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One more enjoyable run of a popular game by one of TASVideos' greatest frame warriors TASers. Easy yes vote for complete brokenness.
Incredible work klmz, the highlights for me were:
* Much smoother damage taking in Underground Reservoir
* Quicksaving as a shortcut after Tasty Meat o_O Amazing Find!
* Cancelling Ronginus Spear without Winged Skeleton soul
* Great luck manipulation (except for manipulating a drop from Stolas, but I guess there was nothing to do because there was a reset shortly before).
I have only one question: why did you manipulate Cloth Tunic? I couldn't figure...
EDIT:
klmz's updated submission wrote:
Tunic (dropped by Zombie)
Getting this didn't cost any time, and instead, it actually saved time by reducing lags mysteriously in the Mermen room.
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mobilisq wrote:
am i getting a desynch or is there a lot of screwery going on in the menu and pc system near the end of this wip?
In Fuschia, right? It's a desync, you must be playing it in the latest v23. Use an older version (except v20) and it'll play fine. This movie was started long before the current v23 was released, in this new version, they apparently changed the way reset input works, so this movie won't sync after a reset occurs.
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I found this extremely boring, I really don't see the entertainment in any of the slow levels, it's just pointless jumping between enemies forever while waiting for the timer to run out. Besides, I don't find SMB suitable for a playaround at all, the fastest completion goal is what this game is about. Voted No.
At Nach's question:
As said before, I disagree with your entertainment rating of this run, but this is just one's opinion and shouldn't be taken in an impartial judging.
Like many other movies, there are arguments that support both acceptance and rejection of this movie. However, despite this run having an explicitly defined goal, it's still arbitrary. There are movies that claim to be 100% when the game doesn't provide a completion counter and others that are glitchfree when it's unclear what a programming error is.
Still, those are entertaining categories that get relevant support. The addition of the lowest score goal, however, reduces the entertainment from a speedrun and reduces the entertainment from an unrestricted playaround, like many pointed out. It can be concluded, then, that this is an inferior category in both sides, and seeing from the initial negative feedback, this movie doesn't stand well to the general viewer in its own, so it's not a good idea to publish it until a playaround is made (we really can't be sure if it will).
So, if I were the judge, my verdict would be the rejection of this run.
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I didn't find this run boring at all. It was an easy watch for me.
There were also some nice tricks used here that made it interesting. Very solid run, Yes vote.
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I'll answer some posts here.
Mukki wrote:
Clever. I liked how counter-intuitive this seemed at first, but the less time required for the new byte change technique paid off better than I thought it would (well...not completely new, but this application could easily have been completely overlooked). Yes vote, of course.
The new warp also surprised me. At first, I only did a small test to see if it'd be faster. I almost fell off my chair when I saw I was seconds ahead.
Mothrayas wrote:
Love it. Pokémon RBY glitching will never get old to me.
Yes vote, obviously.
PS: How's the 100% run going?
Good progress is being made, thanks for asking! The biggest problem is the obscene amount of luck manipulation and the route complexity. I've lost my count of how many times Mukki and I had to redo stuff because of some significant improvements. At the moment, we're almost at Sabrina's battle.
Dragonfangs wrote:
I love how the Podédex Rating seems like some epic plot twist or something because of the music. Is that a glitch or does it do that in the actual game? Doesn't really seem fitting.
It's a glitch, I think the game reads garbage data as audio because there's no valid sound for pokedex completion after 151.
Dunnius wrote:
Throwing away his master balls? I didn't realize becoming a eunuch was necessary to become a pokemon master. Now that sounds painful!
N. Harmonik wrote:
May someone please do a glitchless run now? I want to see Jessie and James' butts kicked on the small screen (Game Boy) instead of the medium screen (TV) and big screen (movie) for once.
I understand why some people would like a glitchless run, but to me, it'd be taking out what made RBY TASing interesting. Long RPG runs aren't very popular here, and this broken one is almost star-worthy. If someone would run the game without bugs, it'd be better to do the FRLG remakes IMO.
