\section{Introduction}
	Given a metric space $(M,d)$, we consider  a finite subset $X$ of $M$.  We denote by  $l_{\operatorname{opt}}(X)$  the minimal length of a path which visits all points of $X$.
	Now assume that  $T$ 
	is a total order on $M$.
	For a finite subset $X \subset M$ we consider the restriction of the order
	$T$ on $X$, and enumerate 
	the points of $X$ accordingly:
	$$
	x_1 \leq_T x_2 \leq_T x_3 \leq_T \dots \leq_T x_{k}
	$$
	where $k = \#X$. Here and in the sequel  $\#X$ denotes the cardinality of the
	set $X$.
	We denote by $l_T(X)$ the length of the corresponding path
	$$
	l_T(X) := d(x_1, x_2) + d(x_2, x_3) + \dots + d(x_{k-1}, x_k).
	$$ 
	
	
	\begin{definition}
		Given an ordered metric space $\operatorname{M}=(M, d,T)$ containing at least two points
		and $k\ge 1$, we define the {\it order ratio function}
		$$
		\operatorname{OR}_{M,T}(k) := \sup_{X\subset M | 2 \le \# X \le k+1} \frac{l_T(X)}{l_{\operatorname{opt}}(X)}.
		$$
	\end{definition}
	
	If $M$ consists of a single point, then the supremum in the definition above is taken over an empty set. 
	We use in this case the convention that $\operatorname{OR}_{M,T}(k)=1$ for all $k\ge 1$.
	
	
	Given an (unordered) metric space $(M,d)$, we also  define the {\it order ratio function} as 
	$$
	\operatorname{OR}_{M}(k) := \inf_{T} \operatorname{OR}_{M,T}(k).
	$$
	
	Given an algorithm to find an approximate solution of an optimization problem, the worst case of the ratio between the value provided by this algorithm
	and the optimal value is called the {\it competitive
		ratio} in computer science literature.
	We will recall below the setting of the universal travelling salesman  problem. 
	In this setting
	the function $\operatorname{OR}(k)$ corresponds to the competitive ratio for $k$-point subset, in the situation when we do not require for a path to return to the starting point.
	
	It can be shown that even in a case of a finite metric space, where the infimum in the definition above is clearly attained, 
	the order that attains this minimum
	might depend on $k$ (see Remark \ref{rem:numberofpoints}).
	
	We say that a metric space $M$ is {\it uniformly discrete}, if there exists $\delta>0$ such that for all pairs of points $x \ne y$ the
	distance between $x$ and $y$ is at least $\delta$. 
	It is not difficult  to see that the asymptotic class of the order ratio function behaves well with respect to quasi-isometries of
	uniformly discrete spaces, see Lemma \ref{le:qi}.
	
	This is no longer true without the assumption of uniform discreteness:  $M_1 = \mathbb{R}$ and 
	$M_2=\mathbb{R}\times [0;1]$ are in the same quasi-isometry class, but $M_1$ has a bounded order ratio function and the order ratio function of $M_2$ is unbounded; 
	indeed, the latter space contains a square. It is shown in \cite{HKL}
	that given  any order on the $n\times n$ square,  there exist a finite subset of this lattice, such that the length of the path, associated to this order, is at least  ${\rm Const} \sqrt[6]{\frac{\log n}{\log \log n}} $ multiplied  by the length of the optimal path. This implies
	an analogous bound in terms of number of points
	of the subset $X$ rather than the size of the square:
	$\operatorname{OR}_{[0,1] \times [0,1]}(k) \ge  {\rm Const}\sqrt[6]{\frac{\log k}{\log \log k}}$.
	
	To avoid non-stability with respect to quasi-isometries, given a metric space $M$,  we can consider  order ratio
	functions for $(\varepsilon, \delta)$-nets of $M$ (see Definition \ref{def:net}).
	It is not difficult to see (see Lemma \ref{le:pullback}) that the function $\operatorname{OR}_{M'}$ is well-defined, up to a multiplicative constant,
	for $M'$ being an $(\varepsilon, \delta)$-net of $M$. 
	This allows us to speak about the {\it order
		ratio invariant} of a metric space, see Definition \ref{def:ORI} in Section \ref{sec:firstexamples}.
	
	
	In contrast with previous works on  competitive ratio of universal travelling salesman  problem (\cites{bartholdiplatzman82, bertsimasgrigni, jiadoubling, HKL, Christodoulou, Schalekamp, bhalgatetal, gorodezkyetal, eades2}),
	we are interested not only in this asymptotic behaviour but also in particular values
	of $\operatorname{OR}(k)$. In the definition below we introduce {\it order breakpoint} of an order :
	
	\begin{definition} \label{def:mins}
		Let $M$ be a metric space, containing at least two points,
		and let $T$ be an order on $M$.  
		We say that the order  
		breakpoint 
		$\operatorname{Br}(M,T) = s$ if $s$ is the smallest integer such that $\operatorname{OR}_{M,T}(s) < s$. 
		If such $s$ does not exist, we say that $\operatorname{Br}(M,T)=\infty$.
		A possible convention for a one-point space $M$ is to define $\operatorname{Br}(M,T)=1$.
	\end{definition}
	
	Given an (unordered)
	metric space $M$, we define $\operatorname{Br}(M)$
	as the minimum over all orders $T$ on $M$:
	$$
	\operatorname{Br}(M) = \min_{T}
	\operatorname{Br}(M,T).
	$$
	
	Given an order on $M$, the order breakpoint describes the minimal value $k$ for which  using this order as an universal order for the travelling salesman problem 
	has some efficiency on $(k+1)$-point subsets.
	
	
	From definition, it is clear that if $M$ has at least two points then $\operatorname{Br}(M,T) \ge 2$  and it is easy to see that $\operatorname{Br}(M,T)\leq 2$
	if $M$ is finite.  In Section \ref{sec:examples}, Corollary \ref{cor:finitegraphs} we evaluate the order breakpoint of finite graphs (containing edges).
	In what concerns infinite graphs and metric spaces, it can be shown (see Lemma \ref{le:notrayline}) that a metric space quasi-isometric to a geodesic one has $\operatorname{Br} \ge 3$ unless it is bounded, quasi-isometric to a ray or to a line;  metric on vertices of an	infinite graph has order breakpoint equal to $2$ if and only if $M$ is quasi-isometric to a ray or a line. 
	
	We characterize finitely generated groups with small ($\le 3$) order breakpoint
	(in other words, groups that admit an order that has some efficiency on $4$-point subsets):
	
	\begin{namedthm*}{Theorem A} (= Thm \ref{thm:girth4})
		Let $G$ be a finitely generated group. 
		Then $G$ admits an order $T$ with  $\operatorname{Br}(G,T) \le 3$
		if and only if $G$ is virtually free.
	\end{namedthm*}
	
	Using Stallings theorem 
	and Dunwoody's result about accessibility of finitely presented groups, we reduce the proof of the Theorem to  Lemma \ref{le:oneended} that provides a lower bound for $\operatorname{Br}$
	for one-ended groups and to Lemma \ref{le:infprcycle} about
	mappings of 
	trimino graphs to infinitely presented groups. 
	
	It seems interesting to better understand  groups and spaces with given order breakpoint, in particular with $\operatorname{Br} =4$.
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.45]{pictures/capture2.png}
		\caption{Bartholdi and Platzman used the order of the first appearance along the space-filling Sierpi\'{n}ski curve.}
		\label{pic:bartholdiplatzman}
	\end{figure}
	
	\bigskip
	
	Now we discuss the asymptotics of order ratio function.	
	The travelling salesman problem is a problem to find a cycle of minimal total length that visits each of $k$ given points.
	Bartholdi and Platzman introduced the idea to order all points of a metric space and then, given a $k$-point subset,  visit its points in the corresponding order
	\cite{bartholdiplatzman82}, \cite{bartholdiplatzman89}.
	Such approach is called {\it universal travelling salesman problem}.
	(One of motivations of  Bartholdi and Platzman was that this approach works
	fast for subsets  of a two-dimensional plane). 
	Their  argument implies a logarithmic upper bound for the function $\operatorname{OR}_{\mathbb{R}^2}(k)$. 
	
	For some metric spaces the function $\operatorname{OR}$ is even better. 
	Namely, it is not difficult to show  that $\operatorname{OR}(k) \leq 2$, for all $k\ge 1$ in case of metric trees, see Thm 2 \cite{Schalekamp}
	(an approriate order is shown on Picture \ref{pic:orderedtree}, see also Lemma \ref{lem:hierarchicaltrees}). 
	
	\bigskip
	
	We prove that this best possible situation ($\operatorname{OR}(k)$ is bounded by above by a constant) holds true for uniformly discrete
	hyperbolic
	spaces.
	
	\begin{namedthm*}{Theorem B} (= Thm \ref{thm:hyperbolicspace}). Let $M$  be a $\delta$-hyperbolic graph of
		bounded degree. 
		Then there exists  an order $T$ and
		a constant $C$
		such that for all $k\ge 1$
		$$
		\operatorname{OR}_{M,T}(k) \le C.
		$$
	\end{namedthm*}
	
	It is essential that constant $C$ depends  on  $M$. 
	In contrast with trees, this constant does not need to be bounded by $2$ and it can be arbitrarily large.
	In particular, for subsets of $\mathbb H^d$ this constant
	depends on the dimension $d$ and the constants of uniform discreteness; 
	if we fix $\varepsilon$ and consider $(\varepsilon, \varepsilon)$-nets, then this constant tends to infinity as dimension $d$ of $\mathbb{H}^d$ tends to infinity. 
	In contrast with trees where the statement is true for any metric tree, it is essential in general to assume that the degree of the graph
	$M$ is bounded.
	
	
	It is not difficult to see that  some not hyperbolic metric spaces also have bounded order ratio function; a particular family that arises from gluing together trees or hyperbolic spaces is studied in
	Subsection \ref{Subsection:nothyperbolic}.
	So far we do not know whether a finitely generated not hyperbolic group can have bounded $\operatorname{OR}(k)$; for more related questions see the discussion at the end
	of the latter section.
	
	
	
	
	
	We continue our study of the order ratio function and the order breakpoint in our subsequent paper \cite{ErschlerMitrofanov2}, where we show  their relation with Assouad-Nagata dimension.
	
	
	\subsection{Plan of the paper}
	
	In Section $2$ we discuss basic properties of the order ratio  function and the order breakpoint, their relation to quasi-isometries,  uniform mappings and the notion of "snakes".
	
	In Section 3 we start 
	with elementary examples of metric trees. We describe the asymptotic of the order ratio
	function for compositions of wedged sums of metric spaces.
	As a particular case we characterise  the values of $\operatorname{Br}$ of finite graphs.
	
	In Section 4 we study metric spaces and groups with small order breakpoint. 
	The already mentioned Lemma \ref{le:notrayline} studies spaces with $\operatorname{Br} \le 2$. Its proof that uses a result of M.Kapovich about what is called $(R,\varepsilon)$ tripods  is given in Appendix $B$.
	The main goal of the section is  to prove  Theorem \ref{thm:girth4} and thus characterise groups with $\operatorname{Br} \le 3$.
	
	The goal of Section 5 is the proof of  Theorem B. Since  by a theorem of Bonk and Schramm \cite{bonkschramm} any $\delta$-hyperbolic space of bounded geometry can be  quasi-isometrically imbedded to $\mathbb H^d$, taking in account  Lemma \ref{le:qi} about quasi-isometric imbeddings,
	the main goal in the proof of Theorem \ref{thm:hyperbolicspace} is to prove it for an $(\varepsilon, \varepsilon)$-net of $\mathbb H^d$. 
	We do it by choosing an appropriate tiling
	of this space and considering a naturally associated tree; there is a family of {\it hierarchical orders} on this tree, and we choose any of these orders. 
	While it is well-known
	that finite subsets of hyperbolic spaces can be well approximated by metric trees, we would like to stress that such approximation does not provide a priori any upper bound for order ratio function.
	We control the total length of piecewise-geodesic paths associated to this order  by bounding the number of tiles that are visited by such paths (here we use  bounds on cones of a path in a hyperbolic space, Lemma \ref{le:cone}), and  bounding the number of visits
	(see Lemma  \ref{lem:multiplicity}). 
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.75]{pictures/tilingwithtree2.png}
		\caption{An order on the tiles of a hyperbolic plane $\mathbb H^2$ 
			corresponding to the first visit of a tile by a green path.
			For $\mathbb H^d$, $d\ge 2$,  an order can be chosen to be 
			hierarchical  on an appropriate tree.
		}
		\label{pic:tilingwithtree2}
	\end{figure}
	
	
	In subsection \ref{Subsection:nothyperbolic} we study (not necessarily hyperbolic) families of metric spaces with bounded $\operatorname{OR}$.
	
	
	
	{\bf Acknowledgements. } 
	We would like to thank 	Chien-Chung Huang and Kevin Schewior
	for references  on travelling salesman  problem,  and Nick Ramsey for conversations on first order logic.
	This project has received funding from the European Research Council (ERC) under
	the European Union's Horizon 2020 
	research and innovation program (grant agreement
	No.725773).
	The work of the second named author is also supported by Russian Science Foundation (grant no.17-11-01377).
	
	\section{Preliminaries and basic properties}\label{sec:firstexamples}
	
	In this section we describe basic properties of the order ratio function and order breakpoint.
	
	\begin{lemma}
		\label{prop:max_min_or}
		For any ordered metric space $(M, d, T)$ and any  $k \ge 1$
		it holds
		$$
		1 \leq \operatorname{OR}_{M,T}(k) \leq k.
		$$
	\end{lemma}
	\begin{proof}
		The first inequality is obvious. 
		Let $X \subset M$, $\# X = k+1$ and let $L$ be the diameter of $X$. 
		Then $l_{T}(X) \leq k L$ and $l_{\operatorname{opt}}(X) \geq L$.   
	\end{proof}
	
	
	\begin{definition} \label{de:qi} Given metric spaces $N$ and $M$, a map $\alpha$  from $N$ to $M$ is a {\it  quasi-isometric imbedding} if
		there exist $C_1$, $C_2$ such that for any $x_1, x_2 \in N$ it holds
		\[
		\frac{1}{C_1}(d_N(x_1, x_2)) -C_2  \le   d_M(\alpha(x_1), \alpha(x_2)) \le C_1(d_N(x_1, x_2)) +C_2
		\]
		
		If $M$ is at bounded distance from $\alpha(N)$, this map is called a quasi-isometry, and the spaces $X$ and $Y$ are called {\it quasi-isometric.}
		If $\alpha$ is bijective and $C_2=0$, then the spaces are said to be  {\it bi-Lipschitz equivalent}.
	\end{definition}
	
	A significantly weaker condition is 
	
	\begin{definition}\label{de:ue}  Given  metric spaces $N$ and $M$, a map $\alpha$  from $N$ to $M$ is an
		{\it uniform imbedding} (also called {\it coarse imbedding})
		if there exist two non-decreasing functions $\rho_1$, $\rho_2:$ $[0, +\infty) \to [0, + \infty)$,
		with $\lim_{r\to \infty} \rho_1(r) = \infty$,  such that
		\[
		\rho_1 (d_X(x_1, x_2)) \leqslant d_M(\alpha(x_1), \alpha(x_2)) \le \rho_2 (d_X(x_1, x_2)).
		\]
	\end{definition}
	
	\begin{definition}\label{def:net}
		We say that for $\varepsilon, \delta > 0$ a subset $U$ in a metric set $M$ is an $(\varepsilon, \delta)${\it -net} if any point of $M$ is at distance at most
		$\varepsilon$ from $U$ and the distance between  any two distinct points of $U$ is at least $\delta$.
	\end{definition}
	
	It is clear that any $\varepsilon$-net of $M$ (in particular, any $(\varepsilon, \delta)$-net of $M$),  endowed with the restriction of the metric of $M$, is quasi-isometric to 
	$M$. 
	Indeed, its imbedding to $M$ satisfies the condition of Definition \ref{de:qi} with $C_1=1$ and $C_2=0$. 
	Observe also  that $M$  is at distance of most $\varepsilon$ from  the image of this imbedding.
	It is also   clear from the definition that any $(\varepsilon, \delta)$-net is uniformly discrete.
	For convenience of the reader, we recall the following well-known observation.
	
	\begin{rem} \label{re:separatednet}
		
		Let $M$ be a metric space. For any  $\varepsilon >0$ there exists an 
		$( \varepsilon, \varepsilon)$-net of $M$.
		
	\end{rem}
	
	\begin{proof}
		Consider a maximal, with respect to inclusion, subset $U$ of $M$ such that the distance between any two points is at least $ \varepsilon$ (in other words $U$ is a maximal with respect to inclusions $\varepsilon$-separated net of $M$).   The maximality of this set guarantees that any point $x \in M$ is at distance at most $\varepsilon$ from $M$.
	\end{proof} 
	
	
	\begin{definition}
		Given (a not necessary injective) map $\phi: N \to M$ and on order $T_M$ on $M$ let us say that an order $T_N$ 
		on $N$ is a pullback of $T_M$ if the following holds: for any $x , x' \in N$ such that  $\phi(x) <_{T_M} \phi(x')$  we have
		$x<_{T_N} x'$. 
	\end{definition}
	
	Note that a pullback of an order always exists: to construct a pullback it is (necessary and) sufficient to fix an order on complete preimage of each $y$ preimage $ \phi^{-1}(y)$.
	
	\begin{lemma}\label{le:pullback}
		Let $\alpha$ be a quasi-isometric imbedding of an uniformly discrete space $N$ to uniformly discrete space $M$, then there is a constant $K$ such that if \: $T$ is an order on $Y$ and $T_X$ is its pullback on $X$, then for all $k \ge 1$ 
		\[
		\operatorname{OR}_{N,T_N} (k) \le K \cdot \operatorname{OR}_{M,T} (k).
		\]
		The constant $K$
		can be chosen depending on only on constants $C_1$, $C_2$ in the definition of quasi-isometries and {\it a constant of uniformly discreteness $\delta$} (that is,
		on $\delta$ such that $\inf_{x\ne y} d_M(x,y) \ge \delta$).
	\end{lemma}
	
	\begin{proof}
		There exists $\delta > 0$ such that all distances between distinct points in $N$ and $M$ are greater than $\delta$. 
		Since $\alpha$ is a quasi-isometry  and  since $N$ and $M$ are  uniformly  	discrete, we know  that there exists $C$ such that
		\[\frac{1}{C} \rho_{M}(\alpha(x_1), \alpha(x_2)) \le  \rho_{N}(x_1, x_2) \le C \rho_{M}(\alpha(x_1), \alpha(x_2))
		\]
		if $\alpha(x_1) \neq \alpha(x_2)$, and $\rho_X(x_1, x_2) < C$ if $\alpha(x_1) = \alpha(x_2)$.
		
