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Warp wrote:
Warp wrote:
Speaking of the Riemann zeta function, I have had for some time now an interest in understanding the Riemann hypothesis. Just out of curiosity, as a little "hobby project". I was wondering if someone could help me with it.
In order to understand the hypothesis, I first need to understand the Riemann zeta function. In order to understand said function, I need to understand analytic continuations. And the topic already becomes really complicated (not to talk about how complicated the zeta function itself is.)
I always like to think of new math concepts as generalizations of old ones. In the case of analytic continuations, I imagine it as related to the geometric series.
I expect you are aware that
1 + x + x^2 + x^3 + ... = 1/(1-x)
for |x|<1, right? Well, consider graphing both sides of the equation. On the left hand side, you get a function that is well-behaved only for |x|<1. Outside this range, it blows up. This is easily seen using complex analysis because there is a pole at x = 1. Using real analysis, we just say that the domain of any series cannot extend past a vertical asymptote. It's left as an exercise to the reader to show that the radius of convergence is a "true radius" in the sense that if the series cannot extend past R to the right, nor can it extend past the same distance R to the left.
Anyway, we look at this graph and lament that its left end "looks" like it's begging for a continuation. In particular, although the graph blows up at x=1 and therefore the series cannot extend past that point (well, actually, it can, but never mind that), nothing seems to be stopping it from continuing leftward, past x = -1. The right hand side of our equation provides exactly such a continuation. The two functions are perfectly identical in all but one aspect: their domain.
Often, it's taught that the left hand side is the Taylor series of the right hand side and the radius of convergence is derived as an exercise. Here, I'm working in the opposite direction and I argue that the left hand side is practically begging to be written as the right hand side. The right hand side is merely a generalization of the left.
If you'd like a slightly better example that doesn't depend on asymptotes on the real domain, consider the function 1/(1+x^2). Famously, its Taylor series,
1 - x^2 + x^4 - x^6 + ...,
only converges for |x|<1. This is rather startling, since the function seems well-behaved throughout and there aren't any asymptotes. Again, learning some complex analysis brings it within your grasp.
thatguy wrote:
Now of course that's not a proof because probability is not certainty and the above assumes that the primes are randomly distributed, which they most certainly are not. But you can see how it's plausible. With the Riemann Hypothesis, it's more "look at the first few zeroes, they all lie on this line" with, for me at least, no deeper intuition as to why that should be so.
It's funny you should mention Goldbach's conjecture. I've heard it said that a single negative result for the Riemann hypothesis would "wreak havoc" on the distribution of prime numbers. While we all know that prime numbers are not distributed truly randomly, apparently the Riemann hypothesis is connected to the idea that the distribution of primes is characterized by certain qualities shared closely with random numbers. In that sense, your post is rather ironic because it boils down to the following argument:
"I believe Goldbach's conjecture to likely be true because it would almost certainly be so if the primes were randomly distributed.
One consequence of the Riemann hypothesis is that the primes are practically randomly distributed.
I'm skeptical of the Riemann hypothesis."
By invoking the apparent randomness of prime numbers, you unfairly favor one theory over the other.
To be perfectly clear, no one has explicitly shown that the Riemann hypothesis is equivalent to Goldbach's conjecture. Apparently, there is some suggestive evidence in its favor, however. It's just funny that the two are intrinsically linked to the idea of prime numbers being "random".
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I didn't really intend to sign up this year since the contest will likely interfere with my work, but I notice that there are 11 individual players and 12 slots available. Therefore, I'm throwing my hat in the ring, albeit with the following caveats:
I'll probably be very busy over the next few months and really shouldn't devote too much time to the contest. I figure that as an extra player, I won't hurt any team, but I probably won't be the primary contributor.
I'm signing up as the lowest priority player. If anyone else signs up in the next 24 hours, count me out to make it an even 12.
