Where are some interesting places where the axiom of choice crops up in category theory?

Question: The two that come to mind are splitting epics in Set and taking the Skel of a category. Surely there are lots of other interesting (and maybe upsetting) places where this comes up.
Comment: I find it slightly odd to say that splitting epics in Set is a place "where the axiom of choice crops up in category theory".  The existence of splittings for epics in Set *is* the axiom of choice.
Comment: I community wiki'd this. It's borderline for mathoverflow anyway (see the FAQ), but for now we're allowing "sorted list" discussin questions. They do have to be wiki'd, however.
Answer: Using the usual definition of "functor," almost any functor
constructed using only universal properties requires the axiom of
choice.  For instance, if a category C has products, then one wants a
"product assigning" functor C×C → C, but in order to define this you
have to choose a product for each pair of objects.  If C is a large
category, then you need an axiom of choice for proper classes.
However, this sort of thing is arguably not a "real" use of the axiom
of choice.  It's more accurate to say that in the absence of the axiom
of choice the usual definition of "functor" is not sufficient, and one
must use anafunctors
instead.  Proving that fully faithful + essentially surjective =
equivalence is the same.  Most often in category theory when we want
to "choose" something, that thing is in fact determined up to unique
isomorphism (though not uniquely on the nose) and in that case using
anafunctors is sufficient to avoid choice.
The axiom of choice does, however, come up in the study of particular
properties of the category Set.  One interesting consequence of the
fact that epics split in Set is that all functors defined on Set
preserve epics.  I think this is an important part of Blass' proof
that the existence of nontrivial (left and right) exact endofunctors
of set is equivalent to the existence of measurable cardinals.
Another interesting consequence is that Set is its own "ex/lex
completion."
Answer: In proving that whenever there is a full, faithful functor F: C to D which is essentially surjective (i.e. hits every isomorphism class of objects in D), C and D are equivalent categories.
(This is pretty close to your example of taking Skel of a category.)
Comment: The nlab page on anafunctors (http://ncatlab.org/nlab/show/anafunctor) says that this is actually equivalent to AC.
Answer: While not the traditional Axiom of choice, Zorn's Lemma makes an appearance on the road to Mitchel's Theorem;
If X is an abelian category with the sup property(Grothendeick's AB5), then an object E is injective iff it does not have a non-trivial essential extension.
Mitchel's theorem is stated in Rotman's Homological Algebra(page 316) as;
If A is a small abelian category, then there is a covariant full faithful exact functor from A to abelian groups.
I believe(but do not have access to the appropriate materials to give details) that Zorn's lemma appears in homotopy limits also.
Answer: If $s:\mathcal{X} \to \mathcal{C}$ is a fibered category and $\varphi:C \to D$ is a morphism in $\mathcal{C}$, for each object $y$ in the fiber $\mathcal{X}_D$ the axiom of choice allows us to specify exactly one pullback $f:y_C \to y$ (i.e., a cartesian arrow $f$ with $s(f)=\varphi$).  The choice of such a collection is called a 'cleavage', and a cleavage always exists by the axiom of choice.
This enables us to define the 'change of base functor' $\varphi^*:\mathcal{X}_D \to \mathcal{X}_C$.
Answer: Choosing fiber products to define the sheafification of a presheaf with respect to a grothendieck pretopology (you later prove that this construction is independent of choice, but it's necessary during the construction).  See stacks-git in the remark immediately preceding proposition 8.10.3.


I've just purchased a VPS to use for developing and managing my clients hosting packages. What necessary steps do I have to take?

Question: The VPS is running CentOS and I have full access.
I only looked around using the cPanel installation and already disabled anonymous FTP access.
What other security risks are there with a default VPS install?
Answer: If you're running cPanel and have root (which it reads as if you do), install this free script, called CSF:  http://www.configserver.com/cp/csf.html
There are several tools to check your servers security and in most cases, this software will tell you where to go, to "harden" your cPanel install.  It's not a end-all-be-all to your cPanel security but it most definitely goes a long way!  Good luck.
Oh... one more thing.  If you don't use CSF (or anything else) the first thing I'd do is change your SSH port and disable root's access to SSH (make sure you have a new user created to join the sudo'ers and wheel groups).  That's basic Linux stuff right there.  You can lock SSH down even further by disabling password auth. into SSH, only allowing keys to gain access to SSH.  Just my 2 cents.  :)
Answer: I'll add some info to @ChristopherHaynes answer.
I also run some server for web purpose mainly and here's what I do :
First of all, I create a new user with non admin rights.
Then, I update the sshd_config file in order to :

Disallow root access
Change the port for SSH
Allow ONLY the user I want (aka, the one I created in the first step)

I say that because if you see the logs, you'll see many tries regarding a lot of users, not only root, in your SSH connexions (www-data, etc). If you allow only a user that is not common, you will reduce the potentiality of a success attack.
Now, for the rest, you use CPanel so it should configure the server with your need and some security checks, but I don't personnaly like to use those "all in one server configuration" since they often install many unnecessary apps and are more subject to threat.
Of course, it depends on your knowledge of managing a server via command line, but for every app you install (http, ftp, etc), I warmly recommand you to search which options you should enable/disable in order to make it more secure.
With Apache for example, you can follow this tutorial, and this one.
And always take in consideration that the more app you install, the less secure your server become. (but this is all relative! It doesn't mean it will be more easily hacked, but it will be more potentially hackable).
Comment: Disallowing `root` is a very good idea.  The whole point of root is of escalation, if common access ( i.e. you log into the server as root daily ), the idea of esclation no longer exists.


Can anyone give me a good example of two interestingly different ordinary cohomology theories?

