Posts for Warp


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This is a very easy problem mathematically speaking, but that's not the challenge. The problem is: What is the area of the red square compared to the area of the green square? The challenge is not finding the answer. The challenge is to find a geometric proof (or, could it perhaps be more accurate to say "geometric argument"?) that requires no math at all. Perhaps put your answer in spoiler tags, if other people would want to think about it as well.
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Famously, lim(x->∞) (1 + a/x)bx = eab This got me thinking what happens if that 1 were something else. Not a very hard problem given that even I could figure it out, but thought you might find it a mildly interesting question.
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Warp wrote:
- Would a TAS made of a VC game theoretically sync, using a real Wii and a tasbot connected to its controller port?
This question still remains unanswered.
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Stovent wrote:
So I miss-understood his question. I'm considering that to reach the credits in the fastest way, you need to go to a point at which no further input is required to get to the credits as fast as possible
Perhaps more precisely (and as has been suggested many times in the past): The input should be ended at the point where no further input can make the ending be reached faster.
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Stovent wrote:
The best answer to this is Ocarina of time, which have a glitch in VC thzt is impossible on a real N64. And I think it's not the only game in this case
Would that constitute abusing emulator inaccuracies? (Question asked only half-seriously.)
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That seems to work. In this example, exp1=2, and exp2=4. On the left the threshold is 0.3, and on the right it's 0.7.
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One argument, and one question: - Wii Virtual Console speedruns seem to be A-ok in the real-time speedrunning community, so is there a reason why they shouldn't be ok in the tool-assisted realm? - Would a TAS made of a VC game theoretically sync, using a real Wii and a tasbot connected to its controller port?
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I was wondering if it would be reasonable to generalize the challenge: What if I wanted for the transition from the first curve to the second one to happen somewhere else than at x=0.5? Could the formula be generalized so that transition happens at any given x within the range (0, 1) (non-inclusive)? I suppose that if I expressed the question as code, it would be like:
double easeInOut(double t, double exp1, double exp2, double threshold)
{
    double thresholdY = ...;
    if(t < threshold) return pow(t / threshold, exp1) * thresholdY;
    else return 1 - pow((1 - t) / (1 - threshold), exp2) * (1 - thresholdY);
}
In my original question threshold was 0.5, and thresholdY = exp2/(exp1+exp2). But what would a more generalized formula be, for any value of threshold between 0.0 and 1.0 (non-inclusive)?
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One think baffles me about the episode The Perfect Pear. (I really think that the spoiler tags are very necessary here. If you follow the show, please watch the episode before reading any spoilers.) It's heavily implied that none of the three siblings know anything at all about their parents, because Granny Smith never told them anything. As in they... disappeared... when they were too young to even remember. But the age difference between Big Mac and Apple Bloom is quite large. As in, something like Big Mac probably was already at least a "teenager" (in pony years, whatever they might be) when Apple Bloom was born. Yet it's still kind of implied that even he doesn't know anything about his parents. It's a bit strange.
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FractalFusion wrote:
That gives y=n/(m+n) at x=0.5.
That seems to indeed give a proper result. For example, if m=2 and n=4, plugging in 4/6 and 1-4/6 as the factors in the equation gives us the nice smooth curve:
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I suppose that my use of the word "smooth" was not what it means in mathematics, as there it means "infinitely differentiable", rather than just "once differentiable". A hint that I referred to the latter might have been given by me saying in my post "there's a visible jump in slope at t=0.5", which refers to the first derivative (and that only).
FractalFusion wrote:
So we can assume that the function f is defined as y=Axm on [0,0.5], y=1-B(1-x)n on [0.5,1]. ... So at x=0.5, y=n2n-m/(m+n).
That doesn't seem correct. Rather obviously (at least for values of m and n greater than 1) y ought to be between 0 and 1. In other words, somewhere in between the starting point (which is at y=0) and the ending point (at y=1). But, for example, if m=2 (the first half is quadratic) and n=4 (the second half is quartic), that formula gives 4*24-2/(2+4) = 16/6, which is > 1. Edit: By the way, your functions are not correct. They should be: y = A(2x)m, x=[0, 0.5] y = 1 - B(2(1-x))n, x=[0.5, 1] I don't know if the error stems from this.
