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I must say I understood close to nothing in this run, but it was mildly funny seeing the player's number go up a lot while his opponents stayed in the negative. I get the feeling that luck manipulation changed gameplay drastically and that's usually very entertaining for those who know the game well. While the movie was repetitive, the other reasons are enough for me to vote Yes.
SD Encode: http://www.mediafire.com/?03w72d18j21kolg
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Fortunately, Eternal SPU plugin syncs for this movie. I'll encode it in SD when the updated movie file is available. It may still be possible to do an eventual publication this year.
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sgrunt wrote:
Although this run doesn't use save game corruption, it's still fairly heavily glitched - notably, getting to the hall of fame seems to use a trick fairly similar in nature to what's ultimately used in the Yellow run.
While I can appreciate the desire of a run to reach the end of the game as quickly as possible, do we really need three runs of the various versions of the first generation, each demonstrating various glitches to get to the end of the game?
Granted, this run is significantly longer and demonstrates quite a few other glitches at this point, but I think we should think very carefully as to how far we should go in this respect.
I'm not sure if I understand what you wrote. Currently, there are three runs of the first generation using glitches to finish the game and it's been like this for a long time.
If you're against publishing three movies of the (U) versions, this run is intended to obsolete the current Blue run. If you look up the history of primo's movie, there have been obsoletions by different versions in the past. This is not a separate category, I'd rather have one more category for the 100% run, which is being worked on (it'd be much different from this one because completing the pokedex doesn't allow the glitch here to be used).
Ah, and also, I put the route in the submission since some people have been confused ;)
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ais523 wrote:
What were the speed/entertainment tradeoffs? Just forgoing the save corruption glitch (in which case they probably shouldn't be listed as that's part of the category), or something else?
Ah, I forgot to put that. It's taking a little time to name some Pokemon with "better" characters, but the biggest one was not to abuse the near death noise to skip pokemon cries in battles, because the sound is very annoying. I'll edit it later.
EDIT: I seem to have trouble encoding this with the border. Here's a temporary (no border, no logo, no subs) encode:
http://www.mediafire.com/?bksrjgzf7mrcod1
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Zowayix wrote:
Remind me again how the Pokedex counter gets overwritten with 152?
There's no pokedex counter. The Seen pokedex and the Own pokedex (they are separated) are stored in binary flags. (0 - wasn't seen/caught, 1 - was seen/caught). Corrupting the save overwrites the entire pokedex region to 0xFF.
Since the game works mostly with bytes, in order to store 151 bits, it needs 19 bytes. That gives 19 x 8 = 152 bits. To evaluate the total amount, it basically counts all 1's in that region. Therefore, it believes you have 152 pokemon because the entire portion was overwritten with 0xFF.
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adelikat wrote:
On another note, that ending level glitch is what? one in a million? and it happens 3 times. It isn't manipulating an RNG specifically, but is it a contender for Lucky TAS of the year? No RPG (or any TAS for that matter) to date has ever manipulated something so rare.
<offtopic>
The case here may be completely different and I have no statistics to show, but in this movie, since the game doesn't allow the RNG to be manipulated during the battle, all manipulation must be done before a duel starts so it leads to pretty rare scenarios.
I remember to have calculated that, for the water master Amy battle, the probability for the starting hand was around 1 in 240000, and it was still necessary to manipulate her hand to lose fast, which was something like 1/50 and some coin flips. Additionally, for this instance, I needed to manipulate all those things at once for the Amy duel and others for a duel before it at the same time, so it's probably rarer than DelayStageClear.
The difference is that, this run is manipulating hardware behavior from what I understand, and the only way is to do a gigantic amount of bot rerecords, where in TCG, knowing how the hands and flips are generated made it possible to run it with a good degree of optimization avoiding massive rerecording.
The manipulation in the 4 player monopoly run might also be of the magnitude in this run. Nevertheless, this is easily the most improbable run of a popular game here and is a favorite for the lucky TAS award.