Question: An answer to the following question would clarify my understanding of what a cohomology theory is. I know it's something that satisfies the Eilenberg-Steenrod axioms, and I know that those axioms allow you to work out quite a lot. But what sort of thing is not determined by the axioms? In particular, can someone give me a simple example of a space that has different cohomology groups with respect to two different theories? Obviously a trivial answer would be to take coefficients in different rings, so let me add the requirement that the coefficient rings should be the same. And if there's some other condition needed to make the question non-trivial, then add that in too.
Answer: For any space that has the homotopy type of a CW complex, its cohomology is determined purely formally by the Eilenberg-Steenrod axioms, so a counterexample is necessarily some reasonably nasty space.  Here's an example you can see with your bare hands: consider the space $X=\{1,1/2,1/3,1/4,...,0\}$.  Now 0th singular cohomology is exactly the group of $\mathbb{Z}$-values functions on your space which are constant on path-components, so $H^0(X)=X^\mathbb{Z}$ (an uncountable group) naturally for singular cohomology.  On the other hand, 0th Cech cohomology computes global sections of the constant $\mathbb{Z}$ sheaf, i.e. locally constant $\mathbb{Z}$-valued functions on your space.  These must be constant in a neighborhood of 0, so the Cech cohomology $H^0(X)$ is actually free of countable rank, generated (for example) by the functions $f_n$ that are $1$ on $1/n$, $-1$ on $1/(n+1)$, and $0$ elsewhere, plus the constant function $1$.
I should add that topologists don't actually care about such examples.  The point of the Eilenberg-Steenrod axioms is to show that cohomology of reasonable spaces is determined by purely formal properties, and these formal properties are actually much more useful than any specific definition you could give (the only point of a definition is to show that the formal properties are consistent!).  What is of interest is when you remove the dimension axiom to get "extraordinary" cohomology theories, which Oscar talks about in his answer.
Comment: I unfortunately don't have anything of value to add, but I couldn't resist pointing out that "an example you can see with your bare hands" is a wonderful (and apparently original!) mixed metaphor. :)
Comment: Here's a question I don't know the answer to (perhaps I should post it, hmm):  What *is* the class of spaces for which the ES axioms determine ordinary cohomology?  We know it includes all spaces with the homotopy type of a CW complex, but is that all, or are there others?
Comment: @Charles: I think this indeed deserves its own question-page; please post it!
Comment: Oh, and @Charles: I'd like to second Benoit here, please ask it!
Answer: The sheaf cohomology with coefficients in the constant sheaf $Z$ coincides with the singular cohomology or the Cech cohomology for all homologically locally connected hausdorff paracompact spaces (this is the case for finite CW-complexes or locally contractible spaces).
Sheaf cohomology is better behaved than singular cohomology when it comes to dimension. Recall that the cohomological dimension of a compact topological space $X$ is the greatest integer $n$ (or $\infty$) such that $H^n(X)\neq 0$.
Theorem: If $F$ is a closed subset of a compact topological space $X$, then its sheaf cohomological dimension is less than the sheaf cohomological dimension of $X$ (see e.g. Bredon "sheaf theory", II.16).
Surprisingly, this result is false for singular cohomology. The following example is due to Barratt and Milnor (1962). Let $X$ be the union in $R^3$ of countably many spheres of radius 1/n, all tangent to the xy-plane at the  origin. The sheaf cohomological dimension of this compact space is 2. But the singular cohomology of this space is non-zero in arbitrarily large degrees.
In other words, its singular cohomological dimension is infinite.
Comment: This is really cool!
Answer: An example I recently came across while learning about shape theory is the closed topologist's sine curve. The zero-th singular cohomology (integer coefficients) is Z^2, while the zero-th Cech cohomology (again integer coefficients) is Z.
I can't give a great reason why this is so as I'm just learning about the technicalities myself, but it seems to be a standard example.
Comment: And, to close the circle: if your cohomology theory does take weak equivalences to isomorphisms, then it *is* uniquely determined (up to isomorphism).
Comment: Yes.  In fact, it actually satisfies a stronger form of excision than singular cohomology.  However Cech cohomology fails to satisfy that a weak equivalence of spaces induces an isomorphism on (ohomology.
Comment: Does Cech cohomology in this setting satisfy all the Eilenberg-Steenrod axioms?
Answer: It might be interesting to note that the examples given by Eric and Justin also show why it is necessary to use Čech cohomology when treating Alexander duality for arbitrary closed subsets of a closed (and orientable) manifold.  If one considers A={1/n | n>=1} as a subset of R/Z=S^1 then its complement is a countable disjoint union of open intervals and its 0-th homology (for any theory satisfying the axioms) is free with a countable basis. So Alexander duality would demand that the 0-th cohomology of A also have a countable basis, which is true for Čech cohomology but not for singular cohomology.
Similarly for for the closed topologist's sine curve as a subset of the two-dimensional sphere.  Its complement has the homology of a point, so it itself should have the cohomology of a point, which again is true for Čech cohomology but not for singular cohomology.
This has indeed lead to a remark in the classical textbook by Seifert and Threllfall that while it is possible to define singular homology for arbitrary topological spaces and not just complexes, it is not very useful to do so.  But that was in 1934.
Comment: Isn't it rather that with Čech cohomology that the closed topologist's sine curve has the homology of $S^1$, and with singular homology it has the homology of a point?
Comment: @ziggurism I think "closed" meant to include the segment $\{0\}\times[-1,1]$ approached  by the curve $\{(x,\sin\frac1x):x>0\}$, not to include another arc joining the segment to the curve.
Answer: The examples mentioned so far, made the point that Cech cohomology differs from singular cohomology in many examples which are, to the eyes of most topologists, quite pathological. I want to concentrate on an example of an ordinary homology theory, which is probably much less well known than Cech cohomology, but quite interesting in geometric contexts: stratifold homology (see Kreck's Differential Algebraic Topology). It is defined as bordism classes of certain stratified spaces (including manifolds, but much more).
An example, where stratifold homology and singular homology differ is the one-point compactification of the surface of infinite genus (i.e. the infinite connected sum of tori), denoted by $F_\infty^+$. This can be given the structure of a stratifold and therefore possesses a fundamental class, which is non-zero, even after applying the induced map of $F_\infty^+\to F_g$ to the surface of genus g. Such a class does not exist in singular homology (see Differential Algebraic Topology, chapter 20.2). The space $F_\infty^+$ is surely not the first thing one considers, but it has still some geometric appeal (and is a compact set of $\mathbb{R}^3$). I think, there are probably similar phenomena for one-point-compactifications of certain other open manifolds.
Answer: An interesting cohomology theory is the continuous singular cohomology theory with real coefficients introduced, as far as I know, in the seventies by R.Bott, G.D.Mostow and others.
Let a singular cochain be called continuous if it restricts to a real continuous map on the space of singular simplices, endowed with the compact-open topology. It is readily seen that continuous cochains provide a subcomplex of the complex of singular cochains. The (co)homology of the complex of continuous cochains is called continuous cohomology.
As one can expect, continuous cohomology is canonically isometric to singular cohomology for reasonable spaces (for example, for locally contractible metrizable spaces). However, one can construct spaces with non-isomorphic continuous and singular cohomology just by looking at subspaces of the Euclidean plane.
Answer: (Some inaccurate info appeared in the Answers above).
Different co/homology theories may behave differently on infinite CW-complexes. If two E-S theories are the same for the singleton then they are equivalent for all spaces homotopically dominated by finite polyhedra. It seems to me that for any other topological space one can construct two E-S theories (equal for singletons) which will give different groups (let's consider completely regular $T_1$-spaces only).
Surely dimension of the Hilbert cube   $C$   is infinite, while   $H^n(C)=0$   for every   $n > 0$. In the case of any Hausdorff compact space   $X$   the (topological) covering dimension, say   $n = \dim(X)$,   is equal to the cohomological dimension, meaning that there exists a closed subset   $A$   of   $X$   such that   $H^n(X\ A) \ne 0$,   but there is no such closed subset   $B$   that   $H^m(X\ B)\ne 0$   for any   $m > n$   (a few more words should be devoted to the infinite-dimensional case).
Intuitively the singular and the Cech co/homology view topological spaces in a drastically different way. What is connected for the cruder Cech theory might be not connected for the more sensitive singular theory; the former one is much more tolerant. One can say that the Cech theory applies the shape theory as its foundation rather than just homotopy theory. More precisely, all co/homology functors split into a composition of the homotopy functor and the rest; while the Cech functor does more, it splits into composition of the shape functor and the rest (while the shape functor is a composition of the homotopy functor and the rest).
Comment: You will not find, Yemon, in what I wrote any suggestion whatsoever to edit this answer... It was expressing my preference for MO not becoming the math-social-site analogue of a kidnapper's letter.
Comment: In the grand scheme of things, it is probably better *not* to tweak the spacing like you did here. Consistently across the site is, in the end, better than getting one's own personal posts to look perfectly *to one's own eye*.
Comment: @MarianoSuárez-Alvarez I disagree - I don't think the spacing here is a big enough deal to be discouraged or altered from the author's own intentions, and I don't have any desire to find some editing of spacing bumping this answer unnecessarily three weeks from now


Reference request: compact objects in R-Mod are precisely the finitely-presented modules?

Question: Let $R$ be a ring. According to this MO question, the modules $M \in R\text{-Mod}$ such that $\text{Hom}(M, -)$ preserves all filtered colimits (the compact objects) are precisely the finitely-presented modules. Where can I find a proof of this?
Comment: By the way: If $X$ is a quasicompact quasiseparated scheme, then I can show that the compact objects of $\mathrm{Qcoh}(X)$ are precisely the quasi-coherent sheaves which are locally of finite presentation.
Comment: Possibly in Adamek and Rosicky, *Locally presentable and accessible categories.*
Comment: @Math Gems: great! In fact Theorem 3.12 gives the general version of this result for categories of algebras. I would prefer not to have to absorb the previous portions of the book to understand this result so I would still appreciate references to a more self-contained proof...
Comment: Note that, if we require $\mathrm{Hom}(M,{-})$ to commute with _all_ colimits, then $M$ is necessarily the zero module. See [_tiny object_ on the nLab](http://ncatlab.org/nlab/show/tiny+object).
Answer: This is easy enough to prove directly:
For the direction finitely presented $\implies$ compact, one can easily
prove it by hand, and there is also the following somewhat slicker argument:
For any module $M$ and any filtered direct limit $N = \varinjlim_i N_i$,
there is a natural transformation $\varinjlim_i Hom(M,N_i) \to Hom(M,N).$
This is certainly an isomorphism when $M = R$ (because it then reduces
to the original equality $\varinjlim_i N_i = N$), and hence when $M = R^n$
for some $n \geq 0$, because both functors $\varinjlim_i Hom(\text{--},N_i)$
and $Hom(\text{--},N)$ commute with finite direct sums.
Since filtered direct limits are exact, both functors are also right
exact.
Now if $M$ is finitely presented, choose a presentation $R^m \to R^n \to M \to 0,$ and apply both functors to it.  By what has already been said together with
the five lemma, we see that $\varinjlim_i Hom(M,N_i) \to Hom(N,M)$ is an isomorphism.
For the other directon, suppose that $M$ is compact.
Like any module, $M$ can be written as a filtered direct limit of finitely presented modules $M_i$.  Thus $Hom(M,M) = Hom(M,\varinjlim_i M_i)
= \varinjlim_i Hom(M,M_i)$
(the last equality following by compactness).  In particular, the identity
$M \to M$ can be factored through the map $M_i \to M$ for some sufficiently
large $i$.  Thus the map $M_i \to M$ can be split, and so $M$ is a direct summand of the finitely presented module $M_i$.   Consequently, $M$ is itself finitely presented.

A related fact, mentioned in the excerpt quoted by Math Gems:
If we restrict to filtered direct limits with injective transition maps
(diret unions in the language of the excerpt quoted by Math Gems), then we similarly find, using the fact that any module can be written as the filtered union of its finitely generated submodules, that finitely generated modules
are precisely  those $M$ for which $Hom(M,\text{--})$ commutes with filtered direct limits having injective transition maps.

