Two Trees in Fall is a canvas print of a colorful and modern Ledent painting showing trees and landscape in autumn. Originally oil on canvas. Pol Ledent was born in 1952 in Belgium. He came to painting in 1989. He started with watercolor but felt rapidly that oil painting would match his way of being. He is a self-taught painter . Nevertheless he took some drawing lessons in a Belgian academy. After taking part into numerous group exhibitions, some galleries in Belgium proposed to him to exhibit his works. Dinant, Bouillon, Brussels , Paris and Moscow in October 2006.


text:
Ríos Rosas – stacja metra w Madrycie, na linii 1. Znajduje się w dzielnicy Chamberí, w Madrycie i zlokalizowana jest pomiędzy stacjami Cuatro Caminos i Iglesia. Została otwarta 17 października 1919.

Linki zewnętrzne

Metro w Madrycie

redpajama_set_name: RedPajamaWikipedia


text:
Q: Grouping by contiguous dates, ignoring weekends in SQL I'm attempting to group contiguous date ranges to show the minimum and maximum date for each range. So far I've used a solution similar to this one: http://www.sqlservercentral.com/articles/T-SQL/71550/ however I'm on SQL 2000 so I had to make some changes. This is my procedure so far:
create table #tmp
(
date smalldatetime,
rownum int identity
)

insert into #tmp
select distinct date from testDates order by date

select
min(date) as dateRangeStart,
max(date) as dateRangeEnd,
count(*) as dates,
dateadd(dd,-1*rownum, date) as GroupID
from #tmp
group by dateadd(dd,-1*rownum, date)

drop table #tmp

It works exactly how I want except for one issue: weekends. My data sets have no records for weekend dates, which means any group found is at most 5 days. For instance, in the results below, I would like the last 3 groups to show up as a single record, with a dateRangeStart of 10/6 and a dateRangeEnd of 10/20:

Is there some way I can set this up to ignore a break in the date range if that break is just a weekend?
Thanks for the help.

A: EDITED
I didn't like my previous idea very much. Here's a better one, I think:

*

*Based on the first and the last dates from the set of those to be grouped, prepare the list of all the intermediate weekend dates.

*Insert the working dates together with weekend dates, ordered, so they would all be assigned rownum values according to their normal order.

*Use your method of finding contiguous ranges with the following modifications:
1) when calculating dateRangeStart, if it's a weekend date, pick the nearest following weekday;
2) accordingly for dateRangeEnd, if it's a weekend date, pick the nearest preceding weekday;
3) when counting dates for the group, pick only weekdays.

*Select from the resulting set only those rows where dates > 0, thus eliminating the groups formed only of the weekends.
And here's an implementation of the method, where it is assumed, that a week starts on Sunday (DATEPART returns 1) and weekend days are Sunday and Saturday:
DECLARE @tmp TABLE (date smalldatetime, rownum int IDENTITY);
DECLARE @weekends TABLE (date smalldatetime);
DECLARE @minDate smalldatetime, @maxDate smalldatetime, @date smalldatetime;
/* #1 */
SELECT @minDate = MIN(date), @maxDate = MAX(date)
FROM testDates;
SET @date = @minDate - DATEPART(dw, @minDate) + 7;
WHILE @date < @maxDate BEGIN
  INSERT INTO @weekends
  SELECT @date UNION ALL
  SELECT @date + 1;
  SET @date = @date + 7;
END;
/* #2 */
INSERT INTO @tmp
SELECT date FROM testDates
UNION
SELECT date FROM @weekends
ORDER BY date;
/* #3 & #4 */
SELECT *
FROM (
  SELECT
    MIN(date + CASE DATEPART(dw, date) WHEN 1 THEN 1 WHEN 7 THEN 2 ELSE 0 END)
      AS dateRangeStart,
    MAX(date - CASE DATEPART(dw, date) WHEN 1 THEN 2 WHEN 7 THEN 1 ELSE 0 END)
      AS dateRangeEnd,
    COUNT(CASE WHEN DATEPART(dw, date) NOT IN (1, 7) THEN date END) AS dates,
    DATEADD(d, -rownum, date) AS GroupID
  FROM @tmp
  GROUP BY DATEADD(d, -rownum, date)
) s
WHERE dates > 0;

redpajama_set_name: RedPajamaStackExchange


Biochimie, 92 (12), p. 1749–1759, 2010, ISSN: 1638-6183.
Proteolytic signalling cascades control a wide range of physiological responses. In order to respond rapidly, protease activity must be maintained at a basal level: the component zymogens must be sequentially activated and actively degraded. At the same time, signalling cascades must respond precisely: high target specificity is required. The insects have a wide range of trapping- and tight-binding protease inhibitors, which can regulate the activity of individual proteases. In addition, the interactions between component proteases of a signalling cascade can be modified by serine protease homologues. The suicide-inhibition mechanism of serpin family inhibitors gives rapid turnover of both protease and inhibitor, but target specificity is inherently broad. Similarly, the TEP/macroglobulins have extremely broad target specificity, which suits them for roles as hormone transport proteins and sensors of pathogenic virulence factors. The tight-binding inhibitors, on the other hand, have a lock-and-key mechanism capable of high target specificity. In addition, proteins containing multiple tight-binding inhibitory domains may act as scaffolds for the assembly of signalling complexes. Proteolytic cascades regulated by combinations of different types of inhibitor could combine the rapidity of suicide-inhibitors with the specificity lock-and-key inhibitors. This would allow precise control of physiological responses and may turn out to be a general rule.
Nature Immunology, 4 (8), p. 794–800, 2003, ISSN: 1529-2908.
The extracellular protein Spätzle is required for activation of the Toll signaling pathway in the embryonic development and innate immune defense of Drosophila. Spätzle is synthesized as a pro-protein and is processed to a functional form by a serine protease. We show here that the mature form of Spätzle triggers a Toll-dependent immune response after injection into the hemolymph of flies. Spätzle specifically bound to Drosophila cells and to Cos-7 cells expressing Toll. Furthermore, in vitro experiments showed that the mature form of Spätzle bound to the Toll ectodomain with high affinity and with a stoichiometry of one Spätzle dimer to two receptors. The Spätzle pro-protein was inactive in all these assays, indicating that the pro-domain sequence, which is natively unstructured, acts to prevent interaction of the cytokine and its receptor Toll. These results show that, in contrast to the human Toll-like receptors, Drosophila Toll requires only an endogenous protein ligand for activation and signaling.
Biochem. J., 345 Pt 3 , p. 653–664, 2000, ISSN: 0264-6021.
Androctonin is a 25-residue non-haemolytic anti-microbial peptide isolated from the scorpion Androctonus australis and contains two disulphide bridges. Androctonin is different from known native anti-microbial peptides, being a relatively hydrophilic and non-amphipathic molecule. This raises the possibility that the target of androctonin might not be the bacterial membrane, shown to be a target for most amphipathic lytic peptides. To shed light on its mode of action on bacteria and its non-haemolytic activity, we synthesized androctonin, its fluorescent derivatives and its all-D-amino acid enantiomer. The enantiomer preserved high activity, suggesting a lipid-peptide interaction between androctonin and bacterial membranes. In Gram-positive and (at higher concentrations) Gram-negative bacteria, androctonin induced an immediate perturbation of the permeability properties of the cytoplasmic membrane of the bacterial energetic state, concomitant with perturbation of the morphology of the cell envelope as revealed by electron microscopy. Androctonin binds only to negatively charged lipid vesicles and induces the leakage of markers at high concentrations and with a slow kinetics, in contrast with amphipathic alpha-helical anti-microbial peptides that bind and permeate negatively charged vesicles, and to a smaller extent also zwitterionic ones. This might explain the selective lytic activity of androctonin towards bacteria but not red blood cells. Polarized attenuated total reflection-Fourier transform infrared spectroscopy revealed that androctonin adopts a beta-sheet structure in membranes and did not affect the lipid acyl chain order, which supports a detergent-like effect. The small size of androctonin, its hydrophilic character and its physicochemical properties are favourable features for its potential application as a replacement for commercially available antibiotics to which bacteria have developed resistance.
Proc. Natl. Acad. Sci. U.S.A., 95 (19), p. 11342–11347, 1998, ISSN: 0027-8424.
We have developed an approach based on a differential mass spectrometric analysis to detect molecules induced during the immune response of Drosophila, regardless of their biological activities. For this, we have applied directly matrix-assisted laser desorption/ionization MS to hemolymph samples from individual flies before and after an immune challenge. This method provided precise information on the molecular masses of immune-induced molecules and allowed the detection, in the molecular range of 1.5-11 kDa, of 24 Drosophila immune-induced molecules (DIMs). These molecules are all peptides, and four correspond to already characterized antimicrobial peptides. We have further analyzed the induction of the various peptides by immune challenge in wild-type flies and in mutants with a compromised antimicrobial response. We also describe a methodology combining matrix-assisted laser desorption ionization time-of-flight MS, HPLC, and Edman degradation, which yielded the peptide sequence of three of the DIMs. Finally, molecular cloning and Northern blot analyses revealed that one of the DIMs is produced as a prepropeptide and is inducible on a bacterial challenge.
Biochimie, 80 (4), p. 343–346, 1998, ISSN: 0300-9084.
By combination of size exclusion and reversed-phase chromatography, we have isolated a novel member of insect defensin-type antimicrobial peptides from the entire bodies of bacteria-challenged Formica rufa (hymenoptera, formicidae). The molecular mass of the purified peptide was estimated to be 4120.42 by matrix-assisted laser desorption/ionization-time of flight/mass spectrometry. Sequence analysis revealed that this peptide consisted of 40 amino acid residues with six cysteines engaged in the formation of three intramolecular disulfide bridges. This peptide is unique among the arthropod defensins in terms of the presence of asparatic acid and alanine at position 33 and as C-terminal residue, respectively. In addition, this novel defensin from Formica rufa has the particularity to have no C-terminal extension in contrast to those reported for other hymenoptera defensins.
J. Biol. Chem., 269 (52), p. 33159–33163, 1994, ISSN: 0021-9258.
In response to a septic injury (pricking with a bacteria-soaked needle) larvae and adults of Drosophila produce considerable amounts of a 44-residue peptide containing 8 cysteines engaged in intramolecular disulfide bridges. The peptide is synthesized in the fat body, a functional homologue of the mammalian liver, and secreted into the blood of the insect. It exhibits potent antifungal activity but is inactive against bacteria. This novel inducible peptide, which we propose to name drosomycin, shows a significant homology with a family of 5-kDa cysteine-rich plant antifungal peptides recently isolated from seeds of Brassicaceae. This finding underlines that plants and insects can rely on similar molecules in their innate host defense.


text:
//
// Created by Matthijs Oosterhoff on 17/01/2017.
//

#ifndef SC_CONTROLLER_H
#define SC_CONTROLLER_H

void wait_for_neutral();

int get_battery_voltage();

int get_temperature_voltage();

void controller_setup();

void controller_loop();

