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Surfing at WAI'AO in Yilan
East Taiwan, Featured, Joe Henley, Northeast and Yilan Coast National Scenic Area, Yilan County On May 29, 2018 No Comments
First-Time Surfing Experience on the Northeast Coast
Text: Joe Henley; Photos: Maggie Song
Taiwan as a surfing destination used to be a well-kept secret, known only to intrepid explorers of waves worldwide. This has changed significantly in recent years, however, and surfing beaches around the island have become hotspots attracting surfers from near and far. Among the hottest is at Wai'ao in Yilan County, a conveniently reached spot that is ideal for novice surfers.
When travelers think of surfing in East Asia it is the wave-riding hotspot of Bali, where everyone from beginner to surfing professional can enjoy the famed breaks at beaches like Kuta, Dreamland, and Medewi, that most likely comes to mind.
While Taiwan does not have the sheer number of surf spots nor the imposing waves of that Southeast Asian haven, the breaks this island does possess, especially those along its eastern shores all the way down to the far south, are slowly making a name for themselves among surfers around the world.
Surfer dude at Wai'ao Beach in Yilan
Surfing in Taiwan
Surfing Experience
One popular location is Donghe Township in Taitung County. This laid-back southeastern locale has excellent waves along its breaks. The waves at Jinzun Fishing Harbor – a 20-minute drive north of the tourist-popular village of Dulan – are said to be the best in the area. Jinzun is venue for the annual Taiwan Open of Surfing (www.taiwanopenofsurfing.org), and the World Longboard Championship was staged there in 2017 as well, both events attracting top surfers from Taiwan and abroad. Jinzun will again be venue for the championship in 2018.
Close to the island's southern tip, on the east side of Hengchun Peninsula, surfers in the know head for Jialeshui. There is a small collection of surf shops and accommodations in the village of Gangkou (also called Chashan), not far from the surf beach. Not known for being the most picturesque, the beach is a welcome out-of-the-way-spot for those who want to enjoy the sun and the waves free of the hustle and crowds encountered at more popular spots further south in Kenting National Park, which takes up a good portion of the peninsula.
Pacific Ocean near Jialeshui in southeastern Taiwan
While Jinzun and Jialeshui are great options for surfers who want to test the Taiwan waters, they are not the easiest to get to if your base is Taipei. Jialeshui, for example, is in one of the most remote areas along the coastline of mainland Taiwan, and getting there without your own means of transport can be a hassle. Luckily, there are many surfing spots that can be reached from the capital far more easily, the most popular of which is Wai'ao Beach, located just to the north of Wushi Harbor in Toucheng Township on the Northeast Coast. It's a 1-hour drive from Taipei (via National Freeway 5) or a 2-hour train ride on a Local Train (the only train type that stops at Wai'ao Railway Station).
The Northeast Coast is known for its varied coastal landscapes. If you are not in a hurry to get to the beach, instead of taking the freeway, the drive along coastal Provincial Highway 2 from the harbor city of Keelung is highly recommended. You take in a breathtaking array of dramatic cliff faces and high-mountain slopes carpeted with tall silvergrass. There are beaches too, most notably Fulong Beach, a fine-golden-sand section of coast that each summer is venue for the Fulong Sand Sculpting Art Festival and the Ho-Hai-Yan Gongliao Rock Festival. If you take the train from Taipei to Wai'ao you won't be able to see the cliff faces, but you'll get close to the coast at Fulong (from the station, it's a short walk to the beach), and after passing through a long tunnel you'll have the ocean in full view almost all the way until you reach Wai'ao.
Black-sand Wai'ao Beach
The beach at Wai'ao is quite different from the one at Fulong; it has dark, almost black, sand, and the waves are much more suitable for surfing. In recent years it has become a popular spot for surfers for a number of reasons. For starters, the water area close to the harbor is "surfers only." Though people are free to play in the fine sand and lounge under a sun umbrella, anyone going into the water had better have a board under their arm. Swimming is prohibited here, to keep the rolling surf open for wave riders and avoid board/swimmer collisions. It is permitted north of the surfing area, however.
Another reason why surfers, both locals and foreigners, are drawn to Wai'ao is the welcoming and friendly atmosphere of the place. Though it can get crowded, especially during summer weekends and holidays, Wai'ao – and surf spots in general around Taiwan, save the "secret" spots you won't hear about – lack the "locals only" mentality that is prevalent in so many other surfing locales in Southeast Asia. So long as you behave respectfully while on the water, don't drop in on anyone's line, and don't cut the queue, there will be no hassle from the local surfers who come to the beach daily to catch their waves.
Summer weekend beach crowd
Not long ago, my own first-time surfing experience was launched at Surffella, a combination surf shop, surf school, and board-rental facility located on the narrow side road fronting Wai'ao Beach, which is lined with a cluster of surf shops and small eateries. It was an unseasonably warm day in early February, temperatures in the mid-twenties Celsius. I was introduced to my coach, 18-year-old Kevin Han, a mature and well-spoken local pro surfer and the son of Surffella's owner, who set up the shop back in 2007.
Surffella surf shop
I was given a wetsuit and a long beginner's board (which is easier to balance on), and we headed down to the beach, dark clouds rolling in over the coastal mountains. A few meters from where the waves rolled onto the sand, my lesson began.
Taking the board to the beach
Kevin imparted to me some of the wisdom he's accrued over his seven years of surfing waves up and down Taiwan's east coast and abroad in Bali. We began with the basics: an intro to the various parts of the board, from the nose to the tail, to the fins, and to the leash. Then we moved on to how to paddle out with your body centered on the board, toes flush with the tail, doing a dry run on the sand.
Dry lesson
After this, while still on land I was shown the proper way to push myself up into a standing position with board pointed toward the beach, knees bent to create a low center of gravity and arms up over my head to avoid that "rolling down the windows" motion that almost always precedes a wipeout. This was practiced a few times, going through the motions from paddling to catching a wave to standing up, correction provided by Kevin at every step when needed.
Thereupon, satisfied with my progress, my teacher motioned toward the sea and the gentle waves rolling in within an area demarcated by buoy lines (the beginners' area), away from the higher waves to our right reserved for more experienced surfers. As we waded slowly in, the water felt cool at first, but soon the chilly sensation turned into one of comfort and calm.
Facing the waves
We went out to a spot where the water reached our waistlines, then pointed the board back toward the shore. Kevin stood before me, looking seaward, watching the waves. Then he gave me the sign. "Paddle, paddle, paddle," he instructed. I applied his teachings, getting in a few strokes before he called out, "OK, stand up!" I stood, but too high, legs straight rather than bent, and fell ingloriously into the water with a large splash and an unquiet yelp. After I resurfaced Kevin pointed out my mistakes, and on my second run I was able to stand, catch, and ride a wave for a few seconds, ending in the shallows just before the beach.
My runs thereafter were a mix of euphoric successes and dismal failures, but as the end of our two-hour lesson neared I was already able to stand and ride the waves consistently – owing more to Kevin's skill as an instructor than to any inborn talent I might possess. Every time I was able to get up on the board and enjoy those few seconds upright, riding a wave, it was pure bliss – a feeling of tranquility normally associated with the endorphin rush at the end of a long run or a particularly intense workout. Perhaps this was the surfer's high I had often read about, but never experienced. Until now.
Master of the board
After our lesson, Kevin and I sat down for a chat at one of the picnic tables set up outside Surffella. Dogs belonging to owners and staff from the various shops lazed in the sunshine, traffic both vehicular and human sparse on this early weekday afternoon. "Wai'ao is perfect for novice surfers, due to a number of factors," said Kevin, explaining that "It's much safer for beginners, because it's a sand beach," alluding to the inherent dangers of getting injured on an outcropping of rock or reef. "At Kending, for example, there are reefs, and the water is much deeper – you can't reach the bottom.
"The crowds at Wai'ao are smaller during the winter," he continued, "with only the more serious surfers coming out during the cooler months for the tall waves driven by distant monsoon winds." Things start to pick up around Wai'ao in March or April. When the weather gets warmer, the waves are among the most consistent around Taiwan, now blown by the strong breezes of the typhoon season. Kevin admits that during the winter Taitung County is the place to be, save for the fact that the waters there are significantly deeper and a degree more dangerous. It is also generally far less crowded there, no matter the time of year.
KLOOK:
Yilan Waiao Surfing Experience
"During the busy season, we might rent out a hundred boards a day," he says of the rush Wai'ao can experience, "so it can get crowded in the water; but the weekdays aren't as bad."
A two-hour Surffella lesson costs NT$1,500, the fee for a full-day board rental included. If you just want to rent the board, it's NT$500 for the day, from sunrise to sunset. If traveling by public transport, Surffella offers free pickup from either Wai'ao Railway Station or the bus stop at nearby Wushi Harbor, as long as you rent at least two boards or book at least one surf lesson. Call at least a day in advance to make a booking. Although there is no English version of Surffella's website, the coaches are able to provide instructions in English.
Surffella (蕃薯衝浪俱樂部)
Add: No. 92-9, Gangkou Rd., Toucheng Township, Yilan County
(宜蘭縣頭城鎮港口路92-9號
Website: blog.surffella.com.tw (Chinese)
Chashan 茶山
Donghe Township 東河鄉
Dulan 都蘭
Fulong 福隆
Gangkou 港口
Jialeshui 佳樂水
Jinzun Fishing Harbor 金樽漁港
Kenting National Park 墾丁國家公園
Toucheng Township 頭城鎮
Wai'ao Beach 外澳沙灘
Wushi Harbor 烏石港
KAYAKING at Fulong on the NORTHEAST COAST
taiwaneverything.cc/2018/01/31/kayaking-fulong/
Visiting TOUCHENG FARM in Yilan
taiwaneverything.cc/2016/08/19/toucheng-farm/
Joe Henley is a Taipei-based freelance writer, journalist, author, and musician. Originally from Saskatoon, Saskatchewan, he moved to Taiwan fresh out of journalism school in 2005. After a seven-year stint as an editor at a Taipei publishing firm, he decided to dedicate himself to freelancing full-time in 2012.
www.jwhenley.com


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Hereford & Worcs
You are in: Birmingham > People > Stories > Dying to meet you
Jane Officer reads Feltus's letter
Dying to meet you
By Scott Faulkner
Grandmother Jane Officer, 74, spends her retirement gardening, visiting her family and writing to her American pen friends - men who have murdered.
It started with a letter written in a Selly Oak living room and ended with a journey to America's Deep South on the eve of a man's execution.
Jane Officer was keeping a promise she'd made to her pen friend Feltus Taylor as he prepared to die for his crimes.
Feltus Taylor always wore leg chains
The last time she'd journeyed to Louisiana he'd eaten his last meal and was 20 minutes away from death when his lawyers had new grounds for appeal accepted by the Supreme Court.
There was to be no reprieve this time and no return to his 6ft by 9ft cell in Angola state penitentiary, America's largest maximum security prison and home to 5,000 inmates.
After five desperate 'stays' of execution, Feltus, 38, was finally killed by lethal injection for the murder of a restaurant worker.
Shot four times
According to the Associated Press, Feltus (or Taylor if using crime reporter parlance) killed Donna Ponsano during a robbery at a fried chicken restaurant in Baton Rouge on March 27, 1991.
One of Feltus's pencil-written letters
Manager Keith Clark was shot four times in the head and left partially paralyzed.
Listen: A letter from death row >
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Prior to his execution on June 6, 2000, Feltus apologized to the slain woman's family and to the wounded restaurant manager.
"I want to tell you, Keith, and the Ponsano family that I always regretted what I've done," he said.
"It was my own doing. After this is over with, I hope you can find the peace to move on."
Guns and money
"Feltus was 30 going on 15 and very immature," adds Jane.
Jane Officer with Feltus Taylor
"His girlfriend started putting pressure on him for money and a lot of people in that area carry guns.
"Just before he was killed, Keith got a message through to him to say he forgave him."
The hundreds of letters she'd received from him in nine years of corresponding are stored in date-marked cardboard crates at her home but the memory of his death still haunts her.
Feltus was one of nine men retired teacher Jane has written to who have been imprisoned for murder.
Terrible crimes
She has been writing to men in prison and on death row in America for 20 years and all but one of her pen friends have now been executed.
Fred Tucker
The last, Fred Tucker, is a 'lifer' and has been told he will die in prison.
They have been corresponding for 18 years and Jane has visited him several times and has vowed to continue until it's no longer possible.
It could be a long time.
Fred is 53 and while Jane is nearly twenty years his senior, she looks much younger.
Jane believes the terrible upbringings of the people she encounters, and their problems with substance abuse, are often at the heart of their terrible crimes and that given the right circumstances "we are all capable of terrible things".
She says Fred was repeatedly sexually abused by his mother's boyfriend as a child and when he was put into a children's home he killed an older boy who did the same.
Jane opposes the death penalty
He was 17 and in custody when he was given the chance of serving two years in the Army in return for his sentence being rescinded.
On being discharged he turned to crime and committed an armed robbery in New Orleans, taking just $20. He has been in prison ever since.
"He's murdered and intimated he's also killed someone inside the prison," she remarks.
Thinking about him
"Once your past's out you're seen as fair game. We don't know a fraction of what goes on.
"He's a very bright man but he's life has been completely ruined."
Capital punishment facts
About 50 women and 3,500 men are on Death Row in USA.
Two thirds of countries have abolished death penalty.
China, Iran, Saudi Arabia, Pakistan and USA carry out 93% of executions.
Nearly 9,000 people were sentenced to death in 52 countries in 2008.
37 executions were carried out in USA in 2008.
Death penalty abolished in Britain for murder in 1965.
She dutifully writes to him once a fortnight but still worries for his state of mind.
"He's depressed and absolutely terrified that I will die before him," she says.
"I know it sounds cynical but I can write more or less the same letter every two weeks.
"What's important, though, is that he knows that I'm here and still thinking about him."
When talking to Jane about her work writing to death row inmates and 'lifers' she does not mention the victims of their crimes.
There's also no mention of the 'm' word either and instead she focuses on her belief in extending kindness to all, regardless of what they have done.
"It's not saying anyone's condoning what's happened but you cannot undo what you've done," she pleads.
Horrendous thing
"It's about coming to terms with it. It's an horrendous thing to live with."
Andrew Lee Jones
Jane started writing letters after seeing the award-winning BBC documentary 14 Days in May, which followed events in the final days of a death row inmate's life.
She responded to an appeal for writers from charity Life Lines in May 1990 and was given her first pen-friend – Andrew Lee Jones.
'Old Sparky'
He had been convicted of murdering and raping his estranged girlfriend's 11-year-old daughter in a trial Jane strongly believes was weighted against him.
He was the last inmate at Angola prison to die in "Old Sparky" – Louisiana's electric chair - in July 1991 – which was also shown in the film The Green Mile.
A saline-soaked sponge is used
After his death she set up the Andrew Lee Jones Fund to train UK lawyers and students to go and work in the US alongside attorneys, under the umbrella of charity Amicus.
"No-one will ever know if Andrew committed the crime," she says.
"You cannot say sorry to someone when you've executed them.
"You only have to look at the number of miscarriages of justice in this country to know the death penalty is wrong."
Do people who commit terrible crimes deserve compassion?
The BBC reserves the right to edit comments submitted.
I say 'never judge a man until you've walked a mile in their shoes'. If we can say another person is bad, that makes us feel good about ourselves. The truth is, there's good and bad in all of us and all of us deserve compassion.
They should all rot in hell. She well messed up writing to them. Some people are just pure evil so why show them compassion ever? Let them die a painful long death I say!
Theresa Heine
I found this a very moving interview. I have also started writing through Lifelines to two prisoners on death row. It is a life changing experience. These are desperate people and a letter to show that someone cares means the world. Yes they deserve our compassion - there but for fortune, go you or I.....
Eirian
'Speak as you find'. That's how I like to form an opinion of someone. I, like Jane, have 2 penfriends on death row. They are both intelligent, caring men, who have become firm friends. They offer me as much support and encouragement as I offer them - which, surely, is the mark of a true friendship. Their past does not come into it.
Jane Macdonald
Of course they deserve compassion. Most of these men who have killed (I write to two) have had horrendous upbringings. I came from a stable family which gave me the tools to cope with frustration and my own anger. Not everyone is so lucky.
There was a teacher at my secondary school who did exactly the same. She wrote frequently to an inmate on death row and she used to tell us often how he was doing and about her yearly visits to Texas to see him. I personally would not have the dedication to keep writing to these people as soon as you had asked the obvious questions such as 'why did you do it', 'did you do it' and, if so, 'do you regret it'? Once incarcerated do you not forego your right to contact with the outside world? These people have been placed in the penal system for punishment, shouldn't having a penpal be classed as a luxury? These people are kept on 'the row' for far too long. It would seem America's system mirrors our own sumptuous, nonsensical sham!
Andrew Trinick
Whilst I am not a big fan of the death penalty for some of the reasons listed in this article I think that we have to remember that 99 percent of these people are in jail because they failed to show any compasion to their victims! I feel that if we should show any compassion it should be to those people left hurt and scarred by the actions of these people. Frankly in the main they do not deserve any sypmathy or thought. They are off the streets and cannot hurt anyone else. Plenty of people have a poor life and background but do not commit terrible, violent crimes. They chose the path they now walk and they only have their choices in life to blame for it.
In a Christian sense perhaps they do deserve compassion and forgiveness. If they have only ever committed one such crime and if there were mitigating circumstances this may be the right way to treat them. But if they are repeat offenders who obviously show no remorse and who gain perverse pleasure from such things then.... Death with no possibility of expensive appeals would be another option, but is justice always carried out fairly? I'm sure that we have all wrongly accused a person of something, later realising that we are wrong. I suppose the question is: If we don't show compassion, then what do we do, bully and abuse the guilty?
Listen: A letter from death row
On bbc.co.uk
Feature: Life Lines founder
On the rest of the web
Life Lines - official site
Office of the Clark County Prosecuting Attorney
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\section{Introduction}
The measure entropy was introduced by Kolmogorov and Sinai in ergodic theory in the mid fifties of the last century.
Some ten years later  Adler, Konheim, and McAndrew \cite{AKM} introduced the notion of topological entropy $h_{top}(T)$ of a continuous self-map $T: X \to X$ of a compact topological space $X$. A prominent example in both cases is provided by the \emph{Bernoulli shifts}. Since these shifts are the core of this paper, we introduce them here in full detail.

