Been playing around with Pixi.js and DroidScript trying to make some stuff for android. Nothing spectacular so far, but fun to play with.

That's definitely a lot clearer. It looks like you normalized the fonts as well, which Lummox'd brought up.

As far as the names go, there's one map in particular that I always use as a what-not-to-do reference. It's basically every blatantly cliche naming convention that's been used in high fantasy.



That's not to say that cliche names or places is a bad thing, but it can certainly help break up the insert-different-name-of-same-fantasy-town-here vibe that I get with way too many fantasy names.

How'd you go about deciding what the map would look like?

I've been too busy to do a whole lot of anything lately, largely because of work but also because of some other things. It's been annoying and I've been feeling very unproductive. To mitigate this, I've slowly been working through John M. Lee's Introduction to Topological Manifolds both to get back into doing math stuff and to get a better understanding of topology, and extremely important field and one of the three fields you're supposed to have a good understanding of before you go to grad school (along with abstract algebra and analysis).

While working on the exercises in the book, I ended up working on the proposition below as I thought I would need it to prove one of those exercises. Turns out, it was a lot of work for nought because when I sat down to prove the exercise in earnest, I figured it out within ten minutes not needing this result. Still, it was a couple days well spent thinking about it while I was at work and scribbling stuff down, and the proof itself is pretty nice I think. So I'm gonna ~~~share~~~ it, because sharing is caring, and explain all the terminology. Proposition first, and then I'll explain everything, though I'm going to assume there's at least some familiarity with what things like sets are. If you have any questions, go ahead and ask.

Proposition. Let A be a non-empty and possibly infinite set. Let L ⊂ R⁺ be indexed by A. Let S ⊂ M—where (M, d) is a metric space—be defined as S = ∪_{α ∈ A} B_rα(c) for some c ∈ M. Then S = B_supL(c).

Now for the plain English version (with some simplification): Imagine you're drawing a bunch of circles centered at the same point, and considering all the stuff inside the circle, not considering the edge as being part of it. Intuitively, all of those circles should be inside the largest one. This is pretty simple to see if you have finitely many circles, but what if there are infinitely many circles? And what if their radii are (just choosing some arbitrary example) L = {1/2, 3/4, 7/8, 15/16, 31/32, ...}? Then things become a little more complicated.

The first two sentences are what allow for the possibility of the radii being finite or infinite. A is what's called an indexing set, and basically it says that for every α ∈ A (read as "for every alpha in A") there is exactly one corresponding r_α ∈ L (generally read as "exactly one corresponding r-sub-alpha in L). If A is finite, then L is finite; if it's infinite, then so is L, whether A is as infinite as the natural numbers or as infinite as the real numbers.

L is also a subset of R⁺, the positive real numbers (and this is what L ⊂ R⁺ means). Thus, we don't have to worry about awkward notions of negative or zero radii, only radii that are strictly positive.

Next, S (the set we're concerned with) is a subset of some larger set M. The notation "(M, d)" means that the set M together with another mathematical object d is a metric space, which is an important mathematical concept. The object d is a binary function from M to the non-negative real numbers (denoted R_≥0) called a metric—binary meaning that it takes two arguments. If x, y, z ∈ M are arbitrary (meaning that we don't pick x, y, or z in any particular manner), then d is a metric if it has the four following properties:

1. d(x, y) ≥ 0. That is, the distance between two objects in M is always positive or 0.
2. d(x, y) = 0 if and only if x = y. That is, the distance between two objects is 0 when they are the same object and only when they are the same object.
3. d(x, y) = d(y, x). That is, the distance from x to y is the same as the distance from y to x.
4. d(x, y) + d(y, z) ≥ d(x, z). This is the triangle inequality, and basically states that "detours should not be shortcuts". That is, if I go from x to y and then from y to z, it should not be a shorter path than if I go from x directly to z.

These generalize what we intuitively think of when we talk about notions of "distance" between two objects (though some particular metrics are weird and unintuitive, even though they satisfy the above criteria). Most people are familiar with, at the very least, the Euclidean metric: d(x, y) = √(x - y)². Metric spaces are important because they tell us what it means to talk about distance, and give a lot of structure to an otherwise-structureless set.

Once we're able to consider distances, we are then able to talk about open balls. Go ahead and laugh now, since non-math people love to laugh at that terminology.

