Let q be any integer ≥ 2. In this paper, we consider the q-ary n-dimensional cube whose vertex set is Z n q and two vertices (x1,..., xn) and (y1,..., yn) are adjacent if their Lee distance is 1. As a natural extension of identifying codes in binary Hamming spaces, we further study identifying codes in the above q-ary hypercube. We let M q t (n) denote the smallest cardinality of t-identifying codes of length n in Z n q. Little is known about ternary or quaternary identifying codes. It is known [2, 14] that M 2 1(n) 

Jon-lark Kim

Seog-jin Kim

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Identifying Codes in q-ary Hypercubes
Jon-Lark Kim
Department of Mathematics
University of Louisville
Louisville, KY 40292, USA
e-mail: jl.kim@louisville.edu
Seog-Jin Kim
Department of Mathematics Education
Konkuk University
Seoul 143-701, Korea
e-mail: skim12@konkuk.ac.kr
4/22/09
Abstract
Let q be any integer ≥ 2. In this paper, we consider the q-ary n-
dimensional cube whose vertex set is Z
n
q and two vertices (x1,...,xn)
and (y1,...,yn) are adjacent if their Lee distance is 1. As a natural
extension of identifying codes in binary Hamming spaces, we further
study identifying codes in the above q-ary hypercube. We let M
q
t (n)
denote the smallest cardinality of t-identifying codes of length n in
Z
n
q. Little is known about ternary or quaternary identifying codes.
It is known [2, 14] that M
2
1(n) ≥
2v
d+1+ 2
n
where v is the number of
vertices of Z
n
2 and d is the degree of any vertex of Z
n
2. In a similar
manner, we show that M
q
1(n) ≥
2v
d+1+ 1
n
, where d is the degree and
v = v(q) is the number of vertices of Z
n
q for q = 3 and q = 4,
respectively. We also give some constructions to show that M
3
1(2) =
4, M
3
1(3) = 9, and M
4
1(2) = 7, deriving some upper bounds on
M
3s
1 (n) and M
4s
1 (n).
Keywords: fault tolerance; identifying codes; parallel processing , q-
ary hypercubes
11 Introduction
Identifying codes were introduced by Karpovsky, et al. [14] in order to
ﬁnd faulty processors in a multiprocessor system. In general, let G be a
graph with vertex set V . Now we place |V | processors in G. We assume
that some processors can check themselves and all the vertices at distance
≤ t, and report if there is a fault. The problem is to choose as few checking
processors as possible so that if we see the reports, we know which processor
is malfunctioning.
Identifying codes have been studied in graphs [3, 4, 5, 6, 14]. Identifying
codes in binary Hamming spaces are related to covering codes from coding
theory (cf. [1, 2, 8, 9, 10, 11, 12]). As a natural extension, we further
study identifying codes in q-ary hypercubes. Moreover, it is remarked [14]
that a q-ary hypercube has several applications in parallel processing (for
example, see [7, 13]). Hence ﬁnding identifying codes in q-ary hypercubes
is of special interest.
We give some basic deﬁnitions. Let q be any integer, q ≥ 2. Denote
the set of integers modulo q by Zq. We deﬁne a q-ary n-dimensional cube
(or hypercube) to be the set Zn
q with the following rule; two vertices x =
(x1,    ,xn) and y = (y1,    ,yn) in Zn
q are adjacent if xi = yi for all i
except one index, say j, and xj −yj = ±1 (mod q). This means that there
is an edge between x and y if and only if their Lee distance dL(x,y) is 1.
From now on, we write dL(x,y) by d(x,y) for simplicity.
As usual, for x ∈ Zn
q, let Bt(x) = {y ∈ Zn
q | d(x,y) ≤ t}. A nonempty
set C of Zn
q is called a code of length n. Now we deﬁne t-identifying codes
as follows.
Deﬁnition 1.1. A code C of length n is called t-identifying if the sets
Bt(x) ∩ C, x ∈ Zn
q, are nonempty and diﬀerent. We let M
q
t (n) denote the
smallest cardinality of t-identifying codes of length n in Zn
q.
In Figure 1, we give two examples. The set {a1,a3,a4,a6} is a 1-
identifying code in Z3
2. On the other hand, the set {b1,b3,b6} is not since
b8 is not 1-identiﬁed by this set.