\begin{example}
Let $K$ be a non-trivial finite group with neutral element $e_K$.
\begin{itemize}
\item[(a)]  The \emph{two-sided Bernoulli shift} $\overline{\beta}_K$ of the group $K^{\mathbb Z}$ is  defined by
$$\overline\beta_K(\ldots, x_0,x_1,x_2, \ldots)=(\ldots, x_{-1}, x_0, x_1, \ldots),\ (\mbox{i.e.},\  \overline\beta_K((x_n)_{n\in\mathbb Z})=(x_{n-1})_{n\in\mathbb Z}, \mbox{ for } (x_n)_{n\in\mathbb Z}\in K^{\mathbb Z}).$$
\item[(b)] The \emph{right  Bernoulli shift} $\beta_K$ \ (respectively, \emph{left Bernoulli shift} $_K\beta$) of the group $K^{\mathbb N_0}$ is  defined by
$$
\beta_K(x_1,x_2,x_3,\ldots)=(e_K,x_1,x_2,\ldots), \; (\mbox{respectively}, \;\; _K\beta(x_0,x_1,x_2,\ldots)=(x_1,x_2,x_3,\ldots).
$$
\end{itemize}

The standard product measure of the compact group $K^{\mathbb Z}$  (respectively, $K^{\mathbb N_0}$) coincides with its Haar measure and $\overline{\beta}_K$ (respectively, $_K\beta$)  is a measure-preserving continuous automorphism (respectively, endomorphism) with topological entropy $\log |K|$ coinciding with the measure entropy. This explains their relevance to both ergodic theory and topological dynamics.

The right Bernoulli shift $\beta_K$ of $K^{\mathbb N_0}$  is less relevant in this respect for two reasons:  it is {\em not} measure-preserving (so not relevant for ergodic theory) and its topological entropy is 0.
\end{example}

A possible definition of algebraic entropy  for endomorphisms of abelian groups was briefly mentioned in \cite{AKM}.
Later on, in 1975 in \cite{W} Weiss defined the algebraic entropy as follows: let $G$ be an abelian group and $F$ be a finite subgroup of $G$; for an endomorphism $\phi:G\rightarrow G$ and $n\in\mathbb N$, let $T_n(\phi,F):=F+\phi(F)+\ldots+\phi^{n-1}(F)$ be the \emph{$n$-trajectory} of $F$ with respect to $\phi$. The \emph{algebraic entropy of $\phi$ with respect to $F$} is $$H(\phi,F):={\lim_{n\to +\infty}\frac{\log|T_n(\phi,F)|}{n}},$$ and the \emph{algebraic entropy} of $\phi:G\to G$ is $$\mathrm{ent}(\phi)=\sup\{H(\phi,F): F\ \text{is a finite subgroup of } G\}.$$

Since the definition is based on finite subgroups $F$, and in particular $F$ is contained in the torsion part $t(G)$ of $G$, the algebraic entropy depends only on the restriction of $\phi$ on $t(G)$, that is $\mathrm{ent}(\phi)=\mathrm{ent}(\phi\restriction_{t(G)})$.
The basic properties of the algebraic entropy can be found in \cite{DGSZ,W}. The most relevant of them, known also as
Addition Theorem, can be found in \S \ref{background} (Theorem \ref{AT}), which collects all relevant properties of the algebraic entropy used in this paper.

As far as the algebraic entropy is concerned,  the right Bernoulli shift restricted to the direct sum $K^{(\mathbb N_0)}$ turned out to be more relevant (while the restriction of the left Bernoulli shift $_K\beta\restriction_{K^{(\mathbb N_0)}}$ has algebraic entropy $0$, see Example \ref{NewExample}). More precisely, for a non-trivial finite abelian group $K$ the restriction $\beta_K\restriction_{K^{(\mathbb N_0)}}$ has entropy  $\log|K|$ \cite[Example 1.9]{DGSZ} and
one can show that every function $f$ defined on all endomorphisms of torsion abelian groups with values in the extended non-negative reals and satisfying $f(\beta_{\mathbb Z(p)}\restriction_{\mathbb Z(p)^{(\mathbb N_0)}})=\log |p|$, the Addition Theorem and
a few other natural properties (namely, Lemmas \ref{conjugation_by_iso}, \ref{log_law} and Remark \ref{lemma1} (b)) must necessarily coincide with the algebraic entropy $\mathrm{ent}(-)$ \cite[Theorem 6.1]{DGSZ}.

With the aim of computing the entropy of other endomorphisms of abelian groups, in this paper we consider a modification of the generalized shifts, introduced in \cite{AKH}.

\begin{deff}\cite{AKH}
For a non-empty set $\Gamma$, an arbitrary map $\lambda:\Gamma\rightarrow\Gamma$  and an abelian group  $K$ the \emph{generalized shift} $\sigma_{\lambda,K}: K^\Gamma\to K^\Gamma$ is defined by $(x_i)_{i\in\Gamma}\mapsto(x_{\lambda(i)})_{i\in\Gamma}$ for every $(x_i)_{i\in\Gamma}\in K^\Gamma$.
\end{deff}

When there is no possibility of confusion we write $\sigma_\lambda$ instead of $\sigma_{\lambda,K}$.
In case $|K|>1$, the subgroup $K^{(\Gamma)}$ of $K^\Gamma$ is $\sigma_{\lambda,K}$-invariant if and only if $\lambda$ has finite fibers (see Lemma \ref{lemma2}), and it is possible to consider the restriction  $\sigma_{\lambda,K}^\oplus=\sigma_{\lambda,K}\restriction_{K^{(\Gamma)}}$ of  $\sigma_{\lambda,K}$ to $K^{(\Gamma)}$.
Again, when there is no possibility of confusion we write $\sigma_{\lambda,K}^\oplus$ simply as $\sigma_{\lambda}^\oplus$, $\sigma_{\lambda,K}$ or just $\sigma_\lambda$.

There is a close relation between the Bernoulli shifts and the generalized shifts. For example, the left Bernoulli shift and the two-sided Bernoulli shift are generalized shifts (see Examples \ref{NewExample} and \ref{theorem4} (d) respectively), while the right Bernoulli shift $\beta_{K}\restriction_{K^{(\mathbb N_0)}}$ cannot be obtained as a  generalized shift $\sigma_{\lambda}^\oplus$ from any function $\lambda:\mathbb N_0\to\mathbb N_0$.
Nevertheless, it can be ``approximated" quite well by the generalized shift $\sigma_{\psi}^\oplus$ of $K^{(\mathbb N_0)}$ induced by the map $\psi: \mathbb N _0 \to \mathbb N_0$ defined by $\psi(i)={i-1}$ for every $i>0$ and $\psi({0})={0}$. Indeed, both $\sigma_{\psi}^\oplus$ and   $\beta_{K}\restriction_{K^{(\mathbb N_0)}}$ leave invariant the finite-index subgroup $H=K^{(\mathbb N)}$  and $\sigma_{\psi}^\oplus\restriction_H =\beta_{K}\restriction_H$ (in particular, they have the same entropy $\log|K|$).

In this paper we compute the entropy of an arbitrary generalized shift $\sigma_{\lambda}^\oplus:K^{(\Gamma)}\to K^{(\Gamma)}$.
More precisely, we show that  $\mathrm{ent}(\sigma_{\lambda}^\oplus)$, depends only on combinatorial properties of the map $\lambda$ and the cardinality of $K$.
To prove this we analyze the structure of the map $\lambda$ and, more specifically, its (iterated) counter-images
(since, in some sense, the iterations of $\lambda$ and the iterations of $\sigma_\lambda$ ``go in opposite directions'').
Roughly speaking we decompose the generalized shift $\sigma_{\lambda}^\oplus$ in ``independent elementary shifts'' (as the generalized shift $\sigma_{\psi}^\oplus$ considered above), which have the same algebraic entropy as the right Bernoulli shift $\beta_K\restriction_{K^{(\mathbb N_0)}}$, and the number $s_\lambda$ of these independent elementary shifts, multiplied by $\log|K|$, gives precisely $\mathrm{ent}(\sigma_{\lambda}^\oplus)$ (see Theorem \ref{theorem6}).

\bigskip
\noindent\textbf{Convention.}
From now on we assume that $\Gamma$ is a non-empty set, $\lambda:\Gamma\rightarrow\Gamma$ is an arbitrary map, and $K$ is a non-trivial finite abelian group.
We denote by $G_\Gamma$ the group $K^{(\Gamma)}$ and for a subset $A$ of $\Gamma$, we identify $K^{(A)}$ with the subgroup $\{x\in G_\Gamma:\mathrm{supp}(x)\subseteq A\}$ of $G_\Gamma$, and we denote $K^{(A)}$ by $G_A$. In case $A=\emptyset$ we assume that $G_\emptyset=\{0\}$.
We denote the generalized shift based on $G_\Gamma$ and $\lambda:\Gamma\to\Gamma$ simply by $\sigma_{\lambda}$, writing it in some cases $\sigma_{\lambda,K}$, when we need to specify the group.

