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Author Microsoft. Updated Over a year ago. Office Microsoft Excel Microsoft Word Microsoft Pffice Ok We use our own and third-party cookies for advertising, session, microsoft office professional plus free 2013 free, and social network purposes. Any action other than blocking them or the express request of the service associated to tree cookie in question, involves providing your microxoft to their use.
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Despite the drastic development of civilization, the introduction of innovation, the application of modernity, freshness and comfort – ecology is still in a state of real danger. Few people realize that the natural environment, that is, everything that surrounds us and what we breathe, where we walk, what we celebrate, watch and thanks to which we live, can transform into a deadly trap. Nearly 40% of the world's population lives in areas with limited access or deliberately decides to give up the use of modern, efficient, non-invasive and, above all, safe energy sources for their life and health. A perspective and open world for everyone – it's a world free of burning high-emission and pathogenic high-carbon fuels. We must be responsible for our planet to create opportunities for ourselves, our families and future generations and a place to do so. It's worth deepening your self-awareness and paying attention to what you use every day, especially when you use energy. Emergency rescue in the form of taking vitamins and minerals – can help, but will not save the environment and humanity, so what to replace with coal?
There is no doubt that it's the heavy and invasive coal, which significantly contributes to air pollution and the formation of many cancers, as well as those that are directly related to the respiratory system – in recent years has been the most popular and the cheapest method of obtaining energy. Today, we have no time or opportunity to wait for a change in the situation – we must act. Coal burning must end when we hear about growing smog, children born with genetic mutations, an increase in disease and infections, fish and animal fever – we do not realize that these are the consequences of our choice or consent to this type of procedure. We are moving away from the fact that something terrible, a personal tragedy caused by burning coal and using it as the main source of energy can happen to us. There are many alternatives that will allow us to live healthier, breathe more cleanly – save the planet and humanity from destruction. Fuel oil or natural gas – slightly more expensive solutions, but still available for many years. An increasing number of households, global corporations, companies, as well as administrative, governmental or diplomatic offices in the world – place priority on the implementation of ecological methods of obtaining energy, and hence, reducing the emission of lethal substances. For modern heating methods, which pose the least threat to the natural environment and for ourselves, we can include, for example, biomass, i.e. pellets, as well as an automatic heat pump. Pellets, which few people know, are nothing more than granular waste from trees, which are not only valuable as wood, but also a safe alternative that protects forests from felling. It should be added that using pellets and boilers intended for its distribution, we do not emit any CO2 gases to the environment, to the atmosphere (this is due to the specificity of construction of this type of equipment, and also due to the elemental composition of substances contained in wood waste).
In addition to biomass, a heat pump can be equally well regarded as a source of energy. Its varieties are diverse and rely on the synthesis as well as the natural processing of what the environment gave us. We can use, for example, an "air-water" solution, where the pumps draw energy from atmospheric air and are able to successfully provide sufficient energy through this operation (also by heating the water). It's impossible to forget about air power plants and hydropower plants. The famous energy producing windmills are now found all over the world and are not only popular with tourists, but also environmentalists and people involved in the process of saving the comfort of their own lives, the condition of the lungs and everything that surrounds us every day.
As you can see – definitely you can act in harmony with nature, let's look at air power plants, which, apart from a one-off investment of funds – now successfully earn on themselves and produce energy in a completely natural, healthy and safe way. Wind gusts, water currents – they cost nothing, are the result of changing weather conditions, and provide the world community with everything that they would not be able to survive without. Let's remember that coal is our curse, it's something that in the near future will annihilate the environment and ourselves. Today's research suggests that the negative effects of using this type of energy can be observed in over 35% of humanity. It's surprising, however, that we start to take care of ourselves and act only when what happens is actually preventing us from everyday life – maybe we should think about it before we touch something terrible, something we will be struggling with for the rest of our lives – maybe it's worth to start working today?


An already booming local construction sector is going to get a lot busier now that the federal government has approved construction of the new Ambassador Bridge, with work slated to begin as early as this year.
The long-sought replacement span for the 90-year-old existing Ambassador — the only bridge linking Windsor and Detroit and the busiest commercial crossing between Canada and the United States — comes at a time when plans are ramping up to build the new Gordie Howe International Bridge, a public-private partnership under government oversight that officials say will get underway late next year.
The Gordie Howe bridge is to be located about three kilometres south of the Ambassador Bridge.


issertation on South Africa's transition to democracy, and recently visited refugee camps and communities in the Middle East. He has been involved with truth and reconciliation processes in Africa, the South Pacific, and Canada. Two of his current avocations are researching church revitalization with the Thriving Christian Communities Project and assisting congregations with renewal as a member of EDGE network of consultants and coaches. His passions include interfaith cooperation, camping in the desert, and pickup games of basketball with his sons, Sam and Will. If you are interested in spiritual exploration or justice seeking that paints outside the lines, he wants to buy you a coffee.
As an educator, minister and lover of life, Susan has lived and worked in the Arctic, East Africa, and Central and Eastern Canada. She is continually amazed by the abundance in self and community, and the many and diverse ways it comes to light. Her joy in St. Andrew's is its concern with real world engagement. Matters of spirit and faith are held with pressing human concerns: of the heart, the household, the community, the planet … What we explore here helps us all to bear more gracefully the realities and responsibilities of being human.
Susan has been involved in the healing work of reconciliation, and is committed to new and open pathways. She has particular interest in the sacred lives and spirituality of those living with illness, dementia, loss, and grief. AND she has a life-long, never-ending heart space for children and youth. She knows they fulsomely contribute to the life, spirit, and mission of any vibrant community – the St. Andrew's community especially – and we can help them to find belonging and grounding that truly helps as they face life's changes and challenges.
Susan's own family is growing and changing, each in their own way. So much to explore … to share … to live with grace.
Jazz, R&B, gospel, pipe organ, spirituals, folk, piano improv, traditional church choral settings, global genres…you'll hear Kevin Parks playing any of these at St. Andrew's–often blending multiple styles in the same service. Kevin's music interests are diverse…his three teenagers even have him enjoying, and talking about, hip-hop and rap. Kevin's passion is to engage people and bring life to the Gospel through a well-selected text and tune; to encourage everyone to "make a joyful noise" regardless of perceived ability or experience. Music ministry has taken him across the country to serve in United Church, Anglican, Presbyterian and Salvation Army congregations, and as a leader in regional conferences and events. Kevin is a resource to musician colleagues, church committees and clergy to develop their music ministry or offer best practices related to employment of church musicians, collegial relationships between clergy, musicians and lay leadership, as well as copyright and intellectual property rights in the church. Beyond St. Andrew's, Kevin is the University Musician at Atlantic School of Theology, and co-music director at St. Luke's Community Players.


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Dying To Divorce (2021)
What would you risk for your freedom?
Directed by Chloe Fairweather, 81 mins
UK premiere
More than one in three Turkish women have experienced domestic violence and the number of femicides is rising. But some Turkish women are fighting back. Ipek Bozkurt, a courageous lawyer, is determined to challenge this misogynistic trend by putting abusive men behind bars.
Working with a group of activists, Ipek is fighting to get justice for two survivors of horrific assaults: Arzu, married off at 14 to a farmer ten years her senior and Kubra, a successful and glamorous TV presenter. Arzu lost both legs and the use of her arms when her husband fired seven shotgun shells into her after she asked for a divorce. She must try to rebuild her life to regain custody of her children, who have been taken into care.
Kubra suffered a brain haemorrhage after being attacked by her husband, two days after giving birth. Her injuries resulted in the loss of her ability to speak and walk. Her husband denies attacking her and has kept their daughter. Kubra must undergo intensive speech therapy in order to testify against him in court. Unless he is convicted, Kubra may not see her daughter again.
During these cases, Ipek must fight not only against a legal system, which regularly gives light sentences to male perpetrators but an increasingly repressive government. After an attempted coup, there is an unprecedented crackdown on dissenting voices leaving Ipek, like thousands of other lawyers, fearing imprisonment.
Filmed over five years, Dying To Divorce takes viewers into the heart of Turkey's gender-based violence crisis and the recent political events that have severely eroded democratic freedoms. Through intimately shot personal stories, the film gives a unique perspective on the struggle to be an independent woman in modern Turkey.
The panel discussion was moderated by documentary journalist Carol Nahra and included Dying To Divorce producer Sinead Kirwan, and protagonist Ipek Bozkurt as well as Professor Heidi Stoeckl, Director of the Gender Violence and Health Centre, London School of Hygiene & Tropical Medicine. The replay can be found below:
Dying To Divorce: additional resources
African Apocalypse (2020)
And Outside Life Goes On (2021)
Be My Voice (2021)
Congo Calling (2019)

redpajama_set_name: RedPajamaCommonCrawl


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Home/World News/Middle East/11 'israeli' pilots stuck in Egypt after entering without visas
11 'israeli' pilots stuck in Egypt after entering without visas
Eleven Israeli pilots have been reportedly stranded in Egypt after landing in the country without having the required visas.
syriatrNovember 25, 2022
Israeli media outlets claimed that the pilots landed in Egypt but did not have sufficient fuel to fly back home.
The reports did not mention whether the pilots were military-affiliated or civil nor the exact type of the aircraft, only saying that the planes were "light".
Citing "national security matters", the Egyptian authorities have reportedly sought to deport the pilots.
"The pilots are out of fuel and are not being allowed to refuel," Israel's Maariv newspaper said, adding that pilots are being "pushed to take other flights but the weather is currently too stormy for them to fly back."
Meanwhile, Israel's foreign ministry has confirmed the news. "From our inspection, the Israelis did receive permission to land the planes but did not receive visas for their entry into the country," an Israeli spokesperson said.
"We are handling the issue both at the embassy in Cairo and at the headquarters in Jerusalem so that the pilots can leave Egypt and return to Israel. They flew to Egypt without a visa. This is a mistake on their part and, as a result, they cannot enter the country," Israel National News online outlet quoted a ministry source as saying.
Both Egyptian median and authorities have declined to report and comment on the development by the time of writing.
Egypt was the first Arab state to recognize Israel and sign a peace treaty with the occupying regime. In 1978, then-Egyptian president Anwar Sadat shook hands with Israeli prime minister Menachim Begin in a meeting hosted by US President Jimmy Carter at Camp David.
At a diplomatic level, the Egyptian government currently treats Israel as a friendly neighbor with which it has strong ties in different fields, particularly security.
US, allies to pay dearly for backing terrorism: Assad
North Korea in final stage of preparing ICBM: Kim
At least 12 killed as bomber targets Pakistan anti-Taliban event
Leader of Hezbollah al-Nujaba: General Soleimani in Iraq
Progress of Iran invalidated logic of liberal democracy


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\section{Introduction}
	When Alan Turing originally defined his $a$-machines, which would later be called Turing machines, he envisioned a machine whose memory was laid out along a one-dimensional tape, inspired by the ticker tapes of the day.  This seemed somewhat arbitrary and perhaps unduly restrictive, and so, very quickly, machines with multiple and multi-dimensional tapes were proposed.  The focus at the time was defining the term ``computable", and as adding tapes and dimensions defined the same class of functions as Turing's original, simpler model, studying alternate tape geometries fell out of favor for some time.