Noob Irdoh's link doesn't contain a glitchless run, Tilus's initial 1:51 run is usually considered glitchfree, despite using the Pokedoll trick, which I consider a programming error (I'm the minority in this matter though). There's also a more optimized glitchfree run of Red (J) by was0x that contains no speed/entertainment tradeoffs, which makes the movie a little hard to watch because of the near death noise (it also uses the pokedoll trick iirc).
If you just want to see Jessie and James beaten, there's an old cancelled Yellow run by FractalFusion using the Lv.100 glitch, it's the most substantial completion of Yellow afaik.
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Submission is updated with relevant information. Thanks to anyone who supported the run without knowing what was happening, and a big thank you for Flygon and mmarks for their encodes.
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Although I'm in love with this game, I'm forced to vote No to this run, sorry.
Blue lags horribly if you mash buttons, don't do it. The fastest way to get through the title screen is by pressing Start/A/Start/A without mashing.
You lost frames naming the rival A, it'd be faster to give him a default name, even if you have to lose time by scrolling the menu with B.
Also, where exactly did you manipulate luck here? You didn't get you trainer ID to match the Hall of Fame map offset, which can be done in Blue like it was done in Yellow.
For anyone wondering how he beat the game, he inserted Lorelei's Map ID in the staircase warp and tricked the game to walk through the doors by changing coordinates, which is even slower than gia's pioneering run.
Anyway, it seems to me that this is not technically on par with the other RBY runs we have here, and I don't see any reason to prefer this type of warp opposed to a faster one.
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mmbossman wrote:
p4wn3r wrote:
Kirkq wrote:
Since the author was new, I made sure to check he wasn't turboing buttons or anything catastrophic like that. =p
Mashing is catastrophic only if it causes lag. If it doesn't, it's a very useful technique that allows the runner to quickly get past the annoying dialogs some RPGs have without any loss of frames. There are also some RPG games where mashing accelerates the closing of text boxes.
Not true, assuming input is polled at 60 fps, "mashing" will lead to a sequence of A_A_A_A_A (obviously). If the text box is able to advance at the first possible time on that second underscore, then you lose one frame. Depending on the amount of text present, that can add up to significant time for an RPG run (not this run so much, but hour+ runs can easily have several seconds lost due to lack of frame precision).
Most RPGs allow two buttons to cancel the text. Just mash ABABABAB for a GBA game, for example, and you don't lose anything.
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Kirkq wrote:
Since the author was new, I made sure to check he wasn't turboing buttons or anything catastrophic like that. =p
Mashing is catastrophic only if it causes lag. If it doesn't, it's a very useful technique that allows the runner to quickly get past the annoying dialogs some RPGs have without any loss of frames. There are also some RPG games where mashing accelerates the closing of text boxes.
About the movie, I liked it a lot. Games of this genre are naturally repetitive, this type of criticism is common to almost all short RPG/strategy game movies.
The glitching is also very nice and unexpected, I agree with klmz that it makes the movie very different from the TAS of the sequel.
Definite Yes vote here, good job!
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Fast rejections are the biggest source of judging controversy in this site, I don't think it's a good idea to quickly reject this. It has beenrepeatedlyproven that casting a verdict solely by the rules is not the ideal way to go.
Also, no matter how sure one can be of the impossibility of an explanation being given, it's a good policy to give the author a chance to explain himself before making a final decision.
That said, I'd vote No because I don't think nfq's reasons for using a savestate are strong enough. However, since he can provide another one that starts from power-on, I'm abstaining from voting until he does it.
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Oh, I missed this submission in the queue and watched it so late that it's already accepted. I'll add my Yes vote anyway.
It's extremely funny how the game cannot handle such a great speed.
I don't know if the rules allow, but I'd rather have the Camhack encode as the main publication video, I find it a lot more entertaining.
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Damn, I thought it was hard to see someone submit a run of MK3 more boring than the tournament mode run...
