34

For question 3, the answer is yes: take a solid disc and excavate half of the Penrose unilluminable room from it. Then, there are boundary arcs which can never be touched, and you can perturb them without altering the bounce behaviour.
It's possible to make this simple closed curve $C^{\infty}$ if you carefully smooth the corners.
Edit: included a rough ...

answered Apr 19 '18 at 13:15

Adam P. Goucher

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33

Gosper showed how to produce the continued fraction for
$$\frac{axy+bx+cy+d}{exy+fx+gy+h}$$
and even more complicated expressions given the continued fractions for $x$ and $y$. As a special case, he covers fractional linear transformations of $x$, which of course includes adding $1/2$ or multiplying by $2$. I haven't digested this yet.
Here is a special ...

16

The sequence $\sum a_n$ is unbounded.
This is a consequence of a general result from Kesten,
On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arithmetica (1966). The proof is not very long but quite computational, using properties of continued fractions.
Theorem
Let $\xi \in [0,1]$, $0\leq a < b \leq 1$, denote by $N(M, \...

11

(This answer has been substantially edited since its original version.)
1.
The surreal number with sign-expansion $+-^{\omega}++-^{\omega}+++-^{\omega}++++-^{\omega}\cdots$ (indexed by $\omega^2$) can be rewritten in a tidy form, which shows that it does lie in the field generated by $\omega$.
As you read off that sign-expansion, the number follows the ...

10

The answer appears to be no.
Consider first the case of the anchor-tile
periodic tiling problem, where we insist that a particular anchor
tile is used. Let's modify the usual Wang tile argument, due to Berger, for the oringal tiling problem. That argument shows that for any Turing machine
program $p$, we can create a set of Wang tile types (square tiles
...

answered Feb 11 '13 at 15:15

Joel David Hamkins

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10

Kind of a late response, but since no one else mentioned it I think it is worth pointing out that the procedure for computing homographies/linear fractional transformations on continued fractions (mentioned in Douglas Zare's answer) was also described explicitly in terms of finite state automata by George N. Raney, in "On continued fractions and finite ...

10

If $X$ is minimal and not a periodic orbit then it cannot contain a periodic orbit and hence in particular cannot contain a Markov shift. A classical construction by Grillenberger shows that one can construct uniquely ergodic (hence in particular minimal) subshifts with arbitrary entropy. (This result also follows from the symbolic version of the Jewett-...

10

Wang tiles are unit squares with edges marked with colors, and the problem of whether a given set of Wang tiles can tile the plane such that edges of adjacent squares match has been studied exhaustively (see https://en.wikipedia.org/wiki/Wang_tile). In particular, there is a set of 11 Wang tiles using four colors which tiles the plane, but there is no ...

10

Yes. By Proposition 2.1.10 in Lothaire, Algebraic Combinatorics on Words, if $u$ is any substring of the Fibonacci word then
$$\left| \frac{\mbox{number of $1$'s in $u$}}{\mbox{length of $u$}} - \frac{1}{\phi^2} \right| \leq \frac{1}{\mbox{length of $u$}}.$$
Any length $n$ substring of any $\omega \in \Omega$ is also a length $n$ substring of the Fibonacci ...

answered Feb 20 '19 at 3:49

David E Speyer

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9

A good way to understand measurable entropy is via the Shannon-McMillan-Breiman Theorem. Roughly speaking it says that there is a constant $c$ so that most atoms $A$ in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$ have measure $m(A)\approx e^{-cn}$, and the value of $c$ is the measure entropy $h(\mathcal{A},T,m)$. More precisely, given $\epsilon>0$, for all ...

8

For odd $k$, the shift in $X^k$ has no square root (and hence does not lie in a 1-parameter subgroup). Indeed, the the set of 2-periodic points can be identified with $(I^k)^2$, where the shift acts as $(x,y)\mapsto (y,x)$. It would then preserve the set of fixed points, namely the diagonal $(x,x)$, so any root should preserve the subset of points of period ...

7

Just as the $\beta$-shift arises as the coding space for the transformation $x\mapsto \beta x$ (mod 1), so the shifts you describe arise as coding spaces for $x\mapsto \alpha + \beta x$ (mod 1). The characteristic sequences $a,b$ correspond to the codings of 0 and 1, respectively.
In particular, you can describe the subshift in terms of a graph -- this was ...

answered Apr 8 '14 at 15:05

Vaughn Climenhaga

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7

Yes. If $X$ is minimal (every orbit is dense) then the only subshift $Y\subset X$ is $X$ itself. The Jewett-Krieger theorem allows the construction of minimal subshifts with positive entropy, which therefore have the property you desire (albeit somewhat vacuously). Googling "minimal subshifts with positive entropy" brings up this paper by Henk Bruin, ...

answered Dec 6 '13 at 19:46

Vaughn Climenhaga

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According to the 2013 paper "Computing the Partial Word Avoidability Indices of
Ternary Patterns" by Blanchet-Sadri, Lohr, and Scott,
The problem of deciding whether a given pattern is avoidable has been
solved [1, 14], but the one of deciding whether it is k-avoidable has remained
open.
So, it seems to be an open problem.

co.combinatorics semigroups-and-monoids algebraic-combinatorics symbolic-dynamics combinatorics-on-words

answered Jul 19 '14 at 0:50

Bjørn Kjos-Hanssen

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7

If $\phi$ is an automorphism of $X$ and $Y$ is the set of points in $X$ of exact period $n$, then $\phi|_Y$ is an automorphism of $Y$. There is a subtle relationship between the sign of the permutation that $\phi$ induces on orbits of lengths $n$ for various $n$, and the value of a transfer function called the nth gyration number of $\phi$, called the sign-...

