For a linear code over GF (q) we consider two kinds of `subcodes&apos; called  residuals and punctures. When does the collection of residuals or punctures  determine the isomorphism class of the code? We call such a code residually   or puncture reconstructible. We investigate these notions of reconstruction  and show that, for instance, selfdual binary codes are puncture  and residually reconstructible. A result akin to the edge reconstruction of  graphs with sufficiently many edges shows that a code whose dimension is  small in relation to its length is puncture reconstructible

Maynard, Philip

Siemons, Johannes

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For a linear code over GF(q) we consider two kinds of “subcodes” called residuals and punctures. When does the collection of residuals or punctures determine the isomorphism class of the code? We call such a code residually or puncture reconstructible. We investigate these notions of reconstruction and show that, for instance, selfdual binary codes are puncture and residually reconstructible. A result akin to the edge reconstruction of graphs with sufficiently many edges shows that a code whose dimension is small in relation to its length is puncture reconstructible

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Discrete Mathematics 293 (2005) 205–211www.elsevier.com/locate/discEfﬁcient reconstruction of partitionsPhilip Maynard1, Johannes SiemonsSchool of Mathematics, University of East Anglia, Norwich NR4 7TJ, UKReceived 2 July 2003; received in revised form 10 May 2004; accepted 18 August 2004Available online 31 March 2005AbstractWe consider the problem of reconstructing a partition x of the integer n from the set of its t-subpartitions. These are the partitions of the integer n − t obtained by deleting a total of t from theparts of x in all possible ways. It was shown (in a forthcoming paper) that all partitions of n can bereconstructed from t-subpartitions if n is sufﬁciently large in relation to t. In this paper we deal withefﬁcient reconstruction, in the following sense: if all partitions of n are t−-reconstructible, what isthe minimum number N =N−(n, t) such that every partition of n can be identiﬁed from any N + 1distinct subpartitions?We determine the functionN−(n, t) and describe the corresponding algorithmfor reconstruction. Superpartitions may be deﬁned in a similar fashion and we determine also themaximum number N+(n, t) of t-superpartitions common to two distinct partitions of n.© 2005 Elsevier B.V. All rights reserved.Keywords: Reconstruction; Partitions1. IntroductionLet P(n) denote the set of all partitions of the integer n. Thus, if x ∈ P(n) then x =[x1, x2, . . . , xl] is amultiset of integersxi with 0xi and∑li=1 xi=n.To avoid cumbersomedistinctions we identify two partitions if they differ by parts of size 0 only. If x ∈ P(n) andif t is an integer with 0 tn then we say that x′ := [x1 − e1, x2 − e2, . . . , xl − el] is at-subpartition if 0eixi and∑li=1 ei = t .E-mail address: j.siemons@uea.ac.uk (J. Siemons).1 The author acknowledges ﬁnancial support through the project on Reconstruction Indices funded by theLeverhulme foundation.0012-365X/$ - see front matter © 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2004.08.037206 P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211The set of all t-subpartitions of x is denoted byDt(x) and we say that x is reconstructiblefrom its t-subpartitions, or that x is t−-reconstructible, if and only if the following is true:Whenever Dt(x) = Dt(y) for some y ∈ P(n), then x = y. For instance, x = [3, 2, 2, 1]has D2(x)= {[3, 2, 1], [3, 1, 1, 1], [2, 2, 