Asymmetric Single Magnitude Four Error Correcting Codes
aa r X i v : . [ c s . I T ] M a r Asymmetric Single Magnitude Four Error Correcting Codes
Derong Xie Jinquan Luo ∗ Abstract − Limited magnitude asymmetric error model is well suited for flash memory. In this paper,we consider the construction of asymmetric codes correcting single error over Z k r and which are basedon so called B [4](2 k r ) set. In fact, we reduce the construction of a maximal size B [4](2 k r ) set for k ≥ B [4](2 k − r ) set. Finally, we give a explicit formula of a maximalsize B [4](4 r ) set and some lower bounds of a maximal size B [4](2 r ) set. By computer searching up to q ≤ Index Terms − Asymmetric error, single error, flash memories, limited magnitude error.
I Introduction
Flash memory is a kind of non-volatile memory which has higher transfer speed, longer life span andless sensitive of vibration than hard disks. But the material of flash memory is expensive and has fixedblocks, which makes it necessary to increase the density of flash memory. At the same time, it faces manychallenges such as how to implement codes correcting asymmetric errors into the flash memories. In [1],the asymmetric channel with limited magnitude errors was introduced and the further results were givenin [2, 3]. An error model with asymmetric errors of limited magnitude is a good model for some multilevelflash memories. In the asymmetric error model, a symbol a over an alphabet Z q = { , , · · · , q − } may be modified during transmission into b , where b ≥ a , and the probability that a is changed to b isconsidered to be the same for all b > a . For some applications, the error magnitude b − a is not likely toexceed a certain level λ. In general, the errors are mostly asymmetric and some classes of construction ofasystematic codes correcting such errors were studied in [5, 6, 7]. Also, several constructions of systematiccodes correcting single errors are given in [6] and the symmetric case is closely related to equi-differenceconflict-avoiding codes see e.g., [10, 13]. In addition, splitter sets can be seen as codes correcting singlelimited magnitude errors in flash memories see e.g., [4, 7, 8, 9, 11, 12, 13].On the other hand, construction of codes correcting t errors can be transformed to B t [ λ ]( q ) sets and theconstruction of a maximal size B [3](2 k r ) set, B [3](3 k r ) set and B [4](3 k r ) set can be found in [7]. In thispaper, we consider the construction of a maximal size B [4](2 k r ) set. In Section II, we briefly introduce B [ λ ]( q ) set and linear codes over the ring Z q . Indeed, we recall some basic results on B [ λ ]( q ) sets. InSection III, we reduce the construction of a maximal size B [4](2 k r ) set for k ≥ B [4](2 k − r ) set. In Section IV, we give an exact formula for calculating a maximal size ∗ The authors are with school of mathematics and statistics & Hubei Key Laboratory of Mathematical Sciences, CentralChina Normal University Wuhan 430079, China. E-mail: [email protected](J.Luo); [email protected]. B [4](4 r ) set. In section V, we consider maximal size B [4](2 r ) set. Finally, we give a short summary ofthis paper in Section VI. II Preliminaries
The following result is almost identical to the introduction given in [6, 7, 8]. But we include it here for thecompleteness of this paper.If H is an h × m matrix over Z q , the corresponding code of length m with parity check matrix H , is C H = { x ∈ Z mq | x H t = } where H t denotes the transposed of H .Let E ⊂ Z mq be the set of error patterns that we want to correct and consider single errors of magnitudeat most λ . If x ∈ C H is a sent codeword and e ∈ E is an error introduced during transmission, then thereceived m -tuple is y = x + e . Therefore y H t = x H t + e H t = e H t . As usual, we call e H t the syndrome of e . Let S H, E = { e H t | e ∈ E} be the set of syndromes. We require these to be all distinct, i.e., |S H, E | = |E| . When this is the case, thecode is able to correct all error patterns in E . Moreover, [ x ∈C H { x + e | e ∈ E} is a disjoint union, and so we get the Hamming type bound |C H | · |E| = |C H | · |S H, E | ≤ q m . For ordinary linear codes (for q a prime power), when h = 1, C H is an [ m.m −
