A Decoding Scheme for the 4-ary Lexicodes with D = 3

We introduce the algorithms for basis and decoding of quaternary lexicographic codes with minimum distance d = 3 for an arbitrary length n. 1. Introduction. In this section, we define some particular operations and discuss q-ary lexicographic codes with minimum distance d. The game-theoretic operations of nim-addition ⊕ and nim-multiplication ⊗ which are used in the Game of Nim are introduced by Definitions 1.1 and 1.2. The Game of Nim is played by two players, with one or more piles of counters. Each player, in turn, removes from one to all counters of a pile. The player taking the last counter wins.


Introduction.
In this section, we define some particular operations and discuss q-ary lexicographic codes with minimum distance d. The game-theoretic operations of nim-addition ⊕ and nim-multiplication ⊗ which are used in the Game of Nim are introduced by Definitions 1.1 and 1.2.
The Game of Nim is played by two players, with one or more piles of counters. Each player, in turn, removes from one to all counters of a pile. The player taking the last counter wins. Definition 1.1. Let (α 1 ···α r ), (β 1 ···β r ) be the binary representation of α, β, respectively. For each i, α ⊕ β has a 0 digit in the position i where α i = β i , and α ⊕ β has a 1 in the position i where α i = β i . In other words, α⊕β is the Exclusive OR (XOR) of each digit in their binary representations.
For example, the nim-addition table for numbers less than 4 is given in Table 1.1. There is a nim-multiplication ⊗ which, together with nim-addition ⊕, converts the integers into a field [1]. With nim-multiplication, we know that 0 ⊗ α must be 0 which is the zero of the field. Also 1⊗α must be α. Since the elements other than 0, 1 satisfy α ⊗ α = α ⊕ 1 in the finite field of order 4, we have 2 ⊗ 2 = 3. Next 2 ⊗ 3 cannot be one of 0, 2, 3 and so must be 1.
In general, using the above value α we can define the following nim-multiplication.
For example, the nim-multiplication table for numbers less than 4 is given in The following is an easy rule enabling us to compute nim-additions: (1) the nim-sum of a number of distinct 2-powers ("2-power" means a power of 2 in the ordinary sense) is their ordinary sum; (2) the nim-sum of two equal numbers is 0. For finite numbers, the nim-multiplication follows from the following rules, analogous to those for nim-addition. We will use the term Fermat 2-power to denote the numbers 2 2 a in the ordinary sense; (3) the nim-product of a number of distinct Fermat 2-powers is their ordinary product; (4) the square of a Fermat 2-power is the number obtained by multiplying it by 3/2 in the ordinary sense.
In [1], ⊕ and ⊗ convert the numbers 0, 1, 2,... into a field of characteristic 2. Also, for all a, the numbers less than 2 2 a form a subfield isomorphic to the Galois field GF(2 2 a ).
Consider the lexicographic codes (for short, lexicodes) with base B = 2 2 a . A word of this code is a sequence x = ···x 3 x 2 x 1 of elements of {0, 1,...,2 2 a − 1}. The set of words is ordered lexicographically, that is, the word x = ···x 3 x 2 x 1 is smaller than y = ···y 3 y 2 y 1 , written x < y, in case of some r we have x r < y r and x s = y s for all s greater than r .
Lexicodes are defined by saying a word in the code in case it does not conflict with any previous codewords. That is, the lexicode with minimum distance d is defined by saying that two words do not conflict in case the Hamming distance between them is not less than d. We write n,d for the 4-ary lexicode consisting of the codewords with length n or less and minimum distance d.
In [2], Conway and Sloane showed that lexicodes with base B = 2 a are closed under nim-addition, and if B = 2 2 a the lexicodes are closed under nim-multiplication by scalars. Therefore if B is of the form 2 2 a , then the lexicode is a linear code over GF(B).

The basis and decoding for 4,3
Lemma 2.1. Let e n be the basis of length n in 4,3 . Then 111 = e 3 , 1012 = e 4 , and 10013 = e 5 .
Proof. Since the weight of e n must be greater than or equal to 3, the first basis has at least 3 nonzero digits, and so the smallest codeword is 111. The second basis e 4 is the type of 10ab, where neither a nor b is zero. Let "ab" n be the first two digits of e n . Since "ab" 3 = "11", "ab" 4 is lexicographically ordered "12", and then d(α ⊗ e 3 , 1012) ≥ 3, for α ∈ GF(4). Therefore, 1012 = e 4 . In a similar way, we obtain 10013 = e 5 .
Proof. Suppose that 1000ab ∈ 4,3 . Let α ∈ GF(4). If neither a nor b is zero, there exists e i (3 ≤ i ≤ 5) such that d(1000ab, αe i ) < 3. This contradicts the hypothesis. In all other cases, the weight of 1000ab is 2, and so the basis of length 6 does not exist.
Consider the basis e 7 of length 7. Then 10000ab of length 7 also conflicts with any smaller basis, for all "ab". Thus 10000ab needs a digit 1 in the 6th position. If "ab" = "0b" (b ≠ 0), then 110000b does not conflict with any smaller codeword. Hence 1100001 is the smallest codeword with more digits than e 5 , that is, 1100001 = e 7 . Therefore, for 7 ≤ n ≤ 21, "ab" n takes ordered digit from "01" to "33".
Suppose that there exists a basis of length 22, that is, 10 ···01000ab ∈ 4,3 . Since there exists e i (7 ≤ i ≤ 21) such that d(10 ···01000ab, e i ) < 3 for any "ab", this is a contradiction to the hypothesis. So the basis of length 22 does not exist.
As we have seen in the proof of Theorem 2.2, the basis e n has digit 1's in the nth, 6th, and (17s + 5)th positions, for all s ∈ N satisfying 6 < 17s + 5 < n.
The following tables give "ab" n corresponding to the length n, where 7 ≤ n ≤ 21 or 17p + 6 ≤ n ≤ 17q + 4, for p ∈ N and q = p + 1. Now we may consider the basis e n satisfying n ≥ 7 and n ≠ 17s + 5, s ∈ N, in the following algorithm.

