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Therealssjlink wrote:
p4wn3r wrote:
arukAdo wrote:
The entertainement is supposly subjective, personally i dont know the games well enought to have a realistic technical comment against the movie, but seriously, how hard is it to admit that it can amuse people to watch this kind of movie in all fairness?
What you are arguing here doesn't even matter since NO ONE is holding you back from voting no.
Regardless, look at this and tell me what you think since it doesn't follow MANY of the reasons on that list and STILL got published (long obsoleted but still)
http://tasvideos.org/forum/viewtopic.php?t=8004
I sincerely don't understand what you're trying to bring up here. Anyway, my point was that it's not hard for me to admit that people can enjoy this run, people can enjoy a sudoku TAS, a minigame TAS or a rhythm game TAS, doesn't mean it should be published. I don't get why my refusal to cast a vote affects anything that I argued.
And no, I don't feel like telling you what I think about a discussion of eleven pages, this has nothing to do with this topic, and wouldn't mean anything useful even if I did. You could bring up ten submissions that were published despite me being against them, it doesn't mean that I should want the site to make a mistake because it made others in the past.
Indeed, no one is forcing me to abstain (despite the recurrent appearance in some submissions of "whoever voted No should be banned", "obviously someone had to vote No so that it doesn't get 100% yes"), I can vote No if you insist, I assure you it won't affect anything. Will you now demand more clarifications of everyone who voted Yes?
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arukAdo wrote:
The entertainement is supposly subjective, personally i dont know the games well enought to have a realistic technical comment against the movie, but seriously, how hard is it to admit that it can amuse people to watch this kind of movie in all fairness?
AI manipulation obviously not, but combos probably, as this demonstrates, I know it's not the same game, but still proves the point.
arukAdo wrote:
I wouldnt like to take a dozen every day for breakfast but once in a while its still fun to me.
Now if there could be less mortal kombat troll submisions it would be a breath thats sure.
AFAIK it's fair to vote No to something that's entertaining but that wouldn't be watched again, I don't even consider this entertaining, and I actually enjoy the troll submissions because they're rejected.
arukAdo wrote:
Did i won a cookie ? :p
Welcome to the dark side...
and we lied about the cookies
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At least this time you made proper use of frame advance...
Still doesn't mean this is any sort of high quality technical achievement to warrant publication as a playaround like thesemovies. There was a time when I thought that under some very restricted conditions, MK TASes could be accepted. Not anymore.
Yes, I know that this game has a lot of combos, nearly all of them pretty doable in real-time, I know what a glitched sprite is and have seen lots of them, it's no justification to publish a movie with an arbitrary goal that doesn't require skillful use of the tools that come with the emulator, it's all about AI manipulation of a single opponent with no obligation to aim for the least amount of frames dropped and not even this I thought it was done well, the 2-player parts don't even have that.
I still doubt the entertainment value of this given the countless MK runs published (in fact I feel more stupid whenever one of them gets published because I see nothing special) and even if I didn't, "to Ceasar that which is Ceasar's", this doesn't seem to differ enough from a skilled player using savestates. I'd vote No, but given what's passed already this'll have an "accepted" status after 72 hours, enjoy your voting unanimity.
I hate to explain a joke, but well, the point was that it's absurd to derive any sort of scientific validation out of a set of data without any theoretical background that would consolidate it as unbiased and accurate enough to arrive at the conclusion.
The amount of experiments in the example is COMPLETELY IRRELEVANT and serves only as a form to evidence such absurdity.
Anyway, this sort of stuff has been around for years, especially in 1999, when many religious and pseudo-scientific movements postulated that the passage to the year 2000 and the arrival of the new millenium would cause severe changes to the world or other weird stuff like that (and the widespread belief that a bug in computer dates would cause economical havoc when changing from 1999 to 2000), even serious areas of science dedicated to studying this but fortunately they realized they didn't have to bother.
