Because of a page limit for a report, I would like to decrease the linespread between individual \items. I know this is generally not a good idea, but I really need the space.  How can I achieve this globally, say, with a couple of lines of code in the preamble?
What is the best way to globally change the amount of vertical space between items in list environments?
Multirow and Multicolumn with missing lines Do you need to take the Hide action when you are Invisible, make no noise and leave no traces? The Situation: Lets say you are under the effect of Invisibility.  An invisible creature is impossible to see without the aid of magic or a special sense. For the purpose of hiding, the creature is heavily obscured. The creature's location can be detected by any noise it makes or any tracks it leaves.  And you are wearing the Boots of Elvenkind  While you wear these boots, your steps make no sound, regardless of the surface you are moving across. You also have advantage on Dexterity (Stealth) checks that rely on moving silently.  Now your groups ranger or druid casts Pass Without Trace on you.  [...] can’t be tracked except by magical means. A creature that receives this bonus leaves behind no tracks or other traces of its passage.  My Question(s): I know that when you are Invisible and you did not hide, you are still detectable because of the noise and the track you leave behind. But the Boots of Elvenkind and Pass Without Trace take care of that. So:  Are you still required to take the hide action in this situation if you don't want enemies to know your location/presence? If so, why? What gives away your location? If not and you are under the effect of greater Invisibility do you get the benefits of &quot;Unseen Attackers and Targets&quot; for every attack you make when you move between attacks? (You would give away your location, but are not detectable, so completely hidden again when you move) Keeping tables/figures close to where they are mentioned Is there any package or a method to force LaTeX to keep floating environments like table and figure closer to where they are declared? Why are local variables not initialized in Java? Looking for illustrator script to swap all objects with another I have a map with small circles for over 200 cities. I need to change all of the circles into map pins. Is there a way to do this? I think it's a script maybe.   I know how to select all objects of the same appearance and give them a new graphic style but that will only let me change the color, stroke, etc. not replace with a new shape. Are \( and \) preferable to dollar signs for math mode? What is the difference in meaning between "proposal" and "proposition"? Why won't vertical-align: middle work?  And yet, vertical-align: top does work.   span{   vertical-align: middle; } &lt;div&gt;   &lt;img src="https://via.placeholder.com/30" alt="small img" /&gt;   &lt;span&gt;Doesn't work.&lt;/span&gt; &lt;/div&gt; Why does Worf want O'Brien to go to Minsk? I've just watched the last 20 minutes of an episode of Deep Space 9, and towards the end Worf is very insistent that O'Brien and his wife move to Minsk. Is there any in or out of universe reason for this? Area of the Limiting Polygon Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.   Continue this process ad infinitum.  What is the area of the limiting "polygon"?  My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.   Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance. How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$, assuming f is continuous and not zero? This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking? We denote the s value of an ECDSA signature $(r, s)$ on a message $m$ as: $s=\frac{H(m)+xr}{k}$  Assume two ECDSA signatures sharing the same nonce $(r, s_1) , (r, s_2)$ on two messages $m_1, m_2$, that verify under two pubkeys $x_1G, x_2G$.    If the two public keys are equal then the secret keys should be equal $x_1 = x_2$ and we can easily recover the $k$ using the standard attack on nonce reuse. Once we know $k$ we can recover the secret key.  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+x_1r - x_2r}$  $x_1 = x_2 \rightarrow x_1r - x_2r = 0$  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)} = k$  My question is can this attack be made to work if the secret keys are not equal i.e. $x_1 \ne x_2$:  $\frac{H(m_1)-H(m_2)}{(s_1 - s_2)} =\frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+x_1r - x_2r} = \frac{k(H(m_1)-H(m_2))}{H(m_1)-H(m_2)+ (x_1 - x_2)r}$  If you know either $x_1 - x2$ or $\frac{x_1}{x_2}$  you should be able to compute $k$ as long as $s_1 \ne s_2$.  