hypergraphs, some selfcomplementary trple systems. Graphs Combnatorcs, 8:259 276, 1992. [Pal73] E.M. Palmer. On the number of n-plexes. Dscrete Math., 6:377 390, 1973. [Rn63] G. Rngel. Selbstkomplementare graphen. Arch. Math., 14:354 358, 1963. [Sac62] H. Sachs. Uber selbstkomplementare graphen. Publ. Math. Debrecen, 9:270 288, 1962. [SW] A. Szymańsk A.P. Wojda. A note on k-unform self-complementary hypergraphs of gven order. submted. [Szy05] A. Szymańsk. A note on self-complementary 4-unform hypergraphs. Opuscula Mathematca, 25/2:319 323, 2005. [Woj06] A.P. Wojda. Self-complementary hypergraphs. Dscussones Mathematcae Graph Theory, 26:217 224, 2006.
124 Artur Szymańsk A. Paweł Wojda