On conflict-avoiding codes ofweight 3 and odd length

Kenneth ShumOct 2011

Definitions

• Optical orthogonal code OOC(n,w,a,c)– Length n– Weight w– Hamming auto-correlation a

– Hamming cross-correlation c

• Conflict-avoiding code (Tsybakov and Rubinov (02))

CAC(n,w) = OOC(n,w,,1)no requirement on Hamming auto-correlation.

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Application to multiple-access collision channel without feedback

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Hello ! message

Scheduler

Packet 1 Packet 2 Packet 3 Packet 4

Channel coding(Reed-Solomon)

Any two packets can recoverthe original message

x Receiverxx

Other user

Other user

Hello !

colli

sionco

llisi

on

colli

sion

Parameters• Number of codewords = T

– Total number of potential users– Each user is statically assigned a unique codeword

• Sequence period = n– maximal delay experience by an active user

• Hamming weight = w– Maximal number of simultaneously active users

• Objective: Given n and w, maximize T

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Outline

• Review of the literature on CAC• Formulation using graph theory• Some new optimal CAC of weight 3 and odd

length

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Maximal number of codewords

• Let M(n,w) be largest number of codewords in a CAC of length n and weight w.

• Levenshtein (07)

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for n = 4t + 2

for odd n, n

CAC of even length and weight 3• For n = 4t,

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• Jimbo, Mishima, Janiszewski, Teymorian and Tonchev (07)• Mishima, Fu and Uruno (09)• Fu, Lin and Mishima (10)

CAC of weight > 3

• Some constructions of optimal CAC of weight 4 and 5– Momihara, Müller, Satoh and Jimbo (07)

• CAC in general– S and Wong

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For w 3,

Outline

• Review of the literature on CAC• Formulation using graph theory

– hypergraph matching

• Some new optimal CAC of weight 3 and odd length

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Terminology

• A binary sequence can be represented by a characteristic set.– Sequence: 0 1 1 0 0 1 0 0 {1,2,5} Indices 0 1 2 3 4 5 6 7

• The set of differences contains the separations between the ones in a sequence– (A) = {x – y mod n: x, y A, x y}– For example ({1,2,5}) = {1,3,4,5,7}

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0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 0

Equivalent definition of CAC

• The characteristic sets of CAC is a collection of subsets of Zn, say A1, A2, …, AM , such that– Each of them has size w.– (Ai) (Ak) = for i k.

• Example: n=15,– 111000000000000 {0,1,2}, ({0,1,2}) = {1,2,13,14}– 100100100000000 {0,3,6}, ({0,3,6}) = {3,6,9,12}– 100010001000000 {0,4,8}, ({0,4,8}) = {4,7,8,11}– 100001000010000 {0,5,10}, ({0,5,10}) = {5,10}

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distinct

(A) = {x – y mod n: x, y A, x y}

Equi-difference codewords• By cyclically shifting the sequence, we can

assume without loss of generality that 0 belongs to the characteristic set.

• For sequence with Hamming weight 3, we can write the characteristic set as {0,a,b} WLOG.– ({0,a,b}) = {a, b, (a – b)}

• In particular, a sequence with characteristic set {0,a,2a} is said to be equi-difference.– The integer a is called the generator of this codeword– ({0,a,2a}) = {a, 2a}

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100010000010000000

a b-a

b

100010001000000000

a a

2a

Formulation using (hyper)graph• Observation: x (A) implies n – x (A) we can identify x and –x mod n.• Assume n odd. Let m = (n – 1)/2.• Undirected graph with vertex set {1,2,…,m}.• Construct hyperedges ({0,a,b}) {1,2,…,m}

– for a and b running over all distinct elements in {1,2,…,n}

• Objective: look for a maximal collection of non-intersecting hyperedges.– A matching problem.

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A greedy method for equi-difference codewords

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1

35

n=15, m=7

2 4

7 6

111000000000000 {0,1,2} ({0,1,2}) = {1, 2}.

101010000000000 {0,2,4} ({0,2,4}) = {2, 4}. (conflict with {0,1,2})

100100100000000 {0,3,6} ({0,3,6}) = {3, 6}.

100010001000000 {0,4,8} ({0,4,8}) = {4, 7}.

100001000010000 {0,5,10} ({0,5,10}) = {5}.