I have a mild preference for working with Team 4 if they'll have me, since I've worked with FatRatKnight before. Team 7 also looks great and working with Tompa would be a privilege. I realize that my previous two points might make me a less-than-ideal partner, so I understand if they'd like to hold out for a better team member. There's some great talent already in the individual player pool. I'll be happy to lend my services to any team that will have me.
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Nach wrote:
BrunoVisnadi wrote:
Nach wrote:
I have no idea what "lefty liberal" means. Is it something many common people classify themselves as? If not, but you are looking to classify yourself as "lefty liberal", then you're asking to be the one who is insulted. Otherwise, he's just insulting some group outside of TASVideos.
If you evaluate what grassini said with a bit of common sense, you will see used the term ''lefty liberals'' as a characteristic of the people he was answering to.
Who is he answering to? He did not quote anyone, he did not mention anyone by name, he just threw it out there.
BrunoVisnadi wrote:
He immediately mentioned them again in ''I pity you guys'', making clear he was not referring to some group outside of TASVideos.
Perhaps it's not some group outside of TASVideos, but it's up to the reader to determine if they are being specified or not, since the specification is anything but clear.
From IRC:
<Niamek> Lefty liberals, isn't those guys who are liberal, but more on the left that regular liberals? LIke left is liberalism and right is communism?
<BrunoVisnadi> It doesn't matter that much what lefty liberal means, grassini was clearly calling other people in the forum of that
Being that he's posting here, it's likely he's referring to 1+ people. But you have to buy into his claim and believe it's about you to be insulted in the first place, since he could just as easily be referring to someone else who posted in this thread who is more extreme in whatever viewpoint you're going with than you are.
Personally, I stick with refusing to be defined, and let unspecified remarks always mean the "other" guy. Unless he specifically calls out a member or members of our site in an insulting manner, I cannot condone just terminating him. If people want to espouse a view which is extremist and exaggerated, as long as not directed at anyone in particular. so be it.
Good grief, Nach, your posts in this thread have been horrendously tone deaf. No one wants grassini "terminated" and, unless I'm sorely mistaken, we are all trying to ignore him. He posted another inflammatory comment and Mothrayas, recognizing that a little more attention was needed, called him out directly and pretty much everyone was satisfied. You, on the other hand, have bent over backwards to say, "Well, maybe he wasn't really insulting members of our community..."
Just bow out of the thread. You're not contributing anything and the rest of us are ready to move on. If things get bad again to the point where an admin needs to step in, I trust that you will act appropriately.
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BrunoVisnadi wrote:
Not only grassini used memes satirizing other people's opinion, he also called them paranoiac and mentioned them twice in not very friendly ways. That sounds pretty personal. I don't want to make from this thread a mess of personal attacks, but it is very unpleasant that moderation ignored him, but warned against the use of the word 'troll'.
This is what I'm really astonished by. "Troll" isn't really much of an insult, it's a characterization of certain behavior and, yes, I think grassini is trolling. I suppose the definition has shifted a little bit so that under certain definitions it means being deliberately provocative specifically without actually holding the beliefs you espouse. What's even more ridiculous is that my statement was, do not engage him because it will be a waste of time. I wasn't trying to stoke any rage.
Anyway, back on topic, I wasn't sure if I should mention Trump in the first place because I come to TASVideos in part to escape politics. It would seem that most others feel approximately the same way because I don't think there's been much mention of the election. Having said that, the election is over and there's no great point in tiptoeing around the topic since the die has been cast. I don't think TASVideos was the right place to discuss the election before it happened, but now that it's over, one of my regrets looking back was that I wasn't vocal enough. I'm liberal, damn proud of it, an unchecked Trump presidency will be bad for everyone, and I want the right wing to know that I and my kind will not just roll over because we lost one election.
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OmnipotentEntity wrote:
grassini,
You unironically believe pizzagate? Really?