		Let $A \subset N$, $\# A = k$, $B = \alpha(A)$, $\# B = n$, $n \le k$.
		Let $L_1= l_{\operatorname{opt}}(B)$, $L_2=l_{T}(B)$.
		We know that $L_2 \leqslant L_1 \cdot \operatorname{OR}_{M,T}(n) \leqslant L_1 \cdot OR_{M,T}(k)$, because the $\operatorname{OR}$ function is non-decreasing.
		If some path goes through all points  of $A$, the corresponding path in $M$ goes through all points of $B$. 
		Hence $l_{\operatorname{opt}}(A) \ge \frac{1}{C}L_1$.
		The path of length $l_{T_N}(A)$ consists of $n-1$ preimages of edges of $l_T(B)$ and $k-n$ edges with one point image.
		The first group gives us $\le C L_2$, and the second group of edges gives $\leq C(k-n) \leq C (k - 1)$.
		We also know that $k < \frac{l_{\operatorname{opt}}(A)}{\delta}$.
		Putting these inequalities together, we obtain
		\[l_{T_N}(A) \le C^2 \cdot \operatorname{OR}_{M,T}(n) l_{\operatorname{opt}}(A) + \frac{C}{\delta} l_{\operatorname{opt}}(A).
		\]
		Therefore
		\[
		\operatorname{OR}_{X,T_N}(k) \leq C^2 \cdot \operatorname{OR}_{M,T}(k) + \frac{C}{\delta},
		\]
		and we can take $K(\alpha) = C^2 + \frac{C}{\delta}$.
	\end{proof}
	
	The Lemma \ref{le:pullback} above  allows  to speak about order ratio invariant of a metric space.
	
	In the definition below we say that two functions $f_1,f_2(r) :\mathbb{N} \to \mathbb{R}$ are equivalent up to a multiplicative constant if  there exists $K_1,K_2>0$ such that $f_1(r) \le K_1 f_2(r)$ and $f_2(r) \le K_1 f_1(r)$  for all $r\ge 1$.
	
	\begin{definition}\label{def:ORI}
		For a metric space $(M, d)$ its {\it order ratio invariant} $\operatorname{ORI}_{M}$ 
		is defined to be the equivalence class, with respect to equivalence up to a multiplicative constant, 
		of functions  $\operatorname{OR}_{M'}$, where $M'$ is some $(\varepsilon, \delta)$-net of $M$.
	\end{definition}
	
	In view of Lemma \ref{le:pullback} we obtain the following 
	
	\begin{lemma} \label{le:qi}
		For any metric space $M$, $\operatorname{ORI}_M$ is well-defined, and it is a quasi-isometric invariant of a metric space.	
	\end{lemma}
	
	\begin{proof}
		If $M_1$ and $M_2$ are $(\varepsilon, \delta)$-nets of $M$, they are quasi-isometric and from Lemma \ref{le:pullback} it follows that $\operatorname{OR}_{M_1} = \Theta(\operatorname{OR}_{M_2})$.
		If two spaces are quasi-isometric then there $(\varepsilon, \delta)$-nets are also quasi-isometric.
		
	\end{proof}
	
	While specific values of the order ratio function can change under quasi-isometries, as can be easily seen already  from finite space examples, 
	the equality  $\operatorname{OR}(s) = s$ for given $s$ (the maximal possible value, see Lemma \ref{prop:max_min_or}) is preserved under quasi-isometries, as we will see in Lemma \ref{le:snakesqi}.
	
	Consider a sequence of  $s$  points: $x_1 <_T x_2 <_T \dots <_T x_s$.
	
	\begin{definition} \label{def:snake}
		We call such sequence to be  {\it a snake} on  $s$ points, of diameter $a$ and of  {\it width} $b$,
		if the diameter of this set is $a$ and the maximal  distance $d(x_i, x_j)$ where $i$ and $j$ are of the same parity, is $b$.
		For a snake on at least two points, 
		we say that the ratio $a/b$ is the {\it elongation} of the snake.
	\end{definition}
	
	In the definition above, if $s=2$, the then the width of the snake is $0$, and we can say in this case that the elongation is infinite.
	
	\begin{lemma}\label{lem:snake} Let $M$ be a metric space, containing at least two points, 
		and let $T$ be an order on $M$ and consider
		$s\geq 1$. 
		The following conditions are equivalent:
		\begin{enumerate}
			\item $\operatorname{OR}_{M,T}(s) = s$
			\item There exists a snake on  $s+1$ points in $(M,T)$ of arbitrary large elongation.
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		$(2)\to (1)$. Let $S = \{x_1, \dots, x_{s+1}\}$ be a snake of diameter  $a$ and width $b$.  Observe that the distance between any consequent points $d(x_i, x_{i+1})$ is greater or equal to  $a - 2b$.  This shows that  $l_T(S) \geqslant s(a - 2b)$. 
		Observe also that $l_{\operatorname{opt}} (S) \leqslant a + (s-1)b$, because we can pass first through
		all points with odd indexes and then all points with even indexes.
		Hence
		\[\operatorname{OR}_{M,T} \geqslant \frac{s(a-2b)}{a + (s-1)b}
		\]
		that tends to $s$ as $b/a$ tends to $0$.
		
		\medskip
		\noindent$(1)\to (2)$. 
		We can assume that $s \geq 2$.
		Let  $S = \{x_1, \dots, x_{s+1}\}$ and
		$l_{T}(S) \ge s(1 - \varepsilon) l_{\operatorname{opt}} (S)$.
		
		Since $l_T(S) \leqslant s\operatorname{diam}(S)$ and 
		$l_{\operatorname{opt}} \geqslant \operatorname{diam}(S)$, we conclude that 
		\begin{itemize}
			\item $s\operatorname{diam}(S)(1-\varepsilon) \leqslant l_T(S)\leqslant s\operatorname{diam}(S)$ and
			\item $\operatorname{diam}(S) \leqslant l_{\operatorname{opt}}(S) \leqslant \operatorname{diam}(S)/(1-\varepsilon)$.
		\end{itemize} 
		The first inequality implies that for all $i$, 
		$d(x_i, x_{i+1})$ is close to the diameter of $S$: $d(x_i, x_{i+1}) \ge \operatorname{diam}(S)(1-s\varepsilon)$.
		
		The second inequality implies that for any triple of points of  $S$, the minimal sum of two pairwise distances  is at most  $\operatorname{diam}(S)/(1-\varepsilon)$.
		If $\varepsilon < 1/2$, then 
		$1/(1-\varepsilon) < 1 + 2\varepsilon$. 
		Applying these inequalities for the triple
		$(x_{i},x_{i+1},x_{i+2})$, we observe that
		$d(x_i, x_{i+2}) < (2+s)\varepsilon \operatorname{diam}(S) \leq 2s\varepsilon \operatorname{diam}(S)$.
		This implies  that the width of the snake $S$ is at most 
		$2s^2\varepsilon \operatorname{diam}(S)$,
		and its elongation is greater of equal than  $1/(2s^2\varepsilon)$.
	\end{proof}
	
	Since a snake on $k$ points contains snakes on lesser number of points, it is clear that 
	$\operatorname{OR}_{M,T}(k) = k$ 
	implies  $\operatorname{OR}_{M,T}(a) = a$ for all integer $a < k$.
	
	\begin{lemma}\label{le:snakesqi}[Snakes, Order Breakpoint and quasi-isometries]
		\begin{enumerate}
			\item
			Let $k$ be a number, let $N$ be an uniformly discrete space with $\operatorname{OR}_N(k) = k$ and let $\alpha$ be a quasi-isometric imbedding of $N$ to a metric space $M$. 
			Then $\operatorname{OR}_M(k) = k$.
			\item In particular, if $N$ and $M$ are quasi-isometric uniformly discrete metric spaces of cardinality $>1$, then 
			$\operatorname{Br}(N) = \operatorname{Br}(M)$
		\end{enumerate}
	\end{lemma}
	
	\begin{proof} 
		(1) Observe that if $\operatorname{OR}_N(k)=k$, then by definition of order ratio function we have $\operatorname{OR}_{N,T}(k)=k$ for any order $T$ on $N$.
		If $\operatorname{OR}_M(k) \ne k$, then there exist an order $T_M$ such that $\operatorname{OR}_{M,T_M}(k) < k$. Let $T$ be a pullback of $T_M$.
		
		Since $\operatorname{OR}_{N,T}(k)=k$, we know that  $(N,T)$ admits a sequence of snakes on $k+1$ points with elongations that tend to infinity. Because of the uniformly discreteness, the widths of snakes are uniformly separated from zero, so their diameters tend to infinity.  Observe that then the diameter of the image of these snakes  tends to infinity. 
		This implies that  the 
		elongation of the  image of the snakes in $(M,T_M)$ tends to infinity, and this is in contradiction with the fact that   $\operatorname{OR}_{M,T_M}(k) < k$.
		
		(2) Follows from (1).
	\end{proof}
	
	
	Now we introduce a strengthening of a notion of a snake. In contrast with the property of admitting
	snakes of large elongation, this property will be inherited by the images 
	of uniform imbeddings.
	\begin{definition} \label{def:strongsnakes}
		We say that an ordered metric space $(M,T)$ admits {\it a sequence of  snakes of bounded width} on $s$ points, if it admits a sequence of
		snakes  on $s$ points of uniformly bounded width and with diameters tending to infinity.
	\end{definition}
	
	This notion allows us to obtain estimates for $\operatorname{Br}$ in case when one metric space is uniformly imbedded in another one.
	
	\begin{lemma} \label{lem:uniform}
		Let $N$ and $M$ be metric spaces.
		Let $\alpha$ be an uniform imbedding of $N$ to $M$, $T_M$ be an order on $M$, and $T_N$ be its pullback on $X$. 
		Then if for some $k\ge 1$ the ordered space $(N,T_N)$ admits a sequence of  snakes of bounded width on $k+1$ points, then $(M,T_M)$ also
		admits such sequence of snakes and
		$\operatorname{OR}_{M,T_M}(k)= k$.
	\end{lemma}
	
	\begin{proof} Consider  a sequence of snakes in $N$, of bounded width with respect to  $T_N$.
		Observe that its image under an uniform imbedding is a sequence of  snakes in $M$, of bounded width with respect to $ T_M$. 
		We can conclude therefore that $M$ admits snakes on $k+1$ points of arbitrary large elongation, and hence by Lemma \ref{lem:snake} we know that 
		$\operatorname{OR}_{M,T_M}(k)= k$.
	\end{proof}
	
	The following lemma implies that the order ratio function of a metric space  is defined by  order ratio functions of all its finite subsets.
	
	\begin{lemma}\label{rem:goedel}
		Let $M$ be a metric space. Consider a  function $F:\mathbb{Z}_+ \to \mathbb{R}_+$ and assume that for any finite subset $M' \subset M$ there exists an order $T'$ satisfying
		$\operatorname{OR}_{M', T'} (k) \le F(k)$ for all $k\ge 1$.
		Then there exists an order $T$ on $M$ satisfying 
		$\operatorname{OR}_{M, T} (k) \le F(k)$ for all $k\ge 1$.
	\end{lemma} 
	
	\begin{proof}
		
		Indeed, we can take one variable for each point of $M$.
		Observe that for binary predicate ``$<$'', the fact that it defines an order  can be described by first order sentences.
		Also for any finite subset $X\subset M$ we forbid all its orders $T_X$ for which $\operatorname{OR}_{X,T_X}(k) \geq F(k)$ for some $k$, this can also be described by a set of first order formulas. 
		Then the statement of this lemma follows from the compactness theorem.
	\end{proof}
	
	Note that for any finite metric space of cardinality $\ge 2$ its $\operatorname{Br}$ is equal to $2$. 
	This means that the inequality $\operatorname{Br}(M) \leq s$ does not follow from inequalities $\operatorname{Br}(M') \leq s$ for finite subsets of $M$.
	To prove that $\operatorname{Br}(M)\leq s$ you  also have to provide for all finite $M'\subset M$ some uniform bound for elongations of snakes on $s-1$ points.
	
	\medskip
	
	Now we give one more definition that will be used in next sections.
	
	\begin{definition}\label{def:convex}
		Let $M$ be a set, $T$ is an order on  $M$ and $A\subseteq M$ is a subset, 
		we say that $A$ is \emph{convex} with respect to  $T$, if for all  $x_1, x_2, x_3 \in M$ such that  
		$x_1 <_T x_2 <_T x_3$ it can not be $x_1, x_3 \in A$ and $x_2 \not \in A$.
	\end{definition}
	
	
	We give a lemma that 
	provides a sufficient condition for existence of an order, where the sets of some given family are "convex".
	
	\begin{lemma} \label{le:convex} 
		Let  $\mathcal{A}$ be a family of subsets of  $M$ such that for
		any two sets $A_1, A_2 \in \mathcal{A}$ either their intersection is empty, 
		or one of the sets is contained in another one.
		Then there exists an order $T$ on $M$ such that all sets in $\mathcal{A}$ are convex with respect to $T$.
	\end{lemma}
	
	
	\begin{proof}
		For finite $M$ the claim can be proved by induction on the cardinality of $M$.
		If $\# M = 0$ (and hence $M=\emptyset$) then the statement is obvious.
		Inductive step: if there are no sets besides $\emptyset$ and $M$ in $\mathcal{A}$, then any order satisfies desired conditions.
		Otherwise, let a set 
		$B_1 $  be a maximal by inclusion among sets
		of $\mathcal A\setminus M$ (we use the same letter $M$ for the space $M$ and here for the set containing all points of $M$)
		and let $B_2 = M\setminus B_1$.
		Our assumption on $\mathcal{A}$ and the maximality of $B_1$ implies that 
		any $A\in \mathcal A$, 
		$A \ne M$, $A \ne \emptyset$  is either a subset of $B_1$ or a subset of $B_2$.
		Since $\#B_1, \#B_2 < \#M$, the induction hypothesis implies the existence of orders $T_1$ on $B_1$ and $T_2$ on $B_2$ such that for $i\in \{1,2\}$ all sets of $\mathcal{A}$ that are subsets of $B_i$ are convex with respect to $T_i$.
		The required order $T$ can be constructed as follows: all elements of $B_1$ are $<_T$ than all elements of $B_2$, and $T = T_i$ on $B_i$ for $i=1$ or $2$.
		
		Now consider the case when $M$ is not necessarily finite. We use again Compactness theorem (similarly to the proof of Lemma \ref{rem:goedel}) to make a reduction to finite case. 
		Consider variables corresponding to points of $M$ 
		and to the  binary relation $<_T$.
		Observe that axioms of a relation  to be an order and condition in the definition of convex sets 
		can be described by sentences of first order. 
		Therefore, since if any finite number of sentences has a model, 
		then the whole set of sentences has a model.
	\end{proof}
	
	Now we mention some examples of families of convex sets for metric spaces which we describe in the following sections.
	\begin{enumerate}
		\item For rooted trees we can choose an order such that all branches of the tree are convex with respect to the order (see Lemma \ref{lem:hierarchicaltrees}.)
		
		\item For the graph $\Gamma_d$ associated to a tiling of a hyperbolic space $\mathbb H^d$ and the order, constructed in Section \ref{sec:hyper}, the sets that correspond to branches of the corresponding tree are convex.
		
		\item In Definition \ref{def:star} we define Star orders for graphs endowed with edges. 
		Star figures (union of a vertex and all the outgoing half-edges) are convex with respect to these orders.
	\end{enumerate}
	
	In these examples we could find the orders by showing  that the above mentioned sets satisfy the  assumption of Lemma \ref{le:convex} and apply this lemma. 
	We do not explain the details, since, as we have mentioned, we will describe these orders more explicitly in the  Sections \ref{sec:examples}, \ref{sec:hyper}.
	
	We already mentioned an order on a square, constructed by Bartholdi and Platzman.  
	Now we describe explicitly another hierarchical  order on the unit square.
	
	This order also uses self-similarity in its construction, 
	and we chose this particular order to guarantee that the squares (discussed below) are convex.
	
	\begin{example} Let $M =[0;1)^2$.
		For each point $x = (x_1,x_2)\in M$ we consider the binary representation $0.a_1a_2\dots$ and $0.b_1b_2\dots$ of  $x_1$ and $x_2$ correspondingly. Given $x$, we define $c(x)$ to be the number with binary representation $0.a_1b_1a_2b_2\dots$.
		We define the order $T$ by setting $x <_T y \Leftrightarrow c(x) < c(y)$.
		Then all parts that are obtained by splitting $M$ into $2^{2k}$ equal squares are convex with respect to $T$.
		
	\end{example}	
	The picture below illustrates the order on these smaller squares.
	
	\begin{figure}[h!]
		
		\begin{tikzpicture}
		\draw[thick](0,0) -- (2,0) -- (2,2) -- (0,2) -- cycle;
		\draw[thick](1,0) -- (1,2);
		\draw[thick](0,1) -- (2,1);
		\draw[thick][-stealth](0.5,0.5) -- (0.5,1.5) -- (1.5,0.5) -- (1.5,1.5);
		
		\begin{scope}[shift={(3,0)}, scale = 0.5]
		\draw[thick](0,0) -- (4,0) -- (4,4) -- (0,4) -- cycle;
		\draw[thick](1,0) -- (1,4);
		\draw[thick](2,0) -- (2,4);
		\draw[thick](3,0) -- (3,4);
		\draw[thick](0,1) -- (4,1);
		\draw[thick](0,2) -- (4,2);
		\draw[thick](0,3) -- (4,3);
		\draw [-stealth] (0.5,0.5) -- (0.5,1.5) -- (1.5,0.5) --
		(1.5,1.5) -- (0.5,2.5) -- (0.5,3.5) -- (1.5,2.5) --
		(1.5,3.5) -- (2.5,0.5)  -- (2.5,1.5) --  (3.5,0.5) -- (3.5,1.5) --
		(2.5,2.5) -- (2.5,3.5) -- (3.5,2.5) -- (3.5,3.5);
		\end{scope}
		\end{tikzpicture}
		\caption{Each of the small squares is convex with respect to $T$. The picture illustrates the induced order on the set of squares.}
	\end{figure}
	
	\section{ Metric trees, acyclic gluings, finite graphs}\label{sec:examples}
	
	In this section we 
	describe first examples of evaluation of the order breakpoint and of asymptotic behaviour  of order ratio function.
	Some of these examples will be used in the following sections.
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.35]{pictures/orderedcircle.png}
		\caption{The clockwise order $T_0$ on the circle.}
		\label{pic:circleorder}
	\end{figure}
	
	\begin{lemma} \label{le:examplecircle} Let $M=S^1$ be a circle (with its inner metric). 
		Then $\operatorname{OR}_M(k)=2$ for all $k \geqslant 2$. More precisely, for any order $T$ of $S^1$ and any $\varepsilon > 0$ there exists a snake on $3$ points such that its 
		points are located in $\varepsilon$-neighborhoods of two antipodal points of $S^1$.
	\end{lemma}
	
	\begin{proof} 
		Denote by $2R$ the length of the circle.
		Let $T_0$ be a natural  order  on this circle (see Picture \ref{pic:circleorder}).
		Suppose that $X \subset M$ and $l_{\operatorname{opt}}(X) = a$. 
		This means that the set $X$ lies on an arc $AB$ of length $a$. 
		If this arc does not contain the point $O$, then 
		$l_{T_0}(X) = a$. If this arc contains $O$, then $l_{T_0}(X) \leqslant 2a$.
		