1] code of minimumHamming distance two that an not correct any errors (without limitation on the magnitude). When weconsider errors of limited magnitude, the situation may be quite different, and it is a nontrivial task to findgood H . Therefore, we consider h = 1, that is H = ( b , b , · · · , b m − ), and the error patterns we considerare E λ,m , the set of sequences ( e , e , · · · , e m − ) ∈ [0 , λ ] m of Hamming weight at most 1. We see that |E λ,m | = mλ + 1 . Permuting the elements of ( b , b , · · · , b m − ), we get another code with the same error correctingcapability. Therefore, from now on we consider sets B = { b , b , · · · , b m − } of distinct positive integers such that the corresponding syndromes S = m − X j =0 e j b j (mod q ) | ( e , e , · · · , e m − ) ∈ E λ,m are distinct. This is called a B [ λ ]( q ) set, see [6]. The corresponding code we denote by C B , that is C B = ( ( x , x , · · · , x m − ) ∈ Z mq (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m − X i =0 x i b i ≡ q ) ) . For any positive integer l coprime to d , let ord d ( l ) be the order of l in Z ∗ d , that is,ord d ( l ) = min { n > | l n ≡ d ) } . For a ∈ Z ∗ d and d a divisor of q , we let β = aq/d. For gcd( al, d ) = 1, define the cyclotomic set σ l ( β ) = { l i β (mod q ) | i ≥ } . Then | σ l ( β ) | = ord d ( l ) .Lemma d = p e p e · · · p e s s with p i distinct primes not diving l , we haveord d = lcm (cid:16) ord p e ( l ) , ord p e ( l ) , · · · , ord p ess ( l ) (cid:17) . b) If p is a prime not dividing l and l ord p ( l ) − p µ p a , where gcd( a, p ) = 1, thenord p k ( l ) = ord p ( l ) if k ≤ µ p ,p k − µ p ord p ( l ) if k > µ p , p = 2 and l ≡ ,p k − µ p ord p ( l ) if k > µ p and p > , if p = 2 , l ≡ k = 2 , , k − if p = 2 , l ≡ k > . III Maximal size B [4](2 k r ) set For q = 2 k r with gcd( r,
6) = 1, we will introduce a result reducing the construction of a maximal size B [4](2 k r ) set for k ≥ B [4](2 k − r ) set in this section. Define M ( q ) to be the maximal size of a B [4]( q ) set.For a positive integer q = 2 k r with gcd( r,
6) = 1 and d | r , let V d = { ar/d (mod q ) | a ∈ Z k d , gcd( a, d ) = 1 } and U i = { x ∈ V d | gcd( x, k ) = 2 i } . We have Z k r = [ d | r V d and V d = [ ≤ i ≤ k U i . (3.1)Let M ′ (2 k d ) = max (cid:8)(cid:12)(cid:12) { B ∩ V d | B ∈ B [4](2 k r ) } (cid:12)(cid:12)(cid:9) . (3.2)Then M (2 k r ) = X d | r M ′ (2 k d ) . If k ≥
3, we consider the following disjoint decomposition Z q \{ } = N ∪ N ∪ N ∪ N , where N i = { a mod q | ≤ a ≤ k − i r, ∤ a } for 0 ≤ i ≤ N = { a mod q | ≤ a < k − r } .T heorem r,
6) = 1 and k ≥
3, then we have M (2 k r ) = M (2 k − r ) + 2 k − r.P roof : For any d | r , write n k = ord k d (3) for short if no confusion occurs. Then n k − is even and n k − = n k − or n k − = 2 n k − . For any a , let α = ar/d and the value of α varies with a . If gcd( a, k d ) = 1 , we have | σ ( α ) | = ord k d (3) = n k . Let Γ k d be a set of coset representatives of the group generated by 3 in Z ∗ k d . It suffices to consider thefollowing cases.(i) If n k − is odd, then n k − = 2 n k − and k = 3. Hence, n k = lcm (ord (3) , ord d (3)) = lcm (ord (3) , ord d (3)) = n k − . We can choose T d = [ a ∈ Γ k − d (cid:8) i α (mod 2 k r ) | ≤ i ≤ n k − − (cid:9) . (ii) If n k − is even and n k − = n k − , then 1 + 2 k − d is not in the cyclic group generated by 3 in Z ∗ k − d .We can choose T d = [ a ∈ Γ k − d (cid:16)n i α (mod 2 k r ) | ≤ i ≤ n k − − o ∪ n i +1 ( α + 2 k − r ) (mod 2 k r ) | ≤ i ≤ n k − − o(cid:17) . (iii) It is easy to see that n k − is even and n k − = 2 n k − is equivalent to k ≥ v (ord d (3)) ≤ k − n k = 2 n k − . In this case, then we can choose T d = [ a ∈ Γ k − d (cid:16)n i α (mod 2 k r ) | ≤ i ≤ n k − − o ∪ n i +1 α (mod 2 k r ) | n k − ≤ i ≤ n k − − o(cid:17) . In total, we choose S = S d | r T d . Let S be a B [4](2 k − r ) set. Define S = S ∪ S ′ with S ′ = { c (mod 2 k r ) | c ∈ S } . Obviously S ′ is a B [4](2 k r ) set. Indeed, it is easy to verify that there do not exist distinct elements x, y ∈ S such that 2 x ≡ y (mod 2 k r ), 3 x ≡ y (mod 2 k r ) or 4 x ≡ y (mod 2 k r ). Hence 2 S , 3 S and 4 S both have the same size as S . Note that S ′ is a B [4](2 k r ) set contained in N . We see that • S ∩ S = ∅ since S ⊂ N , 2 S ⊂ N . • S ∩ S = ∅ since S ⊂ N , 3 S ⊂ N and S ∩ S = ∅ . • S ∩ S = ∅ since S ⊂ N , 4 S ⊂ N . • S ∩ S = ∅ since 2 S ⊂ N , 3 S ⊂ N . • S ∩ S = ∅ since 2 S ⊂ N , 4 S ⊂ N . • S ∩ S = ∅ since 3 S ⊂ N , 4 S ⊂ N .Then S is a B [4](2 k r ) set of size M (2 k − r ) + 2 k − r .On the other hand, firstly we note that at least one of a, a, a, a belongs to N for any a ∈ N ∪ N ∪ N .Therefore, in N ∪ N ∪ N , at most | N | = 2 k − r elements can be chosen in a B [4](2 k r ) set. Also, neitherof a, a, a, a belongs to N ∪ N ∪ N for any a ∈ N . Since( S ∪ S ∪ S ∪ S ) ⊂ ( N ∪ N ∪ N ) , then the set S is a maximal size B [4](2 k r ) set. (cid:4) Example • For q = 40, we have M (40) = M (5) + 5 = 6 . The construction of maximal set in the proof of Theorem 1 is presented as follows. Firstly we have k = 3 and r = 5. Hence, d = 1 or d = 5. − If d = 1, then n = 2, n = 1, and Case (i) applies. We can choose Γ = { } and so T = { } . − If d = 5, then n = 2, n = 4, and Case (ii) applies. Choose Γ = { } . Then T = (cid:8) i (mod 40) | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
11 (mod 40) | ≤ i ≤ (cid:9) = { , , , } . We can choose { } as a perfect B [4](5) set. Then we obtain a maximal size B [4](40) set T ∪ T ∪ · { } = { , , , , , } . • For q = 160, note that { , , , } is a maximal size B [4](20) set (see Example 2). Therefore M (160) = 20 + M (20) = 24 . The construction of maximal set in the proof of Theorem 1 is depicted as follows. Firstly we have k = 5 and r = 5. Hence, d = 1 or d = 5. − If d = 1, then n = 4, n = 2, and Case (iii) applies. We can choose Γ = { , } and so T = (cid:8) · , · (cid:9) ∪ (cid:8) ·
25 (mod 160) , ·
25 (mod 160) (cid:9) = { , , , } . − If d = 5, then n = 4, n = 4, and Case (ii) applies. Choose Γ = { , , , } . Then T = (cid:8) i (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
41 (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i · | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
47 (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i ·
11 (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
51 (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i ·
13 (mod 160) | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
53 (mod 160) | ≤ i ≤ (cid:9) = { , , , , , , , , , , , , , , , } . Hence, T ∪ T ∪ · { , , , } = { , , , , , , , , , , , , , , , , , , , , , , , } is a maximal size B [4](160) set. IV Construction of maximal size B [4](4 r ) sets In this section, we give an explicit construction of maximal size B [4](4 r ) sets. We consider the followingdisjoint decomposition Z r \{ } = N ∪ N ∪ N ∪ N where N = { a mod 4 r | ≤ a ≤ r, ∤ a } N = { a mod 4 r | ≤ a ≤ r, ∤ a } N = { a mod 4 r | ≤ a < r } .T heorem r,
6) = 1, then M (4 r ) = r − .P roof : 1) If r = 1, then M = 0 .
2) If r > r,
6) = 1, for any d | r and d ≥ d be a set of coset representatives of thegroup generated by 3 in Z ∗ d . It suffices to consider the following cases.(i) If n = ord d (3) is odd, then n = n and n = 2 n . We can choose T d = [ a ∈ Γ d (cid:8) i α (mod 4 r ) | ≤ i < n (cid:9) . For distinct elements 3 i α (mod 4 r ) , i ′ α (mod 4 r ) ∈ T d , since n is odd, then 3 i − i ′ ) d ) andso 4 · i α (mod 4 r ) · i α (mod 4 r ) . (ii) If n = ord d (3) is even, then n = n = n and 1 + 2 d is not in the cyclic group generated by 3 in Z ∗ d . We have α (1 + 2 d ) ≡ α + 2 r (mod 4 r ) , · i α ( α + 2 r ) ≡ · i α (mod 4 r ) , · i α ( α + 2 r ) ≡ · i α (mod 4 r ) . We can choose T d = [ a ∈ Γ d (cid:16)n i α (mod 4 r ) | ≤ i < n o ∪ n i +1 ( α + 2 r ) (mod 4 r ) | ≤ i < n o(cid:17) . In total, we choose S = S d> ,d | r T d . It is easy to verify that there do not exist distinct elements x, y ∈ S such that 2 x ≡ y (mod 2 k r ), 3 x ≡ y (mod 2 k r ) or 4 x ≡ y (mod 2 k r ). Hence 2 S , 3 S and 4 S both have the same size as S . Obviously, S ∩ S = ∅ . Since S ⊂ N , 2 S ⊂ N , 3 S ⊂ N and 4 S ⊂ N ,then S ∩ S = ∅ , S ∩ S = ∅ , 2 S ∩ S = ∅ , 2 S ∩ S = ∅ and 3 S ∩ S = ∅ .Therefore, S is a B [4](4 r ) set of size r − M (4 r ) ≤ (cid:22) r − (cid:23) = r − . Hence, the set S is a maximal size B [4](4 r ) set. (cid:4) Remark k ≡ M (2 k r ) = 17 (2 k r + 3 r − .Example • For q = 20, we have M (20) = 5 − . The construction of maximal set in the proof of Theorem 2 is presented as follows. Firstly we have r = 5. Hence, d = 1 or d = 5. If d = 5, then n = 4 is even and Case (ii) applies. We can chooseΓ = { } and so T = (cid:8) i (mod 20) | ≤ i ≤ (cid:9) ∪ (cid:8) i +1 ·