I have a friend who messages me in a semi-daily basis with this conspiracy theory stuff and so far I've failed to convince him that it's stupid. To me, all this pseudo-scientific stuff originates from a widespread misconception that the scientific genius is the guy who stays away from the literature, constantly gets into trouble for rebelling against current paradigms and ad nihilum comes up with an extremely innovative theory that will be ignored until he dies, when everybody suddenly realizes the genius he was and grants him his well deserved praise.
This idea couldn't be more wrong, I can safely tell you that there are no scientists that can generate groundbreaking discoveries through independent research on a regular basis and encourage you to be really skeptical of their claims if they say so. Experiments may not lie, but people do.
Of course, the story of Newton deriving the laws of gravitation or Hamilton discovering the fundamental operations of quaternions while walking across a bridge is much more romantic than a series of letters exchanged between mathematicians and physicists of the time, and for this reason, it will be told for hundreds of years.
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Oh no, not another one of these threads.
I also didn't care about youtube, and only read some excerpts of that paper, personally I'd never have the balls to publish something like that.
It all looks like an anecdote I heard sometime ago, where a scientist has to test the validity of the hypothesis "all odd numbers greater than 1 are primes" and does the set of experiments: "3 is prime, 5 is prime, 7 is prime, 9 is not prime, 11 is prime and 13 is prime".
Conclusion: out of 6 experiments, 5 favor the hypothesis, then it must be true.
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Hi sprocket2005 and welcome here. I'm the author of the movie that sgrunt linked above and unfortunately I find it unlikely that your movie will be accepted, because there are some flaws that can't be overlooked.
First, this game is compatible with SGB and should be run in this mode, the reason is that it looks nicer to who watches it, despite being slower than a run in GB mode. Second, you should terminate input immediately after the game ending sequence has started, so that we can time your movie properly, you kept mashing unnecessarily after the end, I haven't timed it precisely, but the end of your movie seems to be around 4:47, which is still slower than the published movie (here we consider the Red and Green versions as the same category since there's no point in having them both because of their similarities).
Now to the technical part, Pokemon RBY is a game with a high technical level and you should try harder, do not use auto-fire to pass through the dialogs, it'll cause the game to lag. Also, you should pick bulbasaur and manipulate its defense DV so that it gets killed by Gary's charmander in 3 hits. Here you got charmander, didn't manipulate the battle properly and ended up taking damage for no reason, losing some time. Don't take extra steps to manipulate the encounter, the Dokokashira step count only starts after you enter Pallet. I take the extra menu call in Celadon is to avoid a crash but iirc that should be possible to do without it.
Finally, 35 rerecords is not an indication that this was done to acceptable optimization levels, indeed this video beats your submission by a considerable amount.
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Judge guidelines wrote:
Keep the number of different branches per game minimal. A run for a proposed new branch for a game should offer compelling differences relative to previously published runs of that game.
Judge guidelines wrote:
Too many hacks of the same game engine causes the same issues as too many categories of the same game.
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Randil wrote:
For a function f to be analytic at point x0, it has to be infinitely differentiable at x0, and its taylor series at x0 must converge to the function value f(x0). A function is smooth as long as a small change in x causes a small change in f (see this definition). So being analytic is a much stronger property than smoothness - all analytic functions are smooth, but not the other way around (abs(x) is smooth but not analytic at x=0 since the derivative does not exist at this point).
Maybe someone can correct me if I missed something.
"Smooth" is not a precise term. It can mean various things depending on the context, in topology and differential geometry it's usually taken as class C1 (has a derivative everywhere in the domain and this derivative is continuous) and in real analysis as class C-infinity (admits derivatives of any order and all are continuous in the entire domain).
Not a correction, but an addition, whenever we talk about the Taylor series of a function around a point x0, we also have to mention its radius of convergence, or the interval where the series converges. When a function is analytic at x0, its Taylor series around x0 has a positive radius of convergence (for some non degenerate interval, it converges) and for every x in the interval, the value it converges to must equal f(x).
To be analytic, the function has to be C-infinity, but the converse is not true, the most famous smooth non-analytic function is f(x)=e^(-1/x^2), its Taylor series at x=0 is the null function, so despite it converging it does not converge to f(x). The function I presented in my incorrect attempt is also not analytic at x=0, its Taylor series will diverge for x!=0, having a radius of convergence = 0. An adaptation of it can be made to make a function that's smooth and non-analytic in all its domain.
