After a long period of relative malaise, the Games Workshop miniature wargaming company seems to have attained enormous heights of financial success starting around 2017. After spending a while growing in the £3,000 range, they have recently crossed the £9,000 threshold.  I don't follow financial news that much, so maybe it's just my limited perspective, but this is by far the highest stock price per share I have ever seen, for any company in any sector. Given the business they're in and the number of fans their products have, Games Workshop seems to me - as a layman - like it would attract a fair number of 'sympathy investors' (i.e., people who don't otherwise own a stock portfolio going &quot;Well, they're my favourite company, I think I'll buy a share&quot;, with little intent to trade beyond that). But their absurdly high price per share puts them utterly out of the reach of any such casual investors. Is there a good reason to refrain from splitting at such a high stock price? Is the 'casual investor' effect I described above perhaps much smaller than I expect (small enough to be negligible)?
When Apple was in the high $600 or so range, it split 6 times and made it more affordable to purchase more shares for less, even though the value was technically the same. I see companies like Google are extremely high per share and am wondering why or why not they would or wouldn't split their stock right now.
I'm heading to India next week and have a connection in Munich, Germany. I made the mistake of booking a flight with only an hour and a half layover (freaking out!) and I'm wondering if I'll have to go through customs in Munich or if I can pass right on through. Would someone explain wheel sizes.  If I am not mistaken, the wheels on my Trek 520 touring bike are 700C. There are no markings, and of course, no documentation of any kind. I assume 700 means 700mm, so 700 / 25.4 = 27.56 inches.  If I measure the rim however, its diameter is approximately 25.1 inches.  If I buy Continental tires, the size is 28 inch, but I have to buy 27 inch Bontrager tubes.  To make matters worse, per this link, tires come in sizes of 26, 27, and 28 inch, and size 700.     How can I determine my wheel size, and are these various sizes interchangable. Also, is the 'C' suffix important, and what does it mean?  Of course the tires are marked, but I am under the impression (per the dealers comments) that Continental marks their tires differently than everyone else. Also, the tire markings don't help for ordering a new wheel.   With so many tire sizes available, how can I know if I am using the correct size.  After reading the various answers here, I should explain that my confusion on this matter stemmed from the fact that I used to work as an auto mechanic. In the auto industry, car tires are based on wheel size, 13", 14", 15" etc, and there is no possible way to mount a 15" tire on a 13" rim for example.  So when I saw all these different tire sizes, I assumed they all required a different wheel size, but in fact, there is essentially just one wheel diameter for my type of bike, which they refer to as 700C, which is actually 622mm.   I also initially assumed that the C referred to the wheel's width, but that isn't true either. The C is apparently just antiquated terminology.  So at least in theory, I can mount a 26", 27" 28" or 29" tire on my 700C (622mm) rim. I just need to verify the width of the tire is suitable my wheel width, which is what Sheldon Brown's article explains.  Also, I measured my Continental tires, which are marked as 28 inch. They actually measure 27 inches (fully inflated), which must be why my dealer calls their sizing non standard. Binary Search in Array How would I implement a binary search using just an array? What are the differences between Desktop, Server, and Alternate Installs? To illustrate, assume that I have two tables as follows:  VehicleID Name 1         Chuck 2         Larry  LocationID VehicleID City 1          1         New York 2          1         Seattle 3          1         Vancouver 4          2         Los Angeles 5          2         Houston   I want to write a query to return the following results:  VehicleID Name    Locations 1         Chuck   New York, Seattle, Vancouver 2         Larry   Los Angeles, Houston   I know that this can be done using server side cursors, ie:  DECLARE @VehicleID int DECLARE @VehicleName varchar(100) DECLARE @LocationCity varchar(100) DECLARE @Locations varchar(4000) DECLARE @Results TABLE (   VehicleID int   Name varchar(100)   Locations varchar(4000) )  DECLARE VehiclesCursor CURSOR FOR SELECT   [VehicleID] , [Name] FROM [Vehicles]  OPEN VehiclesCursor  FETCH NEXT FROM VehiclesCursor INTO   @VehicleID , @VehicleName WHILE @@FETCH_STATUS = 0 BEGIN    SET @Locations = ''    DECLARE LocationsCursor CURSOR FOR   SELECT     [City]   FROM [Locations]   WHERE [VehicleID] = @VehicleID    OPEN LocationsCursor    FETCH NEXT FROM LocationsCursor INTO     @LocationCity   WHILE @@FETCH_STATUS = 0   BEGIN     SET @Locations = @Locations + @LocationCity      FETCH NEXT FROM LocationsCursor INTO       @LocationCity   END   CLOSE LocationsCursor   DEALLOCATE LocationsCursor    INSERT INTO @Results (VehicleID, Name, Locations) SELECT @VehicleID, @Name, @Locations  END      CLOSE VehiclesCursor DEALLOCATE VehiclesCursor  SELECT * FROM @Results   However, as you can see, this requires a great deal of code.  