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Yuan-Hsun Lo ( 羅 元 勳 ). Optimal Conflict-avoiding Codes of Odd Length Weight Three. Department of Applied Mathematics National Chiao Tung University, Taiwan. A joint work with Kenneth Shum and Hung-Ling Fu. ( 1 0 0 1 0). ( 1 1 0 0 0). Definition. Conflict-avoiding code CAC( n , k )
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Department of Applied Mathematics
National Chiao Tung University, Taiwan
A joint work with Kenneth Shum and Hung-Ling Fu
(11 0 0 0)
DefinitionConflict-avoiding codeCAC(n,k)
Multiple-access collision channelwithout feedback
Guarantee:every active user can transmit at least one packet successfully in a frame of n slots.
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)
M = 4,n = 17, k = 3
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0)
Silence Symbol
M = 4,n = 17, k = 3
Survived Packet
Collided Packet
Senders
Receivers
A
A’
B
B’
C
C’
D
D’
Time Slots
a = (1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0)
CAC(17,3)
b = (0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0)
c = (0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0)
d = (0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0)
Given n and k, maximize M.
Theorem(Levenshtein and Tonchev, 2005)
Mishima et al., 2009 →
Fu, Lin and Mishima, 2010 →
Optimal CAC of weight 3Theorem(Jimbo et al., 2007)
Let n = 4t. Then
±2
±3
Set RepresentationExample (n = 13, k = 3)
x = (11 0 1 0 0 0 0 0 0 0 0 0 )
0 1 2 3 4 5 6 7 8 9 10 11 12
Δ(x) = {±1, ±2, ±3} = {1, 2, 3, 10, 11, 12}
x = {0, 1, 3}
Packing {1, 2, …, n/2}to obtain as many codewords
as possible (optimal CAC).
|Δ(x)| is as small as possible
A codeword of form {0, i ,2i} is said to be
equi-difference.
Example (n = 15, k =3)
equi-difference codewords
x = {0, 5, 10}
y = {0, 4, 8 }
z = {0, 7, 9 }
→Δ(x) = {5}
→ Δ(y) = {4, 7}
→ Δ(z) = {2, 6, 7}
i
0
i
2i
i
i
j
0
i
2i
j
Characterization of ΔLet x be a codeword of a CAC (n, 3).
j
k
i
j
0
i
i+j
0
i
i+j
k
Characterization of ΔLet x be a codeword of a CAC (n, 3).
H(n): a hypergraph (V, E)
(the set of differences arising from codewords)
An optimal CACcorresponds to
a maximum hypergraph matching.
G(n): a graph obtained from H(n) by
dropping all hyperedges with size 3
In G(n), i ~ j iff i ≣ ±2j (mod n).
Each edge of G(n) corresponds to
an equi-difference codeword.
1
2
4
3
5
G(17) :
1
2
4
8
3
6
5
7
1
2
4
8
5
10
G(21) :
3
6
9
7
Graphical Characterizationi ~ j iff i ≣ ±2j (mod n)
Δ = {4, 8} → {0, 4, 8} → 100010001000000000000
Δ ={5, 10} →{0, 5, 10}→ 100001000010000000000
1
2
4
8
5
10
G(21) :
3
6
9
7
Graphical Characterization3
Δ = {7} → {0, 7, 14} → 100000010000001000000
M(21,3) = 5
Δ = {6, 9} →{0, 6, 12}→100000100000100000000
{0,4,8}
{0,15,30}
15
8
4
1
{0,7,14}
{0,2,5}
2
14
3
10
{0,10,20}
5
7
6
13
11
12
{0,9,18}
9
{0,6,12}
Look for a hyperedgewhich intersects three distinct odd cycles
M(31,3) = 7
9a
G(34) :
27
9
18
36
3b
30
21
39
3
6
12
24
33
15
16
17
13
29
35
1
4
c
32
34
26
23
11
2
8
38
31
28
7
22
40
10
20
5
19
25
14
37
There is no hyperedges lying across distinct odd cycles.
M(81,3) = 19
9a
G(34) :
27
9
18
36
3b
30
21
39
3
6
12
24
33
15
16
17
13
29
35
1
4
c
32
34
26
23
11
2
8
38
31
28
7
22
40
10
20
5
19
25
14
37
There is no hyperedges lying across distinct odd cycles.
M(81,3) = 19
9a
G(34) :
27
9
18
36
3b
30
21
39
3
6
12
24
33
15
16
17
13
29
35
1
4
c
32
34
26
23
11
2
8
38
31
28
7
22
40
10
20
5
19
25
14
37
There is no hyperedges lying across distinct odd cycles.
Conjecture.
There are O(p)/ 3 mutually disjoint phyeredges
lying across distinct odd cycles if O(p) ≥ 3.