2], [2, 2, 1, 1]} and by elementary arguments onecan show that x is 2−-reconstructible. On the other hand, D5(x)= {[3], [2, 1], [1, 1, 1]} =D5([5, 2, 1]) and so x is not 5−-reconstructible. Note also that in contrast to other recon-struction problems herewe have no information about themultiplicitywithwhich a partitionoccurs in Dt(x).The problem of reconstructing partitions was proposed in [4,1]. Intuitively, if t is small inrelation to n then the partitions of n should be t−-reconstructible, and this is proved in [5].In this paper we deal with an additional issue: if all partitions of n are t−-reconstructible,is there an N =N−(n, t) such that an arbitrary x ∈ P(n) can be identiﬁed from any N + 1distinct members of Dt(x)? This is the problem of efﬁcient reconstruction ﬁrst consideredin Levenshtein’s seminal paper [2] on the reconstruction of sequences. Here we also answerthe question of efﬁcient reconstruction of partitions. To state these results we need someadditional deﬁnitions.If x = [x1, x2, . . . , xl] ∈ P(n) and if 0 t and 0e1, e2, . . . , el′ are integers with l′ land∑l′i=iei = t then x′ = [x1+ e1, x2+ e2, . . . , xl + el, el+1, . . . , el′ ] is a t-superpartitionof x. The set of all t-superpartitions of x is denoted by Ut(x). The partition x is said to bereconstructible from its t-superpartitions, or t+-reconstructible, if and only if the followingis true: whenever Ut(x)= Ut(y) for some y ∈ P(n) then x = y. For n t0 we deﬁneN−(n, t) := maxx,y∈P(n); x =y|Dt(x) ∩Dt(y)| andD(n, t) := maxx∈P(n)|Dt(x)|,and for arbitrary t0 we deﬁneN+(n, t) := maxx,y∈P(n); x =y|Ut(x) ∩ Ut(y)| ,U(n, t) := maxx∈P(n)|Ut(x)|.Clearly, N−(n, t)D(n, t) and N+(n, t)U(n, t), and if x ∈ P(n) satisﬁes N−(n, t)<|Dt(x)| then x is t−-reconstructible. In fact, the functionN−(n, t)measures efﬁcient recon-structibility in the sense that for an arbitrary partition x ∈ P(n) any N−(n, t)+ 1 or moredistinct members of Dt(x) determine x uniquely, as intended above. A similar statementholds for N+(n, t) and t+-reconstructibility.Theorem 1.1. Let n2 and t1.ThenN−(n, t)=D(n−1, t−1) for n t andN+(n, t)=U(n+ 1, t − 1) for all t.This theorem is the exact analogue of the main result of Levenshtein in [2] on the efﬁcientreconstruction of a sequence from its sub- and supersequences. In particular, every partitionx ∈ P(n) is reconstructible from any two of its 1-subpartitions, as D(n − 1, t − 1) = 1for t = 1. In Section 2 it is shown that D(n, t) = |P(n − t)| for some values of n and t.P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211 207In the same sections we obtain the proof of the theorem above and we give algorithms toreconstruct the partitions in each case.We may regard partitions asYoung diagrams and hence as elements of theYoung lattice.In this lattice the set of elements of rank n isP(n), see [6, Chapter 7] for instance, and forx ∈ P(n) the setD1(x) are the partitions which appear in the Murnaghan–Nakayama rule.Therefore, partition reconstruction can be placed into the larger context of reconstructionproblems in lattices. In a recent paper Stanley [7] determines the number of standardYoungdiagrams of n+ t cells which contain a given partition of n. It may well be possible to usethese techniques to give formulae for the values of D(n, t) and U(n, t) in general.One further general comment should be made.As we have seen, partitions and sequencesboth belong to a class of reconstruction problems where reconstruction is guaranteed, andwhere even the problem of efﬁcient reconstruction has a satisfactory answer. One maytherefore ask which other kinds of reconstruction problems belong to that class. For orbitreconstruction