Added in response to the OP's questions below:
If $A \to B$ is a morphism of rings with $1$, then
$M \mapsto Hom_A(B,M)$ (Homs of left $A$-modules, and $Hom_A(B,M)$
being endowed with the left $B$-module structure coming from the right
$B$-module structure on $B$) is a right adjoint to the forgetful functor
from left $B$-modules to left $A$-modules.   This shows that formation
of colimits commutes with this forgetful functor.
Alternatively, one can check by hand that the formula for filtered direct limits in $R$-mods is the same as in abelian groups.  (Just use the fact that direct sums, kernels, and cokernels are the same.)
Also, any finitely generated $R$-module $M$ is of the form $R^n/N$ for some submodule $N$.  We may write $N = \varinjlim_i N_i$ as the direct limit of its
f.g. submodules.  Then $M = \varinjlim_i R^n/N_i$ is the filtered direct limit
of f.p. modules.  Since any module is the direct limit of its f.g. submodules,
we see that it is also the filtered direct limit of f.p. modules.
Comment: Thanks! I don't know this material well so I need to ask a few naive follow-up questions. 1) I'm not sure why "this is certainly an isomorphism when $M = R$": that is, I'm not sure why the forgetful functor from $R\text{-Mod} \to \text{Ab}$ preserves colimits. Does it have a right adjoint? 2) I see why any module can be written as the filtered union of its finitely generated submodules, but I don't see why any module can be written as a filtered colimit of finitely presented modules. Can you expand on this?
Comment: Thanks! The situation is clearer to me now.
Answer: This is going to be a bit of a ramble, but hopefully relevant to the present question -- probably everything in here is in Adams and Rosicki, though.
Recall that an object $X$ in a category $\mathcal{C}$ is called compact (as Qiaochu states in his question) if $\hom(X, \cdot)$ commutes with filtered colimits: intuitively, a map from $X$ into an ascending union must land in one of the factors.
Compactness is thus a useful version of "smallness,"  and it turns out to be very useful (e.g. in formulating adjoint functor theorems) that in many relevant categories, the class of all objects (which may be gigantic) is controlled by the much more reasonable category of "small objects."
One way of saying this is that $\mathcal{C}$ is cocomplete, and there is a set of compact objects such that every object in $\mathcal{C}$ can be obtained as a colimit of objects in this set. Let me assume that $\mathcal{C}$ is compactly generated in this sense. (This is a bit more restrictive than the notion of locally presentable, where we don't assume that the generating set consists of compact objects but rather of small objects -- those such that homming out of them commutes with sufficiently filtered colimits.)
For instance, in the present question, the category of $R$-modules is generated by the $R$-module $R$ under colimits, and $R$ is compact. As another example, if $\mathcal{D}$ is a test (small) category, then the category of presheaves on $\mathcal{D}$ is generated by the representable ones under colimits, and the representable ones are compact (in fact, by Yoneda's lemma, homming out of a representable presheaf preserves all colimits -- the representables are tiny).
Anyway, the question asks when we can identify precisely what the compact objects are in the category of $R$-modules. I think we can do this in
general for compactly generated categories.
Let $\mathcal{C}$ be a compactly generated category and $S$ the given set of compact objects. Note that $S$ might not consist of all the compact objects: $S$ might just consist of the representables. It is easy to check that a finite colimit of compact objects is compact (this is a consequence of the fact that in sets, finite limits and filtered colimits commute). The claim is that all compact objects are finite colimits of objects in $S$.
In the case of $R$-modules, every module generated under finite colimits from $R$ is finitely presented. So the only compact objects are the finitely presented ones. The same argument works in the category of $R$-algebras (where the compact ones are the finitely presented algebras and a generator is $R[x]$) or in groups (where the compact objects are the finitely presented groups).
To see this, fix any object in $\mathcal{C}$. Since we are assuming that the category is generated by $S$ under colimits, we can form a canonical way of writing $X$ as a colimit of objects in $S$. Namely, we take the  (small) category consisting of pairs $(s, f)$ where $s \in S$ and $f: s \to X$ is a morphism. There is a natural functor from this category to $\mathcal{C}$ sending $(s, f)$ to $s$. (This is a generalization of the "category of elements"  construction, which writes any presheaf as a canonical colimit of representables.)
The claim is that the colimit of this functor gives $X$ again. In fact, if we have maps out of $s$ for every $s \in S$ and $s \to X$, we can get a map out of $X$ just by writing $X$ as a colimit of objects in $S$, and vice versa.
Let's call this generalized category of elements $\mathcal{E}$. We have thus chosen a diagram $D: \mathcal{E} \to \mathcal{C}$, taking values in $S$, such that $X$ is a colimit of $D$. Unfortunately, this needn't be a very good colimit.
Fortunately, if we replace $S$ with the collection $S'$ of objects which are finite colimits of $S$ (rather a small skeleton of this), we can get something better. Let $S'$ be the set in question; we are going to see that it consists of all compact objects. So we can run the same "category of elements"-type construction for $S'$ and, to any $X$, write down $X$ canonically as a colimit of objects of $S'$.
But there's something else you get here: the category of elements $(s' \to X)$ for $s' \in S'$ is actually a filtered category. (This is pretty easy: if we have $s_1' \to X$ and $s_2' \to X$, we can map them both to $s_1' \sqcup s_2' \to X$. If we have two maps $s_1' \rightrightarrows s_2' \to X$, we can form the coequalizer, which is in $S'$ and which maps to $X$.)
So we've thus found that we can write any object of $\mathcal{C}$ as a filtered colimit of objects of $S'$. (This generalizes the observation in Matt E's answer that any module is a filtered colimit of finitely presented ones.) Let $X$ be compact. Since $\hom(X, -)$ commutes with filtered colimits, if take the identity morphism $X \to X$, it must factor through one of the objects in $S'$, which means that $X$ is a retract of an object in $S'$. But $S'$ is closed under retracts (which are finite colimits). So $X \in S'$.

This argument based on "categories of elements" seems to be pretty standard in category theory, when one wants to show that an object with a "good" property (e.g. compactness) can be obtained as a special colimit of certain special good objects.
For instance, the same approach is used to prove that the tiny objects in a presheaf category are precisely the retracts of representables. (A consequence: the presheaf category on a category $\mathcal{C}$ determines the idempotent completion up to equivalence.)
Another application is to Lazard's theorem that a flat $R$-module is  a filtered colimit of free modules. Namely, given an $R$-module $M$, one can consider the category of finite free modules  $F \to M$ together with morphisms over $M$. This expresses $M$ canonically as a colimit of finite free modules. When $M$ is flat, one can show (using the identity $\hom(P, M) \simeq \hom(P, R) \otimes M$ valid for a finitely presented module $P$)  that this category turns out to be filtered.
Comment: Thanks, Akhil! You've given me some good stuff to digest.
Answer: This is a complement to Matt E's nice answer. Let us prove
Lemma A. A direct summand of a finitely presented $R$-module is finitely presented.
Consider the following condition $(*)$ on a finitely generated $R$-module $C$:
$(*)$ If
$$
0\to A\to B\to C\to0
$$
is an exact sequence of $R$-modules, and if $B$ is finitely generated, then so is $A$.
Lemma B. Condition $(*)$ holds if and only if $C$ is finitely presented.
The implication "$(*)\implies C$ finitely presented'' is clear. It is also clear that a direct summand of an $R$-module satisfying $(*)$ satisfies $(*)$. So, we are left with proving that finitely presented $R$-modules satisfy $(*)$.
Let
$$
0\to A\to B\to C\to0,\quad F\to G\to C\to0
$$
be respectively an exact sequence and a finite presentation of $C$. There are morphisms making the diagram
$$
\begin{matrix}
F&\to&G&\to&C&\to&0\\
\downarrow&&\downarrow&&||&&||\\
A&\to&B&\to&C&\to&0
\end{matrix}
$$
commute. By the Snake Lemma, the vertical arrows have isomorphic cokernels. QED
Now let us rephrase Matt E's proof that finitely presented $\implies$ compact. (I'll use the setup of Categories and Sheaves by Kashiwara and Schapira.)
Let $M$ be a finitely presented $R$-module and $F:I\to\mathcal M$ a functor from a small filtrant category to the category of $R$-modules (category which depends on the underlying universe). We must show that the natural map
$$
\lim_{\longrightarrow}\text{Hom}(M,F)\to\text{Hom}\left(M,\lim_{\longrightarrow}F\right)
$$
is bijective.
As $M$ is finitely presented, there is a finite category $J$ and a functor $G:J\to\mathcal M$ such that $G(j)$ is free and finitely generated for all $j$, and
$$
\lim_{\longrightarrow}G\simeq M.
$$
Then there are natural isomorphisms
$$
\lim_{\longrightarrow}\text{Hom}(M,F)
$$
$$
\simeq\lim_{\substack{\longrightarrow\\ i}}\text{Hom}\left(\lim_{\substack{\longrightarrow\\ j}}G(j),F(i)\right)
$$
$$
\simeq\lim_{\substack{\longrightarrow\\ i}}\lim_{\substack{\longleftarrow\\ j}}\text{Hom}(G(j),F(i))
$$
$$
\simeq\lim_{\substack{\longleftarrow\\ j}}\lim_{\substack{\longrightarrow\\ i}}\text{Hom}(G(j),F(i))
$$
$$
\simeq\lim_{\substack{\longleftarrow\\ j}}\text{Hom}\left(G(j),\lim_{\substack{\longrightarrow\\ i}}F(i)\right)
$$
$$
\simeq \text{Hom}\left(\lim_{\substack{\longrightarrow\\ j}}G(j),\lim_{\substack{\longrightarrow\\ i}}F(i)\right)
$$
$$
\simeq \text{Hom}\left(M,\lim_{\longrightarrow}F\right).
$$
The third isomorphism follows from the fact that, in the category of sets, small filtrant limits commute with finite projective limits. The fourth isomorphism results from the assumption that $G(j)$ is free and finitely generated. The other isomorphisms are obvious.
Answer: According to Simion Breaz [1] this result is due to H. Lenzing [2]. Since the introduction of Breaz's paper may be of general interest, I have appended an excerpt below.
[1] S. Breaz. Modules M such that ${\rm Ext}_1^R(M,−)$ commutes with direct limits.
[2] H. Lenzing. Endlich präsentierbare Moduln, Arch. Math. (Basel) 20 (1969), 262–266.

It is well known that commuting properties of some canonical functors (as $\operatorname{Hom}$
or tensor functors induced by a right module) provide important information (about
that module) or some important tools in the study of some subcategories for the
module category. For instance, H. Lenzing proved in [19, Satz 3] that a right $R$-module $M$ is finitely presented if and only if the functor $\operatorname{Hom}_R(M,-)$ preserves
direct limits (i.e. filtered colimits) or the tensor product $-\otimes_R M$ commutes with
direct products, [19, Satz 2]. These theorems had a great influence in modern algebra: the first result is used to define finitely presented objects in various categories,
e.g. [1], while the second is an important ingredient in Chase's characterization of
right coherent rings [11, Theorem 2.1].
The property that the covariant $\operatorname{Hom}$-functor commutes with direct sums provides us with the notion of small (compact) module, [5, p.54]. This notion is useful
in many topics in module theory as generalizations of Morita equivalences [12], almost free modules [24] or the internal structure of the ring, [27]. It is well known
that every finitely generated module (these are the modules such that the induced
covariant $\operatorname{Hom}$-functor commutes with direct unions, [29, 24.10]) is small. Moreover, for some important classes of rings (as right noetherian or right perfect by [20],
[13]) these two conditions are equivalent, but there are also important types of rings
for which there are non-finitely generated small modules (e.g. non-artinian regular
simple rings), [14]. It is proved in [26] that every small module is finitely presented
if and only if the ring is noetherian. A similar problem can be proposed for the
covariant functor $\operatorname{Ext}^1_R$: Identify classes of rings $R$ such that a functor $\operatorname{Ext}^1_R(M, -)$
commutes with direct unions (or direct limits) if and only if it commutes with direct
sums. Corollary 2.5 provides an answer to this problem. Moreover, in Example 2.6
it is shown that these two conditions are not equivalent in general. We mention
here that $R$ is right coherent exactly if for every right $R$-module $M$ the functor
$\operatorname{Ext}^1_R(M, -)$ commutes with direct unions if and only if it commutes with direct
limits, [7, Corollary 7].


Time, what is it?

