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Can I use my laptop as a second monitor? Number of ways to pair off $2n$ points such that no chords intersect For $n \geq 0$ evenly distribute $2n$ points on the circumference of a circle, and label these point cyclically with the numbers $1, 2 . . . , 2n$ Let $h_n$ be the number of ways in which these $2n$ points can be paired off as $n$ chords where no two chords intersect  I want to find a recursive formula for $h_n$.  First I find $h_1,h_2,h_3$ to see the recursive nature.    $h_1 = 1$ as their only one way to make one chord  $h_2 = 2$   Now $\require{enclose}      \enclose{horizontalstrike}{h_3=4}h_3=5$ ((1,4), (2,3), (5,6) case is missed in image)  I found a wikipedia link    and it is very close to my problem, But in this link you don't need to pair off $n$ chords.  But I can't get to come up with a recursive formula for my question.  If I would to guess I would say $h_n = 2 \times h_{n-1}$, And would this means that $h_4 = 8$ ?? System freezes completely with Intel Bay Trail My system freezes completely at random, frequent intervals. I started to have the same problem in Ubuntu 14.04 but after recent upgrade to 16.04 there is no improvement, in fact it seems worse.  When it happens, it's impossible to do anything. I've tried everything in this thread:  but nothing works, I have to hard reset. I have read all the system logs and journalctl but there is never any information that could help diagnose the problem.  This is a dual-boot system with Windows 10 and there's no problem there, so it's not defective hardware.  My laptop has an Intel Bay Trail processor (Pentium N3540) I am creating a query that finds the shortest path from dangerous crossing nodes to hospitals (many-to-many). I have the proper output from pgr_dijkstra, with output as follows  seq | path_seq | start_vid | end_vid | node | edge | cost | agg_cost  -----+----------+-----------+---------+------+------+------+----------    1 |        1 |         2 |       3 |    2 |    4 |    1 |        0    2 |        2 |         2 |       3 |    5 |    8 |    1 |        1    3 |        3 |         2 |       3 |    6 |    9 |    1 |        2    4 |        4 |         2 |       3 |    9 |   16 |    1 |        3    5 |        5 |         2 |       3 |    4 |    3 |    1 |        4    6 |        6 |         2 |       3 |    3 |   -1 |    0 |        5    7 |        1 |         2 |       5 |    2 |    4 |    1 |        0    8 |        2 |         2 |       5 |    5 |   -1 |    0 |        1   I am trying to use st_makeline to create seperate lines for each individual node path, however thus far the function has been creating one big line SELECT  st_transform(st_makeline(t1.the_geom),4326)         FROM pgr_dijkstra(                 'SELECT osm_id as id, source, target, cost FROM spatial.melbourne_edges',                 array(select hospital_id from hospitals_in_vicinity),                  array(select crossing_id from dangerous_crossing_nodes), false) as route                 inner join spatial.melbourne_nodes t1 on route.node = t1.id --);  How am i able to make a column of all the different paths? Let $\mathcal{C}$ be a finite binary code such that for any two sequences $\{a_n\},\{b_n\}\in \mathcal{C}$ if $a_j\not=b_j$ for some $j$ then the concatenation binary sequence $a_1a_2\cdots a_n\not=b_1b_2\cdots b_n$, the Kraft-Mcmillan Inequality says that $$\sum_{i}\frac{N_i}{2^{i}}\le 1,$$where $N_i$ is the number of codes in $\mathcal{C}$ with length of $i$.  If we assume the code is prefix free, we will have an elegant probabilistic proof at  (consider the infinite binary sequence generated by i.i.d. binomial random variable with probability $1/2$, define $E_x$ be the event that the sequence has prefix $x\in \mathcal{C}$, then $\sum_i\frac{N_i}{2^i}\le \sum_i P[E_x]=  P[\cup_i E_x]=1$). But for the general case one will not necessarily have the events $E_i$s disjoint, I'm curious if similar argument exists for the general case. (The problem is an excercise in the probabilistic method of alon, I guess some probabilistic argument should exist)  (My attempt is that consider the product topology $T=\prod_i\mathcal C$ whose open sets generated by $E_x$s where $x$ is the concatenation codes of $\mathcal{C}$, we define a natural product measure $\mu$ over $T$ by $\mu(E_x)=2^{-|x|}$. Clearly $\mu(T)\le 1$ (as $T$ is a sub-topology of $\prod_i\{0,1\}$), I want to show $\mu(E_x\cap E_y)=0$ for $x\not=y\in \mathcal{C}$, if not we can find $E_z$ such that $E_z\subset E_x\cap E_y$ by compactness, if we can show $z$ can be write as concatenation code start with both $x,y$ we will done, but I get stuck at here.) How can one change a coordinate system for a .lasd (Lidar) dataset?  The file is currently in UTM_Zone_15N and I am trying to convert it to IL_East. StackExchange drop down hiding chat, blog and log out, concerns How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?  