Cycle Decompositions of de Bruijn Graphs for Robot

Introduction de Bruijn sequences Results Summary

Cycle Decompositions of de Bruijn Graphsfor Robot Identification and Tracking

Tony Grubman

Joint Supervisors: Y. Ahmet Sekercioglu David R. Wood

Department of Electrical and Computer Systems Engineering

School of Mathematical Sciences

September 23, 2013

Tony Grubman Cycles in de Bruijn graphs 1 / 24


Outline

1 IntroductionMotivationDemonstrationde Bruijn graphs

2 de Bruijn sequencesExistenceConstruction: Linear feedback shift registers

3 ResultsSplitting linear feedback shift register sequencesProduct colouringCombining necklaces



Outline






Outline





Introduction de Bruijn sequences Results Summary Motivation Demonstration de Bruijn graphs

eBugs — colourful robots

Wireless robot network research platform

I ‘Swarm’ of up to 20 robots

Mobile

I Precision controlled stepper motors

16 multicolour LEDs (red, green and blue)

I Can display a sequence of colours aroundits perimeter

Expandable

I Vision capabilities can be provided with acamera

Problem

Can a sequence of colours be assigned to the LEDs of each eBug such thatany observer (camera) can identify the eBug and its orientation?




Wireless robot network research platformI ‘Swarm’ of up to 20 robots

Mobile




Expandable


Problem






Mobile




Expandable


Problem






MobileI Precision controlled stepper motors



Expandable


Problem









Expandable


Problem







16 multicolour LEDs (red, green and blue)I Can display a sequence of colours around

its perimeter

Expandable


Problem








its perimeter

Expandable


Problem








its perimeter

ExpandableI Vision capabilities can be provided with a

camera

Problem








its perimeter

ExpandableI Vision capabilities can be provided with a

camera

Problem




Example (4 eBugs, 8 LEDs, 2 colours)



Preliminary bounds

Definition (eBug number)

Suppose every eBug has k LEDs, each of which can be illuminated in oneof q colours, and that a camera can reliably detect ` adjacent LEDs. Anassignment of colours to the LEDs of all eBugs is valid if the camera candistinguish each eBug in each of the k orientations.

The eBug number E(q, k, `) is the maximum number of eBugs for whichthere exists a valid assignment of colours.

Upper bound: E(q, k, `) ≤⌊q`

k

⌋

I Each of the q` possible sequences cannot appear more than onceI Each eBug will account for k of the sequences

Initial lower bound: Lovasz local lemma gives E ≥ q`

8`kComputation shows that upper bound is achieved in small casesMain problem — when is the upper bound achievable?



Preliminary bounds


Suppose every eBug has k LEDs, each of which can be illuminated in oneof q colours, and that a camera can reliably detect ` adjacent LEDs. Anassignment of colours to the LEDs of all eBugs is valid if the camera candistinguish each eBug in each of the k orientations.The eBug number E(q, k, `) is the maximum number of eBugs for whichthere exists a valid assignment of colours.


k

⌋






Preliminary bounds




k

⌋






Preliminary bounds




k

⌋I Each of the q` possible sequences cannot appear more than once

I Each eBug will account for k of the sequences





Preliminary bounds




k

⌋I Each of the q` possible sequences cannot appear more than onceI Each eBug will account for k of the sequences





Preliminary bounds




k



8`k

Computation shows that upper bound is achieved in small casesMain problem — when is the upper bound achievable?



Preliminary bounds




k



8`kComputation shows that upper bound is achieved in small cases

Main problem — when is the upper bound achievable?



Preliminary bounds




k






Demonstration



What is a de Bruijn graph?

Definition

The `-th order q-ary de Bruijn graph dB(q, `) is the digraph (V,E), whereV = Z`

q and E = {(a0a1 . . . a`−1, a1a2 . . . a`) | ai ∈ Zq}.

