TITLE: Probability of being first, second, or third in a contest
QUESTION [0 upvotes]: Question in the book:
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Michelle wins one of
these prizes if she is one of the contestants?
My work:
Being first: 1/100
Being second: 1/99
Being third: 1/98
So total probablity is 1/100 + 1/99 + 1/98
Book Answer:
But the book says 3/100, I understand why kind of, but I don't understand what I did wrong in my work?

REPLY [0 votes]: $1/99$ is the probability of coming second given that she didn't come first.  The chance of not coming first is $99/100$ so the chance of coming second is $$\frac1{99}\frac{99}{100}=\frac1{100}$$


TITLE: Showing if $f_n \to f$ uniformly and each $f_n$ has at most $10$ discontinuities, then so does $f$
QUESTION [1 upvotes]: Suppose that $f_n:[a,b] \to \Bbb R$ and $f_n$ uniformly converges to $f$ as $n$ goes to infinity. How to prove that if each $f_n$ has at most ten discontinuities (the discontinuities for each $f_n$ might be different) then $f$ also has at most ten discontinuities?

REPLY [3 votes]: If for some $x_0\in [a,b]$, there is a subsequence of $\{f_n\}$ with elements functions which all continuous at $x_0$, then $f$ is continuous at $x_0$. This is due to the following fact:
If $\{g_n\}$ converges uniformly to $g$, and all the $g_n$'s are continuous at $x_0$, then $g$ is also continuous at $x_0$.
So if $f$ is discontinuous at $x_0$, then all but finitely many $f_n$'s are discontinuous at $x_0$.
Assume now that $f$ is discontinuous at the distinct points $x_1,\ldots x_k$, with $k>10$.
Then for the $x_j$, all but finitely many $f_n$'s are discontinuous at it. Say that for $n\ge n_j$, the $f_n$ is discontinuous at $x_j$. Setting $n_0=\max\{n_1,\ldots,n_k\}$, we have that all the functions $f_n$, for $n\ge n_0$, are discontinuous at $x_1,\ldots,x_k$.
This is a contradiction, since these $f_n$'s are discontinuous to more than $10$ points.


TITLE: Prove $ n!$ is divisible by the number of its positive divisors
QUESTION [6 upvotes]: Prove that for all positive integers $ n$ different from
  $ 3$ and $ 5$, $ n!$ is divisible by the number of its positive
  divisors.

I tried some things,such as  the number of divisors of $ n!$ is just $ \prod_p a_p$ where $ a_p=1+\sum_{j\ge 1}\left\lfloor\frac n{p^j}\right\rfloor\le\min\left(1+\frac n{p-1},n\right)$.

