The Covering Excess Method in the Theory of Covering Codes

Transcription

1 The Covering Excess Method in the Theory of Covering Codes Wolfgang Haas Dissertation zur Erlangung des Doktorgrades der Fakultät für Mathematik und Physik der Albert-Ludwigs-Universität Freiburg im Breisgau Oktober 008

2 Dekan: Prof. Dr. Ludger Rüschendorf (Freiburg) 1. Referent: Prof. Dr. Jan-Christoph Schlage-Puchta (Gent). Referent: Prof. Dr. Laurent Habsieger (Lyon) Betzreuung: Prof. Dr. Dieter Wolke (Freiburg) Datum der Promotion:

3 Für Lara

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5 Acknowledgements First of all I wish like to thank Dieter Wolke (Freiburg) for his friendly support over many, many years. Without his permanent encouragement this thesis would not have been possible. I thank Laurent Habsieger (Lyon), whose fine papers on the subject are an important source of inspiration to me. Moreover I wish like to thank Jörn Quistorff (Berlin), who brought me nearer to the world of mathematics. Finally I thank Michael Botzet (Hamburg), who brought me to coding theory.

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7 Contents 1 Introduction A sphere-covering problem Overview of the thesis Preliminaries 5.1 Codes The covering excess A bound for K(n, 1) with even n Bounds for K(n, 1) with odd n The case n 5 (mod 6) The bound of van Wee The Theorem of Blass and Litsyn The new bound The case n 1, 3 (mod 6) Steiner systems Perfect coverings of spheres The general excess bound Habsiegers improvement and the excess matrix Congruence properties for the covering excess function Some lemmas Proof of Theorem Proof of Theorem The main property of the excess matrix Congruence set systems and their resolvability Simple properties of congruence set systems The classification of congruence set systems on small sets

8 3.6.3 Open questions Proof of Theorem The general excess bound improved Bounds for K(n, R) The bounds of van Wee and Honkala An improvement of Honkala s bound The case ɛ = R A Further results 71 A.1 Notes to Chapter and A. Notes to Chapter B The new bounds 73 Bibliography 75

9 Chapter 1 Introduction Let K(n, R) denote the minimal cardinality of a binary code of length n and covering radius R. This thesis is concerned with lower bounds on K(n, R) by the method of covering excess, which was first proposed by Johnson [18] in 197 and later rediscovered by van Wee [9] in A sphere-covering problem Let F = {0, 1} denote the finite field with two elements. To measure the similarity between two words of length n over F, we introduce their Hamming distance, which simply counts the number of places, where the words differ. The space F n of the binary words of length n over F, equipped with the Hamming distance, is one of the most important finite metric spaces. Like in every (finite) metric space there are two fundamental questions, the sphere-packing and the sphere-covering problem. (i) Determine the maximal number A(n, e+1) of non-overlapping spheres of given radius e, that can be placed in the Hamming space F n. (ii) Determine the minimal number K(n, R) of spheres of given radius R needed to cover the whole space F n. The quantity A(n, d) was widely studied during the last six decades in the framework of error-correcting codes. The sphere-covering problem (ii) however was not studied equally extensive, but in the last two or three decades, it attracted much attention to the mathematicians. In generally the determination of K(n, R) is very difficult and we have to be content with lower 3

10 and upper bounds. A trivial lower bound for K(n, R) certainly is the cardinality of the whole space F n divided by the cardinality of a Hamming sphere with radius R. Many methods are known to improve on this lower bound, the so-called sphere-covering bound. One of the most efficient method is the covering excess counting method, which is the object of the present thesis. It relies on the fact, that for arithmetical reasons under certain conditions a sphere-covering is forced to cover words in a small Hamming sphere more than once, the so-called covering excess. 1. Overview of the thesis In chapter we give the necessary background on codes as well as the notations and terminology on covering excess, which are used throughout this work. As a first instance of the covering excess counting method we give Johnson s and van Wee s bound on K(n, 1) for even n, which makes use of a forced covering excess in a sphere of radius one. In chapter 3 we study bounds for K(n, 1) when n is odd. We first use a method of Blass and Litsyn [1] to give the presently best known lower bound for K(n, 1) if n 5 (mod 6). Then we discuss the general excess bound, which follows from the fact, that if n 1 (mod p) for an odd prime p, there exists a covering excess on a sphere with radius p 1. We give a new lower bound for the covering excess, leading to a general improvement of the general excess bound for binary codes with covering radius one. In chapter 4 we study bounds for K(n, R), which rely on a covering excess in spheres with radius one. We describe a classical bound of van Wee for this case from 1988 and a slight improvement of this bound by Honkala. We present a short proof of a further improvement of Honkala s bound. Moreover we give an improvement of the van Wee bound in case that R + 1 divides n, when both bounds coincide and no general further progress of the van Wee bound is known up to now. 4

11 Chapter Preliminaries.1 Codes Let F = {0, 1} denote the finite field with two elements. The Hamming distance d(x, y) between x = (x 1,..., x n ) F n and y = (y 1,..., y n ) F n is defined by d(x, y) = {i {1,..., n} x i y i }. One easily checks, that F n, equipped with the Hamming distance, is a finite metric space, the so-called Hamming space. Any nonempty subset C F n is called a code. For x F n the distance d(x, C) is defined by d(x, C) = min d(x, c). c C We say that x R-covers y, if d(x, y) R holds. C R-covers x means d(x, C) R. In both cases we simply say cover instead of R-cover, if R is clear from the context. If C, then d(c) denotes the minimum distance of C: d(c) = min d(c 1, c ). c 1,c C,c 1 c Let B(x, e) denote the Hamming sphere with radius e centered on x F n, B(x, e) = {y F n : d(x, y) e}. One easily sees, that the cardinality of a sphere B(x, e) does not depend on 5

12 x F n. So we may set V (n, e) = B(x, e) = 0 i e ( ) n i (.1) for any x F n. Like in every (finite) metric space there exists the fundamental spherepacking problem: (i) Determine the maximal number of non-overlapping spheres of given radius e, that can be placed in the Hamming space F n. One easily sees, that the centers of the spheres in such a packing form a code C F n with minimum distance at least e + 1. Thus the number requested in (i) is A(n, e+1), where A(n, d) denotes the maximal cardinality of a code C F n with minimum distance at least d. Assume that a set of messages is encoded by the words of a code C F n with minimum distance at least e + 1. If during the transmission of such a word there occur at most e errors, then these errors can be corrected by the receiver. Thus A(n, d) is the most basic quantity in coding theory, since A(n, e + 1) is the maximal cardinality of a code correcting up to e errors. For a monograph on error-correcting codes see [6]. Updated internet tables for A(n, d) are given by Brouwer []. In this thesis however we are concerned with the dual sphere-covering problem: (ii) Determine the minimal number of spheres of given radius R needed to cover the whole space F n. The covering radius R = R(C) of a code C F n denotes the least integer R, such that the spheres with radius R centered on the codewords cover the whole space. It is clear, that the centers of the spheres of a covering described in (ii) form a code with covering radius at most R. Equivalently, a code C F n has covering radius (at most) R, if x F n c C with d(x, c) R (.) holds or, simply, if each x F n is R-covered by C. K(n, R) denotes the minimal cardinality of a code C F n with covering radius (at most) R. 6