Question: If you ask any person about time, she/he will give you some answer.
I suspect that it is extremely difficult, (if not impossible) to define time.
Is there a definition of what it is in physics? Is it an "axiom" that has to be taken as it is, without explanations? I also noticed that the tag "time" has no pop-ups with comments/definition/explanations.
Comment: Related: http://physics.stackexchange.com/q/15371/2451
Comment: The latest episode of Brian Greene's "The Fabric of the Cosmos" on PBS (TV) was wholly dedicated to your question. Watch it at http://motls.blogspot.com/2011/11/fabric-of-cosmos-ii.html
Comment: Time is a term describing temporal motion. In other words, we time through the temporal dimension. Because the spatial dimension is interlinked with the temporal dimension, the faster we move through the spatial dimension, the slower we time.
Answer: I am not a physicist, so take this with a massive grain of salt. I recently watched "What is Time?" on the new series, The Fabric of the Cosmos. I've also been reading a lot about this from other sources. I will try to summarize what is explained in that show in a way that makes sense.
There are some other aspects to the "what is time" question that I won't cover here much. They involve aspects of how time is measured in different parts of the universe based on the position of observers in relation to each other. The premise is that there isn't a single "now" moment that is universal. If you and an alien are stationary, you can be considered to be in the same "now". However, if you or the alien start moving around then you have to take into account how that speed is measured in relation to the other observer, and the "now" moment can either be in the future or the past of the other observer. It all ties into the relative nature of space and time, and is at the core of special relativity.
So, to the question of what is time (and putting all of the special relativity stuff on a shelf).
In general, what is presented is that time is what we perceive as the increase in entropy in the universe.
Entropy can be described as energy starting in a highly ordered state and moving into an increasingly disordered state.
A physical example, as an illustration, might be an ice cube with a drop of food coloring frozen in the center. As long as the ice cube is cold, all of the molecules of water remain frozen and the ice's crystalline structure is maintained. The molecules of the food color also remain locked into their position. The state of the entire ice cube can be consider a highly ordered state.
When the cube starts to warm up, the molecules in the ice begin moving around. The molecules in the food coloring begin moving too. The density of the food coloring decreases as its molecules spread around and mix with the water. When the energy level of the water and food coloring meet the level of energy outside, in one manner of thinking they have now entered their high-entropy state. They are throughly mixed and all the densities of h2o and dye molecules are evenly distributed. Since there are no more unequal densities, there isn't any lower-order state for them to be in.
Similar examples can be given for gases, or rocks, or plasma, or whatever. Pick any medium and the same principal applies. The biggest difference is simply the rate at which the entropy of a given system increases.
When you look at the equations that govern physics, they are solvable in either "direction". If you drop a glass and it breaks, if you reversed the directions and momentum of all the particles in the glass it should reassemble. The equations work in either direction. This means that there isn't anything really saying that energy has to move from a highly ordered to lesser ordered state. Why this happens, and why we don't spontaneously see things happen in reverse, is still a question that needs solving.
So back to time.
Current thinking is that the universe started in a highly ordered state. This means that at the exact first Planck moment of the Big Bang, all of the energy in the universe existed all together in an extremely highly ordered state. All of the forces were unified, as was the energy that became matter, antimatter, dark energy and dark matter, etc. Everything we are now was just one thing. (I have no idea what the one thing was. Like I said, SALT.)
So the idea is that, in a conceptually similar way the ice cube melted, during the last 13.7 billion years all of the energy from that initial state has been moving from a highly ordered state to a disordered state.
In the process of becoming disordered, all of the initial energy has formed the forces that guided the development of everything from subatomic particles to everything we see and are today.
As living creatures, we perceive the passage of energy through all of the physical systems around us as time. I am not entirely comfortable with this definition, as I think a barren rock in the middle of space between two galaxies would still undergo an increase in entropy regardless of whether there was a living thing to witness it or not. Again, I'm not a physicist and am simply trying to understand this stuff like many others.
So time, under this model of entropy and disorder, is not a thing in itself. Its not a force, its not something that can be pulled apart and studied with a particle accelerator. Its a term we use to describe the passage of energy through physical systems and how those systems change as a result.
So does time end? If we can say that time had a beginning as a perfectly ordered thing of energy, does time have an end? In a way, it might.
If we go back to the ice cube analogy, eventually all of the molecules in the water and dye reach an equilibrium. If left alone for long enough without external influence, the movement of molecules will cease. (Notice that in this analogy, I'm not referencing Brownian motion or movement from quantum particles emerging from the vacuum. Its just an analogy.)
When cosmologists take into account the discovery of dark energy, it paints a very said picture for the fate of the universe. Dark energy is a loose term that was defined in the last decade or so to describe observations that all galaxies are moving further and further apart. The further away a galaxy is, the faster its moving.
Scientists don't know yet what dark energy actually is, but we can see its effects in much the same way as we can see the effects of dark matter. What dark energy appears to be is some universal force that is causing everything in the universe to spread out. It was originally predicted, in a fashion, by Albert Einstein.
In his calculations, Einstein found that the universe should either be expanding or contracting. In his day, astronomers had not yet discovered that galaxies were moving away from each other. It was commonly believed that the universe was static and eternal. As a way to correct his calculations, he added an additional mathematical constant that balanced out his equations to account for what was believed to be that static universe. It was only a few years later that Edwin Hubble made the first red-shift observations that indicated the universe was, indeed, expanding.
Einstein was never very happy with his Cosmological Constant, and said,

"It is to be emphasized, however, that a positive curvature of space is given by our results, even if the supplementary term [cosmological constant] is not introduced. That term is necessary only for the purpose of making possible a quasi-static distribution of matter, as required by the fact of the small velocities of the stars."

Observations of the ever increasing rate of expansion of the universe were only made in the last 15 years or so. Its source, dubbed "dark energy", hasn't been explained yet.
When you take into consideration what dark energy means for the universe and time, here's the scenario. In something like one hundred billion years, all of the galaxies of the universe will be so far away from each other that they will be moving faster than the light they emit can reach each other. If this sounds like it conflicts with the proverbial, "nothing can travel faster than light", remember that we're talking about the expansion of space itself. Space is the fabric in which all energy and matter exist, and it can and has expanded faster than the speed of light in the past. Physicists call it inflation and it helps to explain why the universe looks the way it does now. I won't go into that too much though.
What it means is that to any future intelligent creatures in the far future, when they look out into their skies they won't see the abundance of the cosmos we see now. All they will see is an infinite black sky. The left over radiation from the Big Bang, the Cosmic Background Radiation, will have spread out so much as to be undetectable. They will not be able to observationally deduce the history of the universe as we have. To them, their universe will be just their single galaxy. They may attempt to come up with an explanation of the universe that fits the evidence they can see, but will be hopelessly wrong.
Going even further into the future, trillions of years from now (sorry, I don't have an accurate estimate for this. Its a long, long, long time though) and dark energy will become the pervading force in the universe. It will become stronger than gravity and will force any surviving collections of galaxies, star systems, and planets to fly apart. It will become stronger than the strong and weak nuclear forces and will pry apart atoms at the subatomic level.
Black holes will dominate the late universe for a long time. Eventually, even these will evaporate away their energy into space and cease to be. Look up "Hawking radiation" if you want to learn more about that.
What will be left is an immensely huge void of space filled very sparsely with subatomic particles. Particles will be spread so thinly that they will almost never interact with each other. Even if they happened to be close enough, the nuclear forces would not be able to overcome dark energy to interact with each other. Essentially, the universe will be a cold and empty place.
In many ways, when all of the particles themselves evaporate away into the quantum vacuum, time can be considered to have ended. When the energy of the universe finally has smoothed out to such a low entropy state that nothing can interact, time is done as we perceive it.
On a positive note...
I love to look up at the stars and know that I am lucky to be alive at a time in the universe where everything is still young. Interesting things are still happening and we are able to discover really interesting, important, and beautiful things about the place we exist in.
There are so many things left to be discovered that current ideas about the ultimate fate of the universe could be drastically wrong. Its very hard to say, because even the folks that have dedicated their lives to studying the science involved in these questions are still observing, experimenting, and trying to come up with explanations. Its a great time to be a scientist.
Answer: Time is like colour in that it's difficult to explain to someone who's time-blind: A clock and a mind is to time what an eye and a brain is to colour. To someone that is partially time-understanding to a sufficient degree, I can say that time is that quality which orders identical events at the same location as occuring before, simultaneously or after one another.
Comment: A color-blind person could understand about wavelengths of light, corresponding to energy levels of photons, and how structures in an eye could be receptive to particular wavelengths, and how the visual cortext could summarize the rates of firing of three different types of such structure. This wouldn't cause the color-blind person to experience color, but they could understand the physical process of how someone would. How would you explain to a "time-blind" person the physical process of how some would experience time?
Comment: @JGWeissman The Amondawa people of Brazil don't have a concept of time as developed as in west, so I would explain to them, using what's available from their culture, the idea of at the same location: 1. Two things happening together, 2. One thing happening after another, 3. One thing happening before another. Having got this, I would show how things happening can be ordered further by assigning a symbol to each happening which is called allocating a point in time. How successful I am depends upon mine and their intellect and how sophisticated their culture is.
Answer: It's the universe' way of keeping everything from happening at once. ;-)
Comment: I was serious. Your answer is funny and useless (recursive, it uses the concept of simultaneity i.e. at the same time...).
Comment: "Nature" is strange, whereas 3 dimensions are used to avoid everything being at the same place, one dimension is sufficient to avoid everything happening at once ! :=)
Comment: I'm not sure if Daedelus meant it as such, but in a way, this is a serious answer. It gets at the nature of time in physical theories as a dimension in which different events can be given different coordinates.
Comment: OK, in this sense it is clearer. Thank you for the clarification.
Comment: Hey, Physiks lover said the same thing with more words.
Answer: Unforntunately, for any such definitions, it will depend on the individual situation in which the concept is being used.  Typically physicists use Einstein's definition that "time is what is measured by a clock" but I think this is a bit too generic; after all, it is possible to use the physical dimensions of the clock to measure all sorts of things.  Using this definition you have to then define what you mean by clock and it just gets more complicated from there.
My own personal definition of "time" would run something like:
Time is that property whereby one state of a well-defined system is transformed into another.
Again, the definitions of the words used becomes a complicating factor.  I use "well-defined system" in a sense similar to how mathematicians use "well-behaved function."  I mean a system that is logically coherent and consistent and follows the fundamental laws of nature.
edit
I should probably mention that the word "state" is also very misleading as the phrase "equation of state" for a system typically refers to a property that does not change with time.  For example, the equation of state for an ideal gas is given by the ideal gas law $PV = nRT$.  I would then say that time is the property of a system for which an equation of state can be written whereby the variables in that equation are transformed to different values.
Answer: Neuroscience tells us our perceptions are stored in short term memory and after much processing, are backdated in time to give the perception of "now" at a later time. This can take many many milliseconds later, even close to a second. Or imagine watching a family movie filmed years or decades ago and becoming so engrossed in it so that it seems to be happening all over again in the present. When is now? The neurosurgeon Penfield stimulated parts of the brains of some patients and they relived events from years or decades before as if they were happening in the present. Now is not when you think it is.
In near death experiences, a person reviews memories of their life flashing by. How do you know your now isn't actually happening much later and backdated? Let's say decades in the future, you upload your mind to a computer and a replay of your life memories is made on some playback machine to be observed by some computer mind. It would seem backdated to 2011, would it not, even if it's much later.
This process and go on over and over until the end of time. Maybe "now" is a playback at the end of time at the end of the universe and we have been fooled about the actual date. Whatever it is observing the playback at the end of the world sustains us over and over again by the act of observation. Being, with no becoming.
Comment: .... too difficult....
Answer: Each paradigm shift leads to a change in the conception of time. Newton introduced the notion of absolute time. That was hardly a mainstream view before him. Special relativity overturned the nature of time, and so did general relativity after that. When quantum gravity is cracked, the nature of time will need to be revised again appropriately. Quantum gravity may or may not be the last word in fundamental physics. If not, we might expect yet another overturning of what we think time is.
At any rate, defining what time is prematurely runs the risk of stifling future paradigm shifts.

