Dam, E.R. van

Fon-der-Flaass, D.

English

Research Papers in Economics

Journal of Algebraic Combinatorics 12 (2000), 115–121
c ° 2000 Kluwer Academic Publishers. Manufactured in The Netherlands.
Uniformly Packed Codes and More Distance
Regular Graphs from Crooked Functions
E.R. VAN DAM edwin.vandam@kub.nl
Department of Econometrics, Tilburg University, PO Box 90153, 5000 LE Tilburg, The Netherlands
D. FON-DER-FLAASS¤ d.g.ﬂaass@writeme.com
Institute of Mathematics, Novosibirsk 630090, Russia
Received February 23, 1999; Revised July 27, 1999
Dedicated to Jaap Seidel on his 80th birthday
Abstract. Let V and W be n-dimensional vector spaces over GF.2/. A function Q : V ! W is called crooked
(a notion introduced by Bending and Fon-Der-Flaass) if it satisﬁes the following three properties:
Q.0/ D 0I
Q.x/ C Q.y/ C Q.z/ C Q.x C y C z/ 6D 0 for any three distinctx; y;zI
Q.x/ C Q.y/ C Q.z/ C Q.x C a/ C Q.y C a/ C Q.z C a/ 6D 0i f a 6D 0 .x; y;z arbitrary/:
We show that crooked functions can be used to construct distance regular graphs with parameters of a Kasami
distanceregulargraph,symmetric5-classassociationschemessimilartothoserecentlyconstructedbydeCaenand
vanDamfromKasamigraphs,anduniformlypackedcodeswiththesameparametersasthedoubleerror-correcting
BCH codes and Preparata codes.
Keywords: crooked function, distance-regular graph, association scheme, uniformly packed code
1. Crooked functions
Crooked functions were introduced in [1] as a means to generalise the construction of new
distanceregulargraphsfoundbydeCaen,Mathon,andMoorhouse[4]. Inthisnoteweshow
that crooked functions can similarly be used to generalise the constructions of the distance
regularcosetgraphsoftheKasamicodes(Kasamigraphs)[2,Theorem11.2.1,(13),q D 2],
of symmetric 5-class association schemes related to Kasami graphs which were recently
found by de Caen and van Dam [3], and of the double error-correcting, uniformly packed
BCH (Kasami) codes and Preparata codes.
First we recall from [1] the deﬁnition and some basic properties of crooked functions,
and some useful notations.
¤Partly supported by the grant 96-01-01614 of the Russian Foundation for Fundamental Research. The work was
partly done while the author worked at London School of Economics.116 VAN DAM AND FON-DER-FLAASS
LetV andW ben-dimensionalvectorspacesoverGF.2/,and Q : V ! W anymapping.
We shall use the notation
Q.a1;a2;:::;a m/DQ.a 1/CQ.a 2/C¢¢¢CQ.a m/:
Also, for 0 6D a 2 V, we denote by Ha.Q/, or simply Ha, the set
Ha D Ha.Q/ Df Q . x/CQ . xCa /jx2Vg :
We shall denote the size of a ﬁnite set X by jXj.
Deﬁnition1[1] Amapping Q :V !W iscalledcrookedifitsatisﬁesthefollowingthree
properties:
(1.1) Q.0/ D 0;
(1.2) Q.x; y;z;x C y C z/ 6D 0 for any three distinct x; y;z;
(1.3) Q.x; y;z;x C a; y C a;z C a/ 6D 0i f a 6D 0.
An equivalent but in some situations more useful description of crooked functions is
given in the following proposition which was proved in [1].
Proposition 2 If Q is a crooked mapping then
.2:0/ n D dimV must be odd.
.2:1/ Q is a bijection.
.2:2/ Every set Ha.Q/ is the complement of a hyperplane.
.2:3/ The sets Ha are all distinctI in particular; every complement of a hyperplane
appears among them exactly once.
Moreover;everymapping Qsatisfyingproperty.2:2/;andsuchthat Q.0/D0;iscrooked.
Examples of crooked functions can be constructed as follows. Let V DW DGF.2n/
with n odd. Let k be a natural number coprime to n. Then the function Q.x/ D x1C2k
is
crooked. If, in the constructions to follow, we use these examples, we obtain precisely the
Kasami graphs, the schemes constructed in [3], and the double error-correcting BCH codes
and Preparata codes.