13 In the last two or three decades the theory of covering codes attracted a great amount of attention. A monograph [4] appeared in Updated internet tables for K(n, R) are given by Kéri [19]. An internet bibliography on covering codes is compiled by Lobstein [5]. Let 0 = (0,..., 0) F n denote the all-zero word of length n. Replacing the ith coordinate of the all-zero word by one yields the unit vector e i F n, 1 i n. By the weight w(x) of a word x F n we denote the number of its nonzero coordinates. The support of x F n = (x 1,..., x n ), supp(x), is defined by supp(x) = {1 i n x i 0}. (.3) Binary vectors may be identified with their supports and if x, y F n, then x y, x y and x y refer to the supports of x and y. We further set F n i = {x F n the ith coordinate of x equals 1}. Let C F n be a code with covering radius R and cardinality K(n, R). Then by F n c C B(c, R) we get Fn c C B(c, R) = C V (n, R) and thus K(n, R) n V (n, R). (.4) This is the most basic lower bound on K(n, R), called the sphere-covering bound. In generally this bound cannot be improved, since there are codes attaining equality in (.4). These codes are called perfect. One especially famous family of perfect codes are the so-called Hamming codes. Theorem 1 (Hamming [11]). If n = m 1, then there exists a perfect code C F n with covering radius one. Especially K(n, 1) = n m. In the majority of cases however the sphere-covering bound can be improved. Many different methods are known to do this, see [4]. One of the most efficient method is the covering excess counting method, which is the object of the present thesis. It relies on the fact, that sometimes a covering code is forced to cover words in a small Hamming sphere more than once, the 7

14 so-called covering excess. As a simple example it could be possible, that a codeword covers always an even number of points in a Hamming sphere, but the cardinality of the sphere is odd. Then a covering excess in that sphere is unavoidable. The first who introduced the method of covering excess was Johnson [18] in 197, but his remarkable paper remained unnoticed until the appearance of a monograph on covering codes [4] in The method was rediscovered by van Wee [9] in 1988, so it should be quite fair, to attribute the method both to Johnson and van Wee. After some prerequisites in section. we recall their most prominent example in section.3, a lower bound for K(n, 1) for even n.. The covering excess In this section we give the definition of covering excess, some notations and a few basic equations, which will be used throughout this work. Let C F n be a code with covering radius R. For x F n we define the covering excess E(x) by E(x) = E C (x) = B(x, R) C 1. (.5) By (.) E(x) 0 for every x F n, and E(x) is strictly positive, if x is covered by at least two codewords from C. For V F n, i 0 and x F n we set E(V ) = y V E(y), E i (x) = E(y), y F n,d(x,y)=i Z = {z F n E(z) > 0}, Z i = {z F n E(z) = i}. 8

15 For V F n we have E(V ) = ( B(x, R) C 1) (.6) x V = 1 V x V c B(x,R) C = 1 V c C x B(c,R) V = B(c, R) V V. c C In the special case V = F n we get E(F n ) = C V (n, R) n, (.7) which in case of R = 1 by (.1) further simplifies to E(F n ) = C (n + 1) n. (.8).3 A bound for K(n, 1) with even n The case of binary codes with covering radius one is of special interest. Here the sphere-covering bound (.4) reads K(n, 1) n n + 1. (.9) We now present the argument of Johnson and van Wee to improve on this bound for even n, which is probably the most famous application of the covering excess method. Theorem (Johnson [18], van Wee [9]). Let C F n be a binary code with covering radius one. If x F n, then { 1 (mod ) if n is even and x C, E(B(x, 1)) 0 (mod ) if n is odd. 9

16 Proof. Let δ(x) = 1 if x C, otherwise δ(x) = 0. If c C, then B(c, 1) B(x, 1) equals the number of words in the sphere B(x, 1), which were 1- covered by c. One easily sees, that B(x, 1) = n + 1 if d(c, x) = 0, B(c, 1) B(x, 1) = if d(c, x) = 1 or, 0 if d(c, x) 3. Now (.6) applied to V = B(x, 1) yields E(B(x, 1)) = c C and Theorem follows. B(c, 1) B(x, 1) B(x, 1) = δ(x)(n + 1) + c C, 1 d(c,x) (δ(x) 1)(n + 1) (mod ) 1 (n + 1) Now let n be even and C F n a binary code with covering radius one. Since E(B(x, 1)) is always nonnegative, from Theorem follows E(B(x, 1)) 1 if x C. (.10) Thus there exists a covering excess on every sphere of radius one not centered on a codeword. This was used by Johnson and van Wee to derive the following improvement of the sphere-covering bound (.9), the first excess bound for binary codes with covering radius one (see [4]). Theorem 3 (Johnson [18], van Wee [9]). When n is even we have K(n, 1) n n. Proof. Let C F n be a binary code with covering radius one. Note, that z Z means E(z) 1 and thus B(z, 1) C, implying B(z, 1) \ C B(z, 1) = n 1 if z Z. (.11) 10

17 We now get n C = 1 x F n \C x F n \C = E(B(x, 1)) by (.10) E(z) x F n \C z B(x,1) = z Z E(z) x B(z,1)\C (n 1)E(F n ) by (.11) = (n 1)( C (n + 1) n ) by (.8) 1 implying C n /n and Theorem 3 follows. In general, Theorem 3 cannot be improved, as we see in the next theorem. Theorem 4 (Johnson [18], van Wee [9]). For m 1 we have K( m, 1) = m m. Proof. By Theorem 1 there exists a perfect code C F m 1 of length m 1, covering radius one and cardinality C = m m 1. One easily sees, that the code C F m of length m defined by C = {(0, x) x C} {(1, x) x C} has covering radius one and satisfies C = m m. Therefore K( m, 1) m m. The corresponding lower bound follows from Theorem 3. If n is odd, there is no forced covering excess in a sphere of radius one. For instance the Hamming sphere B(0, 1) centered on the all zero word 0 of length n could be perfectly covered (i.e. without covering excess) by the words e 1, e + e 3, e 4 + e 5,..., e n 1 + e n. What can we do to improve on the sphere-covering bound? We have to look for a covering excess in a sphere with radius greater than one! 11

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19 Chapter 3 Bounds for K(n, 1) with odd n In this chapter we study bounds for K(n, 1), where n is odd. We start with the case n 5 (mod 6), where a covering excess in a sphere of radius two can be found. We describe a classical bound of van Wee [9] for this case as well as an improvement due to Blass and Litsyn [1]. We further improve on this bound, giving the presently best known lower bound for K(n, 1) if n 5 (mod 6). Next we discuss the general excess bound for K(n, 1) (due to Habsieger [9] and Honkala [16]), which follows from the fact, that if n 1 (mod p) for an odd prime p, there exists a covering excess p 1 on a sphere with radius p 1. Answering a question of Habsieger, we prove that this covering excess indeed grows quadratic in p. This yields a general improvement of the general excess bound for binary codes with covering radius one. The proof uses a classification theorem for certain subset systems as well as new congruence properties for the excess function, which were conjectured by Habsieger. 3.1 The case n 5 (mod 6) The bound of van Wee Let C F n be a binary code with covering radius one. As we have already seen, there is no need for a covering excess in a sphere of radius one in case of odd n, so lets hope n 1 (mod 3), when the following theorem of van Wee applies. 13