#endif /* SC_CONTROLLER_H */

redpajama_set_name: RedPajamaGithub


text:
November 24, 2020 - 06:00 AM EST
Trump transition order follows chorus of GOP criticism
By Alexander Bolton
President Trump's decision to tell his administration to start the transition came as GOP criticism of his efforts to overturn the results of President-elect Joe Biden's win reached a crescendo on Monday.
In the hours leading up to Trump's announcement, more Republicans came out with statements saying the election was over, Biden was likely the next president and it was important to begin the transfer of power.
Among those GOP voices were Sens. Shelley Moore Capito (W.Va.) and Rob Portman (Ohio), two advisers to the Senate Republican leadership. Criticism also came from prominent conservatives like talk show host Rush Limbaugh, who said Trump's legal team needed to back up its claims of widespread voter fraud with hard evidence.
"While some irregularities and fraud have been found and should be punished, there is no indication that these are widespread enough to call into question the outcome of the election," Capito said in a statement, while acknowledging Biden is likely to be the next president.
Portman, who co-chaired Trump's campaign in Ohio and is up for reelection in 2022, said the campaign has yet to provide compelling evidence that voting irregularities and fraud denied Trump a second term.
"Based on all the information currently available, neither the final lawful vote counts nor the recounts have led to a different outcome in any state. In other words, the initial determination showing Joe Biden with enough electoral votes to win has not changed," he wrote in an op-ed published Monday in the Cincinnati Enquirer.
Capito and Portman said Biden and Vice President-elect Kamala Harris should begin receiving briefings on national security and the COVID-19 pandemic.
The General Services Administration (GSA) on Monday night issued a letter of ascertainment to officially start the presidential transition, with Trump saying he ordered the agency to take that step while he continues his legal fights.
But fading Republican support for those legal challenges contributed in large part to Trump's GSA move, with the president's lawyers coming under increasing scrutiny for making explosive claims without providing much in the way of hard evidence.
Former New Jersey Gov. Chris Christie (R), a close Trump ally, on Sunday called the conduct of Trump's legal team, led by former New York City Mayor Rudy Giuliani, "a national embarrassment."
A day later, Limbaugh used his radio show to draw attention to the fallout from last week's widely criticized press conference with Giuliani.
"I talked to so many people who were blown away by it, by the very nature of the press conference," Limbaugh said. "They promised blockbuster stuff, and then nothing happened. And that's just not good."
National security officials and business leaders often allied with Republicans also put pressure on Trump before he announced the transition.
More than 100 Republican foreign affairs and national security experts on Monday issued a statement warning that Trump's refusal to begin the transfer of power to Biden "poses significant risks to our national security" at a time the nation faces "a global pandemic" and "serious threats from global adversaries."
The signatories included former Homeland Security Secretary Tom Ridge and former CIA Director Michael Hayden.
Separately, more than 160 CEOs and leaders of major New York-based companies such as Goldman Sachs, Visa, Mastercard, BlackRock and Deloitte published a letter on Monday calling on Trump to agree to an orderly transition.
"Every day that an orderly presidential transition process is delayed, our democracy grows weaker in the eyes of our own citizens and the nation's stature on the global stage is diminished," they wrote.
Republican strategists say statements like the ones from Capito and Portman are a sign that support for Trump's legal challenges are rapidly eroding and that GOP lawmakers want to start preparing to work with what they recognize will be a new administration.
"It shows that members understand the reality is that Trump isn't going to win these cases. The members want to work with a potential Biden administration. They want to get ahead of the curve and not start off on the wrong foot," said Brian Darling, a Republican strategist and former Senate aide.
Up until this past weekend, only four Republican senators had congratulated Biden on winning the presidential election: Sens. Susan Collins (Maine), Lisa Murkowski (Alaska), Mitt Romney (Utah) and Ben Sasse (Neb.). The four are considered Trump's most outspoken GOP critics in the Senate.
Now, more mainstream GOP lawmakers, who are generally less likely to criticize the president, are speaking out.
On Saturday evening, Sen. Pat Toomey (R-Pa.) issued a strong statement declaring that Trump had "exhausted all plausible legal options" and called on the president to concede.
"I congratulate President-elect Biden and Vice President-elect Kamala Harris on their victory," Toomey said.
James Wallner, a former Senate Republican aide, said other GOP senators are likely to make similar statements in the coming days.
"You're going to see more and more senators and congressmen and other elected officials and party officials come out and basically echo what Portman and Toomey have said, which is, 'OK, now it's done,' " he said.
Terry Madonna, director of the Center for Politics and Public Affairs at Franklin and Marshall College, said Biden won Pennsylvania because he won big in the suburbs and that voting anomalies were minor.
"We're not naïve, we know that there was probably some fraud. But widespread? No," he said. "We're probably the most litigated state in the country largely because of the importance of our 20 electoral votes."
Gerald Seib, executive Washington editor of The Wall Street Journal, wrote on Monday that Trump's repeated attacks on the integrity of the election results may have the unintended effect of casting doubt on the legitimacy of Republican officeholders.
"For other Republicans, one of the costs associated with the president's continued attempt to overturn the results of the 2020 presidential election is that he is attacking the legitimacy of the very electoral system that brought all of them into office in the first place - and a system that actually has been quite good to Republicans at every level of government," Seib wrote.
Republicans outperformed expectations in Senate and House races this year - limiting Senate Democrats to a net pickup of only one seat on Election Day and picking up at least eight House seats, with several races yet to be called.
Some Republicans, however, argue there are voting irregularities that still need to be investigated.
"I don't think we have to rush to this. I think there are a lot of voters out there that are saying, 'OK, let's take a look at this. Let's see what happened,' " said Jim McLaughlin, a Republican pollster.
McLaughlin noted the Electoral College will meet on Dec. 14, giving Trump's legal team more time to make its arguments.
But as Trump's legal challenges fail to gain traction in court or significantly change states' vote totals, Republican lawmakers are getting ready to close the book on the 2020 presidential election.
Asked about the growing impatience of senators like Capito, Portman and Toomey, McLaughlin insisted the delay in conceding this year's election won't have any long-term political impact on Trump.
"I think more may come out and say Donald Trump should concede, but I honestly believe some things don't look right in this election and the president feels he owes it to his voters to take a look at it. We'll see what they come up with," he said.
Alex Gangitano contributed.
More in Administration
Trump has talked to associates about forming new political party: report
Howard University's marching band to escort Harris at inauguration

redpajama_set_name: RedPajamaCommonCrawl


We usually cook a pot of black-eyed peas and greens for New Year's but this year we decided to make black-eyed pea salsa instead. I had a bag of lady peas in the freezer and figured those would be perfect to use in a salsa. We enjoyed it so much that it took me less than a month to make it again.
I cooked the peas as I usually do with onions, carrots, a spoonful of brown sugar, black and red pepper and salt in a pot of boiling water until peas are soft. When slightly cooled, I removed 2 cups of the cooked peas from the pot, drained off the pot liquor and mixed with other salsa ingredients.
Black-Eyed Pea Dip can also be made using lady peas or other varieties of peas you have in the freezer from last summer's harvest.
Makes 2 ½ cups. Original recipe is from Judy Jumonville.
1. Mix everything together. Cover and chill for at least 4 hours.
2. Serve with crackers, tortilla chips and gluten-free rice crackers.
Testing note: I made the recipe using cooked lady peas. This is a good dip to serve with gluten-free crackers.


text:
\section{Introduction}\label{intro-sec}

In a foundational paper \cite{shannon-1948-bell} in information theory, Shannon introduced a model of a noisy communication channel, in which the input and output are modeled by stationary probability measures on a space of sequences of symbols. Shannon gave conditions under which the input can be recovered from the output, at least with an acceptable rate of error or ambiguity, in the case of a Bernoulli source, and this work has since been extended to more general sources \cite{kieffer-1981-wahr}.

This paper is motivated by the particular question of when one can ensure zero error, not just almost surely as in information theory but in fact deterministically. A deterministic channel can be modeled by a sliding block code, i.e. a continuous, shift-commuting map on a subshift, on which a stationary process could be supported. In this model, we can apply techniques of symbolic dynamics to investigate the effects of \textit{deterministic noise} \cite{mpw-82-sk}, also called \textit{distortion} \cite{shannon-1948-bell}, which we can interpret as a failure of injectivity of the sliding block code representing the channel, even in the absence of random errors.

The main result of this paper, \Cref{main-thm}, determines the extent to which the non-injectivity of a sliding block code on a mixing shift of finite type (SFT) can be avoided by restricting to a subshift of the domain. Interpreting the sliding block code as a channel with deterministic noise, \Cref{main-thm} characterizes the sources with entropy strictly lower than that of the output which can be transmitted without error or ambiguity.

\begin{thm}\label{main-thm}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $Z$ be a subshift with topological entropy strictly less than that of $Y$. Then there exists a subshift $Z'$ of $X$ conjugate to $Z$ such that $\pi|_{Z'}$ is injective, if and only if for every $n \geq 1$, the number of periodic points of least period $n$ in $Z$ is at most the number of periodic points of least period $n$ in $Y$ with a $\pi$-preimage of equal least period.
\end{thm}

\Cref{main-thm} is a generalization of the following theorem of Krieger in the case of unequal entropy; in particular, \Cref{main-thm} reduces to \Cref{krieger-thm} in the case that $Y=X$ and $\pi$ is the identity.

\begin{thm}[Theorem 2 in \cite{wk-82-etds}]\label{krieger-thm}
Let $Y$ be a mixing shift of finite type and $Z$ a subshift. Then there is a subshift $Z' \subseteq Y$ conjugate to $Z$ if and only if $Z$ and $Y$ are conjugate or the (topological) entropy of $Z$ is less than that of $Y$ and, for every $n \geq 1$, the number of periodic points of least period $n$ in $Z$ is at most the corresponding number in $Y$.
\end{thm}

We note that with $X,Y,Z,\pi$ as in the statement of \Cref{main-thm}, clearly there exists a subshift $Z'$ of $X$ conjugate to $Z$ such that $\pi|_{Z'}$ is injective if and only if there exists a sliding block code $\psi: Z \to X$ such that $\pi \circ \psi$ is injective, in which case $Z' = \psi(Z) \subset X$. To verify the ``only if'' statement in \Cref{main-thm}, suppose that there is a subshift $Z'$ of $X$ conjugate to $Z$ such that $\pi|_{Z'}$ is injective. Let $y \in \pi(Z')$ be periodic. Let $x = \pi|_{Z'}^{-1}(y)$ be the unique preimage of $y$ in $Z'$. Then the orbit of $x$ is in bijection with the orbit of $y$; otherwise, $\pi$ would fail to be injective on the orbit of $x$, which is contained in $Z'$. In particular, $x$ has finite orbit, so $x$ is periodic, moreover with $\mathrm{per}(x) = \mathrm{per}(y)$. Thus, every periodic point in $\pi(Z') \subset Y$ has a periodic preimage in $Z' \subset X$ of equal least period, which shows the necessity of the stated condition.

Both \Cref{main-thm} and \Cref{krieger-thm} give conditions for the existence of an embedding in terms of entropy and a periodic point condition. The following corollary, which we prove in \Cref{sec-conseq}, shows that the periodic point condition can be removed in exchange for a small loss of injectivity.

\begin{cor}\label{mmb-cor}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $Z$ be a subshift with topological entropy strictly less than that of $Y$. Then there exist a subshift $Z'$, a finite-to-one factor code $\chi: Z' \to Z$, and a sliding block code $\psi: Z' \to X$ such that $\pi \circ \chi$ is injective. Moreover, if $Z$ is mixing sofic with positive entropy (i.e. not a single fixed point), then $Z'$ can be taken to be a mixing SFT and $\chi$ can be taken to be almost invertible.
\end{cor}

The code $\chi$ is in fact injective except on points in $Z'$ whose images in $Z$ are backward-asymptotic to one of finitely many periodic points in $Z$. See \Cref{blowup} and \Cref{rmk-blowup-details}. From \Cref{mmb-cor}, we can immediately conclude the following, with $h$ denoting the topological entropy of a subshift.

\begin{cor}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. For any $\varepsilon > 0$, there exists a mixing SFT $Z \subset X$ with $h(Z) > h(Y) - \varepsilon$ such that $\pi|_Z$ is injective.
\end{cor}

The proof of \Cref{main-thm} adapts the strategy used to prove \Cref{krieger-thm} in \cite{wk-82-etds,lm-95-intro} and related results in \cite{mb-1984-etds}. The outline of the proof is as follows. We use a marker set, as in the proof of \Cref{krieger-thm}, to break points in $Z$ into moderate blocks and long periodic blocks, separated by marker coordinates. We code these separately using certain ``data blocks'' in $Y$, some of moderate length and some long and periodic, where the long periodic data blocks come from periodic points with $\pi$-preimages of equal least period in $X$. A block in $Z$ between marker coordinates is coded to a data block in $Y$ which is shorter by an additive constant, so that there are gaps between the data blocks, filled with repetitions of a ``blank'' symbol. We then lift the data blocks from $Y$ to data blocks from $X$, then replace the blanks with a ``stamp'' block from $X$ to form a valid point in $X$. The stamp block is chosen to ensure that once the point in $X$ is coded into $Y$ by $\pi$, the locations of the stamp, and thus of the marker coordinates, can be recognized. These manipulations of markers, blanks, and stamps are presented in detail in \Cref{sec-code}, while the quantitative arguments required to construct the data blocks and stamps are given in \Cref{sec-count}.

The statement of \Cref{krieger-thm} is false for $X$ merely mixing sofic, and to date there is no known characterization of the subshifts that embed into a given mixing sofic shift, though some sufficient conditions are known \cite{mb-1984-etds,thomsen-2004-sofic}. \Cref{main-thm} sheds some light on this problem, without resolving it. Salo-Törmä have answered \cite{vs-it-22-mo} the following related question: let $Y$ be a mixing sofic shift and $Z \subset Y$ a mixing SFT. For which such $Y,Z$ do there exist a mixing SFT extension $\pi: X \to Y$ and a (mixing SFT) $Z' \subset X$ such that $\pi|_{Z'}: Z' \to Z$ is a conjugacy? However, it is unclear how the conditions given in that answer compare to those in \Cref{main-thm}, or to the results given in \cite{thomsen-2004-sofic}. As a final related question, when $Y$ is an SFT and $Z$ is conjugate to $Y$, the existence of an SFT $Z' \subset X$ conjugate to $Z$ such that $\pi|_{Z'}: Z' \to Y$ is a conjugacy, i.e. is surjective as well as injective, has been studied in \cite{hmw-2022}, continuing work from \cite{mpw-82-sk}.

\section{Conventions, definitions, and background on symbolic dynamics}\label{conv-sec}

\subsection{Subshifts and sliding block codes}\label{ss-sbc}

Let $\mathcal{A}$ be a finite set with the discrete topology, which we will call an \textit{alphabet}. The set $\mathcal{A}^{\mathbb{Z}}$ of bi-infinite sequences over $\mathcal{A}$, equipped with the product topology, is called the \textit{full shift} over $\mathcal{A}$, so called because the shift action $\sigma: \mathcal{A}^{\mathbb{Z}} \to \mathcal{A}^{\mathbb{Z}}$, given by $(\sigma x)_i = x_{i+1}$, is a homeomorphism. A closed, shift-invariant subset of the full shift is called a \textit{subshift}. The topology on $\mathcal{A}^{\mathbb{Z}}$ is generated by \textit{cylinders}, which are sets of the form
\[
[w]_i := \{ x \in \mathcal{A}^{\mathbb{Z}} \, | \, x_{i+j} = w_{j}, \, 0 \leq j \leq n -1   \},
\]
where $w \in \mathcal{A}^n$ is a \textit{block} or \textit{word} of length $n \in \mathbb{N}$, and $i \in \mathbb{Z}$. Note that by shift-invariance, for any subshift $X \subset \mathcal{A}^{\mathbb{Z}}$ and any block $w \in \mathcal{A}^*$, we have $X \cap [w]_i \neq \emptyset$ for some $i \in \mathbb{Z}$ if and only if $X \cap [w]_i \neq \emptyset$ for all $i \in \mathbb{Z}$.

