I'm trying to find a formula that's fits the way I want my player to gain experience and level up. I know this is primarily opinion based but I hope someone can help me!  I've used the following formula as a base, I found it :  //Calculate level Level = Math.Max( Math.Floor(6.7f * Math.Log(Experience + 455f) + -40f), 1 );  //Calculate experience for level Experience = (long)(Math.Exp((Level - -40f) / 6.7f) - 455f);   I wrote a script that will show what it will look like, which is this:    I want to find a way to make it very easy for the player at early levels (let's say until level 15~), and then slowly harder. I'm not sure how to change the formula or if it's even possible! Hope you can help!
I am currently developing an RPG game with a classic experience-points and levelup mechanic. The player kills monsters, gains experience points, levels up when they collect enough and become stronger.  How do I balance the amount of Exp points necessary for reaching each level and the amount of Exp awarded by each monster on each level?
Recovering a lost website with no backup? So, when someone is red-green colorblind, the colors appear the same to them, like this:   And if you're totally colorblind, then things presumably just appear like they would in a black-and-white movie.  However, this isn't how ultraviolet patterns seem to work. Compare how we see this flower to the version where ultraviolet is visible:   The UV pattern is completely invisible here. However, unlike with the red and green, this isn't because yellow and UV are colors that appear identical when you can't see UV. Look at these flowers:     This time the flowers are purple, but the UV pattern is still invisible. Why is that? Shouldn't the UV pattern still be apparent on at least one of the flowers, just in a different color? And on some other flowers, the UV does appear as a different color. So:    Why is the UV invisible only sometimes?  Does it have to do with the flower using iridescent structures to produce color, instead of a pigment? Can this happen with red and green, as well? Does changing your Compute Device, change all computes of Blender? So, I have recently moved from using Blender, on my family's Nettop PC, to my personal, gaming PC. And with this move, came to option to use CUDA instead of CPU.  Which leaves the question, if I move to CUDA, does that change all computing, for like physics baking, or only for rendering?  Just a nooby question that I wanted to know, and yes, I know you can specify GPU rendering separately. Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$ It is changing the coordinate from one coordinate to another. There is an angle and radius on the right side. What is it? And why?  I got:  $2\,\mathrm dy\,\mathrm dx = r(\cos^2\alpha-\sin^2\alpha)\,\mathrm d\alpha \,\mathrm dr$,  where $x = r \cos(\alpha)$ and $y = r \sin(\alpha)$.  but cannot understand and get the right side. The problem emerged when trying to integrate $\displaystyle \int_0^\infty e^{\frac{-x^2}{n^2}}\,\mathrm dx$ where I tried to change the problem knowing $r^2=x^2+y^2$ but stuck to this part. What is the change in the title called and why is it so? Javascript Closures and 'this' I have a problem with an object I have created that looks something like this:  var myObject = {      AddChildRowEvents: function(row, p2) {         if(document.attachEvent) {             row.attachEvent('onclick', function(){this.DoSomething();});         } else {             row.addEventListener('click', function(){this.DoSomething();}, false);         }     },      DoSomething: function() {         this.SomethingElse(); //&lt;-- Error here, object 'this' does not support this method.     } }   The problem is that when I am inside the 'DoSomething' function, 'this' does not refer to 'myObject' what am I doing wrong? How do I install the Cinnamon Desktop? I have read that there is a new Gnome-2 like interface called Cinnamon produced by the Linux Mint developers.  Is this available for Ubuntu users - and if so how do I install it? Private IP address in public DNS We have an SMTP only mail server behind a firewall which will have a public A record of mail..  The only way to access this mail server is from another server behind the same firewall.  We do not run our own private DNS server.  Is it a good idea to use the private IP address as an A record in a public DNS server - or is it best to keep these server records in each servers local hosts file? 64 bit enum in C++? Is there a way to have a 64 bit enum in C++?  Whilst refactoring some code I came across bunch of #defines which would be better as an enum, but being greater than 32 bit causes the compiler to error.  For some reason I thought the following might work:  enum MY_ENUM : unsigned __int64   {       LARGE_VALUE = 0x1000000000000000,   }; extended stars-and-bars problem(where the upper limit of the variable is bounded) The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+....a_n=N$$ can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that $a_i\leq r_i$ for certain integers $r_1,\ldots,r_n$?  