Vertices are words of length ` over an alphabet of size q

Edge from u to v if shifting u left and appending any letter gives v

Example (dB(2, 3))

010

100

000

001

111

110

101

011




Definition





Example (dB(2, 3))

010

100

000

001

111

110

101

011




Definition





Example (dB(2, 3))

010

100

000

001

111

110

101

011




Definition





Example (dB(2, 3))

010

100

000

001

111

110

101

011



de Bruijn graphs and eBug numbers

Example (dB(2, 3))

010

100

000

001

111

110

101

011

Every vertex is a sequence of ` colours

I This represents the camera’s view

Rotating the eBug corresponds to following an edge

A cycle of length k represents the whole eBug

E(q, k, `) is the maximum number of disjoint k-cyclesin dB(q, `)




Example (dB(2, 3))

010

100

000

001

111

110

101

011









Example (dB(2, 3))

010

100

000

001

111

110

101

011









Example (dB(2, 3))

010

100

000

001

111

110

101

011

Every vertex is a sequence of ` coloursI This represents the camera’s view







Example (dB(2, 3))

010

100

000

001

111

110

101

011








Example (dB(2, 3))

010

100

000

001

111

110

101

011








Example (dB(2, 3))

010

100

000

001

111

110

101

011








Example (dB(2, 3))

010

100

000

001

111

110

101

011







Construction via line digraphs

Alternate construction

dB(q, 1) =−→Kq

Example (q = 2)





dB(q, 1) =−→Kq

Example (q = 2)

10





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

10





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

10

10

01

00 11





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

10

10

01

00 11





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

11

10

00

01





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

11

10

00

01

000 111

001 011

110100

010 101





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

11

10

00

01

000 111

001 011

110100

010 101





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

010

100

000

001

111

110

101

011





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

010

100

000

001

111

110

101

011

0000 1111

1100

0011

0101

1010

01101001

01001000

0001 0010

11101101

1011 0111





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 2)

1001

0100

0010

1101

0110

1011

0000 1111

1100

0011

1010

0101

1000

0001 0111

1110





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 3)

1

0

2





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 3)

1

0

2

1002

21

0120

12

11

00

22





dB(q, 1) =−→Kq; dB(q, `+ 1) = L(dB(q, `))

Example (q = 3)

1002

21

11

00

22

0120

12


Introduction de Bruijn sequences Results Summary Existence LFSRs

de Bruijn graphs are Hamiltonian

Theorem

If a digraph G is Eulerian, then the line digraph L(G) is Hamiltonian.

An Eulerian circuit in G is equivalent to a Hamiltonian cycle in L(G)

Corollary

Every de Bruijn graph has a Hamiltonian cycle.

Proof

Can assume ` ≥ 2 as dB(q, 1) =−→Kq is Hamiltonian

Every vertex in dB(q, `− 1) has in-degree q and out-degree q

I dB(q, `− 1) is EulerianI dB(q, `) is Hamiltonian




Theorem



Corollary


Proof







Theorem



Corollary


Proof







Theorem



Corollary


Proof







Theorem



Corollary


Proof







Theorem



Corollary


Proof







Theorem



Corollary


Proof


Every vertex in dB(q, `− 1) has in-degree q and out-degree qI dB(q, `− 1) is Eulerian

I dB(q, `) is Hamiltonian




Theorem



Corollary


Proof


Every vertex in dB(q, `− 1) has in-degree q and out-degree qI dB(q, `− 1) is EulerianI dB(q, `) is Hamiltonian



Definition

A q-ary de Bruijn sequence of order ` is a Hamiltonian cycle in dB(q, `).

Theorem

The number of Eulerian circuits in an Eulerian digraph G is

τ(G)∏

v∈V (G)

(d+(v)− 1)!

d+(v) is the out-degree of vertex v

τ(G) is the number of spanning arborescences rooted at some vertex

I Does not depend on choice of root vertex

Corollary

There are exactly(q!)q

`−1

q`q-ary de Bruijn sequences of order `.



Definition


Theorem


τ(G)∏

v∈V (G)

(d+(v)− 1)!




Corollary


`−1




Definition


Theorem


τ(G)∏

v∈V (G)

(d+(v)− 1)!




Corollary


`−1




Definition


Theorem


τ(G)∏

v∈V (G)

(d+(v)− 1)!