21 (mod 20) | ≤ i ≤ (cid:9) = { , , , } is a maximal size B [4](20) set. • For q = 44, we have M (20) = 11 − . The construction of maximal set in the proof of Theorem 2 is depicted as follows. Firstly we have r = 11. Hence, d = 1 or d = 11. If d = 11, then n = 5 is odd and Case (i) applies. We can chooseΓ = { , } and so T = (cid:8) i (mod 44) | ≤ i ≤ (cid:9) ∪ (cid:8) i · | ≤ i ≤ (cid:9) = { , , , , , , , , , } is a maximal size B [4](20) set. V On lower bounds of M ′ (2 d ) Let h i d be the group generated by 3 in Z ∗ k d . Recall V d in 3.1 and M ′ (2 d ) in 3.2. We give a formula or lowerbound for M ′ (2 d ) whether 2 ∈ h i d or not. Firstly, M ′ (2) = 0 . For q = 2 r , we note that | U | = | U | = ϕ ( d )where ϕ is the Euler , s totient function. Let θ : Z r → Z r defined by θ ( x ) = 2 x (mod 2 r ) .Lemma θ to U i of Z r , we have(1) θ ( U ) = U , which is a bijection.(2) θ ( U ) = U , which is a bijection. P roof : For x ∈ Z r , let t = x + r (mod 2 r ), then x = t and θ ( x ) = θ ( t ) = 2 x (mod 2 r ).For any 2 a ∈ U , we have two cases to consider: if a is odd, then a ∈ U , t ∈ U and θ (2 a ) = θ ( t ) =2 a (mod 2 r ); if a is even, then t ∈ U , a (mod 2 r ) ∈ U and θ (2 a ) = θ ( t ) = 2 a (mod 2 r ). Since | U | = | U | ,then θ | U and θ | U are both bijections. (cid:4) V.1 On lower bounds of M ′ (2 d ) with ∈ h i d and d ≥ For brevity, we let • n = ord d (3); • ≡ s (mod d ) with s ∈ [1 , n ]; • m = min { s, n − s } ; • n = 2 k ′ m + r ′ with 0 ≤ r ′ < m ; T heorem n is even and s is odd, then M ′ (2 d ) = 12 ϕ ( d ) .P roof : If n is even and s is odd, we can choose T d = [ a ∈ Γ d n i α (mod 2 r ) | ≤ i < n o . We note that T d ⊂ U . For distinct elements x, y ∈ T d , we have 2 x y (mod 2 r ) and 4 x y (mod 2 r )by Lemma 2. Clearly, 3 x y (mod 2 r ). Hence 2 T d , T d and 4 T d both have the same size as T d . Obviously, T d ∩ T d = ∅ . Checking binary parity we can get T d ∩ T d = ∅ , T d ∩ T d = ∅ , T d ∩ T d = ∅ and 3 T d ∩ T d = ∅ .Since s ∈ [1 , n ] is odd and i, j ∈ [0 , n/ − i · j (mod d ) which implies that 2 T d ∩ T d = ∅ .Hence S is a B [4](2 r ) set. Therefore, M ′ (2 d ) ≥ | T d | = 12 ϕ ( d ) . On the other hand, it is clear that M ′ (2 d ) ≤ | V d | ϕ ( d )and so M ′ (2 d ) = ϕ ( d ). (cid:4) Example q = 2 · ·
19 = 190, we have r = 95. Hence, d = 1, d = 5, d = 19 or 95. − If d = 5, then s = 3 and n = 4. We have | M ′ (10) | = ϕ (5) / = { } and so T = (cid:8) i ·
19 (mod 190) | ≤ i < (cid:9) = { , } . − If d = 19, then s = 7 and n = 18. We have | M ′ (38) | = ϕ (19) / = { } . Then T = (cid:8) i · | ≤ i < (cid:9) = { , , , , , , , , } . − If d = 95, then s = 7 and n = 36. We have | M ′ (190) | = ϕ (95) / = { , } . Then T = (cid:8) i (mod 160) | ≤ i < (cid:9) ∪ (cid:8) i · | ≤ i < (cid:9) = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } . Hence, M (190) = P d | M ′ (2 d ) = 47 and T ∪ T ∪ T = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } is a maximal size B [4](190) set. T heorem n and s are even , then(1) in the case m = 2, M ′ (2 d ) = ϕ ( d ) n · j n k ;(2) in the case m > M ′ (2 d ) ≥ ( m − ϕ ( d )2 m if r ′ = 0 , ( k ′ m − k ′ +1) ϕ ( d ) n if r ′ = 2 , mϕ ( d ) n if 2 < r ′ ≤ m and k ′ = 1 , ( k ′ m − k ′ +2) ϕ ( d ) n if 2 < r ′ ≤ m and k ′ ≥ , ( k ′ m + r ′ − m − k ′ ) ϕ ( d ) n if r ′ > m.P roof : (1) For m = 2, we note that 4 · i α ≡ · i +2 α (mod 2 r ) or 2 · i α ≡ · i +2 α (mod 2 r ). Hence,if 3 i α is chosen in a B [4](2 r ) set, then 3 i +1 α and 3 i +2 α can not be chosen. Therefore, M ′ (2 d ) ≤ ϕ ( d ) n · j n k . By Lemma 2, we know that θ | U and θ | U are both bijections. Hence, for any z , z ∈ [1 , i, i ′ < · (cid:4) n (cid:5) and 3 | ( i − i ′ ), we have z · i α ≡ z · i ′ α (mod 2 r ) if and only if z = z and i = i ′ .Hence, we can choose T d = [ a ∈ Γ d n i α (mod 2 r ) | ≤ i < j n ko . T d ∪ T d ∪ T d ∪ T d and | T d | = ϕ ( d ) n · (cid:4) n (cid:5) , then T d is the maximal size of the elements of B [4](2 r )set in V d .