\section{Background on algebraic entropy}\label{background}

We start collecting basic results on algebraic entropy, mainly from \cite{DGSZ,W}.

\begin{lemma}\label{conjugation_by_iso}\emph{\cite[Proposition 1.2]{W}}
Let $G$, $H$ be abelian groups and $\phi:G\to G$, $\eta:H\to H$ endomorphisms. If there exists an isomorphism $\xi:G\to H$ such that $\phi=\xi^{-1}\eta\xi$, then $\mathrm{ent}(\phi)=\mathrm{ent}(\eta)$.
\end{lemma}

\begin{lemma}\label{log_law}\emph{\cite[Proposition 1.3]{W}}
Let $G$ be an abelian group and $\phi:G\to G$ an endomorphism. Then $\mathrm{ent}(\phi^k) = k\,\mathrm{ent}(\phi)$ for every non-negative integer $k$. If $\phi$ is an
automorphism, then $\mathrm{ent}(\phi^k) = |k|\mathrm{ent}(\phi)$ for every $k\in \mathbb Z$.
\end{lemma}

The following is one of the main results on algebraic entropy.

\begin{theorem}[Addition Theorem]\label{AT}\emph{\cite[Theorem 3.1]{DGSZ}}
Let $G$ be a torsion abelian group, $\phi:G\to G$ an endomorphism and $H$ a $\phi$-invariant subgroup of $G$. If $\overline{\phi}:G/H\to G/H$ is the endomorphism induced on the quotient by $\phi$, then $\mathrm{ent}(\phi)=\mathrm{ent}(\phi\restriction_H)+\mathrm{ent}(\overline\phi)$.
\end{theorem}

\begin{remark}\label{lemma1} Let $G$ be an abelian group and $\phi:G\to G$ an endomorphism.
\begin{itemize}
\item[(a)] A particular case of the above theorem was proved in \cite[Proposition 1.4]{W}: if $n\in\mathbb N$, $G=\bigoplus_{i=1}^nG_i$  and $G_i$ is a $\phi$-invariant subgroup of $G$ for $i= 1,\ldots, n$,
then $\mathrm{ent}(\phi)=\sum_{i=1}^n\mathrm{ent}(\phi\restriction_{G_i})$.
\item[(b)] If the group $G$ is a direct limit of $\phi$-invariant subgroups $\{G_i:i\in I\}$, then ${\mathrm{ent}}(\phi)=\sup_{i\in I}\mathrm{ent}(\phi {\restriction_{G_i}})$ \cite[Proposition 1.6]{DGSZ}.
\item[(c)] Using (b), one can extend (a) to arbitrary direct sums $G=\bigoplus_{i\in I}G_i$.
\end{itemize}
\end{remark}

\begin{lemma}\label{locnilp}\emph{\cite{DGSZ}}
Let $G$ be an abelian group and $\phi:G\to G$ an endomorphism.
\begin{itemize}
	\item[(a)]If $X$ is a set of generators of $G$ and for every $x\in X$ there exists $d_x\in\mathbb N$ such that $\phi^{d_x}(x)=0$, then $\mathrm{ent}(\phi)=0$.
	\item[(b)]If $\phi$ is periodic, then $\mathrm{ent}(\phi)=0$.
\end{itemize}
\end{lemma}

\section{Strings and an effective equivalence relation}

Now we introduce a notion that will play a prominent role in the computation of the algebraic entropy of the generalized shifts.

\begin{deff}
\begin{itemize}
	\item[(a)] A \emph{string} of $\lambda$ (in $\Gamma$) is an infinite sequence of pairwise distinct elements $S=\{m_t\}_{-t\in\mathbb N_0}$ such that $\lambda(m_t)=m_{t+1}$ for every $-t\in\mathbb N$.
	\item[(b)] Let $s_\lambda:=\sup\{|\mathcal F|: \mathcal F\ \text{is a family of pairwise disjoint strings of $\lambda$} \}$, and
	\item[(c)] $\Gamma^+:=\bigcap_{n=1}^\infty\lambda^n(\Gamma)$.
\end{itemize}
\end{deff}

A string $S=\{m_t\}_{-t\in\mathbb N_0}$ of $\lambda$ in $\Gamma$ is said to be \emph{acyclic} if $\lambda^n(m_0)\not\in S$ for every $n\in\mathbb N$. The next claim is easy to prove.

\begin{claim}\label{acyclic}
Each string $S$ of $\lambda$ in $\Gamma$ contains an acyclic string $S'$ of $\lambda$.
\end{claim}

The importance of $\Gamma^+$ consists in the fact that it contains all strings of $\lambda$ as well as all periodic points of $\lambda$. Obviously, $\Gamma^+ = \Gamma$ if and only if $\lambda$ is surjective.  In general, the restriction $\lambda\restriction_{\Gamma^+}: \Gamma^+\to \Gamma^+$ need not be surjective (but this holds true if $\lambda$ has finite fibers).

\medskip
Consider the following equivalence relation: $i\Re_\lambda j$ in $\Gamma$ if and only if there exist $m,n\in\mathbb N_0$ such that $\lambda^n(i)=\lambda^m(j)$.
Let $\alpha_\lambda:=|\{i/\Re_\lambda\in\Gamma/\Re_\lambda:i/\Re_\lambda\ \text{contains at least one string of $\lambda$}\}|$. Obviously, $\alpha_\lambda\leq s_\lambda$.

\begin{example}\label{theorem3-first_part}
Suppose that $\lambda$ is injective.
\begin{itemize}
    \item[(a)] The relation $\Re_\lambda$ in this particular case becomes: $\Re_\lambda=\{(i,j)\in\Gamma\times\Gamma:\exists m\in\mathbb Z:
i=\lambda^m(j)\}$.
    \item[(b)]The relation $\Re_\lambda$ has three types of equivalence classes:
    \begin{itemize}
	\item[(b$_1$)] finite equivalence classes,
	\item[(b$_2$)] infinite equivalence classes contained in $\Gamma^+$ (i.e., containing a string of $\lambda$), of the form
	$$\{\ldots,\lambda^{-1}(i),i,\lambda(i),\lambda^2(i),\ldots\}.$$
	\item[(b$_3$)] infinite equivalence classes non-contained in $\Gamma^+$ (i.e., non-containing a string of $\lambda$), of the form $\{i,\lambda(i),\lambda^2(i),\ldots\}$ with $i\in\Gamma\setminus\lambda(\Gamma)$.
	\end{itemize}
    \item[(c)] Then $\alpha_\lambda$ is the number of the infinite equivalence classes in (b$_2$). Consequently $\alpha_\lambda=s_\lambda$.
\end{itemize}
\end{example}

\begin{example}\label{example5-firstpart}
Let $\Gamma=\mathbb N_0$. For every $n\in\mathbb N$ let $\varphi_n:\Gamma\to\Gamma$ be defined by
\begin{equation*}
\varphi_n(m)=
\begin{cases}
0 & \text{if}\ m=0,1,\ldots,n, \\
m-1 & \text{otherwise}.
\end{cases}
\end{equation*}
The diagram for $\varphi_n$ with $n>1$ is the following:
\begin{equation*}
\xymatrix@-1pc{
  &   &        &     & \vdots \ar@{->}[d] \\
  &   &        &     & n+2 \ar@{->}[d] \\
  &   &        &     & n+1 \ar@{->}[d]\\
1 \ar@{->}[drr] & 2 \ar@{->}[dr] & \ldots & n-1 \ar@{->}[dl] & n \ar@{->}[dll]  \\
  &   & 0\ar@(dl,dr)[]    &  & \\
}
\bigskip
\end{equation*}
In this case $s_{\varphi_n}=\alpha_{\varphi_n}=1$ and $\lambda(\Gamma)=\Gamma^+=\Gamma\setminus\{1,\ldots,n-1\}$.

\medskip
For every $n\in\mathbb N$, let $\psi_n:\Gamma\to\Gamma$ be defined by
\begin{equation*}
\psi_n(m)=
\begin{cases}
0 & \text{if}\ m=0,1,\ldots,n, \\
(k-1)n+i & \text{if}\ m=kn+i\ \text{with}\ 0\leq i<n\ \text{and}\ k\in\mathbb N.
\end{cases}
\end{equation*}
The diagram for $\psi_n$ is the following:
\begin{equation*}
\xymatrix@-1pc{
\vdots \ar@{->}[d]& \vdots \ar@{->}[d] & \ldots & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] \\
2n+1 \ar@{->}[d] & 2n+2 \ar@{->}[d] & \ldots & 3n-1 \ar@{->}[d] & 3n \ar@{->}[d] \\
n+1 \ar@{->}[d]  & n+2 \ar@{->}[d]  & \ldots & 2n-1 \ar@{->}[d] & 2n \ar@{->}[d] \\
1 \ar@{->}[drr]  & 2 \ar@{->}[dr]   & \ldots & n-1 \ar@{->}[dl] & n \ar@{->}[dll] \\
                 &                  & 0\ar@(dl,dr)[]&                  & \\
}
\bigskip
\end{equation*}
For this function $s_{\psi_n}=n$, $\alpha_{\psi_n}=1$ and $\Gamma^+=\Gamma$. Note that $\psi_1=\varphi_1$.

\medskip
Let $\Gamma=\mathbb N_0\times\mathbb N_0$ and $\lambda_0:\Gamma\to\Gamma$ be
defined by
\begin{equation*}
\lambda_0(m,k)=
\begin{cases}
(0,0) & \text{if}\ m=k=0, \\
(m-1,0) & \text{if}\ k=0\ \text{and}\ m\in\mathbb N, \\
(m,k-1) & \text{if}\ m\in\mathbb N_0\ \text{and}\ k\in\mathbb N.
\end{cases}
\end{equation*}
The diagram for $\lambda_0$ is the following:
\begin{equation*}
\xymatrix@-1pc{
\vdots\ar[d] & \vdots\ar[d] & \vdots\ar[d] & \ldots\ar[d] & \vdots\ar[d] & \ldots\ar[d] \\
(0,2)\ar[d] & (1,2)\ar[d] & (2,2)\ar[d] & \ldots\ar[d] & (m,2)\ar[d] & \ldots\ar[d] \\
(0,1) \ar[d] & (1,1)\ar[d] & (2,1)\ar[d] & \ldots\ar[d] & (m,1)\ar[d] & \ldots\ar[d] \\
(0,0)\ar@(dl,dr)[] & (1,0)\ar[l] & (2,0)\ar[l] & \ldots\ar[l] & (m,0)\ar[l] & \ldots\ar[l] \\
}
\bigskip
\end{equation*}
In this case $s_{\lambda_0}=\omega$, $\alpha_{\lambda_0}=1$ and $\Gamma^+=\Gamma$.
\end{example}

\section{The entropy of the generalized shift}

\begin{remark}\label{subgroup}
Let $\mu:\Gamma\to\Gamma$ be a function. If $H$ is a subgroup of an abelian group $L$, then $H^{\Gamma}$ is a $\sigma_{\mu,L}$-invariant subgroup of $L^\Gamma$. Moreover, $\sigma_{\mu,L}\restriction_{H^{\Gamma}}=\sigma_{\mu,H}:H^\Gamma\to H^\Gamma$.
Analogously, if $L^{(\Gamma)}$ is a $\sigma_{\mu,L}^\oplus$-invariant subgroup of $L^\Gamma$, then $H^{(\Gamma)}$ is a $\sigma_{\mu,L}^\oplus$-invariant subgroup of $L^{(\Gamma)}$, and $\sigma_{\mu,L}^\oplus\restriction_{H^{(\Gamma)}}=\sigma_{\mu,H}^\oplus:H^{(\Gamma)}\to H^{(\Gamma)}$.
\end{remark}

\begin{claim}\label{G_F}
Let $x\in G_\Gamma$ and $F=\mathrm{supp}(x)$. Then for every $m\in\mathbb N$:
\begin{itemize}
	\item[(a)]$\mathrm{supp}(\sigma_\lambda^m(x))=\lambda^{-m}(F)$;
	\item[(b)]$\sigma_\lambda^m(G_F)\leq G_{\lambda^{-m}(F)}$;
	\item[(c)]$T_m(\sigma_\lambda,G_F)\leq G_{F\cup\lambda^{-1}(F)\cup\ldots\cup\lambda^{-m+1}(F)}$.
\end{itemize}
\end{claim}
\begin{proof}
(a) If $y=\sigma_\lambda(x)$, then $y_i=x_{\lambda(i)}\neq0$ if and only if $\lambda(i)\in F$, that is, $i\in\lambda^{-1}(F)$, and so $\mathrm{supp}(y)=\lambda^{-1}(F)$. Proceeding by induction it is possible to prove that $\mathrm{supp}(\sigma_\lambda^m(x))=\lambda^{-m}(F)$ for every $m\in\mathbb N$.

(b) Follows from (a).

(c) By (b) $T_m(\sigma_\lambda,G_F)\leq G_F+G_{\lambda^{-1}(F)}+\ldots+ G_{\lambda^{-m+1}(F)}$ and $G_F+G_{\lambda^{-1}(F)}+\ldots+ G_{\lambda^{-m+1}(F)}\subseteq G_{F\cup\lambda^{-1}(F)\cup\ldots\cup\lambda^{-m+1}(F)}$.
\end{proof}

The next lemma shows the relevance of our following assumption on $\lambda$ of having finite fibers.

\begin{lemma}\label{lemma2}
The following conditions are equivalent:
\begin{itemize}
\item[(a)] $\lambda^{-1}(i)$ is finite for each $i\in\Gamma$;
\item[(b)] $\sigma_{\lambda,K}(G_\Gamma)\subseteq G_\Gamma$;
\item[(c)] $\sigma_{\lambda,L}(L^{(\Gamma)})\subseteq L^{(\Gamma)}$ for every non-trivial abelian group $L$.
\end{itemize}
\end{lemma}
\begin{proof}
(b)$\Rightarrow$(a) Let $i\in\Gamma$ and $x\in G_{\{i\}}\setminus\{0\}$. By Claim \ref{G_F}(a) $\mathrm{supp}(\sigma_\lambda(x))=\lambda^{-1}(i)$ and by the assumption $\sigma_\lambda(x)\in G_\Gamma$, hence $\lambda^{-1}(i)$ is finite.

A similar argument shows that (a)$\Rightarrow$(c) and (c)$\Rightarrow$(b) is obvious.
\end{proof}

\bigskip
\noindent\textbf{Convention.}
From now on we suppose that $\lambda$ has finite fibers, that is, $\lambda^{-1}(i)$ is finite for every $i\in\Gamma$.