A subshift $X \subset \mathcal{A}^{\mathbb{Z}}$ is characterized by the set $\mathcal{B}(X)$ of blocks $w \in \mathcal{A}^*$ such that $X \cap [w] \neq \emptyset$, called the \textit{language} of $X$. When the intended subshift $X$ is clear, we write $[w]_i$ for $X \cap [w]_i$. We write $\mathcal{B}_n(X) = \mathcal{B}(X) \cap \mathcal{A}^n$. We can equivalently characterize a subshift by a set of forbidden words $\mathcal{F} \subset \mathcal{A}^*$, writing $\mathsf{X}_{\mathcal{F}} := \overline{\mathcal{A}^{\mathbb{Z}} \setminus \bigcup_{w \in \mathcal{F}} \bigcup_{i \in \mathbb{Z}} [w]_i  }$. Note that in general $\mathcal{F} \subsetneq \mathcal{A}^* \setminus \mathcal{B}(\mathsf{X}_{\mathcal{F}})$. For a given subshift $X \subset \mathcal{A}^{\mathbb{Z}}$, there may be several different sets of forbidden words $\mathcal{F} \subset \mathcal{A}^*$  such that $X = \mathsf{X}_{\mathcal{F}}$. A \textit{shift of finite type} (SFT) is a subshift $X$ such that $X = \mathsf{X}_{\mathcal{F}}$ for some finite set $\mathcal{F}$. A \textit{$k$-step} SFT over $\mathcal{A}$ is an SFT of the form $X = \mathsf{X}_{\mathcal{F}}$ for some set $\mathcal{F} \subset \mathcal{A}^{k+1}$.

It is a theorem (the Curtis-Hedlund-Lyndon theorem) that, for subshifts $X,Y$, a function $\phi: X \to Y$ is continuous and shift-equivariant if and only if it is a \textit{sliding block code}, which means that there exist $m,n \geq 0$ and $\Phi: \mathcal{B}_{m+n+1}(X) \to \mathcal{B}_1(Y)$ such that for every $x \in X$ and every $i \in \mathbb{Z}$, $\phi(x)_i = \Phi(x_{[i-m,i+n]})$. We say that $\phi$ is a $k$-block code if $m+n+1=k$. A \textit{factor code} is a surjective sliding block code, and for a sliding block code $\phi$ defined on a subshift $X$, we say that the image $\phi(X)$ is a \textit{factor} of $X$, and that $X$, or more properly $\phi: X \to \phi(X)$, is an \textit{extension} of $\phi(X)$. A \textit{sofic shift} (from the Hebrew \<swpy>, ``sofi'', meaning ``finite'') is any factor of a shift of finite type. An injective sliding block code is called an \textit{embedding}, and a bijective sliding block code is called a \textit{conjugacy}. The properties of being sofic and of finite type are both invariant under conjugacy.

A subshift $X$ is said to be \textit{irreducible} if for all $u, w \in \mathcal{B}(X)$, there exists $v \in \mathcal{B}(X)$ such that $uvw \in \mathcal{B}(X)$, and \textit{strongly irreducible} with gap $g \geq 1$ if, for any $u,w$, we can take always take $v \in \mathcal{B}_g(X)$. Any factor of an irreducible (resp. strongly irreducible) subshift is irreducible (resp. strongly irreducible). A \textit{periodic point} in a subshift $X$ is a point $x \in X$ with $x = \sigma^n x$ for some $n \geq 1$---we say that $x$ has period $n$. The \textit{least period} $\mathrm{per}(x)$ of a periodic point $x$ is the least $n$ such that $\sigma^n x = x$. Note that $| \{ \sigma^n x \, | \, n \in \mathbb{Z}  \} | = \mathrm{per}(x)$. We write $P(X)$ for the set of periodic points in a subshift $X$, $Q_n(X)$ for the set of periodic points of least period $n$, and $q_n(X) = |Q_n(X)|$. The number of periodic points of a given least period is a conjugacy invariant.

It is a theorem that periodic points are dense in any irreducible shift of finite type. The \textit{period} $\mathrm{per}(X)$ of an irreducible shift of finite type $X$ is the gcd of the periods of the periodic points of $X$. An irreducible SFT with period $1$ is said to be \textit{aperiodic}. An irreducible SFT is strongly irreducible if and only if it is aperiodic, if and only if has periodic points of all sufficiently high periods. For irreducible sofic shifts, strong irreducibility is equivalent to having periodic points of all sufficiently high periods, which clearly implies that the periods have gcd $1$, but the reverse implication fails. For example, consider the odd shift over $\{ 0,1 \}$, in which the block $10^n1$ is permitted only for odd $n$. This is an irreducible sofic shift which contains the fixed point $0^{\infty}$, so the periods of periodic points trivially have gcd $1$. However, the odd shift has no other periodic points of odd period. We follow the convention of the literature in referring to strongly irreducible sofic shifts (in particular SFTs) as mixing sofic shifts (mixing SFTs), because they are also characterized by a topological mixing property, but we will not use that property explicitly, so we do not define it here.

The following definition is new, and we use it extensively.

\begin{defn}
Let $X$ and $Y$ be subshifts and let $\pi: X \to Y$ be a factor code. We write $R_n(\pi)$ for the set of periodic points $y \in Y$ such that $y = \pi(x)$ for some periodic point $x \in X$ with $\mathrm{per}(x) = \mathrm{per}(y)$. We write $r_n(\pi) = |R_n(\pi)|$.
\end{defn}

For a subshift $X$, the (topological) \textit{entropy} of $X$ is the value $h(X) = \inf_{n \geq 1} \frac{1}{n} \log |\mathcal{B}_n(X)|$; in fact, the limit $\lim_{n \to \infty} \frac{1}{n} \log |\mathcal{B}_n(X)|$ exists and is equal to $h(X)$. For a mixing sofic shift (in particular, a mixing SFT) $X$, we also have $h(X) = \lim_{n \to \infty} \frac{1}{n} \log q_n(X)$. Entropy is non-increasing under factor codes and is thus a conjugacy invariant, though certainly not a complete invariant. For any irreducible sofic shift $X$, and any proper subshift $V \subset X$, we have $h(V) < h(X)$. In \Cref{sec-count}, we use the following lemma of Marcus, which allows us to approximate a sofic shift from the inside by SFTs in terms of entropy.

\begin{lemma}[Proposition 3 in \cite{bm-1985-encod}]\label{sofic-approx-inside-sft}
Let $Y$ be a sofic shift. For every $\varepsilon > 0$, there exists an irreducible SFT $U \subseteq Y$ with $h(U) > h(Y) - \varepsilon$.
\end{lemma}

For any subshift $X$ and any $k \geq 1$, we can form the \textit{$k$th higher block shift} $X^{[k]}$ with alphabet $\mathcal{B}_k(X)$, where
\[
w = (a_{1,1} a_{1,2} \dots a_{1,k}) (a_{2,1} a_{2,2} \dots a_{2,k}) \dots (a_{\ell,1} a_{\ell,2} \dots a_{\ell,k}) \in \mathcal{B}(X^{[k]})
\]
if any only if for each $i,j$ we have $a_{i,j} = a_{i+1,j-1}$, so that
\[
w = (a_1 a_2 \dots a_k) (a_2 a_3 \dots a_{k+1}) \dots (a_{\ell+1} a_{\ell+2} \dots a_{\ell+k}),
\]
and $a_1 a_2 \dots a_{k+\ell} \in \mathcal{B}(X)$. Observe that $X$ and $X^{[k]}$ are conjugate for any subshift $X$ and any $k \geq 1$. Moreover, if $X$ is an $m$-step SFT and $k \leq m-1$, then $X^{[k]}$ is an $(m-k)$-step SFT. In particular, every SFT is conjugate to a $1$-step SFT, and every sliding block code on an SFT can be written as a composition of a conjugacy and a $1$-block code. We will therefore frequently assume WLOG that a given sliding block code on an SFT is a $1$-block code on a $1$-step SFT.

For a sliding block code on an irreducible shift of finite type, either every fiber is a finite set (indeed, of bounded cardinality), in which case the code is said to be \textit{finite-to-one} and the entropy of the image is equal to that of the domain, or almost every fiber is uncountable, and the entropy of the image is strictly less than that of the domain. In the finite-to-one case, the minimum fiber cardinality is generic and is known as the \textit{degree}. In particular, a code (on an irreducible SFT) with degree $1$ is said to be \textit{almost invertible}. It is a theorem that every irreducible (resp. mixing) sofic shift is an almost invertible factor of an irreducible (resp. mixing) SFT. We use the following construction of almost invertible codes, known as the ``blowing-up lemma'', in the proof of \Cref{mmb-cor} in \Cref{sec-conseq}.

\begin{lemma}[Lemma 10.3.2, \cite{lm-95-intro}]\label{blowup}
Let $Z$ be a mixing SFT and let $z \in Z$ be a periodic point with least period $p$. Let $M \geq 1$. Then there exist a mixing SFT $Z'$ and an almost invertible factor code $\chi: Z' \to Z$ such that the preimage of the orbit of $z$ under $\chi$ is a single orbit of length $Mp$.
\end{lemma}

\begin{rmk}\label{rmk-blowup-details}
Note that in \cite{lm-95-intro}, the extension $\chi$ in \Cref{blowup} is only stated to be finite-to-one, but the existence of periodic points having unique preimage already implies almost invertibility. Indeed, the construction in \cite{lm-95-intro}, based on work in \cite{mb-1984-etds}, in fact shows that $\chi$ is injective except on the points that are backward-asymptotic to points in the preimage of the orbit of $z$, where we say that two points $z,z'$ are backward-asymptotic if $d(\sigma^n z, \sigma^n z') \to 0$ as $n \to - \infty$.
\end{rmk}

\subsection{Markers and Markov approximations}\label{ss-mark}

We now recall the constructions with markers and long periodic blocks that are at the heart of the proof of \Cref{main-thm}. For an alphabet $\mathcal{A}$, we say that a block $w = w_1 \dots w_n \in \mathcal{A}^n$ is $k$-periodic, or has self-overlap of $n-k$, if $w_{[k+1,n]} = w_{[1,n-k]}$, i.e. for $1 \leq i \leq n-k$ we have $w_i = w_{i+k}$. A given block may be $k$-periodic for several different $k$.

\begin{lemma}[Lemma 2.3 in \cite{mb-1984-etds}]\label{per-lemma}
Let $Z$ be a subshift, let $N \geq 1$, and $a, b \in \mathbb{Z}$ with $b - a \geq 2N$. Let $z \in Z$. If for every $i \in [a+N,b-N]$ there exists $p \leq N-1$ such that $z_{[i-N,i+N]}$ is $p$-periodic, then there is at most one periodic point $\zeta \in Z$ with $\mathrm{per}(\zeta) \leq N-1$ and $\zeta_{[a,b]} = z_{[a,b]}$. If $Z$ is a $1$-step SFT, then such a $\zeta$ exists.
\end{lemma}

\begin{lemma}[Lemma 2 in \cite{wk-82-etds}]\label{marker-lemma}
Let $Z$ be a subshift. For any $N \geq 1$, there exists a subset $F \subset Z$, which can be taken to be a finite union of cylinders, such that:
\begin{enumerate}
    \item the sets $\sigma^i F$, $0 \leq i \leq N-1$, are all disjoint, and

    \item if $z \notin \sigma^i F$ for all $-(N-1) \leq i \leq (N-1)$, then $z_{[-N,N]}$ is $p$-periodic for some $p \leq N-1$.
\end{enumerate}
\end{lemma}

For any subshift $X$ and any $n \geq 1$, we can form the $n$th Markov approximation $X_n$, which is the SFT defined by allowing precisely the blocks of length $n$ which appear in $X$. Clearly $X_{n+1} \subset X_n$. It is an exercise to show that for any $\varepsilon > 0$ and any $N \geq 1$, there exists $N' \geq N$ such that $h(X_{N'}) < h(X) + \varepsilon$ and $q_n(X_{N'}) = q_n(X)$ for all $n \leq N$. In \Cref{wlog-sft}, we use the Markov approximation, together with higher block shifts, to show that in the proof of \Cref{main-thm}, we can assume WLOG that $Z$ is a $1$-step SFT, which allows us to apply \Cref{per-lemma}.

We remark that there are versions of \Cref{marker-lemma} which obviate the need for \Cref{per-lemma}. However, for our purposes in this paper, embedding $Z$ into an SFT has the additional benefit that the rate of convergence of $\frac{1}{n} \log q_n(Z)$ to $h(Z)$ can be easily estimated when $Z$ is an SFT (see e.g. \cite{lm-95-intro}, pp. 349-351), which gives a procedure for deciding whether a given $X,Y,\pi,Z$ satisfy the periodic point condition in \Cref{main-thm}, assuming that $h(Z) < h(Y)$ (namely, compute $N \geq 1$ such that for all $n \geq N$, $q_n(Z) < r_n(\pi)$, then check all $n \leq N$ to determine whether $q_n(Z) \leq r_n(\pi)$).

\begin{lemma}\label{wlog-sft}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $Z$ be a subshift with $h(Z) < h(Y)$ and $q_n(Z) \leq r_n(\pi)$ for all $n \geq 1$. Then there exists a $1$-step SFT $Z'$ such that $Z$ embeds into $Z'$, $h(Z') < h(Y)$, and $q_n(Z') \leq r_n(\pi)$ for all $n \geq 1$.
\end{lemma}

We defer the proof of \Cref{wlog-sft} to \Cref{sec-conseq}.