Corollary


`−1




Definition


Theorem


τ(G)∏

v∈V (G)

(d+(v)− 1)!


τ(G) is the number of spanning arborescences rooted at some vertexI Does not depend on choice of root vertex

Corollary


`−1




Definition


Theorem


τ(G)∏

v∈V (G)

(d+(v)− 1)!


τ(G) is the number of spanning arborescences rooted at some vertexI Does not depend on choice of root vertex

Corollary


`−1




Galois LFSRs

Let q be a prime power, and consider the Galois field GF(q)

Choose a degree ` primitive polynomial p(x) over GF(q)

I The quotient F = GF(q)[x]/〈p(x)〉 is generated by x

Repeatedly multiplying by x gives every non-zero element of F

Easily implemented as a digital logic circuit

Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1xx2x3x4x5x6



Galois LFSRs


Choose a degree ` primitive polynomial p(x) over GF(q)

I The quotient F = GF(q)[x]/〈p(x)〉 is generated by x



Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1xx2x3x4x5x6



Galois LFSRs


Choose a degree ` primitive polynomial p(x) over GF(q)I The quotient F = GF(q)[x]/〈p(x)〉 is generated by x



Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1000000

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0100000

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0010000

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0001000

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0000100

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0000010

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0000001

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

1110010

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0111001

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0111001

1xx2x3x4x5x6



Galois LFSRs





Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1)

0111001

1xx2x3x4x5x61110010



Fibonacci LFSRs

Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1) — Galois

1000000

1xx2x3x4x5x61110010

Different logic configuration

I Additive feedback is many-to-one, instead of one-to-manyI The shift direction is reversed

Same polynomial may be used

Also represents consecutive powers of x, but in a different basis

Produces an identical sequence in the last digit



Fibonacci LFSRs

Example (q = 2, p(x) = x7 + x5 + x2 + x+ 1) — Fibonacci

10000001x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


00000011x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


00000101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


00001011x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


00010101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


00101011x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


01010111x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010








Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010

Different logic configurationI Additive feedback is many-to-one, instead of one-to-many

I The shift direction is reversed






Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010

Different logic configurationI Additive feedback is many-to-one, instead of one-to-manyI The shift direction is reversed






Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010







Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010







Fibonacci LFSRs


10101101x6 + x4 + xx5 + x3x4 + x2x3 + xx2x

1110010







de Bruijn sequences from LFSRs

In the Fibonacci configuration, state transitions correspond to edgesin dB(q, `)

All non-zero states are traversed in a single cycle

I Gives a Hamiltonian cycle in dB(q, `) \ {00 . . . 0}

Can be extended to all of dB(q, `)

I Insert 00 . . . 0 before 00 . . . 1I Previous edge c0 . . . 0→ 00 . . . 1 becomes two edgesc0 . . . 0→ 00 . . . 0→ 00 . . . 1

Every primitive polynomial gives a different de Bruijn sequence

I There areϕ(q` − 1)

`q-ary de Bruijn sequences of order ` arising from

linear feedback shift registersI Most are “non-linear” feedback shift registers





All non-zero states are traversed in a single cycle

I Gives a Hamiltonian cycle in dB(q, `) \ {00 . . . 0}Can be extended to all of dB(q, `)










All non-zero states are traversed in a single cycleI Gives a Hamiltonian cycle in dB(q, `) \ {00 . . . 0}























Can be extended to all of dB(q, `)I Insert 00 . . . 0 before 00 . . . 1

I Previous edge c0 . . . 0→ 00 . . . 1 becomes two edgesc0 . . . 0→ 00 . . . 0→ 00 . . . 1










Can be extended to all of dB(q, `)I Insert 00 . . . 0 before 00 . . . 1I Previous edge c0 . . . 0→ 00 . . . 1 becomes two edgesc0 . . . 0→ 00 . . . 0→ 00 . . . 1
























linear feedback shift registers

I Most are “non-linear” feedback shift registers












Introduction de Bruijn sequences Results Summary LFSR splits Product colouring Combining necklaces

Length k subsequences in de Bruijn sequences

Consider a de Bruijn sequence in dB(q, `) constructed from a LFSR

I This corresponds to an Eulerian circuit in dB(q, `− 1)I State polynomial of LFSR represents an edge

