Another diametric theorem in Hamming spaces: optimal group anticodes

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1 Aother diametric theorem i Hammig spaces: optimal group aticodes Rudolf Ahlswede Departmet of Mathematics Uiversity of Bielefeld POB 003, D-3350 Bielefeld, Germay ahlswede@math.ui-bielefeld.de Abstract I the last cetury together with Levo Khachatria we established a diametric theorem i Hammig space H = (X, d H). Now we cotribute a diametric theorem for such spaces, if they are edowed with the group structure G P = G, the direct sum of a group G o X = {0,,..., q }, ad as cadidates are cosidered subgroups of G. For all fiite groups G, every permitted distace d, ad all d subgroups of G with diameter d have maximal cardiality q d. Other extremal problems ca also be studied i this settig. I. INTRODUCTION As i [2] we study optimal aticodes i Hammig spaces H = (X,d H ) but ow with the additioal costrait that they form a subgroup of G = G, the direct sum of a group G o X = {0,,...,q }. Thus we cosider where AG(,d) = max{ U : U is a subgroup of G is the diameter of U. with D(U) d}, (.) D(U) = max u,u U d H(u,u ) (.2) Farrell [5], see also [8], has itroduced aticodes (,r,d) as subspaces of GF(2) with diameter costrait d ad dimesio r. But eve this special case of our problem (cosistig i maximizig r for give,d) has ot eve bee cosidered. They were actually The author was supported by the Fiite Structures Marie Curie Host Fellowship for the Trasfer of kowledge project carried out by Alfred Reyi Istitute of Mathematics, i the framework of the Europea Commuity s Structurig the Europea Research Area programme. used for a aalysis of codes (see [8]) ad i that coectio words i U were eve cosidered with multiplicities. I [2] we solved the log stadig problem of determiig A(,d) = max{ A : A X with D(A) d} ad we gave up to isomorphy all extremal aticodes: For 0 i d 2 defie K i X as cartesia product of the ball B d+2i i ( 0) with ceter 0 d+2i ad radius i i X d+2i ad X d 2i. Clearly K i has diameter d. Diametric Theorem of [2]. Let r be the largest iteger s.t. { d + 2r < mi +, d + 2 d }, q 2 the A(,d) = K r. Moreover, up to permutatios of {, 2,..., } ad permutatios of the alphabet X = {0,...,q } i the compoets the optimal cofiguratio is uique, uless d >, d + 2 d q 2 ad d q 2 is itegral, i which case we have two optimal cofiguratios: K d q 2 ad K d q 2. Fially we metio that we write groups additive, because we write cocateatio of words multiplicative, for u G : u = u u 2...u. For A G ad a G we write Aa for the set {a = a a 2...a a : a a 2...a A} ad more geerally for B G m ad a l G l we write Ba l for the set {b m a l : b m B}. Furthermore for A G m, B G l AB = {ab : a A,b B}.