For those unaware, "pizzagate" is a conspiracy theory revolving around a supposed child sex ring at a Washington D.C. pizzeria. In leaked emails between Hillary Clinton and her staff were many references to "getting pizza". The minds at 4chan and Twitter presumed these emails were all a code and if they made certain substitutions, such as "cheese pizza" actually referring to "child pornography", a twisted picture emerged where it was apparent that Hillary Clinton's entire staff was implicated in a child prostitution ring. See this video for reference:
Link to video
I've stated the conspiracy. You make up your own mind whether American public discourse is in a healthy state.
Patashu wrote:
And even if you don't think that Trump being sexist/racist/etc is an issue, you must be worried by the fact that he thinks Global Warming is a scam, that he's OK with nuclear weapons escalation, that he sees it fit to promote and negotiate business deals while preparing to become the President, that he doesn't bother to go to intelligence briefings, that he's close friends with Alex Jones, that he'll post things on twitter that are factually false (causing people to get harassment, death threats, company stocks to plummet, etc. for no good reason), etc.
I strongly agree with this comment. We all should have reason to be concerned about Donald Trump's actions. This is not just preemptive hatred based on what we think his presidency will be like.
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BrunoVisnadi wrote:
Grassini, I get you are trying to base your arguments on memes like this little comic you posted. Nowadays terms such as ''racist, sexist and homophobic'' are being used a lot in a sarcastic/satiric way such as you did there, but if you stop to think seriously about it, there is no mistake in claiming Trump is all of these. Just google his name next to this words and you will find a lot of video content that ends any discussion about it.
Everyone should see how serious it is to have a president with this kind of ideology, it's not a risible complaint as you try to make it look like. You must know how disgusting and messy Brazil's politics are since last year, USA is an example of how it could be much worse.
I wholeheartedly endorse this comment.
Mothrayas wrote:
Stop the personal attacks, all of you.
Just because some people have a different political opinion than you does not mean they are trolls.
Grassini entered this thread by characterizing posters as "lefty liberals". While I don't take umbrage with the phrase, he certainly meant it as an insult and it added nothing to the discussion. Since then, he posted a rage comic, a conspiratorial video, reference to an absurd conspiracy, and another comic.
I'm going to abide by my "don't feed the trolls" statement, but he is not raising the level of discourse in this thread.
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Samsara wrote:
grassini wrote:
(pro-trump picture)
Cool, good one, nice meme. Also, completely not at all how anyone rational thinks. Anybody with half a brain* knows that the election wasn't directly hacked by anyone, Russian or otherwise. The real issue is that it has been proven that the Democrats were hacked, which lead to leaked information that completely damned Hillary Clinton, and someone could easily argue that these leaks persuaded a lot of voters over to Trump's side.
*"anyone with half a brain" is still only about 25% of Americans
That someone is not going to be me, because arguing with anyone who even slightly supports Trump would be like bashing my head into a brick wall until I cracked my skull open, but I do firmly believe that these leaks were basically the final nails in the coffin for Clinton.
Don't feed the trolls, Samsara*. Let grassini play with his straw men.
*... which is sort of what your last paragraph says.
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Warp wrote:
Bobo the King wrote:
I believe Donald Trump has a decent chance of getting us all killed within just four years.
More likely the exact opposite.
The relationship between the US and Russia has been getting worse and worse during the last years because of the economic sanctions that the former (and the United Nations in general) has been imposing onto the latter. There's a reason why Russia has been so brash during the last years, and making so much noise. They are not happy.
Trump has indicated that he wants to mend fences with Russia, while Clinton apparently had no such intent (but the opposite). Keeping Russia happier is probably for the good of the entire world, but especially Europe.
As a Finn, it doesn't exactly make me feel relaxed to see time and again how the Russians are talking about invading Finland. If Trump is able to placate Russia in any way, and make them a bit more content, I'm all for it. That in itself trumps (hah) the other things that I strongly disagree with about Trump's views (such as climate change, healthcare, separation of church and state, etc.)
I was avoiding replying because it would be off topic. Now that the thread has been split, I guess I'm free to do so:
Finland is fucked.
Rather than drag out the political discussion, I'll just say enjoy this while you can:
Link to video
You guys are going to need every ounce of defiance you can muster.