20 Theorem 5 (van Wee [9]). Let C F n be a binary code with covering radius one and n 1 (mod 3). If x F n, then E(B(x, )) 1 (mod 3). (3.1) Proof. If c C, then B(c, 1) B(x, ) equals the number of words in the sphere B(x, ), which were 1-covered by c. One easily sees, that B(c, 1) = n + 1 if d(c, x) 1, B(c, 1) B(x, ) = 3 if d(c, x) = or 3, 0 if d(c, x) 4. Now (.6) applied to V = B(x, ) yields E(B(x, )) = c C = (n + 1) B(c, 1) B(x, ) B(x, ) c C, d(c,x) 1 ( ) n (mod 3) 1 (mod 3) by (.1) and n 1 (mod 3) c C, d(c,x) 3 1 ( 1 + n + ( )) n If n 1 (mod 3) with even n, then the better bound in Theorem 3 applies, so we may restrict us to the case n 5 (mod 6). Theorem 6 (van Wee [9]). When n 5 (mod 6), we have ( ) 4 n K(n, 1) 1 + n + n + n + 1. Proof. Let C F n be a binary code with covering radius one and C = K(n, 1). Since E(B(x, )) is always nonnegative for x F n, from (3.1) follows E(B(x, )) for each x F n. (3.) 14

21 We now get n = x F n x F n E(B(x, )) by (3.) = x F n z B(x,) = z F n E(z) E(z) x B(z,) = V (n, )E(F n ) = V (n, )( C (n + 1) n ) by (.8) and Theorem 6 follows by V (n, ) = 1 + n + ( n ) (see (.1)) The Theorem of Blass and Litsyn Theorem 6 was further improved by Blass and Litsyn [1], Habsieger [9], Honkala [15], [16], [17] and van Wee [31]. The best of these bounds is due to Blass and Litsyn [1]: Theorem 7 (Blass and Litsyn [1]). If n 5 (mod 6), then ( ) 40 n K(n, 1) n 44n + 77 n + 1. (3.3) In section we give a simplification of the proof of Theorem 7, which at the same time leads to a better result. We give the complete proof (for n 17), since it is divided on the three papers [1], [7], [16] and it saves the efforts of the reader to find out, which of the material of [1] is needed and which not. For the rest of this section we give the necessary lemmas, which already can be found in [1] and [16], in a slightly modified manner. All improvements on Theorem 6 mentioned above rely on the fact, that not always equality can hold in (3.) and so one has to upper-bound B, where B is defined by 1 B = {x F n E(B(x, )) = }. (3.4) Crucial is a structural property of B described in the next lemma. Assume for the rest of this section, that C F n is a fixed binary code with covering radius one and n 5 (mod 6), n

22 Lemma 8 (van Wee [31]). For every x B there exists a unique z B(x, ) with E(z) > 0. Consequently E(z) = (i.e. z Z ). Proof. Assume to the contrary, there exist distinct z 1, z B(x, ) with E(z 1 ) > 0 and E(z ) > 0. One easily sees, that in all cases there exists y B(x, 1) such that, say z 1 B(y, 1), but z B(y, 1). Now E(B(y, 1)) E(z 1 ) 1. On the other hand by Theorem we have E(B(y, 1)) 0 (mod ), implying E(B(y, 1)). This however gives the contradiction = E(B(x, )) E(B(y, 1)) + E(z ) 3. For x F n we set L = {y F n : E(B(y, )) = 5}, B(x) = B(x, ) B. One easily checks, that if z Z, then E(B(z, )) > and thus E(B(z, )) 5 by Theorem 5. We partition Z into the following subsets (different from [1]): Z (1) = L Z, { ( )} Z () = z Z \ Z (1) n 3 B(z) >, { ( )} Z (3) = z Z \ Z (1) n 3 B(z). (3.5) The next lemmas are from [1]. Lemma 9 (Blass, Litsyn [1]). If z Z, then ( ) n B(z). (3.6) If additionally z Z (1), then there are exactly three distinct words y i B(z, ), i = 1,, 3, such that E(y i ) = 1. For every c C with d(y i, c) 1 for at least one y i we also have d(z, c) 1. Proof. By z Z, z is covered exactly three times. Case I. z C. Without loss z = 0 and C = {0, e 1, e } C. Clearly E(e 1 ) 1, E(e ) 1, E(e 1 + e ) 1. (3.7) 16

23 To show (3.6), simply note that for x B(0) we have d(x, e 1 ) >, since otherwise E(B(x, )) E(0) + E(e 1 ) 3 contradicting E(B(x, )) = for x B(0) B. Similar d(x, e ) >. Thus every word in B(0) has weight two and starts with two zeros, i.e. B(0) {e i + e j 3 i < j n}, (3.8) and (3.6) follows in case I. If z = 0 Z (1), then E(B(0, )) = 5 and by (3.7) in fact we have E(e 1 ) = E(e ) = E(e 1 + e ) = 1. (3.9) Thus we may set y 1 = e 1, y = e and y 3 = e 1 + e. Assume c C with d(y i, c) 1 for at least one y i. By (3.9), y i is covered exactly twice. Since each y i, i = 1,, 3, is covered by two words from C, we find c C and thus d(z, c) = d(0, c) 1. Case II. z C. Without loss z = 0 and {e 1, e, e 3 } C. Clearly E(e 1 + e ) 1, E(e 1 +e 3 ) 1 and like in case I one shows d(x, e 1 +e ) > and d(x, e 1 +e 3 ) > whenever x B(0). Thus every word in B(0) has weight one or two and starts with three zeros, i.e. B(0) {e i 4 i n} {e i + e j 4 i < j n}, (3.10) implying B(0) n 3 + ( ) ( n 3 = n ) and (3.6) follows in case II too. Setting y 1 = e 1 + e, y = e + e 3 and y 3 = e 1 + e 3, the proof of the second half of the Lemma is very similar like in case I. Lemma 10 (Blass, Litsyn [1]). We have Z 1 3 Z (1). (3.11) Proof. By Lemma 9, Z 1 B(z, ) = 3 whenever z Z (1). Thus Lemma 10 follows, if we can show, that no y Z 1 satisfies y B(z 1, ) B(z, ) whenever z 1, z Z (1) with z 1 z. Assume to the contrary, that y Z 1 B(z 1, ) B(z, ). Since y is covered by C, there exists c C with d(c, y) 1. By Lemma 9 we get d(c, z 1 ) 1 as well as d(c, z ) 1, implying d(z 1, z ). But then B(z 1, ) contains two words from Z (z 1 and z ) as well as three words from Z 1 by Lemma 9 and thus E(B(z 1, )) 7 contradicting z 1 Z (1). 17