2 2 II. MORE NOTIONS Def. : The zero word of legth l ad the oe word of legth l is deoted by 0 l ad l, respectively. Def. 2: For U X (or G ) we defie for S X, S, U S = {u...u : u...u s U for all s S Clearly ad u...u s / U for all s X S}. U S U S = if S S. (2.) Def. 3: For U X we defie U ( ) = {u...u : u...u U for some u X }, which equals U S. S Def. 4: For U X we defie U () = {u X : there exists a u...u with u...u U}. Def. 5: For U X we defie for ε X U[ε] = {u = u...u U : u = ε}. Def. 6: For G = C 2, the cyclic group of order 2, we defie the dow-pushig operatio T 0 by settig for ay U = U {0} 0 U {} U {0,} {0,} T 0 (U) = U {0} 0. U {} 0 U {0,} {0,}. For coveiece we also write A = U {0},B = U {}, ad C = U {0,}. (2.2) Obviously, Def. 7: For two sets V, W distace is D(V, W) = U = T 0 (U). (2.3) X their maximal max d H(u,v). u V,v W III. THE BINARY CASE We first cosidered ad settled the case X = {0,} ad G = C 2. Theorem. For d (i) AC 2 (,d) = 2 d (ii) X d 0 d is optimal ad up to isomorphy (permutatios of compoets) uique for d 3. (iii) For d = 2 there is also the additioal solutio {0,0,0,000}0 3. The proof is based o the followig five lemmas. Lemma. For U {0,} D(U) D ( T 0 (U) ). Proof: Sice D(B) = D(B0), it remais to otice that D ( B,A0 C{0,} ) D ( B0,A0 C{0,} ). (3.) Lemma 2. For a subgroup U C2 a subgroup. T 0 (U) is agai Proof: Sice u + u = 0 C 2, it suffices to show that uε,vδ T 0 (U) implies (u + v)(ε + δ) T 0 (U). Sice T 0 (U)[0] U[0], the implicatio is obvious for ε = δ = 0 ad ε = δ =. So u + v0 remais to be checked. That is, u C{0,}, ad therefore also u0 C{0,}. Sice U is a subgroup, (u + v) ad (u + v)0 U ad i particular u + v C ad u + v0 C{0,}. Lemma 3. For a subgroup U C2 (i) C{0,} is a subgroup (ii) A0 is a subgroup (iii) C ad A are subgroups i C2 (iv) Either C = or A =. Proof: Ad (i) If u,v C, the uε,vδ U for all ε,δ {0,} ad therefore (u+v)0,(u+v) U ad (u + v)0,(u + v) C{0,},u + v C. Ad (ii) u0 + v0 = (u + v)0 A0 C{0,}. If ow (u+v)0 / A0, the (u+v) C ad both, (u+v)0 ad (u+v) C{0,}. But the (u+v)+u0 = v U ad sice also v0 U we get v C i cotradictio to v A. Ad (iii) This way it is also show that A is a subgroup i C 2. For C this is show already i (i). Ad (iv) By defiitio C A = ad as subgroups, if ot empty, they cotai both 0, a cotradictio. Lemma 4. For a subgroup U C2 B = ad U = C{0,}. C implies Proof: We kow from Lemma 3 that C implies A =. Now suppose that b B. The b + b = 0 0 C{0,} ad therefore 0 C{0,}, b+ 0 = b0 U, which cotradicts b B ad thus B =. Lemma 5. For a subgroup U C 2 with C =, clearly (i) U = A0 B = A0. A0 + α, U = 2 A (ii) T 0 (U) = A0. (A + g)0. Proof: (i) By Lemma 3 U = A0 B. Sice A0 is a subgroup of U, U = A0 I g i + A0. Necessarily, i= b + b = (b + b )0 A0 b = b + (b + b )0

3 3 ad cosequetly, I = ad U = A0 g + A0. (ii) This is obvious. Key Example: For the subgroup U = {0,0,0,000} = A0 B, we have T 0 (U) = {00,00,0,000} = A0 B0 = C Notice that D(U) = 2 = D ( T 0 (U) ). Fially, these lemmas make it possible, to iteratively apply trasformatios T 0 to a subgroup, keepig the cardiality costat ad ot icreasig the diameter. We keep extractig factors {0,} util i all compoets C = ad Lemma 5 applies, ad we ca extract a factor 0. The procedure eds with a subgroup of the form C d 20 d. We leave it as a exercise to show that the Key Example provides the oly other extremal cofiguratio. IV. A RELATED INTERSECTION RESULT IN THE BINARY CASE