Suppose we want to find a k-subcircuit in this Eulerian circuit

I After k iterations of the LFSR, we should be at the same vertexI Equivalent to changing the constant term in state polynomial

This can be expressed as an equation in the quotient field:

I xkf(x) = f(x) + c for some c ∈ GF(q) \ {0}I Unique solution for each c: f(x) = c

xk−1

We have found q − 1 k-cycles in dB(q, `)

I Requires (q − 1)k ≤ q` − 1 for cycles to be disjointI This is sufficient because cycles are evenly distributed

E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1




Consider a de Bruijn sequence in dB(q, `) constructed from a LFSRI This corresponds to an Eulerian circuit in dB(q, `− 1)

I State polynomial of LFSR represents an edge





xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1




Consider a de Bruijn sequence in dB(q, `) constructed from a LFSRI This corresponds to an Eulerian circuit in dB(q, `− 1)I State polynomial of LFSR represents an edge





xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1









xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1





Suppose we want to find a k-subcircuit in this Eulerian circuitI After k iterations of the LFSR, we should be at the same vertex

I Equivalent to changing the constant term in state polynomial



xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1





Suppose we want to find a k-subcircuit in this Eulerian circuitI After k iterations of the LFSR, we should be at the same vertexI Equivalent to changing the constant term in state polynomial



xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1








xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1






This can be expressed as an equation in the quotient field:I xkf(x) = f(x) + c for some c ∈ GF(q) \ {0}

I Unique solution for each c: f(x) = cxk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1






This can be expressed as an equation in the quotient field:I xkf(x) = f(x) + c for some c ∈ GF(q) \ {0}I Unique solution for each c: f(x) = c

xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1







xk−1



E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1







xk−1

We have found q − 1 k-cycles in dB(q, `)I Requires (q − 1)k ≤ q` − 1 for cycles to be disjoint

I This is sufficient because cycles are evenly distributed

E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1







xk−1

We have found q − 1 k-cycles in dB(q, `)I Requires (q − 1)k ≤ q` − 1 for cycles to be disjointI This is sufficient because cycles are evenly distributed

E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1







xk−1

We have found q − 1 k-cycles in dB(q, `)I Requires (q − 1)k ≤ q` − 1 for cycles to be disjointI This is sufficient because cycles are evenly distributed

E(q, k, `) ≥ q − 1 for k ≤ q` − 1

q − 1



Optimality for k = q`−1

Consider the case when k = q`−1

I q` − k states of the LFSR are covered by the q − 1 k-cycles

Take these cycles out of the LFSR sequence

I The remaining k − 1 states form a cycle C

Suppose one of the states in C is a constant polynomial

I The zero polynomial can be inserted into CI dB(q, `) contains q disjoint k-cyclesI E(q, q`−1, `) = q

C contains a constant iff − logx(xk − 1) mod q`−1q−1 ≤

k−1q−1

I This may be true for some primitive polynomials and false for othersI For a given q and `, we only need one polynomialI Depends on the distribution of the discrete logarithm











k−1q−1












k−1q−1







Take these cycles out of the LFSR sequenceI The remaining k − 1 states form a cycle C




k−1q−1











k−1q−1








Suppose one of the states in C is a constant polynomialI The zero polynomial can be inserted into C

I dB(q, `) contains q disjoint k-cyclesI E(q, q`−1, `) = q


k−1q−1








Suppose one of the states in C is a constant polynomialI The zero polynomial can be inserted into CI dB(q, `) contains q disjoint k-cycles

I E(q, q`−1, `) = q


k−1q−1








Suppose one of the states in C is a constant polynomialI The zero polynomial can be inserted into CI dB(q, `) contains q disjoint k-cyclesI E(q, q`−1, `) = q


k−1q−1










k−1q−1










k−1q−1

I This may be true for some primitive polynomials and false for others

I For a given q and `, we only need one polynomialI Depends on the distribution of the discrete logarithm









k−1q−1

I This may be true for some primitive polynomials and false for othersI For a given q and `, we only need one polynomial

I Depends on the distribution of the discrete logarithm









k−1q−1




Multiplying cycles

Theorem

Fix a value of ` and set E1 = E(q1, k1, `) and E2 = E(q2, k2, `). Then

E(q1q2, lcm(k1, k2), `) ≥ gcd(k1, k2) E1 E2.