</offtopic>
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gia wrote:
p4wn3r wrote:
I made sure to put "Aims for fastest completion of the game"
Maybe I understood you wrong, but first you said the runs have to be entertaining. Now this is another debate, and my opinion may not be the common one, but if you go for fastest completion then you will not compromise even once and will sacrifice entertainment wherever necessary. The problem here is when you compromise too much and overshoot and end up not actually beating the game.
As for the rest of your post I see you also decided to attack just setting flags, so I'll file it under "an ending requires sensory confirmation".
For all we know many of the extreme possibilities could be slower from a tas perspective and as such much debate saved, but I still would like some opinions.
I'll try to make it more clear. Runs with the "aims for fastest time" tag aren't required to have no time-sacrificing additional objectives (for example, they can still avoid glitches, choose not to use death/take damage). It's simply there to indicate that the author completes the game with the speed objective in mind and, therefore, posterior runs that are faster and don't have big losses in entertainment should replace it.
Of course, most of the time, it's redundant to put such a tag, but in Yellow's case, someone could think of it as a glitch demonstration or playaround (trust me, there are people who think this), so I made sure to put the tag to avoid such misunderstanding, I submitted and it was accepted by the judge as a speed-oriented movie, what means that this site recognizes it as a valid completion of Pokemon Yellow.
And I used flag overwriting as an extreme example. The point I meant to defend in the previous post is: defining an ending is not important, because if a considerable amount of people think that a movie doesn't complete the game when it claims it does, that movie will be rejected. Feel free to file this under anything you want.
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This judgment falls on the shoulders of the person who's running the game.
Every TASer is expected not to go blindly to the fastest completion goal, but also make the movie entertaining and even tolerate a loss of time if it's necessary to make the movie entertaining. Since we're talking more about pokemon, one could take 10 minutes off primorial#soup's Blue run without any route change if he decided to ignore entertainment tradeoffs and use the near-death noise all the way (making the movie unwatchable in the process).
I think the same applies for game ending definition. If we evolve to a point where it's possible to overwrite the end flag and someone submits a movie that ends input with the player standing at his house, it's my opinion at least that, for people who will watch it at youtube/dailymotion/nicovideo, it makes little difference if it was written in the submission text "After that, the game behaves like it was finished, I can go to Mewtwo's dungeon and the Hall of Fame option appears in the PC, meaning that the game was beaten". People would still believe the game wasn't completed.
If it happens to fall on a more gray area, we have the voting poll. People are going to vote No if they think the game wasn't beaten, a lot of No votes and the submission will be rejected in favor of a slower option which was validated before. I really don't think defining an ending is a big issue.
I made sure to put "Aims for fastest completion of the game" in my two Yellow submissions so that people vote for it as a speedrun and not a glitch demonstration. IIRC, both received 100% yes votes and received reasonably high technical ratings, so I take it's widely believed that it has indeed completed the game. However, I know that there are people who consider it invalid, but unfortunately they didn't expose this at the time of submission.
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MESHUGGAH wrote:
Thanks, I edited it. By the way, feel free to edit my post. It would be better if someone with native english would revise it.
I'm not a moderator, I can't edit your post.
By the way, I couldn't understand some of the underlined phrases, but assuming you got them from here, I can translate them.
EDIT: Removed the translation because the text was updated.
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MESHUGGAH wrote:
When starting an EMBASSY*** with the "A" button when the ball is high up, press and hold the X button for turning the directional bias. After that, you'll see that the player will kick the air instead of the ball. Then quickly press the "Y" button and choose a direction to run.
"Embaixada" is the name of a soccer trick/maneuver where the player lifts the ball and "juggles" it with his legs and chest. I don't know if it has an English translation (juggle would be more appropriate, I think), but since the most common meaning of the word is embassy, it leads to this translation error.