Algorithm for the basis e n
Step 1. Suppose that 7 ≤ n ≤ 21. The basis e n has digit 1's in the nth and 6th positions. And "ab" n takes the (n − 6)th lexicographically ordered digit from "01" to "33" (see Table 2.1).
The following table gives the basis e n , where n ≥ 7, n ≠ 17s + 5, for s ∈ N: . .
In the following remark we explain a decoding algorithm of 4,3 in more detail.
Remark 2.7. For a given received vector r = a n a n−1 ···a 2 a 1 , we obtain the testing vector t = b n b n−1 ···b 2 b 1 , by Definition 2.6. Let s ∈ N and α ∈ GF(4), and let "f 2 f 1 " i be the first two digits of e i in t.
(A) Certainly, the codeword c is a linear combination of some bases by scalar nimmultiplication. From the given received vector, we can guess the bases which may generate the codeword.
If d(t, r) = 1, we have the following two cases. First, one of a 1 , a 2 is not correct. In the second case, one of the 6th, (17s +5)th digit is not correct. In all cases, t is obtained by bases which do not depend on errored digit. Therefore, we have the desired codeword c = t.
(B) Suppose that d(t, r) > 1. This means that both a 1 and a 2 are correct. Hence, we have to find "d 2 d 1 " (d 1 ,d 2 ∈ GF(4)) such that "b 2 b 1 " ⊕ "d 2 d 1 "= "a 2 a 1 " because t is more added by a component vector (a p ⊗ e p ) with "d 2 d 1 " of t. Therefore, if such a vector exists, we have the desired codeword c = t ⊕ (a p ⊗ e p ).
(D) Suppose that d(t, r) > 1 and there is no component vector (a p ⊗e p ) with "d 2 d 1 " in t. For all α, a q such that q = 6, 17s + 5, if it does not satisfy the equation α ⊗ (a q ⊗ "f 2 f 1 " q ) = "d 2 d 1 ", then r has a nonzero leading digit in the 6th or (17s + 5)th position. If r has a nonzero leading digit in the 6th position, then we have the desired codeword c = t ⊕ (a k ⊗ e k ), for some a k (7 ≤ k ≤ 21). If r has a nonzero leading digit in the (17s + 5)th position, then we have the desired codeword c = t ⊕ (a k ⊗ e k ), for some a k (17s + 6 ≤ k ≤ 17s + 21). In fact, we can obtain a k satisfying (a k ⊗ "f 2 f 1 " k ) = "d 2 d 1 ".
Step 2. Suppose that d(t, r) > 1 and there is (a p ⊗ e p ) with "d 2 d 1 " in t. Then c = t ⊕ (a p ⊗ e p ).

Call for Papers
This subject has been extensively studied in the past years for one-, two-, and three-dimensional space. Additionally, such dynamical systems can exhibit a very important and still unexplained phenomenon, called as the Fermi acceleration phenomenon. Basically, the phenomenon of Fermi acceleration (FA) is a process in which a classical particle can acquire unbounded energy from collisions with a heavy moving wall. This phenomenon was originally proposed by Enrico Fermi in 1949 as a possible explanation of the origin of the large energies of the cosmic particles. His original model was then modified and considered under different approaches and using many versions. Moreover, applications of FA have been of a large broad interest in many different fields of science including plasma physics, astrophysics, atomic physics, optics, and time-dependent billiard problems and they are useful for controlling chaos in Engineering and dynamical systems exhibiting chaos (both conservative and dissipative chaos).
We intend to publish in this special issue papers reporting research on time-dependent billiards. The topic includes both conservative and dissipative dynamics. Papers discussing dynamical properties, statistical and mathematical results, stability investigation of the phase space structure, the phenomenon of Fermi acceleration, conditions for having suppression of Fermi acceleration, and computational and numerical methods for exploring these structures and applications are welcome.
To be acceptable for publication in the special issue of Mathematical Problems in Engineering, papers must make significant, original, and correct contributions to one or more of the topics above mentioned. Mathematical papers regarding the topics above are also welcome.
Authors should follow the Mathematical Problems in Engineering manuscript format described at http://www .hindawi.com/journals/mpe/. Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at http:// mts.hindawi.com/ according to the following timetable: Manuscript Due December 1, 2008 First Round of Reviews March 1, 2009