Preserves optimality w.r.t. upper bound E ≤ q`

k

I If E1 and E2 are optimal, then so is E(q1q2, lcm(k1, k2), `)

When k1 and k2 are coprime, not much is gained

Largest increase in E when k1 = k2

I Easy to get large E when q is divisible by a high power



Multiplying cycles

Theorem




k

I If E1 and E2 are optimal, then so is E(q1q2, lcm(k1, k2), `)






Multiplying cycles

Theorem




kI If E1 and E2 are optimal, then so is E(q1q2, lcm(k1, k2), `)






Multiplying cycles

Theorem










Multiplying cycles

Theorem










Multiplying cycles

Theorem






Largest increase in E when k1 = k2I Easy to get large E when q is divisible by a high power



Construction for k1 = k2



























Proof of correctness (sketch)

Consider any sequence of ` colour pairs in resulting eBugs

I This gives a red/blue sequence and a yellow/cyan sequenceI These had unique positions in original colourings

Sequence corresponds to a particular pair of eBugs in a particularorientation

I Resulting eBugs can still be uniquely identified and oriented

Theorem easily extends to k1 6= k2 case




Consider any sequence of ` colour pairs in resulting eBugsI This gives a red/blue sequence and a yellow/cyan sequence

I These had unique positions in original colourings







Consider any sequence of ` colour pairs in resulting eBugsI This gives a red/blue sequence and a yellow/cyan sequenceI These had unique positions in original colourings



























Necklaces

Definition

A necklace is an equivalence class of words under cyclic rotation. Thelength of a necklace is the length of any word in the class, while the size ofa necklace is the number of words in the class.

Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102

Every necklace gives a cycle in a de Bruijn graph

I The length of the cycle is the size of the necklace

Cycles can be concatenated in the previous de Bruijn graph

I Two necklaces of length ` are mergable if they share a subword oflength `− 1

Construct a graph N(q, `) of necklaces of length ` over q letters

I Edge between two necklaces if they are mergable



Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102









Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102









Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102

Every necklace gives a cycle in a de Bruijn graphI The length of the cycle is the size of the necklace







Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102








Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102


Cycles can be concatenated in the previous de Bruijn graphI Two necklaces of length ` are mergable if they share a subword of

length `− 1





Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102



length `− 1





Necklaces

Definition


Example

001021 ≡ 010210 ≡ 102100 ≡ 021001 ≡ 210010 ≡ 100102



length `− 1

Construct a graph N(q, `) of necklaces of length ` over q lettersI Edge between two necklaces if they are mergable



Merging necklaces

Consider a connected subgraph of N(q, `)

I Edges can be contracted by merging the appropriate cyclesI The whole subgraph will produce one long cycleI The length of the cycle is the sum of the necklace sizes

Consider a partition of N(q, `) into connected pieces of total size k

I Gives a partition of dB(q, `) into k-cyclesI Difficult to find because N(q, `) has necklaces of different sizes

N(q, `) contains necklaces of each size m that divides `

I Moreau’s necklace counting function: M(q,m) = 1m

∑d|m

µ(d)qm/d

If q and ` are coprime, M(q,m) is divisible by q for each m

I Potential to partition N(q, `) into q connected piecesI For each size m, every piece has same number of size m necklacesI Resulting cycles all have the same lengthI Will give E(q, q`−1, `) = q



Merging necklaces

Consider a connected subgraph of N(q, `)I Edges can be contracted by merging the appropriate cycles

I The whole subgraph will produce one long cycleI The length of the cycle is the sum of the necklace sizes





∑d|m

µ(d)qm/d





Merging necklaces

Consider a connected subgraph of N(q, `)I Edges can be contracted by merging the appropriate cyclesI The whole subgraph will produce one long cycle