MESHUGGAH wrote:
When you touch the REDEPOR behind the goal, the announcer will scream non-stop goal.
REDEPOR is a typo, the words should be separated as "rede por".
"When you touch the net from behind the goal, the announcer will scream non-stop goal".
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(a) Let x1 and x2 be the roots of the polynomial P(x) = x^2 - S x + P , S and P real numbers.
Find a polynomial Q(y) of degree 3, with coefficients in terms of S and P, where is a root of Q(y)
(b) Consider the polynomial F(x) = x^3 + a x^2 + b x + c with real coefficients. Find the value of t, so that the coefficient of y^2 in F(y + t) is 0.
(c) Find all roots of the equation x^3 - 3 x^2 + 2 x - 1 = 0
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janus wrote:
There were a few situations where you could have walked to a box/ talked to a person facing them and instead you took the same number of steps
If i understand correctly, I think I corrected it in Scande to fix the elevator. do you have other examples of that?
Sure, I've seen it around frame 127000 when you turn to talk to a guard. Another similar situation near 129500. When you take a chest at 249000 and when you talk to someone near a bed at 563500.
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Whew, marathon is over.
I've never played this RPG before, after watching this movie, I wish I had, that's what I look for in a long RPG TAS, I'm tired after watching it, but that's because of the length of the movie, the run itself was very enjoyable and worth these four hours.
I have a few points:
1) Most of the time you did alternate A-B mashing to pass through text, which is the best way, but there were others you went with A-A-A and B-B-B, this has 50% chance of losing a frame, did you check both parities (one with presses only in even frames, other in odd frames) each time you did this?
2) There were a few situations where you could have walked to a box/ talked to a person facing them and instead you took the same number of steps, but ended up having to turn to the box in the last step, this loses one frame, judging by what I tested (one of the times this happened is around frame 46500).
3) You took some extra steps to manipulate encounters and waited more frames than I expected for some manipulations, but I really don't know how the RNG in this game works, so it may as well be impossible to improve these. Overall, the luck manipulation was good and, because of the long time it took to make this, route planning must be good too.
None of these points are sufficient to doubt the quality of this run, the first two are actually so small that can actually provide no reduction of time, given the amount of luck manipulation here. Yes vote.
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Your criticism was hard to interpret, MrGrunz. Sometimes you call the run "that shit" and its existence being a motivation to finish your run. Other times you say it's not that bad, despite being improvable by minutes.
The scenario may be different in the speedrunning community you attend, but our general audience here most of the time has no idea of the technical aspect of a TAS. You'd be pissed off by roughly 99% of the viewers here (including me ^^).
It's very common that someone who's new to TAS gets excited by a run to the point of calling "best movie ever" without proper knowledge, to them, it's more favorable to watch a reasonably good movie every month than wait years for an "unbeatable" submission (which normally has improvement in the very day it's submitted).
I agree that calling someone jealous is not the most recommended way to express feelings about a run, but replying harshly to such a comment leads nowhere.
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MUGG wrote:
MrGrunz: I understand your situation, but I want to express I'm getting sick of all the hate towards the author of this TAS. The TAS being "shit" or an "abomination", or the author being a "shitty jap who doesn't know how to speedrun the game". This hasn't primarily been expressed by you, but by someone else, still I want to say I'm getting sick of it and don't want to read it here. Thanks for understanding.
These words are the soundest ones so far.
Come on, people, see the situation better. There's no law that says someone is prohibited to TAS a game that's being worked on. I really don't see what's the problem with him uploading his run to nicovideo. He showed no intention of submitting the movie to the site so far and he's already being accused of stealing input (without an emulator movie to prove afaik) and being called a thief who can't optimize a movie.
abeshi helped the site by obsoleting an OoT movie that could be beaten by a kid with a controller, and instead of gratefulness, his movie was hated by almost everyone who's knowledgeable of the game. That's why I think he won't submit it this time.