Definitely, it's indeed necessary to publish every category of every port of every MK game because of all different gameplay aspects they have, isn't it?
Sorry, Thevlackdemonn2294, but this run is getting a strong No vote from me.
TASes of fighting games need to entertain by showing a lot of the game and giving an outstanding impression of tool assistance. The combos here are slow paced and very repetitive. In fact, most of the entertainment is provided by skipping the game with the Cyrax glitch instead of showing it. The bad graphics and music don't help either.
Aside from that, I've already seen movies get rejected for playing an inferior port even when there was no quality submission of a better port. There really is no reason to publish this. Couldn't you MK TASers decide on the best game and do a run that really shows the potential of this series?
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Tub wrote:
Remember, the expected value does not make any predictions about a single game, it's only useful if you're considering a very large sample. And once you do that, you'll notice that both strategies (switch or don't switch) are equal, because you'll have an infinite payoff either way and it doesn't get any better than that.
The expected infinite payoff cannot be used to prove the equality of both strategies. We can prove it by the symmetry of the two envelope problem.
In my example, {(1,2), (2,4), (3,6), (4,8), ... } you're asked if it's better to pick the first or the second number. For every pair you get, it's better to choose the second number, even when you expect infinite payoffs for both sets.
Evaluating gain - loss in a set with infinite mean would yield infinity - infinity, which is undefined, like you said. So, it cannot be used to prove anything, nor that a strategy is better or if they are equally good. The fundamental difference between these cases is that the envelope problem is symmetrical, making it easy to see both strategies are equal.
Taken from the Wikipedia article:
"However, Clark and Shackel argue that this blaming it all on 'the strange behaviour of infinity' doesn't resolve the paradox at all; neither in the single case nor the averaged case. They provide a simple example of a pair of random variables both having infinite mean but where one is always better to choose than the other."
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With the risk of boring everyone else reading this thread, here I go:
To clarify, I do not believe that switching an envelope everytime is better, I'd be crazy if I thought that, I'm just having a conversation about what is wrong in the common conclusion.
I'm skeptical of the infinity expected payoff approach, because:
* it just says "infinity is strange, by comparing things with infinity, we arrive at strange conclusions"
* concluding that two sets are equally better because their mean is infinity is clearly wrong. Take for example the pairs (1,2) , (2,4) , (3,6) , (4,8) ... When given a pair, the second number is always preferable, the fact that the set of the first numbers and the set of the second numbers have infinite means is irrelevant.
My solution is that the conclusion "Every value in infinite unbounded set A is less than a value in infinite unbounded set B implies that set B is preferable to set A" is not valid.
Again, I don't have the background to provide a formal proof of this, so I'll just give an intuitive one:
* The central tendency of a set is determined by its mean, in the unbounded "two envelope problem", the mean of both sets is infinity, thus it's impossible to compare them.
* Saying, in the reformulated problem, that, for every x, the fact that x < 11x/10 reaches no conclusion. Every value x is smaller than the means of the sets, by comparing values smaller than the mean, no conclusion can be made.
Thus, the comparison x < 11x/10 can't be used to say that the other envelope is better than the one you picked.
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Tub wrote:
p4wn3r wrote:
It can be argued that it's impossible to have a uniform infinite distribution according to modern probability theory. While this solves the problem I brought up, it can be changed so that the set in discussion is { (2^n,2^(n+1)), n in Naturals }.
Nope. There is no uniform probability distribution for any discrete infinite set (or for any unbounded interval in continuous distributions), not in any variant of probability theory.
Huh, I guess you missed the point.