of permutation groupswe know that also cyclic, and possibly solvable groupsbelong to this class. In [3] we consider the reconstruction problems associated with ﬁniteprimitive groups. In this generality efﬁcient reconstruction cannot be expected any longer.2. Subpartition reconstructionIf x ∈ P(n) is a partition, we shall always assume that x is in standard form x =[x1, x2, . . . , xl] with xixi+1 for i = 1, 2, . . . , l − 1. Also, we let ‖x‖ :=∑1 i l xi bethe number partitioned by x. If also y=[y1, y2, . . . , ym] is a partition, possibly of a differentinteger, the intersection x ∩ y is the partitionx ∩ y = [min{x1, y1},min{x2, y2}, . . . ,min{xv, yv}],where v =min{l, m}. Similarly, the union x ∪ y is the partitionx ∪ y = [max{x1, y1},max{x2, y2}, . . . ,max{xw, yw}],where w =max{l, m} and where we assume that xl′ = 0 for l′> l and ym′ = 0 for m′>m.These deﬁnitions make sense since we are assuming that our partitions are in standard form.Note that then also x ∪ y and x ∩ y are in standard form and further that x ∩ y = x if andonly if x ∪ y = y. We can now deﬁne a partial order on partitions byx ⊆ y if and only if x ∩ y = x.For any integer m> 0 we deﬁne the partitionS(m) := [m, m/2, m/3, . . . m/(m− 2), m/(m− 1), 1]of s(m) := ∑mi=1m/i. It is easy to see that for any u ∈ P(m) there exist subpartitionsof S(m) that are equal to u and S(m) is the smallest partition with this property. That is tosay, if x is a partition of some integer m′<s(m) then there are partitions u ∈ P(m) whichare not subpartitions of x. We can now proveLemma 2.1. Let x be a partition of n. Then x ∩ S(n− t) determines Dt(x) uniquely. Thatis, if x and y ∈ P(n) then Dt(x)=Dt(y) if and only if x ∩ S(n− t)= y ∩ S(n− t).208 P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211Proof. First we prove that if x∩S(n− t)=y∩S(n− t) thenDt(x)=Dt(y). It is sufﬁcientto show that if x ∩ S(n − t) = u then Dt(x) = Dr(u), where r = ‖u‖ + t − n. Certainly,Dr(u) ⊆ Dt(x). Thus, assume that there is some z ∈ Dt(x) with z /∈Dr(u). However, thenwe would have z ∩ S(n− t) = z for some z ∈ P(n− t), a contradiction.Next assume that x ∩ S(n− t) = y ∩ S(n− t). Let x ∩ S(t)= S = [s1, s2, . . . , sl] andy∩S(t)=T =[t1, t2, . . . , tl′ ] for some l, l′ ∈ N. Let the ﬁrst instance of inequality of thesesequences be, without loss, si > ti for some i ∈ {1, 2, . . . ,min{l, l′}}. Consider the partitionu=[si, si , . . . , si] of the integer isi . Now si(n− t)/i and so ‖u‖ i(n− t)/in− t .Therefore there is some v ∈ Dt(x) such that u ∩ v = u. It is clear that v /∈Dt(y). We can use Lemma 2.1 to manufacture non-reconstructible partitions: take x, y ∈ P(n)with x = y and x ∩ S(m) = y ∩ S(m) for some m ∈ N. Then Dn−m(x) = Dn−m(y) andso x and y are not reconstructible from their (n − m)-subpartitions. For example, take thepartition S(m) for any m ∈ N and consider x, y ∈ Ui(S(m)), different partitions of theinteger s(m)+i for any i ∈ N. By Lemma 2.1 we haveDv(x)=Dv(y)=Du(S(m))=P(m)where u= s(m)−m and v = u+ i.Next we need the following simple lemma.Lemma 2.2. For any n, t, i ∈ Nwith n t we haveD(n, t)D(n+i, t+i).Furthermore,for any n, t, i ∈ N with 1 i < n we have U(n, t)<U(n− i, t + i).Proof. Assume that y ∈ P(n). For the ﬁrst inequality observe that Dt(y) ⊆ Dt+i (x) forany x ∈ Ui(y) for any i ∈ N. This implies that D(n, t)D(n + i, t + i). Secondly, for1 i < n we have Ut(y) ⊂ Ut+i (x) for any x ∈ Di(y) and so U(n, t)<U(n − i, t + i).We note that in general we cannot have strict inequality in the ﬁrst inequality. Indeed, itfollows from Lemma 2.1 thatD(n, t)=D(n+ i, t + i)= |P(n− t)|for any i ∈ N, if ns(n− t).Theorem 2.3. If n2 and n t1, then N−(n, t)=D(n− 1, t − 