MrGrunz wrote:
you fucking idiot, you know my WIP on youtube saves a minute of that TAS?
how retarted are you? you are just too dumb to see the obvious improvements that are obviously there. I can post a comparison video, I'm quicker everywhere
and he even uses my inputs and some places. he hexed my stuff in without asking me!
Please, this is unnecessary. Not everyone who posts in this thread has to know everything about MM speedrunning. Don't call someone retarded because they can't see a 1-minute improvement.
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I encoded FractalFusion's old run, so I encoded this one too. http://www.mediafire.com/?na6voauo6sa7kud
EDIT:
After some effort, a lot of help from sgrunt and two disconnects while uploading, here is an SD encode: http://www.mediafire.com/?k1fiahfd4305lmz
Regarding the run, this game is notable for using both X and Zero, being amazingly fast and, with its challenging level design, it looks very impressive for those who have played the game. The technical quality is obviously present and this game is one of my favorites. Easy Yes vote.
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That's actually a physics question regarding particle kinematics. Anyway, no problem asking here :)
This is easy if you use polar coordinates instead of cartesian.
Let's assume the destination point is at (0,0), the speed of the current is directed upwards in the y-axis direction and has magnitude w. The magnitude of the boat's speed is v at all times and its initial position is (a,0).
Now, what needs to be done is:
If you want to generalize more when the boat doesn't start at the x-axis, it's just a matter of inputting the right boundary conditions in the formula, but I think this is enough for just seeing the shape of the curve.
EDIT:
rhebus wrote:
dy/dx = v + x*dv/dx = v - (f/b)*sqrt(1 + v^2)
x*dv/dx = -(f/b)*sqrt(1+v^2)
\int(dv/sqrt(1+v^2)) = -(f/b)*\int(dx/x)
arctan(v)^2/2 = -(f/b) ln x + c
No, sir. /int(dv/sqrt(1+v^2)) = arcsinh(v) + c = ln (v + sqrt(1 + v^2)) + c
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We're making progress. At the moment, I almost have the full game route written. It's important if we don't want to redo stuff like what we had to redo before. (Well, we've had to redo things that were entirely planned out, however...)
Real life is getting on the way, but vacation is pretty close. The first half was the hardest one. The run should be ready soon enough, unless the RNG decides not to cooperate at all.
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I'll post the solution to the triangle problem here. If anyone wants a picture to understand it, feel free to ask.
Find the point F in AB so that the angle BCF = 20º. The triangle BCF is isosceles, since both base angles measure 80º, so BC=CF. Additionally, the triangle BCD is also isosceles, because CBD=BDC=50º, we conclude that CD=CF. Notice that triangle CDF is isosceles and has an angle of 60º (DCF). From this, we can see that CDF is not only isosceles, but equilateral. So, CD=DF=FC.
Now, consider triangle CEF. Finding its angles (40º, 40º and 100º), we can conclude it's isosceles and FC=FE. Finally, since DF=FE, we can see that triangle DEF is isosceles, so both base angles are equal. More specifically, FDE=FED= (180º - DFE)/2, meaning that FED=a+CEB= (180º - 40º)/2 , a + 40º = 70º, a=30º.
This is the most famous problem in V. Lidski's book "Problems in Elementary Mathematics" (if anyone would like to check it out, it has problems at an advanced level, if you have the skills to solve them, it'll greatly increase your mathematical knowledge, I recommend this book), possibly because at first, it seems easy to solve, but will soon drive people nuts while finding the solution. Today, almost everybody who studies math at olympic level knows this problem, it's used mostly to demonstrate the beauty and simplicity of synthetic solutions to geometry problems.
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"Here at TASVideos we make perfect movies. And then we beat humiliate them."
Bravo! That was the most insane stuff I've ever seen in a TAS. Luck manipulating for a favorable race condition is just crazy, improbable and brilliant, it takes TASing to a whole new level in glitching.