(2) For m > r ′ = 0, we can choose T d = k ′ − [ j =0 [ a ∈ Γ d (cid:16)n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o(cid:17) . For distinct elements x, y ∈ T d , it is easy to verify that 2 x y (mod 2 r ), 3 x y (mod 2 r ) and4 x y (mod 2 r ). Hence 2 T d , T d and 4 T d both have the same size as T d . Obviously, T d ∩ T d = ∅ .Checking binary parity we can get T d ∩ T d = ∅ , T d ∩ T d = ∅ , T d ∩ T d = ∅ and 3 T d ∩ T d = ∅ . For any x ∈ n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o ,y ∈ n i α (mod 2 r ) | j ′ m ≤ i < m j ′ m o ∪ n i +1 α (mod 2 r ) | m j ′ m ≤ i < m + j ′ m − o , we obtain x = 3 e α (mod 2 r ) and y = 3 e α (mod 2 r ). For j = j ′ , if e , e have the same binary parity then0 < | ind α ( x ) − ind α ( y ) ± m | < n and so 2 x y (mod 2 r ); otherwise, we also have 2 x y (mod 2 r )by checking binary parity of e , e .By Lemma 2, we know that θ is a bijection from U to U . If | j − j ′ | >
1, then 2 x y (mod 2 r )since 2 · t ≡ · t + m (mod 2 r ) or 4 · t ≡ · t + m (mod 2 r ).Similarly, 2 x y (mod 2 r ) holds for | j − j ′ | = 1. Therefore, M ′ (2 d ) ≥ | T d | = ( m − ϕ ( d )2 m . (ii)If r ′ = 2, we can choose T d = [ a ∈ Γ d k ′ − [ j =0 (cid:18) n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o (cid:19) ∪ { n − α (mod 2 r ) } ! and so M ′ (2 d ) ≥ ( k ′ m − k ′ + 1) ϕ ( d ) n . (iii)The case 2 < r ′ ≤ m • For k ′ = 1, we can choose T d = [ a ∈ Γ d (cid:16) { i α (mod 2 r ) | ≤ i < m } ∪ { i +1 α (mod 2 r ) | m ≤ i < m } (cid:17) and so M ′ (2 d ) ≥ mϕ ( d ) n . • For k ′ ≥
2, we can choose T d = [ a ∈ Γ d k ′ − [ j =0 (cid:18) n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o (cid:19) ∪ n i α (mod 2 r ) | k ′ m − m + 1 ≤ i < k ′ m − m o ∪ n i +1 α (mod 2 r ) | k ′ m − m ≤ i ≤ k ′ m o ∪ n k ′ m − m − α (mod 2 r ) o ! and so M ′ (2 d ) ≥ ( k ′ m − k ′ + 2) ϕ ( d ) n . If r ′ > m , we can choose T d = [ a ∈ Γ d k ′ − [ j =0 (cid:18) n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o (cid:19) ∪ ( i α (mod 2 r ) | k ′ m − m ≤ i < k ′ m + r ′ − m ) ∪ ( i +1 α (mod 2 r ) | k ′ m − m + 22 ≤ i < k ′ m + r ′ − m − ) ∪ ( i α (mod 2 r ) | k ′ m + r ′ − m ≤ i < k ′ m + r ′ − m ) ∪ ( i +1 α (mod 2 r ) | k ′ m + r ′ − m ≤ i < k ′ m + r ′ ) ! and so M ′ (2 d ) ≥ ( k ′ m + r ′ − m − k ′ ) ϕ ( d ) n . (cid:4) Lemma n is odd, then m = 1 , .P roof : We assume that m = 1 which implies 2 ≡ d ) or 2 ≡ n − (mod d ). Then d = 5 and ord (3) = 4 which contradicts to that n is odd. Similarly, if m = 2, then d = 7 ,
17 andord (3) = 6 , ord (3) = 16. (cid:4) T heorem n being odd.(1) If m is odd, then M ′ (2 d ) ≥ ( n − m ) ϕ ( d )2 n . m is even, then M ′ (2 d ) ≥ ( n + r ′ − m − k ′ ) ϕ ( d )2 n if r ′ ≤ m, (2 k ′ m + m − k ′ +1) ϕ ( d )2 n if r ′ > m.P roof : (1) If m is odd, similar to Theorem 3, we can choose T d = [ a ∈ Γ d (cid:26) i α (mod 2 r ) | ≤ i < n − m (cid:27) and so M ′ (2 d ) ≥ | T d | = ( n − m ) ϕ ( d )2 n . (2) If m is even, similar to Theorem 4, for r ′ ≤ m , we can choose T d = [ a ∈ Γ d k ′ − [ j =0 (cid:18) n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o (cid:19) ∪ n i α (mod 2 r ) | k ′ m − m ≤ i < k ′ m − m o ∪ (cid:26) i +1 α (mod 2 r ) | k ′ m − m ≤ i < n − m − (cid:27) ∪ (cid:26) i α (mod 2 r ) | k ′ m ≤ i < n − (cid:27) ! and so M ′ (2 d ) ≥ | T d | = ( n + r ′ − m − k ′ ) ϕ ( d )2 n . If r ′ > m , we can choose T d = [ a ∈ Γ d k ′ − [ j =0 (cid:18) n i α (mod 2 r ) | jm ≤ i < m jm o ∪ n i +1 α (mod 2 r ) | m jm ≤ i < m + jm − o (cid:19) ∪ n i α (mod 2 r ) | k ′ m − m ≤ i < k ′ m − m o ∪ n i +1 α (mod 2 r ) | k ′ m − m ≤ i < k ′ m o ∪ (cid:26) i α (mod 2 r ) | n − m − ≤ i ≤ n − (cid:27) ! and so M ′ (2 d ) ≥ | T d | = (2 k ′ m + m − k ′ + 1) ϕ ( d )2 n . (cid:4) If r is a prime, then M (2 r ) = M ′ (2 r ). For prime r < M (2 p ) with 2 ∈ h i p p n m k r ′ M (2 p ) T d Thm.5 4 1 2 { , }
37 6 2 1 2 2 { , } { , , , , } { , , , , , , , , }
323 11 4 1 3 ≥ { , , , , , , , } { , , , , , , , , , , , , , }