24 Lemma 11 (Blass, Litsyn [1]). If z Z (), there are exactly three words y i B(z, ) \ {z} (i = 1,, 3) satisfying E(y i ) =. Proof. Case I. z C. Without loss z = 0 and {0, e 1, e } C. We set y 1 = e 1, y = e and y 3 = e 1 + e. By (3.7) we have y i B(z, ) \ {z} and E(y i ) > 0 (i = 1,, 3). We next prove, that there is no other word y B(0, ) \ {0} satisfying E(y) > 0. Without loss it suffices to consider the cases y = e 3 or y = e i + e 3 with i 3. By E(e 3 ) > 0 (resp. E(e i + e 3 ) > 0) and E(0) = we have E(B(e 3 + e j, )) > for j 3, implying e 3 + e j B(0). Therefore by (3.8) each word of B(0) must be of the form e i + e j, 4 i < j n. The number ) (), contradicting the definition of Z of such vectors is at most ( n 3. It remains to show E(y 1 ) = E(y ) = E(y 3 ) =. The case E(y 1 ) = E(y ) = E(y 3 ) = 1 is impossible, since it yields E(B(z, )) = 5 and thus z Z (1), contradicting the definition of Z (). If E(y 1 ) or E(y 3 ), then e 1 + e (with e = 0 or e = e ) must be covered by a codeword c C, C 0, e 1, e. Now c = e 1 + e + e j with j 3 is impossible, since then E(y) > 0 with y = e 1 + e j and y B(z, ) \ {z, y 1, y, y 3 }. There remains the possibility c = e 1 + e C and E(y 1 ) = E(y 3 ) = follows. For reasons of symmetry we also have E(y ) =. Case II. z C. Without loss z = 0 and {e 1, e, e 3 } C. We set y 1 = e 1 +e, y = e +e 3 and y 3 = e 1 + e 3. Clearly E(y i ) > 0 (i = 1,, 3). We show, that there is no other word y B(0, ) \ {0} with E(y) > 0. It suffices to consider the cases y = e i0 or y = e i0 + e 4, 1 i 0 n. Similar like in case I we see, that we then have B(0) B 1 := {e i + e j 4 i < j n} in the first case and B(0) B := {e i 5 i n} {e i + e j 5 i < j n} in the second case (use (3.10)). Since B 1 = B = ( ) n 3 we get a contradiction to the definition of Z (). It remains to show E(y 1 ) = E(y ) = E(y 3 ) =. Like in case I E(y 1 ) = E(y ) = E(y 3 ) = 1 is impossible. For reasons of symmetry it suffices to assume E(y 1 ). Then y 1 = e 1 +e must be covered by a codeword c C, c = e 1 + e + e with e = 0 or e = e i, i 3. Now e = 0 (resp. e = e i, i 4) is impossible, since then E(e 1 ) > 0 (resp. E(e 1 + e i ) > 0). Hence y 1 C, c = e 1 + e + e 3 C and E(y 1 ) = follows. Since c also covers y and y 3, we get E(y ) and E(y 3 ). Now like in the case of y 1 we have 18

25 E(y ) = E(y 3 ) = also. For z Z () we set H(z) = {z, y 1, y, y 3 } Z () Z (3), (3.1) where y 1, y and y 3 are defined in Lemma 11. Clearly H(z) = 4 if z Z (). (3.13) Following the proof of Lemma 11, we notice, that if z C, then for some i, j {1,..., n}, i j If z C, then H(z) = {z, z + e i, z + e j, z + e i + e j } C. (3.14) H(z) = {z, z + e k + e l, z + e k + e m, z + e l + e m } with H(z) C = (3.15) for some k, l, m {1,..., n}, k l m k. {z + e k, z + e l, z + e m, z + e k + e l + e m } C. Moreover in that case Lemma 1 (Blass, Litsyn [1]). If z 1, z Z (), then either H(z 1 ) H(z ) = or H(z 1 ) = H(z ). Proof. Assume y H(z 1 ) H(z ). By definition of H(z) we have E(y) =. Case I. y C. Then z 1, z C (otherwise H(z 1 ) C = ). From (3.14) follows H(z 1 ) = {z 1, z 1 + e i, z 1 + e j, z 1 + e i + e j } = {y, y + e i, y + e j, y + e i + e j } C for some i, j {1,..., n}, i j. Similar by y H(z ) we have H(z ) = {y, y + e k, y + e l, y + e k + e l } C for some k, l {1,..., n}, k l. We now get {i, j} = {k, l} (otherwise E(y) > ), i.e. H(z 1 ) = H(z ). Case II. y C. Then z 1, z C (otherwise H(z 1 ), H(z ) C). From (3.15) follows H(z 1 ) = {z 1, z 1 + e k + e l, z 1 + e k + e m, z 1 + e l + e m } = {y, y + e k + e l, y + e k + e m, y + e l + e m } for some k, l, m {1,..., n}, k l m k and {z 1 + e k, z 1 + e l, z 1 + e m, z 1 + e k + e l + e m } = {y + e k, y + e l, y + e m, y + e k + e l + e m } C. Similar by y H(z ) we have H(z ) = {y, y + e k + e l, y + e k + e m, y + e l + e m } for some k, l, m {1,..., n}, k l m k and {y + e k, y + e l, y + e m, y + e k + e l + e m } C. Like in case I we now get {k, l, m} = {k, l.m }, i.e. H(z 1 ) = H(z ). 19

26 The proof of the case z C of the following Lemma already appeared in Honkala [16]. Lemma 13 (Blass, Litsyn [1], Honkala [16]). If z Z () we have ( ) n 3 B(y) 4. (3.16) y H(z) Proof. By the definition of H(z) we have y Z whenever y H = H(z). Indeed we will show ( ) n 3 B(y) 3 (3.17) y H\{x} whenever x H, since by (3.13) this implies 3 B(y) = ( ) n 3 B(y) 1 y H x H and (3.16) follows. y H\{x} Case I. z C. Without loss z = 0 and H = {0, e 1, e, e 1 + e }, see (3.14). Moreover, it suffices to show (3.17) for x = e 1 + e, i.e. H := H \ {x} = {0, e 1, e }. For y H we have B(y) B (y) := {y + e i + e j 3 i < j n}, (3.18) (compare (3.8)). Note, that the sets B (y), y H are pairwise disjoint. We now consider the covering excess in a sphere of radius two centered on e i, 3 i n. Clearly B(e i, ) H = H and thus E(B(e i, ) H) = 6 by E(y) = whenever y H. By Theorem 5 however we have E(B(e i, )) 1 (mod 3). This implies 3 i n u i B(e i, ) \ H with E(u i ) > 0. (3.19) We set B = y H B (y) and consider two subcases. Subcase A. Each word u i has weight at least two, 3 i n. We first show B(u i, ) B F n i 6 (3 i n). (3.0) 0