Although the glitching is not as funny as in other heavy glitched runs, it still has the awesome garbage blocks everybody loves, and the destruction of the game is incredibly impressive. After getting my head blown so many times during the course of this run, the only thing I can say is "what is this I don't even".
Biggest Yes vote I've ever given anything. Congratulations on a groundbreaking submission.
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The brilliant solution uses no lengths of sides or trigonometric functions, but it's very hard to see. (HINT: Draw auxiliary lines)
A colleague of mine, amazing at geometry, claims to have solved this problem in seven different ways (!). I only know two, the brilliant solution and a trigonometric one, which is easier to see, but requires a little manipulation of trigonometric functions.
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The answer is not 26.9175, but it's close. It's probably due to rounding errors on the calculation of the sines and cosines. I think you can solve this analytically, but it's like killing a fly with a cannon. Even if you do find a geometric method to determine the exact values of the trigonometric functions, there's always the chance you can get with a lot of radicals that you need some amazing factorization to solve.
It's not possible to determine a only adding the angles, that's why the matrix for your system is singular, you need to add more conditions.
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rhebus wrote:
p4wn3r wrote:
The problem is to find the radius of the circunference that contains the vertices of the triangle ITT' (I don't know the term in English, circunscribed?).
Circumscribed. In full, the problem is to find the radius of the circle which is circumscribed around the triangle ITT'.
Nice work.
Ah, thanks! Now I know that I was misspelling "circumference" all this time xD
This question inspired me to post another interesting geometry question here. Let's see if anyone here can crack the infamous Lidski triangle.
Consider the triangle ABC, where AB = AC and angle BAC = 20º. We now mark the point D in AC so that CBD = 50º, and point E in AB so that BCE = 60º. Evaluate the value of the angle CED.
Now with an awesome picture from MS Paint!
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arflech wrote:
I saw this problem on a math forum, and for some reason it got deleted...
Draw two circles that intersect in two distinct points, and draw one of the common tangents and then the circle through both points of tangency and either point of intersection of the circles; prove that the radius of this third circle depends only on the radii of the first two, and not on their position.
I tried using coordinate geometry but ended up with some nested radicals or a quartic or something else scary.
I've seen this problem before with tangent circles, I didn't know the radius remained the same when they intersected in two points. Anyway, here goes.
Let's call O the center of the first circle, O' the center of the second, I one of the intersection points, T and T' the points where the tangent line encounters each circle.
So, we have the pentagon OIO'TT'. Notice that, OI=OT, O'I=O'T' and OI // O'T' (they are both perpendicular to TT'). The problem is to find the radius of the circunference that contains the vertices of the triangle ITT' (I don't know the term in English, circunscribed?).
Using the formula for the area of the triangle: S = abc / 4R, so for this case: R = IT * TT' * IT' / 4S
Using the isosceles triangles and calling the angle ITT' = x and IT'T = y, we can conclude (I need to draw to show this better, I'm too lazy atm, it's not very hard to do):
IT = 2 OT * Sin(x), IT' = 2 O'T' * Sin(y), S = TT' * OT (Sin(x))^2 , Sin(x) = Sin(y) * (OT/O'T')^1/2
Putting all this in the formula, we get R = (OT * O'T' )^1/2. This means that the size of the radius we're looking for is the geometric mean of the radii of the two circles and doesn't depend on their position.
is a root of Q(y)
(b) Consider the polynomial F(x) = x^3 + a x^2 + b x + c with real coefficients. Find the value of t, so that the coefficient of y^2 in F(y + t) is 0.
(c) Find all roots of the equation x^3 - 3 x^2 + 2 x - 1 = 0
If you want to generalize more when the boat doesn't start at the x-axis, it's just a matter of inputting the right boundary conditions in the formula, but I think this is enough for just seeing the shape of the curve.
EDIT:
"Here at TASVideos we make perfect movies. And then we 