The case q = 2 of the Diametric Theorem stated i the Itroductio was first proved much earlier by D. Kleitma [7]. I [] it was show that this theorem ad Katoa s Itersectio Theorem i equivalet formulatio for uios ca be easily trasformed ito each other by usig operatios T 0. Sice these operatios trasform subgroups ito subgroups ad for E(U) = max u,u U W(u u ) with W coutig the umber of s, we have as aalogue to Lemma Lemma. For U {0,} E(U) E ( T 0 (U) ). so we also get as aalogue to Theorem for KC 2 (,d) = max{ U : U is subgroup of C 2 with E(U) d} Theorem. For d ad all d (i) KC 2 (,d) = 2 d (ii) X d 0 d is optimal ad up to isomorphy uique. The case (iii) i Theorem does ot occur because the equivalece holds for dowsets. Remark: The dual problem of itersectio becomes meaigless, because 0 has empty itersectios with other x U. However, we ca do it with a coset of a subgroup + U. This readily follows, because additio of amouts to complemetatio. V. THE CASE G = C 3 AND BEYOND We show ow how the previous approach geeralizes. We assume q = 3 ad the cyclic group C 3 of order 3. For ay subgroup U C 3 we cosider sets U S, that is, U {0}, U {}, U {2}, U {0,}, U {0,2}, U {,2}, U {0,,2}. A simple basic observatio is, that for u U {0,} u0 + u0 + u + u = u2 U (5.) ad therefore U {0,} =. Similarily, for v U {0,2} v0 + v0 + v2 + v2 = v U (5.2) ad therefore U {0,2} =. Fially, for w U {,2} ad therefore U {,2} =. This leaves us with We summarize this. w + w + w2 + w2 = w0 (5.3) U {0}, U {}, U {2}, U {0,,2}. Lemma 6. For a subgroup U C 3 (i) U = U {0} 0 U {} U {2} 2 U {0,,2} {0,,2} (ii) D(U) D ( T α 0 (U) ) for α =,2, where T α 0 is defied aalogously to T 0 i Defiitio 6. Proof: W.l.o.g. cosider ε = ad add to the proof of Lemma that D(U {}, U {2} 2) = D(U {} 0, U {2} 2), completig the preset proof. Lemma 7. For a subgroup U C3 a subgroup. T α 0 (U) is agai We leave the proof ad the establishmet of the aalogues of the other lemmas i Sectio 3. Also it is a ice exercise to fid out how our approach with operatios T α 0 goes. VI. A GENERAL APPROACH NOT USING DOWN PUSHING We have leart that subsets S cotaiig 0 play a basic role i provig Theorem. We deote them by S 0. They are the startig poit of our secod approach. We begi with Lemma 8. For a subgroup U G a o-empty U {0} 0 is a subgroup of U.

4 4 Proof: For u,v U {0}, if u0 + v0 U S S, S {0}, the a x S,x 0 exists with (u + v)x U S S ad (u + v)x u0 = vx U, but this cotradicts v U {0} (because v occurs with extesio 0 oly). Therefore u0 + v0 U {0}. It remais to be see that u0 has a iverse i U {0} 0. Clearly, it has a iverse v0 i U u0 + v0 = 0. (6.) If ow v0 U S S, S {0}, the for some x S,x 0 vx U S S ad u0 + vx = u0 + v0 + 0 x = 0 x U. Cosequetly u0 + 0 x = ux U i cotradictio with u0 U {0} 0. Therefore v0 U {0} 0 ad U {0} 0 is subgroup of U. Lemma 9 (Geeralizatio of Lemma 8). For a subgroup U G a o-empty U S0 0 is subgroup of U. Proof: If for u,v U S0 u0 + v0 U S0 0, the u0 + v0 U S 0 ad u+v U S, where S S 0 ad S S 0, because u0 U ad for all s S 0 vs U ad hece u0 + vs U, (u + v)s = u0 + vs U. Now for x S \ S 0 (u + v)x u0 = vx U, but this cotradicts v U S0 ad hece u0 + v0 U S0 0. It remais to be see that u0 has a iverse i U S0 0. There is a v0 i U with u0 + v0 = 0. (6.2) If v0 U S0 0, the v0 U S 0, where S S 0 ad S S 0, because for all s S 0 us U ad sice u0 + vs = us + v0 U also vs U. Now for x S \ S 0 u0 + vx U ad therefore ux + v0 