REPLY [1 votes]: So there is some challenge in this, but I imagine talented undergraduates can solve it.  However, there are some subtleties that I would not expect a very talented high school student to navigate.
As remarked above, the goal is to show that the product of $\pi(n)$ many terms $1+ v_p(n!)$ divides $n!$. Each of these terms is at most $n$, and if they had distinct values, the high school student (and every one else) would be close to done.  However, for $p \gt n/2$, these values are $2$, and there are other duplications as well.  So we must go further.
As remarked in another post $1+ v_p(n!)$  is bigger than $n/p$.  For small primes $p$, we can bound the quantity by $n/(p-1)$ and $n/p$; as long as $p$ is at most $\sqrt{n}$ this guarantees that different $p$ give different values for $1+v_p(n!)$. For larger $p$ though we have the following. If $p \gt 2$ and $q$ is the next larger prime (so $q \geq p+2$) and $k=v_p(n!)=v_q(n!)$ then for some a and b less than $p$ we have $n=kp+a=kq+b\gt kp +2k+b$ giving that $p \gt 2k$ and $2k^2\lt n$, so we get distinct values for primes $p$ below $\sqrt{2n}$. So of the terms $1+v_p(n!)$, one of them is in $[n/2,n]$, one in $[n/3,n/2]$, and the rest are at most $n/4$ and when $p \lt \sqrt{2n}$ are distinct.
The primes $p$ which share a value $k=1+v_p(n!)$ now lie in the interval $I_k=[n/k,n/(k-1))$.  If we had the following property (B), that the number of primes in $I_k$ was more than the number of primes in $I_{k+1}$ for enough $k$ ($k$ from 1 to about $\sqrt{n/2}$), then we could finish the problem as follows.
Because of (B), we can collect the repeating values and group them into factorial products, so so many groups form $k!$, so many form $(k-1)!$, all the way down to $2!$. How many? Well, there are $\pi(n)- \pi(n/2)$ many instances of $2$, and about $\pi(n)-\pi(\sqrt{2n})$ terms total, so when we sum up all the largest terms in each group we get $t=2\pi(n) - \pi(n/2) - \pi(\sqrt{2n})$.
Why sum up these largest terms?  Because the product of the factorials divides $t!$, as can be seen from multinomial coefficients. If we can get $t $ less than $1+v_3(n!)$, then we have one grouping that divides $(n/2)!$, another (the distinct terms starting with $p=5$) dividing $(n/4)!$, and the term for $p=2$ which is greater than $3n/4$ for large $n$, and the total product then divides $n!$.  For small $n$ other verification is needed, but for large $n$ we can get $t$ small enough to make things work.
So given (B), we can have a talented high schooler prove the result.  Unfortunately (B) is not always true, and it is not clear if (B) is "true enough" to tweak the argument and finish the proof.  We could use some multiples of $1+v_p(n!)$ (even two at a time to use coprimality to an advantage) to get rid of holes and use a (B') assumption, but I am not seeing how weak (B') needs to be, and I am not seeing a nice assignment of more than $O(\sqrt{n})$ values to get around this.  Hopefully someone else will see a nice assignment.
Gerhard "Some Assignments Are Rather Hard" Paseman, 2018.05.24.


TITLE: Interception Relation
QUESTION [0 upvotes]: Given two points $(a, b, c), (x, y, z)\in R^3$, we say that the midpoint of these two points is the point with coordinates $(\frac{a+x}2 , \frac{b+y}2 , \frac{c+z}2)$.
Take any set S of nine points from $R^3$ with integer coordinates. Prove that there must be at least one pair of points in S whose midpoint also has integer coordinates.
Can someone help me with the prove?

REPLY [0 votes]: Given a binary string of length 3, say $\epsilon_1 \epsilon_2 \epsilon_3$, $\epsilon_j \in \{0, 1\}$, we say that a point $(x_1,x_2,x_3)$ with integer coordinates is of type $\epsilon_1 \epsilon_2 \epsilon_3$ if $x_j \equiv \epsilon_j\ (\mathrm{mod}\ 2)$. For example, the point $(3,42,-16)$ is of type $100$.
Since there are only $2^3 = 8$ different types, at least two of your nine numbers must share the same type. The midpoint of these two has integer coordinates.


TITLE: How can I solve for an unknown vector given an otherwise known cross product?
QUESTION [4 upvotes]: Given known vectors $\vec a$ and $\vec b$, is it possible to solve for $\vec u$, given the following equation?
$\vec a \times \vec u = \vec b$
So far I have found that the following must be true
$\vec u = \vec c + \lambda \hat a$
where

$\vec c = \frac{b}{a} (\hat b \times \hat a) = \frac{\vec b \times \vec a}{a^2}$
$\lambda \in \Bbb R $

but I'm unsure whether it is possible to further determine $\vec u$, since any vector $\vec u$ such that
$(\vec u \cdot \hat c) \cdot \hat c = \vec c$
should do the trick, which means that $\lambda$ can take any value.
However, when developing the cross product $\vec a \times \vec u = \vec b$ by hand, one can reach a matricial equation of the form
$ A \cdot (\vec x)^t = (\vec b)^t $
where $A$ is a matrix with coefficients from $ \vec a $ and which is solvable with at least one method, one example thereof being
$ (\vec x)^t = A^{-1} \cdot (\vec b)^t $
Am I doing something wrong? Or is this equation truly not solvable using vectorial math?