At present, no other examples of crooked functions are known. But the simplicity of
Deﬁnition 1 suggests that many more examples should exist. Thus, an alternative title for
this note might have been: “Wanted: Crooked functions. Reward increased”.
2. Kasami graphs
Let Q :V !W be a crooked function; dimV D dimW Dn. Let N D2n DjVj. We deﬁne a
graph K D K.Q/asfollows. Thevertexsetof K isÄDV £W Df.v;w/ j v 2 V;w2Wg.
Vertices .v;w/ and .v0;w0/are adjacent if and only if v 6Dv0 and w C w0 D Q.v C v0/.UNIFORMLY PACKED CODES AND MORE DISTANCE REGULAR GRAPHS 117
Theorem 3 The graph K.Q/ is distance regular with intersection array
½
N ¡ 1; N ¡ 2;
1
2
N C 1I1;2;
1
2
N ¡ 1
¾
:
Proof: The mappings tx;y :.v;w/ ! .v Cx;wCy/for x 2 V; y 2 W form a subgroup
of the automorphism group of K which acts transitively on the vertices. So, it is enough
to check the parameters just for one vertex, say, for v0 D .0;0/. Let, for i D 0;1;2, Ki
denotethesetofverticesatdistancei fromv0 (K0 Df v 0g ); andlet K3 D Än.K0[K1[K2/
(subsequently we shall see that the diameter of K is indeed 3). Also, let us denote by Wv
the set f.v;w/jw2Wg.
It follows from the deﬁnition of K that the Wx are independent sets, and that every
two distinct sets Wx, Wy are joined by a matching. Also, K1 Df.x;Q.x// j x 2Vnf0gg;
jK1jDN¡1, as required.
Consider 2-paths from v0 to Wa, a 6D 0. They all have the form
.0;0/ ¡ .x; Q.x// ¡ .a; Q.x/ C Q.x C a//
for x 6D 0;a. Thus, by Proposition (2.2), for every h 2 Ha.Q/nfQ.a/g there are precisely
two 2-paths from v0 to .a;h/; and this accounts for all 2-paths from v0 to Wa. In particular,
we see that K has no triangles, and that every vertex from K2 is adjacent to precisely two
vertices from K1.
For a 6D 0w eh a v e
K 1\W a Df . a ;Q . a //gI
K2 \ Wa Df . a ;h /jh2H anfQ.a/ggI
K3 \ Wa Df . a ;h /jh= 2H ag :
For a D 0, we have W0nfv0g½K 3.
Let us count the number of neighbours in K3 of an arbitrary vertex .a;w/2W a,w 6D
Q.a/. One such neighbour can be found in W0.
For every b 6D 0;a, the set K3\Wb is joined by a matching to a subset f.a;x/ j x 2 Xbg
of Wa where Xb ½W iseither Hb or WnHb: whicheverofthesetwodoesnotcontain Q.a/.
Let also Xa D WnHa; again Q.a/= 2X a.
By Proposition (2.2), (2.3), the sets Xx are all possible afﬁne hyperplanes in W not
containing Q.a/. Therefore, every point w 6D Q.a/ is contained in exactly N=2 of them.
It follows that every vertex .a;w/2 K2\W a is adjacent to N=2 C 1 vertices in K3, and
every vertex .a;w/2K3\W a is adjacent to .N=2 ¡ 1/ C 1 D N=2 vertices in K3 (recall
that .a;w/2K3if and only if w 2 Xa).
Neighbours of vertices from W0 are considered similarly. Every set K3 \ Wa for a 6D 0
is adjacent to the complement of a hyperplane in W0; and each such complement occurs
exactly once. Therefore each vertex in W0nfv0g is adjacent to N=2 vertices in K3,a s
required.
Thuswehavecheckedenoughentriesoftheintersectionarraytoconcludethatitisindeed
as stated in the theorem; in particular, that K is of diameter three (that is, every vertex from
K3 is adjacent to some vertex from K2). 2118 VAN DAM AND FON-DER-FLAASS
3. Association schemes
For an arbitrary function R :V!W deﬁne a permutation sR of Ä of order 2; sR..v;w// D
.v;w C R.v//. The graph sR.K/ is isomorphic to K; vertices .v;w/ and .v0;w0/in it are
adjacent if w C w0 D Q.v C v0/ C R.v/ C R.v0/.