1). Furthermore, thereexist partitions x and y of n such thatD1(x)∩D1(y) consists of a single partition z of n−1for which |Dt−1(z)| =N−(n, t).Proof. First we show that N−(n, t)D(n− 1, t − 1). Let z be a partition of n− 1 corre-sponding to D(n− 1, t − 1), i.e., |Dt−1(z)| is maximal among the partitions of n− 1. Wenote that for any 1m ∈ N and p ∈ P(m) we have |U1(p)|2. Thus let x, y ∈ U1(z)with x = y. Certainly Dt−1(z) ⊆ Dt(x),Dt (y) and hence it follows thatN−(n, t) |Dt(x) ∩Dt(y)| |Dt−1(z)| =D(n− 1, t − 1). (1)Next we show that N−(n, t)D(n − 1, t − 1). Thus assume that x, y ∈ P(n), x = yare such that |Dt(x)∩Dt(y)| is maximal. Set w := x ∩ y. ThenDt(x)∩Dt(y) ⊆ Dv(w),P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211 209where v = ‖x ∩ y‖ + t − n. In particular,|Dt(x) ∩Dt(y)| |Dv(w)|D(‖x ∩ y‖, v)D(n− 1, t − 1).The last inequality following from Lemma 2.2 upon taking i = n− 1− ‖x ∩ y‖.To prove the ﬁnal part we consider again the partitions x, y constructed at the beginningof the proof. It now follows that the inequalities of (1) are actually equalities and |Dt(x) ∩Dt(y)| = |Dt−1(z)| =D(n− 1, t − 1). In particular, for any x ∈ P(n) it follows that if |Dt(x)|D(n− 1, t − 1)+ 1 then x isreconstructible from its t-subpartitions.Corollary 2.4. Any two different 1-subpartitions of a partition allow its reconstruction.All partitions of n3 are reconstructible from their 1-subpartitions.Proof. Taking t = 1 in Theorem 2.3 we get N−(n, 1)=D(n− 1, 0)= 1. This proves theﬁrst statement. For the second, note that the only partitions of n that have only one differenttype of 1-subpartition are xl = [l, l, l, . . . , l] ∈ P(n) for any l | n. For n3 it is not hardto see that D1(xi) = D1(xj ) for any i, j | n with i = j . For n = 2 the two partitions of 2given by [1, 1] and [2] clearly have the same 1-subpartitions, hence the restriction. Algorithm for recovering a partition from its subpartitions. The following simple pro-cedure reconstructs a partition x ∈ P(n) if one knows n and at leastN−(n, t)+1 membersfrom the set Dt(x).(i) Take any y1, y2, . . . , yl ∈ Dt(x), where l = N−(n, t) + 1 and put them in standardform.(ii) Form the union x =⋃i=1,...,l yi .We claim that x=x, as follows. Since all partitions are in standard form, we have that yi ⊆ xand so⋃i=1,...,l yi ⊆ x. Assume that⋃i=1,...,l yi = x′ ⊂ x with, say, ‖x′‖ = n′<n. ByTheorem2.3 andLemma2.2 it follows that l >D(n−1, t−1)D(n′, n′+t−n) |Dv(x′)|,where v=n′+ t−n. Conversely, since yi ⊆ x′ for i=1, 2, . . . , l it follows that |Dv(x′)| l,a contradiction.No explicit formula for D(n, t) seems to be known for general n and t. For t = 1 andx=[x1, x2, . . . , xl] it is easy to see that |D1(x)| is the number of distinct xi , that is |D1(x)|=|{x1, x2, . . . , xl}|. ThusN−(n, 1) is the greatest integer j for which j (j +1)/2n. Alreadyfor t = 2 it is much more difﬁcult to give an explicit value for D(n, t).3. Superpartition reconstructionIn contrast to reconstruction from subpartitions, partitions are always reconstructible fromsuperpartitions.210 P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211Lemma 3.1. Every partition is reconstructible from its t-superpartitions, for any t ∈ N.In fact, if t is given, then just one suitably chosen member from Ut(x) allows the uniquereconstruction of x. If t is not given then just two suitably chosen members fromUt(x) sufﬁceto reconstruct x uniquely.Proof. Assume that x ∈ P(n) and without loss that n2 since |P(1)|=1. If t is known, se-lect q=[q1, q2, . . . , qr ] ∈ Ut(x)with the property that q1>p1 for anyp=[p1, p2, . . . , pr ′ ]∈ Ut(x)with p = q. It is clear then that x=[(q1− t), q2, . . . , qr ]. To determine