		To prove $\operatorname{OR}_{M}(2) = 2$ we will show that 
		for any order $T$ on $M$ and any $\varepsilon>0$ 
		there exist points $x_1$, $x_2$, $x_3$ $\in X$ such that $x_1 <_{T} x_2 <_{T} x_3$ and $d(x_1, x_2), d(x_2, x_3) \ge R- \varepsilon$, i.e. $(x_1,x_2,x_3)$ is a snake of large elongation.
		
		\noindent Take $\varepsilon = R/n$, $n \in \mathbb{N}$. 
		Consider points $y_1, y_2, \dots y_{2n}$ with $d(y_i, y_{i+1}) = \varepsilon$, numbered with respect to the  order $T_0$.
		Consider a map $\phi$ from $M$ to the two point set $\{0,1\}$ which takes value
		$0$ if the antipodal point to $x$ is $T$-smaller than $x$ and $1$ otherwise. Clearly, the preimages of $0$ and $1$ are non-empty. 
		Therefore, there exists an index $i$ (considered  modulo $2n$) such that $\phi(y_i) = 1$, 
		$\phi(y_{i+1}) = 0$.
		
		
		\[\begin{tikzpicture}[scale = 0.6, every node/.style={scale=0.6}]
		\tikzmath{\mathbb R = 3; \x1 = \mathbb R*cos(100); \y1 =\mathbb R * sin(100);
			\x2 = 3*cos(60); \y2 =3 * sin(60);} 
		\draw[thick] (0, 0) circle [radius = 3cm];
		\draw[thick, ->] (-0.95 *\x1,-0.95*\y1) -- (0.95*\x1,0.95*\y1);
		\draw[thick, <-] (0.95*\x1,-0.95*\y1) -- (-0.95*\x1,0.95*\y1);
		\draw[dashed, thick] (-0.95*\x1, 0.95*\y1) -- (-0.95*\x1, -0.95*\y1);
		\node[above] at (\x1,\y1) {$y_i,1$};
		\node[above] at (-\x1,\y1) {$y_{i+1},0$};
		\node[below] at (-\x1,-\y1) {$y_{i+N}$};
		\node[below] at (\x1,-\y1) {$y_{i+1+N}$};
		\node[right] at (\x2, \y2) {$y_{i+2}$};
		\node[left] at (-\x2, \y2) {$y_{i-1}$};
		\draw (\x1, \y1) circle[radius = 0.075];
		\draw (-\x1, -\y1) circle[radius = 0.075];
		\draw (-\x1, \y1) circle[radius = 0.075];
		\draw (\x1, -\y1) circle[radius = 0.075];
		\draw (\x2, \y2) circle[radius = 0.075];
		\draw (-\x2, \y2) circle[radius = 0.075];
		\draw (\x2, -\y2) circle[radius = 0.075];
		\draw (-\x2, -\y2) circle[radius = 0.075];
		\end{tikzpicture}\]
		
		
		
		This means that $y_i >_T y_{i+n}$ and $y_{i+1} <_T y_{i+1+n}$.
		If $y_{i+1} >_T y_{i+n}$, then the snake 
		$y_{i+n} <_T y_{i+1} < y_{i+n+1}$ has large elongation.
		Otherwise we take the snake $y_{i+1} <_T y_{i+n} <_T y_i$.  
	\end{proof}
	
	
	\noindent 
	This lemma has a generalisation to spheres of dimension $d$. For statements and applications we refer to Section $6$
	in \cite{ErschlerMitrofanov2}.
	
	\medskip
	
	\noindent A {\it tripod} is a metric space that consists of three segments meeting at one vertex. 
	
	\begin{corol}\label{cor:tripod}
		Let $M$ be a tripod. Then $\operatorname{OR}_M(2)=2$. 
		If $l$ is the minimal length of three 
		segments of $M$, then for all $\varepsilon$ there exists a snake on three points, of diameter at least $l- \varepsilon$ and of width at most $2 \varepsilon$.
	\end{corol}
	
	\begin{proof}
		We can assume that all segments of the tripod $M$ have length $1$. 	
		Consider  $N = S^1$ to be a circle of length $6$.
		There is a $1$-Lipschitz map $\varphi: N \to M$ (see picture) such that any two antipodal points map to a pair of points at distance  $1$.
		
		
		\begin{tikzpicture}[scale = 0.6, every node/.style={scale=0.8}]
		\draw (0,1) circle(3);
		\foreach \x in {0,...,5}{
			\node at ({3.2*cos(60 * \x + 30)}, {3.2*sin(60 * \x + 30)+1}) {$A_{\x}$};
		}
		
		\draw[->][thick] (6.5,1.1) -- (8.5,1.1);
		
		\begin{scope}[shift = {(13,0)}]
		\draw[ultra thick] (0,0) -- (0,3);
		\draw[ultra thick] (0,0) -- (2.5,-1);
		\draw[ultra thick] (0,0) -- (-2.5,-1);
		\draw (-0.2,3.2) node[above]{$A_1$} -- (-0.2, 0.2) node[above left]{$A_2$} --
		(-2.6, -0.8) to[out = -130, in = 190] (-2.4, -1.2) node[below left] {$A_3$} -- 
		(0, -0.25)node[below] {$A_4$} -- (2.4, -1.2) node[below right] {$A_5$}  to [out = -20, in = -10] (2.6, -0.8)
		--(0.2, 0.2) node[above right] {$A_0$} -- (0.2, 3.2) to [out = 90, in = 90] cycle;
		\end{scope}
		
		\end{tikzpicture}
		
		Let $T$ be an arbitrary order on $M$ and let $T_N$ be some pullback of $T$ to $N$. 
		From Lemma \ref{le:examplecircle} we know that for any $\varepsilon > 0$ there are two antipodal points $x_1, x_2 \in N$ and points $y_1 <_{T_N} y_2 <_{T_N} y_3$ such that $y_1, y_3 \in B(x_1,\varepsilon)$, $y_2 \in B(x_2,\varepsilon)$. Here  $B(x, \varepsilon)$ denotes an open ball of radius $\varepsilon$ centered at $x$.
		
		We know that $d_{M}(\varphi(A_1), \varphi(A_2)) \geqslant 1$.
		Then $d_M(\varphi(y_1),\varphi(y_2))$ and $d_M(\varphi(y_1),\varphi(y_2)) > 1 - \varepsilon$, and 
		$d_M(\varphi(y_1),\varphi(y_3)) < 2\varepsilon$.
		Since $T_N$ is a pullback of $T$ we can conclude that $\varphi(y_1) <_{T}\varphi(y_2) <_{T}\varphi(y_3)$.
		This implies the second claim of the Corollary, and in particular that
		$\operatorname{OR}(3) \ge 2$.
	\end{proof}
	
	The example  in Remark  \ref{rem:numberofpoints} shows that for some metric spaces the choice of an order that optimizes $\operatorname{OR}(k)$ depends
	on $k$.
	
	\begin{example}
		[Dependence of an optimal order on the number of points]\label{rem:numberofpoints}
		Let $M$ be a six point metric space, with the distance function described by the matrix
		
		
		\[ 
		\left( \begin{array}{ccccccc}
		0 & 1 & 1.5 & 1.7& 1.5 & 2 \\ 
		1& 0& 1.8& 1.6& 1.5& 1.6  \\ 
		1.5& 1.8& 0& 1& 1.7& 2  \\ 
		1.7 & 1.6& 1& 0& 1.3& 1.6  \\ 
		1.5& 1.5& 1.7& 1.3& 0& 1.7  \\ 
		2& 1.6& 2& 1.6& 1.7& 0 
		\end{array} 
		\right) \] 
		
		\noindent Observe that the pairwise distances take values between $1$ and $2$, and hence satisfy the triangular inequality.
		
		\noindent It can be checked that the only orders that minimize $\operatorname{OR}(2)$ are 
		$T_1 = (1,2,3,4,5,6)$ and $T_2 = (6,5,4,3,2,1)$, but these two orders are not optimal for $k=3$.
		It can be shown that the order $T_3 = (3, 4, 5, 1, 2, 6)$ satisfies $\operatorname{OR}_{M, T_3}(3) < \operatorname{OR}_{M, T_1}(3)$.
	\end{example}
	
	
	\subsection{Orders on trees. Hierarchical orders}
	
	Consider a finite directed rooted tree $\Gamma$.
	The root of the tree is denoted by $\mathcal{O}_{\Gamma}$. Any vertex  can be joined with the root by a unique  directed path, and in particular each vertex has at most one
	parent. We assume that the direction on each edge is from the parent to its  child.
	
	We also assign to  each edge  some positive length, 
	and this
	provides a structure of a metric space on
	the vertices of $\Gamma$. 
	We say that $x$ is an {\it ascendant} of $y$, if there exists a sequence of vertices $(x_0 =x, x_1, \dots , x_l=y)$, such that for each $i$ the vertex $x_i$ is a parent of $x_{i+1}$.
	
	Let $T_{\rm rt}$ be an order on vertices of the rooted tree $\Gamma$, defined as follows.
	First for any vertex we fix an arbitrary order on the children of this vertex.
	Now if a vertex
	$x$ is an ascendant of $y$ then $x>_{T_{\rm rt}} y$. Otherwise consider paths of minimal length from $\mathcal{O}_{\Gamma}$ to $x$ and $y$: $(v_0=\mathcal{O}_{\gamma}, v_1, \dots, v_k=x)$ and $(u_0=\mathcal{O}_{\Gamma} , u_1, \dots, u_m=y)$.
	Observe that  if neither $x$ is an ascendant of $y$ nor $y$ is an ascendant
	of  $x$, then there exists $i: i \le k, i\le m$ such that $v_i \ne u_i$. Take minimal $i$ with this property and put
	$x>_{T_{\rm rt}}y$ if $v_i> u_i$ (in our fixed order on the children of $v_{i-1}= u_{i-1}$).
	
	Any order obtained in this manner we call \textit{a rooted order}.
	Let us say that a subset of a rooted tree is a {\it branch}, if it consists of some vertex $x$ and all vertices $y$ such that $x$ is an ascendant of $y$.
	
	Let us say that an order $T$ on vertices of a rooted tree is {\it hierarchical} if the following holds. 
	Suppose that
	$x<_T y<_T z$ and $x$ and $z$ belong to some branch. Then $y$ also belongs to this branch.
	
	The existence of hierarchical orders  can be deduced from Lemma \ref{le:convex}. Without referring to this Lemma, it is also not difficult to see it more directly, in the Lemma
	\ref{lem:hierarchicaltrees}.
	Claim (1) of this Lemma is proven
	in Thm 2, \cite{Schalekamp}.
	For the convenience of the reader we provide the proof.
	
	\begin{lemma} \label{lem:hierarchicaltrees}
		Let $\Gamma$ be a rooted tree as above, then
		\begin{enumerate} 
			\item 
			For any  $k\ge 1$ and any rooted order $T_{\rm rt}$ 
			it holds
			$$
			\operatorname{OR}_{V(\Gamma), T_{\rm rt}} (k) \le 2,
			$$
			where $V(\Gamma)$ is the set of vertices of $\Gamma$.
			
			\item Any rooted order $T_{\rm rt}$ is hierarchical.
			
		\end{enumerate}
	\end{lemma}
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.45]{pictures/orderedtree.png}
		\caption{Vertices of  the tree are ordered with respect to  the first visit by the orange path. 
			In case when the tree is a Cayley graph of a free group $F_n$, one of such orders corresponds
			to a lexicographic order of $F_n$.}
		\label{pic:orderedtree}
	\end{figure}
	
	\begin{proof} Let $S$ be a  subset of
		vertices of $\Gamma$, $\# S \ge 2$. 
		Observe that any (not necessarily not self-intersecting) path  passing through $S$ passes
		through all the vertices of 
		the spanning tree $\Gamma'$ of $S$. 
		Observe also that $l_{\operatorname{opt}}(S)$ is at least the sum of the lengths of all edges in $\Gamma'$. 
		It is clear that the restriction of $T_{\rm rt}$ to $\Gamma'$ is a rooted order on $\Gamma'$.
		
		Now consider a path through vertices of  $S$ with respect to our order $T_{\rm rt}$. Let us show that this path passes at most twice
		through each edge of the spanning tree, this would show that the length of this path is at most twice the length of the optimal tour. Without loss of generality we can assume that $\Gamma=S$.
		We prove the claim by the induction on the number of vertices of the tree $\Gamma$.
		Observe that any subtree $\Gamma$ contains a vertex, which we denote by $\mathcal{O}'$, such that it is an ascendant of all other vertices in $\Gamma'$.  
		For each child of $\mathcal{O}_{\Gamma}$, consider the corresponding branch. The tour with respect to $T_{\rm rt}$ first visits $\mathcal{O}_{\Gamma}$, then one of its children and all points of its branch, then another child and all points of its branch.
		Observe that the edge from $\mathcal{O}_{\Gamma}$ to any of its children is visited at most twice: once passing from $\mathcal{O}_{\Gamma}$ to its child, and, possibly second time, after visiting all points of the branch of this child, before passing to another child of $\mathcal{O}_{\Gamma}$. 
		This implies the second claim of the lemma that the order is hierarchical.
		
		Now observe also that  all edges not adjacent to $\mathcal{O}_{\Gamma}$ are visited at most twice by induction hypothesis, applied to the branches of $\Gamma$.
		So we have proved also the first claim of the lemma.
	\end{proof}
	
	Recall that by one of the definitions, a metric space $M$ is a {\it metric tree} if it is a geodesic
	metric space which is $0$-hyperbolic.
	
	\begin{corol}\label{re:anytree}
		Let $M$ be a  metric tree. By Lemma \ref{rem:goedel} we know that if the statement that $\operatorname{OR}(k) \le 2$ is true for any finite tree, then it is true for any metric tree. We can therefore conclude that $\operatorname{OR}_M(k) \le 2$ for all $k$.
		Orders that are obtained by applying Lemma \ref{rem:goedel} to hierarchical orders on corresponding finite trees will be also called \emph{ hierarchical}.
	\end{corol}
	
	
	\subsection{Acyclic gluing of spaces}
	
	Let us say that a metric space $M$
	is an {\it acyclic gluing} of metric spaces $M_\alpha, \alpha \in \mathcal{A}$ 
	if it is obtained as follows.
	Consider a graph which is a tree (finite or infinite), vertices of which have labellings of two types: 
	each vertex is either labelled by 
	$M_\alpha$, for some  $\alpha \in \mathcal{A} $ or it is labelled by a point $x$, for some $\alpha \in \mathcal{A} $ and some $x \in M_\alpha$. The vertices are joined by an edge if one of the vertices is labelled with $x$, another one with $M_\alpha$ and $x \in M_\alpha$. 
	Our space is obtained from a disjoint union of $M_{\alpha}$ and points $x$ by gluing all $x$ to the points in corresponding $M_\alpha$.
	
	A graph which is an acyclic gluing of circles and intervals  is also called 
	{\it a cactus graph}.
	An example of a space homeomorphic to such cactus graph, with only circles glued together, 
	is shown on Picture \ref{pic:wedgescircles}.
	Observe, that a finite acyclic gluing can be obtained by a finite number of operations which take
	a wedge sum of two metric spaces.
	
	\noindent We will describe a natural way to order points of $M$ if orders on $M_{\alpha}$ are given. 
	
	\noindent For acyclic gluings it will be more natural to consider {\it cyclic order ratio functions}.
	
	\begin{definition}\label{def:cyclicOrderingRatioFunction}
		Let $M$ be a metric space with a total order $T$.
		If $X$ is a finite subset of $M$ we denote by $l^{\circ}_{\operatorname{opt}}(X)$ the minimal length of a closed path visiting all points of $X$.
		We denote by $l_T^{\circ}(X)$ the length of the closed path corresponding to order $T$: if the distance between first and last vertices of $X$ is equal to $r$, then  $l_T^{\circ}(X) = l_T(X) + r$.
		
		\noindent We define the {\it cyclic order ratio function} as
		$$
		\operatorname{OR}_{M,T}^{\circ}(k) := \sup_{X\subset M : 2 \le \#X \le k+1} \frac{l^{\circ}_T(X)}{l^{\circ}_{\operatorname{opt}}(X)}.
		$$
	\end{definition}
	
	\begin{definition}
		Let $M$ be a metric space (or more generally a set) and $T$ be an order on $M$. 
		Suppose that $M$ is a disjoint union $M =  A \bigsqcup B$ and that for any two points $x\in A$, $y \in B$ we have $x <_T y$.
		We define a new order $T'$ on $M$ as follows: if $x, y \in A$ or $x, y \in B$ we put $x <_{T'} y$ if $x <_T y$. 
		For any two points $x \in A$ and $y\in B$ we put $x >_T y$.
		We say that $T'$ is a {\it cyclic shift} of the order $T$.
	\end{definition}
	
	It is clear that for a given metric space the relation to be a cyclic shift is an equivalence relation on orders.
	For example, all clockwise orders of a circle with different starting points are cyclic shifts of each other.
	For a given a metric space $M$, 
	a total order $T$ and a point $x\in M$,
	there exists a unique cyclic shift $T^x$ such that $x$ is the minimal point for this order. 
	