What I would like is a generic function that would allow me to do something like this:  SELECT VehicleID      , Name      , JOIN(SELECT City FROM Locations WHERE VehicleID = Vehicles.VehicleID, ', ') AS Locations FROM Vehicles   Is this possible?  Or something similar? Why is quantum entanglement considered to be an active link between particles? From everything I've read about quantum mechanics and quantum entanglement phenomena, it's not obvious to me why quantum entanglement is considered to be an active link. That is, it's stated every time that measurement of one particle affects the other.  In my head, there is a less magic explanation: the entangling measurement affects both particles in a way which makes their states identical, though unknown. In this case measuring one particle will reveal information about state of the other, but without a magical instant modification of remote entangled particle.  Obviously, I'm not the only one who had this idea. What are the problems associated with this view, and why is the magic view preferred? How are self-replicating mines supposed to work? How to use a special character as a normal one in Unix shells? Let $C_*$ be a chain complex such that each $C_i$ is a torsion-free, finite-range abelian group with $C_i=0$ for all $i&lt;0$. Suppose that $C_i=0$ for all $i$ is large enough. The Euler characteristic of $C_*$ chain complex is defined as $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(C_i)$$ Prove that $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$$  I have to prove that $\sum_{i\geq 0}(-1)^iRank(C_i)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$, I think that one way to do this is by showing that $Rank(C_i)=Rank(H_i(C_*))$ but I do not know if this is true in general, in a nutshell, I want to show that the cardinality of the base of any $C_i$ is the same as the cardinality of the basis of the corresponding homology, how can I do this? Thank you I didn't grow out of it.   or      I haven't grown out of it.    Which sentence is grammatically more correct? I'd rather say the second, but  I saw in the text the first one. How to sort a Google Sheets Query by a custom ordering I have a table of choir members with a “name” and a “voice type” column. How can I write a query that orders them by decreasing pitch level of the voice type (i.e. soprano > contralto > tenor > bass)?  Is there in the  something similar to ?  I also tried the , but it does not seem to have an option for custom ordering, either. How to create a progress bar? Proof of when is $A=X^TX$ invertible? Is it possible to have a rate of change of acceleration? I know this may seem a weird question, but it always bothers me. My physics book (Resnick,Halliday,Walker), and also various sites never say anything beyond acceleration.   But when a moving body is being acted by a variable force , its acceleration will definitely change: it will either increase or decrease. Then there will be rate of change of acceleration with respect to time. So, why don't books mention this? What is the cause for not measuring $\frac{d\vec{a}}{dt}$ ? If it exists, what is the use of it? Image a phone and browse on computer emulator? I have a galaxy nexus (stock OS). During its use I have used WeChat and WhatsApp, and would like to save my chat and voice histories within these programs.  Does a way exist to image my current phone as it is, so I can load it onto some PC emulator and browse these apps on my PC? This would allow me to save everything as it is, but also start fresh on my phone with a format. I do not want to format my phone then load back on WeChat for instance, yet I still want to be able to browse my old WeChat chats and voice clips (using a PC).  Thanks What happens, if a rocket is filled with a vacuum instead of high pressured air? Suppose you put wheels under a compressed air tank    so that it can move horizontally to the right and to the left. Suppose there is a nozzle on the right hand side of the tank (in the picture on the backside, if you like).  If the tank is filled with high pressured air and you open the nozzle, the tank will move to the left. This is how rockets work and one can easily imagine that.   What happens when the tank is filled with a vacuum and you open the nozzle?  My first impression was, that the tank is moving to the right but someone told me, that it would be stand still. Is this true? What is an intuitive explanation for that? Query Activity for ORDER BY xxx DESC LIMIT Quantum computers are known to be able to crack  a broad range of cryptographic algorithms which were previously thought to be solvable only by resources increasing exponentially with the bit size of the key. An example for that is .  But, as far I know, not all problems fall into this category. On , we can read     Researchers have developed a computer algorithm that doesn’t solve problems but instead creates them for the purpose of evaluating quantum computers.   Can we still expect a new cryptographic algorithm which will be hard to crack using even a quantum computer? For clarity: the question refers to specifically to the design of new algorithms. Reading a file used by another process Putting char into a java string for each N characters