	The complexity theory community then reignited interest in alternative tape geometries by considering not the class of functions computable by Turing machines, but time and space complexity of functions on different tape geometries.  This led to a number of results about relative efficiency of machines with one/many tapes and one/two/high-dimensional tapes.  For example, the language of palindromes (those strings that read the same forward and backward) can be computed in $O(n)$ time on a two-tape machine, but requires $\Omega(n^2)$ on a one-tape machine with one read/write head.  Or, an $m$-dimensional Turing machine running in time $T(n)$ can be simulated by a $k$-dimensional Turing machine ($k<m$) in time $T(n)^{1+\frac{1}{m}-\frac{1}{k}+\epsilon}$ for all $\epsilon>0$ \cite{Grigorev1979}.

	Many of the proofs and algorithms used in the study of multidimensional Turing machines make their way into or are inspired by the world of mesh-connected systems.  Mesh-connected systems are arrays of identical, relatively dumb processors (typically with memory logarithmic in the input size) that communicate with their neighbors to perform a computation.  Time use on a Turing machine with a $d$-dimensional tape is intimately tied to the power use of a $d$-dimensional mesh of processors with finite memory, so there is some natural crossover.  Mesh-connected systems constitute an area of very active research now, but since it remains very closely tied to the physical implementation, research is generally restricted to two- and three-dimensional grids.

	In this paper, we go beyond the world of rectangular grids and consider tapes at their most general.  The purpose of this is two-fold.  First, to give the complexity theorists a general framework in which to work, subsuming all current tape-based Turing models.  Second, to provide some evidence that alternative tape geometries are interesting from a recursion-theoretic perspective.  Along the way, we will encounter some questions in combinatorial group theory that aren't directly related to generalized Turing tapes, but are interesting in their own right.

	Any Turing tape can be modeled as a digraph with nodes corresponding to tape cells and edges corresponding to allowable transitions.  Hence, we start by introducing a number of graph-theoretic conditions that ought to be satisfied by a reasonable tape.  It turns out that the criteria we outline are necessary and sufficient conditions for the graph to be the Cayley graph of a finitely generated, infinite group.

	We then turn to the question of whether allowing arbitrary Cayley graphs as tapes is just another equivalent machine model for the class of computable functions.  Interestingly, this will depend on the structure of the group from which the Cayley graph is generated.  For groups with solvable word problem, this does indeed lead to machines that compute the class of computable functions, however for groups with unsolvable word problems, these machines are strictly more powerful than standard Turing machines.  In fact, they can be as powerful as any oracle machine and we end up with an alternative definition of the Turing degrees that is machine based and doesn't rely on oracles.

	The constructions and proofs of these results begin to raise questions about what kind of computable objects we can hope to find in arbitrary Cayley graphs.  In particular, whether we can always find an infinite, computable, non-self-intersecting path.  We call such a path an escape and we construct a group without any escapes.  This construction is non-trivial as any group without an escape must also be a Burnside group, which we also prove.

	Throughout the rest of this piece, all Turing machines will have a single tape and a single head.  This is the easiest case to treat and the generalization to multiple tapes and multiple heads is entirely analogous to the standard Turing picture.

\section{General Tape Geometries}
	Any tape essentially consists of a collection of cells, each of which can hold a single symbol, together with a mechanism for moving from one cell to another.  So underlying any tape is an edge-colored digraph.  Edges in this graph represent allowable transitions between states and the edge coloring encodes the conditions on the stored symbol and control state under which that particular transition occurs.  In order to be a reasonable Turing tape, this digraph should satisfy a few restrictions.
	\begin{enumerate}
		\item Uniqueness of outgoing colors - From any vertex, there should be exactly one outgoing edge of each color.  Since the mechanism by which the tape head moves is encoded in the edge color, outgoing edges should have different colors.  Also, since the transition function is independent of the tape cell, the collection of colors going out from each vertex should be the same.
		\item Homogeneity - Every vertex should ``look like" every other vertex.  Technically, this means that the subgroup of the automorphism group of the graph that preserves edge colors should be vertex transitive.  This is an extension of the assumption that all tape cells are indistinguishable, in this case by the local geometry.
		\item Infinity - The tape should have an infinite number of cells.  Otherwise, it's just a finite automaton.
		\item Connectedness - As the head moves during the course of a computation, it remains on a single connected component.  Inaccessible states are useless, so we can require that every tape cell be accessible from the starting point.  In particular, this means that the graph is connected.
		\item Finitely many colors - The transition function of the TM should be finite, so there should only be finitely many outgoing colors.  Having more colors doesn't change the computational power, since only finitely many of them could be referenced by the transition function anyway.
		\item Backtracking - The Turing machine should be able to return to the cell it just came from.  This assumption is less essential, since in view of homogeneity, any algorithm that called for returning to the previous tape cell could be replaced by a fixed sequence of steps.  However, many algorithms call for the head to return to the previous cell and forcing the head to do so by a circuitous route seems unduly harsh.  Note that in view of homogeneity, the color of an edge determines the color of the reverse edge.
	\end{enumerate}

	Restrictions~1 and~2 imply that our tape is a Cayley graph, restrictions~3,~4, and~5 make it the Cayley graph of a finitely generated infinite group, and restriction~6 forces the generating set to be closed under inverses.  In addition, any Cayley graph of of a finitely generated infinite group with a generating set that is closed under inverses satisfies~1--6.  This suggests that Cayley graphs are, in some sense, the ``right" degree of generality and leads to the following definition:

	\begin{definition}
		Let $G$ be an infinite group and $S\subset G$ be a finite generating set for $G$ that is closed under inverses.  Then the Cayley graph of $G$ generated by $S$ is called the {\em tape graph}, $(G,S)$.
	\end{definition}

	Using this general type of tape, we can then ask questions about the structure of Turing Machines with tapes given by assorted groups and generating sets.

\section{Turing Machines on Cayley Graphs}
	\begin{definition}
		Let $G=\langle g_1,\ldots,g_n\rangle$ be a finitely generated group with the set $\{g_1,\ldots,g_n\}$ closed under inverses.  Then a Turing Machine over $G$ with generating set $\langle g_1,\ldots, g_n\rangle$ is a 7-tuple, $(Q,\Gamma,b,\Sigma,\delta,q_0,F)$ where
		\begin{itemize}
			\item $Q$ is the finite set of states
			\item $\Gamma$ is the finite set of tape symbols
			\item $b\in\Gamma$ is a designated blank symbol
			\item $\Sigma\subset\Gamma\backslash\{b\}$ is the set of input symbols
			\item $\delta:Q\times\Gamma\rightarrow Q\times\Gamma\times\{g_1,\ldots,g_n\}$ is the transition function
			\item $q_0\in Q$ is the initial state
			\item $F\subset Q$ is the set of terminal states (typically one to accept and one to reject)
		\end{itemize}
	\end{definition}
	This definition varies from the standard definition only in the interpretation of the transition function.  Whereas a standard TM has a two-way infinite one-dimensional tape and the transition function includes instructions for moving left or right, a TM over $G$ has as a tape the Cayley graph of $G$ and the transition function has instructions for moving along edges labeled by a particular generator.  For example, a Turing Machine over $\mathbb{Z}$ with generating set $\{-1,+1\}$ is a standard one-dimensional TM and a TM over $\mathbb{Z}^2$ with generating set $\{(0,-1),(-1,0),(0,1),(1,0)\}$ is a standard two-dimensional TM.

	We have intentionally skipped the notion of how to provide input for machines of this type.  Most generally, we could insist that the initial set of non-blank tape cells be connected and contain the initial location of the tape head.  However, we really intend these machines to work like Turing machines, and therefore, to compute on strings of symbols.  It turns out that there will be a canonical way to lay out strings on the Cayley graph, but we need some machinery first.

\subsection{A Well-ordering in Trees}\label{treeOrdering}
	We shall turn aside from the main topic for a moment to discuss a general statement about trees.  There are many ways to define an order on the vertices of a tree, but we are going to be interested in the lexicographic order.  In general, lexicographic orderings on trees have few nice properties, but we show that finitely branching trees have a subtree where the lexicographic order is in fact, a well-order.

	First, some definitions.  Let $T$ be a finitely branching tree.  Denote by $[T]$ the set of all infinite paths through $T$ and by $\sqsubset$ the partial ordering on vertices induced by the tree.  Our convention will be that $v\sqsubset w$ means that $v$ is closer to the root than $w$.

	In order for the lexicographic ordering on $T$ to even make sense, we must have a linear order on the set of children of each node.  Denote the order on the children of $v\in T$ by $<_v$.  Then the lexicographic order, $<$ is defined as follows:
	\begin{itemize}
		\item If $v\sqsubset w$, then $v<w$.
		\item If neither $v\sqsubset w$ nor $w\sqsubset v$, let $u$ be the greatest lower bound of $v$ and $w$ according to $\sqsubset$ and let $u'\sqsubset v$ and $u''\sqsubset w$ be children of $u$.  Then $v<w\iff u'<_u u''$.
	\end{itemize}
	This order can, in fact, be extended to an order on $T\cup [T]$.  Identifying elements of $[T]$ with subsets of $T$ and elements of $T$ with one-element subsets of $T$, we can define
	\begin{equation*}
		x<y\iff (\exists w\in y)(\forall v\in x)v<w
	\end{equation*}

	Defined in this way, $<$ is a linear order, but we can't really hope for any more structure than that.  But, as promised, with a bit of pruning, we can find a subtree with much more structure.

	\begin{theorem}\label{thmWellOrder}
		Let $T$ be a finitely branching tree and let $<$ be the lexicographic ordering on the nodes of and paths through $T$ as given above. Define
		\begin{equation*}
			T'=\{v\in T|\forall w\in[T],v<w\}
		\end{equation*}
		Then $<$ restricted to $T'$ is order isomorphic to an initial segment of $\omega$.  In addition, if $T$ is infinite, $T'$ is infinite as well.
	\end{theorem}
	\begin{proof}
		This follows from the following direct result of K\"onig's Lemma.
		\begin{lemma}
			 Every element of $T'$ has only finitely many $<$-predecessors.
		\end{lemma}
		\begin{proof}
			Suppose $v\in T'$ had infinitely many $<$-predecessors.  Then we could form the tree,
			\begin{equation*}
				S=\{w\in T'|w<v\}
			\end{equation*}
			This is, in fact, a tree since $T'$ is a tree and $x\sqsubset y$ implies $x<y$.  Since $v$ has infinitely many $<$-predecessors, $S$ is infinite.