\bigskip
\begin{proposition}\label{composition}
Let $L$ be an abelian group with at least two elements and let $\mu,\nu:\Gamma\to\Gamma$ be functions with finite fibers. For $\sigma_\mu,\sigma_\nu:L^\Gamma\to L^\Gamma$ and $\sigma_{\mu}^\oplus,\sigma_{\nu}^\oplus:L^{(\Gamma)}\to L^{(\Gamma)}$:
\begin{itemize}
    \item[(a)] $\sigma_\mu\circ\sigma_\nu=\sigma_{\mu\circ\nu}$ and $\sigma_{\mu}^\oplus\circ\sigma_{\nu}^\oplus=\sigma_{\mu\circ\nu}^\oplus$ (hence $\sigma_\mu^m=\sigma_{\mu^m}$ and $(\sigma_{\mu}^\oplus)^m=\sigma_{\mu^m}^\oplus$ for every $m\in\mathbb N$).
    \item[(b)] \emph{\cite{AKH}} The following conditions are equivalent:
\begin{itemize}
	\item[(b$_1$)]$\mu$ is injective (respectively, surjective);
	\item[(b$_2$)]$\sigma_\mu$ is surjective (respectively, injective);
	\item[(b$_3$)]$\sigma_{\mu}^\oplus$ is surjective (respectively, injective).
\end{itemize}
    \item[(c)] In particular, the following conditions are equivalent:
\begin{itemize}
	\item[(c$_1$)]$\mu$ is bijective;
	\item[(c$_2$)]$\sigma_\mu$ is an automorphism of $L^\Gamma$;
	\item[(c$_3$)]$\sigma_{\mu}^\oplus$ is an automorphism of $L^{(\Gamma)}$.
\end{itemize}
    In this case, $(\sigma_\mu)^{-1}=\sigma_{\mu^{-1}}$ and $(\sigma_{\mu}^\oplus)^{-1}=\sigma_{\mu^{-1}}^\oplus$.
\end{itemize}
\end{proposition}

Note that the equivalences (b$_1$)$\Leftrightarrow$(b$_2$) and (c$_1$)$\Leftrightarrow$(c$_2$) hold without any assumption on the fibers of $\mu$ and $\nu$.

\begin{corollary}\label{G_Fd}
For every $m\in\mathbb N$, $\ker\sigma_\lambda^m=G_{\Gamma\setminus\lambda^m(\Gamma)}$.
\end{corollary}
\begin{proof}
It suffices to prove that $\ker\sigma_\lambda=G_{\Gamma\setminus\lambda(\Gamma)}$ and then apply Proposition \ref{composition}(a). If $x\in\ker\sigma_\lambda$, equivalently $\mathrm{supp}(\sigma_\lambda(x))=\emptyset$. By Claim \ref{G_F}(a) $\mathrm{supp}(\sigma_\lambda(x))=\lambda^{-1}(\mathrm{supp}(x))$. Then $\mathrm{supp}(\sigma_\lambda(x))=\lambda^{-1}(\mathrm{supp}(x))=\emptyset$ if and only if $\mathrm{supp}(x)\cap\lambda(\Gamma)=\emptyset$. This is the same as $\mathrm{supp}(x)\subseteq \Gamma\setminus \lambda(\Gamma)$, that is, $x\in G_{\Gamma\setminus \lambda(\Gamma)}$.
\end{proof}

The next lemma gives a characterization (in terms of $\lambda$) of the $\sigma_\lambda$-invariance of the subgroups $G_A$ of $G_\Gamma$.

\begin{lemma}\label{invariance}
If $A\subseteq \Gamma$, then $G_A$ is $\sigma_\lambda$-invariant if and only if $\lambda^{-1}(A)\subseteq A$. If $A$ is also $\lambda$-invariant, then $\sigma_\lambda\restriction_{G_A}=\sigma_{\lambda\restriction_A}$.
\end{lemma}
\begin{proof}
The condition $\sigma_\lambda(G_A)\subseteq G_A$ is equivalent to $\sigma_\lambda(G_i)\subseteq G_A$ for every $i\in A$, that is, $\lambda^{-1}(i)\subseteq A$ for every $i\in A$, which is equivalent to $\lambda^{-1}(A)\subseteq A$.
Assume now that $\lambda^{-1}(A)\cup\lambda(A)\subseteq A$. Then it is possible to consider both $\sigma_\lambda\restriction_{G_A}$ and $\sigma_{\lambda\restriction_A}$. It is clear that they coincide on $G_A$.
\end{proof}

Lemma \ref{invariance} shows that in case $\lambda^{-1}(A)\subseteq A$ for $A\subseteq \Gamma$, it is possible to consider $\sigma_\lambda\restriction_{G_A}:G_A\to G_A$.

\begin{remark}\label{relambda}
We see here that we can assume that for the relation $\Re_\lambda$ there exists only one equivalence class in $\Gamma$ (so coinciding with the whole $\Gamma$).

\smallskip
Indeed, if $i/\Re_\lambda$ is a generic equivalence class, then $i/\Re_\lambda\supseteq \lambda(i/\Re_\lambda)\cup\lambda^{-1}(i/\Re_\lambda)$. By Lemma \ref{invariance}
\begin{equation}\label{Relambda}
\sigma_\lambda\restriction_{i/\Re_\lambda}=\sigma_{\lambda\restriction_{i/\Re_\lambda}}.
\end{equation}
Let now $R$ be a set of representing elements of $\Re_\lambda$ in $\Gamma$. Then $G_\Gamma=\bigoplus_{i\in R}G_{i/\Re_\lambda}$. By Remark \ref{lemma1}(c) and \eqref{Relambda} $$\mathrm{ent}(\sigma_\lambda)=\sum_{i\in R}\mathrm{ent} (\sigma_\lambda\restriction_{G_{i/\Re_\lambda}})=\sum_{i\in R}\mathrm{ent}(\sigma_{\lambda\restriction_{i/\Re_\lambda}}),$$ and so we can assume that $R$ is a singleton.
\end{remark}

The next result gives the very useful formula \eqref{magic}, which is applied in the proof of the main theorem.

\begin{remark}\label{add-rem}
Let $\Gamma=\Gamma'\cup\Gamma''$ be a partition of $\Gamma$. Then $\lambda^{-1}(\Gamma')\subseteq \Gamma'$ if and only if $\lambda(\Gamma'')\subseteq\Gamma''$. Suppose that these equivalent conditions hold. By Lemma \ref{invariance} $G_{\Gamma'}$ is $\sigma_\lambda$-invariant.
\begin{itemize}
    \item[(a)]  Let $p_2:G_\Gamma=G_{\Gamma'}\oplus G_{\Gamma''}\to G_{\Gamma''}$ and $\pi:G_\Gamma\to G_\Gamma/G_{\Gamma'}$ be the canonical projections.
    Denote by $\xi:G_\Gamma/G_{\Gamma'}\to G_{\Gamma''}$ the (unique) isomorphism  such that $p_2=\xi\circ\pi$.   Finally, let
  $\overline{\sigma_\lambda}:G_\Gamma/G_{\Gamma'}\to G_\Gamma/G_{\Gamma'}$ be the homomorphism induced by $\sigma_\lambda$. Then $\overline{\sigma_\lambda}=\xi^{-1}\sigma_{\lambda\restriction_{\Gamma''}}\xi$.
To better explain the situation, this means that the following diagram commutes:
\begin{equation*}
\xymatrix@-1pc{
G_\Gamma \ar@{->}[dr]_{\pi} \ar@{->}[rr]^{p_2} & & G_{\Gamma''} \ar[rr]^{\sigma_{\lambda\restriction_{\Gamma''}}}& & G_{\Gamma''} \\
 & G_\Gamma/G_{\Gamma'} \ar@{->}[ur]_{\xi} \ar[rr]^{\overline{\sigma_\lambda}}& & G_\Gamma/G_{\Gamma'} \ar@{->}[ur]_{\xi}\\
}
\end{equation*}
By Lemma \ref{conjugation_by_iso} $\mathrm{ent}(\overline{\sigma_\lambda})=\mathrm{ent}(\sigma_{\lambda\restriction_{\Gamma''}})$.
   \item[(b)] By (a) and Theorem \ref{AT}
    \begin{equation}\label{magic}
    \mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_{\lambda}\restriction_{G_{\Gamma'}})+\mathrm{ent}({\sigma_{\lambda\restriction_{\Gamma''}}}).
    \end{equation}
     Therefore,
	\begin{equation}\label{magic2}
	\mathrm{ent}(\sigma_\lambda)\geq\mathrm{ent}(\sigma_{\lambda}\restriction_{G_{\Gamma'}})\ \text{and}\ \mathrm{ent}(\sigma_\lambda)\geq\mathrm{ent}({\sigma_{\lambda\restriction_{\Gamma''}}}).
	\end{equation}
\end{itemize}
\end{remark}

The next corollary shows that $\mathrm{ent}(\sigma_\lambda)$ can be computed from its restriction to $\Gamma^+\subseteq\Gamma$.

\begin{corollary}\label{corollary}
\begin{itemize}
	\item[(a)]$\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_{\lambda\restriction_{\Gamma^+}})$.
	\item[(b)]In particular, for each $k\in\mathbb N$, $\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_{\lambda\restriction_{\lambda^k(\Gamma)}})$.
	\item[(c)]If there exists a non-empty finite subset $F$ of $\Gamma$ such that $\Gamma=\bigcup_{n\in\mathbb N_0}\lambda^{n}(F)$, then $\Gamma^+=\mathrm{Per}(\lambda)$ and consequently $\mathrm{ent}(\sigma_\lambda)=0$.
\end{itemize}
\end{corollary}
\begin{proof}
(a) Since $\lambda(\Gamma^+)\subseteq \Gamma^+$, by \eqref{magic} in Remark \ref{add-rem}(b) $\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_\lambda\restriction_{G_{\Gamma\setminus\Gamma^+}})+\mathrm{ent}(\sigma_{\lambda\restriction_{\Gamma^+}})$. We prove that $\mathrm{ent}(\sigma_\lambda\restriction_{G_{\Gamma\setminus\Gamma^+}})=0$. Let $x\in G_{\Gamma\setminus\Gamma^+}$. Then $\mathrm{supp}(x)\subseteq \Gamma\setminus\Gamma^+$. For every $i\in\mathrm{supp}(x)$ there exists $h(i)\in\mathbb N$ such that $i\not\in\lambda^{h(i)}(\Gamma)$, and so $\lambda^{-h(i)}(i)=\emptyset$. Let $h(x)=\max\{h(i):i\in\mathrm{supp}(x)\}$. By Claim \ref{G_F}(a) $\mathrm{supp}(\sigma_\lambda^{h(x)}(x))=\lambda^{-h(x)}(\mathrm{supp}(x))$, which is empty, and so $\sigma_\lambda^{h(x)}(x)=0$. By Lemma \ref{locnilp}(a) $\mathrm{ent}(\sigma_\lambda\restriction_{G_{\Gamma\setminus\Gamma^+}})=0$.

(b) Follows from (a) since $\mathrm{ent}(\sigma_\lambda)\geq\mathrm{ent}(\sigma_{\lambda\restriction_{\lambda^k(\Gamma)}})\geq\mathrm{ent}(\sigma_{\lambda\restriction_{\Gamma^+}})$, where \eqref{magic2} in Remark \ref{add-rem}(b) is applied twice.

(c) Clearly $\Gamma^+\supseteq\mathrm{Per}(\lambda)$. Let $i\in\Gamma^+$. For every $n\in\mathbb N$ there exists $i_n\in\Gamma$ such that $\lambda^n(i_n)=i$. Moreover, for $n\in\mathbb N$ there exist $m_n\in\mathbb N$ and $j_{m_n}\in F$ such that $\lambda^{m_n}(j_{m_n})=i_n$ and so $\lambda^{n+m_n}(j_{m_n})=i$. Since $F$ is finite, there exists $j\in F$
such that for a strictly increasing sequence $({n_k})_{k\in\mathbb N}$  in $\mathbb N$ one has $j_{m_{n_k}}=j$ for all $k\in\mathbb N$.
Then $\lambda^{n_k+m_{n_k}}(j)=i$ for all $k\in\mathbb N$. Choose $k\in\mathbb N$ such that $n_k > n_1 + m_{n_1}$, then $n_k + m_{n_k} > n_1 + m_{n_1}$ as well.
Therefore, $i=\lambda^{ n_1+m_{n_1}}(j)=\lambda^{n_k + m_{n_k}}(j)$ yields  $i\in\mathrm{Per}(\lambda)$.
\end{proof}

\begin{example}\label{NewExample} Let $\lambda: \mathbb N_0 \to \mathbb N_0$ be defined by $\lambda(n) = n+1$. Then $\sigma_\lambda: K^{(\mathbb N_0)}\to K^{(\mathbb N_0)}$ coincides with the left Bernoulli shift $_K\beta$ of $K^{(\mathbb N_0)}$. Since $\mathbb N_0^+=\emptyset$ (in the notation of Corollary \ref{corollary}), we deduce  from Corollary \ref{corollary}(c) that $\mathrm{ent}(\sigma_\lambda)=0$.
\end{example}

Another application of Corollary \ref{corollary} is the following example, in which $\Gamma$ is a compact metric space.

\begin{example}\label{example19}
Let $(\Gamma,d)$ be a compact metric space and $\lambda:\Gamma\rightarrow\Gamma$ be a contraction (i.e., $d(\lambda(x),\lambda(y)) < d(x,y)$  for every pair of distinct points $x,y$ of $\Gamma$) such that for each $i\in\Gamma$, $\lambda^{-1}(i)$ is finite. Then $\mathrm{ent}(\sigma_\lambda)=0$.

Indeed, using a standard compactness argument one proves that $\lambda(\Gamma^+)=\Gamma^+\ne \emptyset$. Moreover, $|\Gamma^+|=1$ since otherwise
there exist $x,y\in \Gamma^+$ such that $d(x,y)=\mathrm{diam}\, \Gamma^+$. By $\lambda(\Gamma^+)=\Gamma^+$ there exist $x',y'\in \Gamma^+$ with $\lambda(x') = x$ and $\lambda(y') =y$.
So $\mathrm{diam}\, \Gamma^+= d(x,y)= d(\lambda(x'),\lambda(y')) < d(x',y')$, a contradiction. Therefore, $|\Gamma^+|=1$, so Corollary \ref{corollary}(a) applies.
\end{example}

\subsection{The main theorem}

For a set $X$, in what follows we use the following notation: $$|X|^*:=
\begin{cases}
|X| & \text{in case}\ |X|\ \text{is finite},\\
+\infty & \text{in case}\ |X|\ \text{is infinite}.
\end{cases}$$

\begin{deff}
Two strings $S$ and $S'$ of $\lambda$ in the set $\Gamma$ are said to be \emph{strongly disjoint} if $S$ and $\lambda^n(S')$ are disjoint for every $n\in\mathbb N_0$ and $S'$ and $\lambda^n(S)$ are disjoint for every $n\in\mathbb N_0$.
\end{deff}

By the definition of string we immediately have the following result.

\begin{claim}\label{strongly-disjoint}
If $s_\lambda = n\in\mathbb N_0$ is finite, then $\Gamma$ contains $n$ strings of $\lambda$ that are pairwise strongly disjoint.
\end{claim}

The next is our main theorem, which calculates the entropy of a generalized shift $\sigma_\lambda:G_\Gamma\to G_\Gamma$, proving that it depends only on the function $\lambda:\Gamma\to\Gamma$ and on the cardinality of $K$.

\begin{theorem}\label{theorem6}
Let $\Gamma$ be a set, $\lambda:\Gamma\to\Gamma$ a function such that $\lambda^{-1}(i)$ is finite for every $i\in\Gamma$, and consider $\sigma_\lambda:G_\Gamma\to G_\Gamma$, where $G_\Gamma=K^{(\Gamma)}$ and $K$ is a non-trivial finite abelian group. Then $$\mathrm{ent}(\sigma_\lambda)=|s_\lambda|^*\log|K|.$$
\end{theorem}
\begin{proof}
By Corollary \ref{corollary}(a) and Remark \ref{relambda} we can assume without loss of generality that $\Gamma=\Gamma^+$ and that there exists only one equivalence class for $\Re_\lambda$.