\section{Coding}\label{sec-code}

In \Cref{ss-blank-stamp}, we introduce two coding constructions, namely subshifts with blanks adjoined, \Cref{blanks-def}, and stamps, \Cref{stamp-def}, then use them to create one side of an interface between $Z$ on the one hand and $\pi: X \to Y$ on the other. In \Cref{ss-marker-blank}, we use markers in $Z$ to construct the other side of this interface. In \Cref{ss-stamp-sft}, we use stamps to give a construction of SFTs analogous to $S$-gap shifts. We use this construction in \Cref{ss-good-per} to construct the shifts that are used in \Cref{ss-blank-stamp} and \Cref{ss-marker-blank}.

\subsection{Blanks and stamps}\label{ss-blank-stamp}

As outlined in \Cref{intro-sec}, the proof of \Cref{main-thm} involves coding $Z$ into $X$ via certain intermediate subshifts which consist of long ``data'' blocks separated by blanks. We now define this construction precisely.

\begin{defn}[subshift with blanks adjoined]\label{blanks-def}
Let $W$ be a subshift and let $N, \ell \geq 1$ with $\ell < N$. Let $*$ be a symbol not appearing in the alphabet of $W$. Let $M \subset \bigcup_{n=1}^{2N} \mathcal{B}(W)$ be a set of blocks and let $Q \subset \cup_{n=1}^{2N-1} Q_n(W)$ be a set of periodic points. Denote by $\mathrm{Blanks}(M,Q,N,*,\ell)$ the subshift in which each point is of the form $\dots w_{-1} *^{\ell} w_0 *^{\ell} w_1 \dots$ where either $w_i \in M$ or $w_i = y_T$ where $y \in Q$ and $T=(-\infty,0], [0, +\infty), (-\infty, \infty)$, or $[0,m]$ with $m \geq 2N$.
\end{defn}

The purpose of the $\mathrm{Blanks}$ construction is to provide an interface between the channel $\pi: X \to Y$ and the subshift $Z$ to be embedded. One side of this interface, namely the embedding of a $\mathrm{Blanks}$ subshift into $X$, is specified in \Cref{blanks-embed-channel}. The construction involves particular blocks, which we call \textit{stamps}, that can be unambiguously recognized in the following sense:

\begin{defn}[stamp]\label{stamp-def}
Let $Y$ be a subshift, $W \subset Y$ a proper subshift, and $k \geq 1$. We say that $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$ is a $(Y,W,k)$ \textit{stamp} if for all $u_1, u_2 \in \mathcal{B}(W)$ and $v_1, v_2 \in \mathcal{B}_k(Y)$, $\mu$ appears exactly once in $u_1 v_1 \mu v_2 u_2$.
\end{defn}

\begin{rmk}
In \Cref{stamp-def}, continuing with the notation there, we do not explicitly require $u_1 v_1 \mu v_2 u_2$ to be legal in $X$. Doing so would neither affect the results nor simplify the proofs. In all of the examples we consider, such blocks will in fact be legal in $X$.
\end{rmk}

\begin{prop}\label{stamp-exist}
Let $Y$ be a strongly irreducible subshift with gap $g$ and $W \subset Y$ a proper subshift. For every $k \geq g$ and every sufficiently large $n$, there exists a $(Y,W,k)$ stamp of length $n$.
\end{prop}

We defer the proof of \Cref{stamp-exist} to \Cref{ss-overlap}, but before applying stamps in \Cref{blanks-embed-channel}, we prove a lemma that expresses how stamps are actually used in our constructions.

\begin{lemma}\label{stamp-usage}
Let $Y$ be a subshift, $W \subset Y$ a proper subshift, $k \geq 1$, and $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$ a $(Y,W,k)$ stamp. Let $N \geq |\mu|$. Then for any $\gamma^{\pm} \in \mathcal{B}_k(Y)$, and any $w \in \mathcal{B}(W)$ with $|w| \geq N$, the stamp $\mu$ appears exactly twice in the block $\mu \gamma^- w \gamma^+ \mu$.
\end{lemma}

\begin{proof}
By the hypotheses on $\mu$, $\gamma^{\pm}$, and $w$, and \Cref{stamp-def}, $\mu$ appears exactly once in each subblock $\mu \gamma^- w$, $w \gamma^+ \mu$. An appearance of $\mu$ other than at the positions explicitly indicated must therefore overlap both of these subblocks. Since $|w| \geq |\mu|$, $\mu$ must therefore be a subblock of $w$, contradicting the hypothesis that $w \in \mathcal{B}(W)$ and $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$.
\end{proof}

We now give one of the main coding constructions (\Cref{blanks-embed-channel}), embedding a subshift with blanks adjoined, and with blocks from a subshift $V \subset X$, into $X$ via a sliding block code $\gamma$, such that $\pi \circ \gamma$ is injective. The large amount of data in the statement is representative of the complexity of the construction and the modular nature of the proof.

\begin{prop}\label{blanks-embed-channel}
Let $X$ be a mixing SFT with gap $g$, let $Y$ be a mixing sofic shift, and let $\pi: X \to Y$ be a $1$-block factor code.

Let $V \subset X$, $W = \pi(V) \subset Y$ be proper subshifts. Let $*$ be a symbol not appearing in the alphabets of $X,Y$. Let $N \geq 1$. Let $M \subset \bigcup_{n=1}^{2N-1} \mathcal{B}_n(W)$ be a collection of blocks, and let $R \subset \bigcup_{n=1}^{N-1} R_n(\pi|_V)$ be a union of finite (i.e. periodic) orbits in $W$ with $\pi$-preimages of equal cardinality in $V$. Let $\kappa: M \to \mathcal{B}(V)$ be an injection such that $\pi \circ \kappa(w) = w$ for each $w \in M$, and let $\Hat{M} = \kappa(M)$. Similarly, let $\lambda: R \to P(V)$ be a shift-commuting injection such that $\pi \circ \lambda(y) = y$ for each $y \in R$, and let $\Hat{R} = \lambda(R)$. Then for any $\ell \geq 1$, $\mathrm{Blanks}(M,R,N,*,\ell)$ and $\mathrm{Blanks}(\Hat{M}, \Hat{R}, N,*,\ell)$ are conjugate.

Moreover, let $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$ be a $(Y,W,g)$ stamp such that $|\mu| \leq N$, and suppose that $M \subset \bigcup_{n=N}^{2N-1} \mathcal{B}_n(W)$, i.e. $M$ contains no blocks of length less than $N$. Then there exists a sliding block code $\gamma: \mathrm{Blanks}(\Hat{M}, \Hat{R}, N,*, |\mu| + 2g) \to X$ such that $\pi \circ \gamma$ is injective.
\end{prop}

\begin{proof}
First, the conjugacy. Let $W[*] = \mathrm{Blanks}(M,R,N,*,\ell)$ and $V[*] = \mathrm{Blanks}(\Hat{M},\Hat{R},N,*,\ell)$. Consider the $1$-block code $\pi[*]$ defined on $V[*]$ by the block map $\pi[*](a) = \pi(a)$ for $a$ in the alphabet of $V$ and $\pi[*](*) = *$. We claim that $W[*] = \pi[*](V[*])$ and that $\pi[*]: V[*] \to W[*]$ is a conjugacy. To see that $W[*] \subseteq \pi[*](V[*])$, note that any $\xi \in V[*]$ is of the form
\[
\xi = \dots w_{-1} *^{\ell} w_0 *^{\ell} w_1 \dots
\]
where either $w_i \in \Hat{M}$ or $w_i = x_T$ for some $x \in \Hat{R}$ and $T$ an interval with $2N+1 \leq |T|$. If $w_i \in \Hat{M}$, then $\pi(w_i) \in M$; if $w_i = x_T$ for some $x \in \Hat{R}$, then $\pi(w_i) = \pi(x)_T$, and $\pi(x) \in R$. Therefore
\[
\pi[*](\xi) = \dots \pi(w_{-1}) *^{\ell} \pi(w_0) *^{\ell} \pi(w_1) \dots \in W[*]
\]
This shows that indeed $W[*] \subset \pi[*](V[*])$. Similarly, any $\eta \in W[*]$ is of the form
\[
\eta = \dots w_{-1} *^{\ell} w_0 *^{\ell} w_1 \dots
\]
where either $w_i \in M$ or $w_i = y_T$ for some $y \in R$ and $T$ an interval with $2N+1 \leq |T| \leq \infty$. For $\eta$ of this form, using \Cref{per-lemma}, we can use the injections $\kappa$, $\lambda$ to reconstruct a unique $\xi \in V[*]$ such that $\pi[*](\xi) = \eta$.

We now suppose that each block in $M$ has length at least $N$ and that we have a $(Y,W,g)$ stamp $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$ such that $|\mu| \leq N$. Under these assumptions, we construct a sliding block code $\gamma: V[*] \to X$ and show that $\pi \circ \gamma$ is injective. Fix a $\pi$-preimage $\Hat{\mu}$ of $\mu$, and let $\ell = |\mu| + 2g$. Using the hypothesis that $X$ is a mixing $1$-step SFT, define maps $\gamma^{\pm}: \mathcal{B}_1(V) \to \mathcal{B}_g(X)$ such that, for $a, b \in \mathcal{B}_1(V)$, we have $\Hat{\mu} \gamma^{-}(a) a, b \gamma^+(b) \Hat{\mu} \in \mathcal{B}(X)$. We then have a sliding block code $\gamma: V[*] \to X$, given by replacing each block $b *^{\ell} a$ by $b \gamma^+(b) \Hat{\mu} \gamma^-(a) a$, and leaving the non-blank symbols unchanged.

Let
\[
\xi = \dots *^{\ell} v_{-1} *^{\ell} v_0 *^{\ell} v_1 *^{\ell} \dots \in V[*]
\]
Then
\[
\gamma(\xi) = \dots \Hat{\mu} \gamma^-(a_0) v_0 \gamma^+(b_0) \Hat{\mu} \dots
\]
where $a_i, b_i$ are, respectively, the initial and terminal symbols of $v_i$. In turn, we have
\[
\pi \circ \gamma(\xi) = \dots \mu (\pi \circ \gamma^-(a_0)) \pi(v_0) (\pi \circ \gamma^+(b_0)) \mu \dots
\]
Moreover, by \Cref{stamp-usage} and the lower bound on lengths of blocks in $M$, it follows that $\mu$ appears in $\pi \circ \gamma(\xi)$ only where $\Hat{\mu}$ appears at the same position in $\gamma(\xi)$. By replacing, in $\pi \circ \gamma(\xi)$, each appearance of $\mu$, and the blocks of length $k$ to the left and right of $\mu$, with $*^{\ell}$, we obtain the point $\dots *^{\ell} \pi(v_0) *^{\ell} \dots = \pi[*](\xi) \in \mathrm{Blanks}(M,R,N,*,\ell)$, from which $\xi$ can be recovered since $\pi[*]$ is a conjugacy.
\end{proof}

\subsection{Blanks and markers}\label{ss-marker-blank}

We now prove a lemma that encapsulates the use of marker constructions in our proof of \Cref{main-thm}.

\begin{lemma}\label{marker-black-box}
Let $Z,W$ be subshifts with $Z$ a $1$-step SFT. Let $N,\ell \geq 1$ be such that $q_n(Z) \leq q_n(W)$ for $n \leq N-1$ and $|\mathcal{B}_n(Z)| \leq |\mathcal{B}_{n-\ell}(W)|$ for $N+\ell \leq n \leq 2N+\ell-1$. Let $M \subset \bigcup_{n=N}^{2N-1} \mathcal{B}_{n}(W)$ and $Q \subset \bigcup_{n=1}^{N-1} Q_n(W)$ be a union of finite (i.e. periodic) orbits such that $|\mathcal{B}_n(Z)| \leq |M \cap \mathcal{B}_{n-\ell}(W)|$ for $N+\ell \leq n \leq 2N+\ell-1$, and $q_n(Z) \leq |Q \cap Q_n(W)|$ for $n \leq N-1$. Then $Z$ embeds into $\mathrm{Blanks}(M,Q,N,*,\ell)$.
\end{lemma}

\begin{rmk}
The lower bound on the length of blocks in $M$ is not in fact needed for \Cref{marker-black-box}, but it is needed in order to apply \Cref{marker-black-box} in conjunction with \Cref{blanks-embed-channel} in the proof of \Cref{main-thm} below.
\end{rmk}

\begin{proof}
Let $F$ be a marker set for $Z$ with parameter $N$. For $z \in Z$, let $A(z) = \{ i \in \mathbb{Z}  \, | \, \sigma^i z \in F  \}$. Enumerate each $A(z)$ as $\{ a_j(z) \}_{j \in J(z)}$ where the index set $J(z)$ may be the empty set, or a finite set, or the integers, or the positive or negative natural numbers, and where $a_j(z) < a_{j+1}(z)$ for each $j$. We refer to the elements of $A(z)$ as marker coordinates for $z$. Say that $T$ is a \textit{marker interval} for $z$ if: $T = [a_j(z), a_{j+1}(z))$ where $a_j(z), a_{j+1}(z)$ are both defined; or $T=[a_0(z), \infty)$ if $a_0(z) = \max A(z) < \infty$; or $T=(-\infty, a_0(z)]$ if $a_0(z) = \min A(z) > -\infty$; or $T=(-\infty, \infty)$ if $A(z) = \emptyset$.

