It was year 5 and only two colours left on the board (myself and my brother). It was the last turn and his turn to go.   He had finished buying all his moon cards (at this time he had all his commanders except the nuke one). His first move was to play all his nuke cards, forgetting he had lost his commander, this meant he couldn’t win (the cards were his only chance).  I called him out after he tried to play the cards and told him he couldn’t play them if he didn’t have his nuke commander and couldn’t go back to buy it.   Should he have been able to go back and buy his commander which would have resulted in him winning the game, or should we have continued playing allowing him no commander?
A generic question across most board games: once a player has taken their turn (taken their finger off the piece they are playing), should they be allowed to change their mind?  Similarly, if a player forgets to take a privilege, such as an occupied building in Puerto Rico giving an extra worker, or plantation, if they remember a few moves later, should they be allowed to take the privilege?  Very few rules explicitly express how these events should be dealt with, so they are generally left up to the group. However, my group's rules will change depending on the personality of the players playing (not all members are available each time we play). I would like to make this consistent, and would appreciate feedback on how others deal with this scenario.
Covering map between orientable compact connected surfaces without boundary with genus $g$ Similar to a previous question I asked, this is the other question that I couldn't answer in an exam I gave for my Topology course:  Suppose $g_1,g_2$ are non-negative integers such that $2-2g_2$ divideds $2-2g_1$. I have to construct a covering map $S_{g_1} \rightarrow S_{g_2}$, where $S_g$ is denotes an orientable compact connected surface without boundary with genus $g$.  Now, I attempted to try this problem for surfaces with smaller genera, for example a 2-torus and a 2-sphere. Clearly $2-2g$ is the Euler Characteristic of the said surface, and by classification of surfaces we have to only consider connected sum of 2-tori. So, I have $\chi(S^2)=2$, and $\chi(T^2) = 0$, here we have $2$ dividing $0$, hence I have to construct a covering map from $T^2 \rightarrow S^2$, and I have no idea how to move forward.  Addendum: I have to got great answers to my question here but I had a doubt when it comes to the points being mapped, specifically in Giulio Bresciani's answer. Clearly the handles of the figure is mapped to the handles in the figure above, but I'm unable to understand the mapping of the twist of the cover (above figure) to the extreme handles of the base (the figure below). Any help? How to add days to current Date using JavaScript. Does JavaScript have a built in function like .Net's AddDay? How to solve following equation: $x^x = C$ as $x, C \in \mathbb{R}_{&gt;0}$. I tried to use exponent and such that: $x\ln x = \ln C$. I am interested in good integer approximation from below and from above for binary Log(N!).  and  provides only a general idea but not exact values. In other words I need integers A and B so that  A &lt;= Log(N!) &lt;= B Sum over all non-evil numbers I'm working on the following contest math problem:     Define an evil number to be any positive integer that contains the   digit $9$. Show that      $$ \sum_{x} \frac{1}{x} &lt; 80 $$      where the sum is over all non-evil positive integers $x$.   I'm very confused on where to begin. Initially, I tried to consider this sum as part of the sum of $1/x$ over all positive integers, namely by noting that  $$ \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{x} \frac{1}{x} + \sum_{y} \frac{1}{y} $$  where the first summation on the right side is the same as in the problem statement and the second is the sum over all evil numbers. However, I can't seem to find a way to use this since the sum of the left diverges.  Could anyone lend a helping hand? Mail App keeps popping up in the background in Mac OS Mojave When I am doing some work on my macbook pro, Mail app's main window keeps popping up in the background, it even pops up when I watch movies by interrupting full screen. This is so frustrating, I can't find out solution to this, googled and went to many discussion forums, no luck, suggestions that I found are very old and outdated. Has anyone faced similar issues or does anyone knows solution to this? Is there such a thing as a Ubuntu Certified Engineer? Why is a vector space equal to its tangent space for any point? I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$, then $T_x(V)=V$ if $x\in V$.  If $x\in V$, then since $V$ is a manifold, there is a local parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$. Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.   An arbitrary element of $T_x(V)$ looks like $$ d\phi_0(v)=\lim_{t\to 0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t} $$  but this doesn't seem very useful to show $T_x(V)=V$. What is the right approach? Installing to SDCard by default When the good ship Serenity is traveling in deep space, the crew experiences what looks like an Earth-normal (or Earth That Was-normal) level of gravity. Is it ever explained how the ship is able to generate artificial gravity? The ship doesn't spin, so  is out.  is a possibility, but I think the gravity in the ship pulls in a direction that is perpendicular to the direction the ship is usually traveling in, not opposite, like this method would require.   So: how does she pull it off?   Extra credit: how does she pull it off in the episode ? Serenity's engine is totally dead, to the extent where not even life support systems are online. Yet Mal still has to valiantly haul himself around on the floor to replace the compression coil, so whatever creates the artificial gravity on Serenity is clearly still going. How? set fontsize for TikZ figure How to reset evolution settings to defaults? I have not used Evolution for long time now and I would like to reset all the settings and start from scratch. How would I be able to do this? $t^3$ is never equal to Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to: (A). $4t+3$ (B). $8t+5$ (C). $10t+3$ (D). $6t+5$  We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction? The number of labelled graphs with all vertices of even degree I have a graph exam tomorrow, and there is a problem that said number of graph with labeled vertices and all of them of even degree is $ 2^{n-1 \choose 2} $.  According to that topic, it means the number of cut spaces is $ 2^{n-1} $. Well, I don't understand what is cut space yet, and why its count is $ 2^{n-1} $ And don't  know why we should divide the total number of graphs by this! Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$ I need your help with evaluating this limit:  $$ \lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{$n$ compositions}}\,n,$$  i.e. we apply the $\sin$ function $n$ times.  Thank you. Why isn't a virus "alive"? The recent  about a new supermassive virus being discovered got me thinking about how we define viruses as non-living organisms whilst they are bigger than bacteria, and much more complex than we first gave them credit for.   What biological differences between viruses and cellular organisms have made viruses be deemed non-living? Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.  So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I  don't know how to proceed since 1000 seems to be a bit high. Child who entered under ESTA/VWP accompanying US citizen mother Non-trivial cyclic group with unique generator Let $G$ be a non-trivial cyclic group with an unique generator. Show that $G$ has exactly two elements.   Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n &gt; 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$. My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n &gt; 2$ the generator cannot be unique. For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach? How can I improve Ubuntu overall system performance?