Basically when I go underwater, the screen goes green or blue. I have tried the method to set watershadows or something to 0, but it negates the blood textures and makes the game look like a bloodless cartoon. How should I solve this? I do not want to mod my game and like I said before, setting depthoffield and radialblur removes blood which I want to retain.
As of patch 1.5, I am completely unable to see underwater in Skyrim.  It's not even that it's murky or dark - it's just a solid grey or green screen.  I can still skim across the surface and see into deep waters, but if I go any deeper than that, nothing is visible.  I can still access items like chests underwater if I figure out their location first (by skimming across the surface) and then point my character there and hope really hard that it gets me to the right place.  I was never able to see much underwater, but I at least used to be able to see small things like wood floating near a shipwreck.  Now, there's just nothing. I am running the Steam version of Skyrim with graphics on Ultra High, with the high res textures DLC.  I've tried disabling the high res textures (along with all my other mods) and it still happens.  In case it's relevant, my GPU is a GTX 560 Ti.  Does anyone have any solutions to this?
Question about proof that $(a + b)^p = a^p+b^p$ in a field of characteristic $p$. Definition charactersitic $p$: a field has characterstic $p$ is its prime field is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.  Proposition: let $K$ be a field of characteristic $p$. Then for any $a, b \in K$ we have $(a + b)^p = a^p + b^p$.  Proof: (from my text) because $p \choose i$ is divisible by $p$ for $0 &lt; 1 &lt; p$ it is zero $\mod p$, so $p \choose i$  $a^{p - i}b^i$ is zero (this is the part I don't understand), so it follows from Newton's binomial law.  I understand that the prime field of $K$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, so if $a, b$ are in the prime field, then $p \choose i$  $a^{p - i}b^i$ is zero, but that is not necessarily the case is it? How to remove a vector QgsVectorLayer from QGIS python plugin I managed to add QgsVectorLayer in QGIS through a python plugin, for example:  vl = self.iface.addVectorLayer(uri.uri(), layerName, self.dbConn.getProviderName())   But how can I remove it? I'm quite confused about the following regular expressions I found in a shell script:  ${0##*/} ${0%/*}   How do they work? I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents: Why is feed-forward mechanism used in hash functions? The compression function of SHA-1 when used in Davies-Meyer mode adds its input to the chaining values at the final step. For the first message block, the IV is used as the input and in the next step, the previous block's output (as specified by Merkle-Damgård) is used. These values are fed to the last round of the compression function.  Why would removing the feed-forward part (i.e.,the final step where the input value is added to the result to get the final output) make the hash function insecure? Apart from the fact that the compression function would not satisfy desired randomness properties (being a permutation for each fixed message). I would like to insert the below mathematical symbol, it looks like a big Delta but different. Is there command to obtain a similar shape? birthday problem - expected number of collisions There are many descriptions of the "birthday problem" on this site — the problem of finding the probability that in a group of $n$ people there will be any (= at least 2) sharing a birthday.  I am wondering how to find instead the expected number of people sharing a birthday in a group of $n$ people. I remember that expectation means the weighted sum of the probabilities of each outcome:  $$E[X]=\sum_{i=0}^{n-1}x_ip_i$$  And here $x$ must mean the number of collisions involving $i+1$ people, which is $n\choose i$. All $n$ people born on different days means no collisions, $i=0$; two people born on the same day means $n$ collisions, $i=1$; all $n$ people born on the same day means $n$ collisions, $i=n-1$.  Since the probabilities of three or more people with the same birthday are vanishingly small compared to two people with the same birthday, and decreases faster than $x$ increases, is it correct to say that this expectation can be approximated by  $$E[X]\approx {n\choose 0}p_{no\ collisions}+{n\choose 1}p_{one\ collision}$$  This doesn't look right to me and I'd appreciate some guidance.    