\medskip
Suppose that $s_\lambda=n$ for some $n\in\mathbb N_0$.
If $n>0$, there exist $n$ pairwise strongly disjoint acyclic strings of $\lambda$ $$S_1:=\{m^1_t\}_{-t\in\mathbb N_0},\ldots,S_n:=\{m^n_t\}_{-t\in\mathbb N_0}$$ in $\Gamma$ by Claims \ref{acyclic} and \ref{strongly-disjoint}.

Let $\Psi=\emptyset$ if $n=0$ and $\Psi:=S_1\cup\ldots\cup S_n$ otherwise. Then $\lambda^{-1}(\Psi)\subseteq\Psi$, since $\Gamma=\Gamma^+$, and so $G_\Psi$ is $\sigma_\lambda$-invariant by Lemma \ref{invariance}. Let $\nu:=\lambda\restriction_{\Gamma\setminus\Psi}$.
By \eqref{magic} in Remark \ref{add-rem}(b) $$\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_\lambda\restriction_{G_\Psi})+\mathrm{ent}(\sigma_{\nu}).$$
For $F=\{\lambda(m^1_0),\ldots,\lambda(m^n_0)\}$, $\Gamma\setminus\Psi=\bigcup_{n\in\mathbb N_0}\nu^n(F)$, since the strings are acyclic. By Corollary \ref{corollary}(c) $\mathrm{ent}(\sigma_{\nu})=0$ and so
\begin{equation}\label{eq2}
\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_\lambda\restriction_{G_\Psi}).
\end{equation}
Now $\Psi$ is disjoint union of $S_1,\ldots, S_n$ and so $G_\Psi=G_{S_1}\oplus\ldots\oplus G_{S_n}$. Moreover, $\lambda^{-1}(S_j)\subseteq S_j$ since $\Gamma=\Gamma^+$, and so by Lemma \ref{invariance} $G_{S_j}$ is $\sigma_\lambda$-invariant for each $j\in\{1,\ldots,n\}$. By Remark \ref{lemma1}(b)
\begin{equation}\label{eq3}
\mathrm{ent}(\sigma_\lambda\restriction_{G_\Psi})=\mathrm{ent}(\sigma_\lambda\restriction_{G_{S_1}})+\ldots+\mathrm{ent}(\sigma_\lambda\restriction_{G_{S_n}}).
\end{equation}
For every $j\in\{1,\ldots,n\}$, since $\sigma_\lambda\restriction_{G_{S_j}}$ is precisely the right Bernoulli shift $\beta_K\restriction_{G_{S_j}}:G_{S_j}\to G_{S_j}$,
\begin{equation}\label{eq4}
\mathrm{ent}(\sigma_\lambda\restriction_{G_{S_j}})=\log|K|.
\end{equation}

By (\ref{eq2}), (\ref{eq3}) and (\ref{eq4}) $\mathrm{ent}(\sigma_\lambda)=n\log|K|=s_\lambda\log|K|$.

\medskip
Assume now that $|s_\lambda|^*=+\infty$. Then $s_\lambda>n$ for every $n\in\mathbb N$. Fix $n\in\mathbb N$. There exist $n$ pairwise disjoint strings of $\lambda$ $$S_1:=\{m^1_t\}_{-t\in\mathbb N_0},\ldots,S_n:=\{m^n_t\}_{-t\in\mathbb N_0}$$ in $\Gamma$. Define
\begin{equation*}
\Lambda_1:=S_1\cup\{\lambda^s(m^1_0):s\in\mathbb N\}, \ldots,\Lambda_n:=S_n\cup\{\lambda^s(m^n_0):s\in\mathbb N\}\ \text{and}\ \Lambda:=\Lambda_1\cup\ldots\cup\Lambda_n.
\end{equation*}
Note that $\lambda(\Lambda)\subseteq \Lambda$, so that we can consider the map $\lambda\restriction_\Lambda: \Lambda \to \Lambda$.
By \eqref{magic2} in Remark \ref{add-rem}(b)
\begin{equation}\label{eq5}
\mathrm{ent}(\sigma_\lambda)\geq\mathrm{ent}(\sigma_{\lambda\restriction_\Lambda}).
\end{equation}
Since $s_{\lambda\restriction_\Lambda}=n$, by the finite case of the proof of the theorem applied to the map $\lambda\restriction_\Lambda$, $\mathrm{ent}(\sigma_{\lambda\restriction_\Lambda})=n\log|K|$. By (\ref{eq5}) $\mathrm{ent}(\sigma_\lambda)\geq n\log|K|$, and this holds true for every $n\in\mathbb N$, so that $\mathrm{ent}(\sigma_\lambda)=+\infty=|s_\lambda|^*\log|K|$.
\end{proof}

We see now a first application of our main theorem.

\begin{example}\label{theorem4}\label{example5}\label{theorem3}
\begin{itemize}
	\item[(a)]In general $\mathrm{ent}(\sigma_\lambda)\geq|\alpha_\lambda|^*\log|K|$. Indeed, by the definitions $s_\lambda\geq\alpha_\lambda$ and so apply Theorem \ref{theorem6}.
	\item[(b)]The inequality in (a) can be strict: consider the functions of Example \ref{example5-firstpart}. In all three cases there exists just one equivalence class and so one equivalent class containing at least one infinite string of $\lambda$. Then $\alpha_{\varphi_n}=\alpha_{\psi_n}=\alpha_{\lambda_0}=1$, but $\mathrm{ent}(\sigma_{\varphi_n})=\log|K|$, $\mathrm{ent}(\sigma_{\psi_n})=n\log|K|$ and $\mathrm{ent}(\sigma_{\lambda_0})=+\infty$.
	\item[(c)] If $\lambda:\Gamma\to \Gamma$ is injective, the inequality in (a) becomes an equality, since in this case $\alpha_\lambda=s_\lambda$: by Theorem \ref{theorem6} and Example \ref{theorem3-first_part} $\mathrm{ent}(\sigma_\lambda)=|\alpha_\lambda|^*\log|K|=|s_\lambda|^*\log|K|$.
	\item[(d)] For $\lambda: \mathbb Z\to \mathbb Z$ defined by $\lambda(n)=n-1$ for every $n\in\mathbb Z$, the generalized shift $\sigma_\lambda$ coincides with the two-sided shift $\overline{\beta}_K\restriction_{K^{(\mathbb Z)}}$
	of $K^{(\mathbb Z)}$. Since $\alpha_\lambda= s_\lambda= 1$, one obtains from  Theorem \ref{theorem6} $\mathrm{ent}(\overline{\beta}_K\restriction_{K^{(\mathbb Z)}}) = \mathrm{ent}(\sigma_\lambda)= \log |K|$.
\end{itemize}
\end{example}

\section{Applications of the main theorem}

We give now other consequences of Theorem \ref{theorem6}.
The first one is an application of Theorem \ref{theorem6} together with Remark \ref{add-rem}(b). It shows that, even if the restriction of $\sigma_\lambda$ to an invariant subgroup is not necessarily a generalized shift, its entropy obeys the same formula as the generalized shift does.

\begin{corollary}\label{lastcor}
Let $\Lambda\subseteq\Gamma$ be such that $\lambda^{-1}(\Lambda)\subseteq\Lambda$. Then $|s_\lambda|^*\log|K|=\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})+|s_{\lambda\restriction_{\Gamma\setminus\Lambda}}|^*\log|K|$.
\end{corollary}
\begin{proof}
By \eqref{magic} in Remark \ref{add-rem}(b) and Theorem \ref{theorem6} $|s_\lambda|^*\log|K|=\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})+\mathrm{ent}(\sigma_{\lambda\restriction_{\Gamma\setminus\Lambda}})=\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})+|s_{\lambda\restriction_{\Gamma\setminus\Lambda}}|^*\log|K|$.
\end{proof}

\begin{remark}\label{lastrem}
If $\Lambda\subseteq\Gamma$ is such that $\lambda^{-1}(\Lambda)\subseteq\Lambda$, and $S$ is a string of $\lambda$ with $S\cap\Lambda\neq\emptyset$, then $S\setminus\Lambda$ is finite and we can assume without loss of generality that $S\subseteq\Lambda$. Hence we can think that either $S\subseteq\Lambda$ or $S\subseteq\Gamma\setminus\Lambda$. This shows that:
\begin{itemize}
	\item[(a)] $s_{\lambda\restriction_{\Gamma\setminus\Lambda}}$ is the number of all strings of $\lambda$ which miss $\Lambda$, and
	\item[(b)] in case $s_\lambda$ is finite, $s_\lambda-s_{\lambda_{\Gamma\setminus\Lambda}}$ is the number of all strings of $\lambda$ contained in $\Lambda$.
\end{itemize}
\end{remark}

\begin{corollary}\label{theorem18}\label{cor7}\label{cor8}
\begin{itemize}
	\item[(a)] Let $\Lambda\subseteq\Gamma$ be such that $\lambda^{-1}(\Lambda)\subseteq\Lambda$.
If $\{\mathrm{ent}(\sigma_\lambda),\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})\}\cap\{0,+\infty\}=\emptyset$,
then $\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})/\mathrm{ent}(\sigma_\lambda)$ is rational.
Moreover, if $0<r\leq 1$ is a rational number such that $r(\mathrm{ent}(\sigma_\lambda)/\log |K|)\in \mathbb Z$, then there exists
$\Lambda\subseteq\Gamma$, such that $\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})/\mathrm{ent}(\sigma_\lambda)=r$.
	\item[(b)] Let $L$ be another finite abelian group and assume that both $K$ and $L$ have at least two elements.
Then $\log|L|\mathrm{ent}(\sigma_{\lambda,K})=\log|K|\mathrm{ent}(\sigma_{\lambda,L})$.
\end{itemize}
\end{corollary}
\begin{proof}
(a) By Theorem \ref{theorem6}, Corollary \ref{lastcor} and our assumption $\{\mathrm{ent}(\sigma_\lambda),\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})\}\cap\{0,+\infty\}=\emptyset$,
$s_\lambda$ is finite and
$\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})/\mathrm{ent}(\sigma_\lambda)=(s_\lambda-s_{\lambda_{\Gamma\setminus\Lambda}})/s_\lambda$ is a rational number.

\smallskip
By hypothesis there exists $m\in\mathbb N$ such that $r s_\lambda=m$.
Let $S_1,\ldots,S_{s_\lambda}$ be the strongly disjoint strings in $\Gamma$ that realize $s_\lambda$ (this is possible by Claim \ref{strongly-disjoint} since $s_\lambda$ is finite by hypothesis).
Since $r\leq1$, it follows that $m\leq s_\lambda$. So it is possible to consider $T:=S_1\cup\ldots\cup S_m$ and define $\Lambda:= \bigcup_{n\in\mathbb N_0}\lambda^{-n}(T)$. Then $\lambda^{-1}(\Lambda)\subseteq\Lambda$. By Remark \ref{lastrem} $s_\lambda-s_{\lambda\restriction_{\Gamma\setminus\Lambda}}=m$, since $\{S_1,\ldots, S_m\}$ is a family of pairwise disjoint strings in $\Lambda$ of the maximal possible cardinality. By Corollary \ref{lastcor} $\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})=m\log|K|$. Then $\mathrm{ent}(\sigma_\lambda\restriction_{G_\Lambda})/\mathrm{ent}(\sigma_\lambda)=m/s_\lambda=r$.

\smallskip
(b) Is a simple application of Theorem \ref{theorem6}.
\end{proof}

\begin{corollary}\label{cor9}
For $M$ a torsion infinite abelian group and $\sigma_\lambda:M^{(\Gamma)}\to M^{(\Gamma)}$, $$\mathrm{ent}(\sigma_\lambda)=\begin{cases}
0 & \text{if}\ s_\lambda=0,\\
+\infty & \text{if}\ s_\lambda>0.
\end{cases}$$
\end{corollary}
\begin{proof}
In view of Remark \ref{subgroup} and Theorem \ref{theorem6}, it is easy to see that
\begin{align*}
\mathrm{ent}(\sigma_\lambda)&=\sup\{\mathrm{ent}(\sigma_\lambda\restriction_{H^{(\Gamma)}}):H\ \text{is a finite subgroup of}\ M\}\\
&=\sup\{\mathrm{ent}(\sigma_{\lambda,H}):H\ \text{is a finite subgroup of}\ M\}\\
&=\sup\{|s_\lambda|^*\log|H|:H\ \text{is a finite subgroup of}\ M\}.
\end{align*}
If $s_\lambda=0$, $\mathrm{ent}(\sigma_\lambda)=0$.
If $s_\lambda>0$, then $\mathrm{ent}(\sigma_\lambda)$ converges to $+\infty$ with $\log|H|$.
\end{proof}