Time, what is it?


Fermentation locations

Question: Where is the ideal place to ferment my beer?
Possible locations for me are:

Extra room inside
Garage
Back porch

Of those three places, which would be the best overall place?
Answer: Whichever place can maintain a steady temp of about 62-65F for ales or 45-50F for lagers.  That's the main concern.  After that, you want to make it someplace that's easily accessible.  Someplace you can keep dark would be great, but you can always just cover the fermenter if the space isn't dark.
Comment: I would also venture to guess that the spare room will maintain the most steady temperature, since weather generally has a stronger effect on garages and back porches.
Answer: The best place would be the one that maintains a relatively constant target temperature and doesn't expose the beer to sunlight/UV.  My 2/3 season porch tends to fluctuate temperature during the day, especially on sunny days.  My garage is underground and holds a good low, cave-like, temperature (lager temps in winter).  Above-ground garages may not be so ideal.  Depending on time of year, spare rooms in my house sit at, roughly, ideal ale temperatures, but my basement is usually my go-to.
Answer: All three can work very well, as others have said the key aspects are consistent temperature and preventing exposure to light.  If none of your locations suit both these concerns then you could consider using an insulated fermentation chamber such as a freezer, fridge, or a large cooler (if your fermenter can fit).
Using a cooler will likely require some extra monitoring and temperature control (adding bottles of ice to offset the exothermic fermentation); using a fridge or freezer is a larger initial investment, but once converted they can be very nice to have (you can purchase a custom controller to keep the temperature perfectly suited to your fermentation type).
If temperature is fine, but light is an issue, then a simple large box can work fine as a chamber to block light.
Comment: Fermentation chamber is definitely the way to go. Lager in the summer!
Answer: During the winter I ferment in a little closet off my office.  It holds a nice 66 degrees which works well for most yeasts.  I have to move down to my basement to get the cooler temps in the summer.

Fermentation locations


Have too many tables in a Mysql database can affect performance?

Question: Having too many tables (200 for example) in a single Mysql database instance can reduce it performance?
Answer: In general, the more of anything you have will reduce performance. However, 200 seems like a fairly small number. 2,000 might be a real performance hit and definitely 20,000. In general, though, you should keep your number of tables small since MySQL can handle very large number of rows in the table.
Comment: A large number of tables can make a huge difference if your application makes use of 'information_schema'. Nothing in 'information_schema' is cached.  Several different ORM systems make heavy use of this.
Comment: See http://bugs.mysql.com/bug.php?id=19588
Answer: In general 200 tables should not be an issue but it depends on a number of things.
Dedicated MySQL server vs Server shared with other software, 128Mb vs 128Gb RAM etc., small tables with few records vs tables with blobs and millions of rows.
MySQL has may settings and multiple engines each effecting tables performance in different ways.
MyISAM normally has three files per table .frm, .MYD, .MYI
INNODB in normal mode has .frm with all the data stored in the central file
INNODB in one file per table mode has .frm, .idb
table_open_cache is the number of tables that can be open at once (Default 64). This may need to be larger than the number of tables in you schema as it is related to how many connections are querying the DB. 100 connections join 3 tables could mean you have 300 tables cached plus any temporary tables. Generally more complex a schema or connections the bigger the number.
open-files-limit I tend to set this to 4x table_open_cache which should be generous rather than bothering to work out an exact value.
The operating systems file handle limit for the mysql user may also be an issue (on linux the default can often be 1024, ulimit -n to show user limits) this can cause an issue with large numbers of tables when it is less that the number mysql requires. This should be at least the same as the open-files-limit.
As with any database there are hundreds of tuning parameters that you can adjust to optimise the database for your particular schema. MySQL is worse for this than most as you can plug in extra engines if you want i.e. NDB cluster engine.
http://dev.mysql.com/doc/refman/5.1/en/table-cache.html
Hope it helps
Answer: The number of tables doesn't matter so much as:

What queries you're running - you aren't likely to be querying all 200 within a single query
The overall query load on the system at any given point in time

Having excess tables does mean memory & hard drive space that could be regained & used for other things.  Keep in mind that denormalizing your tables increases the risk of bad data because you are getting rid of referencial integrity.
Answer: For every table's index MySQL has its own index. Index takes memory to hold and use, and there's global limit for indexes. When there are many tables indexes can't all be in RAM, so they go to disk, which affects performance directly. Try raising this in my.cnf: key_buffer_size=256M: this is the amount of RAM put aside for holding index information.
Answer: Can it?  Sure.  But how much depends a great deal on your app and its access patterns, and if you're using myisam or innodb, and if innodb is in file-per-table mode or not, and the size of the innodb logs.  You'll need to give us more detail than that.
Answer: No - I don't think the number of tables will be the performance bottleneck in your system.  After all, they are just files on your filesystem.  There's nothing unusual about having hundreds of tables in a database.
Far more likely is that your queries aren't properly optimised.  I would advise turning on the log-slow-queries and log-queries-not-using-indexes.  When you identify slow queries, use the explain option to view the query plan for these queries to identify the places where indexes are missing.
See http://dev.mysql.com/doc/refman/5.0/en/slow-query-log.html for more details.


Games with features like Battlestar Galactica's "Traitor" Mechanic?

Question: My local gaming group has been playing the heck out of Battlestar Galactica for some time now.  Are there any other board games that can be recommended that have a similar theme, or similar mechanics?  Obviously, we're not talking about clones or rip-offs here, but games that embody the "Everyone's in it together except for 1 or 2 who are out to get you.  You better find out who they are!" theme.
Answer: There are many games with a "traitor" or "uninformed majority" type mechanics. In some of them, such as Mafia and Are You The Traitor?, that is the main point of the game; you are trying to discover each person's role, with not much else on top of that. In others, it's a more incidental feature, usually involving the win conditions for each player or team, but with more traditional gameplay mechanics for the main duration of the game. In the following descriptions, I sketch only out how the "traitor" mechanics work for each game, not how the whole game works.
There's Mafia, also known as Werewolf or Assassin, which is probably the most well known, and one of the oldest, of the genre. In its most basic form (using Werewolf terminology), you have two Werewolves, one Seer, and the rest are Villagers, though only the Werewolves know who each other are. Each day, everyone (Villagers, Seer, and Werewolves) try to decide who the Werewolves are, and hang one person. Each night, the Werewolves silently and secretly choose one person to kill. Each night, the Seer gets to discover, silently and secret, whether one player is the Werewolf. The humans (Villagers and Seer) win if they kill both of the Werewolves, the Werewolves win if they equal the humans in number. The Seer is trying to help guide the Villagers to kill the Werewolves instead of innocent Villagers, without revealing himself as he will be killed by the Werewolves the next night (unless he identifies both of them openly; but of course, a Werewolf could also claim to be the Seer...).
In Are You The Traitor? from Looney Labs, there is a Key-Holder, trying to deliver a key to the Good Wizard, while the Evil Wizard tries to get the key instead (only the wizards know which is which). The Guards are trying to help the Key-Holder get the key to the Good Wizard, but there's a Traitor in their midst, who's trying to (secretly) reveal to the Evil Wizard who has the key, without being caught by the Guards. The Good team wins if the Key-Holder delivers the key to the Good Wizard, or the Guards catch the Traitor. The evil team wins if the Key-Holder accidentally delivers the key to the Evil Wizard, the Guards attack someone other than the Traitor, or the Evil Wizard catches the Key-Holder is before the Key-Holder delivers the key (the Traitor can never win directly, only by communicating secretly to the Evil Wizard). It's an interesting twist on traitor mechanics, lots of different roles, all with very different goals, but still has the basics of the traitor mechanics.
The multi-player version of Homeworlds, an Icehouse game has a traitor mechanic, though the 2 player version, Binary Homeworlds, is more popular. In the multi-player version, players are secretly assigned Good and Evil roles. The basic mechanics of the game are an abstract strategy game, in which players are trying to eliminate each other (I won't go into the details, as they aren't relevant to the traitor mechanic). The Good players are on a team; they win if all Evil players are eliminated. Each Evil player is playing for themselves; they win if they eliminate any other player, Good or Evil. So, Good players are all cooperating, and Evil players are all competing.
Bang! is a spaghetti western card game, with a Sheriff who is known to everyone, and Deputies, Outlaws, and a Renegade. The Outlaws are trying to kill the Sheriff, the Sheriff and Deputies are trying to kill the Outlaws and Renegade, and the Renegade is trying to be the last man standing.
Napoleon is a trick-taking card game (played with ordinary playing cards) originally from Japan (and apparently over 100 years old, though not in its exact present form). 10 cards are dealt to each player, and two blinds are left. The players bid a number of tricks and a trump suit, similar to Bridge. The winning bidder is Napoleon. He names another card, and the player who has that card becomes his General (but only the General knows that); they together need to make the bid to win the game. Napoleon now gets to take the two blinds, and discard two cards; if the card he named was one of the blinds, he's playing alone, with no General. The game is then played as a trick taking game with the named trump. No one but the General knows who that is, and Napoleon and the General (or Napoleon alone) are playing on a team against the other players, who are also on a team.
Shadows Over Camelot is a cooperative game, but there is a variant in which one player might be a traitor (or there may be no traitor). The players are working together against the game, but the traitor will try and sabotage the progress of the other players while remaining undiscovered. The loyal knights may accuse someone of being the traitor; if they're correct, they get an advantage, if they're wrong, they're set back; of course, depending on the game, there might not even be a traitor. If the traitor remains secret until the end of the game, he can set the players back and turn victory into defeat.
In The Resistance you are a group of resistance fighter running missions against a corrupt and evil government. However, there are government spies in your midst. You need to complete three out of five missions successfully, without a spy sabotaging the mission. Each round, a new player is the leader, and must select a team, trying to choose only loyal resistance fighters and no spies. After selecting the team, the players vote on whether that team should be formed. If the spies sow discord, and no team can be formed after 5 tries, the spies win and the government crushes the resistance. Otherwise, they go on the mission; but any spies on the mission have the chance to sabotage it, causing it to fail. If they do, however, they risk being found out.
Comment: Keep in mind, that Traitorious Knights can also accuse other knights of being traitors to get an advantage (each player only gets one opportunity to accuse someone)
Comment: The "uninformed majority" type of games can be played both live and online.
Also, they can be either short games (less than 2 hours) or long games where the game day-night cycles are in sync with real life days and nights.

You can read more about this game variations on
http://exposemafia.com/about
Answer: In Betrayal at House on the Hill, someone becomes a traitor part of the way through the game. The traitor gets a secret agenda, but the group of remaining players gets a related counter agenda. Each side is unaware of what the other side is trying to accomplish.
Answer: I would recommend you check out Shadows Over Camelot it was the first game I played with a traitor mechanic. It's a cooperative boardgame with (obviously) an Arthurian theme, and if I remember correctly, a traitor is possible in each game, not mandatory.
So you get "Is there a traitor?" questions instead of just "Which one is the traitor?"
Answer: Saboteur is an excellent card-based game with one or more traitors.
You are a group of dwarves tunneling toward gold, but some of you are traitors paid by a rival mining company.  Saboteur is especially great with large groups (7 or 8, the game supports up to 10) as the accusations fly fast and furiously.
Individual games go quickly, and you play 3 games in a row to decide an overall winner.
Answer: La Resistance! This is a great Werewolf/Mafia game that runs without a moderator, and even works for 5 players! It's tense, tactical, emotionally draining, and best of all, everyone plays right up until the end of the game, no-one gets eliminated during play.
The one drawback is that is becomes a little luck-based as the number of players reaches 8 and beyond, so at that point, we tend to switch back to Werewolf.
Comment: I really like this game. I think it's much better than Werewolf, which is fun, but is very random, especially with a lot of people. I disagree that The Resistance becomes more luck based with more players. I've played with all different numbers of players and I think it's just as good at every number. Actually, I tend to prefer more than seven because the first mission has three people and doesn't always go to the resistance.
Answer: Those types of games are normally called "uninformed majority games"; one game I quite like that's in that class is Mafia (Werewolf, Assassins, etc.). The base game is pretty much as simple as uninformed majority games get; every night the mafia players choose someone to eliminate, and every day the townspeople try to figure out who the mafia players are. Mafia is entertaining mostly because there are dozens of extra roles and house rules people have added that can be played in ridiculous numbers of combinations, so you can almost certainly find something that your particular group enjoys
Answer: The expansion for Pandemic includes a 'bio-terrorist' where one of the players is secretly a 'traitor'.
Comment: -1 because the bio-terrorist is not a secret traitor like in BSG, but his identity is known from the beginning. The bio-terrorist is more like Mr. X from Scotland Yard (everyone knows *who* he is, the problem is more *where* he is)
Answer: I've designed and released two free print-and-play games that have traitor mechanics:
Haunted Destinies: All players start as amnesiacs (Each player has a small set of face-down Psyche cards that you're not allowed to look at.).  The goal is to figure out who The Opener is and prevent that player from killing all of the other players.  If you realize that you're The Opener, your goal is to kill all of the other players.  If you die, and you're not The Opener, you become a Ghost.  Special "Haunting" cards come up during play that give the Ghosts special victory conditions.  There's also a novel mechanic that lets you "steal" destinies from other players, so it's possible to steal The Opener role from a player and take it for yourself.
Space Monster: Playable as pure co-op, one-vs-many, or with an "up for grabs" traitor.  The plot of Space Monster parallels the movie Alien – Figure out the space monster's weaknesses and kill it before it eats all of you.  The "up for grabs" traitor mechanic comes into play with the escape pod.  There's one escape pod, with enough room for one person.  If you think that the monster will win, you can bail and leave in the escape pod.  If you do this, and the monster wins, you win instead.  With this set-up, any player can become a "traitor."
Answer: Nostra City is nice, I played it once a few weeks ago.
The story is that the mafia godfather is arrested and faces enough charges to spend the rest of his life in prison.
He makes clear to his men that when they manage to save him from prison, he will retire and the one who did most to save him will take his place.
The players are his lieutenants and try to bribe or threaten witnesses, judges and so on in order to get him found unguilty. Plus, they have to collect victory points (called "respect" in this game).
But some of the players are FBI agents (when I recall it correctly from the one game I played, there were two of them in a five player game).
Unlike in Battlestar Galactica, roles are not assigned at the start of the game. Instead, there is a deck of cards where every player draws from in each round anyway, and the "You are an FBI agent" cards are somewhere in there.
So at the beginning of the game all players are on the same side, but during the game some of the players will secretly change sides and become FBI agents.
In the end, the winner is declared like this:

If the Godfather is not guilty, the FBI agents have lost and the lieutenant with the most respect (victory points) is the winner
If the Godfather is found guilty, the lieutenants have lost and the FBI agent with the most respect is the winner
Answer: Fury of Dracula has a fog-of-war mechanic in which one player plays Dracula, whose moves are generally hidden from the other players, while the other players collude (in plain view of both each other and Dracula) to kill Dracula. Everyone knows up front who is playing Dracula, but the other players have to guess what Dracula is doing.
Comment: That reminds me of the board game Scotland Yard, in which case one player in Mr./Mrs. X and the other players are trying to catch him/her. http://en.wikipedia.org/wiki/Scotland_Yard_%28board_game%29


How does the cyclic-ketogenic diet allow weight loss and muscle gain?

Question: The common notion is that in order to gain muscle you must be a "calorie surplus", and in order to burn fat/lose weight you must be a in a "calorie deficit. However the cyclic-ketogenic diet claims that it allows muscle building AND fat loss? How is this possible and what is the mechanism behind it? Since surely you cannot be in a surplus and deficit at the same time?
Comment: I'm not familiar with that diet, but I wouldn't say that either of those common notions are exactly correct for people who take on a new form of exercise.
Answer: I know someone who recently used the cyclic ketogenic diet to get good results.  Based on his experience, and some of the other experiences I've read who used the approach, it's important to set your expectations appropriately.

The only demographic that can build muscle while losing weight is an overweight beginner.  This is the same regardless of diet used.
Intermediate and advanced lifters will lose some maximum intensity, but their volume work would be unaffected.
The cyclic ketogenic diet is effective at fat loss while preserving the muscle you have acquired.

Ketosis is itself a contentious subject with researchers on both sides of the argument having studies to back up their arguments.  Unfortunately, most studies use readily available sedentary people, and rarely any demographic that would match my own.  I will also say that ketosis is not for everyone.  I did well with it, but I have friends who didn't.  I recommend a trial run without any exercise to see how you adjust.  The principles behind ketosis are fairly simple and straightforward:

The brain needs roughly 125g of energy per day--blood glucose is the primary source.  Lipids can't cross the blood-brain barrier, but glucose and ketones can.
When the blood glucose drops, the pancreas secretes glucagon to pull energy out of fat, muscles, and organs.  The muscles and organs store glucose in the form of glycogen.
When there are no dietary sources of blood glucose for a prolonged period, the liver will start making ketone bodies which process the stored fat in a way that can pass the blood-brain barrier.
Ketone bodies cannot be reabsorbed back into the fat cells.  Unused ketone bodies are urinated out.
About the same time that ketone bodies are created, the body also starts gluconeogenisis which is the process of converting proteins to blood glucose.

All ketogenic diets share a common trait: medium to high protein intake.  The higher protein intake is necessary to provide the gluconeogenisis process a source of protein other than your muscles and organs.  Additionally, if you are lifting weights, the higher protein intake is needed to repair your muscles and supply the necessary amino acids.  Building muscle does take a lot of energy, but thankfully the body can use fat as a source of energy.
Ketosis based diets will completely deplete the glycogen stores in your muscles and organs by design.  This has a negative impact on performance.  Unchecked, high intensity exercise like weight lifting can increase the demands of energy enough where ketosis alone isn't enough to supply the energy requirements.  This is a bad thing.  The cyclic ketogenic diet replenishes the glycogen stores once a week.  This gives you most of the week to burn fat, while keeping your energy up to meet the demands of exercise.
There have been several studies to show that there is a direct correlation between your body's maximum ability and its weight.  As you lose weight, you can expect a loss of your maximum strength.  Part of this is due to any cut causing a loss of lean mass as well as fat mass.  That doesn't mean you are losing muscle, it very well could be some of the energy support systems that feed the muscle.  However, this loss of absolute strength is relatively small compared to your fat loss.  Once you reach your goal, you can rebuild that max strength.
The high fat content during the week in the cyclic ketogenic diet is sufficient to fuel your training.  However, the body does do better with a periodic refeed when you are training hard.
Answer: I don't think there's very much reliable empirical evidence available on the subject. Every mention of this diet that I've seen tends to be either hyperbolic, anecdotal, or very clearly tainted by conflict-of-interest. Wikipedia's page on the subject, for example, uses as its main reference a book which advocates said diet. It's possible that the author is a verified expert who's diligently studied the matter and scrupulously documented all his sources, but my default reaction is extreme skepticism. For what it's worth, ketogenic diets have been used in the treatment of epilepsy for decades, and the medical consensus appears to be that they're generally safe. That said, there's some evidence that ketogenic diets will, if anything, impair performance during anaerobic activities due to more rapid glycogen depletion (q.v., http://www.ncbi.nlm.nih.gov/pmc/articles/PMC524027/), though aerobic performance should not be detrimentally impacted. Given that muscle-building gains are largely dependent on anaerobic types of exercise, and that greater intensity and increased repetition to failure are generally correlated with greater gains, I would expect negative results, if anything.
As far as the hypothetical mechanism for these diets, the idea is that if the body is deprived of carbohydrates it is forced to begin to metabolize fat into fatty acids through lipolysis. Those fatty acids are then further metabolized into ketone bodies (hence, the name of the diet). Whether this actually results in improved fat loss vs. carb-containing diets while holding caloric intake constant isn't known with any certainty.
Answer: The main idea is  to  have two phases.
Short period of carbs restrictions to lower insulin production and thus heightened lipolysys.
At that time high protein content  of  the  nutrition along  with  weight training must  pre-serve muscles from wasting even  despite lower caloric income.
Then in the second phase carb  refeed comes in raising metabolic rate, hormones production and
improving  metabolism.
This is the idea.
detailed discussion may be found in UUD2 book  by  Lyle McDonald.
Personally i found this approach to  be too difficult for the results.
However  the diet (UUD2,  which  is a detailed plan  based on  carb cycling combined with very  special  training) is quite effective to minimize fat gains during  bulking  or maintain muscles during  fat loss.


Witch Doctor build for Inferno?

Question: I am in the beginning of Inferno and I am having a bit of trouble some places, I have been using the zombie bears build (this or a variant of it), while the damage is very nice and it is a very good build for groups I find some champion packs/elites are impossible to take out with it when running solo. So what is a more well rounded build for soloing Inferno as a Witch Doctor?
Answer: So far I've used two builds with a decent amount of success soloing Inferno Acts 1 and 2 (just grinding for gear and killing champion packs and the end boss). Honestly, the best and most powerful thing you can do in Inferno is just gear yourself up. I've heard of a lot of random builds working, as long as you gear yourself in a specific way and have a pretty decent weapon. Also, using the Templar pet while soloing is a good idea, since he will actually tank for you.
As far as a specific build for soloing champions in inferno, I've had pretty good luck with a poison dart build. For this to be successful you'll need to stack a little bit of attack speed and have a pretty decent weapon though. You'll also need to prioritize abilities that'll let you kite. I like the DoTs because they also help with large packs and you don't just get melted, but you can probably change out some of that stuff for your own personal preferences for your own play style.
I like the ZomBears build, but I do agree that it is difficult to solo with because you've got to be fairly close for it to work. I've been playing around with build alternatives and so far I've only come up with the Poison Dart build.
Editing this in since I just tried a new variant of poison dart build that I liked more. This build for darts was much more effective due to the amount of CC you gain from Horrify and Hex. I had a much easier time doing solo Inferno Butcher runs with this build that my previously mentioned one.
Link to ZomBear build I used for a while that also worked well.
This Pet Build has also been said to be pretty effective, but I've not tested it much myself. It looks interesting though, and a couple of people on Blizzard's Witch Doctor forums have said it is good.
Comment: the 2nd build you posted for your darts was extremely ineffective for me...i just died to four spiders (teleport, extra health, mortar) in hell act 2. the CC is a good idea, but it's extremely situational. the giant just dies a lot and stuns maybe 2 times in his lifetime. I have a little under 11k damage and 2.21 attacks/second. The CC is good, but I ended up just running around with almost no health (around 2k out of 25k).
Comment: Hmm. I've not come across any issues like that with it. The situation is different though so would be hard for me to be able to test. Pretty much any WD build is going to have a rough time with Tele/Mort creatures though because they're difficult to kite.
Comment: seems like any poison dart build would require roughly 2.00+as with high amount of dps, else you'll be kiting forever.
Answer: I use spirit walk (mana regen) to go into the pack  then active soul harvest (life steal 9K HP + 650 Int) then go back from a few metters slow with the 80% Area and throwing bears with my Zombies Wall just in front of me so i have these 4 spells in CD ==> 300% regen mana (masterie) ==> unlimited bears ... until the slows stop (8 sec duraction /// 8 sec CD ) then recast the slow few meters back, back a little w8ting spirit walk and harvest (both 15 sec CD) and start over.
with masterie +10% mana/ globe and some range of grabbing the 2nd spirit walk grab 3-4 globes granting +30-40% int + 650 + 9k HP from harvest + globes ==> double my int ==> double dmgs  and full life
if i can t kill the guy before they reach me and they kill me i got masteries that saves me and i back, slow and start again the combo :-D
i advise some grabbing range and some bonus life on globes so u can never die :-D
PS : sry for the terms, i don t realy know the names of the skills in english ;-)
Answer: Here is my build.
As with most WD builds, the main focus here is on mobility. While spamming Grasp and having several Fetishes out at any time, the mobs are generally caught up either being slowed or swinging at Fetishes/Gargantuan.
You can stack high damage by Soul Harvest and the passive Gruesome Feast, enabling some high initial/DoT Acid Clouds (note that it has the Acid Rain rune, hitting more mobs with the initial). Gargantuan and Fetishes also benefit from Int boosts.
The primary, Rain of Toads, is basically a concentrated Acid Cloud with higher damage in smaller radius, and is highly stackable. Use it for champions, elites, and other key mobs like spawners.
If you're low on mana after dropping several Acid Clouds, Spirit Walk will gain you some mana. I mainly Spirit Walk to hunt the area for dropped Globes which will provide additional mana and boost your Int.
Answer: My build
http://us.battle.net/d3/en/calculator/witch-doctor#aZUTjX!ZWV!aaaZbY
Not sure if anyone else still reads this but here's my build.  I've found it pretty useful with DPS from 14k-30k.  ATK% and Regeneration are great for this build.
Basically Grasp where you want to control the field, and let the Garg take the front while you pepper with Darts.  When things get close use your crowd controls starting with Confuse, anything not caught with the Confuse is stunned and Spirit yourself away and place a new Grasp.  Save your Horrify for the armor buff and Confuse cooldown.  Hope this helps.
Answer: Anyone take advantage of stun/immobilize/freeze gear paired with mana regen.  Most of my gear has a chance to do one of the above.  I use Garg and zombie dogs with acid rain and posion darts and spirit walk with mana regen when my mana runs out and big bad voodoo rain dance for large packs and/or elites.  It works great on hell diablo and so far is ok in inferno.
Answer: I've completed inferno with the witchdoctor. There seems to be a lot of BS running around here so I'll clear the air, take it or leave it.
My build: Poison darts (splinters), Bats (Dire bats), Spirit Walk (the one that extends the duration of the walk), soul harvest (increase damage), Big Bad Voodoo (use mana regen one normally, and damage increase one for doing Belial farms You need to kill him quickly), Confuse (increase radius)
Spirit walk into packs to cast soul harvest buff, and spirit walk whenever it's off cooldown will save you a lot. Passives use spirit vessel, vision quest, and the one that decreases damage 20%. I cannot stress the importance of having spirit vessel as it has saved me time and again from a hefty repair bill. I also can't stress enough the importance of dire bats. They chew up a lot of mana, but they have excellent range and go through targets and do area damage twice if not three times as powerful as poison darts. With a decent weapon like a ceremonial blade that does somewhere around 1000 dps and a lot of INT stacked you can mow enemies over all the way through inferno with dire bats.
Also you're not trying very hard with gear farming if you get one shotted by anything in inferno. Stack 600-700 resist all (not hard to do) and with 30k health you can take quite a few shots.
Pets are the worst idea for inferno, lose the pets. bears are to limiting with their piss poor range. An act 3 champ pack will destroy you in a few quick shots if you are relying on bears as a skill. You need to kite with bats.
Comment: Many people do quite well in Act 3 against champ packs with bears. You do not *need* to kite with bats.  Both are perfectly viable, so claiming one is and the other is not is wrong.
Comment: Because people were using vision quest to spam with bears, which is conterproductive enough for blizzard to notice so they will be changing that skill all together. Having 4 skills on CD is just plain stupid, it handicaps you from playing the game an intelligent way. I contend that those so-called "people doing well with bears" could never solo their way through inferno, but rather are co-oping where that build can actually be viable. I've seen and tested this first hand and you cannot solo inferno with a bears build and not spend mucho dinero on repair costs.
Comment: Perhaps you cannot solo inferno with a bears build, but others absolutely have. Your claims that it is impossible are factually incorrect.  The parts about VQ being "stupid" are subjective, and don't really belong here (not saying I disagree; just that answers here are expected to represent fact and/or direct assistance to the main question, rather than tangential discussion of opinion).
Answer: The zombie bear build works great once you time when to hit spirit walk to do the most damage then to run and kite the mobs until you use the bears again.
Answer: I just hit Inferno, and was surprised at just how well my zombear build's been working: http://us.battle.net/d3/en/calculator/witch-doctor#TfUdR!ZUV!ccZZa.  I left the 4th skill blank since I haven't decided what I like best (I'm probably going to go with horrify/frightening aspect for the armor bonus).  The key is that when you have 4 cooldowns with vision quest you can zombear pretty much indefinitely.
I threw the bruiser to left-mouse since it made it easier to move without attacking something instead, but basically when I see enemies or a boss pack, I make sure bruiser/hex are on cooldown, and rush in spirit walk/soulharvest then zombears for a few seconds to do as much dmg as possible.  Then I dodge/kite while casting zombears when I can to take out what didn't die initially.  The bruiser's stun and hex keep enough enemies down that it's usually pretty quick to mop up all but the really nasty boss packs.
I just have cheap gear I picked up at the AH (10k damage base = 30-40k zombears, 40k life or so)
Answer: Vision Quest build is by far the best build. I've solod and strolled through hell barely ever taking a death. im in act 2 inferno now, and it obviously wont matter what build you use, youre getting 1 hit ko'ed
Comment: You are not adding anything that hasn't already been said.  If you're getting 1 hit, why aren't you changing your build to account for that?
Answer: http://us.battle.net/d3/en/calculator/witch-doctor#aUZSdT!ZWV!acacZc
My build - It works ok because of my massive attack speed.
Comment: Please elaborate on exactly why this build works.  Just posting a link is not a good answer.