331 30 6 2 6 ≥ { , , , , , , , , , , , } { , , , , , , , , , , , , , , , , , , , , }
347 23 6 1 11 ≥ { , , , , , , , , , , , , , , , , , } { , , , , , , , , , , , , , , , , , , , , , , , , , }
371 35 11 1 13 ≥ { , , , , , , , , , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , , , ,
77, 81, 87, 93, 95, 105, 109, 111, 113, 117, 125,127 , , , , , , , , , , } { , , , , , , , , , , , , , , ,
61, 67, 73, 75, 77, 81, 83, 91, 93, 101, 103, 111,113, 117, 119, 121, 127, 133, 147, 149, 151, 153,159 , , , , , , , , , , } V.2 On lower bounds of M ′ (2 d ) with
6∈ h i d and d ≥ Let Λ d be a set of coset representatives of h , i d in Z ∗ d with 1 ∈ Λ d . Let Γ , be a set of coset representativesof the group generated by 3 in h , i d and so we can choose Γ , = { , , · · · , t − } where t = |h , i d ||h i d | . Since4 |h i d | = |h i d | , then the natural map Z ∗ d −→ Z ∗ d x x (mod d )induces a group isomorphism φ : Z ∗ d / h i d −→ Z ∗ d / h i d and φ ( d + 2) = 2. Denote by b = d −
2. Then [ a ∈ Λ d a · { , b, · · · , b t − } is a set of coset representatives of h i d in Z ∗ d . For 2
6∈ h i d , there exists a unique integer s ∈ [1 , n − · b t − · s ≡ d ), i.e., b t · s ≡ d ). T heorem n and t + s are even, then M ′ (2 d ) = 12 ϕ ( d ) .P roof : • If both t and s are even, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:16)n i b j α (mod 2 r ) | ≤ i < n o ∪ n i +1 b j +1 α (mod 2 r ) | ≤ i < n o(cid:17) . • If both t and s are odd, we can choose T d = [ a ∈ Λ d (cid:18) t − [ j =0 n i b j α (mod 2 r ) | ≤ i < n o (cid:19) ∪ (cid:18) t − [ j =0 n i +1 b j +1 α (mod 2 r ) | ≤ i < n o (cid:19)! . For i, i ′ ∈ [0 , n −
1] and j, j ′ ∈ [0 , t − i, j ) = ( i ′ , j ′ ), then3 i j i ′ j ′ (mod d ) . Therefore, for distinct elements x, y ∈ T d , it is easy to verify that 2 x y (mod 2 r ), 3 x y (mod 2 r )and 4 x y (mod 2 r ). Hence all of 2 T d , T d and 4 T d have the same size as T d . Obviously, T d ∩ T d = ∅ . Checking binary parity we can get T d ∩ T d = ∅ , T d ∩ T d = ∅ , T d ∩ T d = ∅ and 3 T d ∩ T d = ∅ .Indeed, if both t and s are even, then n · i α (mod 2 r ) | ≤ i < n o ∩ n · i +1 b t − α (mod 2 r ) | ≤ i < n o = ∅ . If both t and s are odd, then n · i α (mod 2 r ) | ≤ i < n o ∩ n · i b t − α (mod 2 r ) | ≤ i < n o = ∅ . T d ∩ T d = ∅ . Therefore, M ′ (2 d ) ≥ | T d | = | Λ d | · n · t ϕ ( d )2 . On the other hand, it is clear that M ′ (2 d ) ≤ | V d | ϕ ( d )2and so M ′ (2 d ) = ϕ ( d )2 . (cid:4) T heorem n being even.(1) If t is odd and s is even, then M ′ (2 d ) ≥ | Λ d | · ( n − · ( t − s if t − < s < n , | Λ d | · ( n − · t if t < s = n , | Λ d | · ( n − · ( t − n − s if n < s < n − t + 1 , | Λ d | · n · ( t − otherwise . (2) If t is even and s is odd, then M ′ (2 d ) ≥ | Λ d | · ( n − · ( t − s +22 if t − < s < n − , | Λ d | · ( n − · t if t < n and s = n or n ± , | Λ d | · ( n − · ( t − n − s +22 if n < s < n − t + 2 , | Λ d | · n · ( t − otherwise .P roof : (1) The case t is odd and s is even. – If t − < s < n , we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) n i b j α (mod 2 r ) | ≤ i < s o ∪ (cid:26) i +1 b j α (mod 2 r ) | s ≤ i < n − (cid:27) ∪ n i +1 b j +1 α (mod 2 r ) | ≤ i < s o ∪ (cid:26) i b j +1 α (mod 2 r ) | s + 22 ≤ i < n (cid:27) (cid:19) ∪ n i b t − α (mod 2 r ) | ≤ i < s o ∪ n i +1 b t − α (mod 2 r ) | s ≤ i < s o ! . a ∈ Λ d and j ∈ [0 , t − ], (cid:18) n · i b j α (mod 2 r ) | ≤ i < s o ∪ (cid:26) · i +1 b j α (mod 2 r ) | s ≤ i < n − (cid:27) (cid:19) \(cid:18)n · i +1 b j +1 α (mod 2 r ) | ≤ i < s o ∪ (cid:26) · i b j +1 α (mod 2 r ) | s + 22 ≤ i < n (cid:27)(cid:19) = ∅ . Indeed, n · i b t − α (mod 2 r ) | ≤ i < s o = (cid:26) · i bα (mod 2 r ) | n − s ≤ i < n (cid:27) and n · i +1 b t − α (mod 2 r ) | s ≤ i < s o = n · i +1 bα (mod 2 r ) | ≤ i < s o . Therefore, M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · ( t −