\begin{example}\label{example10}
\begin{itemize}
\item[(a)]Let $\Gamma$ be a monoid.
\begin{itemize}
  \item[(a$_1$)] For each $s\in \Gamma$ consider $\lambda_s,\rho_s:\Gamma\rightarrow\Gamma$ defined by $\lambda_s(t)=st$ and $\rho_s(t)=ts$ for every $t\in\Gamma$. The element $s$ is invertible if and only if $\lambda_s$ and $\rho_s$ are bijective. By Example \ref{theorem3}, the endomorphisms $\sigma_{\lambda_s}$ and $\sigma_{\rho_s}$ of $G_\Gamma$ have the following properties.
    \begin{itemize}
        \item[(i)]If $s$ is of finite order $n\in\mathbb N$, then $(\lambda_s)^n=\lambda_{s^n}=id_\Gamma$ (i.e., $\lambda_s$ is periodic) and similarly $(\rho_s)^n=\rho_{s^n}=id_\Gamma$ (i.e., $\rho_s$ is periodic). By Proposition \ref{composition}(a) $\sigma_{\lambda_s}^n=\sigma_{\lambda_s^n}=\sigma_{\lambda_{s^n}}=id_{G_\Gamma}$ and $\sigma_{\rho_s}^n=\sigma_{\rho_s^n}=\sigma_{\rho_{s^n}}=id_{G_\Gamma}$. Hence $\mathrm{ent}(\sigma_{\lambda_s})=\mathrm{ent}(\sigma_{\rho_s})=0$ by Lemma \ref{locnilp}(b).
        \item[(ii)]Suppose that $s$ is invertible. If $s$ has infinite order, then $\lambda_s$ and $\rho_s$ are injective and by Example \ref{theorem4} their entropy is positive. So, $\mathrm{ent}(\sigma_{\lambda_s})=\mathrm{ent}(\sigma_{\rho_s})=0$ if and only if $s$ is of finite order.
    \end{itemize}
\item[(a$_2$)] For each invertible $s\in \Gamma$ consider $\mu_s:\Gamma\to\Gamma$ defined by $\mu_s(t)=s t s^{-1}$ for every $t\in\Gamma$. By Example \ref{theorem3}, the endomorphism $\sigma_{\mu_s}$ of $G_\Gamma$ has $\mathrm{ent}(\sigma_{\mu_s})>0$ if and only if there exists $t\in\Gamma$ such that $\{s^n:n\in\mathbb N\}\cap \{v\in\Gamma:vt=tv\}=\emptyset$.
\end{itemize}
	\item[(b)] Suppose now that $\Gamma$ is an abelian group and $\lambda:\Gamma\rightarrow\Gamma$ a group homomorphism such that $\ker\lambda$ is finite (i.e., $\lambda$ has finite fibers).
	\begin{itemize}
		\item[(b$_1$)]If $\Gamma=\mathbb Z$, there exists $n\in\mathbb Z$ such that $\lambda(x)=n x$ for every $x\in\mathbb Z$. If $n\neq\pm 1$, then there exists no string of $\lambda$ and so $\mathrm{ent}(\sigma_\lambda)=0$ by Theorem \ref{theorem6}. If $m=\pm 1$, then $\lambda^2=id_\Gamma$ and so by Proposition \ref{composition}(a) $\sigma_\lambda^2=\sigma_{\lambda^2}=\sigma_{id_\Gamma}=id_{G_\Gamma}$ and by Lemma \ref{locnilp}(b) $\mathrm{ent}(\sigma_\lambda)=0$.
		\item[(b$_2$)]Suppose that $\lambda\in\mathrm{Aut}(\Gamma)$. Then the orbits of $\lambda$ are exactly the equivalence classes of the relation $\Re_\lambda$ (see Example \ref{theorem3-first_part}). Therefore, if $\lambda$ has infinitely many infinite orbits, $\alpha_\lambda$ is infinite and by Example \ref{theorem3-first_part} and Theorem \ref{theorem6} $\mathrm{ent}(\sigma_\lambda)=+\infty$.
		\item[(b$_3$)]Consider $\Gamma=\mathbb Z\times\mathbb Z$ and $\lambda\in\mathrm{Aut}(\Gamma)$ defined by $\lambda(x,y)=(x+y,y)$ for every $(x,y)\in\Gamma$. For every $n\in\mathbb N$ the orbits of $(0,n)$, that is, $$(0,n)/\Re_\lambda=\{\ldots,(-2n,n),(-n,n),(0,n),(n,n),(2n,n),\ldots\},$$ are infinitely many and pairwise disjoint. Then $\mathrm{ent}(\sigma_\lambda)=+\infty$ by (b$_2$).
	\end{itemize}
\end{itemize}
\end{example}

The next example is dedicated to the composition of generalized shifts. Let us mention that from the formulas $\mathrm{ent}(\sigma_\lambda^k)=k \, \mathrm{ent}(\sigma_\lambda)$ (see Lemma \ref{log_law}) and $\sigma_\lambda^k= \sigma_{\lambda^k}$ (see Proposition \ref{composition}), and Theorem  \ref{theorem6} we obtain the useful non-obvious formula $s_{\lambda^k}=ks_{\lambda}$.

\begin{example}
Let $\Gamma=\mathbb N_0$ and let $\mu_1:\Gamma\to\Gamma$ be defined by
\begin{equation*}
\mu_1(m)=
\begin{cases}
p^{2k} & \text{if}\ m=p^{2k+1}\ \text{with}\ p\in\mathbb P\ \text{and}\ k\in\mathbb N, \\
p^{2k+1} & \text{if}\ m=p^{2k}\ \text{with}\ p\in\mathbb P\ \text{and}\ k\in\mathbb N, \\
m & \text{otherwise}.
\end{cases}
\end{equation*}
Hence $\mu_1^2=id_\Gamma$ and by Proposition \ref{composition}(a) $\sigma_{\mu_1}^2=\sigma_{\mu_1^2}=\sigma_{id_\Gamma}=id_{G_\Gamma}$. Then $\mathrm{ent}(\sigma_{\mu_1\circ\mu_1})=0$ and since $\sigma_{\mu_1}$ is periodic, $\mathrm{ent}(\sigma_{\mu_1})=0$ by Lemma \ref{locnilp}(b).
The diagram for $\mu_1$ is the following:
\begin{equation*}
\xymatrix@-1pc{
 & & \vdots & \vdots & \vdots & &\ldots & \vdots & \ldots\\
 &   &  2^{2k+1} \ar@{->}[d] & 3^{2k+1} \ar@{->}[d] & 5^{2k+1} \ar@{->}[d] & & \ldots &p^{2k+1} \ar@{->}[d]& \ldots\\
 &   & 2^{2k} \ar@{->}@/_1.5pc/[u] & 3^{2k} \ar@{->}@/_1.5pc/[u] & 5^{2k} \ar@{->}@/_1.5pc/[u] & & \ldots & p^{2k} \ar@{->}@/_1.5pc/[u]&\ldots \\
 &   & \vdots              & \vdots & \vdots & & \ldots & \vdots & \vdots \\
 &   & 8 \ar@{->}[d] & 27 \ar@{->}[d]& 125 \ar@{->}[d] & & \ldots & p^3 \ar@{->}[d] & \ldots\\
 &   & 4 \ar@{->}@/_1.5pc/[u] & 9 \ar@{->}@/_1.5pc/[u] & 25 \ar@{->}@/_1.5pc/[u] & & \ldots & p^2 \ar@{->}@/_1.5pc/[u]&\ldots\\
0 \ar@(dl,dr)[]& 1\ar@(dl,dr)[] & 2\ar@(dl,dr)[] & 3\ar@(dl,dr)[] & 5\ar@(dl,dr)[] & 6\ar@(dl,dr)[] &\ldots \ar@(dl,dr)[] & p\ar@(dl,dr)[] &\ldots\ar@(dl,dr)[]\\
}
\end{equation*}

\bigskip

\bigskip
Let $\mu_2:\Gamma\to\Gamma$ be defined by
\begin{equation*}
\mu_2(m)=
\begin{cases}
p^{2k-1} & \text{if}\ m=p^{2k}\ \text{with}\ p\in\mathbb P\ \text{and}\ k\in\mathbb N, \\
p^{2k} & \text{if}\ m=p^{2k-1}\ \text{with}\ p\in\mathbb P\ \text{and}\ k\in\mathbb N, \\
m & \text{otherwise (i.e., $m$ is not a prime power)}.
\end{cases}
\end{equation*}
Analogously, $\mu_2^2=id_\Gamma$ and by Proposition \ref{composition}(a) $\sigma_{\mu_2}^2=\sigma_{\mu_2^2}=\sigma_{id_\Gamma}=id_{G_\Gamma}$. Then $\mathrm{ent}(\sigma_{\mu_2\circ\mu_2})=0$ and sice $\sigma_{\mu_2}$ is periodic, so $\mathrm{ent}(\sigma_{\mu_2})=0$ by Lemma \ref{locnilp}(b).
The diagram for $\mu_2$ is the following:
\begin{equation*}
\xymatrix@-1pc{
 & & \vdots & \vdots & \vdots & &\ldots & \vdots & \ldots\\
 &   &  2^{2k} \ar@{->}[d] & 3^{2k} \ar@{->}[d] & 5^{2k} \ar@{->}[d] & & \ldots &p^{2k} \ar@{->}[d]& \ldots\\
 &   & 2^{2k-1} \ar@{->}@/_1.5pc/[u] & 3^{2k-1} \ar@{->}@/_1.5pc/[u] & 5^{2k-1} \ar@{->}@/_1.5pc/[u] & & \ldots & p^{2k-1} \ar@{->}@/_1.5pc/[u]&\ldots \\
 &   & \vdots              & \vdots & \vdots & & \ldots & \vdots & \vdots \\
 &   & 4 \ar@{->}[d]  & 9 \ar@{->}[d] & 25 \ar@{->}[d] & & \ldots & p^2 \ar@{->}[d] &\ldots\\
0 \ar@(dl,dr)[]& 1\ar@(dl,dr)[] & 2\ar@{->}@/_1.5pc/[u] & 3\ar@{->}@/_1.5pc/[u] & 5\ar@{->}@/_1.5pc/[u] & 6\ar@(dl,dr)[] &\ldots \ar@(dl,dr)[] & p\ar@{->}@/_1.5pc/[u] &\ldots\ar@(dl,dr)[]\\
}
\end{equation*}

\bigskip

\bigskip
The diagram for $\mu_1\circ\mu_2:\Gamma\to\Gamma$ is the following:
\begin{equation*}
\xymatrix@-1pc{
 & & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & &\ldots & \vdots\ar@{->}[d] & \ldots\\
 &   &  2^{2k} \ar@{->}[d] & 3^{2k} \ar@{->}[d] & 5^{2k} \ar@{->}[d] & & \ldots & p^{2k} \ar@{->}[d]& \ldots\\
 &   & 2^{2k-2} \ar@{->}[d] & 3^{2k-2} \ar@{->}[d] & 5^{2k-2} \ar@{->}[d] & & \ldots & p^{2k-2} \ar@{->}[d]&\ldots \\
 &   & \vdots \ar@{->}[d] & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & & \ldots & \vdots\ar@{->}[d] & \vdots \\
 &   & 4 \ar@{->}[d]  & 9 \ar@{->}[d] & 25 \ar@{->}[d] & & \ldots & p^2 \ar@{->}[d] &\ldots\\
0 \ar@(dl,dr)[]& 1\ar@(dl,dr)[] & 2\ar@{->}[d] & 3\ar@{->}[d] & 5\ar@{->}[d] & 6\ar@(dl,dr)[] &\ldots \ar@(dl,dr)[] & p\ar@{->}[d] &\ldots\ar@(dl,dr)[]\\
 &   & 8 \ar@{->}[d] & 27 \ar@{->}[d] & 125 \ar@{->}[d] & & \ldots & p^{3} \ar@{->}[d]&\ldots \\
 &   & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & \vdots \ar@{->}[d] & &\ldots & \vdots\ar@{->}[d] & \ldots\\
 &   &  2^{2k-1} \ar@{->}[d] & 3^{2k-1} \ar@{->}[d] & 5^{2k-1} \ar@{->}[d] & & \ldots & p^{2k-1} \ar@{->}[d]& \ldots\\
 &   &  2^{2k+1} \ar@{->}[d] & 3^{2k+1} \ar@{->}[d] & 5^{2k+1} \ar@{->}[d] & & \ldots & p^{2k+1} \ar@{->}[d]& \ldots\\
 & & \vdots & \vdots & \vdots & &\ldots & \vdots & \ldots\\
}
\end{equation*}
In this case $s_{\mu_1\circ\mu_2}=\alpha_{\mu_1\circ\mu_2}=\omega$ (so $|s_{\mu_1\circ\mu_2}|^*=+\infty$) and by Theorem \ref{theorem6} $\mathrm{ent}(\sigma_{\mu_1\circ\mu_2})=+\infty$.
Similarly one can see that also s$_{\mu_2 \circ \mu_1}=\alpha_{\mu_1\circ\mu_2}=\omega$ and so that $\mathrm{ent}(\sigma_{\mu_2\circ\mu_1})=+\infty$.

By Proposition \ref{composition}(a) $\sigma_{\mu_1\circ\mu_2}=\sigma_{\mu_1}\circ\sigma_{\mu_2}$; hence $\sigma_{\mu_1}\circ\sigma_{\mu_2}$ is an example of an endomorphism of infinite entropy with both $\sigma_{\mu_1}$ and $\sigma_{\mu_2}$ of entropy $0$. The same is $\sigma_{\mu_2}\circ\sigma_{\mu_1}$.

\bigskip

Let $\varrho_1:\Gamma\to\Gamma$ be defined by $\varrho_1(0)=1$ and $\varrho_1(m)=m$ for every $m\in\mathbb N$. Then $s_{\varrho_1}=0$ and so $\mathrm{ent}(\sigma_{\varrho_1})=0$ by Theorem \ref{theorem6}.

\medskip
The diagram for $\varrho_1\circ\varphi_1:\Gamma\to\Gamma$ is the following:
\begin{equation*}
\xymatrix@-1pc{
  & \vdots\ar@{->}[d]\\
  & 3\ar@{->}[d]  \\
0 \ar@{->}[dr] & 2 \ar@{->}[d]\\
 & 1\ar@(dl,dr)[]
}
\bigskip
\end{equation*}

\bigskip
For this function $s_{\varrho_1\circ\varphi_1}=1$, and so by Theorem \ref{theorem6} $\mathrm{ent}(\sigma_{\varrho_1\circ\varphi_1})=\log|K|$.