We construct an embedding of $Z$ into $\mathrm{Blanks}(M,Q,N,*,\ell)$ by constructing a function $\Phi$ that maps a block occurring between marker coordinates to a data block padded with $*^{\ell}$. Let $c_n: Q_n(Z) \to Q_n(W)$ be shift-commuting injections for $n \leq N-1$. Let $d_n: \mathcal{B}_n(Z) \to \mathcal{B}_{n-\ell}(W)$ be injections for $N+\ell \leq n \leq 2N+\ell-1$. For a block $w \in \mathcal{B}_n(Z)$ with $N+\ell \leq n \leq 2N+\ell-1$, let $\Phi(w) = *^{\ell} d_n(w)$. For $z \in Z$ periodic with $n = \mathrm{per}(z) \leq N-1$, if $m \geq 2N+\ell$, let $\Phi(z_{[0,m]}) = *^{\ell} c_n(z)_{[\ell,m]}$. Similarly, let $\Phi(z_{[0,\infty)}) = *^{\ell} c_n(z)_{[\ell,\infty)}$. Finally, let $\Phi(z_{(-\infty,0]}) = c_n(z)_{(-\infty,0]}$ and let $\Phi(z) = c_n(z)$. Observe that $\Phi$ is injective, by \Cref{per-lemma}.

Define $\phi: Z \to W$ by declaring that $\phi(z)_T = \Phi(z_T)$ whenever $T$ is a marker interval for $z$. We need to show that $\phi$ is an embedding. Certainly $\phi$ is shift-commuting, since, if $T$ is a marker interval for $z$, then $T-1$ is a marker interval for $\sigma z$, so
\[
\phi(\sigma z)_{T-1} = \Phi((\sigma z)_{T-1}) = \Phi(z_T) = \phi(z)_T = (\sigma \phi(z))_{T-1}
\]
Thus indeed $\phi(\sigma z) = \sigma \phi(z)$. Moreover, $\phi$ is injective because the appearances of $\beta$ in $\phi(z)$ allow us to reconstruct the marker coordinates, and then the injectivity of $\Phi$ allows us to reconstruct $z_T$ for each marker interval $T$ for $z$.

We need to show finally that $\phi$ is continuous, i.e. that for $z \in Z$, $\phi(z)_0$ depends only on $z_{[-L,L]}$ for some finite $L$ independent of $z$. To see this, let $L'$ be such that $F$ is a union of cylinders on $[-L',L']$. Let $L = L'+2N$. By examining $z_{[-L,L]}$, we can determine whether there are marker coordinates for $z$ in $[-2N,0)$ and/or $[0,2N]$. If each of these intervals contains a marker coordinate, then $\phi(z)_0$ is determined by $z_T$ where $T \subset [-2N,2N]$ is the unique marker interval for $z$ containing $0$. If at least one of $[-2N,0), [0,2N]$ has no marker coordinates, then $0$ is in a long marker interval for $z$. If there is a marker coordinate in $(-\ell,0]$, then $\phi(z)_0 = *$. Otherwise, by \Cref{per-lemma}, $\phi(z)_0$ is determined by any subblock $z_{[a,b]}$ where $a <  0 \leq b$, $b-a \geq 2N$, and $[a,b]$ contains no marker coordinate for $z$. This concludes the proof that $\phi$ is continuous.
\end{proof}

The remainder of the proof of \Cref{main-thm} follows the following proposition, the proof of which is taken up in \Cref{sec-count}.

\begin{prop}\label{custom-W-exists}
Let $X$ be a mixing SFT with gap $g$, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $Z$ be a subshift with $h(Z) < h(Y)$ and $q_n(Z) \leq r_n(\pi)$ for every $n \geq 1$. Then there exist: $N \geq 1$, subshifts $V \subset X$, $W = \pi(V) \subset Y$, and a $(Y,W,g)$ stamp $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$, such that $|\mu| \leq N$, $q_n(Z) \leq r_n(\pi|_V)$ for $n \leq N-1$ and $|\mathcal{B}_n(Z)| \leq |\mathcal{B}_{n-\ell}(W)|$ for $N+\ell \leq n \leq 2N+\ell-1$, where $\ell = |\mu| + 2g$.
\end{prop}

\begin{proof}[Proof of \Cref{main-thm}]
By \Cref{wlog-sft}, assume WLOG that $Z$ is a $1$-step SFT. Let $N$, $\ell$, $V \subset X$, $W = \pi(V) \subset Y$, and $\mu$ be as in \Cref{custom-W-exists}. Let $M \subset \bigcup_{n=N}^{2N-1} \mathcal{B}_{n}(W)$ be as in \Cref{marker-black-box}, and let $R \subset \bigcup_{n=1}^{N-1} R_n(\pi|_V)$ be a union of finite orbits, such that $q_n(Z) \leq |R \cap R_n(\pi|_V)|$ for $n \leq N-1$. Each of which the orbits in $R$ is, by the definition of $R_n$, necessarily the image of an orbit with equal cardinality in $V$. Here, $R$ takes the role that $Q$ plays in \Cref{marker-black-box}, but in \Cref{marker-black-box}, there was no channel $\pi$, and thus no preimage requirement, hence the change in notation. By \Cref{marker-black-box}, let $\phi: Z \to \mathrm{Blanks}(M,R,N,*,\ell)$ be an embedding.

Let $\Hat{M}$, $\Hat{R}$ be as in \Cref{blanks-embed-channel}, let $\pi[*] : \mathrm{Blanks}(\Hat{M},\Hat{R},N,*,\ell) \to \mathrm{Blanks}(M,R,N,*,\ell)$ be a conjugacy, and let $\gamma: \mathrm{Blanks}(\Hat{M},\Hat{R},N,*,\ell) \to X$ be an embedding such that $\pi \circ \gamma$ is injective (by \Cref{blanks-embed-channel}, using $\mu$).  Then $\psi = \gamma \circ (\pi[*])^{-1} \circ \phi: Z \to X$ is a sliding block code such that $\pi \circ \psi$ is injective.
\end{proof}

\subsection{Stamps and SFTs}\label{ss-stamp-sft}

In this subsection, we prove \Cref{enlarge-mixing}, which, in conjunction with \Cref{sofic-approx-inside-sft}, allows us, in \Cref{quant-summary}, to construct a mixing SFT $V \subset X$ such that the image $\pi(V) \subset Y$ is a proper subshift of $Y$ but has entropy at least $h(Y) - \varepsilon$ for a given $\varepsilon > 0$. It may be possible to give a more efficient construction of such a $V$, but we have not found one. We first prove \Cref{gap-sft}, which is related to the characterization of SFTs among $S$-gap shifts (Theorem 3.3 in \cite{dad-sj-2012}).

\begin{lemma}\label{gap-sft}
Let $X$ be a mixing SFT with gap $g$ and let $V_0 \subset X$ be an SFT. Let $k \geq g$ and let $\mu \in \mathcal{B}(X) \setminus \mathcal{B}(V_0)$ be an $(X,V_0,k)$ stamp. Let $N \geq |\mu|$ and let $V_1 \subset X$ be the closure of the set of points of the form
\[
\dots v_{-1} \gamma^+_{-1} \mu \gamma^-_0 v_0 \gamma^+_0 \mu \gamma^-_1 v_1 \dots \in X
\]
where, for each $i$, $\gamma^{\pm}_i \in \mathcal{B}_k(X)$ and $v_i \in \mathcal{B}(V_0)$ with $|v_i| \geq N$. Then $V_1$ is a mixing SFT.
\end{lemma}

\begin{proof}
Assume without loss of generality that $X$ is a $1$-step SFT. We first perform a small recoding for convenience, specifically to make it easier to recognize stamps, by replacing $X$ by a conjugate shift $\hat{X}$. For each $x \in X$, define $\hat{x}$ as follows: if $x_{[i,i+|\mu|)} = \mu$, then for each $i \in [-k, |\mu|+k)$, let $a = x_i$ and let $\hat{x}_i = \hat{a}$, where for symbols $a,b$ in the alphabet of $X$, we have $\hat{a} = \hat{b}$ if and only if $a=b$, and the set of symbols with hats is disjoint from the alphabet of $X$. If there is no $j \in (i-(|\mu|+k), i+k]$ with $x_{[j,j+|\mu|)} = \mu$, then let $\hat{x}_i = x_i$. Clearly the map $x \mapsto \hat{x}$ is a sliding block code, and it is just as clearly injective, since we recover $x$ from $\hat{x}$ by dropping hats. Therefore $\hat{X} = \{ \hat{x} \, | \, x \in X \}$ is a mixing SFT, conjugate to $X$.

Denote by $\Hat{V}_1 \subset \Hat{X}$ the image of $V_1$ under the map $x \mapsto \hat{x}$. Let $\ell = |\mu| + 2k$. Since $\mu$ is an $(X,V_0,k)$ stamp, and $N \geq |\mu|$, blocks of the form $\gamma^+_i \mu \gamma^-_{i+1}$ do not overlap in any point in $V$ by \Cref{stamp-usage}, so symbols with hats occur in $\Hat{V}_1$ in blocks of length exactly $\ell$. The blocks of symbols with hats are separated by blocks from $V_0$. Since $\Hat{V}_1$ is the image of $V$ under a conjugacy $X \to \Hat{X}$, $V_1$ is an SFT if and only if $\Hat{V}_1$ is an SFT.

Let $m \geq N$ be such that $\hat{X}$ and $V_0$ are $m$-step SFTs. We claim that if $\Hat{x} \in \hat{X}$ is such that $x_{[i,i+m]} \in \mathcal{B}_{m+1}(\Hat{V}_1)$ for all $i \in \mathbb{Z}$, then $\hat{x} \in \Hat{V}_1$, which means precisely that $V$ is an $m$-step SFT. To prove this claim, let $\mathcal{F} \subset \mathcal{B}_{m+1}(X)$ be the set of blocks of length $m+1$ which contain at least one of the following: a block of length greater than $\ell$ in which all symbols have hats; a block without hats that is not in $\mathcal{B}(V_0)$; or a block of symbols without hats, of length less than $N$, bounded on both sides by symbols with hats. Note that $\mathcal{F}$ is disjoint from $\mathcal{B}_{m+1}(\Hat{V}_1)$. Suppose that $\hat{x}_{[i,i+m]} \notin \mathcal{F}$ for all $i \in \mathbb{Z}$. Then any block of symbols with hats in $\hat{x}$ has length exactly $\ell$, and is thus of the form $\gamma^+ \mu \gamma^-$, where $\gamma^{\pm} \in \mathcal{B}_g(X)$ (with hats added). Furthermore, the blocks separating the blocks with hats must have length at least $N$ and must be in $\mathcal{B}(V_0)$, since every subblock of length $m+1$ is in $\mathcal{B}(V_0)$ and $V_0$ is an $m$-step SFT. Thus indeed $\hat{x} \in \hat{V}_1$, so $\Hat{V}_1$ is indeed an SFT.

To see that $V_1$ is irreducible, let $u_-, u_+ \in \mathcal{B}(V_1)$. We need to construct $u_0 \in \mathcal{B}(V)$ such that $u_- u_0 u_+ \in \mathcal{B}(V_1)$. We do so as follows. Extend $u_-$ on the right to form a block $v_- \in \mathcal{B}(V_1)$, which begins with $u_-$ and ends with $\gamma^+_{-1} \mu \gamma^-_0$ where $\gamma^+_{-1}, \gamma^-_0 \in \mathcal{B}_k(V_1)$. (It is possible that $u_-$ overlaps $\gamma^+_{-1} \mu \gamma^-_0$.) Let $v_0 \in \mathcal{B}_N(V_0)$ be such that $v_- v_0 \in \mathcal{B}(X)$. Similarly, extend $u_+$ on the left to form a block $v_+ \in \mathcal{B}(V_1)$ which ends with $u_+$ and begins with $\gamma^+_0 \mu \gamma^-_1$, where $\gamma^+_0, \gamma^-_1 \in \mathcal{B}_k(V_1)$ and $v_0 \gamma^+_0 \in \mathcal{B}(X)$. Let $x^{\pm} \in \mathcal{B}(V_1)$ be such that $x^-_{[0,\infty)}$ begins with $v_-$ and $x^+_{(-\infty,-1]}$ ends with $v^+$. Let $x = x^-_{(-\infty,-1]} v_- v_0 v_+ x^+_{[0,\infty)}$. Then $x \in X$ since $X$ is a $1$-step SFT. Moreover, $x \in V_1$, since the tails $x^-_{(-\infty,-1]} v_-$ and $v_+ x^+_{[0,\infty)}$ appear in $V_1$ and are joined together in a way that creates no violations of the restrictions defining $V_1$. Letting $u_0$ be the block appearing between $u_-, u_+$, such that $v_- v_0 v_+ = u_- u_0 u_+ \in \mathcal{B}(V)$, the construction is complete, showing that $V_1$ is indeed irreducible.