			By K\"onig's Lemma, there must be a path through $S$, call it $P$.  But $T'$ is a subtree of $T$ so $P\in [T]$.  By definition of $<$, $P<v$, but $v\in T'$ so $v<P$.  This is a contradiction, so $v$ must have only finitely many predecessors.
		\end{proof}
		Any linear order in which every element has only finitely many predecessors clearly cannot have an infinite descending chain, so must be a well-order.  As $\omega+1$ has an element with infinitely many predecessors, the order type must be an initial segment of $\omega$.

		For the second part of Theorem~\ref{thmWellOrder}, we need an additional lemma.
		\begin{lemma}\label{lmMinPath}
			If $[T]$ is non-empty, then $[T]$ has a minimal element in the $<$ ordering.
		\end{lemma}
		\begin{proof}
			We can inductively construct the minimal element of $[T]$.  For any $v\in T$, define $s(v)$ to be the minimal (according to $<_v$) child of $v$ that is a member of some element of $[T]$ if such a vertex exists.  Note that if there is a path through $v$, $s(v)$ is defined and there is a path through $s(v)$.  If $v_0$ is the root, then $P=\{s^{(n)}(v_0)\}_{n\in\mathbb{N}}$ is the desired minimal element.

			Since $[T]$ is non-empty, there is a path through the root and so, by induction $s^{(n)}(v_0)$ is defined for all $n$.  Therefore $P$ is indeed a path.

			To see that $P$ is minimal, let $P\neq P'\in[T]$.  Let $v=s^{(m)}(v_0)$ be the largest element (according to $\sqsubset$) of $P\cap P'$ and let $w\in P'$ be a child of $v$.  By maximality of $v$, $w\neq s(v)$ and by construction, $s(v)\leq_v w$.  Therefore, $s(v)<_v w$.  By the lexicographic ordering, $w$ is greater than all descendants of $s(v)$ and also greater than all ancestors of $s(v)$ (since ancestors of $s(v)$ are also ancestors of $w$).  Therefore, $w>u$ for all $u\in P$ and $P'>P$.
		\end{proof}

		Now, let $T$ be infinite.  Then, by K\"onig's Lemma again, $[T]$ is non-empty.  Let $P$ be the minimal path in $[T]$ according to Lemma~\ref{lmMinPath}.  Then $P\subset T'$ since for any $v\in P$ and $P'\in [T]$, $v<P\leq P'$.  $P$ is infinite, so $T'$ is infinite as well.
	\end{proof}

\subsection{Power of Turing Machines on Cayley Graphs}
	One of the first questions to be asked about any new model of computation is whether the class of functions computable by the new model is different from the class of computable functions.  For Turing machines on Cayley graphs, this depends rather sensitively on properties of the group producing the tape graph.  For example,
	\begin{lemma}\label{lemWordProblem}
		Let $(G,S)$ be a tape graph.  There is a Turing machine over $(G,S)$ that can solve the word problem for $G$.
	\end{lemma}
	This will not be proved rigorously, since we have not yet defined how input is to be provided, but we will provide an argument that can be made rigorous in an obvious fashion by the end of this section.

	Given two sequences of generators, the machine can simply follow the first sequence of generators, leaving a pointer at each cell along the way pointing to its predecessor.  Marking the end, it can follow the sequence of pointers back to the origin.  Now, it can follow the second sequence of generators and check to see whether the end point was marked in the first step.  Clearly, if this algorithm ends on the marked cell, the two sequences of generators correspond to the same group element.

	Boone and Novikov \cite{Boone1958,Novikov1958} independently showed that there exist groups with undecidable word problems, so this leads us to believe that Turing machines on a given group with unsolvable word problem are strictly more powerful than standard Turing machines.  However, this requires that Turing machines over said group also be able to compute all computable functions.  Fortunately, this is the case.

	\begin{theorem}\label{thmCayleySimulation}
		Let
		\begin{equation*}
			M=(Q_M,\Gamma_M,b_M,\Sigma_M,\delta_M,{q_0}_M,F_M)
		\end{equation*}
		be a standard one-dimensional one-tape Turing Machine and let $(G,S)$ be a tape graph.  Then there is a Turing Machine over $(G,S)$ that can simulate $M$.
	\end{theorem}

	The simulation itself is very straight-forward.  The only difficulty stems from the question of how to arrange the tape contents of the simulated machine on the Cayley graph.  If we could compute an infinite non-self-intersecting path through the Cayley graph, we could use this as a standard one-dimensional tape and do the simulation there.  However, as we will see in Section~\ref{secInescapable}, such a path need not exist.

	Fortunately, we can do the simulation anyway, in this case, by a variant of the ``always turn left" algorithm for solving mazes.  By putting an ordering on the generators of our tape graph, ``always turn left" becomes ``always follow the lexicographically minimal edge".  So, in an appropriate tree, Section~\ref{treeOrdering} gives us a well-ordered subtree where we can do the simulation with the $n$th vertex in the well-ordering storing the contents of the $n$th tape cell.

\subsection{Proof of Theorem \ref{thmCayleySimulation}}
	We will start with a description of the tree where we will store the tape of the simulated machine.  After we describe the operation of the machine, we will prove that this tree can be constructed on-line and navigated effectively.

	We will routinely use the natural correspondence between sequences of generators and group elements given by forming the product of the generators in the given sequence and evaluating in the group.  Henceforth, sequences and group elements will be used interchangeably.  Of course, multiple sequences will correspond to the same group element, but the sequence should always be clear from context.

	\begin{definition}
		Let $G$ be a group.  A {\em super-reduced word} is a finite sequence of elements of $G$ such that no subword, taken as a product in $G$ is equal to the identity.  More precisely, it is a sequence, $g_1,\ldots,g_n$ such that for all $1\leq i<j\leq n$, $\prod_{k=i}^jg_k\neq e$ in $G$.
	\end{definition}
	In the context of tape graphs, super-reduced words with symbols from the generating set correspond exactly to non-intersecting finite paths through the tape graph.  Note that a word is super-reduced if and only if all of its prefixes are super-reduced.

	Form the tree, $T$, of super-reduced words with symbols from $S$.  Since every group element has at least one super-reduced word corresponding to it and $G$ is infinite, $T$ is a spanning tree for the Cayley graph of $G$, hence, infinite.  Therefore, we can construct an infinite $T'$ as in Section~\ref{treeOrdering} where the lexicographic ordering is a well-ordering.  We will not do the computation on $T'$, but on a subtree, which we will call $R$.  To define $R$ we will want another definition.

	\begin{definition}
		We say that two sequences of generators, $v$ and $w$, are equivalent in $G$, or $v\equiv_G w$ if they correspond to the same group element.  In other words, $v\equiv_G w$ if $v=(s_1,\ldots,s_n)$, $w=(r_1,\ldots,r_m)$ and
		\begin{equation*}
			\prod_{i=1}^ns_i=\prod_{i=1}^mr_i\,\,\,\text{(in }G\text{)}
		\end{equation*}
	\end{definition}

	Now we can define $R$ as follows,
	\begin{equation*}
		R=\{v\in T'|(\forall w\in T')\,v\equiv_G w\implies v<w\}
	\end{equation*}
	That is, $R$ is the set of vertices in $T'$ that are lexicographically minimal among sequences that represent the same group element.

	It's not obvious at first glance, but $R$ is a tree.  Suppose to the contrary that $uv= w\in R$ but $u\notin R$.  Then there is some $u'<u$ in $R$ corresponding to the same group element.  Since we began with the tree of super-reduced words, $u$ and $u'$ are incomparable in the tree order.  Therefore, $u$ and $u'$ must differ at some first location.  So, $u'v\equiv_G w$ but $uv$ and $u'v$ differ for the first time at the same location and since $u'<u$, $u'v<uv=w$.  This is a contradiction since $w$ was supposed to be minimal among words corresponding to the same group element, so $R$ is indeed a tree.

	Also non-obvious is the fact that $R$ is infinite.  By the proof of Theorem~\ref{thmWellOrder}, $T'$ contains a path.  As was noted earlier, since we began with the tree of super-reduced words, elements that are comparable in the tree order cannot correspond to the same group element.  Therefore, the path in $T'$ corresponds to an infinite collection of group elements.  $R$ represents the same group elements since we only pruned redundant representations, so $R$ also represents an infinite number of group elements and is therefore infinite.

	Now for any tape graph, we have a tree that is well-ordered lexicographically, represents infinitely many group elements and represents each individual element at most once.  This is where we will do the simulation.

	Let $(G,S)$ be the tape graph given in the statement of the theorem.  Denote the elements of $S$ by $g_1,\ldots,g_n$.  It will be convenient to define a set $S'=S\cup\{g_0,g_{n+1}\}$ with the ordering $g_0<g_1<\cdots<g_n<g_{n+1}$.  We will consider both $g_0$ and $g_{n+1}$ equal to $e$ in the group for the purposes of moving from node to node.  Define a Turing Machine, $N$, over $(G,S)$ as follows:
	\begin{itemize}
		\item $Q_N=Q_M\sqcup(Q_M\times S')\sqcup(Q_M\times S')\sqcup(Q_M\times S)\sqcup(Q_M\times S)$
		\item $\Gamma_N=\Gamma_M\times S'\times\mathcal{P}(S)\times\mathcal{P}(S)$
		\item $b_N=(b_M,g_0,\emptyset,\emptyset)$
		\item $\Sigma_N=\Sigma_M\times S'\times\mathcal{P}(S)\times\mathcal{P}(S)$
		\item ${q_0}_N={q_0}_M$
		\item $F_N=F_M\times S'\times\mathcal{P}(S)\times\mathcal{P}(S)$
	\end{itemize}

	Each state in the tape alphabet will have an intended meaning.  Remember that we are going to do the computation on a tree, so we have to encode in each node not just the symbol of the simulated machine, but also auxiliary information about the structure of the tree.  In particular, if $(\gamma,\sigma,A,B)\in\Gamma_N$, $\gamma$ is the symbol of the simulated machine stored at the node, $\sigma$ is the generator to follow to reach the ancestor of this node in the tree, $A$ is the set of generators corresponding to edges pointing away from the root in the tree, and $B$ is the set of generators defining non-edges of the tree.  Remember that we are going to be constructing this tree on the fly and so we don't have complete information about which generators correspond to edges of the tree at every step.  Thus, elements of $S\backslash(A\cup B)$ are the edges whose membership in the tree has not yet been determined.