By Proposition \ref{composition}(a) $\sigma_{\varrho_1\circ\varphi_1}=\sigma_{\varrho_1}\circ\sigma_{\varphi_1}$; consequently $\mathrm{ent}(\sigma_{\varrho_1}\circ\sigma_{\varphi_1})=\mathrm{ent}(\sigma_{\varphi_1})=\log|K|$, while $\mathrm{ent}(\sigma_{\varrho_1})=0$.
\end{example}

\begin{theorem}\label{theorem14}
If $\lambda:\Gamma\to\Gamma$, $\mu:\Upsilon\to\Upsilon$ are such that for each
$(i,j)\in\Gamma\times\Upsilon$, $\lambda^{-1}(i)$ and
$\mu^{-1}(j)$ are finite, then for the endomorphism $\sigma_{\lambda\times\mu}:G_{\Gamma\times\Upsilon}\to G_{\Gamma\times\Upsilon}$ we have:
\begin{equation*}
\mathrm{ent}(\sigma_{\lambda\times\mu})=0\ \text{\emph{if}}\  \begin{cases}
\mathrm{ent}(\sigma_{\lambda})=\mathrm{ent}(\sigma_{\mu})=0, & \text{\emph{(a$_0$)}}\\
\mathrm{ent}(\sigma_\mu)=|\mathrm{Per}(\mu)|=0\ \text{and}\ \mathrm{ent}(\sigma_\lambda)>0, & \text{\emph{(a$_1$)}}\\
\mathrm{ent}(\sigma_\lambda)=|\mathrm{Per}(\lambda)|=0\ \text{and}\ \mathrm{ent}(\sigma_\mu)>0. &  \text{\emph{(a$_2$)}}
\end{cases}\eqno{(a)}
\end{equation*}
Moreover,
\begin{equation*}
\mathrm{ent}(\sigma_{\lambda\times\mu})=
\begin{cases}
+\infty & \text{\emph{if}}\ \mathrm{ent}(\sigma_{\lambda})>0\ \text{and} \ \mathrm{ent}(\sigma_{\mu})>0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{\emph{(b$_0$)}}\\
|\mathrm{Per}(\mu)|^*\mathrm{ent}(\sigma_\lambda) & \text{\emph{if}}\ \mathrm{ent}(\sigma_\mu)=0,\ |\mathrm{Per}(\mu)|>0\ \text{and}\ \mathrm{ent}(\sigma_\lambda)>0,\ \ \ \text{\emph{(b$_1$)}}\\
|\mathrm{Per}(\lambda)|^*\mathrm{ent}(\sigma_\mu) & \text{\emph{if}}\ \mathrm{ent}(\sigma_\lambda)=0,\ |\mathrm{Per}(\lambda)|>0\ \text{and}\ \mathrm{ent}(\sigma_\mu)>0.\ \ \ \text{\emph{(b$_2$)}}
\end{cases}\eqno{(b)}
\end{equation*}
\end{theorem}
\begin{proof}
If $\mathrm{ent}(\sigma_{\lambda\times\mu})>0$, then by Theorem \ref{theorem6}, there exists a string
$\{(m_t,n_t)\}_{-t\in\mathbb N_0}$ of $\lambda\times\mu$ (for each $-t\in\mathbb N_0$, $(m_{t+1},n_{t+1})=(\lambda(m_t),\mu(n_t))$),
therefore at least one of the sequences $\{m_t\}_{-t\in\mathbb N_0}$ or $\{n_t\}_{-t\in\mathbb N_0}$ is a string, which shows that either
$\mathrm{ent}(\sigma_\lambda)>0$ or $\mathrm{ent}(\sigma_\mu)>0$, in view of Theorem \ref{theorem6}. This proves (a$_0$).

\smallskip
Assume that $\mathrm{ent}(\sigma_\mu)=0$ and $\mathrm{ent}(\sigma_\lambda)>0$. We prove that $\mathrm{ent}(\sigma_{\lambda\times\mu})=0$ if $|\mathrm{Per}(\mu)|=0$. To this end, suppose that $\mathrm{ent}(\sigma_{\lambda\times\mu})>0$. By Theorem \ref{theorem6} $s_{\lambda\times\mu}>0$, so let $\{(m_t,n_t)\}_{-t\in\mathbb N_0}$ be a string of $\lambda\times\mu$. Since $\mathrm{ent}(\sigma_\mu)=0$, $s_\mu=0$ by Theorem \ref{theorem6}, and so $\{m_t\}_{-t\in\mathbb N_0}$ has to be a string  of $\lambda$ and $n_0$ is a periodic point of $\mu$. In particular $|\mathrm{Per}(\mu)|>0$. This proves (a$_1$).

Reverting the roles of $\lambda$ and $\mu$ one can prove (a$_2$).

\smallskip
Now let $\mathrm{ent}(\sigma_\lambda)>0$ and $\mathrm{ent}(\sigma_\mu)>0$. By Theorem \ref{theorem6} there exist strings $\{m_t\}_{-t\in\mathbb N_0}$ and $\{n_t\}_{-t\in\mathbb N_0}$ respectively of $\lambda$ and $\mu$. For each $-l\in\mathbb N_0$, let $z_{l}:=(m_0,n_l)$. Then $\{(m_t,n_{l+t})\}_{-t\in\mathbb N_0}$ is a string of $\lambda\times\mu$ for every $l\in\mathbb N$, and these strings are pairwise disjoint. This means that $|s_{\lambda\times\mu}|^*=+\infty$ and by Theorem \ref{theorem6} $\mathrm{ent}(\sigma_{\lambda\times\mu})=+\infty$. This proves (b$_0$).

\smallskip
Assume that $\mathrm{ent}(\sigma_\mu)=0$, $|\mathrm{Per}(\mu)|>0$ and $\mathrm{ent}(\sigma_\lambda)>0$. We prove that
\begin{equation}\label{sigmaXmu}
\mathrm{ent}(\sigma_{\lambda\times\mu})\leq |\mathrm{Per}(\mu)|^* \mathrm{ent}(\sigma_\lambda).
\end{equation}
If $\mathrm{ent}(\sigma_{\lambda\times\mu})=0$ the inequality in \eqref{sigmaXmu} is trivially satisfied. So we can assume that $\mathrm{ent}(\sigma_{\lambda\times\mu})>0$. By Theorem \ref{theorem6} $s_{\lambda\times\mu}>0$. Let $\{(m_t,n_t)\}_{-t\in\mathbb N_0}$ be a string  of $\lambda\times\mu$. Since $s_\mu=0$ by Theorem \ref{theorem6}, $\{m_t\}_{-t\in\mathbb N_0}$ has to be a string  of $\lambda$ and each $n_t$ is a periodic point of $\mu$. If $\{(m_t,n_t)\}_{-t\in\mathbb N_0}$ and $\{(m'_t,n'_t)\}_{-t\in\mathbb N_0}$ are disjoint strings of $\lambda\times\mu$, then either $\{m_t\}_{-t\in\mathbb N_0}$ and $\{m'_t\}_{-t\in\mathbb N_0}$ are disjoint strings of $\lambda$ or $n_0$ and $n'_0$ are distinct periodic points of $\mu$. This proves that $s_{\lambda\times\mu}\leq |\mathrm{Per}(\mu)| s_\lambda$. In particular \eqref{sigmaXmu} holds by Theorem \ref{theorem6}.

We show now that under the same hypotheses, also the converse implication holds true, that is, we prove that
\begin{equation}\label{sigmaXmu2}
\mathrm{ent}(\sigma_{\lambda\times\mu})\geq|\mathrm{Per}(\mu)|^*\mathrm{ent}(\sigma_\lambda).
\end{equation}
If $\{m_t\}_{-t\in\mathbb N_0}$ is a string of $\lambda$, and $j\in\mathrm{Per}(\mu)$ with $\{j=j_0,j_1,j_2,\ldots, j_s\}$ the finite orbit of $j$ (i.e., $\mu(j_k)=j_{k+1}$ for every $k\in\{0,\ldots,s-1\}$ and $\mu(j_s)=j$), then $\{(m_t,j_{[t]_{s+1}})\}_{-t\in\mathbb N_0}$ is a string of $\lambda\times\mu$ (where, for $a\in\mathbb Z,\ b\in\mathbb N$, $[a]_{b}$ denotes the remainder class of $a$ modulo $b$). In case $i,j$ are distinct elements of $\mathrm{Per}(\mu)$, the strings $\{(m_t,j_{[t]_{s+1}})\}_{-t\in\mathbb N_0}$ and $\{(m_t,i_{[t]_{r+1}})\}_{-t\in\mathbb N_0}$ (where $\{i=i_0,i_1,i_2,\ldots, i_r\}$ is the finite orbit of $i$, i.e., $\mu(i_k)=i_{k+1}$ for every $k\in\{0,\ldots,r-1\}$ and $\mu(i_r)=i$) are pairwise disjoint. This proves that $s_{\lambda\times\mu}\geq|\mathrm{Per}(\mu)|s_\lambda$, and by Theorem \ref{theorem6} \eqref{sigmaXmu2} holds.

By \eqref{sigmaXmu} and \eqref{sigmaXmu2} $\mathrm{ent}(\sigma_{\lambda\times\mu})=|\mathrm{Per}(\mu)|^*\mathrm{ent}(\sigma_\lambda)$. This concludes the proof of (b$_1$).

Reverting the roles of $\lambda$ and $\mu$ one can prove (b$_2$).
\end{proof}

In the notations of this theorem, the following example shows that in the case where $\mathrm{ent}(\sigma_\mu)=0$ and $\mathrm{ent}(\sigma_\lambda)=\log|K|$, it is possible that $\mathrm{ent}(\sigma_{\lambda\times \mu})$ is positive and also infinite, depending on the cardinality of $\mathrm{Per}(\mu)$.

\begin{example}\label{example15}
Let $t\in\mathbb N$, $\Gamma_t=\{1,\ldots,t\}$ and $\theta_t=(123\dots t)\in S_{\Gamma_t}$. Let $\Lambda=\mathbb N_0\times\Gamma_t$. Then $\varphi_1\times\theta_t:\Lambda\to\Lambda$ and its diagram is the following:
\begin{equation*}
\xymatrix@!R@!C@-1pc{
\vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] \\
(m,1) \ar@{->}[d] & (m,2) \ar@{->}[d] & \ldots \ar@{->}[d] & (m,t-1) \ar@{->}[d] & (m,t) \ar@{->}[d] \\
(m-1,2) \ar@{->}[d] & (m-1,3) \ar@{->}[d] & \ldots \ar@{->}[d] & (m-1,t) \ar@{->}[d] & (m-1,1) \ar@{->}[d] \\
\vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] \\
(m-t,1) \ar@{->}[d] & (m-t,2) \ar@{->}[d] & \ldots \ar@{->}[d] & (m-t,t-1) \ar@{->}[d] & (m-t,t) \ar@{->}[d] \\
(m-t-1,2)\ar@{->}[d]&(m-t-1,3)\ar@{->}[d]&\ldots\ar@{->}[d]&(m-t-1,t)\ar@{->}[d]&(m-t-1,1)\ar@{->}[d] \\
\vdots\ar@{->}[d] & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] & \vdots\ar@{->}[d] \\
(2,t-1) \ar@{->}[d] & (2,t) \ar@{->}[d] & \ldots \ar@{->}[d] & (2,t-3)\ar@{->}[d] & (2,t-2) \ar@{->}[d] \\
(1,t) \ar@{->}[d] & (1,1) \ar@{->}[d] & \ldots \ar@{->}[d] & (1,t-2) \ar@{->}[d] & (1,t-1) \ar@{->}[d] \\
(0,1) \ar@{->}[r] & (0,2) \ar@{->}[r] & \ldots \ar@{->}[r] & (0, t-1)\ar@{->}[r] & (0,t) \ar@/^1.5pc/@{->}[llll] \\
}
\end{equation*}

\bigskip
Let $\theta=(12)\in S_\mathbb N$ and let $\Lambda=\mathbb N\times \mathbb N_0$. The diagram for $\theta\times \varphi_1:\Lambda\to\Lambda$ is the following:
\begin{equation*}
\xymatrix@!R@!C@-1pc{
\vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] & \ldots \ar@{->}[d] & \vdots \ar@{->}[d] & \vdots \ar@{->}[d] \\
(1,2) \ar@{->}[d] & (2,2) \ar@{->}[d] & (3,2) \ar@{->}[d] & \ldots \ar@{->}[d] & (n,2) \ar@{->}[d] & \ldots \ar@{->}[d] \\
(2,1) \ar@{->}[d] & (1,1) \ar@{->}[d] & (3,1) \ar@{->}[d] & \ldots \ar@{->}[d] & (n,1) \ar@{->}[d] & \ldots\ar@{->}[d] \\
(1,0) \ar@{->}[r] & (2,0) \ar@{->}@/^1.5pc/[l] & (3,0)\ar@(dl,dr)[] & \ldots\ar@(dl,dr)[] & (n,0)\ar@(dl,dr)[] & \ldots \ar@(dl,dr)[]\\
}
\bigskip
\end{equation*}

Then by Theorem \ref{theorem6} $\mathrm{ent}(\sigma_{\varphi_1\times\theta_t})=t\log|K|$ and $\mathrm{ent}(\sigma_{\varphi_1\times\theta})=+\infty$, while $\mathrm{ent}(\sigma_{\varphi_1})=\log|K|$ and $\mathrm{ent}(\sigma_{\theta})=\mathrm{ent}(\sigma_{\theta_t})=0$, since $|\mathrm{Per}(\theta_t)|=t$, $|\mathrm{Per}(\theta)|^*=+\infty$.
\end{example}

The next is an application of Theorem \ref{theorem14} and Corollary \ref{cor7}. Indeed considering the product of two finite abelian groups $K\times L$ instead of only one finite abelian group $K$ (as in Theorem \ref{theorem14}) is not a more general situation, since Theorem \ref{theorem6}, but also Corollary \ref{cor7}, shows that the entropy of a generalized shift depends mainly on its string number $s_\lambda$.

\begin{corollary}\label{cor16}
Let $K$ and $L$ be finite non-trivial abelian groups and let $\lambda:\Gamma\rightarrow\Gamma$ and
$\mu:\Upsilon\rightarrow\Upsilon$ be such that for each $(i,j)\in\Gamma\times\Upsilon$, $\lambda^{-1}(i)$ and
$\mu^{-1}(j)$ are finite. Then for $\sigma_{\lambda\times\mu}:(K\times L)^{(\Gamma\times\Upsilon)}\to(K\times L)^{(\Gamma\times\Upsilon)}$ we have:
\begin{equation*}
\mathrm{ent}(\sigma_{\lambda\times\mu})=0\ \text{\emph{if}}\ \begin{cases}
\mathrm{ent}(\sigma_{\lambda})=\mathrm{ent}(\sigma_{\mu})=0, & \text{\emph{(a$_0$)}}\\
\mathrm{ent}(\sigma_\mu)=|\mathrm{Per}(\mu)|=0\ \text{and}\ \mathrm{ent}(\sigma_\lambda)>0, & \text{\emph{(a$_1$)}}\\
\mathrm{ent}(\sigma_\lambda)=|\mathrm{Per}(\lambda)|=0\ \text{and}\ \mathrm{ent}(\sigma_\mu)>0. & \text{\emph{(a$_2$)}}
\end{cases}
\eqno{(a)}\end{equation*}
Moreover,
\begin{equation*}
\mathrm{ent}(\sigma_{\lambda\times\mu})=
\begin{cases}
+\infty & \text{\emph{if}}\ \mathrm{ent}(\sigma_{\lambda})>0\ \text{and} \ \mathrm{ent}(\sigma_{\mu})>0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{\emph{(b$_0$)}}\\
\frac{\log|K\times L|}{\log|K|}|\mathrm{Per}(\mu)|^*\mathrm{ent}(\sigma_\lambda) & \text{\emph{if}}\ \mathrm{ent}(\sigma_\mu)=0,\ |\mathrm{Per}(\mu)|>0\ \text{and}\ \mathrm{ent}(\sigma_\lambda)>0,\ \ \ \text{\emph{(b$_1$)}}\\
\frac{\log|K\times L|}{\log|K|}|\mathrm{Per}(\lambda)|^*\mathrm{ent}(\sigma_\mu) & \text{\emph{if}}\ \mathrm{ent}(\sigma_\lambda)=0,\ |\mathrm{Per}(\lambda)|>0\ \text{and}\ \mathrm{ent}(\sigma_\mu)>0.\ \ \ \text{\emph{(b$_2$)}}
\end{cases}\eqno{(b)}
\end{equation*}
\end{corollary}
\begin{proof}
By Corollary \ref{cor7} $\mathrm{ent}(\sigma_{\lambda\times\mu,K\times L})=\frac{\log|K\times L|}{\log|K|}\mathrm{ent}(\sigma_{\lambda\times\mu,K})$. Now apply Theorem \ref{theorem14}.
\end{proof}