27 To see this, fix i with 3 i n. Note that in subcase A by u i B(e i, ) we have u i F n i. If u i F n 1, then B(u i, ) B (e1) F n i = B (e1) F n i implying B(u i, ) B (e1) F n i = B (e1) F n i = n 3 6 and (3.0) follows. If u i F n or w(u i ) =, then the argument is very similar. There remains the case, that u i starts with two zeros and satisfies w(u i ) = 3. Then for each of the three elements y H we have B(u i, ) B (y) F n i = and (3.0) follows again. Each word in B occurs in exactly two of the sets B F n i, 3 i n. Therefore (B(u i, ) B ) (3.1) 3 i n (B(u i, ) B F n i ) 1 B(u i, ) B F n i 3(n 3). 3 i n 3 i n by (3.0). Now, if x B(u i, ) B, (3 i n), then there exists y H with x B (y ) and thus y B(x, ). Therefore E(B(x, )) E(y ) + E(u i ) >, which implies x y H B(y) whenever x B(u i, ) B, 3 i n. Using (3.18) and (3.1) we now get B(y) B (B(u i, ) B ) y H 3 i n ( ) ( ) n n 3 3 3(n 3) = 3 and (3.17) follows, completing the proof of Lemma 13 in subcase A. Subcase B. There is a word u i with w(u i ) = 1. If e j0 with a suitable j 0, 3 j 0 n occurs among the words u i, then (3.1) follows again, since then each of the 3(n 3) words of the form y + e i + e j0, y H, 3 i n, i j 0 lies in B(e j0, ) B. Since we have already seen, that (3.1) implies (3.17), the proof of Lemma 13 is complete in case I. Case II. z C. In this case by (3.15) we may assume z = 0 and H = {0, e 1 + e, e 1 + e 3, e + e 3 }. It suffices to show (3.17) for x = e + e 3, i.e. H = H \ {x} = {0, e 1 + e, e 1 + e 3 }. For y H we have B(y) B (y) := {y + e i 4 i n} {y + e i + e j 4 i < j n}, 1

28 see (3.10). The excess argument from case I leading to (3.19) applies equally well to the spheres B(e 1 + e i, ), 4 i n and yields 4 i n v i B(e 1 + e i, ) \ H with E(v i ) > 0. Again we set B = y H B (y) and consider two subcases. Subcase A. Neither e 1 nor one of the words e 1 + e j occurs among the words v i, 4 i, j n. We show B(v i, ) B F n i 6 (4 i n). (3.) To see this, fix i with 4 i n. Again in subcase A we have v i F n i. If v i does not start with (100), then B(v i, ) B (y) F n i = n 3 6 for a suitable y H and (3.) follows. If v i starts with (100) then B(v i, ) B (y) F n i = for every y H and (3.) follows again. Now (3.1) (with u i replaced by v i ) follows like in case I. The rest of the proof of Lemma 13 in subcase A is the same. Subcase B. Either e 1 or one of the words e 1 + e j, 4 j n occurs among the words v i, 4 i n. If e 1 occurs among the words v i, then each of the 3(n 3) words of the form y+e i, y H, 4 i n lies in B(e 1, ) B. If e 1 +e j0 with a suitable j 0, 4 j 0 n occurs among the words v i, then each of the 3(n 3) words of the form either y + e j0 or y + e i + e j0, y H, 4 i n, i j 0 lies in B(e 1 + e j0, ) B. In both cases the proof proceeds exactly like in case I The new bound In this section we further improve on (3.3). The bound given in the next theorem appears to be the presently best lower bound on K(n, 1) when n 5 (mod 6). Theorem 14 (Haas [7]). If n 11 and n 5 (mod 6), then ( ) 8 n K(n, 1) 1 + n 10n + 19 n + 1. As a corollary we get the following new bounds on K(n, 1). records in parenthesis follow from (3.3). The old

29 Corollary 15. K(17, 1) 7419 (7414 [1]), K(3, 1) 3587 (35736 [1]), K(9, 1) ( [1]). Proof of Theorem 14. We use the notations of the previous section. Let n be an integer with n 11 and n 5 (mod 6). By (3.3) Theorem 14 does hold for n = 11, so we may assume n 17. Let C F n be a binary code of length n with covering radius one. By M Z () we denote a set of maximal cardinality satisfying H(z 1 ) H(z ) = whenever z 1, z M with z 1 z. Moreover we set H = H(z). Then by Lemma 1, (3.1), (3.13) we have H = 4 M and z M Lemma 16. If z L, then B(z, ) Z 1. Z () Z (3) = H (Z (3) \ H). (3.3) Proof. By E(B(z, )) = 5 there exists u B(z, ) with E(u) odd. Fix any x B(u, 1) B(z, 1). By Theorem we have E(B(x, 1)) 0 (mod ). Thus there exists v B(x, 1) B(z, ) with v u and E(v) odd. Now, since E(B(z, )) is odd, there must exist w B(z, ) with E(w) odd, w u and w v. Since E(u), E(v) and E(w) are odd, nonnegative integers adding to at most 5, at least two of them must equal one and Lemma 16 follows. We now may complete the proof of Theorem 14. From Lemma 8 follows B z Z B(z) (remember, that B is defined in (3.4)). Thus by (3.3) we get B z Z B(z) = z Z (1) B(z) z M y H(z) B(y) z Z (3) \H B(z), 3

30 which by (3.5), (3.3), Lemma 9 and Lemma 13 implies B B(z) + B(y) + B(z) (3.4) z Z (1) z M y H(z) z Z (3) ( ) ( ) ( \H ) n Z (1) n 3 n M + (Z () Z (3) ) \ H ( ) ( ) n = Z (1) n 3 + Z () Z (3) ( ) n 3 = Z + (n 3) Z (1). Moreover E(F n ) = i 0 i Z i Z + Z 1. Thus By Theorem 5 we have ( ( )) n 1 + n + E(F n ) = V (n, )E(F n ) Z E(F n ) Z 1. (3.5) = x F n E(B(x, )) B + 5 L + 8( n B L ) = 8 n 6 B 3 L. Substituting (3.11), (3.4) and (3.5) on the right-hand side yields ( ( ) ( )) n n n E(F n ) (3.6) ( ( ) ) n 3 8 n + 3 (n 3) Z 1 3 L. By Lemma 16 for n 17 we have ( ( ) ) n 3 3 (n 3) Z 1 3 = ( 1 + n + z Z 1 x B(z,) ( )) n Z x L z B(x,) Z 1 3 L.