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rhebus wrote:
I'd probably use a B-spline
But in the context of this thread I was posing the challenge of how to calculate that factor in the formula (which is now 0.5) in order to make the curve smooth. Why wouldn't that problem be well-defined?
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In many situations, especially when it concerns animating things, it's useful to have easing functions that take a linearly-changing parameter t, which gets values from 0.0 to 1.0, and return a curve (that likewise starts from 0.0 and ends in 1.0, but does something else than increase linearly). One useful such easing function is one that starts slow and accelerates up to t=0.5, and then decelerates and ends slow in t=1.0. (This is useful eg. when animating an object on screen, to have it perform that kind of smooth accelerating/decelerating motion instead of moving at constant speed.) One way of creating such a function is to make it polynomial, like this:
double easeInOut(double t, double exponent)
{
    if(t < 0.5) return pow(2 * t, exponent) * 0.5;
    else return 1 - pow(2 * (1 - t), exponent) * 0.5;
}
What this does is that when t < 0.5, it's scaled to the range 0-1, the exponent is applied to it, and then scaled back to the range 0-0.5. And when t >= 0.5, the same is done, but mirroring t about 0.5 (thus getting a mirrored curve). This allows choosing the speed at which the acceleration/deceleration changes, so that with exponent values that are close to 1.0 the acceleration is not very pronounced, and with larger values it gets more pronounced. For example with an exponent of 2 we get this nice easing curve: With an exponent of 4 the curve is more pronounced: The important thing is that the curve is smooth over the entire range. Now, what if we wanted the starting acceleration to be less pronounced than the ending deceleration? In other words, what if we wanted to use two exponents rather than one: The first exponent for the first half, and the second exponent for the second half. We can't simply plug the two exponents in the above code because the resulting curve is not smooth: There's a visible jump in slope at t=0.5 (and this jump becomes very obvious if eg. used in an animation). So my question is: How to get a smooth curve like this, where the starting half uses one exponent and the ending half uses another? One possibility would be to change those 0.5 factors in the formula, so that the joint point at t=0.5 goes up or down (in this example it would have to go up). But by how much, exactly?
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Water temperature can have an effect on how fast you get tired swimming. The warmer the water, the less tiring it is (well, up to a point, obviously). Swimming in really cold water for an even moderate period of time can be really hard.
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I don't think any rule should be absolutely 100% rigid, imposed completely blindly in an absolutely totalitarian manner, with absolutely no possibility of special exceptions. Certain rules should be stronger than others, for certain, but there ought to always be at least the possibility of granting an exception, if there are good reasons for it. And exceptions should never create precedents.
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Mothrayas wrote:
This category is known in the real-time speedrunning community as MST (Medallions, Stones & Trials)
Ah. "All Dungeons, Temples & Ganon Trials" was confusing.
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Before I can make a judgment, I would like to know if this is one of the common categories in real-time OoT speedrunning, or if it's a custom one. I'm looking at http://www.speedrun.com/oot and either I'm blind, or I can't find any category that fits this. ("All Dungeons" would at first glance sound closest, but given that its record is 1h 25m, it can't be it. This TAS does more.) If this is a completely customary objective that's not common in the OoT speedrunning community, it raises all kinds of questions about arbitrary goals. Such arbitrary goals are a hot debate topic, but one could argue that if you were to choose one, at least choose one that's popular in the speedrunning community. But I'm jumping ahead of myself before knowing the full story.
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No bonus!