	In the following two remarks we state straightforward properties of cyclic shifts and cyclic order ratio functions.
	
	
	\begin{rem}\label{rem:ORcyclicshift1}
		If $M$ is a metric space endowed with an order $T$, then for any finite subset $X$ of $M$ it holds
		
		$$
		1 \leq \frac{l^{\circ}_T(X)}{l_T(X)} \leq 2; \:\:\:\: 
		1 \leq \frac{l^{\circ}_{\operatorname{opt}}(X)}{l_{\operatorname{opt}}(X)} \leq 2.
		$$
		
		It follows that  for any $k$ 
		$$
		\frac{1}{2} \operatorname{OR}_{M,T}(k) \leq \operatorname{OR}^{\circ}_{M,T}(k) \le 2 \operatorname{OR}_{M,T}(k).
		$$
		
		For any $m$, $\operatorname{Br}(M,T) \geq 2m$ if and only if $\operatorname{OR}^{\circ}_{M,T}(2m-1) = m$.
	\end{rem}
	
	\begin{rem}\label{rem:ORcyclicshfift2}
		If $T'$ is a cyclic shift of $T$, then for any $k$ it holds
		$$
		\operatorname{OR}^{\circ}_{M,T'}(k) = \operatorname{OR}^{\circ}_{M,T}(k),
		\: 
		\operatorname{OR}_{M,T'}(k) \le 2 \operatorname{OR}_{M,T}(k)
		$$
		
		In particular, taking in account the last claim of the previous remark 
		$\operatorname{OR}_{M,T}(2m-1) = 2m-1$ if and only if 
		$\operatorname{OR}_{M,T'}(2m-1) = 2m-1$.
	\end{rem}
	
	
	Now given a family $M_\alpha$ and a family of orders
	on these spaces $T_\alpha$, we consider a
	metric space $M$ 
	which is an acyclic gluing of ordered spaces $M_{\alpha}$, $\alpha \in \mathcal{A}$. We define  a family of  {\it  clockwise orders} on $M$ as follows.
	We assume below that the tree in the definition of acyclic gluing is finite (this is assumption is not essential).
	One way  is to define such orders recursively:
	having already defined  orders $T_A$ and $T_B$ on two metric spaces $A$ and $B$, we will define an order $T_C$ on the wedge sum $C$ of $A$ and $B$ over a point $x$.
	We 
	choose any of the two ways to enumerate $A$ and $B$, for example $(A, T_A)=(M_1, T_1)$ and $(B, T_B)=(M_2, T_2)$. 
	We use the notation that denotes by $x$ both a point in $M_1$ and a one in $M_2$.
	
	Recall that for any point in any space, there is a cyclic shift that makes this point minimal.
	Let $T^x_1$ and $T^x_2$ be  cyclic shifts of $T_1$ and $T_2$ respectively such that $x$ is the minimal point in $(M_1,T^x_1) $ and $(M_2, T^x_2)$. 
	Consider the order $T_C$ on a wedge sum of $M_1$ and $M_2$, such that $x$ is a minimal  point, then come all points of $M_1 \setminus \{ x \}$ ordered as in  $T^x_1$ and
	then all points of $M_2 \setminus \{ x \}$
	ordered as in  $T^x_2$.
	In such a way  we construct recursively  an order on an acyclic gluing. 
	We say that such orders and any cyclic shift of these orders are  {\it clockwise orders} on $M$.
	Finally, if the tree in the definition of an acyclic gluing is infinite, we can say that an order on this space is a clockwise order, if the restriction of this order to any finite (connected) acyclic gluing
	is a clockwise order.
	
	There is another way to formulate this definition and to visualize clockwise orders. 
	Given an acyclic gluing $M$ of $M_\alpha$, $\alpha \in \mathcal{A}$, we consider circles $N_\alpha$, $\alpha \in A$ and choose a direction (clockwise order) on each $N_{\alpha}$.
	For all joint points on $M_\alpha$ we consider points on $N_\alpha$ enumerated in the same order, up to a cyclic shift, and we consider the corresponding acyclic gluing of $N_\alpha$. 
	The clockwise order on an acyclic gluing $N$ of $N_\alpha$ can be obtained in the following way.
	We construct a continuous imbedding of $N$ in the plane in such a way that 
	the image of $N$ is equal to its outer boundary, and the clockwise orientation of the boundary  of the image of $N$  (in the plane) coincides with clockwise orientations of $N_{\alpha}$.
	
	We choose a joint point $O$ (among joint points joining our circles)
	as a base point  and  consider an order corresponding to the fist visit of a clockwise path in the plane, see
	Picture \ref{pic:wedgescircles}.
	For each point of $M_\alpha$ we can associate an arc in $N_\alpha$. If $m$ is a joint point in $M$, then it corresponds
	to a joint point in $N$. We will use a convention that the initial point of an arc (with respect to the first visit of the clockwise path starting from $O$) is included in the arc and the last point of the arc is not included.
	
	A clockwise order on $M$ can be described as follows. Take $m, m' \in M$. First suppose
	that there exists $\alpha$ such that $m, m' \in M_\alpha$. Observe that our choice of $O$ fixes a
	choice of a cyclic shift on the acyclic gluing of our circles (and in particular on  $N_\alpha$). This induces a choice of an acyclic shift also 
	on $M_\alpha$.
	In case when there exists $\alpha \in \mathcal{A}$ such that $m, m' \in M_\alpha$ we compare $m, m'$ with respect to the above mentioned cyclic shift of $T_\alpha$ (on $M_\alpha$).
	Finally, if there is no $\alpha$ such that $m, m'$ belong to the same $M_\alpha$, we consider corresponding arcs $\gamma$, $\gamma'$, choose any point $n\in \gamma$, $n' \in \gamma'$ and compare
	$n$ and $n'$ in $N$. If $n<n'$ with respect the order in $N$,
	then we say that $m<m'$
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.45]{pictures/wedgedcircles.png}
		\caption{An acyclic gluing of circles. A clockwise order corresponds to a first visit by an orange path. More precisely,
			to choose one among cyclic shifts of this clockwise order, we fix any point of the gluing and consider the first visit by an orange path starting at this point.}
		\label{pic:wedgescircles}
	\end{figure}
	
	Now observe that a metric tree is an acyclic gluing of intervals, and recall 
	that $\operatorname{OR}(k) \le 2$ for all $k$ in this case. 
	Here we prove a statement
	about general acyclic gluings.
	
	\begin{lemma} \label{le:ORacyclicunion}
		Let $M$ be an acyclic gluing of ordered spaces $(M_\alpha, T_{\alpha})$, $\alpha \in \mathcal{A}$. 
		Let $T$ be a clockwise order on $M$. 
		For all $k$ it holds
		$$
		\operatorname{OR}^{\circ}_{M,T}(k) =  \sup_\alpha \operatorname{OR}^{\circ}_{M_\alpha, T_\alpha} (k).
		$$
	\end{lemma}
	
	\noindent If we allow metric spaces with possible infinite
	distances between some points, there is a way to define $\operatorname{OR}$ also in this more general setting (considering subsets $S$ with finite pairwise distances)
	and in this setting the lemma can be reformulated to claim that order ratio function of an acyclic gluing $M$ is the same as for the disjoint union of $M_\alpha$.
	
	
	\begin{proof}
		It is obvious that the left hand side in the equation in  the lemma  is at least
		as large as the right hand side, so it is enough to prove that
		$$
		\operatorname{OR}^{\circ}_{M,T}(k) \leq  \sup_\alpha \operatorname{OR}^{\circ}_{M_\alpha, T_\alpha} (k).
		$$
		
		To prove this inequality, 
		we first observe that it is enough to
		prove this claim for finite acyclic gluings.
		We consider a finite set $X$ consisting of $k+1$ points $x_1, x_2, \dots x_{k+1}$.
		Consider a minimal (with respect to inclusion) acyclic gluing $M'$ of $M_\alpha$, 
		$\alpha \in \mathcal{A'} \subset \mathcal{A}$, containing $X$.
		If $M_\alpha$ are geodesic metric spaces,
		then observe that any path passing through $X$ visits each $M_\alpha$,
		for $\alpha \in A'$. 
		Even though $M_\alpha$ are not assumed to be geodesic in general, for any two points 
		$x, y$ we can consider a finite sequence of points $x=z_1$, $z_2$ \dots $z_p=y$, where $p$ depends on $x$ and $y$, such that 
		for any $i$ the points $z_i$ and $z_{i+1}$ belong to the same component $M_{\alpha}$, and
		the distance between $x$ and $y$ is
		$$
		\sum_{i: 1\le i \le r-1} d_M(z_i, z_{i+1}).
		$$
		It is clear that such $z_i$ exist and that for $1<i <r$ these points are joint points in the acyclic gluing. 
		
		Now we choose a component $M_\alpha$, $\alpha \in \mathcal{A}'$ and consider 
		the retract mapping $\rho_{\alpha}: M\to M_{\alpha}$ that maps every point of $M$ to the nearest point of $M_{\alpha}$. 
		It is clear that if $x\not\in M_{\alpha}$ then $\rho(x)$ is a joint point.
		Observe, that the retract $\rho_{\alpha}$  sends the clockwise order on $M$ to some (cyclic shift) of $T_\alpha$.
		More precisely, if we choose arbitrary point $y\in M_{\alpha}$ that is not a joint point and consider the cyclic shift $T^{y}$ of $T$, then
		for any two points $x_1 <_{T^y}x_2$ holds
		$\rho_{\alpha}(x_1) \leq_{T^y} \rho_{\alpha}(x_1)$.
		
		\begin{figure}[!htb] 
			\centering
			\includegraphics[scale=.45]{pictures/wedgessumOR.png}
			\caption{A retract of the gluing to s component 
				and the images of $x_i$.}
			\label{pic:wedgessumOR}
		\end{figure}
		
		
		Observe also that any closed path
		passing through  $x_i$ ($1 \le i \le k+1$) passes through all their images $\rho_{\alpha}(x_i)$ ($1 \le i \le k+1$).
		To be more precise, if $M_\alpha$ are not necessarily geodesic, we use the same convention as before: having two points in $M$ at distance $d$, by a path we mean a sequence of points $z_i$, such that consecutive points belong to the same component and such that the total sum of the distances is $d$.
		
		For the shortest closed path visiting all points of $X$ we consider all its ``jumps''
		inside $M_\alpha$
		(shortest paths with endpoint lying in the union of $X$ and joint points of 
		$M_{\alpha}$). 
		These jumps form a cyclic tour visiting all the points of $\rho_{\alpha}(X)$, and their total length is at least $l^{\circ}_{\operatorname{opt}}(\rho_{\alpha}(X))$.
		Now we consider a tour of $X$ with respect to a clockwise order $T$, and take all its ``jumps'' inside  $M_{\alpha}$.
		They form a tour of $\rho_{\alpha}(X)$ with respect to $T$, and its length can be bounded from above as follows
		
		$$
		l^{\circ}_{T}(\rho_{\alpha}(X)) 
		\leq \operatorname{OR}^{\circ}_{M_{\alpha}, T_{\alpha}}\left(\#\rho_{\alpha}(X)\right)
		l^{\circ}_{\operatorname{opt}}(\rho_{\alpha}(X)) \leq
		\operatorname{OR}^{\circ}_{M_{\alpha}, T_{\alpha}}(k)
		l^{\circ}_{\operatorname{opt}}(\rho_{\alpha}(X)).
		$$
		
		Summing these inequalities over all $\alpha\in \mathcal{A}'$, we prove the claim of the lemma.
	\end{proof}
	
	In Remark \ref{rem:ORcyclicshift1} we mention that $\operatorname{OR}$ and $\operatorname{OR}^{\circ}$ are the same up to the multiplicative constant $2$.
	So we get the following
	
	\begin{corol}
		Let $M$ be an acyclic gluing of ordered spaces $(M_\alpha, T_{\alpha})$, $\alpha \in \mathcal{A}$. 
		Let $T$ be a clockwise order on $M$. 
		Then for any $k \ge 1$
		$$
		\operatorname{OR}_{M,T}(k) \le 4 \sup_\alpha \operatorname{OR}_{M_\alpha, T_\alpha} (k).
		$$
	\end{corol}
	
	As another  corollary  we obtain
	\begin{corol}\label{cor:acyclicgirth}
		
		For a finite acyclic gluing 
		$\operatorname{OR}_{M,T}(2k-1)=2k-1$ if and only if
		for some $\alpha$ it holds $\operatorname{OR}_{M_\alpha,T}(2k-1)=2k-1$.
		More generally, for a not necessary finite gluing it holds
		$\operatorname{OR}_{M,T}(2k-1)=2k-1$ if and only if 
		there is no uniform upper bound for all $M_{\alpha}$ on elongations of snakes on $2k$ points.
	\end{corol}
	
	The following corollary is about $\operatorname{Br}$ for free product of groups. 
	In a particular case of virtually free groups we will show later that they can be characterized as groups with small $\operatorname{Br}$.
	\begin{corol} Let $G=A*B$ be a free product of groups $A$ and $B$. 
		If maximum of $\operatorname{Br}(A)$, $\operatorname{Br}(B)$ is odd, then $\operatorname{Br}(G)$ is equal to this maximum.
		If the maximum is even, then it is at most this maximum plus one.
	\end{corol}
	If we do not assume that the maximum is odd  (take for example $A=B=\mathbb{Z}$), it is clear that the "plus one" in the formulation is essential.
	
	
	
	\subsection{Order breakpoint of finite graphs.}
	
	Let $\Gamma$ be a graph, finite or infinite, with edges of length $1$. 
	We consider $\Gamma$ as a metric space with edges included. 
	Given an order on the set of vertices of $\Gamma$, we define an order on all points of $\Gamma$ as follows.
	
	\begin{definition}
		\label{def:star}
		Given an order $T$ on the set of vertices $V$ of $\Gamma$, 
		we define the order ${\rm Star}(T)$ on $\Gamma$. We subdivide each edge $AB$ into two halves of length $1/2$, and we assume
		that the middle point belongs to the half of $A$ if $A<_T B$. For each point $x$ in $\Gamma$ we have defined therefore the vertex $v(x)$ (to the half-interval of which it belongs).
		If $v(x)<_T v(y)$ we put $x <_{{\rm Star}(T)} y$. On points with the 
		same $v(x)$ we consider an hierarchical order on the corresponding star figure, assuming that the vertex is larger than the other points. 
		The half-edges starting from this point are numbered with respect to the order of their non-common endpoints, and on each interval we consider the usual order of this interval, from the vertex to the middle point.
	\end{definition}
	A comment on the definition above: the choice to which half edge the middle point belongs, and the order  we have chosen for half edges with a common vertex (the corresponding star figure) is not important.
	We have made this choice to fix an order on star figures which we will study in the sequel.
	
	The  Lemma \ref{lem:edgesnotedges} below shows that for  the asymptotic behaviour of order ratio function it does not matter whether we consider the graph together with the edges or only the metric
	on the vertices. In a more general setting,  the asymptotic behavior of order ratio function can, under certain assumption, be 
	reduced to studying local geometry of the metric space and its global
	geometry.
	
	
	\begin{lemma}\label{lem:edgesnotedges}
		There exist positive constants $A$ and $B$  such that for any graph $\Gamma$ and all
		$k\ge 2$  it holds
		$$
		\operatorname{OR}_{\Gamma, \operatorname{Star}(T)}(k) \leq A\cdot \operatorname{OR}_{V,T}(k) + B.
		$$
		In particular, one can take $A=8$ and $B=4$.
	\end{lemma}
	This lemma is proven by  associating to a path in $(\Gamma, \operatorname{Star}(T))$ a path that passes through the centers of corresponding figures. For details of the proof see Appendix \ref{app:edgesnotedges}.
	We do not use this lemma in the sequel, but we will use some other properties of the $\operatorname{Star}$ orders.
	
	
	It is clear that order breakpoint can change
	by applying  ${\rm Star}$ operation: for example, $\operatorname{Br}$ of the tripod is $3$ while $\operatorname{Br}$ of the set of its four vertices is $2$. However, this operation provides a bound on $\operatorname{Br}$ of the graph.
	
	\begin{lemma}\label{lem:stargirth}
		If $\Gamma$  is any (finite or infinite)
		graph, then $\operatorname{Br}(\Gamma, {\rm Star}(T))$ is at most $\max (\operatorname{Br}({V, T}), 4)$.
		In particular, if  $\Gamma$ is a finite graph, then $\operatorname{Br}(\Gamma, {\rm Star}(T)) \le 4$.
	\end{lemma}
	
	\begin{proof}
		Let constant $C$ be such that any Star figure does not
		admit snakes on $4$ points of elongation greater
		than $C$ (from the proof of Lemma \ref{lem:snake} it follows that for any tree we can take for example $C = 12$). 
		Let $k\ge 5$ be such that
		$(V,T)$ does not admit snakes of arbitrarily large elongation on $k$ points.
		Let us  show that 
		$(\Gamma, {\rm Star}(T))$ does not admit snakes of arbitrarily large elongation on $k$ points.
		Suppose that this is not the case and $\Gamma$ admits such snakes.
		For each point $x_i$ of the snake $(x_i)$, $1\le i \le k$,  consider the vertex $v_i\in V$
		such that $x_i$ belongs to the Star figure of $v_i$.
		If elongation of the snake is 
		$> C$,
		then it is clear that among the vertices $v_i$ there are at least  two distinct points.
		
		Let $D$ and $\delta$ be the diameter and the width of the snake $(x_i)$, $1 \le i \le k$.
		First suppose that all $v_i$, $1\le i \le k$ are distinct.
		Observe  that $\delta \ge 1/2$. Indeed, $x_1$, $x_3$, $x_5$ belong to three distinct Star figures,
		and then the distance between two of these point is
		at least $1/2$.
		Observe that in this case  the points $v_i$, $1 \le i \le k$,   form a snake, of  diameter $D'$ satisfying  $D'\ge D-1 \ge 1/2 D$ and
		of width $\delta'$ at most $\delta+1 \le 3 \delta$. We see that under our assumption the elongation of the snake $(v_i)$ 
		is at least $1/6$ of the elongation of $(x_i)$.
		
		Now observe that if there exist at least two points of the snake belonging to the same Star figure $S$, then (since the Star figure is convex with respect to the order on $\Gamma$) there exist two points of indices of different parity belonging to $S$.  We see that $D\le 1+2 \delta$. Assume that the elongation $D/\delta \ge 6$. We have $6 \delta  \le D \le 1+ 2\delta$, and
		$\delta \le 1/4$.
		We conclude that all points of our snake are in the "extended" Star figure around $v_i$ with edges of length $1$.
		We know that not  all $x_j$ belong to the Star figure of $v_i$, and that this Star figure is convex with respect to the Star order. Thus, reverting if  necessary our order we can assume that $x_1$ does not belong to this Star figure.
		We know that the distance between $x_1$ and $x_3$ and between $x_3$ and $x_5$ is at most $1/4$.
		This implies in particular that these points belong to the same edge  of $\Gamma$ as $x_1$.
		
		If $x_3$ belongs to the same  half-edge as $x_1$, then,
		taking in account convexity of this half-edge, we conclude that  $x_2$ also belong to the same half-edge. 
		In this case the elongation of the 
		$3$-point snake $(x_1,x_2,x_3)$ is $1$. 
		This means that the elongation of the initial snake is at most $3$.
		
		Finally, if $x_3$ does not belong to the same (external) half-edge as $x_1$, 
		then $x_5$ can not belong to the same interval as $x_1$ because the half edge is convex with respect to the order. 
		Then $x_3$ and $x_5$ belong to the same half-edge.
		In this case we obtain a snake 
		$(x_3, x_4, x_5)$ on $3$ points with elongation $1$ and again get a contradiction.
		