REPLY [2 votes]: Cross product can be written as matrix multiplication:
$$\pmatrix{b_1 \\ b_2 \\ b_3} = \vec{b} = \vec{a} \times \vec{u} = \pmatrix{a_1 \\ a_2 \\ a_3} \times \pmatrix{u_1 \\ u_2 \\ u_3} = \pmatrix{a_2u_3 - a_3u_2 \\ a_3u_1 - a_1u_3 \\ a_1u_2 - a_2u_1} = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0\end{bmatrix}\pmatrix{u_1 \\ u_2 \\ u_3} $$
Now this can be solved for $\vec{u}$ like any linear system, e.g. with row reduction. WLOG assume $a_3 \ne 0$.
$$\left[
\begin{array}{ccc|c}
  0 & -a_3 & a_2 & b_1 \\ a_3 & 0 & -a_1 & b_2 \\ -a_2 & a_1 & 0 & b_3
\end{array}
\right] \sim \left[
\begin{array}{ccc|c}
  1 & 0 & -\frac{a_1}{a_3} & \frac{b_2}{a_3} \\ 0 & 1 & -\frac{a_2}{a_3} & -\frac{b_1}{a_3}\\ 0 & 0 & 0 & \frac{a_1b_1 + a_2b_2}{a_3} + b_3
\end{array}
\right] $$
We see that a solution exists if and only if $b_3 = -\frac{a_1b_1 + a_2b_2}{a_3}$ and in that case the it is given by:
$$\vec{u} = \pmatrix{u_1 \\ u_2 \\ u_3} = \pmatrix{\frac{b_2}{a_3} \\ -\frac{b_1}{a_3} \\ 0} + \lambda \pmatrix{a_1 \\ a_2 \\ a_3} = \pmatrix{\frac{b_2}{a_3} \\ -\frac{b_1}{a_3} \\ 0} +\lambda \vec{a}$$
for some $\lambda \in \mathbb{R}$.


TITLE: A pattern involving '8' and square roots
QUESTION [4 upvotes]: If we take the square root of 012345, we get 111.108055514. The square root of 012345678901234567 approximates to 111111110.611. We're taking the square roots of numbers following a pattern, the numbers 0-9 repeating with an even number of digits before any decimal (I'm counting it starting with 0). Note that we get very close to 111.111 repeating, or a variation with decimal moved.
Now let's omit '8' from the pattern. square root(123456790123456) = 11111111.1111. That's a far closer approximation. I observe, ((10)^nth/9)^2, where n is positive will be represented by the digits 0-9 repeating omitting 8.
This pattern will only appear significant in a number base 10 system, without a number base 10 1234567890123 doesn't appear to be a number with a pattern. What is the significance of '8' in 100/81? in 1000/81?

REPLY [5 votes]: Consider what happens when you square $1111\cdots1.$ Near the beginning (where the most-significant digits are) you are adding the string of ones to itself shifted by every place:
\begin{align*}
111111111111\cdots &\\
+ 11111111111\cdots &\\
+ 1111111111\cdots &\\
+ 111111111\cdots &\\
+ 11111111\cdots &\\
\vdots&\\
= 123456790123 \cdots &
\end{align*}
Notice that after the 10th digit, the sum carries; the carry causes the 9 in the previous digit of the sum to carry, so that instead of
$$789(10)(11)(12)\cdots$$
you get
$$790123\cdots.$$
After another ten digits the sum starts carrying twice instead of once, so that the pattern repeats, etc.


TITLE: Examples of Hyperbolic Groups
QUESTION [2 upvotes]: We already have the Milnor-Svarc Lemma, which tells us that if a group acts "nicely" on a space, then the Cayley graph of the group is quasi-isometric to the given space.  This gives us a lot of hyperbolic groups, because hyperbolic manifolds are ubiquitous.
Edit: For example, the MS-lemma tells us that the surface groups are hyperbolic.
Edit:
Of course, in the large-scale geometric setting here, we consider two metric spaces equivalent if they are quasi-isometric.
There is a result of Kharlampovich and Myasnikov, in "Hyperbolic groups and free constructions" which allows us to construct a "separated HNN extension" of two hyperbolic groups to yield a hyperbolic group.
What are some other ways of constructing word-hyperbolic groups?

REPLY [0 votes]: A similar question was asked on MathOverflow recently. The answer of Genevois is pretty comprehensive.