	We will also give each state a name and an intended interpretation,
	\begin{itemize}
		\item $Cq$ for $q\in Q_M$: We are simulating the computation of $M$ and the current state of $M$ is $q$.
		\item $Rqx$ for $q\in Q_M$ and $x\in S'$: The simulated tape head is moving to the right and $M$ is currently in state $q$.  The argument $x$ encodes the edge we followed to reach our current location.
		\item $Lqx$ for $q\in Q_M$ and $x\in S'$: The simulated tape head is moving to the left and $M$ is currently in state $q$.  The argument $x$ encodes the edge we followed to reach our current location.
		\item $Eqx$ for $q\in Q_M$ and $x\in S$: The simulated tape head is moving to the right and $M$ is in state $q$, but we have run out of tape and are attempting to extend the tree along edge $x$.
		\item $Bqx$ for $q\in Q_M$ and $x\in S$: $M$ is currently in state $q$ and we just failed to extend the tree along edge $x$, so we are backtracking.
	\end{itemize}
	Note that the starting state of $N$ is named $C{q_0}_M$.

	It will also be convenient to talk about the component functions of the transition function of $M$, \begin{eqnarray*}
		\delta_1:Q_M\times\Gamma_M&\rightarrow& Q_M\\
		\delta_2:Q_M\times\Gamma_M&\rightarrow&\Gamma_M\\
		\delta_3:Q_M\times\Gamma_M&\rightarrow&\{L,R\}
	\end{eqnarray*}

	We can now write down the action of the transition function.  If the current symbol being read is $(\gamma,\sigma,A,B)$, then the value of the transition function on each type of state is given in Table~\ref{tblSimTransition}.  Entries in the table are triples in the order, new state, symbol written, direction the tape head moves.
	\begin{table}[htbp]
		\begin{tabular}[htb]{l|c|c}
			$\digamma\in\Gamma_N$ & $\delta_N(\digamma,(\gamma,\sigma,A,B))$ \\ \hline
			$Cq$ & $\begin{cases}\left(R\delta_1(q,\gamma)g_0,(\delta_2(q,\gamma),\sigma,A,B),g_0\right)\text{ if }\delta_3(q,\gamma)=R\\ \left(L\delta_1(q,\gamma)\sigma,(\delta_2(q,\gamma),\sigma,A,B),\sigma\right)\text{ if }\delta_3(q,\gamma)=L\end{cases}$\\
			$Lqx$ & $\begin{cases}\left(Cq,(\gamma,\sigma,A,B),g_0\right)\text{ if }\forall y\in A,y\geq x\\ \left(Lqg_{n+1},(\gamma,\sigma,A,B),\max_{y\in A,y<x}y\right)\text{ otherwise}\end{cases}$\\
			$Rqx$ & $\begin{cases}\left(Cq,(\gamma,\sigma,A,B),\min_{y\in A,y>x}y\right)\text{ if }\exists y\in A,y>x\\ \left(Eqy,(\gamma,\sigma,A\cup\{y\},B),y\right)\text{ where }y=\min_{z\in S\backslash B,z>x}z\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{ if }\exists z\in S\backslash B,z>x\\ \left(Rq\sigma,(\gamma,\sigma,A,B),\sigma\right)\text{ otherwise}\end{cases}$\\
			$Eqx$ & $\begin{cases}\left(Cq,(\gamma,x^{-1},\emptyset,\{x^{-1}\}),g_0\right)\text{ if }A=B=\emptyset\\ \left(Bqy,(\gamma,\sigma,A,B),x^{-1}\right)\text{ otherwise}\end{cases}$\\
			$Bqx$ & $\left(Rqx,(\gamma,\sigma,A\backslash\{x\},B\cup\{x\}),g_0\right)$
		\end{tabular}
		\caption{Transition Table for $N$}
		\label{tblSimTransition}
	\end{table}

	Most of these transitions are pretty opaque, so some explanation is warranted.

	If we are in a $C$-state, then we perform one step of the computation and then transition into either an $R$-state or and $L$-state depending on whether the computation tries to move left or right.  A leftward step in the simulated machine always begins with a step toward the root for the simulating machine, so we take that step immediately.  Rightward steps are more complicated, so we leave the tape head where it is on a rightward step.

	Taking a step to the left, we want to find the lexicographic immediate predecessor of our current vertex.  We begin by taking one step toward the root, which we did when we transitioned out of the $C$-state.  Then, either there is a branch all of whose elements are less than where we started, or not.  If not, then we are at the immediate predecessor of our origin, and we continue computing.  Otherwise, follow the greatest such branch and always move to the greatest child until we reach a dead-end.  This dead-end is the immediate predecessor we were looking for, so we can continue the computation.

	Taking a step to the right is significantly more complicated, as we are most likely going to have to extend the tree as we go.  If we have already built some tree above us ($A$ is non-empty), then we simply use that part of the tree, moving along the minimal edge in the tree and continuing the computation.  This is the reason for using the ``false" generator $g_0$ when we transition out of the $C$-state.  Otherwise, try to extend the tree along the least edge that we have not already ruled out.  We add that edge to $A$ and switch to an $E$-state.

	If $A$ and $B$ are both empty, then this is a new vertex that we have not visited before, so continue with the computation.  Otherwise, we are at a vertex that has already been added elsewhere in the tree.  So, we back up and transition to a $B$-state.  In the $B$-state, we rule out the edge we just took by removing it from $A$ and adding it to $B$ and switch to an $R$-state to try extending the tree again.

	If we can't extend the tree at all ($B=\{g_1,\ldots,g_n\}$), then take a step toward the root and try again, eschewing anything less than or equal to the edge we backtracked along.  Since there is an infinite tree for us to use, we will eventually be able to extend the tree.

	We are now in a position to be more explicit about input.  Take a machine with one tape given by a tape graph, $(G,S)$ and another, read-only one-way standard input tape.  Then, in view of the construction above, it is clear how this machine would transcribe its input from the standard tape onto the tape graph.  Once the transcription is done, the computation can proceed according to the above construction.  Thus, it seems reasonable to require the input to a machine be the result of a transcription of this type.  This does require some (potentially) non-recursive manipulation of a string to produce the appropriate input, but if this offends you, it is always possible to go back to the formalism with an auxiliary read-only input tape.

\subsection{An Alternative Characterization of the Turing Degrees}\label{secTDegrees}
	We have demonstrated that Turing Machines on arbitrary Cayley graphs are strictly more powerful than standard Turing Machines, so the next question to ask is, ``how much more powerful?"  The short answer is ``as powerful as we want".  In \cite{Boone1966a,Boone1966b} Boone showed that for any r.e. Turing degree, there is a finitely presented group whose word problem is complete for that degree.  Using such a group, we can produce a machine (more precisely a class of machines) that computes exactly the functions in or below the given degree.  More precisely,
	\begin{theorem}
		Let $T$ be an r.e. Turing degree.  There is a group, $G$, such that the class of functions computable by a Turing Machine over $G$ is exactly the class of functions in or below $T$.
	\end{theorem}
	\begin{proof}
		By Boone, let $G$ be a group whose word problem is complete for $T$ and let $f$ be a function in or below $T$.  Since $f\leq T$, $f$ is Turing reducible to the word problem for $G$.  Turing machines over $G$ can perform this reduction since they can compute all recursive functions and they can solve the word problem for $G$ by Lemma~\ref{lemWordProblem}.  Therefore, they can compute $f$.

		To show that a Turing Machine over $G$ cannot compute a class larger than $T$, observe that a Turing Machine with an oracle for the word problem for $G$ can easily simulate a Turing Machine over $G$.  It simply maintains a list of nodes written as a sequence of generators for the address together with whichever tape symbol is written there.  When the simulated tape head moves, the machine simply appends the generator to the current address and consults the oracle to determine which node this corresponds to, adding a new entry if the new node isn't in the list.
	\end{proof}

\section{Inescapable Groups}\label{secInescapable}
\subsection{Construction of an Inescapable Group}
	Theorem~\ref{thmCayleySimulation} raises the question of whether or not a Turing Machine can always walk off to infinity on the Cayley graph of an infinite group without retracing its steps.  This suggests the following definition:
	\begin{definition}
		An {\em inescapable group} is a tape graph, $(G,S)$, such that any infinite computable sequence, $s$, of elements from $S$ corresponds to a self-intersecting path.
	\end{definition}

	This definition leads to a number of questions.  Do such things exist?  What do they look like?  Is inescapability a group invariant?  Very little is known in response.  For example, as we will see in Section~\ref{secBurnside}, an inescapable group must be a Burnside group; that is, every element must have finite order.  The purpose of this section is to answer the more fundamental question of whether or not inescapable groups exist.

\subsubsection{Definitions}
	We will need to make a few definitions.  Let $A$ be a finite set.  Let $A^*=A^{<\omega}$ be the set of all finite sequences with elements from $A$, called words over $A$.  Let $\epsilon$ denote the empty word and let $A^+=A^*\backslash\{\epsilon\}$ be the set of non-empty words.  If $w\in A^*$, denote by $|w|$ the length of~$w$, by $w(i)$ the $i$th symbol in~$w$ (indexed from~0), and by $w(i,j)$ the word $w(i)w(i+1)\ldots w(j)$.  Also, let $A^{\leq n}=\{w\in A^*,|w|\leq n\}$ be the set of words of length no greater than~$n$.

	For $w,w'\in A^*$ define the following relations:
	\begin{itemize}
		\item $w'$ is a subword of $w$ if there exist $0\leq i,j<|w|$ such that $w'=w(i,j)$.
		\item $w'$ is a subsequence of $w$ if there exist $0\leq i_1<i_2<\ldots <i_{|w'|}<|w|$ such that $w'=w(i_1)w(i_2)\ldots w(i_{|w'|})$.
		\item $\#(w,w')=\left|\{x\in\mathbb{N}^{|w'|}|x_0<\ldots<x_{|w'|-1}\text{ and }w(x_i)=w'(i)\text{ for all }i\}\right|$ is the number of ways in which $w'$ is a subsequence of $w$.
	\end{itemize}
	Notice that $\epsilon$ is both a subword and subsequence of every word and that $\#(w,\epsilon)=1$ for all $w$.  As an example, note that while $aabbaa$ does not contain $aba$ as a subword, it does contain $aba$ as a subsequence in 8 different ways.

	We will say that an infinite sequence of symbols from $A$, call it $s$, is computable if there is a Turing Machine which, on input $n$, produces $s(n)$.

	When dealing with sets, we will use $\mathcal{P}(X)$ to denote the collection of all subsets of $X$ and $\mathcal{P}_r(X)$ to denote all subsets of $X$ of cardinality $r$.

	We will also be dealing with polynomials in a non-commutative polynomial ring, so let $f\in K\langle x_1,\ldots,x_d\rangle$ be a polynomial over a field, $K$, with non-commuting indeterminates.  We say that $f$ is homogeneous of degree~$n$ if $f$ is a $K$-linear combination of monomials of the form
	\begin{equation*}
		x_{i_1}^{n_1}x_{i_2}^{n_2}\cdots x_{i_k}^{n_k}
	\end{equation*}
	with $n_1+n_2+\ldots+n_k=n$.  In this case, we write $\partial(f)=n$.