Witch Doctor build for Inferno?


Defining 'HTML5 Interface'

Question: I've seen people talk about sites upgrading from Flash to HTML5, but there's no sign of HTML5 anywhere in their source. So what constitutes as an "HTML5 interface"? Is this term becoming the next "Web 2.0"?
Answer: I agree with previous answers (Christopher and user5514) as they state HTML5 is becoming a new buzzword and usually misenterpreted.
As a WHATWG member I feel the urge to point what HTML5 is and what isn't in short.
HTML5 is a new technology definition, intended to turn into a standard (perhaps when this happen for real, I mean, HTML5 proposal becomes a W3C recommendation, too much time passed).
In practice, HTML5 is more than new tags, properties and Doctype. These things (Doctype, tags, properties) are just part of it. HTML5 also defines a lot of new interfaces on Document Object Model, and those can be used with any scripting language that works on or produce HTML5 documents (javascript, vbscript, PHP DOM extension...).
Another important feature is the web related technologies interoperability (SVG, canvas, MathML, Ruby Notation...).
When analizing a new "HTML5" website, despite it contains only "old" tags, it can use a new interface on scripting part.
Ah, another point that worths to mention is: we can define new standards. On the other hand, it should worth nothing if there is no browser support - what makes the standards work in the wild.
Keep an eye into this, and have nice HTML5 experiences.
Answer: You will always have people who misinterpret or misuse a term, though from what I've seen out on the web an HTML5 interface still contains HTML5 markup.
For it to become another "Web 2.0" term would mean that there wouldn't be an actual difference between HTML5 and other web languages.
This is mostly up for your own interpretation.
Answer: Agree with Christopher. Strictly it just refers to the next standard of HTML (HTML 5.0). The feeling of HTML5 becoming a buzzword is because it supports and simplifies many previously complicated/convoluted issues like embedding Flash, audio, and other media.
Here is a great primer:
http://diveintohtml5.org/

Defining 'HTML5 Interface'


XNA: Perform Collision Detection Against 3D Scene Geometry

Question: How would one perform collision detection between a simple scene like the one below and a simple object like a sphere?
I assume the best solution would involve convex hulls in some way but I have no idea how that works (still learning :)
Answer: If the terrain is fixed you can do some precalculations to cut down on the problem size.  For example, for each sector (pick reasonable sizes to work with) you can define a guaranteed-clear altitude.  Each object also has a maximum object height variable.
If the object position - object height is greater than the known safe altitude then no further checks need to be done.  This won't do anything about your worst time but if you have a lot of moving objects that are mostly clear of the terrain then it will do a lot for your average time.
Answer: For simple scenes, use an aggregate of geometric primitives for collision testing. Your example can be broken down into a finite plane, box and rectangular prism. Since the only object interacting is a single sphere, your testing can be kept very simple. If your scene is considerably larger, an quadtree or octree approach to organizing collision shapes is sufficient enough and can ensure that you only test what is close to the sphere.
Convex hulls would apply to the moving sphere, for representing the space that the sphere moves within a given time frame. This space is best envisioned as a short capsule with round ends. For collision detection, you can break this into one cylinder and two half-spheres and test for those.
Going back to the quadtree, you will first recursively check which leaves/nodes the capsule shape is occupying, not the sphere (capsule-box intersection). Then iterate through the collision shapes that are in them and do a test case for each.
It's probably not the most efficient method for collision checking (sweep and prune handle many more scenarios) but for one sphere this is good enough.
Answer: I use Mogre3D, but I think this could work in XNA as well.
Use a raycast that is cast downward, check where the ray hits the poly or face so that your character does not fall through the level.
Don't start the ray for very high, just something above your character, so the character can still walk under a bridge or something.
If you are using high detailed scenes, try creating a very low poly mesh that you can use with your ray, so that the performance is improved greatly.
Then for doodads, or scene objects,  use a collision sphere, some kind of fixed radius  or bounding box when running around your character.
For speeding up the loading of your game, you could already pre-calculate the triangles of the scene mesh.
Hope this helps.
T.


What area of Los Angeles does this historical map depict?

Question: What area of Los Angeles does this historical map depict?
Comment: Please note, some of the existing street names are on your map.
Answer: That appears to be the survey of LA drawn out by Lieutenant Edward Ord in 1849.
If so, then the Fort Street in the grid-like area in the upper left area of the map later became Broadway street. Hill Street, the next over from Fort Street on the survey, is still the next street NW from Broadway today.
As for where on Broadway that is, I'd say its almost certainly centered around 1'st St. For one thing, LA's grid bends a bit there much like the bend on this map. But more importantly, cities tend to grow outward, with the center of their downtowns typically occupying the area of original settlement.


94' Dodge Spirit 3.0l v6 Throttle position sensor over 4.96V

Question: I pulled up the codes from my 94 spirit that's been running sluggish and unresponsive lately. I got a 24* Throttle position sensor over 4.96V. I put a new TPS in and I'm still getting this code. Do I have to unplug the battery and let my computer reset? I keep seeing videos saying you can adjust your TPS but I'm not sure how they do it by moving it. Thoughts?
Answer: If you made sure to clear the codes and it came back, then you probably have high resistance or a broken wire. Most sensors on a car use voltage on sensor wire at 5V, and then ground is applied by what ever sensing mechanism and brings the voltage down... So if the sensor wire is broken or the return wire (aka ground) , the voltage can't be dropped and sticks high.
Keep in mind high Resistance could be an actual break in the wire, corrosion, or connector terminals spread open. Need to check it all.


Why is the water in my toilets orange?

Question: My wife and I just moved in to a brand new house and have started to notice that the water in the toilets is an orange color, and is starting to leave stains in the bowl. After looking online briefly, I have seen the suggestion that perhaps there is rust in the plumbing. I know that all of the plumbing in our house consists of what appears to be plastic piping... is there any other explanation? What steps should be taken to fix the problem?
Comment: do you have a well or municipal water source?
Comment: Our water comes from the city.
Comment: If it's sediment, like clay in the pipes from a messy installer, then clean your aerators and toilet fill valves. But I'm guessing woodchips is right on the bacteria suggestion.
Answer: If the house was vacant for a long period of time, there might be build up in the pipes. Try running a number of facuets for a while and flush the toilet over and over.  Assuming your incoming water isn't orange, it should go away after you flush out the pipes.
Also open the tank and have a look in there - maybe there is a component of the toilet that is rusting, or maybe there is some type of cleaning tablet present (usually they are blue but you never know).
EDIT: You mentioned in your comment that there is orange sediment on the bottom of the tank.  I can't really guess at what this is (maybe orange soil as someone in the comments suggested), but I would suggest that you turn off the water to the toilet, flush it to empty the tank, and then using a sponge, scrub it off, or at least disturb it enough that it will wash away when the tank refills.  Turn the water back on and flush it away.
If it comes back after some time then you'd need to investigate further..
Comment: +1 for looking in the tank. Is the water in the tank orange? Is it orange when the tank is refilling itself?
Comment: The house is brand new, but my wife and I have lived in it for about two weeks now. During the last two months or so of construction though, the water was running and was used fairly frequently. Opening the tank reveals water that appears mostly clear, but a layer of some sort of orange sediment at the bottom of the tank. Could this be from when the water was first turned on?
Comment: Any chance you have red dirt in your area?
Answer: This is clearly not rust in this case, nor is it probably sediment, but bacteria. They are probably living on something in the water, and often that is iron.
Does your hot water also smell a bit, with a bit of rotten egg smell?
If so, then you probably have either iron or manganese in your water supply to the house. Bacteria live in your hot water tank, and your toilet tank.
Get rid of the iron to get rid of the things it causes. To do that, you will need to remove it somehow, and there are several treatments available once you figure out how much iron is there. Start with a home test kit, which you can buy inexpensively from your local home store.
Answer: If you have potassium thiocyanate you can test for ferric (Iron-3 aka rust) by dropping a few drops into the toilet. If it's rust, it will turn blood red. Ferric isn't terribly damaging as far as health-effects go, but it is definitely gross. I'm testing one of the drinking fountains at my school right now with the AP Chem class. Based on my research, it could also be Chromate or Pervanadyl, which could cause potential health risks. Good Luck.
Answer: the stain is coming from the water supply. county water would not tell me what they use but it started at my house three years after i had moved in. at that time the county change chemical vendor and the orange stain appeared. it has been there now for 12 years. it easily wipes away but you must wipe it.
Answer: My water ran orange or red like our dirt after the fire department tapped the water lines to fight a house fire in our neighborhood. I called the water department and that's the first thing they checked. They told me that it would run clear after the lines were flushed and recapped. What color is the dirt in your region?

Why is the water in my toilets orange?