TITLE: Is there an example to show the Hodge decomposition fails on non-compact case?
QUESTION [0 upvotes]: The theorem of Hodge decomposition is on the compact Kahler manifold, is it generally true for the non-compact kahler manifold or are there examples to show the failure?
Here is my Hodge decomposition theorem:

REPLY [3 votes]: Now that it's clear what you want, let me turn my (and various other) comment(s) into an answer.  Any manifold with odd first Betti number will give a counterexample; $\mathbb{C}^*$ is the simplest such. Let me add that the reason why there was some confusion initially, is that "Hodge decomposition" can mean different things to different people. It could mean the $(p,q)$ decomposition, or it could simply mean the fact that on a compact manifold every cohomology class has a unique harmonic representative. BM's answer seems to be using this last interpretation. For noncompact manifolds, things can go badly wrong with this as well.
I'm just leaving this for the record.


TITLE: positive values of $a$ for which the equation $\lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution
QUESTION [2 upvotes]: The  positive values of $a$ for which the equation $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution is,
where  $\lfloor x \rfloor = $ floor function of $x$
Options
$(a)\;\;(0,1)$
$(b)\;\; (1,2)$
$(c)\;\; (0,2)$
$(d)$ None of these
$\bf{My\; Try::}$ Using the formula $x-1<\lfloor x \rfloor \leq x.,$ we get $\displaystyle (x+a)-1< \lfloor x+a \rfloor \leq (x+a)$
Now Given $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right).$ So $\displaystyle (x+a) - 1<\sin \left(\frac{\pi x}{2}\right)\leq (x+a)$
Now we will solve $\displaystyle (x+a)-1<\sin \left(\frac{\pi x}{2}\right)$ and $\displaystyle (x+a)\leq \sin \left(\frac{\pi x}{2}\right)$ for Common values of $x$
But I Did Understand How can I solve it
Help me
Thanks

REPLY [0 votes]: observe that $\sin(\frac{\pi x}{2})$ only takes integer values for integer $x$'s, hence we can restrict our attention to $x \in \mathbb{Z}$ and so the floor function becomes obsolete. Now, the situation is even better, since there aren't plenty of integer values $\sin$ can take - it's just $-1,0$ and $1$, therefore, since it's a test question the simplest way would be to check all the posibilities. Let me do point a) and I hope you'll finish the rest by yourself following what I do.
first let $a = 0$. we want an integer $x$ such that $x = \sin(\frac{\pi x}{2})$. all values we need to check are $1,0,-1$. $x=1$ works, so we're done, that can't be the answer.
we could now cross answer a) as a wrong one, but nevertheless let me do the case of $a=1$ so that its more instructive
if $a = 1$ then we're looking for $x$ s.t.
$$ x+ 1 = \sin(\frac{\pi x}{2})$$
because of what've said about the range of $\sin$ we only need to check $x = 0$, $x = -1$ and $x = -2$.
$x = 0$ won't work since it leads to
$$1 = 0$$
$ x = -1$ doesn't work either cause we get
$$ 0 = -1$$
finally $x = -2$ gives
$$-1 = 0$$ which isn't true either, so for $a=1$ there are no solutions to your equation.


text:
\section{Values of the $(\frakg_2,V)$-invariant on closed planar 3-nets}
In this section, we show how the value of \It on a closed planar 3-net can
be calculated recursively.

The following lemma assures that it is sufficient to consider connected
3-nets.

\begin{lem}
If a closed 3-net $N$ is equivalent to a 3-net consisting of two closed
3-nets $N_1$ and $N_2$ with $N_1 \subset \R^-\times \R^2, N_2\subset
\R^+\times \R^2$, then
$$\I(N) = \I(N_1)\cdot \I(N_2).$$    \hfill $\Box$
\end{lem}

As, by definition, every planar 3-net is  equivalent to a 3-net contained in
$\R^2\times \{0\}$ with only upward pointing vectors assigned, it is enough
to calculate \It for these. Therefore, we assume that all planar 3-nets in this
section are of this type.

\begin{df}
Let $N$ be a planar 3-net. A {\em mesh} of $N$ is the closure of a
connected component of $(\R^2\times \{0\})\backslash N$. A {\em $n$-mesh}
is a mesh with $n$ trivalent vertices in the boundary.
\end{df}

We will show how in any non-empty connected closed planar 3-net the number
of meshes can be reduced without changing the value of the invariant. As
we know that $\I$(empty 3-net) = 1 (the empty 3-net is the unity in \topf
\hspace{0mm} and $\I$ is a monoidal functor),
this will allow us to calculate the
invariant of a closed planar 3-net recursively.

\begin{prop}
Let $N$ be a non-empty connected planar 3-net with $m$ meshes. Then there
exist closed planar 3-nets $N_1,\ldots , N_r$  (not necessarily connected)
with fewer than $m$ meshes and coefficients $\lambda_1,\ldots,\lambda_r \in
\C$ such that $\I(N)=\sum_{i=1}^r\lambda_i\I(N_i)$.
\end{prop}

{\bf Proof:} The idea is to cut out a mesh and replace it by a linear
combination of pieces that lead to 3-nets with fewer meshes.

Thanks to the following lemma, the mesh we want to cut out can always be
chosen to be a simply connected $n$-mesh with $n\leq 5$.

\begin{lem}\label{km}
Let $N$ be a planar non-empty closed connected 3-net. Then $N$ has at least
one simply connected $n$-mesh with $n\leq 5$.
\end{lem}

{\bf Proof of the lemma:} If $N$ is an embedded $S^1$, $N$ has a bounded
0-mesh, and so in this case, the lemma holds.\newline
Let $\sharp$ denote ``number of\," and let $N$ be a 3-net without 0-mesh.

\underline{Observation 1}: $\sharp$ vertices of $N - \sharp$ edges of $N +
\sharp$ meshes of $N =2$ (Euler characteristic of $S^2$).

\underline{Observation 2}: If we assign to each mesh $M$ the appropriate
part of the contributions of its vertices, its edges, and  its region, i.e.\
$$\chi_M:=\left\{\begin{array}{l}\quad\frac{1}{3}\: \sharp\mbox{ vertices of
}M\\ +\frac{1}{3}\: \sharp\mbox{ vertices of }M\mbox{ for  which all adjacent
edges belong to } M\\ +\frac{1}{3}\: \sharp\mbox{ vertices of }M\mbox{ that
belong only to }M\\ -\frac{1}{2}\: \sharp\mbox{ edges of }M\\
 -\frac{1}{2}\: \sharp\mbox{ edges of }M\mbox{ that belong only to } M\\
 +1,
\end{array}\right.$$ $$\mbox{then } \sum_{M \mbox{ {\scriptsize mesh of} } N}\chi_M =
2.$$

\underline{Observation 3}: For a simply connected $n$-mesh $M$ of $N$, we
have $\chi_M= 1-\frac{1}{6}n$.

\underline{Observation 4}: If the unbounded mesh $M'$ of $N$ does not
contain an edge belonging to $M'$ only, then $\chi_{M'}\leq 1$.

If $N$ does not contain an edge that belongs to one mesh only, the lemma
is a consequence of observations 2, 3 and 4\footnote{Note that any bounded mesh $M$ that does not
contain an edge belonging only to $M$ is simply connected.}.

Now suppose that there are edges $e_1,\ldots, e_k$ such that $e_j$ belongs
 to only one mesh $M_j$. Note that forgetting such an edge $e_j$ would split
 the 3-net $N$ into two connected components $N_{j1}$ and $N_{j2}$. As there are only
finitely many edges $e_j$, there is  an edge $e_i$ for which at least for one
$k\in \{1,2\}$ $N_{ik}$ satisfies the following properties:\newline
($i$)$\,\quad$ $N_{ik}$ does not contain an edge that belongs to only one mesh;\newline
($ii$)\quad any mesh $M$ of $N$ that is also a mesh of $N_{ik}$ is bounded.\newline
The sum $\sum \chi_M$ over all meshes $M$ of $N$ mentioned in ($ii$) is greater than
0 (look at $\sum \chi_M$ with $M$ considered as mesh of $N_{ik}$, use observation 4, and subtract
$\frac{1}{6}$ for the influence of $e_i$),
and thus observation 3 guarantees that at least one of
these meshes has fewer than 6 vertices.\hfill $\Box$ of lemma.\\