\subsubsection{A Combinatorial Lemma}
	When we get to the actual construction, we are going to need to find a subsequence of each computable sequence that satisfies certain properties.  In particular, we need the following lemma.

	\begin{lemma}\label{lmEvenSubstrings}
		Let $A=\{x_1,\ldots,x_m\}$ and let $n\geq 1$.  There is some $C(m,n)$ such that for all words, $s$, over alphabet $A$ with $|s|\geq C(m,n)$, $s$ contains a non-empty subword, $w$, with the following property.  For each $w'\in A^{\leq n}\cap A^+$, $w$ contains $w'$ as a subsequence an even number of times.
	\end{lemma}

	In fact, the number $C(m,n)$ is a Ramsey number, $R(2,3,2^\frac{m^{n+1}-1}{m-1})$ where the function $R(r,k,n)$ is given in Ramsey's Theorem,
	\begin{theorem}[Ramsey]
		Let $r,k,n$ be positive integers with $1\leq r\leq k$.  Then there exists an integer, denoted $R(r,k,n)$, such that for each set $X$ with $|X|=R(r,k,n)$ and each partition of $\mathcal{P}_r(X)$, $Y_1,\ldots,Y_n$, there exists a $k$-element subset $Y$ of $X$ and a set $Y_i$ with $\mathcal{P}_r(Y)\subset Y_i$.
	\end{theorem}

	To prove the lemma, we need a theorem from the combinatorial theory of words.
	\begin{theorem}[Pirillo \cite{Pirillo1983}]\label{thmPirillo}
		Let $\phi:A^+\rightarrow E$ be a mapping from $A^+$ to a set $E$ with $|E|=n$.  For each $k\geq 1$, each word $w\in A^+$ of length $R(2,k+1,n)$ contains a subword $w_1w_2\ldots w_k$ with $w_i\in A^+$ and
		\begin{equation*}
			\phi(w(i,i'))=\phi(w(j,j'))
		\end{equation*}
		for all pairs $(i,i')$, $(j,j')$ ($1\leq i\leq i'\leq k$ and $1\leq j\leq j'\leq k$).
	\end{theorem}

	\begin{proof}[Proof of Lemma \ref{lmEvenSubstrings}]
		Consider the function $\phi:A^+\rightarrow \mathbb{Z}_2^{A^{\leq n}}$ defined as follows:
		\begin{equation*}
			\left(\phi(w)\right)(w')=\#(w,w')\bmod{2}
		\end{equation*}
		Since
		\begin{equation*}
			\left|\mathbb{Z}_2^{A^{\leq n}}\right|=2^{\frac{m^{n+1}-1}{m-1}}
		\end{equation*}
		We can apply Theorem~\ref{thmPirillo} with $k=2$ to $s$ to get $w_1w_2$, a subword of $s$ such that
		\begin{equation*}
			\phi(w_1)=\phi(w_2)=\phi(w_1w_2)
		\end{equation*}
		Then the word $w_1w_2$ contains any non-empty sequence over $A$ of length no greater than $n$ as a subsequence an even number of times.

		We will prove something slightly stronger by induction on the length of the contained subsequence.  Specifically, we will show that not only does $w_1w_2$ satisfy the theorem, but so do $w_1$ and $w_2$ individually.

		For the base case, let $x_i$ be a subsequence of length one.  Then
		\begin{equation*}
			\phi(w_1w_2)(x_i)=\phi(w_1)(x_i)+\phi(w_2)(x_i)=2\phi(w_1)(x_i)=0\bmod{2}
		\end{equation*}
		since the number of occurrences of a single symbol in a concatenation of words is simply the sum of the number of occurrences in each factor.  Using the fact that $\phi(w_1w_2)=\phi(w_1)=\phi(w_2)$, we have established the base.

		The induction step isn't significantly more difficult; the only difficulty arises from the fact that the number of occurrences of a substring of length greater than one isn't additive.  However, we do have the formula,
		\begin{equation*}
			\#(w_1w_2,w')=\sum_{i=0}^{|w'|}\#(w_1,w'(0,i-1))\#(w_2,w'(i,|w'|-1))
		\end{equation*}
		We really only care about the parity of this expression and if we assume the induction hypothesis for all strings shorter than $w'$, most of the terms are even.  So by reducing,
		\begin{equation*}
			\#(w_1w_2,w')=\#(w_1,w')+\#(w_2,w')\bmod{2}
		\end{equation*}
		Now, in an argument analogous to the base case, we get
		\begin{equation*}
			\phi(w_1w_2)=\phi(w_1)=\phi(w_2)=0\bmod{2}
		\end{equation*}
	\end{proof}

\subsubsection{The Golod-Shafarevich Theorem}
	The Golod-Shafarevich Theorem is a powerful tool from algebra that gives a sufficient condition for a particular quotient algebra to be infinite dimensional.  For our purposes, it is a tool that will ensure that as we start adding relations to a free group, we don't collapse the group to something finite.

	The power of the theorem comes from the fact that the criterion it presents is based only on the number of relations of certain types, and not on the relations themselves.  This gives us a large amount of freedom to choose the relations we want without having to worry about bad interactions between them.

	\begin{theorem}[Golod-Shafarevich \cite{GolodShafarevich}]\label{thmGS}
		Let $R_d=K\langle x_1,\ldots, x_d\rangle$ be the polynomial ring over a field, $K$, in the non-commuting indeterminates $x_1,\ldots,x_d$.  Let $f_1,f_2,\ldots\in F$ be a set of homogeneous polynomials of $R_d$, and let the number of polynomials of degree $i$ be $r_i$.  Let $2\leq\partial(f_i)\leq\partial(f_{i+1})$ and let $I$ be the ideal generated by $F$.  Let $R_d/I=A$.  If all the coefficients in the power series,
		\begin{equation*}
			\left(1-dt+\sum_{i=2}^\infty r_it^i\right)^{-1}
		\end{equation*}
		are non-negative, then $A$ is infinite dimensional.
	\end{theorem}

	In a subsequent paper, Golod \cite{Golod1964} proves the following corollary,
	\begin{corollary}\label{corDimBound}
		In Theorem~\ref{thmGS}, if
		\begin{equation*}
			r_i\leq\epsilon^2(d-2\epsilon)^{i-2}
		\end{equation*}
		where $0<\epsilon<\frac{d}{2}$, then $A$ is infinite-dimensional.
	\end{corollary}
	For example, taking $d=2$ and $\epsilon=\frac{1}{4}$ in the corollary, we see that if $r_i\leq 2$ for all $i\geq 11$ and $r_i=0$ for all $i<11$, $A$ is infinite dimensional.

	Golod used this fact to establish the existence of a Burnside group, that is, an infinite group in which every element has finite order.  He did more than this, in fact, and produced an infinite $p$-group for each prime, $p$.  The diagonalization in Section~\ref{secDiag} will follow the same general ideas as Golod's construction as simplified for countable fields in Fischer and Struik \cite{FischerStruik}.

\subsubsection{The Construction}\label{secDiag}
	In Fischer and Struik \cite{FischerStruik}, there is a construction of a nil-algebra over finite and countable fields.  Although not expressly stated there, the construction is essentially a diagonalization over all polynomials.  In fact, even Golod's original construction of a nil-algebra can be viewed as a diagonalization over all polynomials, but the fact that the collection of all polynomials in the general case is uncountable makes it more difficult to see.

	The presentation given here will follow the construction in Fischer and Struik, since we will be diagonalizing against the set of computable sequences, which is countable.  Thus, we will use the more straightforward construction.

	\begin{theorem}\label{thmInescGroup}
		There exists an inescapable group.
	\end{theorem}

	\begin{proof}
		Let $d\geq 2$ and $A=\{x_1,\ldots,x_d\}$.  Consider the algebra, $\mathcal{A}=\mathbb{F}_2\langle A\rangle$ in non-commuting indeterminates, $x_i\in A$.  Let
		\begin{equation*}
			S=\bigcup_{i=1}^d\{(1+x_i),(1+x_i)^{15}\}
		\end{equation*}
		and enumerate all c.e. sequences over $S$: $s_0,s_1,\ldots$.  We will use the elements of $S$ interchangeably as characters in an alphabet and as polynomials in $\mathcal{A}$.

		We will construct a set of homogeneous polynomials, $F$, such that the following conditions hold:
		\begin{itemize}
			\item $x_i^{16}\in F$ for $1\leq i\leq d$.  (This will ensure that $(1+x_i)$ has order~16 and is therefore multiplicatively invertible in the quotient algebra)
			\item For each $i\in\mathbb{N}$, $s_i$ has a subword, $w$ such that
				\begin{equation*}
					\left(\prod_{j=0}^{|w|-1}w(j)\right)-1
				\end{equation*}
				is in the ideal generated by elements of $F$.
			\item The number of elements of $F$ of degree $i$ is~0 for $i<16$, $d$ for $i=16$, and either~0 or~1 for every $i>16$.
		\end{itemize}

		Define the following sequence recursively:
		\begin{eqnarray*}
			r_0&=&16\\
			r_{n+1}&=&15\cdot R\left(2,3,2^{\frac{|S|^{r_{n}+1}-1}{|S|-1}}\right)
		\end{eqnarray*}
		We can now enumerate the elements of $F$ as follows.

		Start with $F=\{x_1^{16},\ldots,x_d^{16}\}$ and begin enumerating the elements of all $s_i$ in parallel.  If, at any point, we have enumerated a contiguous subsequence of elements of some $s_i$ of length $r_{i+1}$, call it $v$ and do the following.

		By Lemma~\ref{lmEvenSubstrings}, we can find a non-empty subword, $w$, of $v$ such that $|w|\leq \frac{1}{15}r_{i+1}$ and every sequence of length $\leq r_i$ occurs an even number of times in $w$.  Then, by multiplying out,
		\begin{equation*}
			p(x_1,\ldots,x_d):=\prod_{k=0}^{|w|-1}w(k)=\sum_{w'\in S^*}\#(w,w')\prod_{k=0}^{|w'|-1}(w'(k)-1)
		\end{equation*}
		Note that if $0<|w'|\leq r_i$, $\#(w,w')\equiv 0\bmod{2}$ and that the constant term of $p$ is~1.  Note also that $w'(k)-1$ always has constant term~0, so $p-1$ has no non-zero terms with degree $\leq r_i$.  Since $p$ is a polynomial, we can write $p-1$ as a sum of homogeneous components, $f_1,\ldots,f_m$.  Note that for all $1\leq i\leq m$, $r_i<\partial(f_i)\leq 15|w|\leq r_{i+1}$.  Add all of these to $F$, stop enumerating $s_i$ but continue enumerating everything else that hasn't been similarly halted.

		The $F$ that is so constructed clearly contains $x_i^{16}$ for $1\leq i\leq d$ and for each $s_i$, if $s_i$ is total, it has a subword whose corresponding product is equivalent to~1 mod the ideal generated by $F$.  Also, each $s_i$ only adds polynomials to $F$ that have degree in the interval $(r_i,r_{i+1}]$ and adds at most one polynomial of any given degree.  Since we start with $d$ polynomials of degree~16 and $r_0=16$, the number of elements of $F$ of degree $i$ must be~0 for all $i<16$ and either~0 or~1 for every $i>16$.

		We would then like to show that the multiplicative semigroup of $\mathcal{A}/(F)$ generated by elements of $S$, call it $G$, is an inescapable group.  First, note that by the binomial theorem,
		\begin{equation*}
			(1+x_i)(1+x_i)^{15}=1+x_i^{16}=1
		\end{equation*}
		so $G$ is a genuine group and $S$ is closed under inverses.  $S$ trivially generates $G$, so we only need $G$ to be infinite for $(G,S)$ to be a tape graph.

		By elementary calculus,
		\begin{equation*}
			2\leq d\leq\frac{1}{16}(d-.5)^{i-2}
		\end{equation*}
		for all $i\geq 11$.  Therefore, $F$ satisfies the hypotheses of Corollary~\ref{corDimBound} and $\mathcal{A}/(F)$ is infinite-dimensional.

		Now, for any positive integer, $d$, there must be a monomial of degree $d$ that does not lie in the ideal generated by $F$.  Otherwise, every monomial of degree~$\geq d$ would be in the ideal and the quotient algebra would be finite-dimensional.  So, consider two generic such monomials,
		\begin{equation*}
			x_{i_1}x_{i_2}\ldots x_{i_M}\text{ and }x_{j_1}x_{j_2}\ldots x_{j_N}
		\end{equation*}
		with $M>N$ and the group elements,
		\begin{equation*}
			u=(1+x_{i_1})(1+x_{i_2})\ldots(1+x_{i_M})\text{ and }v=(1+x_{j_1})(1+x_{j_2})\ldots(1+x_{j_N})
		\end{equation*}
		Then,
		\begin{equation*}
			u-v=x_{i_1}x_{i_2}\ldots x_{i_M}+\ldots
		\end{equation*}
		where the remaining terms all have degree $\leq M$.  Since the ideal $(F)$ is generated only by homogeneous polynomials, $u-v\in (F)$ if and only if every homogeneous component of $u-v$ is in $(F)$.  However, the degree $M$ component of $u-v$ is clearly not in $(F)$, so neither is $u-v$.  Therefore, $u$ and $v$ are different elements in $G$.

		Thus, we can find infinitely many distinct elements of $G$, one for each degree.  Therefore, $G$ is infinite and $(G,S)$ is a tape graph.

		In addition, any computable sequence of generators is one of the $s_i$, so we have ensured that it has a subword such that the corresponding product is equal to~1 in the quotient algebra.  This is the same as having product~1 in the group, so all computable sequences of generators must correspond to self-intersecting paths.
	\end{proof}

	The construction given above in fact does better than producing an inescapable group.  Since every step of the construction can be done recursively, the set of relations in the group is r.e.  It is a standard result that a group with an r.e. set of relations is recursively presentable, so the construction produces a {\em recursively presentable} inescapable group.  That being said, the question of whether there exists a finitely presentable inescapable group remains open.  It is also unlikely that the word problem for the group constructed above is solvable, so there also remains the question of whether there exists an inescapable group with solvable word problem.  It should also be noted that the only property of computable sequences that the construction used was that there are countably many of them.  Thus, the argument relativizes.  In particular, the construction will produce a group with no escape in Turing degree $T$, but with a presentation in $T$.

	Since the presentation is in $T$, the set of escapes is $\Pi^0_1$ in $T$.  You can see this by observing that if an infinite sequence has a self-intersection, we will eventually know about it.  Trivially, there are escapes, so by the low basis theorem, there must be a low escape.  So, for any Turing degree, we have a group with no escapes in or below the given degree, but with an escape low relative to it.

\subsection{Inescapable Groups are Burnside Groups}\label{secBurnside}
	It is clear that in an inescapable group, all generators must have finite order, but must every group element have finite order?  An infinite group in which every element has finite order is called a Burnside group and the existence of Burnside groups was an open problem for some time before E. Golod constructed one in 1964~\cite{GolodShafarevich,Golod1964}.

	So our question can be rephrased as, must every inescapable group be a Burnside group?  It turns out the answer is yes, but it's not as obvious as it appears.  Suppose your group did have an element of infinite order.  The obvious thing to do would be to write this element of infinite order as a product of generators and simply repeat that sequence to produce an escape.  This certainly gives a computable sequence of generators that hits infinitely many elements of the group, but there is no reason this is a {\em non-self-intersecting} path.  Fortunately, we can find some other element of infinite order and an expression of it as a product of generators such that this naive construction does give a non-self-intersecting path.

	The general idea is to start with the obvious construction and cut out the loops.  Since our original element has infinite order, the constructed walk can only return to a given point a fixed, finite number of times.  So, we wait at each point of the walk until it returns to our current position for the last time, then we follow for one step, and repeat.  The only difficulty is knowing when the last return will be.  Fortunately, this is invariant under shifting by our infinite order element, so we can just record it in a finite table indexed by the generators in the expression of our infinite order element.

\begin{theorem}\label{thmInescBurn}
	Let $G=\langle g_0,\ldots,g_n\rangle$ be a group and $a\in G$ have infinite order.  Then there exists $b=\prod_{j=0}^{k-1}{g_{i_j}}\in G$ such that $\prod_{j=0}^N g_{i_{j\bmod{k}}}$ is distinct for each $N$.
\end{theorem}
\begin{proof}
	Let $a=\prod_{i=0}^{m-1} h_i$ with $h_i\in\{g_0,\ldots,g_n\}$ be an expression for $a$ of minimal length.  Define
	\begin{equation*}
	\delta(r,s,M)=\left(\prod_{i=r+1}^{m-1}h_i\right)a^M\left(\prod_{j=0}^{s-1}h_j\right)
	\end{equation*}
	and consider relations of the form $\delta(r,s,M)=e$ with $M\geq 0$.  Fixing $r$ and $s$, there is at most one $M$ for which $\delta(r,s,M)=e$ since $a$ has infinite order.  Similarly, fixing $r$ and $M$, there is at most one $s$ for which $\delta(r,s,M)=e$ since we chose an expression for $a$ of minimal length.  This allows us to define the following functions, $\alpha:[0,m-1]\rightarrow\mathcal{P}(\mathbb{N}\times[0,m-1])$, $\beta:[0,m-1]\rightarrow\mathbb{N}$, and $\gamma:[0,m-1]\rightarrow[0,m-1]$ by
	\begin{eqnarray*}
		\alpha(r)&=&\{(M,s)|\delta(r,s,M)=e\}\\
		\beta(r)&=&\begin{cases}0\text{ if }\alpha(r)=\emptyset\\
			\max_{(M,s)\in\alpha(r)}M\text{ otherwise}\end{cases}\\
		\gamma(r)&=&\begin{cases}s\text{ if }(\beta(r),s)\in\alpha(r)\\
			(r+1)\bmod{m}\text{ if }\alpha(r)=\emptyset\end{cases}
	\end{eqnarray*}

	Notice that $\alpha(r)$ is always a finite set since $s$ can take on at most finitely many values and for each value, there is at most one $M$ such that $(M,s)\in\alpha(r)$.  Therefore, whenever $\alpha(r)$ is non-empty, $\beta(r)$ exists and is attained by exactly one element, $(\beta(r),s')$ of $\alpha(r)$.  By definition, $\gamma(r)=s'$, so $\gamma$ is well-defined.

	\begin{lemma}\label{prefixes}
		For all $n$, there is a $k_n$ such that
		\begin{equation*}
			\prod_{i=1}^n h_{\gamma^{(i)}(m-1)}=a^{k_n}h_0\ldots h_{\gamma^{(n)}(m-1)}
		\end{equation*}
		where the sequence $\{k_n\}$ is defined by
		\begin{eqnarray*}
			k_0&=&-1\\
			k_{n+1}&=&k_n+\begin{cases}0\text{ if }\alpha(\gamma^{(n)}(m-1))=\emptyset\text{ and }\gamma^{(n)}(m-1)\neq m-1\\
				1\text{ if }\alpha(\gamma^{(n)}(m-1))=\emptyset\text{ and }\gamma^{(n)}(m-1)=m-1\\
				\beta(\gamma^{(n)}(m-1))+1\text{ otherwise}\end{cases}
		\end{eqnarray*}
	\end{lemma}
	\begin{proof}
		We proceed by induction.  If $n=0$, the product on the left side is empty, so we can take $k_0=-1$.

		For the induction step, suppose the lemma holds for $n-1$.  Then we can apply the induction hypothesis to reduce the problem to
		\begin{equation*}
			a^{k_{n-1}}h_0\ldots h_{\gamma^{(n-1)}(m-1)}h_{\gamma^{(n)}(m-1)}=a^{k_n}h_0\ldots h_{\gamma^{(n)}(m-1)}
		\end{equation*}
		or more simply,
		\begin{equation*}
			h_0\ldots h_{\gamma^{(n-1)}(m-1)}=a^{k_n-k_{n-1}}h_0\ldots h_{\gamma^{(n)}(m-1)-1}
		\end{equation*}
		If $\alpha(\gamma^{(n-1)}(m-1))$ is empty, then
		 \begin{equation*}
			\gamma^{(n)}(m-1)=(\gamma^{(n-1)}(m-1)+1)\bmod{m}
		\end{equation*}
		If $\gamma^{(n-1)}(m-1)=m-1$, then the product on the left is $a$ and the product on the right is $a^{k_n-k_{n-1}}$, so take $k_n=k_{n-1}+1$.  Otherwise, take $k_n=k_{n-1}$.

