What are the polygon count standards for game assets? I'm thinking about games that are lower poly but aim for the best graphics possible, I'm thinking like L4D / L4D2 or Insurgency level of detail.
Recently, I am trying to create 3D models for my own game project. However, when I do some researches on game models, I come up with the term - game-ready models and I could not find any clear definition online. Could anyone tell me about what exactly is game-ready models and how to know if my 3D models (created by Blender mainly) is suitable for games.  From what I know by far, quad topology is preferred in Blender. However, for most of the game engine, triangle topology is preferred (Please correct me if I am wrong). Does it mean I have to manually change the models in Blender from quad topology to triangle topology before importing to the game engine?  Thanks for any helps.
When using an Int- or FloatProperty, is there any way of changing their min and max values dynamically?  I'm currently loading images on the fly and would like to limit a frame property to run from 0 to len(loaded_images) every time I call the loading operator. However, the property seems to be read-only and re-assigning gives the following error:  bpy.context.scene.project_props.bl_rna.properties['myIntProp'] = bpy.props.IntProperty(...) TypeError: bpy_prop_collection[key] = value: invalid, expected a StructRNA type or None, not a tuple   I understand the bpy.props constructor uses an intermediary tuple, but I'm not sure how to proceed. Is there any way around this? Is the 2DS region locked? Will I be able to play NA games on a Japanese System? If I purchase a 2DS from Japan, will I be able to play American games on it? Should we `enable` the User Systemd service `pulseaudio.service`? Checking if a key exists in a JavaScript object? How to remove garlic smell from refrigerator/ice maker Why is the product of all units of a finite field equal to $-1$? Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as  $$ a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. $$ However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks! Almost surely convergence of the sequence Let ${X_n}$ be a sequence of independent and identically distributed, square integrable random variables.   Write $ u = E(X_n)$. Study the almost sure convergence, as $n \rightarrow \infty$, $$S_n = (X_1X_2 + X_2X_3 + ... + X_{n-1}X_{n})/n$$   Since $X_iX_{i+1}$ are not independent, it seems we cannot directly use law of large number for that, so anyone can give me some idea? Integral $ \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx $ I am trying to solve this integral $$ \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx $$ A closed form does exist despite the looks of the integrand.  This problem is from some old high school IMO training courses.  I am not sure how to solve it. The only information that may be of help is $$ \sin x=\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x=\sum_{n=1}^\infty \frac{ x^{2n}}{(2n)!} \quad \forall \ x $$ Thanks, I am looking for a complete solution, not a description as of what to do. I work sometimes till 4 am and my friends and the rest of the world obviously don't know in advance that I don't want to be called at 8 am.  Is there an app or setting that can allow me to have an emergency override list of people I need to be woken up for if a possible emergency arises but ignore all other calls until I wake up?    I'm familiar with the Do Not Disturb option, but it has limitations that it blocks all calls. What if there is an emergency at night?  What if I want to allow certain callers during the Do not Disturb time? Is there a way to have an emergency list that allows calls during silent? What is the name of an anime from 80s-90s movie/series in which there is a scene, maybe the end, where a girl (who has an​ armor) exchanges age with a dying man in a beach, so the man transforms into a baby and the girl grows to a woman? SFDX Pull error: Entity of type 'ListView' named XXX cannot be found While trying to pull changes from a scratch org (sfdx force:source:pull, tried with and without -f), I'm getting this error:     Entity of type 'ListView' named 'Note_Relation__c.My_Notes' cannot be found.   I do have a custom object Note_Relation__c, and it exists both locally and in the scratch org.  However, it's not an object that is normally accessed directly.  It has no tab, and should have no list views.  Nonetheless, I've checked the object's list view (via prefix, e.g., /a0D in both Classic and LEX view), and I can find no list views at all.  I've also tried adding an entry to .forceignore:  force-app/main/default/objects/Note_Relation__c/listViews/My_Notes.listView   This has no effect either.  At this point, I'm stuck with a bunch of config changes in a scratch org that I cannot pull.  I can find no references to this error message with scratch orgs.  Has anyone seen this issue?  