To replace $n$-meshes for $n\leq 5$, we will use the following lemma:

\begin{lem}
The following equations hold:
\begin{tabbing}
(o)\quad \=$\I (\psdiag{1}{3}{0mesh})$\quad\quad\= =\quad\quad \=$7c\I(\quad\quad)=7c$,\\[5pt]
(i)\>$\I(\psdiag{1}{3}{1mesh})$\>=\> 0,\\[5pt]
(ii)\>$\I(\psdiag{1}{3}{2mesh})$\>=\>$r\I(\pstext{ek33})$,\\[5pt]
(iii)\>$\I(\pstext{3mesh})$\>=\>$t\cdot\I(\pstext{tril})$ with
$t:=\frac{1}{\delta}(-q^3+\alpha q^{-3}-\gamma)$,\\[5pt]
(iv)\>$\I(\pstext{4mesh})$\>=\>$\frac{r^2q^5}{g(q^4+1)}(\I(\pstext{wbog})
+\I(\pstext{sbog}))$\\[5pt]
\>\>\>$+\frac{rq^2(q^2+1)}{g}(\I(\pstext{h})+
\I(\pstext{i}))$\\[0em]
\>\>\>with $g:=q^6+q^5+q^4+q^2+q+1$, \\[5pt]
(v)\>$\I(\psdiag{3}{9}{5mesh})$\>=\>$-d(\I({\mbox{\psdiag{3}{9}{5k1}}})+
\I({\mbox{\psdiag{3}{9}{5k2}}})+\I({\mbox{\psdiag{3}{9}{5k3}}})$\\[5pt]
\>\>\>$\quad\quad+\I({\mbox{\psdiag{3}{9}{5k4}}})+
\I({\mbox{\psdiag{3}{9}{5k5}}}))$ \\[5pt]
\>\>\>$-d^2(\I({\mbox{\psdiag{3}{9}{5k6}}})+ \I({\mbox{\psdiag{3}{9}{5k7}}})+
\I({\mbox{\psdiag{3}{9}{5k8}}})$\\[5pt]
\>\>\>$\quad\quad+\I({\mbox{\psdiag{3}{9}{5k9}}})+
\I({\mbox{\psdiag{3}{9}{5k10}}}))$ \\[5pt]
\>\>\>with $d:=\frac{rq^3}{g}$.
\end{tabbing}
\end{lem}

{\bf Proof of the Lemma:} Equation ($o$) is proved in remark
\ref{unknot}.

Equation ($i$) is true because $\I(\psdiag{1}{3}{1mesh})$ is a $\frakg_2$-linear
map from $\C$ to $V$ and must therefore be 0.\newline
Equation ($ii$), we get from
$\I(\pstext{5k11})=\Psi(\pstext{ab3box})\stackrel{\mbox{{\scriptsize Sec.\ 4}}}{=}
r\Psi(\pstext{ek33})$.\newline
To get equation ($iii$), we use the skein relation given in theorem \ref{thm4}
(rotated by $90^\circ$):
$$\I(\psdiag{1}{3}{3mesh})=\frac{1}{\delta}(\I(\pst{5k13})-\alpha\I(
\pst{5k14})-\beta\I(\pst{5k15})-\gamma\I(\pst{tril}).$$
Note that
$\I(\pst{5k13})=\I(\pst{tril})\circ\I(\pst{nkreuz})=-q^3\I(\pst{tril})$
(only the eigenvalue of $\I(\pst{nkreuz})$ on $V$ matters).\newline
Equation ($iv$) and ($v$) can be obtained in a similar way. For ($v$), it may
help to use that $\pstext{5k16}\sim\pstext{5k17}$.\hfill
$\Box$of lemma.\\

As \It is a monoidal functor, equations ($o$)-($v$) will still hold if the
3-nets depicted in the arguments of \It are parts of  bigger 3-nets that
are identical outside the depicted region for all arguments in the same
equation. The observation that for all equations, the 3-nets appearing on
the right-hand side have fewer meshes than the one on the left-hand side
concludes the proof of the proposition.\hspace*{\fill} $\Box$\\

Now we are able to compute the value of the $\I$-invariant for every
oriented closed planar 3-net recursively. Planar 3-tangles that are not closed can
often be reduced by the same technique, but because in this case, lemma
  \ref{km} is no longer true, it may happen that we get stuck before reaching
the empty 3-net (example:\pstext{6mesh}).\\

config: arxiv