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Ready Steady Yeti wrote:
Bobo the King wrote:
Ready Steady Yeti wrote:
What are the reasons to believe or to not believe that the singularity will happen?
I believe Donald Trump has a decent chance of getting us all killed within just four years. Even after dodging that bullet, it seems like only a matter of time before we manage to kill ourselves. I've never been so pessimistic about the human race as I've been this last month.
I meant that whole "when technology becomes superintelligent and makes everything perfect" thing. Updates itself faster than we can handle, has its own superconsciousness, etc. etc.
I know what the singularity is. I mostly agree with Patashu, but I was merely pointing out a practical hurdle immediately in front of us...
What are the reasons to believe or to not believe that the singularity will happen?
I believe Donald Trump has a decent chance of getting us all killed within just four years. Even after dodging that bullet, it seems like only a matter of time before we manage to kill ourselves. I've never been so pessimistic about the human race as I've been this last month.
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Derakon wrote:
If anyone's interested in redoing an old game, this one should be a straightforward obsoletion; there's a glitch involving arranging your consumables inventory and then buying 0 seeds that lets you skip all of the Fire chapter, including three dungeons. It's presumably common knowledge among speedrunners of the game; I just happened to stumble across it while watching a race.
For reference, the relevant skip and brief explanation appears to happen around 45:51:
Link to video
Give me a few days and I may look into this. I'm very curious about what's going on underneath and whether it can be further exploited.
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Hey, here's something you didn't ask for:
The top image is the average of the two images. The bottom is the difference.
I think the average looks best, but whatever.
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OmnipotentEntity wrote:
Bobo the King wrote:
I was about to throw down a healthy dose of skepticism, but a quick analysis below shows that your answer is plausible. I'm still troubled by the fact that you very quickly state that the area of the cross-section is f*x*dy. Can you go into more detail on how you got this? I don't see where you took into account that the villi exhibits the same ridges along the y direction.
f*x being the length is plausible. So, I believe you're taking exception with the width being dy.
So here's my logic: if you take that cross section and pull it straight you get a line with a length of f*x. But the width does not change with this action. Also, this action smooths out the undulations in both the x and the y directions for that slice. Then we just add up all of the slices.
I could, of course, be overlooking something critical. It's early now, and late when I wrote that. And I need more coffee.
I followed that logic and was pretty convinced for a little while, but now I'm skeptical again. If we start with that same slice and extrude it directly along the y direction, your logic perfectly checks out.
But, as a simple example, suppose that we extrude back along a diagonal line. In more mathematical terms, f(x,0) is our slice, and f(x,y) = f(x,0) + C*y, where C is some constant. Because it is upward-sloping, it clearly results in a greater surface area over the same bounds on y. (If you're at all bothered by the fact that the slice no longer "lays flat", just play the same game with the height zigzagging as y varies.)
Based on this, I argue that we need some sort of approximation to the function's "average derivative" along the y direction. From isotropy, the derivative along the x direction is suitable and we just need to carry out the integration. I find this path to the solution inelegant, however.
I'll carry out the derivation at some point, but for now, I'm dead tired (and the day is only starting).
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OmnipotentEntity wrote:
We can assume that the villi are spaced randomly and that our single cross section represents the mean cross section.
Because this is our mean, all cross sections will have on average, the same arc length.
Let's call the arc length / length = f, our constant of proportionality.
Let's consider the area of this cross section to be f*x*dy. x = length.
Therefore, the area is just integral of this value, or f*x*y. So f*A rather than f^2*A.
I was about to throw down a healthy dose of skepticism, but a quick analysis below shows that your answer is plausible. I'm still troubled by the fact that you very quickly state that the area of the cross-section is f*x*dy. Can you go into more detail on how you got this? I don't see where you took into account that the villi exhibits the same ridges along the y direction. This all seems too easy...