However, smooth non-analytic functions are usually taken as monsters of real analysis and have little importance outside of the theoretical realm.
Warp wrote:
Randil wrote:
abs(x) is smooth but not analytic at x=0 since the derivative does not exist at this point
How can it be smooth at x=0 since there's a clear discontinuity in the derivative of the function at that point? (The derivative is -1 when x<0>=0, hence by definition it's not a smooth function, AFAIK.)
I think he may be using another definition of the term or confusing smooth with continuous. Anyway, |x| is indeed not smooth, but its derivative doesn't equal 1 at x=0, it doesn't exist.
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Nitrodon wrote:
That said, you only have to differentiate twice to get rid of the indeterminacy, since f''(0) is nonzero. You will obtain L=(-1/2)0/f''(0)=0, and hence this is not a counterexample.
ROFL I fail!!!! *facepalms at himself*
That's right, I was careless at that step...
To make it even more embarassing, I ended up actually proving it when I tried to find another counterexample o_O, I think my intuition is not very good. (Or also my math, if the following turns out to be wrong).
If for some n, f(n)(c)!=0, that means limx->c f(x)/(x-c)n != 0, where n is the smallest order non-zero derivative.
I prove that with l'Hopital's rule, it can be made more rigorous with induction, but well, if you keep differentiating the numerator and the denominator, you get:
limx->c f'(x)/n(x-c)n-1 , limx->c f''(x)/n(n-1)(x-c)n-2, ... , limx->c f(n)(x)/n!
Now, if n is indeed the smallest non-zero derivative, all fractions will have a 0/0 indeterminacy, except the last one, which will be different from zero because fn(c)!=0, this is enough to say that the original limit is not zero. Also, assuming that f doesn't get negative for a neighborhood of c, its smallest order non-zero derivative can never be an odd number, since in this case c wouldn't be a local minimum. So, the exponentiation of the two functions can be written without loss of generality as:
(f(x)/(x-c)n)g(x) * ((x-c)n)g(x)
The base in the first factor goes to something other than 0 when x->c and g(x)->0, so that factor has limit 1, and the other one is equivalent to exp(n g(x) ln (x-c)).
For the exponent: n g(x) ln x = n (g(x)/(x-c)) * (x-c)ln (x-c)
Since g(c)=0, the limit of g(x)/(x-c) is g'(c), which exists, since g is smooth. An application of l'Hopital's rule in (x-c)ln(x-c) will show that it goes to 0, so the limit of the exponent is 0 and thus the limit is exp(0)=1.
I'm surprised by this, since it's much weaker than the conditions I read (f and g analytic), in fact that only requires f to be smooth and have derivatives that don't vanish and g to be differentiable at x=c, leaving me to wonder why they put those much stronger conditions there, perhaps because the most common counterexample to that is f(x)= exp(-1/x^2) and g(x)=x^2, a non-analytic function.
If what I wrote is right, that's an interesting problem, it's been a while since I got beat up by a limit, it took me a long time to figure out the division by (x-c)^n, perhaps I'll put it in a test some time.
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I finally got around to doing this problem:
Warp wrote:
Let's assume we have two functions f(x) and g(x), and a constant c, such that:
1) Both f(x) and g(x) are smooth functions (around x=c).
2) f(c) = 0 and g(c) = 0.
3) The nth derivative of f(x) is non-zero at c. (I think this is a more formal way of stating that f(x) is not simply the function "0". Is there a better way?)
Conjecture: limx->c f(x)g(x) = 1
We can verify that f(0)=0, because when x=0 all cosines turn to 1 and the series becomes a geometric progression. For x in (]-2pi,2pi[ - {0}), at least one of the cosines is less than 1, so the series will sum to less than e/(e-1) and thus, f(x)>0 for that interval, so the exponentiation is defined there.
Using the fact that cos(nx) is at most 1, and taking Mn = e-n, we can apply the Weierstrass M-test to the series of functions in f, since the series with terms Mn converges absolutely, the series of functions in f will converge uniformly and thus can be derived term by term.
