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Automatic Sequences and why I am interested Wieb Bosma Radboud Universiteit Nijmegen Streams Seminar 25 January 2011 Nijmegen, 25 January 2011 0
Transcript

Automatic Sequences

and why I am interested

Wieb Bosma

Radboud Universiteit Nijmegen

Streams Seminar

25 January 2011

Nijmegen, 25 January 2011 0

disclaimer

Most material taken from book Automatic Sequences by Jean-Paul Allouche and

Jeffrey Shallit (Cambridge University Press, 2003), and paper Number theory and

formal languages by J.O. Shallit, (Springer 1999).

Automatic sequences form a class of sequences somewhere between simpleorder and chaotic disorder.

Nijmegen, 25 January 2011 1

PART I

automata and sequences

Nijmegen, 25 January 2011 2

automata

A deterministic finite automaton starts in a distinguished initial state, reads letters

from a finite input word, which determine transitions to finite number of states, to

end in one of the final states, which is either accepting or rejecting for the input word.

A = (Q,Σ, δ, q0, F ):

Q finite set of states,

Σ finite alphabet,

δ : Q× Σ→ Q transitions,

q0 ∈ Q initial state,

F ⊂ Q accepting states.

Nijmegen, 25 January 2011 3

example

Nijmegen, 25 January 2011 4

answers

Theorem 1 A language is accepted by a deterministic finite state automaton if and

only if it can be specified by a regular expression.

Theorem 2 There is a unique deterministic finite state automaton with minimal

number of states accepting a given regular language.

Nijmegen, 25 January 2011 5

generalizations

A non-deterministic finite state automaton allows any finite number (including

zero) of transitions from a state for a given input letter. A word is accepted if and

only if there exists a choice of transitions leading to an accepting state.

A deterministic finite state automaton with output produces an output symbol at

final states (instead of a Boolean), and hence defines a finite state function from Σ∗

to (a possibly different output alphabet) ∆.

A transducer produces a word over the output alphabet ∆ for every transition

(q, a) ∈ Q× Σ. It is uniform if all output words have the same length.

Nijmegen, 25 January 2011 6

examples

Nijmegen, 25 January 2011 7

examples

Nijmegen, 25 January 2011 8

sequences

A sequence (an)n≥0 is a k-automatic sequence if there exists a deterministic finite

state automaton which produces for all n ≥ 0 on input the base-k representation [n]kof n, the element an as output.

Nijmegen, 25 January 2011 9

example

Define the Thue-Morse sequence (tn)n≥0 by tn =∑∞

i=0 di mod 2 if n =∑∞

i=0 di2i

is the binary expansion of n. It starts 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, . . ..

Theorem 3 The Thue-Morse sequence is 2-automatic.

Nijmegen, 25 January 2011 10

example

Define the Rudin-Shapiro sequence (rn)n≥0 by rn = 1 or −1 according to whether

the number of pairs of consecutive 1’s in the binary expansion of n is even or odd. It

starts 1, 1, 1,−1, 1, 1,−1, 1, 1, 1, 1,−1, . . ..

Theorem 4 The Rudin-Shapiro sequence is 2-automatic.

Nijmegen, 25 January 2011 11

example

Define the Baum-Sweet sequence (bn)n≥0 by bn = 1 or 0 according to whether the

number of blocks of odd length of consecutive 0’s in the binary expansion of n is

zero or positive. It starts 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1 . . ..

Theorem 5 The Baum-Sweet sequence is 2-automatic.

Nijmegen, 25 January 2011 12

characterisations

For positive integer k and output letter d, define the fiber Ik(a, d) of the sequence a

as the set of input words [n]k such that an = d.

Theorem 6 a is k-automatic if and only if every Ik(a, d) is a regular language.

Define the k-kernel of a to be the set of subsequences

{(aki·n+j)n≥0 : i ≥ 0, 0 ≤ j < ki}.

Theorem 7 a is k-automatic if and only if the k-kernel is finite.

Nijmegen, 25 January 2011 13

bases

Theorem 8 If (an)n≥0 is ultimately periodic, then it is k-automatic for all k ≥ 2.

Theorem 9 (an)n≥0 is ultimately periodic if and only if it is 1-automatic.

Theorem 10 (Cobham) (an)n≥0 is ultimately periodic if it is k- and l-automatic,

for multiplicatively independent k, l ≥ 2.

Nijmegen, 25 January 2011 14

morphisms

A k-uniform morphism is a map φ : Σ∗ → Σ∗ such that φ(xy) = φ(x)φ(y) for all

words x, y and such that the length of φ(a) is k for all letters a ∈ Σ.

If φ(a) = ax for some letter a then

φω(a) = axφ(x)φ2(x) · · ·

is the unique fixed point of φ starting with a.

Theorem 11 (Cobham) a is a k-automatic sequence if and only if it is fixed point

of a k-uniform morphism (followed by a coding).

Nijmegen, 25 January 2011 15

example

The Thue-Morse sequence is the fixed point of φ defined by φ(0) = 01 and

φ(1) = 10. Indeed,

01φ(1)φ2(1)φ3(1) · · · = 0110100110010110 · · · .

The Baum-Sweet sequence is obtained from the fixed point of

φ(a) = ab, φ(b) = cb, φ(c) = bd, φ(d) = dd,

followed by replacing a, b by 1 and c, d by 0. Indeed, we obtain

abcbbdcbcbddbdcb · · · ,

which becomes

1101100101001001 · · · .