I The length of the cycle is the sum of the necklace sizes





∑d|m

µ(d)qm/d





Merging necklaces

Consider a connected subgraph of N(q, `)I Edges can be contracted by merging the appropriate cyclesI The whole subgraph will produce one long cycleI The length of the cycle is the sum of the necklace sizes





∑d|m

µ(d)qm/d





Merging necklaces






∑d|m

µ(d)qm/d





Merging necklaces


Consider a partition of N(q, `) into connected pieces of total size kI Gives a partition of dB(q, `) into k-cycles

I Difficult to find because N(q, `) has necklaces of different sizes



∑d|m

µ(d)qm/d





Merging necklaces


Consider a partition of N(q, `) into connected pieces of total size kI Gives a partition of dB(q, `) into k-cyclesI Difficult to find because N(q, `) has necklaces of different sizes



∑d|m

µ(d)qm/d





Merging necklaces





∑d|m

µ(d)qm/d





Merging necklaces



N(q, `) contains necklaces of each size m that divides `I Moreau’s necklace counting function: M(q,m) = 1

m

∑d|m

µ(d)qm/d





Merging necklaces




m

∑d|m

µ(d)qm/d





Merging necklaces




m

∑d|m

µ(d)qm/d

If q and ` are coprime, M(q,m) is divisible by q for each mI Potential to partition N(q, `) into q connected pieces

I For each size m, every piece has same number of size m necklacesI Resulting cycles all have the same lengthI Will give E(q, q`−1, `) = q



Merging necklaces




m

∑d|m

µ(d)qm/d

If q and ` are coprime, M(q,m) is divisible by q for each mI Potential to partition N(q, `) into q connected piecesI For each size m, every piece has same number of size m necklaces

I Resulting cycles all have the same lengthI Will give E(q, q`−1, `) = q



Merging necklaces




m

∑d|m

µ(d)qm/d

If q and ` are coprime, M(q,m) is divisible by q for each mI Potential to partition N(q, `) into q connected piecesI For each size m, every piece has same number of size m necklacesI Resulting cycles all have the same length

I Will give E(q, q`−1, `) = q



Merging necklaces




m

∑d|m

µ(d)qm/d

If q and ` are coprime, M(q,m) is divisible by q for each mI Potential to partition N(q, `) into q connected piecesI For each size m, every piece has same number of size m necklacesI Resulting cycles all have the same lengthI Will give E(q, q`−1, `) = q



Partitioning N(q, `)

Example (q = 2)

As ` must be odd, each necklace contains a majority of some letter (0 or1). Let N0 (N1) be the set of necklaces with more 0s (1s).

Let n0 ∈ N0 be the all zeroes necklace

Pick a necklace n in N0 \ {n0}

I Change the first 1 in n to a 0, and call this necklace nI There is an edge in N(q, `) between n and nI Note that n ∈ N0

By induction, there is a path in N0 from n to n0

I N0 is connectedI By symmetry, N1 is connected

Each component gives a cycle of length 2`−1 in dB(q, `)

Similar methods give a partition for general q whenever gcd(q, `) = 1

I E(q, q`−1, `) = q




Example (q = 2)









I E(q, q`−1, `) = q




Example (q = 2)









I E(q, q`−1, `) = q




Example (q = 2)



Pick a necklace n in N0 \ {n0}I Change the first 1 in n to a 0, and call this necklace n

I There is an edge in N(q, `) between n and nI Note that n ∈ N0





I E(q, q`−1, `) = q




Example (q = 2)



Pick a necklace n in N0 \ {n0}I Change the first 1 in n to a 0, and call this necklace nI There is an edge in N(q, `) between n and n

I Note that n ∈ N0





I E(q, q`−1, `) = q




Example (q = 2)



Pick a necklace n in N0 \ {n0}I Change the first 1 in n to a 0, and call this necklace nI There is an edge in N(q, `) between n and nI Note that n ∈ N0





I E(q, q`−1, `) = q




Example (q = 2)








I E(q, q`−1, `) = q




Example (q = 2)




By induction, there is a path in N0 from n to n0I N0 is connected

I By symmetry, N1 is connected



I E(q, q`−1, `) = q




Example (q = 2)