Done? Okay, moving on.

An open ball is basically what I mean by saying "everything inside the circle without the circle itself": it's the set of all points less than a certain distance away from a central point. More formally, the notion B_r(c) for a positive real number r and a center point c ∈ M is define as: B_r(c) := {p ∈ M : d(p, c) < r} ("the set of all points p in M such that the distance between p and the center point c is less than r"). A sort of analogy is that if you have a basketball, from a mathematical viewpoint the ball is gonna be the air inside but not the rubber and leather at the edge of it. This notion is often quite useful, even though the distinction probably doesn't seem very important to a non-mathematician.

And now we can move on to the definition of what this set S that we talked about five or so paragraphs ago is. Remember that S is a subset of M (S ⊂ M). Specifically, it's defined as the union of every open ball of a given radius r_α ∈ L with an arbitrary center point c ∈ M. That is, using the notation from the statement of the proposition: S = ∪_{α ∈ A} B_rα(c) ("S is the union of all open balls of radius r_α centered at c for every α ∈ A").

To make it clear, this is what the proposition assumes that S is defined as. So far, I have not yet covered the thing that we're actually trying to prove. Mathematical notion and terminology is very dense.

So, what I'm actually trying to prove is that, given all the preceding things, is that assuming S is defined as above, then S = B_supL(c). That is, S is just an open ball centered at c with a radius equal to supL. A simple notion (assuming you were able to follow all the preceding paragraphs), except that we just have to explain what supL is. Quite plainly, supL (in this case) captures the idea of the "largest circle" from the plain English statement of the proposition. In particular, supL is the supremum of the set L. Supremums are sort of like a slight generalization of the notion of a maximum element. For example, let's say you have some set T = {1, 2, 3, 4, 5}. The maximum element in T is simply the largest element in T: maxT = 5. If, for a given set, the maximum element exists, then the supremum is equal to it. Thus, supT = 5.

But, there are some sets without a maximum. Recall the example choice of L from earlier? L = {1/2, 3/4, 7/8, 15/16, 31/32, ...}. This is an infinite set, whose elements are all numbers (2ⁿ - 1)/2ⁿ for all positive natural numbers (i.e., n = 1, 2, 3, 4, 5, ...). This set has infinitely many elements, but with two important properties:

1. For any element of the set, there is another element larger than it.
2. No element of the set is greater than or equal to 1.

This first property clearly shows it can't have a maximum: there is no element in the list that is larger than every other element. But, the second property is suggestive. Clearly, 1 is an upper bound on the set: every element in the set is less than 1. But, there are infinitely many upper bounds: 37, 149, π, 6/5, and so forth. 1 is special, though: it is the least upper bound. No other upper bound is smaller than 1, meaning that if you choose any number smaller than 1, you'll find an element in the set larger than that element. The supremum is exactly this: the least upper bound. Thus, for our example choice of L we have that supL = 1.

An extremely important property of the real numbers is that every set of real numbers with an upper bound has a supremum (a property called Dedekind completeness).

So, by saying that S = B_supL(c), we're basically saying that in the way we defined S (as the union of those particular open balls), we can show that S is actually just the open ball with the smallest radius possible so that it's still at least as larger as anything in the open ball. While this sounds obvious, actually proving it is another matter.

We're basically ready to move on to trying to prove this. All that we need is the following: for consistency's sake, I will define B_∞(c) = M for any c ∈ M. Basically, an 'infinitely large' open ball in any metric space is just going to be every point in the metric space (i.e., the space itself).

With that, we can begin!

Proof. I shall break the proof into two parts. First, I'll assume that L is unbounded, and then I'll assume that L is bounded (I'll explain these notions in the relevant portion). For any subset of the real numbers (which L is), the set is either bounded or unbounded. Thus, we can consider these two disjoint cases on their own, and the assumptions are valid.

First, let us assume that L is unbounded. That is, the elements of L get 'arbitrarily large' (e.g., L = {1, 2, 3, 4, 5, ...} is unbounded). Formally, this means that for any arbitrary ρ ∈ R ("any arbitrary rho in the real numbers") there exists some α ∈ A such that ρ < r_α, where r_α ∈ L. In plainer language: pick any number, and I can find a larger one in L.