twe selectin addition the (unique) partition in Ut(x) which has the largest number of parts. If thisnumber is l then clearly t = l − r . This completes the proof. The above lemma shows that for any x ∈ P(x) some two suitably chosen membersfrom Ut(x) allows the reconstruction of x. The next theorem gives the minimum number ofarbitrarily chosen members from Ut(x) needed to reconstruct x.Theorem 3.2. If n2 and t1 thenN+(n, t)=U(n+ 1, t − 1). Furthermore, there existpartitions x and y of n such that U1(x) ∩U1(y) consists of a single partition z of n− 1 forwhich |Ut−1(z)| =N+(n, t).Proof. First we show thatN+(n, t)U(n+ 1, t − 1). Thus let z ∈ P(n+ 1) be a partitioncorresponding toU(n+1, t−1). Since n2 there are at least two different 1-subpartitionsof z unless possibly z = zv = [v, v, . . . , v] for any v | n. However, consider the partitionsof n+ 1 deﬁned bywv =[v+ 1, v, v, . . . , v, v− 1] for v <n andwn=[n, 1]. It is not hardto see that Um(wv)Um(zv) for any v | n and m ∈ N. In particular, we may assume thatthere exist x, y ∈ D1(z) with x = y. Then we have Ut−1(z) ⊆ Ut(x), Ut (y). HenceN+(n, t) |Ut(x) ∩ Ut(y)| |Ut−1(z)| = U(n+ 1, t − 1). (2)Next we prove that N+(n, t)U(n + 1, t − 1). So let x and y be distinct partitions of nsuch that |Ut(x) ∩ Ut(y)| is maximal. Clearly, any z ∈ Ut(x) ∩ Ut(y) must contain both xand y, i.e. p = (x ∪ y) ⊂ z. It follows that |Ut(x) ∩Ut(y)| |Uv(p)|, where v = n+ t − land l = ‖x ∪ y‖. In particular,N+(n, t)= |Ut(x) ∩ Ut(y)| |Uv(p)|U(l, n+ t − l).Now if l > n+1, then by Lemma 2.2 it follows thatN+(n, t)<U(n+1, t−1), in contradic-tion to the ﬁrst part of the proof. It follows that l=n+1 and thenN+(n, t)U(n+1, t−1).To prove the ﬁnal part, we consider again the partitions x, y constructed at the beginningof the proof. It now follows that the inequalities of (2) are equalities and |Ut(x)∩Ut(y)| =|Ut−1(z)| = U(n+ 1, t − 1). Algorithm for recovering a partition from its superpartitions. The following simpleprocedure reconstructs a partition x ∈ P(n) if one knows n and at least N+(n, t) + 1members from the set Ut(x).(i) Take any y1, y2, . . . , yl ∈ Ut(x)where l=N+(n, t)+1 and put them in standard form.(ii) Form the intersection x =⋂i=1,...,l yi .P. Maynard, J. Siemons / Discrete Mathematics 293 (2005) 205–211 211We claim that x=x, as follows. Since all partitions are in standard formwe have that x ⊆ yiand so x ⊆⋂i=1,...,l yi . Assume that⋂i=1,...,l yi = x′ with x ⊂ x′. Say, ‖x′‖ = n′>n. ByTheorem3.2 and Lemma 2.2 it follows that l >U(n+1, t−1)U(n′, n=n′+t) |Uv(x′)|,where v= n+ t − n′. On the other hand, since x′ ⊆ yi for i = 1, 2, . . . , l we also have that|Uv(x′)| l, a contradiction. References[1] P. Cameron, Stories from the age of reconstruction, Congr. Numer. 113 (1996) 31–41.[2] V. Levenshtein, Efﬁcient reconstruction of sequences from their subsequences and supersequences, J. Combin.Theory Ser. A 93 (2001) 310–332.[3] P. Maynard, J. Siemons, On the reconstruction of permutation groups: general bounds, A equationes Mathe-maticae, 2005, in press.[4] V. Mnukhin, Combinatorial Properties of Partially Ordered Sets and GroupActions, TEMPUS Lecture Notes:Discrete Mathematics and Applications, vol. 8, J. Siemons (Ed.), 1993.[5] O. Pretzel, J. Siemons, On the reconstruction of partitions and applications, to appear.[6] R.P. Stanley, Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge, 1999.[7] R.P. Stanley, On the enumeration of skewYoung tableaux, Adv. Appl. Math. 30 (2003) 283–294.

On the reconstruction of linear codes

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On the reconstruction of linear codes

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