One distribution of (2^n,2^(n+1)) has probability P(n) = 2^n/3^(n+1). The sum of this series converges to 1, so it's clearly a valid distribution. (I never said that set had uniform distribution either, please read that again)
So, for a value 2^n for n larger than 0, if I calculate the gain like I did before, I get
( ( 2^(n+1) - 2^n )p(n) - ( 2^n - 2^(n-1) ) p(n-1) )/( p(n) + p(n-1) ) =
( 2^n ( 2^n/3^(n+1) ) - 2^(n-1) ( 2^(n-1)/3^n) )/( 2^n/3^(n+1) + 2^(n-1)/3^n ) =
( 2^n/3 - 2^(n-2) )/( 1/3 + 1/2 ) = 2^n(1/3 – 1/4 )/( 5/6 ) = 2^n(1/12)(6/5) = 2^n/10
So, using non-uniform distribution, I'd still get 1/10th more of the sum if I switch the envelope.
While this analysis is not trivial, I didn't point it out because rhebus had already linked it before. It's usually good to read some posts before so that people don't need to spend time explaining the idea when it isn't necessary, and that you don't need to teach me things I already know.
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rhebus wrote:
Bertrand's Paradox is no paradox -- just an underspecified problem with three solutions which complete the specification in different ways. They are measuring different distributions and get different results.
It was. I think it was this problem that took down the old theory and made people aware that they had to consider the distribution. I put it as a form of saying how mathematical proofs can be non-intuitive, even when they're clear. Looking back at my post, I seemed to say that the measuring had to be unique, thanks for clarifying.
Warp wrote:
Tub wrote:
Note that if you pick an envelope with $1000, you're absolutely correct to say that switching is advantageous.
I don't understand why switching would be advantageous. It being advantageous would result in a contradiction (because once you switch, switching again would also be advantageous by the same reasoning).
Well, I only teach math up to calculus and linear algebra, so I'm far from being a specialist in this area, but I had a deeper look into this and I'll try to answer this (I may say something stupid, be warned).
Old/classic theory, in a nutshell, considered that the probability of an event is computed by dividing the amount of events favorable divided by the total events (or, in the case of a continuous domain, the length/area/volume of the favorable domain divided by the total length/area/volume).
However, Bertrand's paradox shows that this fails in some cases. Like rhebus pointed out, there must be a distribution of probability so that we can evaluate it. I can consider that the chords in a circumference magically appeared out of nowhere or that the two envelopes were conceived by a mysterious entity, but there's no way to tackle this problem mathematically, it would be the same as asking "what's the chance that adelikat is thinking about redoing the Gradius run right now?"
To properly compute the probability, we have to take into account all possible distributions. In Tub's case, switching is favorable because you looked at the value of the envelope and can compare it to the possible range.
As an example, consider the uniform distribution of pairs (1,2) , (2,4) , (4,8) ... (2^99,2^100), which is bounded (my explanation can be made for an arithmetic distribution, instead of a geometric one, but it's more tiresome).
See that, when you look at the envelope and see 1, it's obvious that you must switch, while if you have 2^100, a switch will only reduce your money. For other values n, switching is overall better, we can compute the expected gain by calculating (2^(n+1)-2^n)-(2^n-2^(n-1)) = 2^(n-1)
The contradiction arises if it's better to switch without looking at the envelope, i.e, for any value there. For that, we need to add the gains for all values you can open.
1 = 2^0: Gain is (2-1) = 1
2 = 2^1: Gain is 2^(1-1) = 1
4 = 2^2: Gain is 2^(2-1) = 2
... ... ...
2^99: Gain is 2^(99-1) = 2^98
2^100: Gain is -2^99
Adding everything, S = 1 + (1 + 2 + 4 + ... + 2^98) - 2^99 = 1 + 2^99 - 1 - 2^99 = 0
Thus, it makes no difference to switch the envelope without looking at its value first. See the problem now? When the value is not bounded, there is no last term to compensate the other gains, and doing this procedure would result that it's better to switch in all cases.
It can be argued that it's impossible to have a uniform infinite distribution according to modern probability theory. While this solves the problem I brought up, it can be changed so that the set in discussion is { (2^n,2^(n+1)), n in Naturals }.