Lemma 4 Let R :V !W be a mapping such that
.8a 2 V/ Ha.R/ µ Ha.Q/:
Then the graphs K and L D sR.K/ satisfy the following properties:
.4:1/ They are edge-disjoint.
.4:2/ The graph K [ L has no triangles.
.4:3/ There are no 4-tuples of vertices .x1;x2;x3;x4/ such that .x1;x2/ and .x2;x3/ are
edges of K;. x 1;x 4/and .x4;x3/ are edges of L; and x1 6D x3.
Proof: The hypothesis implies that R is a bijection. Indeed, none of the sets Ha.Q/
contains 0; therefore R.x/ C R.y/ 6D 0 when x 6D y. This proves (4.1): the equalities
wCw0 D Q.vCv0/ and wCw0 D Q.vCv0/C R.v/C R.v0/ cannot hold simultaneously.
Suppose that vertices x D .a;w 1/; y D .b;w 2/;z D .c;w 3/form a triangle. Neither
K nor L contain triangles; so let the edges xy, yz be in K, and xz in L (the other case is
similar). We have
Q.a C b/ C Q.b C c/ C Q.a C c/ D R.a/ C R.c/:
Thisisimpossible, since Q.aCb/CQ.bCc/ 2 HaCc.Q/, Q.aCc/ D Q.0/CQ.aCc/ 2
HaCc.Q/, and R.a/ C R.c/ 2 HaCc.R/ µ HaCc.Q/;b u tH a C c.Q/is sum-free.
A similar easy calculation proves (4.3). 2
We shall call a mapping R satisfying the hypothesis of Lemma 4 for a crooked function
Q an accomplice of Q. Trivially, every crooked function is an accomplice of itself.
Let R be an accomplice of a crooked function Q, and L D sR.K/.
Following [3], we shall now deﬁne 5 symmetric relations A1;:::;A 5 on Ä which will
be shown to form, together with the identity relation A0, an association scheme.
Let A1 and A3 be, respectively, the relations of being at distance 1 and at distance 2 in
K; and A2 and A4, similarly, the relations of being at distance 1 and at distance 2 in L. The
relation A5 holds for vertices .v;w/ and .v0;w0/if and only if v D v0 and w 6D w0; that is,
when they lie in the same class Wv.
Theorem 5 The relations A0;:::;A 5deﬁned above form a 5-class association scheme.
Proof: As usual, we shall identify relations with subsets of Ä £ Ä, and with their char-
acteristic vectors viewed as .N2 £ N2/-matrices.
First let us show that A0 C A1 C¢¢¢CA 5 D J, the trivial relation (that is, the all-one
matrix). To do this, it is sufﬁcient to check that no pair of vertices can be in more than one
of these relations—then comparing sizes does the job.UNIFORMLY PACKED CODES AND MORE DISTANCE REGULAR GRAPHS 119
Obviously, A0 \ Ai D;for i 6D 0. It is just as easy to see that A5 \ Ai D;for i 6D 5.
Also, we already know from the previous section that A1 \ A3 D A2 \ A4 D; . The
remaining cases follow from Lemma 4: A1 \ A2 D;from (4.1), A1 \ A4 D A2 \ A3 D;
from (4.2), and A3 \ A4 D;from (4.3).
Now, following the lines of [3, Theorem 2], we prove that A1A2 D A2A1 D A3 C A4C
A5. Consider all walks of length 2, starting at some vertex x and going ﬁrst along an edge
of K, and then along an edge of L. There are .N ¡ 1/2 of them. By (4.1), none of them
returns to x; by (4.2), none of them ends in a vertex adjacent to x; and by (4.3), they all end
in distinct vertices. Since jÄjDN 2D1C2 . N¡1 /C. N¡1 / 2, the claim is proved.
The relation A5 has a very simple structure, so it is not difﬁcult to check that, for every
i, Ai A5 is a linear combination of Ai’s.
FromTheorem3wealreadyknowthat.A0; A1; A3; A2CA4CA5/and.A0; A2; A4; A1C
A3 C A5/ are association schemes of distance regular graphs. Together with the equation
A1A2 D A2A1 D A3 C A4 C A5 this sufﬁces to check that every product Ai Aj is a linear
combination of Ai’s with integer nonnegative coefﬁcients (without any further resorting to
the actual deﬁnition of the relations Ai). We leave this exercise to the reader. 2
Itistemptingtolookforlinearaccomplicesofcrookedfunctions. Firstly,becausethesets
Ha.R/ are particularly small for a linear function: Ha.R/ Df R . a / g . The second reason
is that, as was mentioned in [1] just before Proposition 11, ﬁnding such a linear function
would immediately give us a new example of a closed bent Kerdock set of functions, and a
new Kerdock code.
Unfortunately,forknowncrookedfunctionsindimensionsupto9therearenosuchlinear
accomplices R, as was shown by an exhaustive computer search.
4. Uniformly packed codes
Let Q :V !W be a crooked function; dimV DdimW Dn >1. Let N D2n DjVj.W e
deﬁne the code C D C.Q/ as the set of characteristic vectors of all subsets S of Vnf0g such
that
P
r2S r D 0 and
P
r2S Q.r/D0. Clearly, C is a binary linear code of length N ¡1.
In fact, C is a generalization of the double error-correcting BCH codes. These codes are
uniformly packed, i.e., the number of codewords at distance 3.De C 1/ from a word X
which is at distance 2 from the code is constant, and the number of codewords at distance
3 from a word X which is at distance greater than 2 from the code is also constant.
Theorem 6 For n 6D 3; the code C.Q/ is a double error-correcting uniformly packed
code. For n D 3;C.Q/ is the perfect repetition code.
Proof: First, suppose that there is a codeword of weight at most 4. Then there are
r1;r2;r3;r4 2 V such that r1 Cr2 Cr3 Cr4 D 0 and Q.r1/C Q.r2/C Q.r3/C Q.r4/ D 0.
This contradicts the fact that Q is a crooked function unless all ri are zero, so C.Q/ has
minimum distance at least 5. Since the zero word and the all-ones word are codewords this
implies that for n D 3, C.Q/ is the repetition code of length 7, and this code is perfect.120 VAN DAM AND FON-DER-FLAASS
Next, consider a word X which is at distance 2 from the code C.Q/. We want to show
that for any such X the number of codewords S at distance 3 from X is the same. Now let
T be a codeword at distance 2 from X, say X and T differ in coordinates indexed by e1 and
e2. Then
P
r2X r D e1 C e2 and
P
r2X Q.r/ D Q.e1/ C Q.e2/.
Suppose that S and X differ in coordinates indexed by x1;x2;x3, then it becomes clear
that we want to count the number of triples fx1;x2;x3g of nonzero elements of V, such
that x1 C x2 C x3 D e1 C e2 and Q.x1/ C Q.x2/ C Q.x3/ D Q.e1/ C Q.e2/. Substituting
x3 D x1 C x2 C e1 C e2 in the second equation, and substituting y D x1 C e1 C e2;z D
e1 C e2;wD Q.e 1/CQ.e 2/, we obtain that Q.y C z/ C w D Q.x2/ C Q.x2 C y/. This
equation has precisely two solutions for x2 if Q.y C z/ C w2 Hy.Q/, and otherwise it
has none. Note that z and w are given. Since Q.y C z/ C Q.z/2 Hy.Q/, we have that
Q.y Cz/Cw 2 Hy.Q/ if both Q.z/ and w are in Hy.Q/ or if both are not in Hy.Q/ (here
we use that Hy.Q/ is the complement of a hyperplane).
Since w and Q.z/ are distinct and nonzero (by the properties of Q), the number of
hyperplanes containing w and Q.z/ equals 1
4N ¡1, and the number of hyperplanes not
containing w and Q.z/ equals 1
4N (this follows easily by counting). Hence by Proposi-
tion 2 it follows that the number of y such that Q.y C z/ C w2 Hy.Q/ equals 1
2N ¡1,
and consequently the number of triples fx1;x2;x3g with the required properties equals
1
3.1
2N ¡1/¡1 D N¡8
6 (each triple occurs 3! times as a solution, and the solution f0;e1;e2g
is not allowed).
Note that the integrality of the above number of triples forces n to be odd. Also, we may
now assume that n > 3, so that the number of triples is greater than zero, which shows that
C.Q/ has minimum distance exactly 5.
Similarly, one can show that the number of codewords at distance 3 from a word which
is at distance at least 3 from the code equals N¡2
6 , which completes the proof. 2
Note that the proof that C.Q/ is a uniformly packed code goes along the same lines
as the proof in [7, p. 45] that the double error-correcting BCH codes (Kasami codes) are
uniformly packed. Note also that it now follows from counting that the dimension of C.Q/
equals N ¡ 1 ¡ 2n (cf. [7, Thm. 1.3]).
An important consequence of the theorem is that C.Q/ is a double error-correcting lin-
ear code with dual degree 3 (cf. [7, Thm. 3.11]) (or is perfect in case n D 3), and hence
it follows by the work of Delsarte (cf. [2, Chapter 11]) that the coset graph of C.Q/ is
distance regular. Following Proposition 1 in [3] this coset graph can be reformulated as
follows. Its vertex set is V £W, and two distinct vertices .v;w/ and .v0;w0/are adjacent
if w C w0 D Q.v C v0/. Hence the coset graph is precisely the Kasami graph of Section 2.
Closely related to the double error-correcting BCH codes are the Preparata codes. These
are binary, non-linear, double error-correcting, nearly perfect codes, that is, each word at
distance at least 2 from the code has distance 2 or 3 to exactly L
3 codewords, where L is
the length of the code (clearly such a code is also uniformly packed). Also here we give
a generalization: by adapting the Baker-van Lint-Wilson description (cf. [6, Def. 7.4.4])
of the original Preparata code (note that other Preparata-like codes have been constructed
over the ring of integers modulo four (cf. [6, Chap. 7])).UNIFORMLY PACKED CODES AND MORE DISTANCE REGULAR GRAPHS 121
Let P.Q/ be the code consisting of characteristic vectors of pairs .S;T/ with S µ
Vnf0g;T µ V, such that jTj is even,
P
s2S s D
P
t2T t, and
P
s2S Q.s/ D
P
t2T Q.t/ C
Q.
P
t2T t/.
Theorem 7 The code P.Q/ is a double error-correcting; nearly perfect code of size
22N¡2¡2n; and length L D 2N ¡ 1.
Proof: First, note that for every choice of T, jTj even, there are jC.Q/jD2 N¡ 1 ¡ 2 nsets S
such that .S;T/ is a codeword (this follows by counting, and the observation that if .S;T/
is a codeword, then so is .S ¥ R;T/ for every R 2C.Q/, where S ¥ R stands for the
symmetric difference of S and R). Thus P.Q/ has 22N¡2¡2n codewords.
Next, suppose that P.Q/ has mimimum distance at most 4, say the two codewords
.S1;T1/ and .S2;T2/ have distance at most 4. Then it follows that S1 and S2 differ in 1
or 2 elements, and T1 and T2 differ in 2 elements (since C.Q/ has minimum distance
5, and T1 and T2 differ in an even number of elements). Without loss of generality we
assume that S1 and S2 differ in s1;s2 (where we allow s1 to be zero to cover the case
where S1 and S2 differ in only one element), and that T1 and T2 differ in t1;t2.N o w
it follows that s1 C s2 D t1 C t2, and Q.s1/ C Q.s2/ D Q.t1/C Q.t2/ C Q.
P
t2T1 t/ C
Q.
P
t2T1 t C t1 C t2/. But Q.s1/ C Q.s2/, Q.t1/ C Q.t2/, and Q.
P
t2T1 t/ C Q.
P
t2T1 t C
t1 C t2/ 2 Ht1Ct2.Q/, which is sum-free. Hence we have a contradiction, and P.Q/ has
minimumdistance5.Itnowfollowsfromtheobtainedparametersthat P.Q/isnearlyperfect
(cf. [6, p. 122]). 2
Added in proof. After writing this paper, we discovered the paper [5]. In this paper so-
calledalmostbentfunctionsarerelatedtouniformlypackedcodes. Inasense, theapproach
in [5] is dual to ours. It follows from the results in [5] and this paper that a crooked function
is almost bent. D. de Caen [private communication] showed us an easy, direct argument
that this is indeed the case.
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Uniformly packed codes and more distance regular graphs from crooked functions.

Uniformly packed codes and more distance regular graphs from crooked functions.

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