1) + 2 s . – If t < s = n , we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) n i b j α (mod 2 r ) | ≤ i < s o ∪ (cid:26) i +1 b j α (mod 2 r ) | s ≤ i < n − (cid:27) ∪ n i +1 b j +1 α (mod 2 r ) | ≤ i < s o ∪ (cid:26) i b j +1 α (mod 2 r ) | s + 22 ≤ i < n (cid:27) (cid:19) ∪ n i b t − α (mod 2 r ) | ≤ i < s o ∪ n i +1 b t − α (mod 2 r ) | s ≤ i < s − o ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . – If n < s < n − t + 1, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) (cid:26) i b j α (mod 2 r ) | ≤ i < n − s (cid:27) ∪ (cid:26) i +1 b j α (mod 2 r ) | n − s ≤ i < n − (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − s (cid:27) ∪ (cid:26) i b j +1 α (mod 2 r ) | n − s + 22 ≤ i < n (cid:27) (cid:19) ∪ (cid:26) i b t − α (mod 2 r ) | ≤ i < n − s (cid:27) ∪ (cid:26) i +1 b t − α (mod 2 r ) | s − ≤ i < n − (cid:27) ! M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · ( t − n − s . – Otherwise, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) n i b j α (mod 2 r ) | ≤ i < n o ∪ n i +1 b j +1 α (mod 2 r ) | ≤ i < n o (cid:19) and M ′ (2 d ) ≥ | T d | = | Λ d | · n · ( t − . (2) The case t is even and s is odd. – If t − < s < n − , then we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) (cid:26) i +1 b j α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i b j α (mod 2 r ) | s + 32 ≤ i < n (cid:27) (cid:19)! ∪ t − [ j =0 (cid:18) (cid:26) i b j +1 α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | s + 12 ≤ i < n − (cid:27) (cid:19)! ∪ (cid:26) i b t − α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i +1 b t − α (mod 2 r ) | s + 12 ≤ i < s + 1 (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · ( t − n − s +22 . – If t < n and s = n − , we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) n i +1 b j α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i b j α (mod 2 r ) | n + 44 ≤ i < n (cid:27) (cid:19)! ∪ t − [ j =0 (cid:18) n i b j +1 α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | n ≤ i < n − (cid:27) (cid:19)! ∪ n i b t − α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i +1 b t − α (mod 2 r ) | n ≤ i < n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . – If t < s = n , we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) (cid:26) i +1 b j α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i b j α (mod 2 r ) | s + 32 ≤ i < n (cid:27) (cid:19)! ∪ t − [ j =0 (cid:18) (cid:26) i b j +1 α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | s + 12 ≤ i < n − (cid:27) (cid:19)! ∪ (cid:26) i b t − α (mod 2 r ) | ≤ i < s + 12 (cid:27) ∪ (cid:26) i +1 b t − α (mod 2 r ) | s + 12 ≤ i < n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . – If t < n and s = n +22 , we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) n i b j α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i +1 b j α (mod 2 r ) | n + 44 ≤ i < n (cid:27) (cid:19)! ∪ t − [ j =0 (cid:18) n i +1 b j +1 α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i b j +1 α (mod 2 r ) | n ≤ i < n − (cid:27) (cid:19)! ∪ n i +1 b t − α (mod 2 r ) | ≤ i < n o ∪ (cid:26) i b t − α (mod 2 r ) | n ≤ i < n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . – If n +22 < s < n − t + 1, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) (cid:26) i b j α (mod 2 r ) | ≤ i < n − s + 12 (cid:27) ∪ (cid:26) i +1 b j α (mod 2 r ) | n − s + 12 ≤ i < n − (cid:27) (cid:19)! ∪ t − [ j =0 (cid:18) (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − s + 12 (cid:27) ∪ (cid:26) i b j α (mod 2 r ) | n − s + 32 ≤ i < n (cid:27) (cid:19)! ∪ (cid:26) i +1 b t − α (mod 2 r ) | ≤ i < n − s + 12 (cid:27) ∪ (cid:26) i b t − α (mod 2 r ) | s − ≤ i < n (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · ( t − n − s +22 . – Otherwise, we can choose T d = [ a ∈ Λ d (cid:18) t − [ j =0 n i b j α (mod 2 r ) | ≤ i < n o (cid:19) ∪ (cid:18) t − [ j =0 n i +1 b j +1 α (mod 2 r ) | ≤ i < n o (cid:19) and M ′ (2 d ) ≥ | T d | = | Λ d | · n · ( t − . (cid:4) T heorem n being odd.(1) If t is odd, then M ′ (2 d ) ≥ | Λ d | · ( n − · t if s = 1 , | Λ d | · ( n − t − n − s +22 if s is odd and s > , | Λ d | · ( n − · ( t − if s = 0 , | Λ d | · ( n − t − s if s is even and s > . (2) If t is even, then M ′ (2 d ) ≥ | Λ d | · ( n − t − s +12 if s is odd , | Λ d | · ( n − · t if s = 0 , | Λ d | · ( n − t − n − s +12 if s is even and s > .P roof : It is similar to the proof of Theorems 6 and 7.(1) The case t is odd.0 • If s = 1, we can choose T d = [ a ∈ Λ d (cid:18) t − [ j =0 (cid:26) i b j α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:18) t − [ j =0 (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19)! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . • If s is odd and s >