For this case, by computing the distribution of each value and evaluating the probability, we still come to the conclusion that it's better to switch for every value. Here, my knowledge fails me and I can't really say more than what I already have. Hope this was helpful.
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Tub wrote:
Now you're saying: any amount goes. To which I reply: there is no uniform distribution over all numbers, your premise is flawed.
p4wn3r, in his original post wrote:
Basically, in an infinite distribution, the probability of the numbers appearing is not the same.
p4wn3r, later wrote:
To completely solve a paradox, you have to point out a mistake on the calculation of the final value 5A/4 (which is not the case here, the value is gotten using correct premises of classical probability) OR say why the theory you're using to calculate it is mistaken, that's the difficulty of the problem.
Where exactly did I say that any amount goes?
I never said my premises weren't flawed either, I just cited that problem to explain why classical probability theory can't be used in flagitious's problem when there's no range determined.
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Oops, I miscalculated before, 1/2*(A/2)+1/2*2A = 5A/4, not 3A/2
It really makes no difference whether you switch or not, I implied this when I said that conclusion was absurd.
The paradox rises when you reach a false conclusion using classical probability theory. To completely solve a paradox, you have to point out a mistake on the calculation of the final value 5A/4 (which is not the case here, the value is gotten using correct premises of classical probability) OR say why the theory you're using to calculate it is mistaken, that's the difficulty of the problem.
In order words, it's easy to understand why your proof by contradiction is correct and the previous one is wrong, because human intuition gives the answer.
However, take Bertrand's Paradox, there are at least three proofs leading to different results that don't have any flaws when you consider classical theory. Determining which value is the correct one would need a counter-argument to the other proofs.
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Decision problems involving probability distributions are usually very hard to solve and, IIRC, no consensus has been reached yet.
An example that illustrates the non-triviality of these puzzles is the "Two envelopes paradox".
Suppose that you have to choose between two envelopes, one contains X dollars and the other 2X dollars. After choosing an envelope with A dollars, you're given the chance to switch to the other one. To see if it's worthy, you assume that there's 50% chance the other one will have A/2, and another 50% that it'll have 2A. Summing, your expected gain will be 5A/4, thus it's better to switch. However, since that's true for every value, switching before even choosing the envelope would rise your expected value, and this is absurd.
These problems introduce a concept called "expectation", which is hard to formalize in mathematics. By picking a large value, you may think "it's unlikely that the other envelope contains an even larger sum". Basically, in an infinite distribution, the probability of the numbers appearing is not the same. This may influence your players' problem.
I lack the mathematical background to offer a reasonable solution to this problem, I suggest that you look at other paradoxes of probability and decision theories and their proposed solutions and see if it helps you.
EDIT: Wow, it seems Tub posted something similar while I was writing, lol
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Wrong, sorry.
ln (sin x) is non-positive in the given interval. This comparison criteria only works if the function considered is non-negative.
Try proving the convergence of \int -ln (sin x)dx in the given interval, where most methods will apply.
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Very good! You're correct.
More rigorously though, the second integral at (b) is an improper integral of the second kind, since lim x->0+ (ln (sin x)) = -infinity. Thus, you have to prove this integral converges before calculating it.
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Guidelines wrote:
The game should give an impression of complexity; it should not be overly easy or straightforward.
Based on this, I vote No for this submission.
This is not the first time, and probably not the last, that a run of an extremely simple game with no interesting gameplay is submitted. As far as I'm concerned, all have been rejected.
The tricks used in this improvement don't contribute in any way to properly show tool-assistance. The gameplay continues obvious and straight-forward. This game is not well suited for TASing and that's unlikely to change no matter how much time is saved.
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No one? Hmm, that integral is not really hard, I think people here haven't seen the method to solve it. I'll reformulate the question pointing the right direction, but also putting one that's a little harder ;)
(a) Prove that, for every real function f, integrable in the interval [a,b], the following equality holds:
(b) Evaluate:
Come on, guys! You can do it!
(b) Evaluate:
Come on, guys! You can do it!