John Rickert Tom Langley Ralph Grimaldi April 28,...
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The Jacobsthal Subcube of the Hypercube John Rickert Tom Langley Ralph Grimaldi April 28, 2012
Transcript of John Rickert Tom Langley Ralph Grimaldi April 28,...
The Jacobsthal Subcube of the Hypercube
John Rickert
Tom Langley
Ralph Grimaldi
April 28, 2012
Cubes using the alphabet Σ = {0, 1} with languageQ = {0, 1}
Q2
00
01 10
11
Q3
000
001 010 100
011 101 110
111
Q4
0000
0001 0010 0100 1000
0011 0101 0110 1001 1010 1100
0111 1011 1101 1110
1111
Vertices: 2n Edges: n2n−1
Restrict to the language A = {0, 01, 11}
A0 = {∅}A1 = {0}A2 = {00, 01, 11}A3 = {000, 010, 110, 001, 011}A4 = {0000, 0100, 1100, 0010, 0110,
0001, 0101, 1101, 0011, 0111, 1111}A4 = A30 ∪ A201 ∪ A211|An| = |An−1|+ 2|An−2|
Jacobsthal numbers
J0 = 1 J1 = 1Jn = Jn−1 + 2Jn−2
1, 1, 3, 5, 11, 21, 43, 85, 171, 341, . . .
|A0| = 1 |A1| = 1|An| = |An−1|+ 2|An−2|
Jn = 23 · 2
n + 13 · (−1)
n
Jacobsthal subcube Hn
A2 = {00, 01, 11}
00
01 10
11
00
01 10
11
H3 Q3
000
001 010 100
011 101 110
111
000
001 010
011 110
A3 = {000, 010, 110, 001, 011}
H4 Q4
0000
0001 0010 0100 1000
0011 0101 0110 1001 1010 1100
0111 1011 1101 1110
1111
0000
0001 0010 0100
0011 0101 0110 1100
0111 1101
1111
Subcube recursion
An = An−10 ∪ An−201 ∪ An−211
000
110010
001011
0000
0010
0100
0110
11000001
01011101
00110111
1111
00000
00100
01000
01100
1100000010
0101011010
0011001110
11110
00001
00101
01001
01101
11001
00011
00111
01011
01111
11011
A string starts with an even number of 1s
00111 0 H5
11 0111 11
00000
00100
01000
01100
1100000010
0101011010
001100111011110
00001
00101
01001
01101
11001
00011
00111
01011
01111
11011
A = {0, 01, 11}
Subcube recursion with cubes
An = 0Qn−1 ∪ 11An−2
00000
00001 00010 00100 01000
00011 00101 00110 01001 01010 01100
00111 01011 01101 01110
0111111011
11110 11001
1101011000
Counting the edges
0 Q7 7 · 261 1 0 Q5 5 · 241 1 1 1 0 Q3 3 · 221 1 1 1 1 1 0 Q1 1 · 20
Count of edges between rows 2 and 3
0 Q7 7 · 261 1 0 1 0 Q5 5 · 241 1 1 1 0 23 Q3 3 · 221 1 1 1 1 1 0 Q1 1 · 20
Count of edges between rows
0 Q7 7 · 261 1 0 1 · 25 Q5 5 · 241 1 1 1 0 2 · 23 Q3 3 · 221 1 1 1 1 1 0 3 · 21 Q1 1 · 20
Total number of edges
0 Q7 7 · 26 7 · 261 1 0 1 · 25 Q5 5 · 24 7 · 241 1 1 1 0 2 · 23 Q3 3 · 22 7 · 221 1 1 1 1 1 0 3 · 21 Q1 1 · 20 7 · 201 1 1 1 1 1 1 1 4∑b(n+1)/2c
k=1 (n − 1)2n−2k =(n−13
)2n[1−
(14
)b n+12c].
en =(n4
)[1 + (−1)n] +
(n−13
)2n[1−
(14
)b n+12c].
Direct count of edges in H5
00000
00001 00010 00100 01000
00011 00101 00110 01001 01010 01100
00111 01011 01101 01110
0111111011
11110 11001
1101011000
from \ to (start 0) (start 11) (start 1111)(start 0) 4 · 23 1 · 22 1 · 20(start 11) 2 · 21 1 · 20
(start 1111) 0Total to 4 · 23 4 · 21 4 · 2−1
32 8 242 edges in H5
Does Hn have a Hamiltonian path?
000000
001000
010000
011000
110000
000100
010100
110100
001100011100
111100
000010
001010
010010
011010
110010
000110
001110
010110
011110
110110
000001
001001
010001
011001
110001
000101
010101
110101
001101011101
111101
000011
001011
010011
011011
110011
000111
010111
110111
001111011111
111111
H6
Take the path through 0Qn−1 then move to 11Hn−2
A2 = {00, 01, 11}
00
01 10
11
00
01 10
11
H3
000
001 010 100
011 101 110
111
000
001 010
011 110
A3 = {000, 010, 110, 001, 011}
H4
0000
0001 0010 0100 1000
0011 0101 0110 1001 1010 1100
0111 1011 1101 1110
1111
0000
0001 0010 0100
0011 0101 0110 1100
0111 1101
1111
Summary
000
110010
001011
Verticies: |Hn| = Jn = 13(−1)
n + 23(2)
n
Edges: en =(n4
)[1 + (−1)n] +
(n−13
)2n[1−
(14
)b n+12c]
Hamiltonian Paths!
000000
001000
010000
011000
110000
000100
010100
110100
001100011100
111100
000010
001010
010010
011010
110010
000110
001110
010110
011110
110110
000001
001001
010001
011001
110001
000101
010101
110101
001101011101
111101
000011
001011
010011
011011
110011
000111
010111
110111
001111011111
111111