We have .
The absolute convergence of the series of functions isn't true just for the geometric series of 1/en, it also holds for n/en, n2/en and nk/en in general. There are many ways to prove this, the one that comes to my mind right away is to use the inequality:
And use a comparison test with the absolute convergent series 1/n^2.
Because of this, the Weierstrass M-test will guarantee uniform convergence for all of f's derivatives, which will always exist for x=0 and will be differentiable term by term. This is enough to say that f is smooth, and to satisfy another condition:
For g, it's fairly obvious that g(0)=0 and that g is smooth.
Now, for the limit, we use an idea similar to FractalFusion's: fg = exp(g ln f) and find the limit of the exponent using l'Hopital:
This time, it's significantly more annoying because we have to derive the numerator and the denominator four times to get away from the indetermination, eventually:
Just substitute x=0 and:
(the minus sign shouldn't be in the last member, oh well)
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Warp wrote:
3) The nth derivative of f(x) is non-zero at c. (I think this is a more formal way of stating that f(x) is not simply the function "0". Is there a better way?)
Yes, just saying "f is not the null function for any neighborhood of c".
It's not required that a derivative be different from zero for the function not to be constant. Take f(x)=e^(-1/x^2) for x!=0 and f(0)=0.
The nth derivative at x=0 is always zero and f is not the constant function for any neighborhood of x=0. Of course, having a nonzero derivative will guarantee that it's not constant around that point, but the converse is not true.
That said, I know that result holds if f and g are both analytic and f differs from the null function at a neighborhood of c. I also know that the condition requiring f and g to be analytic can be weakened, but probably not so much as just f(n)!=0 or they wouldn't even mention this requirement on f.
However, all examples I know of smooth non-analytic functions that have a non-zero derivative involve Fourier series or other long stuff like that, which I find way too boring to be interested in or to verify if they hold here. Perhaps you can find something in "Counterexamples in analysis".
Felipe: The Hedgehog
My little hedgehog, Felipe. He hates the lights, so he only goes out at nights, and he's very active. That's why it's very difficult to do him a good photo.
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I disagree with the above post for many reasons. DarkKobold's decision was completely fitting and was in accordance with the audience, that gave feedback and voted based in the site's guidelines.
Notice that rejecting a movie doesn't mean that most people disliked it, in many occasions we vote No to movies that we like and Yes to those we don't. The question presented is "Should this movie be published?". The site is dedicated to tool-assisted movies of a wide variety of games and publishing every category and hack of SM64 may not be the best way to represent the community's desire in general.
I have nothing against complaining about a judge's decision, the gruefood forum is for this anyway, but I recommend people to refrain from attacking well estabilished guidelines with no argument as this is often misinterpreted.
That said, I'll go along with most users and say that a higher completion run would show this hack's potential better and be more worthy of publication and that the authors should submit it when/if they finish it.
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FODA wrote:
That friday music has been hand-crafted by the devil I say. I came to this topic and watched every spin off of the original song you guys posted, and the music stayed in my head for the whole day and then the night, tormenting me like khan's worms except less pleasant to the point I seriously thought I'd go nuts. I swear I'm never gonna click anything related to this monstrosity again (and any "so bad you must watch it!" thing for that matter).
After this, I swear I tried really hard not to post this, but I can no longer help it. Dedicated to our friend FODA:
Link to video
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@Warp:
Using heuristic arguments for constructing the proof :P
Invariants are the best method you have at tackling problems like that, (in my previous example, if you found out that the square sum always doubles, it'd be enough to say it diverges), but the arguments are usually ad hoc and will work only for a very limited class of numbers (I think that's the reason there are a lot of classes of primes...).
I see it could be done by taking the number, finding an invariant of the sequence it generates (it doesn't have to be valid for the entire sequence, if it doesn't work for a finite number of terms, it still goes) and prove that this invariant causes the sequence to diverge.