100000100000100 {0,6,12} ({0,6,12}) = {3, 6} (conflict with {0,3,6})

100000010000001 {0,7,14} ({0,7,14}) = {1, 4} (conflict with {0,1,2}) and {0,4,8}

A perfect matchingEach vertex is covered by a red edge,and all red edges are disjoint.

Another example: n = 31, m = 15equi-difference codewords only

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8 15

41

12

214 3 5 10

13 117 6

9

{0,1,2}

{0,4,8}

{0,5,10}

{0,9,18}{0,3,6}

{0,7,14}

Theorem (Levenshtein (07))The graph with edgesfrom the equi-difference codewordsare decomposed into cycles.

Find a maximalmatching

Six equi-differencecodewords

The optimal solution with hyperedges

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8 15

41

12

214 3 5 10

13 117 6

9

{0,15,30}{0,4,8}

{0,10,20}

{0,9,18}{0,6,12}

{0,7,14}

Seven codewords

{0,6,13}

An example of hyperedge

{0,2,6}

Another example of hyperedge

{0,2,5}

Look for a hyperedgewhich intersects three distinct odd cycles

M(31,3) = 7• n=31

– {0,4,8} 1000100010000000000000000000000– {0,6,12} 1000001000001000000000000000000– {0,7,14} 1000000100000010000000000000000– {0,9,18} 1000000001000000001000000000000– {0,10,20}1000000000100000000010000000000– {0,15,30}1000000000000001000000000000001– {0,2,5} 1010010000000000000000000000000

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The cycle graph for n=99.

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1

2

48

163235

29

41

17

34 49

253731

7

14

28

431326

47

5

10

20

40 46

233819

11

2244

48

24

9

12

6

3 18

36

2745

42

15

30

392133

{0,1,11}

{0,6,15}

M(99,3) = 24

• Two non-equi-difference codewords: {0,1,11}, {0,6,15}.

• Twenty two equi-difference codewords generated by 2,7,8,12,13,17,18,19,20,21,22,23,25,27,28,29,30,31,32,33,47,48.

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A sufficient condition for being an optimal CAC

• Theorem 1: – Let n be an odd integer, and let Nodd(n) be the

number of odd-cycle in the graph.

– If we can find Nodd(n) / 3 mutually disjoint hyperedge of size 3 lying across 3 Nodd(n) / 3 cycles of odd length, then equality holds.

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The upper bound in Thm 1 is not tight

• Theorem 2: for e 1,

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For n= powers of 3 or 7, M(n,3) is strictly less than the upper bound in Theorem 1.

(because in these cases, non-equi-difference codewords are not useful in constructing optimal CAC.)

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n 11 13 15 17 19 21 23 25 27 29

M(n,3) 2 3 4 4 4 5 5 6 6 7Thm 2

n 31 33 35 37 39 41 43 45 47 49

M(n,3) 7* 8* 8 9 10 10 10* 11 11 11new new new Thm 2

n 51 53 55 57 59 61 63 65 67 69

M(n,3) 13 13 13 14* 14 15 15 16 16 17new

n 71 73 75 77 79 81 83 85 87 89

M(n,3) 17 17* 19 18 19 19 20 21 22 21*new Thm 2 new

n 91 93 95 97 99 101 103 105 107 109

M(n,3) 22 23* 23 24 24* 25 25 26 26 27new new

* non-equiv-difference codewords are required to construct optimal CAC

Conclusion

• Numerical results:– For all odd n <500, except n=81, 189, 243, 343,

405, 441,

– M(81,3) = 19, M(189,3) = 47– M(243,3) = 60, M(343,3) = 85– M(405,3) = 101, M(441,3) = 110

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81=34, 189 = 33 7 243=35, 343=73,405 = 34 5, 441 = 32 72

.

References• Tsybakov and Rubinov, Some constructions of conflict-avoiding codes,

Problems of Inf. Trans., 2002.• V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems,

Probems of Inf. Trans., 2007.• M. Jimbo et al., On conflict-avoiding codes of length n=4m for three active

users, IEEE Trans. Inf. Theory, 2007.• M. Mishima, H.-L. Fu and S. Uruno, Optimal conflict-avoiding codes of

length n0(mod16) and weight 3, Des. Codes Cryptogr., 2009.• H.-L. Fu, Y.-H. Lin and M. Mishima, Optimal conflict-avoiding codes of even

length and weight 3, IEEE Trans. Inf. Theory, 2010.• K. W. Shum and W. S. Wong, A tight asymptotic bound on the size of

constant-weight conflict-avoiding codes, Des. Codes Cryptogr., 2010.

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n 111 113 115 117 119 121 123 125 127 129

M(n,3) 28 28 28 29 29 29 31 31 30* 31*Thm 1 Thm 1

n 131 133 135 137 139 141 143 145 147 149

M(n,3) 32 32 33 34 34 35 35 36 36 37

n 151 153 155 157 159 161 163 165 167 169

M(n,3) 36* 38 38* 39 40 39* 40 41* 41 42Thm 1 Thm 1 Thm 1 Thm 1

n 171 173 175 177 179 181 183 185 187 189

M(n,3) 41* 43 43 44* 44 45 46 46 46 47Thm 1 Thm 1 Similar

to Thm 2

n 191 193 195 197 199 201 203 205 207 209

M(n,3) 47 48 49 49 49 50* 50 51 51 51*Thm 1 Thm 1

* non-equiv-difference codewords are required to construct optimal CAC

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n 211 213 215 217 219 221 223 225 227 229

M(n,3) 52 53 53* 52* 54* 55 55* 56 56 57Thm 1 Thm 1 Thm 1 Thm 1

n 231 233 235 237 239 241 243 245 247 249

M(n,3) 57* 57* 58 59 59 60 60 60 61 62*Thm 1 Thm 1 Thm 2 Thm 1

n 251 253 255 257 259 261 263 265 267 269

M(n,3) 60* 62 64 64 64 65 65 66 65* 67Thm 1 Thm 1

n 271 273 275 277 279 281 283 285 287 289

M(n,3) 67 68 68 69 69* 69* 70* 71* 71 72Thm 1 Thm 1 Thm 1 Thm 1

n 291 293 295 297 299 301 303 305 307 309

M(n,3) 73 73 73 73* 74 74* 76 76 76* 77Thm 1 Thm 1 Thm 1

* non-equiv-difference codewords are required to construct optimal CAC

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n 311 313 315 317 319 321 323 325 327 329

M(n,3) 77 78 78 79 79 80* 80 81 82 81*Thm 1 Thm 1

n 331 333 335 337 339 341 343 345 347 349

M(n,3) 80* 83 83 82* 85 84* 85 86 86 87Thm 1 Thm 1 Thm 1 Thm 2

n 351 353 355 357 359 361 363 365 367 369

M(n,3) 87 88 88 89 89 89 88* 89* 91 92Thm 1 Thm 1

n 371 373 375 377 379 381 383 385 387 389

M(n,3) 92 93 94 94 94 94* 95 95 94* 97Thm 1 Thm 1

n 391 393 395 397 399 401 403 405 407 409

M(n,3) 97 98* 98 99 99* 100 100* 101 101 102Thm 1 Thm 1 Thm 1 Similar to

Thm 2

* non-equiv-difference codewords are required to construct optimal CAC

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n 411 413 415 417 419 421 423 425 427 429

M(n,3) 103 102 103 104* 104 105 105 106 106 107*Thm 1 Thm 1

n 431 433 435 437 439 441 443 445 447 449

M(n,3) 106* 108 109 108 109* 110 110 110* 112 112Thm 1 Thm 1 Similar

to Thm 2Thm 1

n 451 453 455 457 459 461 463 465 467 469

M(n,3) 112 112* 113 114 114 115 115 116* 116 116Thm 1 Thm 1

n 471 473 475 477 479 481 483 485 487 489

M(n,3) 118 116* 118 119 119 120 120* 121 121 122*Thm 1 Thm 1 Thm 1

n 491 493 495 497 499 501 503 505 507 509

M(n,3) 122 123 123* 123* 124* 125 125 126 127 127Thm 1 Thm 1 Thm 1

* non-equiv-difference codewords are required to construct optimal CAC

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n 511 513 515 517 519 521 523 525 527 529

M(n,3) 103 102 103 104* 104 105 105 106 106 107*

n 531 533 535 537 539 541 543 545 547 549

M(n,3) 106* 108 109 108 109* 110 110 110* 112 112

n 451 453 455 457 459 461 463 465 467 469

M(n,3) 112 112* 113 114 114 115 115 116* 116 116

n 471 473 475 477 479 481 483 485 487 489

M(n,3) 118 116* 118 119 119 120 120* 121 121 122*

n 491 493 495 497 499 501 503 505 507 509

M(n,3) 122 123 123* 123* 124* 125 125 126 127 127

* non-equiv-difference codewords