Báo cáo toán học: "On Mixed Codes with Covering Radius 1 and Minimum Distance 2" pot

... combinatorics 14 (2007), #R 51 10 Table II. Bounds on K(r, s, s; 2) for r ≤ 16 and 9 ≤ s ≤ 16 . s = 9 s = 10 s = 11 s = 12 s = 13 s = 14 s = 15 s = 16 r = 1 9 10 11 12 13 14 15 16 r = 2 16 B 18 B 20B 22B ... 71F 72F a85E a96 − 98E a103 − 11 3E 11 0 − 12 6G r = 13 60 − 65C 66 − 72C 73 − 81F 79 − 84C 85F a98E 10 5 − 11 3E a 117 − 12 8E r = 14 63 − 67C 70 − 76C...

Ngày tải lên: 07/08/2014, 15:22

... 0 1 LF 0 2 1 X 1 0 X 3 2 3 0 1 3 1 2 0 LG 0 2 1 X 1 3 X 0 2 1 0 3 3 0 2 1 LH 0 2 1 X 1 3 X 0 2 0 3 1 3 1 0 2 LI 0 2 1 X 1 3 X 2 2 0 3 1 3 1 2 0 LJ 0 2 1 X 1 3 X 2 2 1 0 3 3 0 2 1 LK 0 2 3 X 1 ... combinatorics 15 (2008), #R68 10 LA 0 1 2 X 1 0 X 3 2 3 1 0 3 2 0 1 LB 0 1 2 X 1 0 X 3 2 3 0 1 3 2 1 0 LC 0 1 2 X 1 2 X 3 2 3 0 1...

Ngày tải lên: 07/08/2014, 15:23

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Ngày tải lên: 05/08/2014, 09:46

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Ngày tải lên: 05/08/2014, 15:20

... adja- cent. Suppose 1 ≤ i<j≤ 2δ − 1withi and j odd and v i adjacent to v j .Then the vertices x, v i 1 ,v i ,v j ,v j 1 ,v j−2 , ···,v i +1 ,y,v j +1 ,v j+2 , ···,v 2δ ,v 1 ,v 2 , ···,v i−2 form ... answered by Erd¨os and Gallai [4] and again by Andrasfai [1] . Our formulation of their answer is Theorem 1 Let G be a connected graph with minimum degree δ and at least...

Ngày tải lên: 07/08/2014, 06:20

... =1andn = 2 are produced by hand. 1 10 2 10 01 3 10 0 010 1 4 10 00 010 10 01 5 10 000 010 1 010 010 0 01 6 10 0000 010 1 010 010 100 010 010 00 01 7 10 00000 010 1 010 10 010 100 010 1000 010 010 010 0 010 000 01 5 Remarks on generalizations It ... 5 (19 98), #R35 4 00000 000 01 00 010 0 010 1 010 010 1 010 10 000 10 0 01 10 010 10 10 010 1 01 0 010 0 010 00 V 0 V V V V...

Ngày tải lên: 07/08/2014, 06:22

... spanned by M 1 into A 1 and B 1 ,soM 1 is a matching between A 1 and B 1 . Halve the vertices of A arbitrarily and denote one of the halves by A  1 . Denote by B  1 the set of vertices in B 1 which ... any 1 ≤ k ≤ l p k  j =1 a j ≥ a 0 − a k − 1= p − a k − 1. Proof: k  j =1 a j = l p 1  j=0 a l p −j − l p −k 1  j=0 a l p −j = a 0 − δ 0 − a k + δ k ≥ a 0...

Ngày tải lên: 07/08/2014, 07:21

... ,W 8 } where W 1 = 000, 10 0, 10 1, 11 1, 11 0 W 5 = 10 0, 11 0, 11 1, 10 1, 10 0 W 2 = 0 01, 011 , 010 , 000, 0 01 W 6 = 10 1, 11 1, 11 0, 10 0, 10 1 W 3 = 010 , 000, 0 01, 011 , 010 W 7 = 11 0, 010 , 011 , 0 01, 000 W 4 = 011 , ... exact. n 2 3 4 5 6 7 8 910 111 213 1 415 1 617 1 819 20 21 (n) 2 ∗ 2 ∗ 2 ∗ 4 ∗ 4 ∗ 5 ∗ 568889 911 111 212 1 415 16 We now describe a hitherto...

Ngày tải lên: 07/08/2014, 07:21

... 1) T , c 1 = (1, 1, 0) T ,c 2 = (1, 0, 1) T ,c 3 =(0, 1, 1) T , d 1 =( 1, 1, 1) T ,d 2 = (1, 1, 1) T ,d 3 = (1, 1, 1) T , e 1 = (1, 0, 0) T ,e 2 =(0, 1, 0) T ,e 3 =(0, 0, 1) T . Then orthogonality-preserving ... {a, b 1 ,b 2 ,b 3 ,c 1 ,c 2 ,c 3 ,d 1 ,d 2 ,d 3 ,e 1 ,e 2 ,e 3 }, be a system of vectors from R 3 , where a = (1, 1, 1) T , b 1 =( 1, 1, 0) T ,b...

Ngày tải lên: 07/08/2014, 08:20

... edges from (0, 0) to ( 1, 1) , ( 1, 0) and ( 1, 1) , and the three edges from (1, 0) to (2, 1) , (2, 0) and (2, 1) . Then I r (K  ; p) = I r (K  ; q). 3.2 Codes with r = 1, 1 ≤ t ≤ 5 The following ... is 1 2r 1 . Proof. To obtain the lower bound, take p = (x, y) and q = (x + 1, y), and let K  be the graph obtained by deleting the three edges from p to p + ( 1, 1) , p...

Ngày tải lên: 07/08/2014, 15:22