		On the other hand, if $\alpha(\gamma^{(n-1)}(m-1))$ is non-empty, $\gamma^{(n)}(m-1)=s$ for $(M,s)\in\alpha(\gamma^{(n-1)}(m-1))$ where $M=\beta(\gamma^{(n)}(m-1))$.  Since $(M,s)\in\alpha(\gamma^{(n-1)}(m-1))$,
		\begin{eqnarray*}
			h_0\ldots h_{\gamma^{(n-1)}(m-1)}&=&h_0\ldots h_{\gamma^{(n-1)}(m-1)}\delta(\gamma^{(n-1)}(m-1),s,M)\\
			&=&h_0\ldots h_{\gamma^{(n-1)}(m-1)}h_{\gamma^{(n-1)}(m-1)+1}\ldots h_{m-1}a^Mh_0\ldots h_{s-1}\\
			&=&a^{M+1}h_0\ldots h_{\gamma^{(n)}(m-1)-1}
		\end{eqnarray*}
		so we can take $k_n=M+1+k_{n-1}=\beta(\gamma^{(n)}(m-1))+1+k_{n-1}$.
	\end{proof}

	\begin{lemma}
		The sequence $h_{\gamma(m-1)},h_{\gamma(\gamma(m-1))},h_{\gamma^{(3)}(m-1)},\ldots$ corresponds to a non-self-intersecting path
	\end{lemma}
	\begin{proof}
		Suppose $\prod_{i=1}^{l_1}h_{\gamma^{(i)}(m-1)}=\prod_{i=1}^{l_2}h_{\gamma^{(i)}(m-1)}$ with $l_2>l_1$.  Then, by Lemma~\ref{prefixes},
		\begin{equation*}
			a^{k_{l_1}}h_0\ldots h_{\gamma^{(l_1)}(m-1)}=a^{k_{l_2}}h_0\ldots h_{\gamma^{(l_2)}(m-1)}
		\end{equation*}
		or,
		\begin{equation*}
			e=h_{\gamma^{(l_1)}(m-1)+1}\ldots h_{m-1}a^{k_{l_2}-k_{l_1}-1}h_0\ldots h_{\gamma^{(l_2)}(m-1)}
		\end{equation*}
		Then there are two cases.

		First, if $k_{l_1}=k_{l_2}$, then $\gamma^{(l_1)}(m-1)=\gamma^{(l_2)}(m-1)$ by the minimality of the expression for $a$.  In addition, for all $i$ with $l_1\leq i<l_2$, $\alpha(\gamma^{(i)}(m-1))=\emptyset$ and $\gamma^{(i)}(m-1)\neq m-1$.  Therefore, for all such $i$, \begin{equation*}
			\gamma^{(i+1)}(m-1)=\gamma^{(i)}(m-1)+1\bmod{m}
		\end{equation*}
		Since $\gamma^{(l_1)}(m-1)=\gamma^{(l_2)}(m-1)$, $l_2-l_1\geq m$.  But then for some $l_1\leq j<l_2$, $\gamma^{(j)}(m-1)=m-1$, which is a contradiction.

		Second, if $k_{l_1}>k_{l_2}$, then $(k_{l_2}-k_{l_1}-1,\gamma^{(l_2)}(m-1)+1)\in\alpha(\gamma^{(l_1)}(m-1))$.  Then $k_{l_1+1}\geq k_{l_1}+(k_{l_2}-k_{l-1}-1)+1=k_{l_2}$.  Since $l_2\geq l_1+1$ and the sequence of $k$'s is non-decreasing, $l_2=l_1+1$.  We can then calculate $\gamma^{(l_1+1)}(m-1)$ directly.  Note that
		\begin{eqnarray*}
			(\beta(\gamma^{(l_1)}(m-1)),\gamma^{(l_2)}(m-1)+1)&=&(k_{l_1+1}-k_{l_1}-1,\gamma^{(l_2)}(m-1)+1)\\
			&=&(k_{l_2}-k_{l_1}-1,\gamma^{(l_2)}(m-1)+1)\\
			&\in&\alpha(\gamma^{(l_1)}(m-1))
		\end{eqnarray*}
		So
		\begin{equation*}
			\gamma^{(l_1+1)}(m-1)=\gamma^{(l_2)}(m-1)+1=\gamma^{(l_1+1)}(m-1)+1
		\end{equation*}
		which is again a contradiction.
	\end{proof}

	Since $\gamma$ is a function from a finite set to itself, the sequence,
	\begin{equation*}
		\gamma(m-1),\gamma(\gamma(m-1)),\gamma^{(3)}(m-1)\ldots
	\end{equation*}
	is eventually periodic.  Let $\gamma^{(j_1)}(m-1),\ldots,\gamma^{(j_2)}(m-1)$ be a single period and take $b=\prod_{i=j_1}^{j_2}h_{\gamma^{(i)}(m-1)}$.  This expression of $b$ as a product of generators clearly fits the theorem.
\end{proof}

\begin{corollary}
	Inescapable groups are Burnside groups
\end{corollary}
\begin{proof}
	Suppose $(G,S)$ is a tape graph and $G$ is not a Burnside group.  Then $G$ has an element of infinite order.  By Theorem~\ref{thmInescBurn}, we can find some $i_0,\ldots,i_{k-1}$ such that
	\begin{equation*}
		\prod_{j=0}^N g_{i_{j\bmod{k}}}
	\end{equation*}
	is distinct for each $N$.  The sequence $\{g_{i_{j\bmod{k}}}\}_{j\in\mathbb{N}}$ is computable (in fact, it's regular), so $G$ is not inescapable.
\end{proof}

\subsection{Further Remarks}
	As was previously mentioned, generalizing to multiple tapes and multiple heads in this context is done in the same way as in the context of standard Turing machines.  For example, a one-tape machine with multiple heads can be simulated by a one-tape machine with one head.  The locations of all the heads can simply be marked on the tape with special symbols and for each step of the simulated machine, the simulating machine can search through the entire tape for all the head symbols and then update the tape accordingly.  Similarly, a machine with multiple, identical tapes can be simulated by a machine with a single tape of the same type.  Simply enlarge the tape alphabet to tuples of tape symbols together with special symbols for the head on each tape and follow the procedure for multiple heads.

	The only major difference is in a machine with multiple, different tapes.  In this case, the class of functions computable by the machine is the class of functions in or below the join of the r.e. degrees of the word problems of the tapes.  Just as in Section~\ref{secTDegrees}, this machine is mutually simulatable with a standard Turing machine with oracles for the word problem of each tape.  In fact, this case subsumes the multiple-head, single-tape and multiple-head, multiple-tape cases.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
Q: Accessing array of from another view controller class I am new in iPhone development and Objective-C. I have RootViewController and DetailViewController. I don't wanna go back to table view and select other row to show data. I want to check next data directly from the DetailView. But i dont know how to access that array which is defined in RootViewController. Can anyone please help me with that? Please give some example code because i am very new to Objective-C and iPhone development.
Thanx in advance.

A: There are a couple ways you might pursue this. You could pass the array to the detail view controller. Alternately, and more my recommendation is that you can have your next button handler send a message back to the root controller, to have it run the detailView showing routine on the next item instead of restoring the rootview.
Assuming you have a situation where the delegate of the detailViewController is the rootViewController, you'd have a method in the root viewController like:
- (void)detailViewControllerDidAskForNext:(detailViewController *)controller {
     [self showNextDetail]; //you'd supply the showNextDetail method
}

And a method in the detailController like this, hooked to your next button.
- (IBAction)next {
    [self.delegate detailViewControllerDidAskForNext:self];
}

Sorry if these are too vague.

redpajama_set_name: RedPajamaStackExchange


Planned Business Growth We plan to increase our citrus packing facility and continue our very busy squash seasonal work.
What is your hourly rate for seasonal staff?
Four Seasons Packhouse was developed by growers to provide a service that involves the harvesting and packing of buttercup squash ready for export.
We also assist growers in the packing, marketing and distribution of Fejioa's and citrus to local and export markets.


text:
Q: Double vertical spacing : how to fix this? Consider the following :
\documentclass[12pt]{report}
\usepackage{amsmath, amssymb, amsthm}
\usepackage{lipsum}

\usepackage{tikz}
\usetikzlibrary{shadows}
\newcommand*\squared[1]{\tikz[baseline=(char.base)]{\node[shape=rectangle, draw, inner color = white, drop shadow = {opaque, black}, inner sep=3pt, text justified] (char) {#1};}}

\newtheoremstyle{exercise}
  {\topsep}
  {\topsep}
  {\hangindent=2em}
  {}
  {}
  {}
  {.8em}
  {\squared{\thmnumber{#2}}}

\theoremstyle{exercise}
\newtheorem{exercise}{}

\renewcommand{\arraystretch}{1.75}
\usepackage{tabularx}

\begin{document}

\begin{exercise}
    \begin{tabularx}{\textwidth}[t]{r @{\hskip .4em} X}
        a) & This is a test. \\
        b) & I say
        $$
            \pi \approx 3.14.
        $$ \\
        c) & Do you agree?
    \end{tabularx}
\end{exercise}

\end{document}

It seems like the fact that b) finishes with math mode adds extra vertical spacing before c). But I would like the spacing to be uniform. That is, I want the same vertical space before b) and before c). How to do it ?

A: You have few options.
First one is to add the equation in a row of its own:
\documentclass[12pt]{report}
\usepackage{amsmath, amssymb, amsthm}
\usepackage{lipsum}

\usepackage{tikz}
\usetikzlibrary{shadows}
\newcommand*\squared[1]{\tikz[baseline=(char.base)]{\node[shape=rectangle, draw, inner color = white, drop shadow = {opaque, black}, inner sep=3pt, text justified] (char) {#1};}}

\newtheoremstyle{exercise}
  {\topsep}
  {\topsep}
  {\hangindent=2em}
  {}
  {}
  {}
  {.8em}
  {\squared{\thmnumber{#2}}}

\theoremstyle{exercise}
\newtheorem{exercise}{}

\renewcommand{\arraystretch}{1.75}
\usepackage{tabularx}

\begin{document}

\begin{exercise}
    \begin{tabularx}{\textwidth}[t]{r @{\hskip .4em} X}
        a) & This is a test. \\
        b) & I say \\
           &
            \hfil $\pi \approx 3.14$.
            \\
        c) & Do you agree?
    \end{tabularx}
\end{exercise}
\end{document}

However, I feel that a list enumerate would be the better way to go after some modifications to the exercise environment.
Here is some more additions. These are not perfect as the \arraystretch is modified and the widths of \parbox and \minipage are not accurately calculated. Nevertheless, it may serve as a starting point.
\documentclass[12pt]{report}
\usepackage{amsmath, amssymb, amsthm}
\usepackage{lipsum}

\usepackage{tikz}
\usetikzlibrary{shadows}
\newcommand*\squared[1]{\tikz[baseline=(char.base)]{\node[shape=rectangle, draw, inner color = white, drop shadow = {opaque, black}, inner sep=3pt, text justified] (char) {#1};}}

\newtheoremstyle{exercise}
  {\topsep}
  {\topsep}
  {\hangindent=2em}
  {}
  {}
  {}
  {.8em}
  {\squared{\thmnumber{#2}}}