Nijmegen, 25 January 2011 16

morphic

Theorem 12 The image of an automatic sequence under a uniform transducer is

again automatic.

A morphic sequence is the fixed point of a morphism (after a coding).

The infinite Fibonacci word is the morphic sequence that is the fixed point of

φ(0) = 01 and φ(1) = 0. We get

f = 010010100100101001010 · · · .

Theorem 13 The image of a morphic sequence under a transducer is finite or again

a morphic sequence.

Nijmegen, 25 January 2011 17

PART II

numbers and functions

Nijmegen, 25 January 2011 18

conway

Conway’s prime-producing machine

Nijmegen, 25 January 2011 19

fractions

Regular continued fractions for real numbers are of the form

a0 +1

a1 +1

a2 +1

a3 +1

. . .

with positive integers ai. Denoted [a0, a1, a2, a3, . . .].

(In)finite continued fractions represent (ir)rational numbers.

For irrational x the infinite sequence of ai is obtained by putting x0 = x and

repeating

ai = bxic, xi+1 =1

xi − ai.

Alternative (semi-regular) continued fraction expansions are obtained by rounding

differently.

Nijmegen, 25 January 2011 20

approximation

Important because this furnishes infinite sequence of (best) rational approximations

by the convergentspn

qn= [a0, a1, a2, . . . , an],

each satisfying ∣∣∣∣x− pn

qn

∣∣∣∣ < 1q2n.

Theorem 14 The regular continued fraction expansion of x is ultimately periodic if

and only if x is a quadratic irrational number.

For example, √67 = [8, 5, 2, 1, 1, 7, 1, 1, 2, 5, 16].

Nijmegen, 25 January 2011 21

transcendental

This automaton generates the continued fraction expansion of

f(B) =∞∑

k=0

1B2k ,

a transcendental number, for any integer B ≥ 3.

Nijmegen, 25 January 2011 22

Thue-Morse

The Thue-Morse number∞∑

i=0

ti2−i

is also transcendental.

Nijmegen, 25 January 2011 23

conjecture

Thus we find two classes and lots of examples of real numbers having bounded

partial quotients: the rational numbers (finite) and the quadratic irrationalities

(ultimately periodic) and some transcendental numbers.

Conjecture 1 If a real number x has bounded partial quotients then it is either

rational, quadratic irrational, or transcendental.

Note: this is no longer true for complex continued fractions!

There are also interesting questions related to sums of real numbers with bounded

partial quotients: M. Hall proved that every real number in the unit interval is the

sum of two reals with partial quotients at most 4.

Nijmegen, 25 January 2011 24

arithmetic

Unfortunately, continued fraction representations are hopeless for doing arithmetic.

The main result is by Raney.

Theorem 15 For integers a, b, c, d with ad− bc 6= 0, there exits a finite transducer

that, on input the regular continued fraction representation of x, produces the

continued fraction representation of

y =ax+ b

cx+ d.

Nijmegen, 25 January 2011 25

example

Nijmegen, 25 January 2011 26

Farey

Raney uses the LR-representation, which is closely related to Farey-fractions and the

Stern-Brocot representation for reals.

The numerators pn and denominators qn for the convergents of x = [a0, a1, a2, . . .]satisfy the same recursion

pn = anpn−1 + pn−2, qn = anqn−1 + qn−2,

if we put

p−1 = 0, p−2 = 1, q−1 = 1, q−2 = 0.

Nijmegen, 25 January 2011 27

LR

It turns out that matrices like pn+1 pn

qn+1 qn

can be written as product

La0Ra1La2 · · ·Lan

(or a slight variant, depending on the parity of n) where

L =

1 0

1 1

, R =

1 1

0 1

,

so

Lai =

1 0

ai 1

, Raj =

1 aj

0 1

.

Nijmegen, 25 January 2011 28

seven

Nijmegen, 25 January 2011 29

Christol

Let (an)n≥0 be a sequence over the finite alphabet ∆, let p be a prime number, and

Fpm the finite field of pm elements..

Theorem 16 (Christol) (an)n≥0 is p-automatic if and only if there exist a positive

integer m and an injection ι of ∆ in Fpm such that

∞∑k=0

ι(ak)Xk

is algebraic over Fpm(X).

Nijmegen, 25 January 2011 30

example

Let (tn)n≥0 be the Thue-Morse sequence again, and

T (X) =∞∑

k=0

tkXk ∈ F2[[X]].

Since t2n+1 = tn + 1 we find

T (X) =∞∑

k=0

t2kX2k +X

∞∑k=0

(tk + 1)X2k

and hence

T (X) = T (X)2 +XT (X)2 +X1

1−X2,

so T satisfies

(1 +X)3T 2 + (1 +X)2T +X = 0.

Nijmegen, 25 January 2011 31

references

Jean-Paul Allouche, Jeffrey Shallit, Automatic Sequences, Cambridge University

Press, 2003

J.O. Shallit, Number theory and formal languages, in: Emerging applicationsof number theory, pp. 547–570, Springer, 1999.

Richard K. Guy, Conway’s prime producing machine, Math. Mag. 56 (1983),

26–33.

Doug Hensley, Continued Fractions, World Scientific, 2006.

Marshall Hall, Jr., On the sum and product of continued fractions, Ann. Math.

48 (1947), 966–993.

George N. Raney, On continued fractions and finite automata, Math. Ann.

206 (1973), 265–283.

Nijmegen, 25 January 2011 32


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