Let p ∈ M be arbitrary. Because we can pick ρ to be any number, we can also pick ρ = d(p, c). From the definition of d, it is a real number and so a valid choice. But, by assumption we know that L is unbounded meaning that there is some r_α ∈ L with ρ = d(p, c) < r_α, and so p is in some open ball that was contained in S and therefore p ∈ S. Because p can be any point in M, this result must be true of all points in M, and so all points of M are in S, which is to say that M ⊂ S. But, we know from the proposition that S ⊂ M. If two sets are subsets of one another, we say they are equal and thus S = M.

Next, when a set is unbounded we say that its supremum is infinite, and therefore supL = ∞. But this means that S = M = B_∞(c) = B_supL(c), that is S = B_supL(c), as we were to show. Thus, the result holds when L is unbounded.

Now, assume that L is bounded. Then, as stated before, supL must exist. Let us consider some set Σ = B_supL(c). Because it's true for every p ∈ S that d(p, c) < supL, then it must be that p ∈ Σ. That is, S ⊂ Σ.

Let us now assume that there is some q ∈ Σ such that q ∉ S. That is, we assume that S and Σ are, in fact, different. Then we know that for every α ∈ A, the inequality r_α ≤ d(p, c) < supL must hold.

It is at this point we will use an equivalent definition of supremum to make some progress. Earlier, I mentioned that the supremum is the number such that if you choose any number smaller than it, you'll find a number in the set larger than the number you chose. There's a way to make this notion precise, and it is as follows: For every positive number ε (that is, every ε > 0) there exists some r_α ∈ L such that supL - r_α < ε. In plain English: pick any positive number, no matter how small, and I can find a number in L closer to supL than the number you chose. This definition is equivalent to the one that I mentioned earlier, about the supremum being the smallest upper bound (equivalent here meaning that if you assume one is true, you can prove the other and vice-versa).

Thus, not only is the above inequality satisfied (r_α ≤ d(p, c) < supL) but it is also satisfied that for every ε > 0 there exists some β ∈ A with supL - r_β < ε. Because we can choose ε to be any positive real number, we can also make a particular choice: let ε = supL - d(p, c), recalling that p is the point we assume is in the set Σ but not in S. Because we know that d(p, c) < supL, supL - d(p, c) > 0 and is then a valid choice. We then require that for every β ∈ A that supL - r_β < ε = supL - d(p, c) ⇒ -r_β < -d(p, c) ⇒ r_β > d(p, c).

Thus, for every β ∈ A we have that r_β > d(p, c). But, from the previous inequality we also had that r_α ≤ d(p, c) for every α ∈ A, including the particular choice of α = β. This is a contradiction! Going back, every step we made was logically justified, until we reach the step where we assume that p exists, that is where we assume that there is some element in Σ that is not in S. Therefore, this step must be false, and it must actually be the case that S = Σ, but Σ = B_supL(c) and therefore we have that S = B_supL(c) and the result holds when L is bounded.

Thus, we have that for any indexed set of positive real numbers L, S = B_supL(c), as was to be shown. ∎

And there you have it, a proof of the proposition! Even though I ended up not needing the proof, I feel it was a very nice proof and wanted to share. If any of you are mathematically-inclined or simply curious, here is a "homework" problem of a point I sort of glossed over in the proof. See if you can work out why it's true:

• I stated that the inequality r_α ≤ d(p, c) < supL must hold. Why?

If anyone has any questions, I'll do my best to answer them.

Popisfizzy wrote:

I quit my last job because it was stressful and filled with bullshit,

If you're talking about the job that required you to cut your hair, I mean, I could've told you that job was going to be bullshit. That was red flag #1.

Been a while since I participated in this thread. If I have anything to talk about, it's my virtual machine again. Been working on plans for new system calls.

Looks very nice Flick. I only have some basic grasps of UE4 and Blender myself.

EmpirezTeam wrote:

You could say he's a... hexcellent game developer!

Lol.

The blueprint system is definitely awesome. Heck, that sort of system is starting to get more common in games that support modding. Space Engineers being one recent example.

If I ever work on a UE4 game, I would definitely use C++ and blueprints together (since the former can provide support for the latter). I was supposed to be working on a game with someone who used to be on BYOND, but haven't done anything yet (due to like being caught up on my own projects).