\theoremstyle{exercise}
\newtheorem{exercise}{}

\renewcommand{\arraystretch}{1.75}
\usepackage{tabularx,enumitem,showframe}

\begin{document}

\begin{exercise}
    \begin{tabularx}{\dimexpr\textwidth-2.8em\relax}[t]{@{\,}r @{\hskip .4em} X}
        a) & This is a test. \\
        b) & I say \parbox[t]{0.88\linewidth}{%
           \[
            π\approx 3.14.
           \]
           }\\
        c) & Do you agree?
    \end{tabularx}
\end{exercise}
\begin{exercise}
    \begin{minipage}[t]{0.928\linewidth}
     \begin{enumerate}[label = \alph*),leftmargin=*,topsep=0pt,partopsep=0pt]
        \item This is a test.
        \item I say
           \[
            π\approx 3.14.
            \]
        \item Do you agree?
    \end{enumerate}
    \end{minipage}%
\end{exercise}
\end{document}


text:
A 19-year-old Vancouver resident has been charged in a human trafficking investigation. (Black Press Media file photo)
Vancouver man, 19, charged in human trafficking case involving teen girl
The 16-year-old girl was reported missing and later discovered in Vancouver
A 19-year-old Vancouver man has been charged in a human trafficking investigation involving a 16-year-old girl.
What began as a missing person's investigation on Jan. 6 evolved after police learned the girl was allegedly being held against her will in Vancouver.
Police from Surrey and Vancouver later found her, without the suspect there, but RCMP did not provide further details in a release on Thursday.
Raimon Geday has been charged with 14 offences, including recruiting, transporting, harbouring and exploiting a minor; production and possession of child pornography; knowingly advertising sexual services; and possession of a gun.
His next court appearance is on April 29 in Vancouver.
VIDEO: Are you concerned about Facebook leaking personal information?
Housing for Penticton's homeless to be ready September 1


The discovery of Bluetooth-enabled credit card skimming devices at a Madison County gas station has raised the alarm on a growing crime trend.
Law enforcement said thieves are increasingly using technology at gas pumps to steal credit or debit card information. The thieves place the devices - known as skimmers - behind gas pump card readers. The fake reader stores your card number and, if a phony keypad is placed over the real one, can also collect your PIN.
If the skimmer is Bluetooth enabled, the thief can download the stolen information remotely, avoiding the risk of collecting skimming devices later.
According to the National Association of Convenience Stores, a single compromised pump can capture data from 30 to 100 cards per day.
There are ways to protect yourself from skimmers, both at gas pumps and ATMs. Here are suggestions from security company LifeLock.
If you can pay for your gas inside a station or use an ATM inside a bank lobby, take the extra minute to go inside for the transaction. When you pay at a point-of-sale terminal that sits right next to a clerk, you're much less likely to encounter a skimmer.
Do business at places that make it harder for thieves to put skimmers on pumps or ATMs. Thieves look for undetected, uninterrupted access to point-of-sale terminals. Make sure the machine is in a brightly lit area where lots of people walk past it often. It's also best to choose the pump closest to the cashier.
Use a credit card--preferably with a chip--if you have one. Using a credit card with chip technology will make it harder for thieves to skim your data in the first place. But if a thief still gets your credit card info, he can use only that existing account to make unauthorized charges.
Check your financial statements to catch fraudulent charges--and do it regularly. If you see fraudulent charges on your statement, report it to the card company immediately and shut down the account.


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West Midlands Police to be given short term funding to fight violent crime
The government has confirmed West Midlands Police will be given a one-off payment of £6 million this year to boost police resources.
The Home Secretary announced in March that he was going to parachute in extra money to the forces which have been hardest hit by the national rise in violent crime.
As the extra money is a one-off payment it will largely be spent paying current officers to do more overtime, rather than recruiting new PCs.
The additional funding will be used by our force to ensure extra officers are available to respond to crimes and be on patrol in communities.
Knife crime has risen in the West Midlands area by 85% since 2012. A similar pattern has been seen in other cities across the country.
The Home Office has also promised extra funding for projects which prevent crimes, but the Commissioner is yet to be told whether the West Midlands has been successful with its bid.
To put the additional £6 million funding into context West Midlands Police has a total budget of just under £570 million in 2019/20.
The West Midlands Police and Crime Commissioner, David Jamieson, said: "Since 2010 our force has been hit by funding cuts of £175 million, leading to the loss of some 2,000 officers, despite the huge efficiencies we have driven.
"I have been saying over the last few years that unless we are given adequate resources to tackle the crimes being committed in our region lives will be lost.
"That has sadly proved to be true.
"Whilst this does not make up for the funding we have lost, I do welcome even a small reversal in the cuts we've endured. This money will be directed straight to the front line, ensuring more officers are available to respond to emergency calls and there is a greater police presence on our streets.
"I hope the government has now realised that the thin blue line has become too thin and more officers are needed in the long term to keep us all safe.
"The government now needs to come forward with a long term funding solution for West Midlands Police and deliver extra funding to support young people and prevent violent crime.
The Deputy Chief Constable of West Midlands Police, Louisa Rolfe said: "The additional funding is a welcome boost which will support our commitment to deploy more officers to tackle knife crime in our region, working with communities to understand what motivates young people to carry knives and how we can protect them.
"Boosted by efforts to improve efficiency and get more from our budget we have also recently announced our plans to recruit a further 200 officers over the next two years to improve our ability to police priority areas and investigate crimes that most concern our communities in the West Midlands."
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Who is Dr. Bryn Cooper?
Dr. Bryn Vaughan Cooper, a board-certified orthodontist, works for more than just straighter teeth – she wants patients to love their smile.
Dr. Bryn Vaughan Cooper has been a Bear, a Longhorn, and a Billiken on her long path to become a board-certified orthodontist. She earned Bachelor's and Master's degrees in Biology from Baylor University in 2005 and 2007, respectfully. While pursuing her Master's degree at Baylor, she researched water quality controls on the Central Texas Watershed. That research led to presentations to the national and Texas branches of the American Society of Microbiology, and the Texas Academy of Science.
After her stint as a Baylor Bear, she became a Texas Longhorn by attending the University of Texas Health Science Center San Antonio completing her Doctorate of Dental Surgery in 2011. During dental school, she authored or co-authored nine article reviews.
Saint Louis University's Center for Advanced Dental Education (CADE) – where Dr. Cooper spent her residency learning the ins-and-outs of creating beautiful smiles.
Finally, Dr. Cooper began the last leg of her orthodontic journey by joining Saint Louis University's orthodontic residency program, where she became a Billiken. Upon completing the program, she earned an orthodontic certificate and a Masters of Science. During residency, she presented to the American Association of Orthodontists and collected and analyzed genetic data as part of an orthodontic thesis project. The resulting thesis, titled "Genes Influencing Mandibular Size," was some part of the genetic expression research in orthodontia and led to a pending patent submission.
Dr. Cooper proved her qualifications to a board of her peers in both written and oral form to earn her American Board of Orthodontics Certification. She serves on the College of Diplomates American Board of Orthodontics. And since moving to Houston, she has gotten involved with the local dental and orthodontic communities, including joining the Southwestern Society of Orthodontists, Greater Houston Dental Society, and the Texas Dental Association. She looks forward to serving patients and networking with doctors throughout the Houston area.
Dr. Cooper lives in the Greater Heights Area of Houston with her husband, daughter and golden retrievers. When she isn't walking the trails with her family and friends, you can find her sampling the new restaurants' fares, reading a best-seller, and working on her business. After growing up in the DFW area, she is excited to be building deep roots in the city of bayous.


In the past, a foreign bank's use of correspondent bank accounts in the United States to facilitate wire transfers has not necessarily given New York courts a sufficient basis for jurisdiction over the bank. But a recent 4-3 Court of Appeals decision may change that. In Rushaid et al. v. Pictet & Cie et al., No. 180, 2016 BL 387923 (N.Y. Nov. 22, 2016), Judge Jenny Rivera, writing for the four-person majority (and overturning decisions of both the First Department and the Commercial Division), ruled that a foreign bank's "repeated, deliberate" use of correspondent bank accounts in the United States is enough to establish New York jurisdiction.
To continue reading Stephen Younger and Sarah Ferguson's article from Law360 please click here.


THE scheme to create a national stock exchange through the 'integration' of the three existing exchanges might prove detrimental to the development of the capital market and for interests of up-country investors.
It will also establish the complete monopoly of a single bourse over all key areas of the market, such as depository, clearing and settlement, and commodity services.
The Security and Exchange Commission of Pakistan (SECP) is executing the project in spite of strong opposition by a committee comprising the chairmen of the three bourses that the apex regulator had constituted earlier this year.
Some Lahore Stock Exchange brokers claim that the regulator had actually twisted their arms to compel them to agree to the proposal for the creation of the Pakistan Stock Exchange (PSE). For all practical purposes, it will be the 'renaming' of the Karachi Stock Exchange (KSE) as the two other bourses in Lahore and Islamabad will be de-licensed, they add.
But the regulator insists that the creation of a single national stock exchange is crucial for an efficient and transparent capital market and for attracting a foreign strategic partner for investment in latest technology and products.
The plan, an SECP official told Dawn, is being implemented to complete the demutualisation of the exchanges initiated under the Stock Exchanges (Corporatisation, Demutualisation and Integration) Act 2012.
The law, which segregated the majority ownership of each bourse from their right to trade on it, bound the three exchanges to sell 40pc of their shares to foreign strategic investors by this August, which they failed to comply with because of the overall investment conditions in the country.
The committee of the chairmen had recommended extending the disinvestment deadline in view of the overall investment climate conditions.
Meanwhile, SECP officials privately say the integration of the bourses into one exchange makes sense as over 95pc of the trading is commanded by the KSE and the Lahore and Islamabad brokers have for long been trading through their peers in Karachi.
Former SECP chairman Khalid Mirza is opposed to the creation of a national bourse, saying the regulator's scheme will "eliminate competition for the KSE and help the big Karachi brokers establish their monopoly over the capital market".
"Competition is important because demutualisation makes stock exchanges 'for profit' entities. It is also important because the quality of the market in one region [for example Punjab] is quite different from another [like Karachi]," he argues.
He says countries like Malaysia and Singapore with single stock exchanges are regionally connected and compete with each other for investors and company listings.
Mirza, who currently teaches at LUMS, believes that the merger of the LSE and the ISE would have made sense, as recommended by the SECP committee.
It is pertinent to note that the demutualisation law provides for a framework for the SECP to facilitate the integration of the stock exchanges that are inclined to merge, but this is not 'mandated' by the law; nor does the law specifically encourage stock exchange mergers.
The commission is only obliged to regulate merger proposals in accordance with the law and not 'manage' them.
Asim Zafar, a leading Lahore broker, also doesn't agree with the concept of a single bourse. "Investors from Punjab have a very important role to play in the capital market. The Lahore bourse could have taken care of the needs of the off-market (trade), bonds market and small companies, as well as prevented the establishment of Karachi's monopoly.
"All exchanges should have been given an opportunity to target potential investors based on their own realities rather than forced to integrate," he says.
He agrees that the merger of the Lahore and Islamabad bourses would have created a competitive environment for the KSE and contributed to the growth of the capital market in the country.
But he does not favour the heavy regulatory regime, including capital requirements for TREC holders, that the apex regulator is introducing to reduce the number of brokers, and believes that they are detrimental to the market.