In [2, 14], identifying codes in Zn
2 and in Zn
q with q > 4, q even, were
considered. The remaining cases are open. Among them, this paper consid-
ers 1-identifying codes in Zn
q with q = 3 or q = 4. For q = 2, it was proved
in [2, 14] that M2
1(n) ≥ 2v
d+1+ 2
n
where v is the number of vertices of Zn
2 and
d is the degree of any vertex of Zn
2. As a generalization to q = 3 and q = 4,
we show that M
q
1(n) ≥ 2v
d+1+ 1
n
, where d is the degree and v = v(q) is the
number of vertices of Zn
q for q = 3 and q = 4, respectively. Furthermore in
Section 3, we give some constructions to show that M3
1(2) = 4, M3
1(3) = 9,
and M4
1(2) = 7, deriving upper bounds on M3s
1 (n) and M4s
1 (n).
2a1 a2
a3 a4
a5 a6
a7 a8
b1 b2
b3 b4
b5 b6
b7 b8
Case 1 Case 2
Figure 1: Case 1 is a 1-identifying code and Case 2 is not.
2 Improved Lower Bounds
If C is a 1-identifying code for a d-regular graph G with v vertices, then C
satisﬁes the following lower bound [14, Theorem 2].
|C| ≥
2v
d + 2
. (2.1)
For the special case of the q-ary hypercube, when q = 2, a better lower
bound is obtained in [2, 14] as follows.
Theorem 2.1. [2, 14]
M
2
1(n) ≥
2v
d + 1 + 2
n
where v is the number of vertices of Zn
2 and d is the degree of any vertex of
Zn
2.
Here we generalize the results on the lower bound for q = 3 and q = 4 by
modifying the argument of Theorem 9 in [2]. When the distance between a
vertex x and codeword z is at most 1, then x is called 1-covered by z. Note
that a codeword covers itself.
Theorem 2.2. For q = 3 in Lee metric,
M3
1(n) ≥
2v
d + 1 + 1
n
=
2   3n
2n + 1 + 1
n
,
where v is the number of vertices of Zn
3 and d is the degree of any vertex
in Zn
3.
3Proof. Suppose that C is a 1-identifying code with K = M3
1(n). If a vector
x ∈ Zn
3 is 1-covered by exactly i + 1 codewords of C, the excess E(x) on
x is deﬁned to be i as in [2]. In general, if V is a subset of Zn
3, then
E(V ) =
P
x∈V E(x). Every point x with E(x) = 1 is called an employee,
and every point with E(x) > 1 is called an employer. We are going to show
that, in some sense that we shall deﬁne below, every employee x has an
employer and that sometimes an employee may have two employers.
Let T be the set of employees and employers. Let x be an employee.
Then x is 1-covered by two codewords y1 and y2. We have the following
three cases, which exclude each other.
Case 1: x ∈ {y1,y2}.
If x = y1, then E(y2) > 1 by the deﬁnition of the 1-identifying code.
Hence y2 is called the employer of x and it is its only employer.
Case 2: x / ∈ {y1,y2} and y1 and y2 are not adjacent.
Then y1 and y2 have another unique common neighbor z. In this case,
z must be 1-covered by at least a third codeword, z is called the employer
of x, and z is its only employer.
Case 3: x / ∈ {y1,y2} and y1 and y2 are adjacent.
In this case, x,y1, and y2 diﬀer at only one coordinate. Here if E(y1) =
1, then B1(x)∩C = B1(y1)∩C. This contradicts the deﬁnition of C. Hence
E(y1) ≥ 2, and similarly E(y2) ≥ 2. We call both of y1 and y2 the
employers of x.
We call a subset of points of all employees with a common employer a
company. Note that the employer also belongs to the company. We next
show that we can distribute the excesses of employers to their employees
so that each member of T has average excess at least f(2n) where
f(2n) =
￿￿
2n
2
￿
+ 2n − 1
￿
/
￿￿
2n
2
￿
+ 1
￿
.
Here, in Case 3, the employee x receives excess from both of its employers,
but this does not reduce its average excess. Hence we do not need to worry
about this situation. Note that no point is 1-covered by more than 2n
codewords in the minimum 1-identifying code. Hence the employer of each
company is 1-covered by at most 2n codewords.
If an employer is 1-covered by exactly i ≥ 3 codewords and has exactly
t employees (t ≥ 0), then the average excess of the company is at least
g(t) = (t + i − 1)/(t + 1).
Note that each employee is covered by two codewords among the i code-
words that cover the employer. Hence t ≤
￿i
2
￿
. Since the function g(t) is
4decreasing function as t increases, we have
g(t) ≥ g
￿￿
i
2
￿￿
=
￿￿
i
2
￿
+ i − 1
￿
/
￿￿
i
2
￿
+ 1
￿
= f(i).
Recall that there is no point 1-covered by 2n + 1 codewords in the
minimum 1-identifying code. Hence the average excess of each member of
the company is at least f(2n), since f(i) is decreasing in each company and
i ≤ 2n.
The total excess E(Zn
3) trivially equals K(2n + 1) − 3n. Since each
member of any company has the average excess of at least f(2n) and there
are at most K vertices that have excess zero, we have
K(2n + 1) − 3n ≥ (3n − K)f(2n).
If we simplify the above inequality, we have
K ≥
2   3n
2n + 1 + 1
n
=
2v
d + 1 + 1
n
where d is the degree of each vertex and v is the number of vertices.
Remark In the proof of Theorem 9 of [2], the authors assume that the
families (companies in our term) are disjoint. However in our above proof
the companies do not need to be disjoint.
Similarly, we improve the lower bound given in (2.1) when q = 4.
Theorem 2.3. For q = 4 in Lee metric,
M
4
1(n) ≥
2v
d + 1 + 1
n
=
2   4n
2n + 1 + 1
n
,
where v is the number of vertices of Zn
4 and d is the degree of any vertex
in Zn
4.
Proof. The proof is almost identical with the proof of Theorem 2.2. Hence
suppose that C is a 1-identifying code with K = M4
1(n). Since Case 3 does
not happen, we just need to consider Case 1 and Case 2. With the same
argument as in the proof of Theorem 2.2, we can show that the average
excess of each point of T is at least f(2n).
The total excess E(Zn
4) trivially equals K(2n+1)−4n. Since the average
excess is at least f(2n), we have
K(2n + 1) − 4
n ≥ (4
n − K)f(2n).
If we simplify the above inequality, we have
K ≥
2   4n
2n + 1 + 1
n
=
2v
2n + 1 + 1
n
,
5where d is the degree of each vertex and v is the number of vertices.
Remark We note that the above arguments do not work if q ≥ 5 because
we cannot deﬁne an employer like Case 1 if q ≥ 5.
3 Upper Bounds
In this section, we extend certain results on 1-identifying codes in binary
Hamming spaces (see Section 1 of [2]) to 1-identifying codes in q-ary hy-
percubes.
Theorem 3.1. Let q ≥ 2. Assume that C is 1-identifying in Zn
q. The direct
sum Zq ⊕ C is 1-identifying if and only if d(c,C \ {c}) ≤ 1 for all c ∈ C.
Proof. We omit the proof as it is basically the same as that of Theorem 1
of [2].
Theorem 3.2. Let q ≥ 4. If C is 1-identifying in Zn
q, then so is C ⊕ Zq.
Proof. We denote the code C⊕Zq by D. Let z1 = (x,W1) and z2 = (y,W2),
where W1,W2 ∈ Zq and x,y ∈ Zn
q.
• Case 1: W1 = W2 and x  = y.
Since C is a 1-identifying code, B1(x) ∩ C  = B1(y) ∩ C. Without loss
of generality, we may assume that there is a codeword c ∈ C such
that c ∈ (B1(x) ∩ C) \ (B1(y) ∩ C). Then (c,W1) ∈ (B1(z1) ∩ D) \
(B1(z2) ∩ D). Hence B1(z1) ∩ D  = B1(z2) ∩ D.
• Case 2: W1  = W2 and x  = y.
Subcase 2.1: x ∈ C or y ∈ C.
Without loss of generality, we may assume that x ∈ C. Since d(z1,z2) ≥
2, (x,W1) ∈ (B1(z1) ∩ D) \ (B1(z2) ∩ D).
Subcase 2.2: x,y / ∈ C.
All codewords in B1(z1) ∩ D end with W1 and all codewords in
B1(z2) ∩ D end with W2. Hence we have B1(z1) ∩ D  = B1(z2) ∩ D.
• Case 3: W1  = W2 and x = y.
Since W1  = W2 and q ≥ 4, we can ﬁnd W3 ∈ Zq such that d(W3,W2) ≥
2 and d(W3,W1) ≤ 1. Then (x,W3) ∈ (B1(z1)∩D)\(B1(z2)∩D).