1, then (cid:26) · i α (mod 2 r ) | ≤ i < n − (cid:27) ∩ (cid:26) · i b t − α (mod 2 r ) | s − ≤ i ≤ n − (cid:27) = ∅ . We can choose T d = [ a ∈ Λ d t − [ j =0 (cid:18) (cid:26) i b j α (mod 2 r ) | ≤ i < n − (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:26) i b t − α (mod 2 r ) | s − ≤ i ≤ n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − t − n − s +22 . • If s = 0, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:26) i b j α (mod 2 r ) | ≤ i < n − (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) ∪ (cid:8) n − b t − α (mod 2 r ) (cid:9) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · ( t − . • If s is even and s >
0, then (cid:26) · i α (mod 2 r ) | ≤ i < n − (cid:27) ∩ n · i b t − α (mod 2 r ) | ≤ i < s o = ∅ . We can choose T d = [ a ∈ Λ d t − [ j =0 (cid:26) i b j α (mod 2 r ) | ≤ i < n − (cid:27) ∪ (cid:26) i +1 b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) ∪ n i b t − α (mod 2 r ) | ≤ i < s o ! M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − t − s . (2) The case t is even. • If s is odd, then (cid:26) · i +1 α (mod 2 r ) | ≤ i < n − (cid:27) ∩ (cid:26) · i b t − α (mod 2 r ) | ≤ i ≤ s − (cid:27) = ∅ . We can choose T d = [ a ∈ Λ d (cid:18) t − [ j =0 (cid:26) i +1 b j α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:18) t − [ j =0 (cid:26) i b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:26) i b t − α (mod 2 r ) | ≤ i < s + 12 (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − t − s +12 . • If s = 0, we can choose T d = [ a ∈ Λ d t − [ j =0 (cid:26) i +1 b j α (mod 2 r ) | ≤ i < n − (cid:27) ∪ (cid:26) i b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − · t . • If s is even and s >
0, then (cid:26) · i +1 α (mod 2 r ) | ≤ i < n − (cid:27) ∩ (cid:26) · i b t − α (mod 2 r ) | s ≤ i ≤ n − (cid:27) = ∅ . We can choose T d = [ a ∈ Λ d (cid:18) t − [ j =0 (cid:26) i +1 b j α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:18) t − [ j =0 (cid:26) i b j +1 α (mod 2 r ) | ≤ i < n − (cid:27) (cid:19) ∪ (cid:26) i b t − α (mod 2 r ) | s ≤ i ≤ n − (cid:27) ! and M ′ (2 d ) ≥ | T d | = | Λ d | · ( n − t − n − s +12 . (cid:4) M (2 p ) with 2
6∈ h i p p n t s | Λ d | M (2 p ) T d Thm.11 5 2 1 1 ≥ { , , } ≥ { , , , } ≥ { , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , , , , , , , , , , , , } { , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , }
673 12 3 10 2 ≥ { , , , , , , , , , , , , , , , , , , , , , , , } ≥ { , , , , , , , , , , , , , ,
45, 47, 49, 59, 67, 73, 77, 79, 85, 91, 95, 97, 101,103 , , , , , , , , , } r < M (2 r ) = P d | r M ′ (2 d ), the lower bounds on M ′ (2 d ) can imply a lower bound for M (2 r ).Numerical result indicate that all the lower bounds on M (2 r ) deduced from Theorems 3-8 are tight for r ≤
53 and gcd( r,
6) = 1 . Conjecture : The lower bound of M (2 r ) deduced from Theorems 3-8 are tight for all gcd( r,
6) = 1 . VI Summary
In this paper, we are mainly consider the constructions of a maximal size B [4](2 k r ) set with gcd( r,
6) = 1 . It can be applied to error correction for single asymmetric error of limited magnitude since all the syndromesare distinct. For k ≥
3, we first reduce the construction of a maximal size B [4](2 k r ) set to a maximalsize B [4](2 k − r ) set which implies that we only need to determine a maximal size B [4](4 r ), B [4](2 r )and B [4]( r ) set. The construction of a maximal size B [4](4 r ) set has been completely solved in Theorem2. Furthermore, we discuss maximal size B [4](2 r ) set and given the calculation formula or lower boundof M ′ (2 r ). On the other hand, for q ≤
106 we can determine all maximal size B [4]( q ) sets by computersearch. In all these examples, M ′ (2 r ) is equal to the lower bound in our result. It is reasonable toconjecture that all the lower bounds are tight. We invite the readers to attack these open problems.3 References [1] R. Ahlswede, H. Aydinian and L.H. Khachatrian, “Unidirectional errors control codes and relatedcombinatorial problems,” in
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