31 This inserted in (3.6) yields E(F n ) and Theorem 14 follows by (.8). 8 ( ) n ( ) n n n The case n 1, 3 (mod 6) We have seen, that in the case n 5 (mod 6) a binary code of length n and covering radius one forces a covering excess in every sphere of radius two (Theorem 5). In this section we have a look on the remaining odd cases n 1, 3 (mod 6) and show, that then there are coverings of spheres with radius three indeed without covering excess. For the proof we have to review shortly some basic facts on Steiner systems Steiner systems We shortly give the definition and a few basic facts for Steiner systems, which are basic objects in design theory. They are related to the construction of perfect coverings of Hamming spheres, i.e. covering codes without covering excess on certain spheres. Definition 17. A Steiner system S(t, k, n) is a collection of k-subsets (called blocks) of {1,..., n}, such that every t-subset of {1,..., n} is contained in exactly one of the blocks. The connection to coding theory is given by the quantity supp defined in (.3), which yields a natural one to one correspondence between the words x F n and the subsets of {1,..., n}. The history of Steiner systems goes back until 1847, when Kirkman considered the special case t = and k = 3. A Steiner system S(, 3, n) is called a Steiner triple system. Theorem 18 (Kirkman [0] 1847). A Steiner triple system S(, 3, n) exists if and only if n 1, 3 (mod 6). A Steiner system S(3, 4, n) is called a Steiner quadrupel system. Theorem 19 (Hanani [1] 1960). A Steiner quadrupel system S(3, 4, n) exists if and only if n, 4 (mod 6). 5

32 3.. Perfect coverings of spheres Definition 0. A perfect R-covering of a Hamming sphere B(x, r) with x F n is a code C F n, such that each y B(x, r) is R-covered exactly once by C. Clearly a perfect code C F n with covering radius R is a perfect R- covering of any sphere B(x, r), x F n, r 0. Perfect coverings of spheres are related to Steiner systems as we will show in the next theorem. Theorem 1. A perfect 1-covering of a Hamming sphere B F n with radius three exists, if and only if there exists a Steiner quadruple system S(3, 4, n + 1), i.e. if n 1, 3 (mod 6). Proof. It suffices to consider the case B = B(0, 3) with the all-zero word 0 F n. Assume D 1,..., D r {1,..., n+1} are the blocks of a Steiner quadruple system S(3, 4, n + 1) on the set {1,..., n + 1}. Then one easily sees, that the code C = {0, supp 1 (D 1 \ {n + 1}),..., supp 1 (D r \ {n + 1})} F n is a perfect 1-covering of B(0, 3). Conversely assume that C F n is a perfect 1-covering of B(0, 3). Let C i C be the words of weight i in C, i 0. If 0 C, then C 1 = C =. One easily sees, that in this case {supp(u) {n+1} u C 3 } {supp(u) u C 4 } is a Steiner quadruple system S(3, 4, n + 1) on the set {1,..., n + 1}. Consider the case 0 C. Then C 1 = 1, say C 1 = {e n }. Then supp(u) {1,..., n 1} and u v = 4 whenever u, v C, u v. One checks, that {u v u, v C, u v} {supp(u) {n, n + 1} u C } {supp(u) {n + 1} u C 3 } {supp(u) u C 4 } is a Steiner quadruple system S(3, 4, n + 1) on the set {1,..., n + 1}. The last part of the claim now follows from Theorem 19. Similar (somewhat easier) one can show, that a perfect 1-covering of a Hamming sphere B F n with radius two exists if and only if a Steiner triple system S(, 3, n) exists, which by Theorem 18 is the case, if n 1, 3 (mod 6). One might try to extend the construction from Theorem 1 to spheres of radius four, thereby obtaining the statement, that a perfect 1-covering of 6

33 a Hamming sphere B F n with radius four exists if and only if a Steiner quintuple system S(4, 5, n + ) exists. This however in generally is incorrect, since for instance by Theorem 1 there exists a perfect 1-covering of a Hamming sphere B F 15 with radius four, but there is no Steiner system S(4, 5, 17) (Östergård and Pottonen [7]). At the present nothing seems to be known concerning perfect 1-coverings of spheres with radius greater than three. Question. For which values of n there exists a perfect 1-covering of a Hamming sphere B F n with radius four? By Theorem 1 there is no forced covering excess on a sphere with radius three if n 1, 3 (mod 6). In order to improve on the sphere-covering bound (.9) we will have to look on a covering excess in a sphere with radius greater than three! 3.3 The general excess bound Again let C F n be a code with covering radius one. Theorem 5 was independently generalized by Habsieger [9] and Honkala [16], who showed that there is a covering excess in a sphere of radius p 1 whenever n 1 (mod p) for an odd prime p. For the proof we need a simple lemma, which was first stated in Cohen et al. [5]. For x F n and an integer i we define A i (x) by Note that A i (x) = 0, if i is negative. A i (x) = {c C d(c, x) = i}. (3.7) Lemma 3 (Cohen et al. [5]). For x F n and i 0 we have ( ) n E i (x) = (n i + 1)A i 1 (x) + A i (x) + (i + 1)A i+1 (x). (3.8) i Proof. It suffices to consider the case x = 0. Then A j (0) is the number of codewords of weight j, j 0. Let S(0, i) denote the set of words with weight i. One easily sees, that for c C B(c, 1) S(0, i) = n i + 1 if w(c) = i 1, 1 if w(c) = i, i + 1 if w(c) = i + 1, 0 otherwise. 7

34 Therefore by (.6) applied to V = S(0, i) and R = 1 we have E i (0) = E(S(0, i)) = c C = j B(c, 1) S(0, i) S(0, i) c C w(c)=j B(c, 1) S(0, i) S(0, i) = (n i + 1)A i 1 (0) + A i (0) + (i + 1)A i+1 (0) ( ) n. i Theorem 4 (Habsieger [9], Honkala [16]). Let C F n be a binary code with covering radius one and n 1 (mod p) for an odd prime p. If x F n, then E(B(x, p 1)) = E 0 (x) + E 1 (x) E p 1 (x) 1 (mod p). (3.9) Proof. Summing (3.8) with 0 i p 1 gives E i (x) = 0 i p 1 0 i p 1 0 i p 1 = (n + 1) 0 i p 1 0 i p 1 (n i + 1)A i 1 (x) + ( ) n i 0 i p ( ) n i 1 (mod p) by n 1 (mod p). 0 i p 1 A i (x) + A i (x) + p(a p 1 (x) + A p (x)) (mod p) ( 1) i (mod p) 0 i p 1 0 i p 1 (i + 1)A i+1 (x) ( ) n i 8

35 As a consequence we get the following bound on K(n, 1), the general excess bound for binary codes with covering radius one. Theorem 5 (Habsieger [9], Honkala [16]). When n 1 (mod p) for an odd prime p, then ( ) p 1 n K(n, 1) 1 + V (n, p 1) n + 1. (3.30) Proof. Let C F n be a binary code with covering radius one and C = K(n, 1). From (3.9) follows E(B(x, p 1)) p 1, since E(B(x, p 1)) is always nonnegative for x F n. We now get (p 1) n = x F n (p 1) and Theorem 5 follows. x F n E(B(x, p 1)) = x F n z B(x,p 1) = z F n E(z) E(z) x B(z,p 1) = V (n, p 1)E(F n ) = V (n, p 1)( C (n + 1) n ) by (.8) The condition n 1 (mod p) is always satisfied for a suitable odd prime p, except if n = m 1, but then the value of K(n, 1) is known by Theorem 1. If we define the residue classes R p N for an odd prime p by we have R p = {n N n 1 (mod p)}, N \ { m 1 m N} = R 3 R 5... R p.... Then, as a major statement, the general excess bound says K(n, 1) n n + 1 if n, n R p for a fixed odd prime p. (3.31) 1 9