		We have proved the first claim of the lemma. The second claim follows from the fist one, since it is clear that finite sets do not admit snakes on three points of large elongation.
		
	\end{proof}
	
	For the rest of this section, we assume that our graph is finite.
	As before, we consider a metric space of a finite graph $\Gamma$, with edges (not only vertices) included. 
	
	
	Now we compute the order breakpoint for such graphs. It is clear that $\operatorname{Br}$ is equal to $2$ if $\Gamma$ is homeomorphic to an interval.
	We have seen already that $\operatorname{Br}$ is equal to $3$ for a circle or a tripod. We start with an example with $\operatorname{Br}(\Gamma) =4$.
	
	
	
	\begin{figure}[!htb] 
		\centering
		\begin{tikzpicture}[scale = 2]
		\draw (0,0) -- (0,1) -- (1,1) -- (1,0) -- (-1,0) -- (-1,1) -- (0,1);
		\draw[ultra thick] (-0.33,0) -- (0.33,0);
		\draw[ultra thick] (-0.33,1) -- (0.33,1);
		\draw[ultra thick] (0,0) -- (0,0.33);
		\draw[ultra thick] (0,1) -- (0,0.66);
		\node[above] at (0, 1) {$A$};
		\node[below] at (0, 0) {$B$};
		\end{tikzpicture}
		\caption{Two tripods in a domino graph.}
		\label{pic:domino}
	\end{figure}
	
	\begin{lemma}\label{lem:domino}[Order breakpoint of a domino graph]
		Let $\Gamma$ be a domino graph (with its inner metric, see Picture \ref{pic:domino}). Then for any order $T$
		on $\Gamma$, this graph admits snakes on $4$ points of large elongation. More precisely, assume that the lengths of the paths between the vertices ($A$ and $B$ on the picture) are at least $1$.
		For any $\varepsilon>0$ there exists a snake on 4 points of diameter $\ge 1/3-\varepsilon$ and of width at most $2 \varepsilon$.
	\end{lemma}
	
	\begin{proof} Consider two tripods ${\rm Tr}_1$ and ${\rm Tr}_2$, with segments  of length $1/3$, with their joint vertices at two joints vertices $A$ and $B$ of the domino graph (see Picture \ref{pic:domino}).
		By Corollary \ref{cor:tripod}
		we know that in ${\rm Tr}_1$  there exists a snake on  three points of diameter at least $1/3-\varepsilon$ and of width at most $2 \varepsilon$. 
		Denote its points $x_1$,$x_2$, $x_3$, here $x_1<_T< x_2<_T x_3$.
		Observe that there is a path from $x_2$ to any point of another tripod ${\rm Tr}_2$ which stays at distance at least $1/3-\varepsilon$ from $x_1$ and $x_3$. If $\Gamma$ equipped with 
		our order $T$ does not admit
		snakes on $4$ points of width at most $2\varepsilon$ and diameter at least $1/3-\varepsilon$, we would see that any point on this path stays (with respect to the order $T$) between $x_1$ and $x_3$. 
		This would show that all points of ${\rm Tr}_2$ are between some two points of the first tripod ${\rm Tr}_1$.
		The same argument shows that all points of the first tripod are between some two points of the second one, and this contradiction implies the claim of the lemma.
	\end{proof}
	
	An argument  similar to the argument above about the order breakpoint of the domino graph will be used later to estimate the order breakpoint of infinitely presented groups.
	
	Now we formulate a corollary of Lemmas \ref{lem:stargirth} and \ref{lem:domino}.
	
	\begin{corol}\label{cor:finitegraphs}
		Let $\Gamma$ be a finite connected graph, with edges included. Then there are $3$ possibilities
		\begin{enumerate}
			\item $\Gamma$ is homeomorphic to an interval, then  $\operatorname{Br}(\Gamma) =2$.
			
			\item $\Gamma$ is homeomorphic to a cactus graph (acyclic gluing of circles and intervals), but not homeomorphic to an interval, then $\operatorname{Br}(\Gamma) =3$.
			
			\item Otherwise, $\operatorname{Br}(\Gamma) =4$.
		\end{enumerate}
	\end{corol}
	
	\begin{proof}
		If $\Gamma$ is homeomorphic to an interval, then the claim is straightforward.
		
		\noindent If $\Gamma$ is homeomorphic to an acyclic gluing of circles and intervals, observe that either there is at least one circle in this wedge sum decomposition, or $\Gamma$ is a tree which contains an isometric  copy of a tripod.
		In both  cases, we know that for a subset of $\Gamma$  it holds $\operatorname{Br} \ge 3$, and hence $\operatorname{Br}(\Gamma) \ge 3$. Also observe that $\operatorname{Br}(\Gamma) \le 3$, as follows from Corollary \ref{cor:acyclicgirth}. 
		
		Finally observe that if $\Gamma$ does not contain an isometric copy of a domino graph (with edges of some positive length), then $\Gamma$ is homeomorphic to an acyclic gluing of circles and intervals. 
		Indeed, if $\Gamma$ is not a tree, then $\Gamma$ contains a simple cycle $c$.  Observe that either
		any two vertices $A,B$ of this cycle belong to distinct connected components
		of $(\Gamma \setminus  c) \bigcup \{A,  B\}$, in this case we can argue by induction claiming that connected components of $\Gamma \setminus c$ are acyclic gluings.
		Or there exist  $A,B$ on $c$, and a path from $A$ to $B$ not passing through $c$, and this provides a homeomorphic image of a domino.
		Therefore,  Lemma \ref{lem:domino} implies that $\operatorname{Br}(\Gamma) \ge 4$.  By Lemma \ref{lem:stargirth} we know that $\operatorname{Br}(\Gamma) \le 4$, and we can conclude that
		$\operatorname{Br}(\Gamma) = 4$
		
	\end{proof}
	
	\section{Spaces and Groups with small order breakpoint}
	
	We recall that given a metric
	space $M$
	of cardinality $>1$
	and an order $T$ on $M$, $\operatorname{Br}({M,T})$ is the minimal $s$ such that $\operatorname{OR}_{M,T}(s)<s$. One can define
	$\operatorname{Br}(M)$ as the minimum of $\operatorname{Br}({M,T})$, where the minimum is taken over all orders $T$ on $M$.
	
	In this section we will be mostly interested in uniformly discrete metric spaces. 
	Given a graph $\Gamma = (V,E)$, we consider the metric space $V$ with graph metric (all the edges have length $1$). 
	For a group $G$ with generating set $S$ we consider the word metric on $G$. 
	
	It is straightforward that for any metric space $M$ of cardinality at least $2$ it holds $\operatorname{Br}(M) \ge 2$.
	We know (see Lemma \ref{le:snakesqi}) that quasi-isometric uniformly discrete metric spaces have the same value of $\operatorname{Br}$.
	In particular, 
	for any finite metric space $M$, $\operatorname{Br}(M, T) = 2$ for any order $T$, because there are finitely many triples of points and elongation of snakes on $3$ points is bounded.  
	
	The following lemma characterizes spaces (quasi-isometric to geodesic ones) with this minimal possible value of $\operatorname{Br}$.
	
	\begin{lemma} \label{le:notrayline}
		Let $M$ be quasi-isometric to  a geodesic metric space, and not quasi-isometric to a point, a ray or a line. Then $\operatorname{Br} \ge 3$. If we assume moreover that $M$ is
		uniformly discrete, then $\operatorname{Br} \leq 2$ 
		if and only if $M$ is either bounded or quasi-isometric to a ray or to a line.
		($\operatorname{Br} \le 2$ means  $\operatorname{Br}=2$ unless $M$ is a one-point set).
	\end{lemma}
	A particular case 
	for the last claim of  Lemma
	\ref{le:notrayline}
	above is when  $\Gamma = (V,E)$ is a connected graph, finite or infinite,
	of bounded degree (as we have mentioned, a convention of this section is that the length of edges equal to $1$).
	Then the order breakpoint $\operatorname{Br}(\Gamma) \leq 2$ if and only if the graph is quasi-isometric to a point, a ray or a line.  
	Moreover, it can be shown in this case that if  $\Gamma$ is not quasi-isometric to a point, a ray or a line, then for any order $T$ the space $(\Gamma, T)$ admits a sequence of snakes on $3$ points of width $1$.
	
	Before we prove this lemma, we formulate a discrete version
	of Lemma \ref{le:examplecircle} and Corollary \ref{cor:tripod}, which we will use also later in this section.
	\begin{lemma}\label{lem:discretecircletripod}
		\begin{enumerate}
			\item Let $M$ be a set of $n$ points of a
			circle of length $L$, with distance $L/n$ between consecutive points.
			Let $T$ be an order on $M$. Then
			$(M,T)$ admits a snake $(x_1, x_2, x_3)$ on three points,
			of diameter at least $L/2 -L/n$ and of
			width at most $L/n$.
			
			\item Consider a tripod with segments of length $L$. Let $N$ be a set of $3n+1$ points, containing the center of the tripod and $n$ points on each segment of the tripod, with distance between consecutive points $L/n$.
			Let $T$ be an order on $N$. Then $(N,T)$ admits a snake on $3$ points, of diameter at least $L-L/n$ and of width at most $L/n$.
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		The claims of the lemma can be proven in the same way as Lemma \ref{le:examplecircle} and Corollary \ref{cor:tripod}. To avoid referring to these proves, we observe that it follows from the claims of this lemma
		and this corollary. Indeed,
		if $T_1$ and $T_2$ are some orders on $M$ and $N$, consider the corresponding $\operatorname{Star}$ orders on the circle (which is the union of points of $M$ and edges between neighbouring points) and on the tripod.
		Let us call these star orders $T'_1$ and $T'_2$. By Lemma \ref{le:examplecircle} we know that $(M,T'_1)$ admits a snake on three points in $\varepsilon$-neighborhoods of two antipodal points, and by Corollary \ref{cor:tripod} we know that $T'_2$ admits a snake on $3$
		points of large diameter and of width at most $L/n$. The claims for $T_1$ and $T_2$ follow.
	\end{proof}
	
	The statement of Lemma \ref{lem:containstripod} below can be obtained
	as a corollary of a result of
	M.~Kapovich  \cite{kapovich}, see
	Appendix \ref{sec:tripods}.
	
	\begin{lemma}\label{lem:containstripod}
		There exists $C: 0<C < 1$
		such that the following holds. If $M$ is a geodesic metric
		space which is not quasi-isometric to a point, a ray or a line, then
		for any $\varepsilon>0$
		$M$ admits
		a  quasi-isometric imbedding of a
		tripod of arbitrarily large size. More precisely,
		for each $n\ge 1$ there exists $\rho_n: T_n \to M$ such that for any $x,y \in T_n$
		$$
		C d_{T_n}(x,y) - \varepsilon \le d_M(\rho_n(x), \rho_n(y)) \le 
		d_{T_n}(x,y) +\varepsilon
		$$
		One can choose $C=1/3$.
	\end{lemma}
	
	
	Now we prove Lemma  \ref{le:notrayline}.
	
	\begin{proof}
		The ``if'' part is straightforward because $\operatorname{Br}$ of a line or a ray  is $2$ and $\operatorname{Br}$ is a quasi-isometric invariant of a metric space, see Lemma \ref{le:snakesqi}.
		Now observe that
		if a geodesic metric space $M$ is not quasi-isometric  to a point, a ray or a line, then  
		the claim of Lemma \ref{le:notrayline}
		follows from Lemma \ref{lem:containstripod} and
		the second claim of Lemma
		\ref{lem:discretecircletripod}.
		Moreover, if $M'$ is not necessarily geodesic, and quasi-isometric to a geodesic space $M$, then by Lemmas
		\ref{lem:containstripod} and
		the second claim of Lemma
		\ref{lem:discretecircletripod} we know that  $M$, and hence  $M'$, admit snakes on $3$ points.
		
		
	\end{proof}
	
	
	
	A graph with $\operatorname{Br} \leq 3$ does not need to be quasi-isometric to a tree.
	Indeed, consider a ray and for each $i$ glue a circle of length $i$ at the point at distance $i$ from  the base point.
	This is a particular case of an acyclic
	gluing constructions
	(in this case applied to circles and intervals),
	considered in Section $2$ (see  Lemma  \ref{le:ORacyclicunion}). We know that in this example $\operatorname{Br} \leq 3$.
	However, groups with the word metric on their vertices satisfying  $\operatorname{Br} \le 3$ are virtually free and, hence, quasi-isometric to trees. We start with the following observation.
	
	\begin{lemma} \label{le: treesmins} 
		Let $M$ be uniformly discrete metric space which is  quasi-isometric to a tree. Then there exists an order $T$ such that
		$\operatorname{OR}_{M,T}(3) <3$.
	\end{lemma}
	
	\begin{proof}
		It follows immediately from Corollary \ref{re:anytree} and second claim of Lemma \ref{le:snakesqi}.
	\end{proof}
	
	Let $G$ be a group with a finite generating set
	$S \subset G$, and let $\Gamma(G,S)$ be the Cayley graph of $G$ with respect to $S$. 
	The {\it number of ends} of a group is the supremum, taken over all finite
	sets $V$, of the  number of infinite connected component of $\Gamma \setminus V$.
	It is not difficult to see that the number of ends does not depend on the choice of the generating set. Moreover, it is a quasi-isometric invariant
	of groups. The number of ends of an infinite  group can be equal to $1$,
	$2$ or to be infinite. The notion of ends and their numbers can be more generally defined for metric, or even more generally for topological spaces,
	but this notion has particular importance in group theory due to Stallings theorem that we will recall in the proof of
	Corollary \ref{cor:oneendedsubgroup}.
	For generalities about number of ends see for example
	\cite{drutukapovichbook}.
	
	
	\begin{lemma}[One-ended groups have snakes on $4$ points] \label{le:oneended}
		Let $G$ be a one-ended finitely generated group.
		Then for any order $T$ and any integer $N$ in $(G, T)$ there exists a snake on $4$ points  of width $1$ and of diameter at least $N$. 
		In particular, $\operatorname{Br}(G) \ge 4$.
	\end{lemma}
	
	\begin{proof}
		Assume the contrary: for some order $T$ and integer $N_0$ there are no snakes on $4$ points of width $1$ and diameter at least $N_0$.
		Take a point $x \in G$ and consider a ball $B(x,2N_0)$ of radius $2N_0$, centered at $x$. 
		Since $G$ is one-ended,
		the complement of this ball has one infinite connected component, which we denote by $Y$, and finitely many finite connected components.
		Take $N_1$ such that all these finite connected components belong to the ball $B(x,N_1)$, and let $N_2$ be the number of points in a ball of radius $N_1$.
		Note that for any two points $a, b\in G$ if 
		$d_G(a,b) > N_1$ then $b$ lies in the unique infinite connected component of $\Gamma(G,S) \setminus B(a, 2 N_0)$ (this unique connected component is a translation of $Y$).
		
		Observe that all groups of linear growth are virtually $\mathbb Z$, and thus have
		two ends. We know therefore that  the  growth function of $G$
		satisfies  $v_{G,S}(n)/n  \to \infty$,
		as follows from an elementary case of Polynomial Growth theorem, due to Justin, see  \cite{justin71}, see also \cite{MannHowgroupsgrow}).
		
		Now we fix a sufficiently large $R$. It is sufficient to assume that
		$v_{G,S}(R)/R > 2 N_2 +2$, $R>N_2$.
		Consider any two points $x, y$ of $G$ at distance  larger than $3R$.
		Observe that the ball $B(x,R)$ contains at least  $M \geq R (2N_2+2)$ points. 
		Enumerate these points with respect to order $T$:  $A_1,A_2,\dots, A_M$;
		$A_1 <_T A_2 <_T A_3 \dots <_T A_M$.
		It is clear that  $A_1$ and $A_M$ can be connected with a path of length not
		greater than $2R$.
		There are two neighbouring points  along this path such that their indices differ at least by 
		$
		\frac{M-1}{2R}> N_2$.
		We denote these points by $z$ and $w$, assuming
		that $z<_T w$.
		Observe that $G$ (in fact the ball
		$B(x,R)$) contains  at least $N_2$ elements $t$ such that $z <_T t <_T w$. 
		The infinite connected component of $\Gamma(G,S) \setminus B(z,2N_0)$ contains at least one such $t$.
		
		Observe that in this case  all elements in this connected  component are between $z$ and $w$ with respect to  $T$. 
		Indeed, otherwise in $\Gamma(G,S) \setminus B(z,2N_0)$ there exist two vertices at distance $1$ such that one of them is $T$-between $z$ and $w$ and the second is either $<_T z$ or  $>_T w$.
		This would imply that  in $G$ there is a snake on $4$ points with width $1$ and diameter $\geq N_0$, this is in  a contradiction with our assumption.
		In particular, for any point $t$ in $B(y,R)$ it holds 
		$z <_T t <_T w$.
		
		But in the same way we can prove that in $B(y,R)$ there are two points $z',w'$ at distance $1$ such that for any $t \in B(x,R)$ it holds $z' <_T t <_T w'$. 
		So we see that all points of $B(y,R)$ are between some two points
		of $B(x,R)$, and all points of $B(x,R)$ are between some two points
		of $B(y,R)$. Since these two balls do no intersect, we obtain a contradiction.
	\end{proof}
	
	
	\begin{corol}\label{cor:oneendedsubgroup}
		If a finitely generated group $G$ contains a one-ended (finitely generated) subgroup, then $\operatorname{Br}(G) \ge 4$. In particular, this inequality 
		holds for any finitely presented not virtually free group.
	\end{corol}
	\begin{proof}
		Let $H$ be a finitely generated one-ended subgroup of $G$ and $T$ an order on $G$.
		By Lemma \ref{le:oneended} we know that
		$H$ admits a sequence of snakes on $4$ points of width $1$.
		Observe that the imbedding map of any finitely generated subgroup into an ambient group is uniform.
		We can therefore apply Lemma \ref{lem:uniform} to conclude that for any order $T$ the space $G$ admits a sequence of snakes of 
		bounded width on $4$ points, and therefore that  $\operatorname{Br}(G) \ge 4$.
		