To see that $V_1$ is mixing, let $u_1, u_2 \in \mathcal{B}(V_0)$ with $|u_1| > m$, where $m$ is as above, and $|u_2| = |u_1| + 1$. Let $\gamma^{\pm}_i \in \mathcal{B}(X)$, $i = 1,2$, be such that $u_i \gamma^+_i \mu \gamma^-_i u_i \in \mathcal{B}(X)$. Then $x_i = (u_i \gamma^+_i \mu \gamma^-_i)^{\infty} \in V_1$ for both $i=1,2$. Indeed, certainly $x_i \in X$, since $u_i \gamma^+_i \mu \gamma^-_i u_i \in \mathcal{B}(X)$ and $X$ is a $1$-step SFT. Moreover, $\mathrm{per}(x_i)$ divides $\ell + |u_i|$, and $\gcd(\ell + |u_1|, \ell + |u_2|) = \gcd(\ell + |u_1|, \ell + |u_1| + 1) = 1$, so $\gcd( \mathrm{per}(x_1), \mathrm{per}(x_2) ) = 1$. Since $V_1$ is an irreducible SFT with periodic points of coprime periods, $V_1$ is mixing.
\end{proof}

As advertised, we now use \Cref{gap-sft} to prove the following lemma, which is applied in the proof of \Cref{quant-summary}, which in turn is the main input to the proof of \Cref{custom-W-exists}.

\begin{lemma}\label{enlarge-mixing}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $W_0 \subsetneq Y$ be an SFT. Then there exists a mixing SFT $V_1 \subset X$ with $W_0 \subset \pi(V_1) \subsetneq Y$.
\end{lemma}

\begin{proof}
Let $V_0 = \pi^{-1}(W_0) \subset X$. Note that $V_0$ is an SFT since $W_0$ is an SFT. Let $g$ be the mixing gap of $X$. Let $y \in Y \setminus W_0$ be a periodic point with least period $k \geq g$. Such a $y$ certainly exists because periodic points are dense in $Y$ and $W_0$ is a proper subshift. Let $k'$ be such that $y_{[0,k')} \notin \mathcal{B}_{k'}(W_0)$. Let $\ell = k+k'$. Then every $\ell$-block in $y$ is forbidden in $W_0$. In particular, for any $x \in \pi^{-1}(\{ y \})$ and any $i \in \mathbb{Z}$, we have $x_{[i,i+\ell)} \notin \mathcal{B}(V_0)$.

By \Cref{stamp-exist}, let $\mu$ be an $(X,V_0,g)$ stamp. Let $V_1$ consist of the closure of the set of points of the form $\dots v_{-1} \gamma^+_{-1} \mu \gamma^-_0 v_0 \gamma^+_0 \mu \gamma^-_1 v_1 \dots \in X$ where each $v_i \in \mathcal{B}(V_0)$ with $|v_i| \geq \ell$ and each $\gamma^{\pm}_i \in \mathcal{B}_g(X)$. By \Cref{gap-sft}, $V_1$ is indeed a mixing SFT. Note that every point in $V_1$ contains $\ell$-blocks permitted in $V_0$, so $V_1$ is disjoint from $\pi^{-1}(\{ y \})$, and therefore $\pi(V_1) \subsetneq Y$.
\end{proof}

\section{Counting}\label{sec-count}

In this section, we prove \Cref{stamp-exist} and \Cref{custom-W-exists}, which state the existence and properties respectively of the stamps and the shifts $V \subset X, W \subset Y$ used in \Cref{sec-code}. \Cref{ss-overlap} contains two results required for the proof of \Cref{stamp-exist}, one (\Cref{self-overlap}) showing that most blocks in a subshift with positive entropy have little self-overlap, and the other (\Cref{synd}) showing that one can assume, at the cost of a small loss of entropy, that a given sufficiently long block appears syndetically in a mixing sofic shift. \Cref{ss-good-per} then gives a crucial asymptotic result on the number of periodic points in $Y$ with a preimage of equal least period in $X$, and applies the results from \Cref{ss-stamp-sft} to construct the shifts $V$ and $W$.

\subsection{Self-overlap and stamps}\label{ss-overlap}

We begin by showing that most blocks have very little self-overlap, which we use both to construct stamps and to determine the asymptotic number of periodic points in $Y$ with a $\pi$-preimage of equal least period.

\begin{lemma}\label{self-overlap}
Let $Y$ be a subshift with $h(Y) > 0$. For every $\alpha \in (0,1)$, there exist $N \geq 1$ and $b > 0$ such that for every $n \geq N$, there are at least $(1-\exp(-bn)) \exp(n h(Y))$ blocks $w \in \mathcal{B}_n(Y)$ with no self-overlap of more than $\alpha n$.
\end{lemma}

\begin{proof}
Let $\varepsilon = \frac{1}{2} (\alpha^{-1} -1 )h(Y)$, so that $\alpha (h(Y) + \varepsilon) < h(Y)$. Let $r = \exp(h(Y))$ and $s = \exp(h(Y) + \varepsilon)$. Note that $s^{\alpha} < r < s$ and that $r^n \leq |\mathcal{B}_n(Y)|$ for every $n$. Let $N_0$ be large enough that for all $n \geq N_0$, we have $|\mathcal{B}_n(Y)| \leq s^n$. Let $C_1 = \sum_{k=1}^{N_0 -1} |\mathcal{B}_k(Y)|$. Then the number of blocks in $X$ of length $n$ with self-overlap of more than $\alpha n$ is at most
\begin{align*}
    \sum_{k=1}^{\lceil \alpha n \rceil } |\mathcal{B}_k(Y)| &\leq \sum_{k=1}^{N_0 -1} |\mathcal{B}_k(Y)| + \sum_{k=N_0}^{\lceil \alpha n \rceil } |\mathcal{B}_k(Y)| \\
    %
    &\leq C_1 + \sum_{k=N_0}^{\lceil \alpha n \rceil } s^n \\
    %
    &\leq C_1 + \frac{ s^{\alpha n + 2} - s^{N_0} }{s-1} \\
    %
    &\leq C_2 s^{\alpha n}
\end{align*}
where
\[
C_2 = C_1 + \frac{s^2}{s-1}
\]

Let $N > \frac{ (1-\alpha)h(Y) - \alpha \varepsilon }{\log C_2}$. Then, for $n \geq N$, the number of blocks in $Y$ of length $n$ with no self-overlap by more than $\alpha n$ is at least
\begin{align*}
    |\mathcal{B}_n(Y)| -  \sum_{k=1}^{\lceil \alpha n \rceil } |\mathcal{B}_k(Y)| &\geq r^n - C_2  s^{\alpha n} \\
    %
    &= r^n \left( 1 - C_2 \left( \frac{s^{\alpha}}{r} \right)^n     \right) \\
    %
    &> (1- \exp(-bn)) \exp(nh(Y))
\end{align*}
where we can take
\begin{align*}
    b &= \frac{1}{2} \log \left(  C_2 \left( \frac{r}{s^{\alpha}} \right)^N    \right) \\
    %
    &= (1-\alpha)h(Y) - \alpha \varepsilon - \frac{1}{N} \log C_2
\end{align*}
which is positive by the choice of $N$.
\end{proof}

We now control the entropy loss incurred by requiring a given long block to appear syndetically.

\begin{lemma}\label{synd}
Let $Y$ be a strongly irreducible subshift with $h(Y) > 0$. For every $\varepsilon > 0$, there exist $\beta \in (0,1)$ and $N \geq 1$ such that for every $n \geq N$ and every $\theta \in \mathcal{B}_{\lfloor \beta n \rfloor}(Y)$, the subshift $S \subset Y$ consisting of points $y \in Y$ in which $\theta$ appears at least once in $y_{[i,i+n)}$ for every $i \in \mathbb{Z}$ has entropy at least $h(Y) - \varepsilon$.
\end{lemma}

\begin{proof}
Let $g$ be the gap for $Y$. Let $\beta = \min \{ \varepsilon/(8 h(Y)), 1/2 \}$ and let $N = \lceil 4(2g+1)h(Y)/\varepsilon \rceil$. Let $n \geq N$ and fix $\theta \in \mathcal{B}_{\lfloor \beta n \rfloor}(Y)$. For $m \geq n$, and for all $u_1, \dots, u_k \in \mathcal{B}_{(1-2\beta)n-2g}(Y)$, where $k = \lfloor m/n \rfloor$, there exist $v^{\pm}_1, \dots, v^{\pm}_k \in \mathcal{B}_g(Y)$ and $v_0 \in \mathcal{B}_{m-kn}(Y)$ such that
\[
v_0 \theta v^-_1 u_1 v^+_1 \theta v^-_2 u_2 v^+_2 \dots \theta v^-_k u_k v^+_k \in \mathcal{B}_m(Y)
\]
Therefore, by manipulation of logarithms and the fact that $h(Y) = \inf_{\ell \geq 1} \frac{1}{\ell} \log |\mathcal{B}_{\ell}(Y) |$ by definition,
\begin{align*}
    |\mathcal{B}_m(S)| &\geq |\mathcal{B}_{\lfloor (1-2\beta)n\rfloor - 2g}(Y)|^{\lfloor m/n \rfloor} \\
    %
    \frac{1}{m} \log |\mathcal{B}_m(S)| &\geq \frac{1}{m} \log \left(  |\mathcal{B}_{\lfloor (1-2\beta)n\rfloor - 2g}(Y)|^{(m-1)/n} \right) \\
    %
    &= \left( 1 - \frac{1}{m}  \right) \frac{1}{n} \log |\mathcal{B}_{\lfloor (1-2\beta)n\rfloor - 2g}(Y)| \\
    %
    &\geq \left( 1 - \frac{1}{m}  \right) \frac{1}{n} ( \lfloor(1-2\beta)n\rfloor - 2g) h(Y) \\
    %
    &> h(Y) - \varepsilon/2
\end{align*}
for large enough $m$, where the final inequality follows from the choices of $\beta$ and $N$. We conclude that $h(S) = \liminf_{m \to \infty} \frac{1}{m} |\mathcal{B}_n(S)| > h(Y) - \varepsilon$.
\end{proof}

\begin{proof}[Proof of \Cref{stamp-exist}]
It is clearly enough to prove the result for $u_1, u_2$ sufficiently long, since we can then pass to subwords of $u_1,u_2$. By \Cref{synd}, let $\beta \in (0,1)$, $m$ sufficiently large, and $\theta \in \mathcal{B}_{\beta m}(Y) \setminus \mathcal{B}(W)$ be such that the subshift $S \subset Y$ defined by requiring at least one appearance of $\theta$ in any block of length $m$ has $h(S) > 0$. Let $\alpha \in (0,1)$ be arbitrary, and let $n > (m+k)/(1-\alpha)$ be large enough that, by \Cref{self-overlap}, there exists $\mu \in \mathcal{B}_n(S)$ such that $\mu$ has no self-overlap by more than $\alpha n$, in particular by more than $n-(m+k)$.

Let $u_1 \in \mathcal{B}_{k_1}(U)$, $u_2 \in \mathcal{B}_{k_2}(U)$ with $k_1, k_2 \geq m$ and let $v_1, v_2 \in \mathcal{B}_{k}(Y)$. Then $\mu$ cannot appear in $u_1 v_2 \mu v_2 u_2$ except at the position explicitly indicated. Indeed, $\mu$ cannot appear at a position shifted by at most $m+k$---otherwise, $\mu$ would overlap itself by too much---and it cannot appear at a position shifted by more than $m+k$, as it would then overlap with $u_1$ or $u_2$ in a block of length at least $m$, contradicting the fact that $\mu \in \mathcal{B}(S)$, and thus has $\theta$ as a subword.
\end{proof}

\subsection{Entropy and periodic points}\label{ss-good-per}

We first show that at least a positive fraction of periodic points in $Y$ of sufficient least period have a preimage of equal least period, and in particular that their growth is exponential with rate $h(Y)$.

\begin{prop}\label{growth-rn}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Then $\lim_{n \to \infty} \frac{1}{n} \log r_n(\pi) = h(Y)$.
\end{prop}

\begin{proof}
Let $g$ be the mixing gap of $X$. By \Cref{self-overlap}, let $b > 0$ and $N > 3g$ be such that, for all $n \geq N$, the number of blocks in $Y$ of length $n-g$ with no self-overlap by more than $n/3$ is at least $c \exp(nh(Y))$, where we may take $c = \frac{1}{2} \exp(-g h(X))$. For each block $v \in \mathcal{B}_{n-g}(Y)$, there exists a periodic point $x \in X$ with $\pi(x)_{[0,n-g)} = v$ such that $\mathrm{per}(x)$ divides $n$. Thus $\pi(x)$ is also periodic with least period dividing $n$. Moreover, if $v$ has no self-overlap by more than $n/3$, then in fact $\mathrm{per}(\pi(x)) = n$. Therefore $r_n(\pi) \geq c \exp(n h(Y))$, so $\liminf_{n \to \infty} \frac{1}{n} \log r_n(\pi) \geq h(Y)$, matching $\limsup_{n \to \infty} \frac{1}{n} \log r_n(\pi) \leq \lim_{n \to \infty} \frac{1}{n} \log q_n(Y) = h(Y)$.
\end{proof}

We now assemble the quantitative results proven so far.

\begin{prop}\label{quant-summary}
Let $X$ be a mixing SFT, $Y$ a mixing sofic shift, and $\pi: X \to Y$ a factor code. Let $\varepsilon > 0$ and $N_0 \geq 1$. Then there exist $N_1 \geq N_0$ and proper subshifts $W \subsetneq Y$, $V = \pi^{-1}(W) \subset X$, such that: $h(W) > h(Y) - \varepsilon$; for $n \leq N_1$, $r_n(\pi|_V) = r_n(\pi)$; and for $n \geq N_1$, $r_n(\pi|_V) > \exp(n (h(Y) - \varepsilon))$.
\end{prop}

\begin{proof}
By \Cref{sofic-approx-inside-sft} and \Cref{enlarge-mixing}, let $V_1 \subset X$ be a mixing SFT such that $h(Y) - \varepsilon/2 < h(\pi(V_0)) < h(Y)$. Let $W_1 = \pi(V_1)$. By \Cref{growth-rn}, let $N_1 \geq N_0$ be such that for any $n \geq N_1$, we have $\frac{1}{n} \log r_n(\pi|_{V_1}) > h(W_1) - \varepsilon/2 > h(Y) - \varepsilon$. Let $W = W_1 \cup \bigcup_{n=1}^{N_1} R_n(\pi)$ and $V = \pi^{-1}(W)$. Then $r_n(\pi|_V) = r_n(\pi)$ for all $n \leq N_1$.