Sorry - edited to change ${n\choose 1}$ to ${n\choose 0}$ in second equation. Sloppy of me. How to calculate $$\int_{0}^{\pi/2}{\{\tan(x)\}dx}$$ where $\{x\}$ means the fractional part of $x$. I mentioned the zero points of $\{\tan(x)\}$. $\{\tan(x)\}$ equals to 0 iff $x=\arctan(k)$, $k$ is an integer. I tried to separate it into$$\sum_{k=0}^{\infty}{\int_{\arctan(k)}^{\arctan(k+1)}{(\tan(x)}-k)dx}$$ Then I found it hard to continue. I don't think I am on the right way. Definition of textbook RSA Different results between individual and batch clip of rasters in QGIS? I am given that $V$ is n-dimensional vector space over $\mathbb{C}$ and $T \in L(V)$. And $T$ has least $m$ distinct nonzero eigenvalues. How do I show that $\text{null}(T^{n-m}) = \text{null}(T^{n-m+1})$?  I'm really not sure how to start here so just the first couple steps would be very helpful. Thanks!! How to redirect output to a file and stdout In bash, calling foo would display any output from that command on the stdout.  Calling foo &gt; output would redirect any output from that command to the file specified (in this case 'output').  Is there a way to redirect output to a file and have it display on stdout? Circular duplicates strike again It looks like this issue has , but it just happened again on Travel.SE. Perhaps this is another corner case? These two questions are closed as duplicates of each other.           I had the last vote on each, although in both cases, my vote was for "Unclear what you're asking", but the weight of votes in both cases were for the duplicate status. So if the existing checks are only in place at the time someone casts a close vote, that would explain how this one got in under the radar. Perhaps another check needs to be done when the question is actually closed, not just when close votes are cast?  And as @gnat points out, low-rep/logged-out users get a Redirect loop when trying to load either question.  NOTE: A moderator has already fixed this particular circular duplicate situation by closing one of the questions with a different close reason. Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map?   I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles.   Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then   1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same  2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$   3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$  Or is there a different/better way to think about this problem? What is the main difference between an inner class and a static nested class in Java? Does design / implementation play a role in choosing one of these? How to flash a BIOS when only a Windows binary is provided? I need to show that for prime numbers of the form $p=4n+1$, $x=(2n)!$ solves the congruence $x^2 \equiv-1\pmod p$. I then need to show this implies p isn't a gaussian prime.   I have started to solve this using Wilson's theorem that a number $z$ is prime iff $(z-1)!\equiv-1\pmod z$. Therefore the endpoint of my proof should be that $(p-1)!\equiv-1\pmod p$.  As $p$ is of the form $p=4n+1$, I only need to prove that $4n!$ is congruent to $-1$ modulo $p$.   Here is my working so far: Starting with the congruence $x^2 \equiv -1\pmod p$:    $$\eqalign{x^2 = -1\pmod p&amp;\implies   x^2 + 1 = kp \implies (x-i)(x+i) = kp\cr &amp;\implies ((2n)!-i)((2n!)+i))=kp \implies 4n!- 1 =kp\cr}$$   This is where I start to run out of any ideas that seem to get me anywhere. Any tips would be greatly appreciated! Basis for Skew Symmetric Matrix I'm trying to find a basis for the kernel for the following mapping:  Considering the linear transformation T: $M_{33} \rightarrow M_{33}  $ defined by $T(A) = .5(A + A^T)$. I know that this is basically asking for the basis under the condition that $T(A)=0$ which means that $A+A^T=0$ so $A^T = -A$.  I found that matrices that fit this condition are Skew Symmetric Matrices.  However, I'm not sure how to find the basis for the kernel of these matrices. There was a book (it actually had a second book as well) where a guy gets paid with a group of others to go to a planet that is enclosed in a metallic sphere.   There are many species in this planet. I don't recall if they actually met the races in the first or the second book but there an incident where an alien in the first book ate fruit from a tree that removed any signs of gender from it. Context: I was looking at . I didn't know the result referred to in Brian M. Scott's answer is hard enough to require a separate proof   ― .  ― .  (1)  ― What's the intuition? Is Bezout Identity supposed to be intuitive?  (2) — Why doesn't this try work? Let $c$ be any common divisor of $a, b$. Therefore $c\mid a, c\mid b \iff cj_1 = a, cj_2 = b$ for some integers $j_1, j_2$.     Call $g = \gcd$. By definition $\gcd(a,b)\mid a$ and $\gcd(a,b)\mid b$ and $c \le \gcd(a,b)$. Therefore $g\mid a, g\mid b \iff gk_1 = a, gk_2 = b$ for some integers $k_1, k_2$.   The two paragraphs together precipitate $cj_1 = a = gk_1,  cj_2 = b = gk_2$ but this only proves $c\mid gk_1, c\mid gk_2$?    Origin - Elementary Number Theory, Jones,  p9, Exercise 1.8