		Now we explain the second claim of the corollary about finitely presented
		groups.
		Stallings theorem (see e.g.
		\cite{drutukapovichbook})
		shows that if a finitely generated group has at least  two ends, then it is an amalgamated free product over a finite group or an HNN extension over a finite subgroup.
		For any group which we obtain in this way, if it has at least two ends, one can again write it as an amalgamated free product or an HNN extension over a finite subgroup.
		We recall that a group is said to be {\it accessible} if this process terminates. That is, if the group is accessible, then it is a fundamental group of a finite graph of
		group such that each of the vertex groups is either finite or has one end, see \cite{dunwoody}, \cite{serreTrees}.
		If all vertex groups
		are finite, the group is virtually free (see Proposition 11,
		Section 2.6 \cite{serreTrees}). 
		Since the vertex groups are subgroups of the fundamental group of a graph of groups, it is clear that if a group is accessible and not virtually free, then
		it contains a one-ended subgroup. By a result of Dunwoody any finitely presented group is accessible
		(see \cite{dunwoody}, see also \cite{drutukapovichbook}). Therefore, any finitely presented not virtually free group admits a one-ended subgroup.
	\end{proof}
	
	
	Any  infinitely presented group
	admits isometric imbeddings of arbitrary large
	cycles. 
	See Theorem A in \cite{Nima}, which states that shortcut groups
	(they are by definitions those that do not admit isometric
	imbeddings of large cycles)  are finitely presented.
	For the convenience of the reader we provide the proof of this statement (see the  first part of the following lemma and its proof).
	
	\begin{lemma}[Cycles and Trimino graphs in infinitely presented groups]\label{le:infprcycle}
		Let $G$ be a group with a finite generating set $S$. Assume that  $G$ is not finitely presented.
		
		\begin{enumerate}
			\item Then for any $N>0$ there is an isometric imbedding of a cycle with length more than $N$ into the Cayley graph $\Gamma(G,H)$.
			\item For any $N_0, N_1, N_2>0$ there exist  three isometric imbeddings of cycles $\omega_1$, $\omega_2$, $\omega_3$ in such a way that $\omega_1$ and $\omega_2$ have a common path
			$AB$ of length $\ge N_0$; $\omega_2$ and $\omega_3$ have a common path
			$CD$ of length $\ge N_0$; the ratio of the length of $\omega_1$ and that of $AB$ is $\ge N_1$; 
			the ratio of the length of $\omega_3$ and that of $CD$ is $\ge N_1$; the ratio of the lengths of $BC$ and $AD$ to that of $\omega_1$ and $\omega_2$ is $\ge N_2$.
			(see Picture \ref{pic:figureeight}).
		\end{enumerate}
		
	\end{lemma}
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.75]{pictures/figureeight2.png}
		\caption{Both black closed paths ($\omega_1$ and $\omega_3$) and the blue closed path (that is $\omega_2$) are isometrically imbedded cycles. The length of both grey-and-blue intervals are much smaller than the lengths of both black intervals, which are in its turn are much smaller than the length of both blue intervals.}
		\label{pic:figureeight}
	\end{figure}
	
	\begin{proof}
		$(1)$ Take in $G$ a relation $u = s_1s_2\dots s_n$, $s_i\in S$ such that $u$ is not a consequence of shorter relations of $G$ (since $G$ is not finitely presented, there are infinitely many such $u$).
		Consider in $\Gamma(G,S)$ a path starting at arbitrary vertex and having labelling $u$.
		This is a cycle $c_u$ with length $n$, assume it is not an isometric imbedding.
		Then some two vertices $v_i$ and $v_j$ are connected with a path $p$ such that $p$ has less edges than both parts of $c_u$ between $v_i$ and $v_j$. 
		If we keep one of these parts and replace another one by $p$, we obtain two relations in $G$ that are shorter than $u$ and imply $u$. This contradiction
		implies the first claim.
		
		$(2)$ To prove the second part of the lemma, it is sufficient to take an isometric imbedding of a long cycle, choose another, much longer isometrically imbedded cycle that has a long common word with  the first one, and finally take a third, long but much shorter than the second one isometrically imbedded cycle which has a long almost antipodal common subword with the second one.
		
		More precisely, we argue as follows. For a word $u$ in the alphabet
		$S$ we denote by $l_S(u)$ its length.
		From $(1)$ we know that there exists an infinite sequence of words $(u_i)$ in
		the alphabet $S$ such that $\lim_{i\to \infty} l_{S}(u_i) = \infty$ and for any $i$ and any $g\in G$, the path in $\Gamma(G,S)$, starting at $g$ and
		labelled $u_i$
		is an isometric image of a cycle of length $l_{S} (u_i)$.
		
		Any word $u_i$  of length greater than $2N_1N_0$ can be decomposed as follows:
		\[
		u_i = x_i a_i y_i b_i,
		\]
		where $l_S(x_i) = l_S(y_i) = N_0$ and $|l_{S}(a_i) - l_{S}(b_i)| \leq 1$. 
		In other words, we chose this decomposition in such a way that two segments labeled by $x_i$ and $y_i$ are almost opposite in the cycle labeled by $u_i$.
		
		\noindent Since the number of such pairs $(x_i, y_i)$ is bounded by $(\#S)^{2N_0}$, there exist $i$ and $j$ such that $x_i = x_j$, $y_i = y_j$,  and $l_S(u_j) > 4N_2l_S(u_i)$.
		
		\noindent In $\Gamma(G,S)$ there are vertices $A, B, C, D$ and six paths (lower indices denote starting and ending points) $p_{AB}$, $p_{BA}$, $p_{BC}$, $p_{CD}$, $p_{DC}$, $p_{DA}$ with label words 
		$x_i$, $a_iy_ib_i$, $a_j$, $y_i$, $b_ix_ia_i$, $b_j$ correspondingly (see Picture \ref{pic:figureeight}).
		
		The cycles $p_{AB}p_{BA}$, $p_{AB}p_{BC}p_{CD}p_{DA}$ and $p_{CD}p_{DC}$ have labels $u_i$, $u_j$ and a cyclic shift of  $u_i$ correspondingly, they are isometrically imbedded,
		$p_{BC}$ and $p_{DA}$ are geodesics, 
		and all the required inequalities hold.
	\end{proof}
	
	\begin{thm}\label{thm:girth4}
		Let $G$ be a finitely generated group.
		Then $\operatorname{Br}(G) \leq 3$ if and only if $G$ is virtually free.
	\end{thm}
	
	\begin{proof}
		We know that  free groups have  $\operatorname{Br}({G,T}) \leq 3$,
		for some choice of a generating set $S$
		and of order $T$. In fact, this breakpoint
		for a lexicographic order with
		respect to this generating set $S$
		(= hierarchical order on a tree) is $3$ if
		the number of generators is at least $2$, and is $2$ otherwise (if $G$ is virtually $\mathbb Z$).
		Since $\operatorname{Br}$ is a quasi-isometric invariant of groups (see Lemma \ref{le:snakesqi}), we can conclude that any virtually free group $G$ satisfies $\operatorname{Br}(G) \leq 3$.
		
		Now we have to prove the non-trivial claim of the theorem. That is, to prove that if $G$ is not virtually free and $S$  is a finite generating set, then  for any order $T$ it holds $\operatorname{Br}({G,T}) \ge 4$.
		
		\noindent The second claim of Corollary \ref{cor:oneendedsubgroup} says that if $G$ is finitely presented but not virtually free, then $\operatorname{Br}({G,T}) \ge 4$ for any order $T$. 
		
		Therefore, it is enough to assume now  that $G$ is not finitely presented. In this case we consider a mapping of the "trimino figure", guaranteed by Lemma \ref{le:infprcycle}, and search an appropriate snake in the image of this figure. (If we would have a quasi-isometric imbedding of a domino figure, 
		with  length of its edges large enough, in our group
		then we could use directly a discrete version of the domino Lemma \ref{lem:domino}. Since we do not have this information, it is more convenient for us to work with the trimino figure).
		
		Suppose that the claim we want to prove is not satisfied: $G$ is a group which is not finitely presented, $S$ is a finite generating set, $T$ is an order on $G$ and
		there are no snakes on $4$ points
		with large elongation in $(G,S)$, with respect to the order $T$. In particular, this means
		that for  some $K$ there are no snakes on $4$ points  in $(G,T)$ with width $1$ and of length greater  or equal to $K$.
		
		Apply the  second part of Lemma \ref{le:infprcycle} and consider  
		the image of the  "trimino figure" satisfying the claim of this lemma 
		for $N_0 = 10K$, $N_1 = N_2 = 10$.
		
		The cycles $p_{AB}p_{BA}$, $p_{AB}p_{BC}p_{CD}p_{DA}$ and $p_{CD}p_{DA}$,
		constructed in this lemma, are isometrically imbedded, 
		$p_{BC}$ and $p_{DA}$ are geodesics, $l_{G,S}(p_{AB}) = l_{G,S}(p_{CD}) = 10K$, $l_{G,S}(p_{BA}) = l_{G,S}(p_{DC}) > 90K$, $\min(l_{G,S}(p_{BC}),l_{G,S}(p_{DA})) > 9 l_{G,S}(p_{BA})$.
		
		\noindent From the first claim of Lemma \ref{lem:discretecircletripod}
		we know that 
		the cycle $p_{AB}p_{BA}$ contains points $v_1, v_2, v_3$ 
		such that $v_1 <_T v_2 <_T v_3$, $d(v_1, v_3) = 1$, $d(v_1, v_2) > 4 K$.
		
		\noindent We claim that in $\Gamma(G, S)$ there is a (not necessarily directed) path $p$  from $v_2$ to the vertex $C$ such that it does not intersect with the ball $B(v_1, 2K)$.
		
		\noindent First observe that $B(v_1, 2K)$ does not intersect with $p_{CD}p_{DC}$, so reaching any point of $p_{CD}p_{DC}$ is enough for our purpose.
		Since the cycle $p_{AB}p_{BA}$ is isometrically imbedded, 
		its intersection with $B(v_1, 2K)$ is a geodesic segment of length $2K$, denote this segment by $p_s$.
		It is  clear that $v_2\not\in p_s$.
		
		\noindent We also observe that $B(v_1,2K)$ can not intersect with both $p_{BC}$ and $p_{DA}$. 
		Indeed, since the cycle $p_{AB}p_{BC}p_{CD}p_{DA}$ is
		isometrically imbedded, the distance in $\Gamma(G,S)$ 
		between the subsets $p_{BC}$ and $p_{DA}$ is equal to $10K > 4K$.
		Without loss of the generality we assume that the set $B(v_1, 2K)$ does not intersects with $p_{BC}$.
		In particular, $B \not \in p_s$. 
		Then there is a path $p_{v_2B}$ from $v_3$ to $B$ such that 
		vertices of it belong to $p_{AB}p_{BA} - p_s$.
		Denote by $p$ the path $p_{g_2B}p_{BC}$, this path satisfies the property we were looking for.
		
		Consider some points
		$x_1,x_2 \in G \setminus B(v_1,2K)$. If $d(x_1, x_2) = 1$ and $v_1 <_T x_1 <_T v_3$ 
		then we claim that $v_1 <_T x_2 <_T v_3$, otherwise we find a snake on $4$ points with width $1$ and length greater than $K$.
		So, all points of the path $p$ are between $v_1$ and $v_3$, and all points of the circle $p_{CD}p_{DC}$ are greater than $v_1$ (in this sentence "between" and "greater" refers to the order $T$).
		
		But in the same way we can prove that for some point $v_4\in p_{CD}p_{DC}$ and any point $v$ of $p_{AB}p_{BA}$ it holds 
		$v_4 <_T v$, this leads to a contradiction.
	\end{proof}
	
	
	
	\section{Hyperbolic spaces} \label{sec:hyper}
	
	It is known that $\delta$-hyperbolic spaces behave nicely with respect to some questions related to travelling
	salesman problem. See Krauthgamer and Lee \cite{KrauthgamerLee}  who show that under a natural local assumption on such spaces there exists an
	approximate randomized algorithm for the travelling salesman problem. 
	
	The goal of this section is to prove Theorem \ref{thm:hyperbolicspace}, showing that there is a very efficient order on these spaces to solve universal travelling salesman  problem.
	In contrast with the context of \cite{KrauthgamerLee},  the constant in the bound for order ratio function can not be close to one even in the case of metric trees (unless the tree is homeomorphic to a ray, line or an interval, otherwise it clearly admits an imbedded tripod).
	
	
	
	As we have already mentioned, in view of the theorem of  Bonk and Schramm, our main step in the proof of Theorem \ref{thm:hyperbolicspace} is to prove it for uniformly discrete spaces that are quasi-isometric to hyperbolic spaces $\mathbb H^d$. 
	We work with a graph that corresponds to a specific tiling of $\mathbb H^d$ (also shown on Picture 
	\ref{pic:tilingwithtree2}  for this tiling in the case of $\mathbb H^2$).  
	The order we define on this graph comes from a  hierarchical order on a certain associated tree (see Picture \ref{pic:tilingwithtree2}). It is well known that the metric of any $\delta$-hyperbolic space can be well approximated by a metric of a tree. A comparison lemma (see for example \cite{ghysdelaharpe}) states that given a subset of cardinality $k$ in a $\delta$-hyperbolic space, its metric can be approximated by a metric of a finite tree
	up to an additive logarithmic error $ \delta O(\ln k)$.
	As we have already mentioned in the introduction, this metric approximation is not enough for our purposes.
	We will work with some trees,  which in some other sense approximate  our space,
	associate an order to these trees and we need to check some properties of these trees and associated orders.
	
	
	\subsection{Definitions and basic properties of $\delta$-hyperbolic spaces}
	
	A metric space is said to be  {\it $\delta$-hyperbolic} if there exists $\delta\ge 0$ such that the following holds. For any points $X_1$, $X_2$, $X_3$, $X_4$ consider three numbers
	$|X_1X_2| +|X_3X_4|$, $|X_1X_3|+|X_2X_4|$, $|X_1X_4|+|X_2X_3|$.  Here we use the notation $|X_iX_j|=
	d_M(X_i,X_j)$.
	We require that the difference between the largest of these three numbers and the middle one is at most $\delta$.
	It is well-known that for geodesic metric spaces  this definition (with appropriate choice of $\delta$) is equivalent to several other possible definitions, one of them is in terms of thin triangles. A geodesic triangle is said to be {\it $\delta$-thin} if any of its side is contained in a $\delta$-neighborhood of the union of two other sides.
	A geodesic metric space is $\delta$-hyperbolic if there exists $\delta$ such that any geodesic triangle is $\delta$-thin.
	Any Hadamard manifold (complete simply connected Riemannian manifold of sectional curvature $\le \kappa <0$) is $\delta$-hyperbolic for some $\delta$ depending on $\kappa$ (see e.g. Chapter $3$ in \cite{ghysdelaharpe}).
	In particular, a hyperbolic space $\mathbb{H}^d$ is $\delta$-hyperbolic for some $\delta>0$.
	We recall that a quasi-geodesic is an image of an interval with respect to quasi-isometric imbedding.
	If a metric space is geodesic, a basic property of a $\delta$-hyperbolic space is
	that all quasi-geodesics between points $X$ and  $Y$  lie at a bounded distance from any geodesic between $X, Y$, where this distance is estimated in terms of $\delta$ and quasi-isometry constants. This statement is sometimes called Morse Lemma.
	
	We also recall that if $M_1$ and $M_2$ are quasi-isometric geodesic  metric spaces and $M_1$ is $\delta$-hyperbolic, then there exists $\delta'$ such that $M_2$
	is $\delta'$-hyperbolic.
	For these and other basic properties of $\delta$-hyperbolic spaces see e.g. \cites{ghysdelaharpe,  BridsonHaefliger}. For optimal estimates of constants in Morse Lemma see
	\cites{schur, gouezelschur}.
	
	
	\subsection{Binary tiling}\label{subsec:bin}
	
	There is a well-known tiling of a hyperbolic plane $\mathbb H^2$, which is called
	{\it binary tiling} and also sometimes called B{\o}r{\o}czky tiling. A version of this tiling that we describe below
	exists in $(d+1)$-dimensional
	hyperbolic spaces $\mathbb H^{d+1}$, $d \geq 0.$
	
	Consider the space  $\mathbb R^{d+1}$, the coordinates of this space we denote by $(x_0, x_1, \dots, x_d)$.
	Denote by  $H^d$ its half-space with  $x_0 > 0$.
	We will use words {\it up} and {\it down} referring to the change of the coordinate
	$x_0$. We also call  changes of this coordinate {\it vertical}, and changes of all the other coordinates {\it horizontal}.
	We subdivide $H^d$ (viewed as a subset of Euclidean space $\mathbb R^{d+1})$ into cubes: for integers $k, a_1, a_2, \dots, a_d$ consider the set of points satisfying
	\[
	\begin{cases} 2^{k} \leqslant x_0 \leqslant 2^{k+1} 
	\\ 2^k a_1 \leqslant x_1 \leqslant 2^k (a_1 + 1) \\
	\dots \\
	2^k a_d \leqslant x_d \leqslant 2^k (a_d + 1) \end{cases}
	\]
	
	We call such cube  {\it a tile with coordinates} $(k,a_1,\dots, a_d)$,  and we speak about   {\it faces} of this cube
	considering to its  $d$-dimensional faces. (This tiling of  $\mathbb{R}\times\mathbb{R}_+$, that is for  $\mathbb H^2$ and   $d = 1$, is shown on Picture \ref{pic:tilingwithtree2}).
	
	We call the subset  $2^{k} \leqslant x_0 \leqslant 2^{k+1}$  {\it a layer of the level $k$}. The layer of level $k$ consists
	of $(d+1)$-dimensional cubes  of the same size, with the sides of length $2^k$. The centers of these cubes form a standard
	Euclidean $d$-dimensional lattice. 
	A tile  with coordinates $(k,a_1,\dots,a_d)$ is adjacent to  $2d$ tiles of the same level  with coordinates $(k, a_1,\dots, a_{i-1}, a_i\pm 1, a_{i+1}, \dots, a_d)$, it is also adjacent to the  one of the upper level, namely to $(k+1, [a_1/2],\dots, [a_{d}/2])$ and it is adjacent to  $2^d$ tiles of the lower level of the form $(k-1, 2a_1 + e_1, \dots, 2a_d + e_d)$, where each  $e_i$ is equal to $0$ or $1$.
	
	
	Consider the graph $\Gamma_{d+1}$, the vertices of which correspond to (centers of) tiles, and
	two vertices are joined by an edge if the tiles have an adjacent face. 
	We know that the degree of each vertex
	of  $\Gamma_{d+1}$ is equal to  $2d + 2^d + 1$. 
	