e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3&gt;2=r_3$. How can I disable Ctrl+Q for all applications (system wide)? A lot of applications like Firefox, Chrome etc, get closed when I accidentally press Ctrl+Q instead of Ctrl+W because of the proximity of the Q and W keys on the keyboard. Is there a way this shortcut can be removed or reset on a system wide basis? if the curvature is constant and positive, then it is on the circunference I'm trying to prove that if $\alpha(t)=(x(t),y(t))$ is a $C^2$ regular curve $(\alpha'\neq0)$ with constant and positive curvature, then $\alpha$ is on the circunference and if $\alpha$ is the circunference, then its curvature is constant.  I solved the second part since $\alpha$ is the circunference, then $\alpha(t)=(r\cos\theta,r\sin\theta)$, so $|\alpha'(t)|=r$, I need help in the first part.  Thanks a lot. Local file access with JavaScript How do I get the next element in HTML using JavaScript?  Suppose I have three &lt;div&gt;s and I get a reference to one in JavaScript code, I want to get which is the next &lt;div&gt; and which is the previous. How to modify an invalid '/etc/sudoers' file? How can anything ever fall into a black hole as seen from an outside observer? The event horizon of a black hole is where gravity is such that not even light can escape. This is also the point I understand that according to Einstein time dilation will be infinite for a far-away-observer.  If this is the case how can anything ever fall into a black hole.  In my thought experiment I am in a spaceship with a powerful telescope that can detect light at a wide range of wavelengths.  I have it focused on the black hole and watch as a large rock approaches the event horizon.  Am I correct in saying that from my far-away-position the rock would freeze outside the event horizon and would never pass it? If this is the case how can a black hole ever consume any material, let alone grow to millions of solar masses.  If I was able to train the telescope onto the black hole for millions of years would I still see the rock at the edge of the event horizon?  I am getting ready for the response of the object would slowly fade.  Why would it slowly fade and if it would how long would this fading take?  If it is going to red shift at some point would the red shifting not slow down to a standstill? This question has been bugging me for years!  OK - just an edit based on responses so far.  Again, please keep thinking from an observers point of view.  If observers see objects slowly fade and slowly disappear as they approach the event horizon would that mean that over time the event horizon would be "lumpy" with objects invisible, but not passed through?  We should be able to detect the "lumpiness" should we not through? complex polynomial satisfying inequality listing files with pattern with starting and ending Slow Network Printing over OpenVPN / Windows Server 2008 Our setup is as follows:  Two offices (A and B) connected with OpenVPN Access Server.  One Windows Server 2008 R2 Standard    Housed in remote datacenter. Terminals at both offices connect through RDP in order to use a particular piece of software concurrently. Network Connected printers (Deployed with GPO) that are connected to separate terminals at each office (and shared) are used with this software to print things throughout the day.   We are experiencing extremely slow printing times.  Opening printer properties can take 30-60 seconds and printing times range from 30-180 seconds with no apparent reason.  File sizes remain &lt; 1mb.  This happens with each of our printers.  All different models.  The server massively underutilized with a 6-core AMD, 16GB ram, 4x 120GB SSDs, and a 100 megabit port.  Our VPN does not seem to be a bottleneck either.  I've been attempting to find ways to increase print speed.  So far no success.  What can you advise? When updating, I get the following error message:   W: The repository 'http://ppa.launchpad.net/mc3man/trusty-media/ubuntu xenial Release' does not have a Release file.   Here, I find another statement on this error:   This recommends removing certain PPAs; and, I'm not sure if I should do that since it might mean not getting the updates that I need.    Is this what I should do? I am an Indian student in France on a 6 month visa for an Internship. Currently I don't have my passport with me as I have applied for UK visa and am not hoping to receive it back in next 10 days. I was wondering if I can travel to Germany without my passport in hand.   The documents I can carry are a photocopy of my passport's identity page, Residence proof here, proof of Internship in France,Indian national identity card, receipt from UK embassy stating my visa application.   Please note that I do NOT have a photocopy of French visa page.