U ad ux U i cotradictio to u U S0. Thus U S0 0 is a subgroup. Lemma 0. For a subgroup U G (i) There is exactly oe S 0 with U S0 (ii) S 0 is a group (iii) U S0 S 0 is a subgroup of U Proof: Ad (i) Sice 0 U S0 S 0 for all sets of type S 0 (by Lemma 9), disjoitess of these sets gives the result. Ad (ii) Sice 0 U S0 S 0, also 0 s U S0 S 0 for all s S 0, ad for all s,s S 0 0 s + 0 s = 0 s U. If s / S 0, the this cotradicts that 0 U S0. Therefore s+s = s S 0. Cocerig the iverse of s i S 0 use that 0 s has a iverse 0 ( s) U. Agai by defiitio of U S0 s S 0. Ad (iii) U S0 S 0 is subgroup because it is a direct sum of groups ad cotaied i U. VII. MORE ON THE STRUCTURE OF SUBGROUPS U We have leart that S 0 is a subgroup i G, that so is U S0 i G, ad fially U S0 S 0 i U. We ca decompose U ito cosets of U S0 S 0 ad begi with coset leaders of the form 0 α i (i =,2,...,I), elemets of U, such that S 0 + α i are disjoit for i =,2,...,I. (7.) This gives cosets i U: U S0 (S 0 + α i ) for i =,2,...,I. However, ecessarily I =, because otherwise we have a cotradictio with the defiitio of U S. So we may choose also α = 0. Next we cosider U S0+ϕ(α i)(s 0 + α i ), α i / S For example for U =, S 0 0 = {0}, U S0 = 0 {00,} we have. U S0 S 0 US0 S = U with α 2 = ad ϕ(α 2 ) = 0. Geerally usig U S0 S 0 we ca make a decompositio. U = (U S0 + γ)(s 0 + ψ(γ)) (7.2) for suitable ψ. Now comes a ew idea. Remember that γ U =. U S S. (7.3) By Lemma 0 there exists exactly oe S 0 with U S0. Lemma. If for a subgroup U G S 0 2, the the trasformatio L : ( ) U S S U S G S S

5 5 results i a group of diameter d ad a ot decreased cardiality. Proof: Use the decompositio i (7.2). Cosequetly every u occurig i some U S0 + γ has multiplicity = S 0 + ψ(γ) = S 0 ad gets by the trasformatio multiplicity G S 0. So the cardiality does ot decrease. Furthermore by (7.2) ad also D ( U S0 + γ ) d D ( U S0 + γ, U S0 + γ ) d ad the trasformatio L is appropriate. It remais to aalyse the case S 0 = {0}. Here, due to the defiitio of U S0, the decompositio U = ( US0 + ϕ(α) ) α holds with U() from α U() Defiitio 4. The terms ( U S0 + ϕ(α) ) α are disjoit or equal. If U S0 + ϕ(α) = U S0 + ϕ(β), ϕ(α) ϕ(β) U S0, sice d(α,β) = D ( U S0 + ϕ(α) ) = D(U S0 ) d, whe D ( U S0 +ϕ(α) ) d for all α we ca replace α by G. So it remais U S0 + ϕ(α) U S0 + ϕ(β) for all α β. [5 ] P.G. Farrell, Liear biary aticodes, Electroics Letters 6, 49-42, 970. [6 ] G. Katoa, Itersectio theorems for systems of fiite sets, Acta Math. Acad. Sci. Hug. 5, , 964. [7 ] D. Kleitma, O a combiatorial cojecture of Erdos, J. Combi. Theory, , 966. [8 ] F.J. MacWilliams ad N.J.A. Sloae, The Theory of Error- Correctig Codes, North-Hollad. Here replace all α by 0. As a illustratio look at the Example: U = 0 0 Theorem 3. For ay fiite abelia group G ad d AG(,d) = G d. As a further problem oe ca try to exted Theorem to the o-biary case with the costrait a b d. REFERENCES [ ] R. Ahlswede ad G. Katoa, Cotributios to the geometry of Hammig spaces, Discrete Mathematics 7, -22, 977. [2 ] R. Ahlswede ad L.H. Khachatria, The diametric theorem i Hammig spaces - optimal aticodes, Advaces i Applied Mathematics 20, , 998. [3 ] R. Ahlswede ad L.H. Khachatria, The complete itersectio theorem for systems of fiite sets, Europea J. Combiatorics, 8, 25-36, 997. [4 ] R. Ahlswede ad L.H. Khachatria, A pushig-pullig method: ew proofs of itersectio theorems, Combiatorica 9(), -5, 999.