36 Note, that (3.31) (and (3.30)) does not imply K(n, 1) n n + 1 if n, n N \ {m 1 m N}, (3.3) since (3.30) cannot improve on the sphere covering bound, if p is too large compared to n. The quality of the general excess bound depends on the least odd prime divisor of n + 1. A proof of (3.3) could be found in [6]. The rest of this chapter is devoted to the possibility for further improvements of the general excess bound. 3.4 Habsiegers improvement and the excess matrix Let C F n be a code with covering radius one, n odd. Theorem 4 states, that whenever n 1 (mod p) for an odd prime p, then there exists a covering excess in a sphere with radius p 1. This gives rise to the natural question, under which circumstances there exists a perfect 1-covering of a sphere with radius p. A necessary condition is, that p is the least odd prime satisfying n 1 (mod p), since if n 1 (mod p ) with an odd prime p < p, then by Theorem 4 there exists a covering excess in a sphere of radius p 1 p. This condition indeed is sufficient in the cases p = 3 (see the last paragraph of section.3) and p = 5 (see Theorem 1). At the present state of knowledge the existence of a perfect 1-covering of a sphere with radius p, where n is odd and p is the least odd prime divisor of n + 1 cannot proved or refuted when p 7, see section 3... It is perhaps worth noting, that another necessary condition for E i (x) = 0 whenever 0 i p arising from (3.8) is always satisfied. Theorem 6. Let n m 1 be odd and p the least odd prime satisfying n 1 (mod p). Then there exist nonnegative integers A 0,..., A p 1 satisfying ( ) n E i := (n i + 1)A i 1 + A i + (i + 1)A i+1 = 0 (3.33) i for 0 i p (assume A 1 = 0). 30

37 The proof of Theorem 6 is postponed to section 3.5, where the necessary tools are developed. If it cannot be excluded, that there is no covering excess in a sphere of radius p, what can be done to further improve on the general excess bound? Let us note the following immediate consequence of Theorem 4. Theorem 7 (Habsieger [9], Honkala [16]). Assume C F n is a binary code of length n and covering radius one with n 1 (mod p) for an odd prime p. Then for every x F n E 0 (x) = E 1 (x) =... = E p (x) = 0 (3.34) implies E p 1 (x) p 1. (3.35) Habsiegers idea was, that if (3.34) cannot excluded, it could be possible to improve on (3.35). His result is conditional and uses another set of congruence properties of the excess function. Definition 8. Assume C F n is a binary code of length n and covering radius one with odd n. We say, that C satisfies condition (q), if there exist integers β 0,..., β q, such that 0 i q holds whenever x F n. β i E i (x) + E q 1 (x) 0 (mod q) (3.36) Theorem 9 (Habsieger [9]). If p 5, then in Theorem 7, (3.35) may be replaced by E p 1 (x) p 1, (3.37) if additionally C satisfies condition (q) for q = p 1 and q = p. We give the proof of Theorem 9 in a framework different from the original, since we develop material needed later. We first introduce the notion of the excess matrix. For the study of covering excess it suffices to consider spheres centered on 0 and we use the abbreviation E i = E i (0) for i 0. 31

38 Definition 30. Let C F n be a binary code of length n and covering radius one with n 1 (mod p) for an odd prime p. Assume E 0 = E 1 =... = E p = 0. (3.38) Let d j (j A = {1,..., E p 1 }) run over the set Z B(0, p 1), such that every d Z B(0, p 1) occurs exactly E(d) times. Then the E p 1 n- matrix D, whose rows consists of the words d j (j A) is called the excess matrix of C. The column vectors of D are the incident vectors of a subset system B 1,..., B n A, i.e. for 1 i n Note that B i = {j A : the ith coordinate of d j is nonzero} A. w(d j ) = p 1 for j A (3.39) by (3.38). Moreover E p 1 > 0 by (3.35). Thus the excess matrix is nonempty. We next show, that condition (q) implies a strong structure on the column sets B 1,..., B n A of the excess matrix. Theorem 31. Assume the propositions of Definition 30 are satisfied. Let s, r 1,..., r s be any integers satisfying 1 s p 1 and 1 r 1 <... < r s n. We set q = p s. If C satisfies condition (q), then the column sets B i, 1 i n of the excess matrix satisfy B ri 0 (mod q). (3.40) 1 i s Proof. Let y F n denote the unique vector, whose ith coordinate is nonzero if and only if i {r 1,..., r s } (1 i n). Now if z B(y, q ), then d(z, 0) d(z, y) + d(y, 0) p and thus z B(0, p ) implying E(z) = 0 by (3.38). Therefore E i (y) = 0 for 0 i q, which yields E q 1 (y) 0 (mod q) (3.41) by (3.36) (applied to x = y). If z Z B(y, q 1), then d(z, 0) d(z, y) + 3

39 d(y, 0) p 1 and thus z Z B(0, p 1). Therefore E q 1 (y) = E(z) (3.4) We next show = z Z d(z,y)=q 1 z Z B(0,p 1) d(z,y)=q 1 E(z) = {j A d(d j, y) = q 1}. {j A : d(d j, y) = q 1} = 1 i s B ri. (3.43) Assume first d(d j, y) = q 1 for an integer j with 1 j E p 1. To show j 1 i s B r i it suffices to show, that the ith coordinates of d j are nonzero whenever i {r 1,..., r s }. Assume to the contrary, that, say the r 1 th coordinate of d j is zero. This means that there are at most s 1 nonzero coordinates among the set {r 1,..., r s }. Since d j has exactly p 1 nonzero coordinates, d j must have at least (p 1) (s 1) = q nonzero coordinates outside of {r 1,..., r s }. In these coordinates y is zero however, implying d(d j, y) q, contradicting d(d j, y) = q 1. Conversely assume j 1 i s B r i. This means that the ith coordinates of d j are nonzero whenever i {r 1,..., r s }. Moreover d j has exactly q 1 nonzero coordinates outside the set {r 1,..., r s }. Using the definition of y we now see, that y and d j exactly differ in these q 1 coordinates, i.e. d(d j, y) = q 1. This completes the proof of (3.43). Now (3.40) follows by (3.41), (3.4) and (3.43). Proof of Theorem 9. It suffices to consider the case x = 0. Assume equality holds in (3.35), i.e. E p 1 = A = p 1. Since any row of the excess matrix D has weight p 1, D has at least two nonempty column sets, say B 1, B A. Since C satisfies condition (p 1) we may apply Theorem 31 with s = 1 and r 1 = 1 (resp. r 1 = ) and get B 1 B 0 (mod p 1), implying B 1 = B = A by A = p 1. Again by Theorem 31 (with s = and r 1 = 1, r = ) however we find B 1 B 0 (mod p ), since C satisfies condition (p ). This gives the contradiction p 1 = A = B 1 = B 1 B 0 (mod p ). Therefore E p 1 > p 1 and (3.37) follows by (3.34) and Theorem 4. 33