Here's the analysis that shows you might be correct. As a very simple example, suppose we have a single cone of base radius r and lateral distance from base to vertex L (not the height of the cone). This fails the homogeneity condition but passes the isotropy condition as long as our cross-section goes through the cone's vertex. As such, it might not be suitable for building an analogy, but I'll explore it anyway. The arc-length is s = 2*L. It can be shown that the lateral surface area is
A = 2*pi^2*r*L
The characteristic dimension of the cross-section is 2*r while the area of the cone's base is pi*r^2. Therefore, your function f becomes 2*L/(2*r) = L/r while the ratio of the areas is 2*pi^2*r*L/(pi*r^2) = 2*pi*L/r. As you predicted, the conversion between them scales linearly, with a constant factor of 2*pi, which is probably unique to the cone.
If we play the same game with a hemisphere, we get an arc-length of pi*r and a surface area of 2*pi*r^2. The ratio of the lengths is pi*r/(2*r) = pi/2 while the ratio of the areas is 2*pi*r^2/(pi*r^2) = 2. Once again, it scales proportionately, with a new factor of 4/pi.
If I'm reading OmnipotentEntity's post correctly, it is implied that the constant of proportionality is one. This seems far off for the cone, but is close to what we observe for the hemisphere.
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Here's a problem I don't know the answer to, so best of luck to everyone.
Suppose we have a membrane in two dimensions that is both homogeneous and isotropic. What immediately comes to my mind is villi in the small intestines. We are supplied only a vertical cross-section of the membrane, like so:
From this, we can determine the arc-length along the cross-section. What is the surface area on, say, one square-centimeter of small intestine? (By which I mean one centimeter by one centimeter punched out of the small intestines. The one square-centimeter measurement is macroscopic while the surface area I'm asking for comes from adding up all the microscopic ridges and folds and should therefore be much larger.)
I am almost certain the conversion factor should be proportional to the square of the arc-length divided by the horizontal distance spanned. Perhaps the proportionality constant is just one. Can anyone succinctly prove what it should be?
I'm really disappointed in the answer they gave to last week's problem. I thought they wanted a closed-form solution and I was somehow missing something. I also think my analysis is more thorough than what they cover in the written solution. It basically just says, "Take the derivatives and set them equal to zero." Well, duh!
Anyway, I noticed the comments at the bottom indicate for the camel problem that you can drop bananas off in the desert (something I wouldn't have considered, as it seems like a bad idea...) and pick them up later. That makes the problem non-trivial.
I like this problem a lot. I started working on it, then realized that I'd already posted a more general math challenge.
Let yi be the y coordinate associated with the top of the ith layer (or said another way, where the boundary is between the ith and ith plus one layer). A little work will show that the volume of this layer is pi*R2*(1-yi/h)2*(yi-yi-1). Let's factor out the constants of pi*R2/h2 to leave the "meat" of the expression: (h-yi)2*Δyi. I've re-written yi-yi-1 as Δyi for convenience.
If you haven't already done so, check out my earlier post, linked above, as well as the post with the solution. We wish to maximize the sum over i of (h-yi)2*Δyi. That means our discrete Lagrangian is (h-yi)2*Δyi. The discrete Euler-Lagrange equation is:
Δ(dL/d(Δyi+1)) - dL/dyi
Plugging in our Lagrangian and simplifying leads to the recurrence relation
yi+12 - 2hyi+1 - 3yi2 + 4hyi + 2yiyi-1 - 2hyi-1 = 0
We have y0 = 0, so for i=1, this becomes
y22 - 2hy2 - 3y12 + 4hy1 = 0
On the other end, this Lagrangian does not depend on yn+1, so bookkeeping is a little tricky here. There is a yn dependence as well as a Δyn dependence, but no Δyn+1 dependence. The net effect is that we're left with a modified Euler-Lagrange equation for the end point:
dL/d(Δyn) + dL/dyn = 0
Which eventually gives us:
3yn2 - 4hyn + 2hyn-1 - 2ynyn-1 + h2 = 0
So in summary...
For i = 1:
y22 - 2hy2 - 3y12 + 4hy1 = 0
For 1 < i < n:
yi+12 - 2hyi+1 - 3yi2 + 4hyi + 2yiyi-1 - 2hyi-1 = 0
For i = n:
3yn2 - 4hyn + 2hyn-1 - 2ynyn-1 +h2 = 0
I believe all of those equations are correct, giving us n equations and n variables. As for using them to solve for the yis explicitly, well, that's left as an exercise to the reader...
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As you know by now, FatRatKnight has done some work on Yoshi's Cookie. For future reference, if you're looking for all information on a game, start by checking the forums (as you now know) and go to the front page of the website and do a search for the game. After executing the search, check the "Include submissions" checkbox, then do it again. When you do so, this submission by FatRatKnight pops up near the top. By including submissions, you can find movies that weren't published for whatever reason.
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FractalFusion wrote:
Bobo the King wrote:
cos(2t) = 2cos2(t) - 1
...
Is there a simple proof without the use of trigonometry
Here is one I came up with that uses only geometry and the definition of cos x. The diagram shows that cos(2x) = 2 cos2(x) - 1
It helps to remember the geometric proof of cos(a+b)=cos(a)cos(b)-sin(a)sin(b).
A wonderful proof without words! There's hardly even much geometry in it! Thanks, FractalFusion!
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Here's a "come up with the most elegant proof" problem.
I noticed by application of the law of cosines and the law cos2(t) + sin2(t) = 1 that cos(2t) = 2cos2(t) - 1. If we substitute in x = cos(t) and x' = cos(2t), then this takes the very compact and elegant form
x' = 2x2 - 1.
Now pretend you don't know any trigonometry but still want to prove this fact. At first I tried drawing the two relevant triangles on the unit circle (I'll provide a figure if requested), but I could find no simple geometric relationship between x and x'. I think that this strategy may not be fruitful because x is a length and x' is written in terms of x2, which we traditionally associate with an area. I was finally able to prove the identity by algebraically defining both the unit circle and a circle centered at (x, y) which intersects points (1, 0) and (x', y'). You're free to do so yourself, but there's a decent amount of algebra involved, including the definitions of those two circles, two applications of the Pythagorean theorem, and culminating in the quadratic formula, which fortunately simplifies nicely.
I tend to believe that elegant identities conceal elegant proofs. Is there a simple proof without the use of trigonometry that x' = 2x2 - 1, where x' is associated with twice the angle of x? Bonus points if you can prove it geometrically with minimal algebra.
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RGamma wrote:
(Note: I don't know how messy this might get, and will look into this tomorrow myself if there are no answers then. If it's too messy, ignore it).
I noticed a problem when playing L.A. Noire: Cole Phelps needs to interrogate a bus driver on duty whose route he inquires at the bus depot. Driving along the path in his car his partner remarks (all paraphrased) "This is taking forever." to which Phelps replies: "Chances are we're going to catch him half-way". Is this true?
You can complicate this question a lot, so for simplification state the problem like this: Given a closed non-self-intersecting path C of length L > 0 in the two-dimensional reals, a fixed point d (the bus depot, Phelps's car's starting point), a point p = d moving at speed |v_p| <= l_p representing Phelps's car and a point b /= p representing the bus moving at speed |v_b| <= l_b and /= 0 (all points on the path; a (positive) negative value for the speed denotes (counter-) clockwise direction, both speeds are constant and bounded).
What is the probability P(v_p, b, v_b) that Phelps encounters the bus b (that is p meets b) after having travelled at most L/2 distance from d? (oh, and assume they're never travelling in the same direction at the same speed if it helps)
Edit: Added bounds on speeds.
I'll admit that I'm not keeping up with the symbols you're using or how you're modeling the problem, but if I'm not mistaken, if the bus is on a continuous cycle and last left the depot at a completely indeterminate time, wouldn't Phelps and the bus meet up, on average, in one-quarter the time it takes to complete the route? If they're traveling in opposite directions at roughly the same rate, they can be assured of meeting up at no later than half the time it takes to complete the route (corresponding to both Phelps and the bus leaving the depot at the same time). If the bus is just pulling into the depot, they meet instantaneously. The bus could be anywhere in its cycle and the distribution should be uniform, so the average length of time it takes for the two to meet up is one-quarter the length of the cycle (e.g., if it takes an hour to complete the route, they'll meet up in 15 minutes on average).
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Andrew wants his Super Mario Bros. record back.
I'm just joking. I don't have anything to contribute to this thread except to say that I agree that there's likely some funny business in the RTA community. Best of luck to you.
Edit: I think it's likely that a scandal will eventually rock the RTA community. Someone will submit an emulated run, then reveal it to be faked (or it will be discovered, though I think that's far less likely). People will argue about it for a few weeks or months and then come up with a two tiered system. They'll still accept runs from talented players who are known to be good at the game in question but there will also be something like a "directly verified" tier that includes only runs that were completed in front of judges and on known hardware. That could be frustrating because that means runs can only be completed at events like AGDQ or perhaps at special stations around the country.
I kind of stopped following SpeedDemosArchive, so let me know if anything like that has already been discussed.
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Piling on with what Samsara said, we have a decent reason to at least sometimes reject a functionally identical run on another console. People are (rightfully) very protective of their work. Creating a high-quality TAS can take years of analyzing game mechanics, glitch-hunting, and even disassembly. We're a pretty friendly community and are open to sharing our findings as we make new discoveries with the understanding that someone won't come along and try to "scoop" the run and make their own before the main authors are done with their own. For the most part, that isn't even something to worry about as a lot of tricks are very technical and whoever is hunting for them should have a sizable head-start.
But it could easily be a problem in multi-platform games. If the Genesis port of an SNES game is virtually identical, there is nothing stopping me from taking all of the tricks found by someone else in the SNES version and making my own Genesis version. I get sole authorship credit despite having discovered nothing. The SNES run's authors can't really do anything about it because they're busy with the SNES run and probably have little interest in duplicating their work on another console.
I don't think any of this is too likely to occur regardless, but I think it's good that we have a cautionary policy in place that effectively says, "Let's take a look at the latest run and make sure it has something new to contribute." I can think of some instances where gameplay mechanics may make two runs on separate consoles both publishable. For example, if I remember correctly, Cool Spot for Genesis allows you to run while shooting whereas the SNES version has fixed shooting, as well as some other differences. Barring these kinds of differences, however, I think the safe thing to do is reject a run on an inferior console, especially when the authorship between the two runs differ.
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Invariel wrote:
But the question says (ambiguously, I suppose) "You hate the crust so much that you’ll only eat the portion of the sandwich that is closer to its center than to its edges", which suggests to me that your graphical interpretation is correct, Flip. Anything closer to the crust than (r/2, r/2) is "closer to an edge than the center" so you get that smaller square that is 1/4 the area of the sandwich.
Which suggests to me that the more general case is that you get to eat 1/n of an n-sided sandwich.
I don't get it. Taking the width of the bread to be 2 units, the corners of the inner boundary are a distance of 1 unit from the outer edge and a distance of sqrt(2) units from the origin. I came up with a picture similar to Flip's when I started this problem, but quickly saw that it couldn't work. In what reasonable sense are those edges equidistant from the center and the crust?
The top image is the average of the two images. The bottom is the difference.
I think the average looks best, but whatever.
From this, we can determine the arc-length along the cross-section. What is the surface area on, say, one square-centimeter of small intestine? (By which I mean one centimeter by one centimeter punched out of the small intestines. The one square-centimeter measurement is macroscopic while the surface area I'm asking for comes from adding up all the microscopic ridges and folds and should therefore be much larger.)
I am almost certain the conversion factor should be proportional to the square of the arc-length divided by the horizontal distance spanned. Perhaps the proportionality constant is just one. Can anyone succinctly prove what it should be?, which I don't know how to do.
It helps to remember the geometric proof of cos(a+b)=cos(a)cos(b)-sin(a)sin(b).