In the next example we associate to a given map $\lambda$ a natural extension map $\Lambda$ such that $\mathrm{ent}(\sigma_\Lambda)$ is either infinite or $0$, depending on whether $\mathrm{ent}(\sigma_\lambda)$ is positive or $0$.

\begin{example}\label{example17}
Define $\Lambda:\mathcal P_{fin}(\Gamma)\rightarrow{\mathcal P}_{fin}(\Gamma)$ by $\Lambda(A)=\{\lambda(i):i\in A\}=\lambda(A)$ for every $A\in {\mathcal P}_{fin}(\Gamma)$.
(Since $\Gamma$ embeds into ${\mathcal P}_{fin}(\Gamma)$ in a natural way via the singletons, $\Lambda$ can be considered as an extension of $\lambda$.)
For each $A\in\mathcal P_{fin}(\Gamma)$, $\Lambda^{-1}(A)$ is a finite subset of ${\mathcal P}_{fin}(\Gamma)$. Consider $\sigma_\Lambda:G_{\mathcal P_{fin}(\Gamma)}\to G_{\mathcal P_{fin}(\Gamma)}$. Then:
\begin{equation*}
\mathrm{ent}(\sigma_{\Lambda})=
\begin{cases}
    0 & \text{if}\ \mathrm{ent}(\sigma_\lambda)=0, \\
    +\infty & \text{if}\ \mathrm{ent}(\sigma_\lambda)>0.
\end{cases}
\end{equation*}
If $\mathrm{ent}(\sigma_\Lambda)>0$, by Theorem \ref{theorem6} there exists at least one string $\{A_t\}_{-t\in\mathbb N_0}$ of $\Lambda$ in $\mathcal P_{fin}(\Gamma)$. In particular there exists a string $\{m_t\}_{-t\in\mathbb N_0}$ of $\lambda$ in $\Gamma$: indeed, there exists $-t\in\mathbb N_0$ such that not all elements of $A_t$ are periodic for $\lambda$. Suppose without loss of generality that $t=0$ and let $m_{0}\in A_{0}\setminus \mathrm{Per}(\lambda)$. Then there exists an infinite sequence $\{m_t\}_{-t\in\mathbb N_0}$ of elements of $\Gamma$ such that $m_t\in A_t$ and $\lambda(m_t)=m_{t+1}$ for every $-t\in\mathbb N$. The elements $m_t$ have to be distinct because $m_{0}$ is not periodic. So $\{m_t\}_{-t\in\mathbb N_0}$ is a string of $\lambda$ and by Theorem \ref{theorem6} $\mathrm{ent}(\sigma_\lambda)>0$. This proves that if $\mathrm{ent}(\sigma_\lambda)=0$ then $\mathrm{ent}(\sigma_\Lambda)=0$.

If $\mathrm{ent}(\sigma_\lambda)>0$, by Theorem \ref{theorem6} there exists a string $\{m_t\}_{-t\in\mathbb N_0}$ of $\lambda$ in $\Gamma$. For each $l\in\mathbb N_0$,
$S_l:=\{\{m_t,m_{t-1},\ldots,m_{t-l}\}\}_{-t\in\mathbb N_0}$ is a string of $\Lambda$ in $\mathcal P_{fin}(\Gamma)$ and obviously $S_1,\ldots,S_l,\ldots$ are pairwise disjoint strings of $\Lambda$ in $\mathcal P_{fin}(\Gamma)$. So Theorem \ref{theorem6} leads us to the desired result, that is $\mathrm{ent}(\sigma_\Lambda)=+\infty$.
\end{example}

\section{Final remarks and open problems}

We consider here the generalized shift $\sigma_\lambda$  on $K^\Gamma$ and, in case $\lambda$ has finite fibers,  its restriction $\sigma_{\lambda}^\oplus$ on $K^{(\Gamma)}$.
Theorem \ref{theorem6} calculates precisely the value of the entropy of $\sigma_{\lambda}^\oplus:K^{(\Gamma)}\to K^{(\Gamma)}$. The necessary property on $\lambda$ to have finite fibers helps us in finding the explicit formula for the entropy of $\sigma_{\lambda}^\oplus$. In the general case of $\sigma_\lambda:K^\Gamma\to K^\Gamma$ we
leave open the following problem.

\begin{problem}
Calculate the entropy of $\sigma_\lambda:K^\Gamma\to K^\Gamma$. Is $\mathrm{ent}(\sigma_\lambda)=\mathrm{ent}(\sigma_{\lambda}^\oplus)$ in case $\lambda$ has finite fibers?
\end{problem}

Note that $\mathrm{ent}(\sigma_\lambda)\geq\mathrm{ent}(\sigma_{\lambda}^\oplus)$ in the latter case, since then $K^{(\Gamma)}$ is $\sigma_\lambda$-invariant in $K^\Gamma$.

\begin{problem}
\begin{itemize}
	\item[(a)] Is it possible to have $\mathrm{ent}(\sigma_{\lambda}^\oplus)=0$, but $\mathrm{ent}(\sigma_\lambda)>0$?
	\item[(b)] Is it possible to have $0<\mathrm{ent}(\sigma_\lambda)<+\infty$?
\end{itemize}
\end{problem}

Moreover, Theorem \ref{theorem6} concerns only a single abelian group $K$. If $\{K_i:i\in\Gamma\}$ is a family of abelian groups and $\rho_i:K_{\lambda(i)}\to K_i$ is a homomorphism for every $i\in I$, define $\widetilde\sigma_\lambda:\prod_{i\in\Gamma}K_i\to\prod_{i\in\Gamma}K_i$ by $\widetilde\sigma_\lambda(x)=(\rho_i(x_{\lambda(i)}))_{i\in\Gamma}$ for every $x=(x_i)_{i\in\Gamma}\in\prod_{i\in\Gamma}K_i$. It is possible to consider also $\widetilde\sigma_\lambda$ restricted to the direct sum, that is, $\widetilde\sigma_{\lambda}^\oplus:\bigoplus_{i\in\Gamma}K_i\to\bigoplus_{i\in\Gamma}K_i$.

\begin{problem}
Suppose that for every $i\in\Gamma$ $K_i$ is a finite abelian group.
\begin{itemize}
	\item[(a)]Calculate the entropy of $\widetilde\sigma_\lambda:\prod_{i\in\Gamma}K_i\to\prod_{i\in\Gamma}K_i$.
	\item[(b)]Calculate the entropy of $\widetilde\sigma_{\lambda}^\oplus:\bigoplus_{i\in\Gamma}K_i\to\bigoplus_{i\in\Gamma}K_i$.
\end{itemize}
\end{problem}

A particular case of this problem is when for each $i\in\Gamma$ $K_{\lambda(i)}\leq K_i$, that is, $\rho_i:K_{\lambda(i)}\to K_i$ is an injective homomorphism. So one can consider first the problem in this case.

\medskip
Let $K$ be a finite field and $R=K[x]$. For $r\in R$, let $m_r:R\to R$ be defined by $m_r(s)=r s$ for every $s\in R$. It is easy to see that for the
natural isomorphism $j:K[x]\to K^{(\mathbb N_0)}$ the conjugated isomorphism $j \circ m_x \circ  j^{-1}$ coincides with the right Bernoulli shift $\beta_K$ of $K^{(\mathbb N_0)}$ (and consequently,
$j \circ m_{x^n} \circ  j^{-1}=\beta^n_K$).
 Therefore, the endomorphism $m_r$ is (conjugated to) a linear combination of powers of the Bernoulli shift  $\beta_K$.

\begin{problem}
Calculate the entropy of $m_r:K[x]\to K[x]$. What about the ring $K[x_1,\ldots,x_n]$ of polynomials in more variables?
\end{problem}

This problem can be generalized for graded rings. (Let us recall that a \emph{graded ring} is a ring $R$ with a family $\{R_i:i\in\mathbb N_0\}$ of subgroups of $(R,+)$ such that $R=\bigoplus_{i=0}^\infty R_i$ and $R_i R_j\subseteq R_{i+j}$ for all $i,j\in \mathbb N_0$ \cite[Chapter 10]{AM}.)
For $r\in R$, let $m_r:R\to R$ be defined by $m_r(s)=r s$ for every $s\in R$.

\begin{problem}\label{Last_problem}
Compute the entropy $\mathrm{ent}(m_r)$ in case $R$ is a graded ring, $r\in R$ and $m_r:R\to R$.
\end{problem}

Problem \ref{Last_problem} can be extended also to graded $R$-modules $M$ and the endomorphism $m_r$ of $M$ defined by the multiplication, in $M$,  by  a fixed element $r\in R$ as above.

\medskip
We conclude with a problem suggested by Example \ref{example10}(b).

\begin{problem}
Let $\Gamma$ be an abelian group and $\lambda:\Gamma\to\Gamma$ an endomorphism.
\begin{itemize}
\item[(a)] Is it true that $s_\lambda>0$ implies $s_\lambda$ infinite?
\item[(b)] Describe in which cases $s_\lambda=0$ and in which cases $s_\lambda$ is infinite.
\end{itemize}
\end{problem}

redpajama_set_name: RedPajamaArXiv


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MOHIT HISSARIA
Mohit started his career with Sunita Minechem Industries as an International Marketing Executive in 2006. Sunita Minechem Industries converted into Sunita Hydrocolloids Pvt. Ltd. (a JV partner of Weatherford Intl Ltd), where Mohit managed oil and gas business development as the Product Line Manager for Weatherford International from 2008-2010. In 2013, he accepted a role as the President of Sunita Hydrocolloids Inc. Houston, TX (USA) developing fully integrated business line in oil and gas chemicals expanding into the major US Basins that include the Permian (TX), Bakken (ND), and the Mid-Continent (OK).
During his career at Sunita, Mohit managed the construction of our new facility in Jodhpur, India, implemented process enhancements to increase production and managed multiple accounts directly related to the growth of Sunita in establishing backward integration. This multi-faceted skill set is uniquely suited for new business development in the United States. Mohit demonstrates leadership abilities, technical knowledge, a dedicated work ethic and the ability to work well with all levels in an organization. He has respect for the people he works with, courage to lead change and shows a soundness of moral character.
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Rahul Bhati, Associate Vice-President of Sunita Hydrocolloids Inc., is a well-qualified and distinguished Chartered Accountant from ICAI with high level of integrity, competence and employment experience. Rahul started his professional career in 2007 as an auditor with Deloitte Haskin & Sales and continued his service with Deloitte through February, 2013. Since March 2013, Rahul became an integral part of the Sunita Hydrocolloids management team. He served as a General Manager in India for a year and relocated to the Sunita USA office in the year 2014 to manage US operations and business development. As a result, Rahul cultivated a knack for leading teams in building the most effective workflow possible to ensure ease in budget due diligence, long-term viability, and compliance in various energy industries such as Oil & Gas, power, chemicals, manufacturing, automobile and mining.
By combining data-driven decision making, interpretation of long/short-term business needs, and an in-depth understanding of how factors M&A, reorganization, outsourcing, and capital projects-impact operations, Rahul offers a track record of mitigating risk and developing opportunities into solid business cases. With a strong understanding of the unique requirements in adapting operations for new development initiatives including M&A, global expansion and capital projects.
Rahul earned a Masters in Commerce and B-Com Honors graduate from Jai Narayan Vyas University in Jodhpur, India.
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Danny Wilson, who majored in Agriculture Education at East Texas State University, joined Sunita Hydrocolloids in 2016 as Associate VP-Sales after spending 32 years with Benchmark Performance Group in various sales roles that included Account manager, VP, Sr. VP and on the board of directors, which was acquired by SCF (Rockwater Energy Solutions) in 2011.
Danny started his career in the oilfield chemical business with BJ Hughes in Odessa, Texas as a warehouseman for their North America distribution Center in 1977. Over a four year period moved into a chemical buyer's role responsible for maintaining inventory for the North America pumping services business line. In 1981 he accepted a role as plant superintendent for Delta Solvents and Chemicals in Odessa where his responsibilities included inventory management, logistics, distribution, and a limited sales role. In 1984 he joined Benchmark Performance Group as a sales representative responsible for all sales that included the pressure pumping service companies in the Permian Basin. Danny was promoted to VP-sales in 1988 and transferred to Houston in 1989 when Benchmark was expanding and then promoted to Sr. VP-Sales in 2010.
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Farhan Siddiqui serves Sunita Hydrocolloids Inc as Associate Vice President of Research and Development; she is responsible for leading the Sunita organization in research and development of new products, product enhancements and product redesign. The company relies on her extensive experience and judgement to plan and accomplish goals. Under her leadership, Sunita Hydrocolloids Inc has filed their first patent in 2018.
Farhan started her career as a graduate intern from University of Houston Department of Chemistry with the Shell Development Company in 1984 and served as key member in different departments. Discovering herself as an oilfield chemist, Farhan joined Texaco Inc. in 1987 to support and perform test on their new microwave water cut monitor where she achieved five patents and presented papers to both SPE and ACS. In 1994, she joined Westport Technology center as a project manager where she managed and worked on several joint industry EOR projects. Her time with Westport, Farhan served as a SPE treasurer for reservoir study group. In 1997, Farhan joined Benchmark Research and Technology as a senior chemist and in 2001, promoted to director of stimulation research, where for she continued her career for 18 years as she directed and managed the research operations department for oil field applications.
Over her 30 years of professional experience, Farhan has earned the respect and recognition of oil and gas industry. Farhan holds a Master's Degree in Organic Synthesis from The Central University of Hyderabad.
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