EDIT: And there's this cute problem: http://en.wikipedia.org/wiki/Lychrel_number :D
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Warp wrote:
I'm not sure I understand correctly. Unless I'm mistaken, you seem to talk about methods that could in some cases be used to disprove or prove the counter-example as valid, rather than a sureproof way of telling if it is.
And indeed that's the case.
Warp wrote:
As far as I understand, a theoretical counter-example to the Collatz conjecture doesn't need to cause a loop (which doesn't contain 1), but could cause the series to grow forever.
Of course.
When you asked that, I understood that you were asking how one could come to the conclusion that the number 1 would never be reached while doing only a finite number of computations, but you seem to be looking for an algorithm that determines whether it terminates or not.
I don't know if there's one, it may be possible that one doesn't exist (Hilbert's tenth problem is the most famous case where it was proved that no algorithm can solve a certain class of number theoretical problems).
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Things like this are proved mostly with invariants. I find it easier to understand if an example is given. Take the triple (3,4,12) and follow an algorithm: at each iteration, pick two numbers a and b from the triple and replace them with 0.6a - 0.8b and 0.8a + 0.6b.
You're faced with the problem "Will this eventually get to (4,6,12)?" the answer is no, we can see this by summing the squares of each number in the triple after an iteration:
(a,b,c) -> (0.6a - 0.8b , 0.8a + 0.6b, c)
(0.6a - 0.8b)^2 + (0.8a + 0.6b)^2 + c^2 = 0.36a^2 - 0.96ab + 0.64b^2 + 0.96ab + 0.36b^2 + c^2 = a^2 + b^2 + c^2
See that, wherever you apply it, the sum of the squares doesn't change. It's an invariant of the process. Notice that the square sum of (3,4,12) is 169, while in (4,6,12) it's 196. Thus, we'll never reach (4,6,12) since it requires the sum to change.
The problem is that afaik nobody has found a strong enough invariant to disprove the Collatz conjecture, that's the reason it's believed to be true. And also, one way to attempt a computational proof is to find a number that reaches itself (without going to 1, of course) the existence of a cycle would disprove the conjecture.
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<Raiscan> (19:32:33) (+TASVideoAgent) Page NitroGenesis edited by NitroGenesis (Dumped Smurfs): http://tasvideos.org/NitroGenesis.html << That must have hurt.
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Tub wrote:
How about, instead of hitting a (yet to be implemented) "mark as screenshot" key, you just hit the "screenshot" key and submit the screenshot together with your movie later?
I agree, many authors won't care about screenshots and for the ones that do, it suffices to put the frame number in the submission. I don't see the need for this feature, it'd make publishing work easier, but it doesn't seem screenshot selection is so hard to demand such a change.
EDIT: Properly quoted because Warp posted before me.
We can verify that f(0)=0, because when x=0 all cosines turn to 1 and the series becomes a geometric progression. For x in (]-2pi,2pi[ - {0}), at least one of the cosines is less than 1, so the series will sum to less than e/(e-1) and thus, f(x)>0 for that interval, so the exponentiation is defined there.
Using the fact that cos(nx) is at most 1, and taking Mn = e-n, we can apply the Weierstrass M-test to the series of functions in f, since the series with terms Mn converges absolutely, the series of functions in f will converge uniformly and thus can be derived term by term.
We have 
.
The absolute convergence of the series of functions isn't true just for the geometric series of 1/en, it also holds for n/en, n2/en and nk/en in general. There are many ways to prove this, the one that comes to my mind right away is to use the inequality:
And use a comparison test with the absolute convergent series 1/n^2.
Because of this, the Weierstrass M-test will guarantee uniform convergence for all of f's derivatives, which will always exist for x=0 and will be differentiable term by term. This is enough to say that f is smooth, and to satisfy another condition:
For g, it's fairly obvious that g(0)=0 and that g is smooth.
Now, for the limit, we use an idea similar to FractalFusion's: fg = exp(g ln f) and find the limit of the exponent using l'Hopital:
This time, it's significantly more annoying because we have to derive the numerator and the denominator four times to get away from the indetermination, eventually:
Just substitute x=0 and:
(the minus sign shouldn't be in the last member, oh well)
The best, however, is this one:
