How do you become 'stylish' enough to access Luminose stores: Pokemon X and Y? So upon entering Luminose city, I immediate sought out the clothing store to look at what was offered in this city. I mean, c'mon, it's pokemon with customization? Nuts! But when I entered the store, my character was told she wasn't 'as stylish as she could be', and asked to visit other clothing stores in Luminose before returning. Any help on how to access this particular store?
Clothing store access in Pokémon X & Y Once the player obtains access to Lumiose City for the first time, they'll be able to visit Boutique Couture from the city's Vernal Avenue (South fork, connecting from South Boulevard). Upon entering the store without performing any other actions, the player will be turned away and informed they "aren't stylish enough" and that they should visit more places in Lumiose City before visiting again.  This description is dubious, and doesn't inform the player of what they need to accomplish to gain access. Specifically:  Even beating the game (completing the Pokémon League) fails to allow access, as does modifying the player's look, visiting most shops in all areas of Lumiose City, and performing all hair stylist, PR studio, and associated actions. Completing a set of triple battles in Restaurant Le Wow did set the event flag that allowed access in my copy of Y, but it is not clear if this is the only requirement for access.  So other players may benefit, and so I know how to repeat this process if I obtain a copy of X or the future release of Z, what exact requirements yield access to this store for the first time?
Proof that $\mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $ Where $ f: V \rightarrow V  $ is a linear operator on an inner product space $V$, I'm trying to show that $$ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $$  i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f.  I taking the inner product of $f(v)$, where $v \in V$  and an arbitary element of the kernel, $k \in \mathrm{ker}(f^*) $  $$\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0$$  And then taking the inner product of $f(v)$ with $w$, where $w \in  \mathrm{im}(f)^{\perp}$  By definition  of the orthogonal complement to the image of f,   $$ \langle f(v), w\rangle = 0 $$  My question is, now that we have   $$ \langle f(v), k\rangle = \langle f(v), w\rangle = 0 $$  What can we conclude? I would like to conclude that because $w$ and $k$ are arbitary elements of their respective subspaces, that these must be equal, i.e. $ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $  However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight? Inequality Proof in Real Numbers: If $p+q=1$ then $pq\le\frac14$ Let $p,q \in \mathbb R$, and $p+q=1$. Prove that $$pq \le \frac{1}{4}.$$   The first thing I did was define 3 possibilities that we can have from $p+q=1$:  1º  $p$ or $q$ is negative. Example:   $p=-10$, $q=11$   $-10+11=1$  $(-10)(11)=-110$  So if $p$ or $q$ are negative the inequality $pq$ $\le$ $\frac14$ holds.  2º $p$ or $q$ is $0$, that makes the inequality $pq$ $\le$ $\frac14$ be $0$ $\le$ $\frac14$, the inequality holds.  3º The final possibility I saw  was $0$ $\lt$ $p,q$ $\lt 1$, here i just see that the max value from $p,q$ can be $p,q = 1/2$ making $\frac12 + \frac12 = 1$ and $(\frac12) (\frac12) \le \frac14$, but i do not see the elegant way of proving that with other values the inequality holds.  I apologize in advance if I made any mistake or missed something. My router can do port-forwarding based on MAC addresses. That is, a specific MAC will get a specific IP, for which I can configure a set of ports to be forwarded. In order to easily change that set of ports, I'd like to have different connections in the Network manager.  How do I change the MAC address for a network connection? How to show wifi password? I am trying to prove the existence of the square root of $2$.  I have some steps with a very vague explanation and I would like to clarify.  The proof: Let  $$S=\{x\in\mathbb R\mid x\geqslant 0 \text{ and } x^2&lt;2\}.$$  I understand the proof of LUB, ∝   and so I am at the step where $\alpha^2=2$.  I know that we are to prove by contradiction so we state let $\alpha^2 &lt;2$ and $\alpha^2 &gt;2$.  Now my instructor wants us to use the Archimedean Axiom $1/n = \varepsilon$.    $(\alpha^2 + 1/n)^2$  then what..... Does the verb &quot;provide&quot; always have to be used with &quot;with&quot;? For example,  Can you provide me with some good examples? Can you provide me some good examples? Can you provide some good examples?  I suppose it's a transitive verb, isn't it? Moreover, is the following type of usage correct?  You should provide food for your dog before you go on vacation.  If you could give me some good examples regarding &quot;provide&quot; with different types of usage, I'd appreciate that. How do I select elements of an array given condition? When, if ever, should voting override answer acceptance for answer sorting? The accepted answer always appears at the top of the list, regardless of sort order. This can be misleading if the question author accepted an invalid solution and a better solution was proposed later, and eventually got more votes. I have seen somewhat obscure questions where the accepted answer has a negative score and the next answer has 10 or more up-votes. After update to OS X 10.11 I started having these bugs:   Finder sidebar text flashing  Very high CPU load:      I have Macbook Pro Retina 13" Mid 2014  I reset NVRAM.  I deleted com.apple.finder.plist, it didn't help.  Also I did this: , no longer flashes. Scoring when two triple word squares are covered in one play The word QUAGMIRE is on the board vertically, with the Q on the top row, second column. The next player creates SQUEEZED horizontally, landing the S on the triple word square in the top row, first column, and remaining UEEZED landing to the right of the existing Q, ending with the D on the triple word square in the eighth column. Since two triple word squares are covered in this one placement, how would it be scored?  Letters:   S: 1 Q: 10 U: 1 E: 1 x 2 (double letter) E: 1 Z: 10 E: 1 D: 1   27 Total for letters. Do you then multiply by 3 for the first triple word square, and then multiply that result by 3?  27 X 3 = 81, 81 X 3 = 243  Bonus for all 7 tiles placed:  243 + 50 = 293 Why do prisms work (why is refraction frequency dependent)? There's a part in my apps that displays the file path loaded by the user through OpenFileDialog. It's taking up too much space to display the whole path, but I don't want to display only the filename as it might be ambiguous. So I would prefer to show the file path relative to the assembly/exe directory.  For example, the assembly resides at C:\Program Files\Dummy Folder\MyProgram and the file at C:\Program Files\Dummy Folder\MyProgram\Data\datafile1.dat then I would like it to show .\Data\datafile1.dat. If the file is in C:\Program Files\Dummy Folder\datafile1.dat, then I would want ..\datafile1.dat. But if the file is at the root directory or 1 directory below root, then display the full path.   What solution would you recommend? Regex?  Basically I want to display useful file path info without taking too much screen space.  EDIT: Just to clarify a little bit more. The purpose of this solution is to help user or myself knowing which file did I loaded last and roughly from which directory was it from. I'm using a readonly textbox to display the path. Most of the time, the file path is much longer than the display space of the textbox. The path is supposed to be informative but not important enough as to take up more screen space.  Alex Brault comment was good, so is Jonathan Leffler. The Win32 function provided by DavidK only help with part of the problem, not the whole of it, but thanks anyway. As for James Newton-King solution, I'll give it a try later when I'm free. Recommend a statistics fundamentals book To give you some background, I have a grasp on the basics of statistics and probability theory and even remember touching Bayes theorem at the university data mining course. But being a few years away from the university made my math got extremely rusty (so much for last-minute pre-exam cramming). While I remember various random basic concepts, there are a lot of gaps in my understanding of them.  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I want to create a scratch-card-like effect in new Unity UI's new Image tag.  I know it's possible with setPixels, but that's prohibitively expensive on mobile devices.  Something like this:    I'd like the user to be able to drag on any part of the image to "scratch off" the top layer and reveal the content underneath it.  How can I do this? Inconsistent line spacing If I type something like this:  {\itshape\LARGE   Dunt venim dolorerosto do odit   ametum vulla conum dolore   conulput in vero od el   Dunt venim dolorerosto do odit   ametum vulla conum dolore   conulput in vero od el }   it results in a block of 4 lines with narrow spacing between the 1st and 2nd line and between the 3rd and 4th line. Line spacing between the 2nd and 3rd line is normal.  I can't fully explaing it to myself so I am looking for a short description for this behaviour. I know for example that with a \par at the end of the text (inside the group), the calculation would correct. But I have no idea why...