To see that $W \neq Y$, observe that the only $n$-blocks in $W$ that may not be in $W_1$ are those in the low-order periodic points that have been adjoined, which are bounded in number by a constant. That is, $|\mathcal{B}_n(W)| \leq |\mathcal{B}_n(W_1)| + C$ for all $n \geq N_1$, where we can take $C = \sum_{k=1}^{N_1} k |R_k(\pi)|$. Thus $h(W) = h(W_1) < h(Y)$.
\end{proof}

\Cref{quant-summary} is the final input to the proof of \Cref{custom-W-exists}, and thus of \Cref{main-thm}.

\begin{proof}[Proof of \Cref{custom-W-exists}]
Let $\varepsilon = h(Y) - h(Z)$. Let $N_0 \geq 1$ be large enough that for all $n \geq N_0$,
\[
\frac{1}{n} \log \max \{ q_n(Z), |\mathcal{B}_n(Z)|  \} < h(Z)+ \frac{\varepsilon}{4}
\]
By \Cref{quant-summary}, let $W \subset Y$, $V = \pi^{-1}(W) \subset X$, and $N_1 \geq N_0$ be such that $h(W) > h(Y) - \frac{\varepsilon}{4}$, $r_n(\pi|_V) = r_n(\pi)$ for all $n \leq N_1$, and $\frac{1}{n} \log  r_n(\pi|_V) > h(Y) - \frac{\varepsilon}{4}$ for all $n \geq N_1$. Note that $h(W) > h(Z) + \frac{\varepsilon}{2}$ and that $q_n(Z) \leq r_n(\pi|_V)$ for all $n \geq 1$.

Let $g$ be the mixing gap of $X$. By \Cref{stamp-exist}, let $\mu \in \mathcal{B}(Y) \setminus \mathcal{B}(W)$ be a $(Y,W,g)$ stamp. Let $\ell = |\mu| + 2g$. Then since $h(Z) < h(W)$, there exists $N$ sufficiently large so that for all $n \geq N$, in particular for $N+\ell \leq n \leq 2N+\ell-1$, we have $|\mathcal{B}_n(Z)| < |\mathcal{B}_{n-\ell}(W)|$.
\end{proof}

\section{Proofs of \Cref{wlog-sft} and \Cref{mmb-cor}}\label{sec-conseq}

We first use \Cref{growth-rn}, along with facts about Markov approximations in \Cref{ss-mark}, to prove \Cref{wlog-sft}, which reduces \Cref{main-thm} to the case where $Z$ is a $1$-step SFT.

\begin{proof}[Proof of \Cref{wlog-sft}]
We use the properties of Markov approximations mentioned in \Cref{conv-sec}. Let $\varepsilon = h(Y)-h(Z)$. Let $m_0$ be such that $h(Z_{m_0}) < h(Z) + \varepsilon/3$, where $Z_{m_0}$ is the $m_0$th Markov approximation to $Z$, and such that, by \Cref{growth-rn}, for all $n \geq m_0$, $\frac{1}{n} \log r_n(\pi) > h(Y) - \varepsilon/3$. Let $m_1 \geq m_0$ be such that $\frac{1}{n} \log q_n(Z_{m_0}) < h(Z_{m_0}) + \varepsilon/3$ for all $n \geq m_1$. Let $m_2 \geq m_1$ be such that for all periodic points $z \in P(Z_{m_1}) \setminus Z$ (under the natural embedding $Z \hookrightarrow Z_{m_1}$) with $\mathrm{per}(z) \leq m_1$, we have $z_{[0,m_2)} \notin \mathcal{B}_{m_2}(Z)$. Then $Z_{m_2}$ satisfies $q_n(Z_{m_2}) = q_n(Z) \leq r_n(\pi)$ for all $n \leq m_1$. Moreover, since $Z_{m_2} \subset Z_{m_0}$, $\frac{1}{n} \log q_n(Z_{m_2}) \leq \frac{1}{n} \log q_n(Z_{m_0})$ for all $n$; in particular,
\begin{align*}
\frac{1}{n} \log q_n(Z_{m_2}) &< h(Z_{m_0}) + \varepsilon/3 \\
&< h(Z) + 2\varepsilon/3 \\
&< h(Y) - \varepsilon/3 \\
&< \frac{1}{n} \log r_n(\pi)
\end{align*}
for all $n \geq m_1$. Taking $Z' = Z_{m_2}^{[m_2]}$ to be the $m_2$th higher block shift, the lemma is proved.
\end{proof}

To prove \Cref{mmb-cor}, in the mixing sofic case, we use \Cref{blowup} to handle low-order periodic point obstructions, with periodic points of sufficiently high order controlled by \Cref{growth-rn}. To handle the arbitrary case, we first give an improved Markov approximation (\Cref{embed-mixing}), embedding an arbitrary subshift into a mixing SFT with only slightly greater entropy. The construction uses \Cref{gap-sft}; in \Cref{gap-sft-entropy} we estimate the entropy of the mixing SFT constructed in \Cref{gap-sft}.

\begin{lemma}\label{gap-sft-entropy}
Let $X$ be a mixing SFT with gap $g$ and let $V_0 \subset X$ be an SFT. Let $k \geq g$ and let $\mu \in \mathcal{B}(X) \setminus \mathcal{B}(V_0)$ be an $(X,V_0,k)$ stamp. For any $\varepsilon > 0$, there exists $N \geq |\mu|$ such that $h(V_1) < h(V_0) + \varepsilon$, where $V_1$ (depending on $N$) is as in \Cref{gap-sft}.
\end{lemma}

\begin{proof}
Let $N_0 \geq 1$ be such that for all $n \geq N_0$ we have $\frac{1}{n} \log |\mathcal{B}_n(V_0)| < h(V_0) + \varepsilon/4$. Let $N > 2N_0$ be such that
\[
\frac{1}{N} \max \{ \log N,  \log |\mathcal{B}_k(X)|^2, \log |\mathcal{B}_{N_0}(V_0)|\} < \frac{\varepsilon}{4}
\]

We will show that $\frac{1}{N} \log |\mathcal{B}_N(V_1)| < h(V_0) + \varepsilon$. Consider a block of length $N$ in $V_1$. Such a block can contain at most one full or partial block of the form $\gamma^+ \mu \gamma^-$ where $\gamma^{\pm} \in \mathcal{B}_k(X)$. The $\mu$, if present, can begin at any of the $N$ positions. The rest of the block of length $N$, outside the block $\gamma^+ \mu \gamma^-$, consists of one or two blocks from $V_0$, with length totalling at most $N$. We thus have
\[
|\mathcal{B}_N(V_1)| \leq N |\mathcal{B}_k(X)|^2 \max_{0 \leq \ell \leq N/2} |\mathcal{B}_{\ell}(V_0)| |\mathcal{B}_{N-\ell}(V_0)|
\]
If $0 \leq \ell \leq N_0$, then $|\mathcal{B}_{\ell}(V_0)| |\mathcal{B}_{N-\ell}(V_0)|  \leq |\mathcal{B}_{N_0}(V_0)| |\mathcal{B}_{N}(V_0)|$, so
\begin{align*}
    \frac{1}{N} \log \left( |\mathcal{B}_{\ell}(V_0)| |\mathcal{B}_{N-\ell}(V_0)| \right) &\leq \frac{1}{N} \log |\mathcal{B}_{N_0}(V_0)| + \frac{1}{N} \log |\mathcal{B}_{N}(V_0)| \\
    %
    &< h(V_0) + \frac{\varepsilon}{2}
\end{align*}
If $N_0 \leq \ell \leq N/2$, then
\begin{align*}
    \frac{1}{N} \log \left( |\mathcal{B}_{\ell}(V_0)| |\mathcal{B}_{N-\ell}(V_0)| \right) &< \frac{1}{N}  \left( \ell \left(h(V_0) + \frac{\varepsilon}{4} \right)  +  (N-\ell) \left(h(V_0) + \frac{\varepsilon}{4} \right)  \right) \\
    %
    &= h(V_0) + \frac{\varepsilon}{4}
\end{align*}
Therefore
\begin{align*}
    \frac{1}{N} \log |\mathcal{B}_N(V_1)| &< \frac{1}{N} \log N + \frac{1}{N} \log |\mathcal{B}_k(X)|^2 + h(V_0) + \frac{\varepsilon}{2} \\
    %
    &< h(V_0) + \varepsilon
\end{align*}
by the above choice of $N$.
\end{proof}

along with the following standard lemma improving the construction of the Markov approximation. For completeness, we include a proof using \Cref{gap-sft}.

\begin{lemma}\label{embed-mixing}
Let $Z$ be a subshift and let $\varepsilon > 0$. Then there exists a mixing SFT $V$ containing $Z$ with $h(V) < h(Z) + \varepsilon$.
\end{lemma}

\begin{proof}
Let $m$ be large enough that $h(Z_m) < h(Z) + \varepsilon/2$, where $Z_m$ is the $m$th Markov approximation to $Z$. Let $X$ be the full shift on the alphabet of $Z$. Certainly $Z_m \subseteq X$. If $Z_m = X$, then we can take $V=X$. If $Z_m \neq X$, then by \Cref{stamp-exist}, let $\mu$ be an $(X,Z_m,k)$ stamp for some $k \geq 0$. Let $V_0 = Z_m$ and $V=V_1$ as in \Cref{gap-sft} where $N$ is large enough that, by \Cref{gap-sft-entropy}, we have $h(V) < \varepsilon/2$. Thus $V$ is indeed a mixing SFT containing $Z$ with $h(V) < h(Z) + \varepsilon$.
\end{proof}

\begin{proof}[Proof of \Cref{mmb-cor}] We first consider the case in which $Z$ is mixing sofic. Let $\Tilde{Z}$ be a mixing SFT and $\chi_0: \Tilde{Z} \to Z$ an almost invertible factor code. If already $q_n(\Tilde{Z}) \leq r_n(\pi)$ for all $n \geq 1$, then we can take $Z' = \Tilde{Z}$ and apply \Cref{main-thm} immediately to construct the claimed embedding $\psi: Z' \to X$. However, if $q_n(\Tilde{Z}) > r_n(\pi)$ for some $n$, so that $X,Y,\pi, \Tilde{Z}$ violate the hypotheses of \Cref{main-thm}, then we need to construct a further extension of $\Tilde{Z}$ which satisfies the hypotheses of \Cref{main-thm}. The construction, consisting a tower of extensions via \Cref{blowup}, is as follows.

By \Cref{growth-rn}, since $h(\Tilde{Z}) = h(Z) < h(Y)$, there are at most finitely many $n$ such that $q_n(\Tilde{Z}) > r_n(\pi)$. Let $N$ denote the greatest such $n$. Let $C = \sum_{k=1}^{N} \max\{0, q_k(\Tilde{Z}) - r_k(\pi)\}$. That is, $C$ is the number of periodic points by which $X,Y,\pi, \Tilde{Z}$ violate the hypotheses of \Cref{main-thm}. For $1 \leq k \leq N$ and $1 \leq \ell \leq k^{-1} \max\{0, q_k(\Tilde{Z}) - r_k(\pi)\}$, let $z_{k,\ell}$ be periodic points with pairwise disjoint orbits, such that $\mathrm{per}(z_{k,\ell}) = k$. For a given $k$, the union of the orbits of the points $z_{k,\ell}$ has cardinality $\max\{0, q_k(\Tilde{Z}) - r_k(\pi)\}$. Let $C' = \sum_{k=1}^{N} k^{-1} \max\{0, q_k(\Tilde{Z}) - r_k(\pi)\}$ (counting orbits, rather than points), and let $\{ z^{(j)}  \}_{j=1}^{C'} = \{ z_{k,\ell} \}_{k,\ell}$ be an enumeration of the points $z_{k,\ell}$.