	
	Recall that the metric on $\mathbb H^{d+1}$ (the upper half-space model) is defined by putting 
	$$(ds)^2 = \frac{(dx_0)^2 + (dx_1)^2 + \dots + (dx_d)^2}{x_0^2}.
	$$
	The distance between two points
	$x = (x_0, x_1, \dots, x_d)$ and 
	$x' = (x'_0, x'_1, \dots, x'_d)$ 
	is then 
	$$
	d_{\mathbb H^{d+1}}(x,x') =2\ln
	\frac{\sqrt{\sum_{i=0}^d(x_i-x'_i)^2} +\sqrt{(x_0+x'_0)^2 + \sum_{i=1}^d(x_i-x'_i)^2}}
	{2\sqrt{x_0 x'_0}}.
	$$
	
	Observe that the   graph $\Gamma_{d+1}$ is quasi-isometric to  $\mathbb H^{d+1}$. To see this observe that
	the mapping from  $\mathbb H^{d+1}$ to $\mathbb H^{d+1}$ 
	\[(x_0,\dots,x_d) \to (2^k x_0, 2^k x_1 + 2^ka_1, \dots, 2^k x_d + 2^ka_d)
	\]
	is an isometry of $\mathbb H^{d+1}$, as follows from the above mentioned definition of the hyperbolic metric.
	Hence any tile  can be mapped to any other tile of our tiling by an
	isometry of $\mathbb H^{d+1}$. 
	Observe also that each tile has a bounded number of faces, and that each point is adjacent to a bounded number of tiles.
	Therefore, our graph is quasi-isometric to $\mathbb H^{d+1}$.
	
	\begin{definition}
		For given two vertices $\alpha$ and $\omega$ of the graph $\Gamma_{d+1}$ we consider a path of the following form.
		This path makes from $\alpha$ several (possibly none) steps   upwards to some vertex $z$.
		After this it makes several horizontal steps to some vertex $t$ that is close to $z$: close in sense that the tiles corresponding to $z$ and $t$ have a common point. 
		We require that  this segment of path has minimal length among all horizontal paths connecting $z$ and $t$ in $\Gamma_{d+1}$.  
		After that the geodesic makes
		several steps (possible none) downwards from $t$ to $\omega$. 
		We will call paths of the form described above
		{\emph{up-and-down paths}}. 
		The first the first group of edges of the path is called its \emph{upwards component}, the last group of edges is called \emph{downwards component}.
		We call an up-and-down path  which has minimal number of vertical edges \emph{  standard up-and-down path.}
		
	\end{definition}
	
	Such standard up-and-down paths are essentially unique, that is, their "upwards" and "downwards" paths
	are uniquely defined, and the only freedom is the choice of at most horizontal $d$ steps.
	
	\begin{lemma}[Standard quasi-geodesics]\label{le:up_and_down}
		There exist $A, B>0$, depending on dimension $d$, such that for any two vertices $\alpha$ and  $\omega$ of the graph $\Gamma_{d+1}$ a standard up-and-down path between them is a $(A,B)$-quasi-geodesic in $\Gamma_{d+1}$.
	\end{lemma}
	\begin{proof}
		Observe that a subpath of an up-and-down path is also an up-and-down path, and that this subpath is clearly standard if the initial one is standard. Therefore in order to show that standard up-and-down paths are quasi-geodesics,
		it is sufficient to show that for some $C$ for all $\alpha$ and $\omega$ at distance $L$ in $\Gamma_{d+1}$ such paths
		between $\alpha$ and $\omega$
		have length at most $C L$. 
		
		
		Observe that a standard quasi-geodesics between two points can be obtained in the following way:
		given two points   $\alpha$ and  $\omega$, if the level of $\omega$ is greater or equal than that of $\alpha$,  we first move
		from $\alpha$ upwards  to some point $\alpha'$ of the same level as $\omega$. Then we start moving upwards (simultaneously) from $\alpha'$ and $\omega$ until we reach for the first time the points of the same level that are close in the sense described in the definition of
		up-and-down paths, see Picture \ref{pic:path}.
		
		Let $m$ be the shortest length of a path among paths that move only horizontally from $\alpha'$ to $\omega$.
		Observe that the length of a standard up-and-down path between $\omega$ and $\alpha'$  
		is at most $d + 2\log_2 m$.
		The length of a standard up-and-down path connecting $\alpha$ and $\alpha '$ is clearly the difference of their levels $k_{\alpha'} - k_{\alpha }$.  
		This shows that, for some positive $C_1$, the length of a standard up-and-down path 
		between $\alpha$ and $\omega$ in $\Gamma_{d+1}$ is  at most $C_1 \ln (m+1) +k_{\alpha '} -k_\alpha=
		C_1 \ln (m+1) + k_\omega - k_\alpha$. 
		
		On the other hand, for any path between $\omega$ and $\alpha$ in $\Gamma_{d+1}$, observe that its length $L$ satisfies $L \ge  k_\omega - k_\alpha$.
		Observe also that in $\mathbb H^{d+1}$
		(as follows from the distance formula we have mentioned) and hence in our graph
		$L \ge C_2 \ln (m+1)$, for some $C_2>0$.
		As we have already mentioned, $\Gamma_{d+1}$ and $\mathbb{H}^{d+1}$ are quasi-isometric.
		
	\end{proof}
	
	We will also call standard up-and-down paths 
	{\it  up-and-down quasi-geodesics} and {\it standard quasi-geodesics}.
	If we relax closeness condition in the definition of up-and-down paths by saying that $z$ and $t$ are of the same vertical level ($z_0=t_0$) and at bounded distance, one can moreover show that a geodesics between any two points
	can be chosen in this standard form. But we do not need this statement.
	
	
	
	Let us say that an edge in $\Gamma_{d+1}$ is {\it vertical}, if this edge has a non-zero change of $x_0$ coordinate.
	Now consider a graph which  has the same vertex set as $\Gamma_{d+1}$ and the edges are vertical edges of
	$\Gamma_{d+1}$.
	
	\begin{rem} The obtained graph is a forest. Indeed, observe that any point has only one vertical ascendant, hence this graph does not have cycles.
		This forest consists of $2^{d}$ disjoint trees. For example, in case $d=1$, these two trees correspond to two quadrants $x<0$,
		and $x>0$, in the upper half plane $y>0$.
	\end{rem}
	
	
	Denote by $\mathcal{F}_{d+1}$ one of connected components of this graph. 
	We know that $\mathcal{F}_{d+1}$ is a tree.  We can introduce an orientation on the edges of $\mathcal{F}_{d+1}$, saying that the edges are oriented downwards. It is clear that we obtain an oriented tree. 
	\begin{rem}
		Any finite set of vertices of $\Gamma_{d+1}$ can be moved to the  component $\mathcal{F}_{d+1}$ by an isometry of $\mathbb H^{d+1}$. 
	\end{rem}
	
	By Lemma \ref{rem:goedel} if we want to  find an order on  $\Gamma_{d+1}$ with bounded order ratio function, 
	it is sufficient to find an order with this property on the connected component $\mathcal{F}_{d+1}$. 
	We will check this claim for  orders on $\mathcal{F}_{d+1}$ with convex
	branches.
	
	In this section when we discuss branches of $F_d$ we mean branches in the sense of an oriented tree:
	a branch of $\mathcal{F}_{d+1}$ corresponds to the set of tiles belonging to 
	\[
	\begin{cases}  x_0 \leqslant 2^{k+1} 
	\\ 2^k a_1 \leqslant x_1 \leqslant 2^k (a_1 + 1) \\
	\dots \\
	2^k a_d \leqslant x_d \leqslant 2^k (a_d + 1) \end{cases}.
	\]
	for some choice of integers $a_1, \dots, a_d$ and $k$.
	
	Observe that   $\mathcal{F}_{d+1}$ admits  orders $T$ such that 
	branches (that are discussed above) are convex
	with respect $T$
	(we know that such orders exist 
	by Lemma \ref{le:convex};  we also discussed such order 
	in case of finite  trees (where orientation could be chosen as direction from the root) in 
	Lemma \ref{lem:hierarchicaltrees}, and  a general can be alternatively obtained from that lemma in combination Lemma \ref{rem:goedel}).
	
	\begin{figure}\label{pic:path}
		\centering
		\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
		\newcommand{\bintile}[5]{
			\begin{scope}[shift={(#1,#2)}, scale = 0.5*#3]
			\draw[fill=#5] (0,0) -- (0,1) -- (1,1) -- (1,0) -- (1,0) -- (0,0) -- cycle ;
			\draw (0.5,0.5)
			node{\scalebox{#3}{#4}};
			\end{scope}
		}
		
		\foreach \x in {0,...,3}{
			\bintile{4*\x}{4}{8}{}{white}
		}
		
		\foreach \x in {0,...,7}{
			\bintile{2*\x}{2}{4}{}{white}
		}
		
		\foreach \x in {0,...,15}{
			\bintile{\x}{1}{2}{}{white}
		}
		
		\foreach \x in {1,...,30}{
			\bintile{0.5*\x}{0.5}{1}{}{white}
		}
		\foreach \x in {2,...,61}{
			\bintile{0.25*\x}{0.25}{0.5}{}{white}
		}
		\bintile{2}{2}{4}{$x$}{gray!40!white}
		\bintile{2}{1}{2}{}{gray!40!white}
		\bintile{3}{1}{2}{}{gray!40!white}
		\foreach \x in {0,...,3}{
			\bintile{2 + 0.5*\x}{0.5}{1}{}{gray!40!white}
		}
		\foreach \x in {0,...,7}{
			\bintile{2 + 0.25*\x}{0.25}{0.5}{}{gray!40!white}
		}
		\filldraw [fill=gray!40!white] (2,0.) rectangle (4,0.25);
		
		\draw[dotted] (0,0) -- (16,0);
		\bintile{8}{2}{4}{}{green!20!white}
		\bintile{12}{2}{4}{}{green!20!white}
		\bintile{8}{4}{8}{$z$}{green!20!white}
		\bintile{12}{4}{8}{$t$}{green!20!white}
		\bintile{9}{1}{2}{}{green!20!white}
		\bintile{12}{1}{2}{}{green!20!white}
		\bintile{9}{0.5}{1}{$\alpha'$}{green!20!white}
		\bintile{12.5}{0.5}{1}{$\omega$}{green!20!white}
		\bintile{9}{0.25}{0.5}{$\alpha$}{green!20!white}
		\draw (0,8) -- (0,9);
		\draw (8,8) -- (8,9);
		\draw (16,8) -- (16,9);
		
		\end{tikzpicture}
		\caption{Branch of a tile $x$ and standard up-and-down path between tiles $\alpha$ and $\omega$.} \label{fig:box}
	\end{figure}
	
	\begin{lemma}\label{lem:insidethebox}[Standard paths are inside branches]
		Suppose that  $\alpha$ and $\omega$ belong to some branch of the tree $\mathcal{F}_{d+1}$. 
		Then the vertices of  the standard
		up-and-down path between $\alpha$ and $\omega$ also belong to this branch.
	\end{lemma}
	\begin{proof}
		Suppose that not.
		Let $x$ be a vertex of $\mathcal{F}_{d+1}$ such that $\alpha$ and $\omega$ belong to the branch of $x$.
		Consider a standard up-and-down path between $\alpha$ and $\omega$. 
		First observe that if the first and  the last vertex  of the corresponding 
		horizontal edges (their number is at most $d$) belong to the branch of $x$, then all the others among these vertices also belong to this branch.
		Also observe that if some vertices of horizontal steps belong to the branch of $x$ and some vertices do not belong to this branch, then $\alpha$ or $\omega$ is not in the branch of $x$.
		
		Hence, the first and the last vertex of the horizontal path do not belong to the branch of $x$. 
		Then the standard up-and-down path between $\alpha$ and $\omega$ contains two vertical edges $(u_1, v_1)$ and $(v_2, u_2)$ such that $u_1$ and $u_2$ are in the branch of $x$, but $v_1$ and $v_2$ are not in this branch.
		
		We see that $u_1$ and $u_2$ are equal to $x$.  
		This means that there is a path up-and-down path from $\alpha$ to $\beta$ with fewer vertical steps than one containing $v_1$ and $v_2$, and this is a contradiction.
	\end{proof}
	
	
	\begin{lemma}[Multiplicity upper bound on paths with respect to hierarchical orders] \label{lem:multiplicity}
		Let  $T$ be an  order on  $\mathcal{F}_{d+1}$
		such that all branches of this oriented tree are convex with respect to $T$. 
		Consider distinct vertices
		$\alpha_0 >_T \alpha_1 >_T \dots >_T \alpha_n$
		and for each  $0\leqslant i < n$ assume that the sequence of tiles
		\[(u^i_0, u^i_1, u^i_2, \dots, u^i_{n_i}),
		\]
		with $u^i_0= \alpha_i$ and $u^i_{n_i} = \alpha_{i+1}$
		is a standard up-and-down quasi-geodesic, we denote it by $p_{i+1}$.
		Then each vertical edge   appears at most once among upwards edges of the above mentioned paths $p_1, \dots, p_n$
		and at most
		once among downwards edges.
		There exists $C$ depending only on $d$ such that any vertex belongs to at most $C$ paths among  $p_1, \dots, p_n$. 
	\end{lemma}
	
	\begin{proof}
		Let $e=(v,w)$ be some edge in $\mathcal{F}_{d+1}$, $w$ is one level lower than $v$.
		Suppose that this edge $e$  belongs to the upward part of the the standard up-and-down path $p_{i+1}$ between
		$\alpha_i$ and $\alpha_{i+1}$. Take $j>i$ and let us show that $e$ can not belong to $p_{j+1}$. 
		
		Indeed, $\alpha_i$ belongs to the branch of $w$. We know that the level of $v$ is higher than that of $w$, and hence
		$v$ does not belong to the branch of $w$. We can therefore
		conclude
		by Lemma \ref{lem:insidethebox} that $\alpha_{i+1}$ is also not inside the branch of $u$. 
		Since branches of the tree are convex with respect to our order, this implies that for  any $j\ge i+1$ the vertex $\alpha_j$ does not
		belong to the branch of $w$ (since by assumption of the lemma $\alpha_{i+1}$ is between $\alpha_i$ and $\alpha_j$, and since we know that $\alpha_i$ is in the branch, and $\alpha_{i+1}$ is not in the branch).
		This implies that
		any path going upward from $\alpha_{j}$ does not enter this branch. In particular, the edge $e$ is not on the standard up-and-down path from $\alpha_{i+1}$ to $\alpha_{i+2}$.
		The case of downwards paths is analogous.
		
		Now to prove the last claim of the lemma observe that it is sufficient to bound multiplicity
		of all (not necessarily  vertical) edges on our path. (Indeed, any but the very last vertex of our path belongs to an edge of our path
		starting from this vertex).
		
		Now consider some horizontal edge $f$. Observe that for each passing through this edge $f$
		there exists either some vertex $\alpha_i$ or some vertical upward edge or downward of our path at distance of at most $d$ steps,
		for this passing through $f$, from both vertices of $f$.
		Now recall that all $\alpha_i$ are distinct and that by already proven part of the Lemma the multiplicities of upwards edges is at most one.
		Since our graph has bounded degree, we can conclude that multiplicity of $f$ is bounded by some
		constant depending on $d$.
	\end{proof}
	
	
	
	Take a subset $X$ in $\mathbb H^d$ and a point $x\in \mathbb H^d$.
	Let us call {\it a cone} the union of all geodesics between
	$x$ and $y$, union is taken over all $y\in X$. We denote this cone by   $\operatorname{Cone} (x, X)$.
	This definition can be also given for an arbitrary metric space $M$
	if we consider the union of all geodesics between $x$ and $y$, but we will use  this notion only in $M = \mathbb H^d$.
	
	For a set $X$ we denote by  $X_{\varepsilon}$ its $\varepsilon$-neighborhood.
	The standard volume in the space $\mathbb H^d$ is denoted by  $\operatorname{Vol}$,
	the volume of a ball or radius $\varepsilon$ is denoted by $\operatorname{Vol}(B(\cdot, \varepsilon))$.
	Clearly, this volume does not depend on the center of the ball, and in the notation above we write $\cdot$ without introducing the notation for the center. 
	
	\begin{lemma}[Linear bound for the  volume of neighborhoods of a cone]\label{le:cone}
		There exists  $C$, depending on $d$ and $\varepsilon$,
		such that if  $\gamma\in \mathbb H^d$  is a curve of length $l$ and $x$ is a point
		of $\gamma$, 
		then the volume of the $\varepsilon$-neighborhood of the cone satisfies  ${\rm Vol}(\operatorname{Cone} (x, \gamma)_{\varepsilon}) \leqslant C(l+1)$.
	\end{lemma}
	
	\begin{proof}
		Note that adding a geodesic interval to our curve, we obtain a closed curve of length at most $2l$. 
		Therefore
		we can assume that our curve in the assumption of lemma is closed.
		
		Note also that it is enough to show the claim when $\gamma$ is a union of geodesics. 
		Indeed, given $\varepsilon'>0$ we can find a finite set of points on $\gamma$ such the following holds.
		The curve $\gamma$ belongs to the $\varepsilon'$-neighborhood
		of the union of these points and  the difference between the length of our curve and  that of the piecewise geodesic curve passing
		through these points is at most $\varepsilon'$. 
		Then the $\varepsilon$-neighborhood of $\gamma$ belongs to $(\varepsilon +\varepsilon')$-neighborhood of the piecewise geodesic closed path.
		Observe that if we bound the volume of the cone of this piecewise geodesic curve, then we can bound the volume of the cone of the initial curve $\gamma$.
		
		In the sequel we use the following notation. Given a path $
		\kappa$, its length is denoted  by  $|\kappa|$.
		Given a metric space $M$ and $\delta > 0$, we 
		will use a notation
		$$
		N(\delta, M)= \min n : \exists M'\subseteq M, \# M' = n, M \subseteq M'_{\delta},
		$$
		that is $N(\delta,M)$ is the minimal cardinality  of  $\delta$-net of $M$.
		
		Observe that for  $M \subset \mathbb{H}^d$ it holds
		$$
		\operatorname{Vol}(M_{\varepsilon}) \leq \operatorname{Vol}(B(\cdot, 2\varepsilon)) N(\varepsilon,M).
		$$
		The volume $\operatorname{Vol}(B(\cdot, 2\varepsilon))$ depends clearly only on  $\varepsilon$ and $d$.
		Therefore, is is sufficient to prove a linear upper bound for $N(\varepsilon, \operatorname{Cone}(\gamma, x))$.
		
		Replace  the condition  $x\in \gamma$ by a weaker one: $d(\gamma, x) \leq |\gamma|$.
		Analogously to the choice of a piecewise geodesic
		curve in the beginning of the proof, we 
		can choose a finite number of points in the neighborhood of $\gamma$ and assume
		that  $\gamma$ is a piecewise geodesic curve and that $x \notin \gamma$.
		In fact, note that under after such  replacement the cone can be assumed to be  in an arbitrarily  small neighborhood of the newly obtained cone. 
		Again note that if we can bound the volume of the newly obtained cone, then we can obtain bounds for the initial one.
		