40 In [9] Habsieger explicitly gave congruence properties of type (3.36) for q 5 and raised the question, if the conditions (p 1) and (p ) are always satisfied for C under the propositions of Theorem 9. He wrote: It is probably true, that a necessary and sufficient condition for the existence of (3.36) (for q = p 1 and q = p ) is that p divides n + 1. Since the applications of... require explicit congruence properties, we shall not try to prove this characterization. In the next section we prove the sufficient part of Habsiegers conjecture. Motivated by important structural properties of the excess matrix (see Theorem 31), we show more generally, that condition (q) is satisfied for any binary code of length n and covering radius one, if q and n + 1 do not have a common odd prime divisor. More precisely we prove the following theorem. Theorem 3. Let n, q be integers with 1 q n and odd n. Assume q and n + 1 do not have a common odd prime divisor. Then there are integers β 0,..., β q depending only on q and the residue class of n (mod q), such that for every binary code C F n with covering radius one and each x F n we have β i E i (x) + E q 1 (x) 0 (mod q). (3.44) 0 i q Note that Habsiegers conjecture follows from Theorem 3 applied to q = p resp. q = p 1, since by (3.9) and (3.34) p is the least odd prime satisfying n 1 (mod p) and therefore q and n + 1 do not have a common odd prime divisor. As the second major question concerning Theorem 9, Habsieger [9] asked whether the estimation E p 1 p 1 could be improved further. Indeed Honkala [17] showed E p 1 3p 1. In Habsiegers words:... the inequality E p 1 p 1 can probably sharpened. One might even get a bound which is quadratic in p. As a main goal for the rest of this chapter we will answer this question in the affirmative. Under essential use of the structural properties of the excess matrix stated in Theorem 31 together with the congruence properties described in Theorem 3 we will give a proof of the following unconditional result. 34

41 Theorem 33. Assume C F n is a binary code of length n and covering radius one with n 1 (mod p) for an odd prime p. Then for every x F n E 0 (x) = E 1 (x) =... = E p (x) = 0 (3.45) implies E p 1 (x) (p )p 1. (3.46) In section 3.7 we will use Theorem 33 to give a general improvement of the general excess bound for binary codes with covering radius one. 3.5 Congruence properties for the covering excess function In this section we prove Theorem 3 and discuss some of its consequences including the outstanding proof of Theorem 6. In this section we always assume, that variables with negative index equal zero Some lemmas For the proof of Theorem 3 we need some lemmas. Lemma 34. Let m, n be nonnegative integers with n odd. Then there are integers γ i (0 i n+1) satisfying ( ) n + 1 (n i + 1)γ i + (i + )γ i+1 (mod m ) (3.47) i + 1 for 0 i n 1. Proof. Recursively choose integers u i (0 i n+1 ) satisfying (n i)u i+1 1 (i + 1)u i (mod m ) for 0 i n 1 (here u 0 may be chosen arbitrarily). This is possible, since n i is odd (0 i n 1) and therefore n i and the modul m do not have a common prime divisor. We now set ( ) ( n + 1 γ i = u i 0 i n + 1 ) i 35

42 and find (0 i n 1 ) (n i + 1)γ i + (i + )γ ( ) i+1 ( ) n + 1 n + 1 = (n i + 1)u i + (i + )u i+1 i i + ( ) ( ) n + 1 n + 1 = (n i + 1)u i + (n i)u i+1 i i + 1 ( ) ( ) n + 1 n + 1 (n i + 1)u i + (1 (i + 1)u i ) i i + 1 ( ) n + 1 =. i + 1 (mod m ) Lemma 35. Let m, n be nonnegative integers with n odd. Then there are integers α i (0 i n) satisfying ( ) n (n i + 1)α i 1 + α i + (i + 1)α i+1 (mod m ) (3.48) i for 0 i n 1. Proof. We use Lemma 34 to get integers γ i (0 i n+1 ) satisfying (3.47). The integers α i (0 i n) are defined recursively. When i is even with 0 i n we set α i = γ i α i 1 (0 i n, i even). (3.49) When i is odd with 0 i n we choose any integer α i satisfying ( ) n iα i (n i + )α i α i 1 i 1 (mod m ) (0 i n, i odd). (3.50) This is possible, since i and m do not have a common prime divisor if i is odd. When i is even with 0 i n 1, then (3.48) follows from (3.50) 36

43 (with i + 1 instead of i). When i is odd with 0 i n 1 we have (n i + 1)α i 1 + α i + (i + 1)α i+1 = (n i + )(α i + α i 1) + (i + 1)(α i + α i+1) = = and (3.48) holds again. ((n i + )α i + α i 1 + iα i) ( n i 1 ) ( + 1 γ i 1 + i 1 ) + γ i+1 by (3.49) ((n i + )α i + α i 1 + iα i) ( ) ( ) n + 1 n (mod m ) by (3.47), (3.50) i i 1 ( ) n i Lemma 36. Assume n, u are nonnegative odd integers, such that n+1 and u do not have a common prime divisor. Then there are integers α i (0 i n) satisfying ( ) n (n i + 1)α i 1 + α i + (i + 1)α i+1 (mod u) (3.51) i for 0 i n 1. Proof. Choose an integer α satisfying α(n + 1) 1 (mod u). (3.5) This is possible, since u and n + 1 are assumed to be without common prime divisor. We set ( ) n α i = α (0 i n) i 37

44 and have (0 i n 1) (n i + 1)α i 1 + α i + (i + 1)α i+1 ( ( ) ( ) ( )) n n n = α (n i + 1) + + (i + 1) i 1 i i + 1 ( ( ) ( ) ( )) n n n = α i + + (n i) i i i ( ) n = α(n + 1) i ( ) n (mod u) by (3.5). i Lemma 37. Let n, q be positive integers with n odd. Moreover assume, that n+1 and q do not have a common odd prime divisor. Then there are integers α i (0 i n) satisfying ( ) n (n i + 1)α i 1 + α i + (i + 1)α i+1 (mod q) (3.53) i for 0 i n 1. Proof. Write q = m u (3.54) with m 0 and odd u. Then the propositions of Lemma 35 and Lemma 36 are satisfied and we get integers α i, α i (0 i n) satisfying (3.48) and (3.51). Now by the well-known chinese remainder theorem there are integers α i (0 i n) satisfying α i α i (mod m ), α i α i (mod u) for 0 i n, since m and u do not have a common prime divisor. By (3.48) and (3.51) this implies ( ) n (n i + 1)α i 1 + α i + (i + 1)α i+1 (mod m ), i ( ) n (n i + 1)α i 1 + α i + (i + 1)α i+1 (mod u) i for 0 i n 1 and (3.53) follows by (3.54). 38