By induction, there is a path in N0 from n to n0I N0 is connectedI By symmetry, N1 is connected



I E(q, q`−1, `) = q




Example (q = 2)







I E(q, q`−1, `) = q




Example (q = 2)







I E(q, q`−1, `) = q




Example (q = 2)






Similar methods give a partition for general q whenever gcd(q, `) = 1I E(q, q`−1, `) = q



Summary

Overall goal: investigate behaviour of eBug number E(q, k, `)

I Each eBug has k LEDs with q available coloursI Camera can see ` consecutive LEDs

Equivalent to finding k-cycles in the de Bruijn graph dB(q, `)

I E(q, k, `) maximised when dB(q, `) is partitioned into k-cycles

de Bruijn graphs are Hamiltonian, so E(q, q`, `) = 1

I Called de Bruijn sequencesI One big eBug is not very useful

In some cases, partition into q q`−1-cycles exists

I Guaranteed when q and ` are coprimeI Likely when q is a prime powerI Requires primitive polynomial with certain properties

Two eBug colourings can be multiplied to give many eBugs

I Must have the same ` valueI Resulting eBugs have q1q2 possible coloursI Resulting cycle length is lcm(k1, k2)I Each pair of eBugs gives gcd(k1, k2) new eBugs



Summary

Overall goal: investigate behaviour of eBug number E(q, k, `)I Each eBug has k LEDs with q available colours

I Camera can see ` consecutive LEDs











Summary

Overall goal: investigate behaviour of eBug number E(q, k, `)I Each eBug has k LEDs with q available coloursI Camera can see ` consecutive LEDs











Summary












Summary


Equivalent to finding k-cycles in the de Bruijn graph dB(q, `)I E(q, k, `) maximised when dB(q, `) is partitioned into k-cycles









Summary











Summary



de Bruijn graphs are Hamiltonian, so E(q, q`, `) = 1I Called de Bruijn sequences

I One big eBug is not very useful







Summary



de Bruijn graphs are Hamiltonian, so E(q, q`, `) = 1I Called de Bruijn sequencesI One big eBug is not very useful







Summary










Summary




In some cases, partition into q q`−1-cycles existsI Guaranteed when q and ` are coprime

I Likely when q is a prime powerI Requires primitive polynomial with certain properties





Summary




In some cases, partition into q q`−1-cycles existsI Guaranteed when q and ` are coprimeI Likely when q is a prime power

I Requires primitive polynomial with certain properties





Summary




In some cases, partition into q q`−1-cycles existsI Guaranteed when q and ` are coprimeI Likely when q is a prime powerI Requires primitive polynomial with certain properties





Summary









Summary





Two eBug colourings can be multiplied to give many eBugsI Must have the same ` value

I Resulting eBugs have q1q2 possible coloursI Resulting cycle length is lcm(k1, k2)I Each pair of eBugs gives gcd(k1, k2) new eBugs



Summary





Two eBug colourings can be multiplied to give many eBugsI Must have the same ` valueI Resulting eBugs have q1q2 possible colours

I Resulting cycle length is lcm(k1, k2)I Each pair of eBugs gives gcd(k1, k2) new eBugs



Summary





Two eBug colourings can be multiplied to give many eBugsI Must have the same ` valueI Resulting eBugs have q1q2 possible coloursI Resulting cycle length is lcm(k1, k2)

I Each pair of eBugs gives gcd(k1, k2) new eBugs



Summary





Two eBug colourings can be multiplied to give many eBugsI Must have the same ` valueI Resulting eBugs have q1q2 possible coloursI Resulting cycle length is lcm(k1, k2)I Each pair of eBugs gives gcd(k1, k2) new eBugs



Summary

Minimum k for which E(q, k, `) = q`

kguaranteed

` q = 2 q = 3 q = 4 q = 6 q = 12

1 1 1 1 1 12 4 3 4 12 123 4 27 4 108 1084 16 27 16 432 4325 16 81 16 1296 12966 64 729 64 46656 466567 64 729 64 46656 46656


Documents

Cycle Decompositions of de Bruijn Graphs for Robot