Edit to clarify: I tried the .forceignore line based on what I think the file name should be; I have no force-app/main/default/objects/Note_Relation__c/listViews directory locally.  Edit 2: Workaround:  I found a workaround that doesn't make sense to me, but it works.  I'm not adding it as an answer yet because I'm not sure what's happening.  The workaround is to add the following line to .forceignore  Note_Relation__c.My_Notes   Even though this isn't a DX filepath, it's the name from the error message, and it seems to work. Disable white background of pasted text in PowerPoint When I paste text from another source with "Keep source formatting option" into PowerPoint 2010 I receive a white background, that I don't want.    How can I make this text background transparent, so there would be just red text on purple? Why not use development provisioning instead of ad hoc? \documentclass[11pt]{article}  \begin{document}   The above equations on application of Laplace Transform become:  $$sL_1\hat{i}_{L1}(s) = D\cdot \hat{v}_{C1}(s) + D\cdot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sL_2\hat{i}_{L2}(s) = -D\prime \cdot \hat{v}_{C1}(s) + D\cdot \hat{v}+{C2}(s) + D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sC_1\hat{v}_{C1}(s) = -D\cdot \hat{i}_{L1}(s)+ D\prime \cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sC_1\hat{v}_{C1}(s) = D\prime \cdot \hat{i}_{L1}(s)- D\cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sL_l\hat{i}_l(s) = D\prime \cdot \hat{c}_{C1}(s) + D\prime \cot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) +(V_g - V_D - V_{C1} - V_{C2})\cdot \hat{d}(s) - R_l \cdot \hat{i}_l(s)$$  \end{document}   Error:   181 Underfull \hbox (badness 10000) in paragraph at lines 181--181 181 Overfull \hbox (22.63293pt too wide) detected at line 181 Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. Does Java Connection.close rollback? Does Java Connection.close rollback into a finally block?.  I know .Net SqlConnection.close does it.  With this I could make try/finally blocks without catch...  Example:  try {     conn.setAutoCommit(false);     ResultSet rs = executeQuery(conn, ...);     ....     executeNonQuery(conn, ...);     ....      conn.commit(); } finally {    conn.close(); } What info does Google backup? Does anyone know what exactly does Google backup (and restore) when you enable that option?  Also, are we able to see this list somewhere? (Maybe hidden n-levels deep in Google Account information?)  Reason I ask is I've had the unfortunate experience of my SD Card dying on me, put in a temporary SD Card and allowed Google to do the auto-restore.  Some things that were restored:   Saved wireless networks Background Some apps - strangely enough only a subset of my apps were restored (some with data, some without). All previous apps were installed via Android Market.   Then finally got my new SD Card, put that in and the following were restored   Saved wireless networks Some apps - again, same problem... only a subset (different subset, even fewer apps). This time no data seemed to have been restored either.   I think it would be very helpful to be able to view/manage a list of what is backed up, and thus would be restored (but I have yet to find this list... if it even exists). Inexact measurement and wavefunction collapse As is usually said, measurement of an observable $q$ leads to collapse of wavefunction to an eigenstate of the corresponding operator $\hat q$. That is, now the wavefunction in $q$ representation is $\psi(q)=\delta(q-q_0)$ where $q_0$ is result of measurement.  Now, in reality measurements are never exact, so the wavefunction should not have that sharp peak. Instead it should be broadened depending on precision of measurement. In this case, can we still introduce an operator, an eigenstate of which would the new wavefunction appear? Is it useful in any way? Or does the new wavefunction depend too much on the way it was measured so that each instrument would have its own operator? How would such opeartor look e.g. for a single-slit experiment? I would like to show that  $$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$$  I'm interested in more ways of proofing this.  My method :  suppose that $\sqrt{\frac{n}{n+1}}\in  \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}{n+1}}=\frac{p}{q}$ thus   $$\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2$$ since $p\neq q\implies p^2\neq q^2$ then  $$n=\dfrac{p^2}{(q^2-p^2)}$$  since $n\in \mathbb{N}^*$ then $n\in  \mathbb{Q}$   I'm stuck here and I would like to see different ways to prove $\sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$ Is it possible to reintegrate a Horcrux? As anyone who is familiar with Harry Potter knows, a horcrux is made when someone commits murder and stores a piece of their soul in something. Would it be possible for this someone to later recover and reintegrate their soul piece(s), should they want to?