Again by \Cref{growth-rn}, let $M > N$ be large enough that for all $n \geq M$, we have $q_n(\Tilde{Z}) + Cn \leq r_n(\pi)$. We now repeatedly apply \Cref{blowup}. Let $Z^{(0)} = \Tilde{Z}$. For $1 \leq j \leq C'$, let $Z^{(j)}$ be a mixing SFT and $\chi^{(j)}: Z^{(j)} \to Z^{(j-1)}$ an almost invertible factor code such that the preimage of the orbit of $z_j$ under $\chi^{(j)}$ is a single orbit of length $M \mathrm{per}(z_j)$, and such that every periodic point in $Z^{(j-1)}$ not in the orbit of $z_j$ has a unique preimage under $\chi^{(j)}$. Let $\eta^{(1)} = \chi^{(1)}$ and $\eta^{(j+1)} = \eta^{(j)} \circ \chi^{(j+1)}$. Let $Z' = Z^{(C')}$ and $\eta = \eta^{(C')} : Z' \to \Tilde{Z}$. Certainly $\eta$ is almost invertible, so $h(Z') = h(\Tilde{Z}) < h(Y)$. We claim that $q_n(Z') \leq r_n(\pi)$ for all $n \geq 1$. Indeed, for each $j$, if $\mathrm{per}(z_j) = k$, then we have $q_k(Z^{(j)}) = q_k(Z^{(j-1)}) - k$, $q_{Mk}(Z^{(j)}) = q_{Mk}(Z^{(j-1)}) + Mk$, and $q_n(Z^{(j)}) = q_n(Z^{(j-1)})$ for all $n \notin \{  k, Mk \}$. Therefore $q_k( Z' ) = r_k(\pi)$, and
\begin{align*}
    q_{Mk}(Z') &= q_{Mk}(\Tilde{Z}) + M \max\{0, q_k(\Tilde{Z}) - r_k(\pi)\} \\
    %
    &\leq q_{Mk}(\Tilde{Z}) + CM \\
    %
    &\leq r_{Mk}(\pi)
\end{align*}
where the last inequality follows from the choice of $M$. Therefore $X,Y,\pi,Z'$ satisfy the hypotheses of \Cref{main-thm}, so there exists a sliding block code $\psi: Z' \to X$ such that $\pi \circ \psi$ is injective. This concludes the proof in the case that $Z$ is mixing sofic.

We now handle the general case, where $Z$ is an arbitrary subshift with $h(Z) < h(Y)$. By \Cref{embed-mixing}, let $V$ be a mixing SFT containing $Z$ with $h(V) < h(Y)$. By the mixing sofic case, let $V'$ be a mixing SFT such that $X,Y,\pi,V'$ satisfy the hypotheses of \Cref{main-thm}, and let $\chi: V' \to V$ be an almost invertible factor code. Let $Z' = \chi^{-1}(Z)$. Then $\chi|_{Z'}$ is still finite-to-one, which concludes the proof.
\end{proof}

\section*{Acknowledgments}

I thank Tom Meyerovitch for posing the question that led to this work and suggesting the argument for \Cref{growth-rn}, and more generally for offering, with Brian Marcus, very generous advice and supervision. I also thank Mike Boyle for a helpful conversation by email about the relationship between my question \cite{vs-it-22-mo} and the embedding problem for sofic shifts.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Q: Sorting variables after importing Excel information into Matlab Hello I'm learning how to use Matlab for the first time and I am stuck on an assignment. Its a two part assignment the first one is to write an Matlab script and have it output an Excel file which I have done.
M = {'City' 'Population'; 'Chandler' 249146; 'Gilbert' 229972;
'Glendale' 234632; 'Mesa' 457587; 'Peoria' 162592;
'Phoenix' 1513367; 'Scottsdale' 226918; 'Surprise' 123546;
'Tempe' 168228; 'Tucson' 526116};

M = xlswrite('largest_cities_in_arizona.xls',M)

The second part is to create a Matlab script that reads the Excel file, creates two variables named city_population and city_name and then sorts both in descending order based upon the population. I have most of this figured out.
[A, B] = xlsread('largest_cities_in_arizona.xls', 1, 'A2:B11');

city_population = sort(A, 'descend');

city_name = sort(B);

The problem I am having is trying to sort the city_name by the city_population. When importing the data Matlab treats city_name as a 10x1 cell and city_population as a matrix, and gives me dimension errors if I try to sort B based upon city_population. Is there some other way to do this? Any help or information would be greatly appreciated.

A: Try this:
[city_population, index] = sort(A, 'descend');
city_name = B(index);

The key is to use the second output of sort as an index which you apply to the other array to get the same order.
In your example, this gives
>> city_population
city_population =
     1513367
      526116
      457587
      249146
      234632
      229972
      226918
      168228
      162592
      123546

>> city_name
city_name =
    'Phoenix'
    'Tucson'
    'Mesa'
    'Chandler'
    'Glendale'
    'Gilbert'
    'Scottsdale'
    'Tempe'
    'Peoria'
    'Surprise'


Website design can be tricky to get just right, so we at Studio Culture have compiled a list of the "4 C's" – clarity, colour, content, and call to action – to help you get started. These ke...	y features are things you can always keep in mind when planning to design a website. Be sure to also contact us for expert help on web design, SEO, branding, digital marketing, and more.


In any job interview, there are inevitably going to be questions that will stump you. They'll stump you not because you don't know the answer or because you're not knowledgeable, but because they seem like obvious questions that don't even need to be asked. It's like asking a race athlete, "Why do you want to win?" Um…? Why WOULDN'T they want to win? Why would they train and prepare for so long if their goal was to lose? It seems obvious, so why bother even asking the question?
However, in an interview, you can't be so blunt when a similar question arises, such as "Why do you want this job?" Well, you're unemployed and on the job market – why on earth WOULDN'T you want the job??? Another common question that gets asked in interviews is, "Why should we hire you?" Well, buddy, you invited me to this interview, so you're obviously interested in hiring me – why don't you tell me what's so appealing about me instead?
Clearly, these responses are not what you want to say out loud even though you're probably thinking them in your head. Regardless, we still have to answer the question, so what should we actually say in response to these types of seemingly obvious questions?
Our natural tendency when asked, "Why should we hire you," is to start talking about what makes us so great. This seems appropriate since the question emphasizes YOU. However, shifting the focus from ourselves to a team-centered vision that merely includes ourselves is the way to go.
When you think about the question rephrased in this manner, it becomes easier to answer because it's not as vague. It's clear that they want you to articulate your assets in the context of their company.
So when you formulate your answer to the question, highlight your strengths and how they align with the company. For example, let's say you're an RNA biochemist interviewing for a pharmaceutical company who has burgeoning interests in RNA therapeutics – and you know this because you've done your homework on them – but they haven't yet launched this research division yet.
Rather than highlighting how skilled you are at the bench and stating that you're a hard worker who would bring your superior intellect to the company's research and development (R&D) team, which is a very "me" centric answer, instead focus on how you've developed a keen sense of new and emerging trends in RNA biology in recent years and you're very interested in the company's vision to move in that direction for future projects. Then highlight how your knowledge base and background would be a good fit for the existing R&D team as the company transitions into this new field of therapeutics.
This is a much more "we" focused answer that is sure to make the interviewer think you are indeed the best candidate for the position over other applicants because you can help the company expand and become more lucrative.
By answering in such a way that demonstrates the match would be beneficial for both parties, you are much more likely to seal the deal. So, although the question seemingly asks about you, assume that it's asking about them and how they would be better if you joined their team.
Similar to highlighting the "we," not the "me," in your answer to the question, "Why should we hire you," you should also focus on how you can solve any existing problems for the company. This is different from the above-mentioned example in which you could help the pharmaceutical company pave a new path into RNA therapeutics. What this perspective gets at is how you can fix something that is currently amiss, not something that has yet to occur, in the company.
For example, let's say the same pharmaceutical company is experiencing a high rate of turnover for their R&D scientists, which is inherently slowing down the team's overall progress. Additionally, let's say that throughout the course of your interview day, you've gathered that there aren't a lot of career development or enrichment programs in place for the scientists at this company.
Given your background working with CNSPY and/or other campus organizations at Yale that work to provide career development opportunities for scientists, you could highlight that, in addition to the skills you'd bring to the bench, you'd be interested in taking the initiative to form a similar organization at this company to provide enrichment programs for the R&D scientists to improve job satisfaction and overall team morale, which should decrease the rate of employee turnover.
This response is sure to catch the attention of the interviewer for reasons other than your technical skills. As a result, you will become a much more desirable candidate over others who merely have the ability to help the company's R&D team at the benchtop.
So, when asked, "Why should we hire you," take this as an opportunity to address how you can solve any existing problems the company may be experiencing. Doing so is selfless and puts the company first and you second. Again, keep in mind that the company has one goal – to improve itself.
Thus, by focusing on how your skills and your background can help the company move forward, either by solving their existing problems or paving the way for them to accomplish their desired future goals, you will have given the interviewer very little reason why they shouldn't hire you.
This is why it's so important to 1) Address the "we," not "me," and 2) Solve problems for the company. Leave no doubt in their minds that you are THE best candidate for the position, and THAT is why they should hire you!
Yes, you nailed this point Vickie! Our qualifications are not the reason we are being interviewed, it's how our qualifications can solve a problem and fit a need in the company. Thanks for the reminder and tips on how to keep this perspective.


Home News Russia downgraded by S&P to near-junk status – good for Russia FX volumes?
The leading credit rating agency has injected more volatility into an already volatile situation.
Global credit rating agency Standard & Poor's (or S&P for short) has downgraded its ratings of Russian foreign debt to BBB-, just one notch above 'junk' status, keeping its outlook negative. S&P backed its decision by continued capital outflows from Russia, as well as the country's financial account deficit.
So what does this mean for the markets?
Well, in short, more negativity. On a usually-quiet Friday the markets have reacted by pushing the Ruble lower against the dollar another 0.3%, after the Ruble had already dropped 8% so far in 2014. And Russian stock exchanges were off between 1.0-1.5%.
But what's bad for the Russian government and Russia's stock market might not be bad for the Russian forex sector – which seems to like volatility. In what's already been a somber first half of the year for Russia's capital markets, we've seen healthy growth in FX trading volume metrics at places like Alpari Russia, the leading retail FX broker in the country, and at the Moscow Exchange.
Lowered credit ratings usually brings with it more currency volatility, and that could be good for business for forex brokers in the country.
To see a copy of S&P's Russia downgrade click here.


text:
Upper Canada College has a long tradition of excellence in the art of writing.
Here are UCC's Upper School student publications:
Blazer is UCC's humour publication for people who think they're way funnier than they actually are. Blazer spoons Convergence every issue in its last two pages and is widely known as the most important part of the paper. It works toward bringing a drop of sunshine, rainbows and unicorns into the dark, dreary and scary world of school. Laughter is the best medicine — that is, aside from actual medicine.
The Blue and White (Tbaw)
The Blue and White is UCC's online arts and life publication. Its focus extends from critiques of Friday assemblies to movie reviews to non-fiction pieces about art festivals happening off campus. It's not only a school newspaper but also a guide to the Toronto arts and life scenes, offering reviews, photo essays and social commentaries. Visit tbaw.ca.
UCC's yearbook, College Times, specializes in capturing the essence of school life each year under the stewardship of art teacher Anne Kaye. It's published every May.
Convergence is tackling the year with all the gusto that comes with new writers, new leadership and new stories. With tons of material, ranging from controversial interviews with Derek Poon about the dismal state of UCC food to in-depth coverage of Association Day, Convergence has undoubtedly become the premier source of school news for Upper Canada College students.
Under the stewardship of Marshall Webb, the editors have introduced innovative sections for stories on international events, comedy and poetry. Convergence endeavours to provide material that appeals to students (and parents) of all interests. Visit Convergence online at uccconvergence.com.
IDiv Times
IDiv Times has been an aspiration of the Intermediate Division since its original publication in 2010-2011 was met with great enthusiasm and an appetite for further growth. The Times is a student-driven newsletter that was originally born out of a desire to fill the gap that existed for Year 1 and Year 2 writers to publish their work. A platform for students to report and reflect on events relevant to peers, the club raises awareness about the Intermediate Division, spotlights new boys and faculty, and offers competing perspectives on school-related concerns such as bullying and appropriate technology use. Beyond garnering journalistic and creative skills, the organization offers excellent leadership opportunities. Students are elected and fulfill a myriad of roles, from editors to section heads to layout designers. The Times has partnered with Convergence so students benefit from the expertise of senior student mentors. This relationship helps support the growth of Intermediate Division writers and empowers these students to take creative risks and assert innovative thought. IDiv Times celebrates student initiative, collaboration and, above all, encourages engagement in the growth and sustenance of the invaluable contributions of UCC student publications. Gillian Levene is the faculty advisor.
This exquisite book includes the best poetry, prose and art created during the school year. It's issued in time for the leaving class graduation in May and receives faculty assistance from Julian Bauld and Dale Churchward.
The Blue Page
The Blue Page is the student-run editorial newspaper and a vehicle for opinion relating to school matters. Now more than a quarter-century old, The Blue Page prints articles written by students and faculty members that demand change and appraisal. Heated arguments, like the removal of boarding a few years ago and the teaching of global warming as fact, have scrolled across The Blue Page. In essence, The Blue Page is a tool for expressing one's perspective, which is exposed to a large-scale audience (the paper prints bi-weekly and is distributed to the entire Upper School). The Blue Page is intriguing, entertaining and effective — a paper for the long term created with the assistance of advisor Marshall Webb.
Literature Advisors
Julian Bauld
Quiddity and
Poetry In-Voice
Dale Churchward
Terence Dick
The Blue and White
Anne Kaye
Gillian Levene
Marshall Webb
Convergence,
Blazer and


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