You can calculate $x_1 - x_2 = \frac{H(m_2) - H(m_1)}{r}$ in case where $s_1 - s_2 = 0$. However this case seems to reduce to the hardness of ECDSA since anyone can compute the pubkey for a new message $m_2$ that verifies under first signature $(s, r)$ using public key recovery.  If $s_1 \ne s_2$ you can compute $\frac{x_1 - x_2}{k}$ which allows you to convert $s_1$ into $s_2$ and vice versa. I know that $S_n$ is generated by a number of things, like all transpositions, all transpositions of the form $(1\,a)$, the transpositions $(1\,2),(2\,3),\dots,(n-1\, n)$, and  just the two elements $(123\dots n),(1\,2)$.      Suppose $n$ is prime. If you just have $(123\dots n)$ and some arbitrary transposition $(a\, b)$, how does this also generate $S_n$?    Can you somehow get to $(1\,2)$ or reduce it to some other previous case? I have joined corresponding Local Enterprise Partnerships table data into a polygon shapefile containing UK Local Authorities.      How do I Dissolve or Merge the the Local Authorities polygons by LEP_NAME as shown in the screenshot?   Ideally, I would like to retain the Local Authority attributes with the resultant merged polygons. I have zero scripting knowledge so am looking for a tool based solution. Any help or pointers much appreciated.    Additional Note: When attempting to use the dissolve tool the LEP_NAME field is not available as shown above. The LEP attribute data was created in Excel, saved as CSV format and joined to the Local Authority polygon shapefile. There are duplicate IDENTIFI0 fields as shown in the other screenshot - could this be the root of the issue? The locker problem - why squares? There are 1000 lockers in a high school with 1000 students.  The    problem begins with the first student opening all 1000 lockers; next    the second student closes lockers 2,4,6,8,10 and so on to locker 1000;   the third student changes the state (opens lockers closed, closes    lockers open) on lockers 3,6,9,12,15 and so on; the fourth student    changes the state of lockers 4,8,12,16 and so on. This goes on until    every student has had a turn.      How many lockers will be open at the end?   I worked out this problem by doing it for 10 lockers. The pattern was obvious (1,4,9 remained open).It's easy to then say that 31 lockers remain open at the end. I don't quite understand why only square numbers remain open at the end. Is there some deeper reason? Prove $\lim\limits_{n\to\infty}n^{-1/(p+1)}(1^{1^p}\cdot 2^{2^p}\cdots n^{n^p})^{n^{-p-1}}=e^{-1/(p+1)^2}$ Prove that for real number $p\geq  0$ we have   $$\lim_{n\to\infty}\frac{(1^{1^p} \cdot2^{2^p}\cdots n^{n^p})^\frac{1}{n^{p+1}}}{n^\frac{1}{(p+1)}} = \exp\left(\dfrac{-1}{(p+1)^2}\right).$$ Javascript: negative lookbehind equivalent? Is there a way to achieve the equivalent of a  in javascript regular expressions?  I need to match a string that does not start with a specific set of characters.  It seems I am unable to find a regex that does this without failing if the matched part is found at the beginning of the string.  Negative lookbehinds seem to be the only answer, but javascript doesn't have one.  EDIT: This is the regex that I would like to work, but it doesn't:  (?&lt;!([abcdefg]))m  So it would match the 'm' in 'jim' or 'm', but not 'jam' Prove that $p \in \mathbb{R}[x]$ can be represented as a sum of squares of polynomials from $\mathbb{R}[x]$ I'm using biblatex for my references.  Normally, if I use \cite{source1, source2, source3}, I get a citation like  [3, 45, 98]   If I now want to give a page number for one of them, it looks confusing imho, as I should get (did not try it out up to now, honestly)   [3, 45, 78, p. 23f, 98]   So imho it would make sense to change one of the separator from a comma to e. g. a semicolon like:  [3; 45; 78, p. 23f; 98]   Question 1: Does that make sense, or are there other conventions for citing sources with and without page numbers together?  (Up to now I collected the citations without page number in one cite command and added singular \cite[][p. xy]{} commands for those with page numbers., so I'd get:   [3, 45, 98][78, p. 23f]   )  the \cites[]{}[]{}[]{} could also do that, but it also uses the same separators for entries and postnotes.  Question 2: How could I change the separator globally for those number-citations from comma to semicolon in biblatex? I am not sure what caused this error, but here is what the whole error says, and also this is sitting as a notification, and preventing me from updating any software using Update Manager - Please provide some assistance or tell me how to figure out what to do to fix it.  Could not calculate the upgrade  An unresolvable problem occurred while calculating the upgrade.  Please report this bug against the 'update-manager' package and include the following error message: 'E:Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages.'