Theorem 3.3. Let q ≥ 3. If C is 1-identifying in Zn
q, then so is C ⊕ Z2
q.
6Proof. Let z1 = (x,W1) and z2 = (y,W2), where x,y ∈ Zn
q and W1,W2 ∈
Z2
q. We consider three cases as in the proof of Theorem 3.2. The ﬁrst
two cases are proved similarly. We only consider the third case, that is,
W1  = W2 and x = y.
Since W1  = W2, we can ﬁnd W3 ∈ Z2
q such that d(W3,W2) ≥ 2 and
d(W3,W1) ≤ 1. Then (x,W3) ∈ (B1(z1) ∩ D) \ (B1(z2) ∩ D).
In the case of q = 3, we can compute exact values of M3
1(2) and M3
1(3)
as follows.
Theorem 3.4. M3
1(2) = 4.
Proof. If we choose randomly four vectors in Z2
3, there are 126 possible sets
of four vectors. We check that exactly 72 such sets become 1-identifying
codes in Z2
3. For example,
{(00),(10),(21),(02)}
is a 1-identifying code (see Figure 2). But it can be checked that there
is no ternary 1-identifying code of length 2 of size 3 by hand. In fact,
Theorem 2.2 shows that M3
1(2) ≥ 4.
Figure 2: A 1-identifying code in Z2
3
Theorem 3.5. M3
1(3) = 9.
Proof. First note that M3
1(3) ≥ 8 by Theorem 2.2. We check that there are
exactly 13384 1-identifying codes of length 3 with size 9 containing (000)
in Z3
3. An example of a ternary 1-identifying code of length 3 is given:
{(000),(012),(201),(100),(011),(122),(222),(220),(001)}.
But it is checked that there is no ternary 1-identifying code of length 3 of
size 8 by computer.
Next we consider the case when q = 4 and n = 2.
Theorem 3.6. M4
1(2) = 7.
7Proof. Note that M4
1(2) ≥ 6 by Theorem 2.3. We check that there are
exactly 960 1-identifying codes of size 7 in Z2
4. We give two quaternary
1-identifying codes of length 2 as follows.
{(22),(00),(32),(13),(30),(33),(11)}
{(00),(32),(10),(03),(20),(23),(02)}.
These are drawn in Figure 3.
Figure 3: Two 1-identifying codes in Z2
4
But there is no quaternary 1-identifying code of length 2 of size 6 by
computer.
In what follows, we give upper bounds on M
p
1(n) when p = 3s and
p = 4s. Using Theorems 3.2, 3.3, 3.4, 3.5, and 3.6, we obtain the following.
Theorem 3.7. (i) M3
1(2k) ≤ 4   32(k−1) for k ≥ 1,
(ii) M3
1(1 + 2k) ≤ 32k for k ≥ 1,
(iii) M4
1(n) ≤ 7   4n−2 for n ≥ 2.
Corollary 3.8. M3
1(4) ≤ 36, M3
1(5) ≤ 81, M3
1(6) ≤ 324, M4
1(3) ≤ 28,
M4
1(4) ≤ 112.
Corollary 3.9. For p = 3s where s ∈ N,
(i) when n = 2k where k ∈ N, M
p
1(2k) ≤
4
9
pn;
(ii) when n = 2k + 1 where k ∈ N, M
p
1(2k + 1) ≤
pn
3
.
Similarly, for p = 4s where s ∈ N, M
p
1(n) ≤
7
16
pn for n ≥ 2.
Proof. (i) Let n = 2k. Theorem 3.7 (i) proves the case when p = 3, i.e.,
s = 1 as p = 3s. In general, for p = 3s, we copy ternary 1-identifying codes
8sn times (that is, shift these copies in every direction). Then it follows from
Theorem 3.7 (i) that
M
p
1(2k) ≤ 4   32(k−1)   sn =
4
9
pn.
(ii) If n = 2k + 1, then by Theorem 3.7 (ii)
M
p
1(2k + 1) ≤ 32k   sn =
pn
3
.
Similarly, if p = 4s, then by Theorem 3.7 (iii)
M
p
1(n) ≤ 7   4n−2   sn =
7
16
pn.
Remark The results of Corollary 3.9 for n = 3 and p = 3s is an analogue
of Corollary 6 in [14], where n = 2 and p = 13s were considered.
Acknowledgments. We thank an anonymous referee for helpful com-
ments.
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Identifying Codes in q-ary Hypercubes

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