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I suppose that I could try to give some advice to younger people here. Perhaps if I could save even one person from being chronically lonely, it would be definitely worth it. If you have a rich social life, and romantic relationships, you regularly hang out with people and interact with people, and you have a busy life socially speaking, this post is not for you. You can safely skip it and go do something else. If you are, however, a young person (late teens, early twenties), and you are quite a nerd, you eg. play a lot of video games, and you don't socialize much, and you don't have many hobbies that you do regularly and which involve socializing with people in real life, and you don't have many or even any romantic relationships at all, I would like to tell you something. There's nothing inherently wrong in being a nerd, and there's nothing inherently wrong in liking video games, but for your own sake, try not to do it at the expense of your social life. Seize the social opportunities that present themselves. If somebody invites you to an event, say yes, even if you are not interested at all and you would be bored out of your skull. Try to enjoy yourself. (If drugs are involved then I wouldn't recommend it; stay out of those. If alcohol is involved, personally I wouldn't recommend it, but it's your choice. But there are so many other possibilities as well.) It doesn't matter if you are bad at smalltalk, and you don't know what to say, or perhaps even constantly make a fool of yourself when you try to. It happens, and it is a skill that can be learned. The more you are exposed to social interaction, the more you learn. With time, you'll learn how to interact, what to say, and how to talk with people. When you witness it happening frequently, you start catching the little tidbits and the little ideas, and with practice you get better at it yourself. The important thing is that you stop saying "no" and start saying "yes". Maybe not to every single possible event under the sun, but in general to most of them. Don't let the opportunities go by. If you let them pass you for long enough, they will stop coming. People will stop asking you. Don't let that happen. If you don't seize the opportunities now, it's not going to get any easier as you get older; on the very contrary, it's only going to get harder and harder. You don't need to go to social events specifically looking for something (such as making friends or romantic relationships). Just being there frequently is enough. Be there, interact with people in person, participate. Try to find the fun in it. Shove aside your prejudices and try to see the events as enjoyable. Look for the enjoyable things. You want to want to be in these events, rather than them feeling like a chore. Start doing it now, while you still can. Or one day you might realize that there are no social events where you can participate. They are gone.
Post subject: Re: #5533: Sniq, Total, Aran Jaeger's SNES Super Metroid "game end glitch" in 07:09.68
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TASVideoAgent wrote:
Executes arbitrary code
Meh.
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Ferret Warlord wrote:
MICHELLE CREBER IS A FERRET OWNER[/URL] OH MY GOODNESS
I have to remove some brony points from myself for not knowing who Michelle Creber is.
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rhebus wrote:
zn+1* = (zn2 + c)* [by definition] = zn2* + c* [conjugation distributes over addition] = zn*2 + c* [conjugation commutes with squaring] = z*n2 + c* [using induction hypothesis]
I don't know what "induction hypothesis" means here.
= z*n+1 [by definition]
This seems to show why it's symmetric on the imaginary axis, but it does not show why it's not symmetric on the real axis.
EDIT: The key reason why this argument doesn't work on the other axis is that the corresponding operation (negation or "unary minus") doesn't commute with multiplication/squaring: (-x)(-y) ≠ -(xy) in general, while x*y* = (xy)*.
I'm not sure I understand that. As an additional question: The zn+1=zn3+c fractal is (or, more precisely, appears to be) symmetric on both the imaginary and the real axes. Why is that the case? zn+1=zn4+c once again is not (even though it has a more complicated three-axis symmetry, where each axis is separated by 120 degrees.) In general, with even powers there is no real-axis symmetry, while with odd powers there is. Can this be shown to be true?
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I was wondering why the Mandelbrot set is symmetric on the y-axis but not on the x-axis. (If you calculate for a value a+bi whether it belongs to the set, or how many steps it takes to bail out, you get the exact same value as with a-bi. However, the same is not true for -a+bi.) It's not obvious directly from the formula zn+1 = zn2+c why it would be symmetric on one axis but not the other.
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thatguy wrote:
Here's a more interesting twin prime problem: prove that the only Fibonacci numbers that are also twin primes are 3, 5 and 13.
I think you mean numbers that are one of the members of a twin prime pair? Btw, in a related note, is it hard to prove how many Fibonacci numbers are prime (as in, are there infinitely many of them)?
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James Grime of Numberphile fame presented this problem in one of his videos: Prove that when you multiply a pair of twin primes you get a number that has remainder 8 after division by 9. With one exception.