This should be easy, but I'm not seeing how.  A hint would be appreciated. Integral of $\frac{1}{2x}$ Cannot get grub menu to timeout (or go away) ls: cannot access .gvfs: Permission denied Taking screenshot of a specific area from the command line Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$ It's been said that .  It the corresponding statement true for typical cryptographic hashes, like MD5, SHA-1, SHA-2 or SHA-3?  For example, would SHA-512 be bijective when hashing a single 512 bit block? How do trees lift water higher than 10 meters? The atmosphere pressure is 10 meters of water (approx). This means that it is impossible to lift water higher than 10 meters with vacuum or сapillary action (on Earth, under normal conditions).  There are trees higher than 10 meters.  How do they lift water to their tops?  UPDATE  In other words: how cohesion-tension theory can be true if it apparently contradicts the laws of physics?   UPDATE 2  Atmospheric pressure helps to rise the water, not resists rising. What is resisting is water weight. When water column is 10 meters high, then atmospheric pressure can't help anymore.   Any adhesion/cohesion mechanism can't help here too, because it acts only in thin molecular layer. To transfer action force further the pressure is required, which is insufficient at 10 meters.  UPDATE 3  If we had capillary small enough to rise water to 10 meters and then we will build smaller capillary which we expect will rise water higher, we will fail. Water column will break and does not climb higher than 10 meters.     Menisci acts like small piston and can't help rising water higher than 10 meters.  UPDATE 4  Common pressure distribution in capillary is follows:    $P_0$ is atmospheric pressure. As you see, right under menisci, the pressure is lowered by $2 \sigma / R$ where $R$ is the radius of menisci and $\sigma$ is surface tension. The entire term is called "Laplace pressure". As you see, it can't supersede atmospheric pressure, because water continuity will be broken in the case.  I.e. no any menisci can rise water higher than 10 meters.  The existence of higher trees PROVES that there are some other significant mechanisms, not adhesion/cohesion, not capillary.  UPDATE 5  Current version, as I understood it, is based on a declaration, that a water, if put into thin capillary, can behave like a solid body. Particularly, it can resist tension up to minus 15 atmospheres.   This is a tensile strength of concrete, so I don't believe that without additional proofs.  I think it is just not hard to make thin tube, put water into it and check, how high it can climb.  Was it done ever? I was wondering if having less than 5-10GB of free space on an SSD would affect performance. Is it specific to SSD or is it more of an operating system aspect?  For example, on an Intel X-25 G2 that has TRIM support (on Windows 7). How to type en-dash and em-dash in Windows 7/8? I write blog posts, and I often need to type en-dash (–) and sometimes em-dash (—) in my blog posts. Currently I keep these characters saved in a text file, and copy-paste them whenever I need them. Unnecessary to say, it is a headache to move between keyboard and mouse when you type large texts.  It would be better for me if there ware a keyboard shortcuts available for these two characters. Is there a way to type en-dash and em-dash in Windows? I use Windows 7 and Windows 8 in different computers.    Here's an aligned comparison of dashes, if you need to see the difference for some reason.   Here's-dash Here's–en-dash Here's—em-dash How can I create a /dev/null-like "blackhole" directory? I would like to create a "/dev/null" directory (or a "blackhole" directory) such that any files written to it are not really written, but just disappear.   I have an application that writes out large temporary files to a directory. I have no control over the name of the files and I don't really care about the content of these files. I could write a script that periodically clobbers these files, but the files are written out very quickly and fill my disk. I'm looking for something cleverer. I want the application to "think" that it is writing out these files, when in fact, the writes are just being discarded at the other end.  Also see  old related thread. I know that for any nonzero $x,y\in\mathbf{R}$,  $$ax^2+by^2+cxy &gt; 0,$$  where $a,b,c\in\mathbf{R}$. What can I deduce about $a$, $b$, and $c$?    For example, letting $x=1$ and $y=0$, I know that $\boxed{a&gt;0}$. Letting $x=0$ and $y=1$, I know that $\boxed{b&gt;0}$. Letting $x=y=1$, I know that $\boxed{c&gt;-(a+b)}$.  What else can I deduce about $a$, $b$, and $c$? How will I know when it's time to stop looking? (Is that last question even answerable?) Gauss circle problem : a simple asymptotic estimation.