7

I've decided to upgrade my comments and make an answer out of them, even though I'm just addressing the (easier) variant suggested by the OP at the end of the post, where we replace the golden ratio by other numbers $b$.
If $b$ can be well approximated by rationals, then it's easy to see that the sum cannot stay bounded: More precisely, suppose there are ...

answered Apr 11 '17 at 20:03

Christian Remling

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6

The answer is no. It's based on a (un?)published example of Crannell, Rudolph and Weiss.
The example is the following shift: $X$ is the subset of $\lbrace 0,\pm 1\rbrace ^{\mathbb Z}$ with the property that $x_k\cdot x_{k+2^n}$ is not allowed to be $-1$ for any values of $k$ and $n$.
What they prove is that there are 2 measures of maximal entropy for $X$: ...

6

Every topologically transitive shift space, whether intrinsically ergodic or not, can be approximated from above by intrinsically ergodic systems.
Indeed, given a finite alphabet $A=\{1,2,\dots,p\}$ and a closed $\sigma$-invariant set $X\subset A^\mathbb{Z}$ (everything works just the same for one-sided shifts), let $\mathcal{L}=\mathcal{L}(X) \subset A^* =...

answered Jul 9 '12 at 15:36

Vaughn Climenhaga

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6

Here are some results which suggest that perhaps what happens is neither very simple nor very random. But see the end for a better result.
These are some of the simplest continued fractions and what they lead to. This also tells you what results in simple continued fractions because $r-\frac12$ and $r+\frac12$ have the same continued fraction after the ...

answered Apr 28 '13 at 21:06

Aaron Meyerowitz

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6

There are some examples related to your third question in "Renewal Systems, Sharp-Eyed Snakes, and Shifts of Finite Type" by Johnson and Madden, Amer. Math. Monthly 109 (2002), 258-272. A long time ago Goldberger, Smorodinsky, and I showed that for every possible entropy of a shift of finite type (or, what amounts to the same thing, for the logarithm of ...

6

The link that you refer to does not describe the boundary as the attractor of a simple IFS, rather it describes a collection of portions of the boundary as the invariant list of a digraph IFS - or directed graph iterated function system. I am not an expert on sofic systems, but I believe that an analysis of that digraph IFS yields the type of description ...

6

(05/05/2015) If $T$ is a permutation of any set $X$ with a (finite) odd number of 2-cycles, then $T$ is not a square in the group of permutations of $X$ (because if $T=U^2$ then $T$ commutes with $U$ --this assumption in the question is thus redundant--, hence $U$ preserves the union $J$ of 2-cycles of $T$ and acts on $J$ so that $U^2$ has no fixed point, ...

6

The most intuitive explanation I know is the following: suppose that you have a certain amount of mass (I usually picture a pile of sand) that is distributed over $\Sigma_A^+$ according to the density $g\,d\nu$, where $g$ is an arbitrary continuous function and $\nu$ is the eigenmeasure for the transfer operator $\mathcal{L}$. Then you move this mass around ...

answered Aug 2 '19 at 19:21

Vaughn Climenhaga

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5

The complexity of an infinite sequence $x$ is a sequence $C(n)$, where $C(n)$ is the number of distinct blocks of length $n$ in $x$. For an interval exchange with $k$ symbols, it's not hard to show that $C(n)=(k-1)n+1$. If your word has more complexity than this, it can never appear as the coding sequence of an IET.

5

To prevent any further beating around the bush let me explain Anthony's answer - as it is in fact much easier than what he actually wrote.
Take any weakly mixing zero entropy transformation $T$ on a probability space $(X,m)$, and take a subset $A\subset X$ with $m(A)=1/2$. Then the symbolic coding of $T$ determined by the partition $(A,X\setminus A)$ ...

5

I you use the definition: $$\mathcal T=\left\{\sum_{n=0}^\infty a_n\left(\frac{1-i}2\right)^n : a_n\in\{0,1\}\right\}$$
it seems that the words on the boundary are exactly those recognized by the automaton (e.g.) in [https://www.ricam.oeaw.ac.at/publications/reports/06/rep06-02.pdf] (Figure 2). For your definition, it seems that we have to complement every ...

5

While I do not know if there is a
"right" surreal generalization of the fact that a real number $r$ is
rational if and only if its sign-expansion is eventually periodic
it is perhaps worth noting that Conway has identified what he believes "are perhaps the closest analogue in No of the ordinary rational numbers" (ONAG, p.47). These surreal "fractional ...

5

No.
Your condition is called being a (topological) subshift. If $(C,f)$ is a topological subshift, then there exists a finite clopen partition $P$ of $C$ such that the family $(f^{-n}P)_{n\ge 0}$ separates the points of $C$. (Indeed this is obvious of the shift on $k$ letters: take the partition into $k$ clopen subsets defined by evaluation at 0).
An ...

5

In the setting you describe, for each $\alpha \in (0,1)$ the $(1-\alpha,\alpha)$-Bernoulli measure is the unique measure achieving the maximum. The function $\alpha \mapsto \eta(\alpha)$ is the Legendre transform of the function $t\mapsto P(t\phi)$ where $\phi(x) = x_0$ and $P$ is topological pressure. This is all part of the "multifractal analysis" of the ...

answered May 17 '18 at 15:23

Vaughn Climenhaga

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5

Let $f$ be a sublinear function that tends to infinity, such as $f(n) = \sqrt{n}$. Define $X \subset \{0,1,2\}^{\mathbb{N}}$ by forbidding all long enough words $w$ with more than $f(|w|)$ occurrences of $2$. Then $X$ has entropy $\log 2$ and is mixing, and properly contains the binary full shift, which likewise has entropy $\log 2$.
If you need a binary ...

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