		For a piecewise geodesic curve $\theta = y_1y_2\dots y_m$, (with $y_iy_{i+1}$ geodesic for all $i<m$) and $x \notin \theta$,  define {\it angular length} of $\theta$  with respect to  $x$ as
		$$
		\operatorname{Angle}_x(\theta) = \angle y_1xy_2 +\dots + \angle y_{m-1}xy_m.
		$$
		
		Decompose the curve $\gamma$ into pieces $\gamma_0, \gamma_1, \gamma_2, \dots, \gamma_n$ in such a way that  $\operatorname{Angle}_x(\gamma_0) \leq \pi$ and
		$\operatorname{Angle}_x(\gamma_i) = \pi$ for  $ 1 \leq i \leq n$.
		
		It is clear that  $\operatorname{Cone}(\gamma, x)$ is the union  of cones $\cup_i \operatorname{Cone}(x, \gamma_i)$.
		Let us show that for some  $C$, 
		depending on $d$ and $\varepsilon$,  it holds
		$$
		N(\varepsilon, \operatorname{Cone}(x, \gamma_0)) \leq C\cdot |\gamma| + 1,
		$$
		and
		for $1 \leq i \leq n, $
		$$
		N(\varepsilon, \operatorname{Cone}(x, \gamma_i)) \leq C\cdot |\gamma_i| + 1.
		$$
		Moreover, we will show that the $\varepsilon$-nets that assure the inequalities 
		above (for each $\operatorname{Cone}(x, \gamma_i)$ and hence for  their union which  is a net in $\operatorname{Cone}(x, \gamma)$) can be chosen among $\varepsilon$-nets, containing $x$. This would imply the claim of the lemma.
		
		Now we will prove our statement for $\gamma_i$, we assume that a piecewise geodesic  $\gamma_i = y^i_0\dots y^i_{k^i}$.
		To simplify the notation, in the sequel we omit sometimes the  index $i$
		and write $y_j$ and not $y_j^i$.
		The cone  $\operatorname{Cone}(x, \gamma_i)$ is a union of  $k$ triangles.
		In the hyperbolic plane  $\mathbb{H}^2$  consider a polygon (possibly degenerate) consisting of the same triangles:
		$\Phi_i = x'y'_0y'_1\dots y'_k$ is such that  the triangle $y_jxy_{j+1}$ is isometric to the triangle $y'_jx'y'_{j+1}$ for all  $0 \leq j < k$.
		(Such $\Phi_i$ is what is called a {\it net}, not to be confused with $\varepsilon$-nets).
		The  perimeter of  $\Phi_0$ can be  estimated by $4|\gamma|$.
		For all  $i\ge 1$ observe that
		$\angle {y'}^i_0x'{y'}^i_{k^i}=\pi$, 
		this implies that the union of  ${y'}^i_0x'$ and $x'{y'}^i_{k^i}$ forms a geodesic,
		and hence  $|{y'}^i_0x{y'}^i_{k^i}| \leq |\gamma_i|$, and therefore,  the perimeter $p_i$ of $\Phi_i$ is at most $2 |\gamma_i|$.
		Again to simplify the notation we omit $i$ and write $p=p_i$. 
		
		By a linear isoperimetric inequality in $\mathbb H^2$ (see e. g. \cite{teufel} for a more general version), we know that a loop  $K$, of length $L$,  in a hyperbolic plane of constant curvature $-c$ is a boundary of  a  figure of the volume (area in $\mathbb H^2$) at most  $L/\sqrt{c}$.
		
		Denote by  $\Phi'_{i}$ the set of points of  $\Phi_i$  that are at distance  greater or equal to
		$\varepsilon/2$ from its boundary.
		Consider  an $(\varepsilon,\varepsilon)$-net $\mathcal{U}_i$ of $\Phi'_i$ (such net exists by Remark \ref{re:separatednet}).
		Its cardinality $\#\mathcal{U}_i$ is at most  $\frac{p_i}{S\sqrt c}$, where  $S$ is the volume (the area in $\mathbb H^2$)  of a ball of radius $\varepsilon/2$.
		We can also find an $(\varepsilon/2)$-net  $\mathcal{W}_i$ of the boundary $\partial \Phi_i$ of  $\Phi_i$ such that $x'\in \mathcal{W}_i$ and $\#\mathcal{W}_i \leq \frac{p_i}{\varepsilon/2} +1$.
		Observe that $\mathcal{W}_i$ is a $\varepsilon$-net of $\Phi_i \setminus \Phi'_i$.
		
		We get therefore a $\varepsilon$-net $\mathcal{U}_i \cup \mathcal{W}_i$ of $\Phi_i$ of cardinality at most  $2p_i/\varepsilon + \frac{p_i}{S\sqrt c} + 1$ points (including $x'$).
		To show that there exists a  $\varepsilon$-net of $\operatorname{Cone}(x,\gamma_i)$ of the same cardinality,  observe the following.
		If $\Phi_i$ is obtained by unfolding a  cone of a piecewise geodesic curve of an angular length at most $\pi$,
		then $\Phi_i$ maps to this cone by a map that does not increase distances.
		
	\end{proof}
	
	
	
	In the claim of the Lemma below the word "convex" refers to adding paths between two points, not to the Definition \ref{def:convex} of convexity
	with respect to an order. 
	Note that in contrast to the result of \cite{benjaminironen} the number of the points (on our path) is not bounded.
	Given a set $X$ in a metric space $M$, define its partial convex hull  $\rm{PCH}(X)$ as a union of geodesics between $x$ and $y$, the union
	is taken over all $x,y \in X$ and all geodesics between  them.
	
	\begin{lemma}[Bound on the partial convex hull]\label{lem:PCH}
		\label{lm_union_geodes}
		There exists  $C$, depending on $d$ and $\varepsilon$,
		such that if  $\gamma \in \mathbb H^d$  is a curve of  length $l$, then
		the volume of $\varepsilon$-neighborhood of its partial convex hull satisfies
		${\rm Vol} ({\rm PCH} ({\gamma})_{\varepsilon}) \leqslant C l+C$.	
	\end{lemma}
	
	
	\begin{proof}
		Fix some point $x\in \gamma$. Since  $\mathbb H^d$ is $\delta$-hyperbolic, there exists $\delta>0$
		Observe that by $\delta$-hyperbolicity of $\mathbb H^d$ we know that the ${\rm PCH}(\gamma)$ lies in the $\delta'$ neighborhood
		of the cone $\operatorname{Cone} (\gamma,x)$, for some $\delta'>0$. Therefore, its $\varepsilon$-neighborhood belongs to $\delta'+\varepsilon$-neighborhood
		of the cone, and the claim follows therefore from the previous lemma.
	\end{proof}
	
	Now to prove Proposition \ref{prop:hyperbolic} below, 
	we take a finite subset $X \subset \mathcal{F}_d \subset \Gamma_d \subset \mathbb H^d$, of cardinality  $n+1$,  and assume that this subset admits a path of length $L$ in $\Gamma_d$
	visiting its points.
	Let number the  points of the set $X$ with respect to  an order
	on $\mathcal{F}_d$, satisfying the assumption of the proposition: 
	$\alpha_0 >_T \alpha_1 >_T \dots >_T \alpha_n$. We want to show that
	$$
	\sum_{i=1}^{n} d_{\Gamma_d}(\alpha_{i-1}, \alpha_i) \le CL,
	$$
	for some constant $C$ depending only on $d$.
	
	Connect $\alpha_i$ with $\alpha_{i+1}$, $0 \le i \le n$ by geodesic in
	$\Gamma_d$, and consider the union of these
	geodesics.
	We estimate the total number of vertices on 
	the obtained path
	using Lemma \ref{le:cone}
	about cones and its corollary (Lemma \ref{lem:PCH}) to partial convex hulls. 
	
	Now for each $i: 0 \le i \le n-1$ consider a standard up-and-down path $\gamma_i$ between $\alpha_{i}$ and $\alpha_{i+1}$:
	\[\alpha_i = u^i_0, u^i_1, u^i_2, \dots, u^i_{n_i} = \alpha_{i+1}.
	\]
	From Lemma \ref{le:up_and_down} we know that these paths are quasi-geodesics in $\Gamma_d$ with some constants $A,B$. 
	Hence they are also  quasi-geodesics in $\mathbb H^d$ with some constants $A'$ and $B'$ (depending only on  $d$).
	
	As we have already mentioned, Morse Lemma for $\delta$-hyperbolic spaces implies that there exist $C_1$ depending on $A'$ and $B'$ (and hyperbolicity constants of the space $\mathbb H^d$), see e.g. Chapter 5 of \cite{ghysdelaharpe}
	such that any $(A',B')$-quasi-geodesic between two points
	belongs to a $C_1$-neighborhood of any geodesic between these points. As we have mentioned, $A',B'$, and therefore also $C_1$ depend only on $d$.
	
	Recall that we are discussing a finite set $X$, which admits a path of length $L$ in $\Gamma_d$ visiting all its points. 
	Since the vertices of $\Gamma_d$ form a uniformly discrete space,  quasi-isometrically imbedded in $\mathbb H^d$, there is a piecewise geodesic path $\gamma_*$ of length at most $C_2 L$ in $\mathbb H^d$, visiting all points of $X$. Here the constant $C_2$ depend only on $d$. The constants $C_3,C_4 \dots$
	that we will choose later in the proof will also depend only on $d$.
	
	The geodesics joining in $\mathbb{H}^d$ points $a_i$ and $a_{i+1}$ lie in ${\rm PCH}(\gamma_*)$. 
	All tiles corresponding to $u^i_j$ are in the $C_3$-neighborhood of ${\rm PCH}(\gamma_*)$, for some $C_3$. Constant  $C_3$ can be chosen as the sum of $C_1$ and the diameter of a tile.
	
	We can therefore apply Lemma \ref{lem:PCH} for $\gamma_*$ and $\varepsilon = C_3$ and find a constant $C_4$
	depending only on $d$ such that all tiles containing corresponding points $u^i_j$
	belong to a subset of volume at most $C_4(L+1)$ of $\mathbb H^d$. 
	
	Let $C_5$ be the volume of a tile in $\mathbb H^d$ (since  the tiles are isometric they have the same volume). 
	We can conclude that
	the number of distinct vertices in $\Gamma_d$ in the union of quasi-geodesics $\gamma_i$ is at most  $\frac{C_4}{C_5}(L+1)$.
	
	The observation above holds for any order, we did not use that branches are convex with respect to $T$. However, in the following observation this assumption  for $T$ is essential. 
	We recall that by Lemma \ref{lem:multiplicity} there exists a constant $C_6$ such that any point appears
	at most $C_6$ times among vertices of quasi-geodesics $\gamma_i$. This implies that  $l_{T}(X)$ 
	is at most
	$\frac{C_4C_5}{C_5}(L+1)$. 
	
	Since $\Gamma_d$ is a disjoint union of trees, it is sufficient to check the claim below for these trees, 
	so we have proved
	
	\begin{prop}\label{prop:hyperbolic}
		There exists a constant $K>0$ depending on $d$  such that
		for any
		order $T$ on $\Gamma_d$, with branches of all trees
		of $\Gamma_d$
		convex with respect to $T$,
		it holds  $\operatorname{OR}_{\Gamma_d, T}(k) \leqslant K$,  for any $k$.
	\end{prop}
	
	
	As we have already mentioned, the result of Bonk and Schramm \cite{bonkschramm}
	implies  that any  $\delta$-hyperbolic metric space of bounded growth  imbeds quasi-isometrically to $\mathbb H^d$ for some $d$.
	We do not explain the definition of bounded growth but recall that any graph of bounded degree is a particular case of this definition.
	
	Observe that if a space imbeds  quasi-isometrically into $\mathbb H^d$, this space also imbeds quasi-isometrically to $\Gamma_d$. Therefore, combining   Proposition \ref{prop:hyperbolic}  with
	Lemma \ref{le:pullback} we obtain
	
	\begin{thm} \label {thm:hyperbolicspace}
		Let $M$  be a $\delta$-hyperbolic graph of
		bounded degree. 
		Then there exists  an order $T$ and
		a constant $C$
		such that for all $k$
		$$
		\operatorname{OR}_{M,T}(k) \le C.
		$$
	\end{thm}
	
	In the formulation of the Theorem above we use the convention that edges of the graph
	are of length $1$, and we can assume that they are included in the metric space $M$.
	
	\begin{rem}\label{rem:graphsqiHd}
		The order breakpoint and the constant $C$ in Theorem \ref {thm:hyperbolicspace} of $\Gamma_d$ tend to infinity as $d$ tends to infinity. 
		To see this it is enough to consider uniform imbeddings
		of a cube of dimension $d-1$ into $\Gamma_d$.
		For more on this see Section $6$
		in \cite{ErschlerMitrofanov2}.		
		
		The same claim is clearly true for any uniformly discrete metric space, quasi-isometric to $\mathbb H^d$. We remind that
		for each $d\ge 1$ there exist finitely generated groups quasi-isometric to $\mathbb H^d$ \cites{Borel63, gromovpiatetski}, but it seems to be a challenging open problem  \cite{gromovMarkov} to understand 
		whether one can find more elementary constructions of such groups (for
		$d\ge 3$).
	\end{rem}
	
	\begin{remark}\label{rem:ahmedov}
		The properties of $OR$ function we consider are not to be confused with "travelling salesman property"  of a very different nature,  introduced for finitely generated groups by Akhmedov 
		\cite{akhmedov1}, see also \cite{akhmedov2}, \cite{guba}. 
		While  hyperbolic groups (and spaces) are best possible cases for our question about $\operatorname{OR}(k)$, 
		hyperbolic groups are among worst possible case ("TS" groups)
		for the question of Akhmedov, see \cite{akhmedov1}.
		We also  mention that all amenable groups are not so bad for that problem (they are not "TS" groups, see \cite{guba} ,
		where this observation is attributed to Thurston), while for the questions of our paper there is no visible  relation with amenability. Both amenable and not amenable groups can  have finite $AN$-dimension (and thus satisfy logarithmic bound for $OR(k)$, as we will show in   in \cite{ErschlerMitrofanov2}, see Theorem I)  and there are numerous classes of amenable groups satisfying  $OR(k)=k$ ( \cite{ErschlerMitrofanov2}, Corollary 6.8).	
	\end{remark}
	
	
	\subsection{Further examples with bounded order ratio functions} \label{Subsection:nothyperbolic}
	
	A metric space with bounded order ratio function does not need to be hyperbolic. We have seen already in Section $4$ that cactus graphs satisfy $\operatorname{OR}(k)\le 2$. Choosing circles of unbounded length we clearly can obtain not-hyperbolic examples (in particular among graphs of bounded degree). 
	
	A wider family of examples is provided by the following lemma.
	\begin{lemma} \label{lem:severalcopies}
		Let $M$ be a metric space that admits an order $T$ such that $\operatorname{OR}_{M,T}(k) \le C$ for all $k\ge 1$.  Let $\Omega$ be a subset of $M$.  Given $s\ge 1$, denote by $M_s(\Omega)$ a space which is obtained from $s$ copies of $M$ by gluing them over $\Omega$. 
		For any $s\ge 1$ the space $M_s(\Omega)$ admits an order $T_s$ such that $\operatorname{OR}_{M_s(\Omega),T_s}(k) \le sC + s-1$.
	\end{lemma}
	
	
	\begin{proof}
		For $1 \leq i \leq s$ we denote by $M_i$ the subset of $M_s(\Omega)$ that corresponds to the $i$-th copy of $M$.
		It is clear that $\Omega$ is the intersection of all $M_i$.
		We define the order $T_s$ on $M_s(\Omega)$ as follows: 
		\begin{enumerate}
			\item if $x\in M_1$, $y\not \in M_1$, then $x<_{T_s}y$;
			\item if $x, y \not \in \Omega$, $x\in M_i$, $y\in M_j$ and $i < j$, then $x<_{T_s}y$;
			\item if $x,y\in M_i$ for some $i$ and $x<_T y$ (we compare these points as points of $M$), then $x<_{T_s}y$.
		\end{enumerate}
		
		Consider a finite set $X\subset M_s(\Omega)$ and denote by $L$ the length $l_{\operatorname{opt}}(X)$ of the shortest path  in $M_s(\Omega)$.
		We want to estimate $l_{T_s}(X)$ from above.  
		Let $X_1$ be the intersection $X\cap M_1$ and let $X_i$ for $i > 1$ be the intersection $X\cap (X_i \setminus \Omega)$.
		Without loss of generality we assume that all $X_i$ are non empty. 
		
		It is clear that 
		$$
		l_{T_s}(X) \leq (l_{T_s}(X_1) +\dots +l_{T_s}(X_s)) + (t_1 + \dots + t_{s-1}),
		$$
		where $t_i$ is the distance between the maximal point of $X_i$
		with respect to $T_s$ and the minimal  point of $X_{i+1}$
		with respect to $T_s$.
		
		For any $i$ we can estimate $l_{\operatorname{opt}}(X_i) \leq L$, $l_{T_s}(X_i) \leq C l_{\operatorname{opt}}(X_i) \leq CL$, $t_i \leq L$.
		Hence $l_{T_s}(X) \leq (sC + s-1)L$.
	\end{proof}
	
	We know that the assumption of Lemma \ref{lem:severalcopies} is in particular
	verified  when $M$ is uniformly discrete hyperbolic space of bounded geometry. This assumption holds also  for any metric tree (not necessarily uniformly discrete).
	Picture  \ref{pic:nothyperbolic} shows an example of the construction above, where $M$ is a tree, $s=2$. Observe that the space in this example is not hyperbolic. For all $r \in \mathbb{N}$ it contains an isometric copy of a square of side $2^r$.
	
	\begin{figure}[!htb] 
		\centering
		\includegraphics[scale=.55]{pictures/nothyperbolic.png}
		\caption{An example of a not hyperbolic space with bounded order ratio function.
		}
		\label{pic:nothyperbolic}
	\end{figure}
	
	Other particular cases of the spaces from the claim of Lemma \ref{lem:severalcopies} are spaces, obtained by gluing together two (or several copies) of graphs $\Gamma^d$ (which is quasi-isometric to $\mathbb H^{d}$), by a horizontal  subspace in
	the half-space model
	of dimension $d-1$. If we glue two copies of $\mathbb H^2$ by this horizontal subspace, we obtain a space $M$, satisfying $\operatorname{ORI}_M \le {\rm Const}$, but admitting contractible quasi-isometrically imbedded circles of
	arbitrarily large length.
	
	We have discussed not hyperbolic
	spaces with bounded function $\operatorname{OR}(k)$. It seems interesting to understand whether this can happen
	for finitely generated not hyperbolic groups. In this context we ask the following questions: what is the asymptotics of $\operatorname{OR}(k)$
	for solvable Baumslag-Solitar groups? The same question can be asked for $\mathbb{Z}\wr \mathbb{Z}/2\mathbb{Z}$.
	
	