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Spikestuff is correct. If the thumbnail is black and white and says
視聴
できません
or if the bottom right has text that says "Foo" and it is a 40-second clip of text and music, then the video has unfortunately been deleted. There is no way to retrieve it.
The "black screen" problem that sometimes occurs is a different problem associated with loading errors. It may solve itself randomly, or if you clear your cache.
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The Wiki page linked to web archive showing what the page with the interview(?) looked like:
http://web.archive.org/web/20040403201057/http://soramimi.egoism.jp/emu.htm
The way the page is colored and structured, it appears as if there was more than one person involved in the production of the movie (Morimoto the TASer + someone who encoded it). Assuming so, Morimoto's comments are in red and the other person's comments are in white.
PoochyEXE wrote:
オレはデスモードクリアできるんですわ。
っていうか結構初期のほうのマスターＧｍとかなんですね。
I can clear death mode.
Saying like this, it's the master Gm of the way of the splendid beginning and such. (?)
I'm not entirely sure if I've gotten this right, but my interpretation is: "It's not like I said I can clear Death Mode, or that I was able to get Master Gm very quickly or something."
I've been trying to translate everything again but I went over
of which what you quoted was the last two lines. I find that those are very difficult for me to understand but are absolutely crucial to the context of the passage, to the point where I cannot tell if Morimoto (assuming he wrote this) is praising someone, or criticizing someone, or if he is actually part of the "Tetris group" and how much qualification he has in that regard, or if he is being sarcastic, or... (you get it). I am not a Japanese person.
Most of everything else on that page is straightforward.
The only wild encounters for Level 10+ Zigzagoon are Route 118 and 119, both of which are too far ahead in the game. (Ruby and Sapphire have such encounters earlier in the game, on route 110).
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Fortranm wrote:
Totally impressive.
Using Swellow is not the fastest route for Emerald? Is it because of the double battles so the main battle Pokemon would better be able to learn Surf?
I don't know if GoddessMaria tested it or not.
There are a few things that make the use of Swellow difficult, the biggest of which is probably Wattson. In Emerald, all his Pokemon have Shock Wave. There's also no getting King's Rock from Pickup like in Ruby/Sapphire, or any Rare Candies for that matter.
Note that Swellow learns Fly, so it won't cost an extra Pokemon to catch.
In any case, it is worth exploring which other Pokemon are good candidates.
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On GoddessMaria's twitch.tv stream, I saw the making of the last part of the TAS (where all the glitching was). Interesting glitch going on there. I find it interesting how one can use the game's "anti-cheat" corruption tool to corrupt RAM and, well, cheat one's way to victory. I looked at how the glitch works a little bit, but I am not familiar with it. Might take some more time to figure out. It seems though that it is faster to scroll the glitched Pokemon menu selector (the one where "cancel" is highlighted) by alternating up instead of holding up.
I was also impressed by the manipulation for Mudkip's IV and the Ralts tutorial glitch, without needing my help.
That being said, it's probably just me being too experienced with TASing this generation of games, but I wasn't impressed with much before the glitch, or much afterward. Afterward is unavoidable, but once you see one battle, it's pretty obvious what is going to happen in the rest of them.
There are a few things that I noticed could be improved:
- The path picking up the Ether and leaving in Petalburg Woods is not the shortest possible.
- In the first battle in Norman's gym, Mud Shot is used, and then Water Gun. Water Gun should be used first, because Mud Shot makes an effect if it does not KO the Pokemon.
- In the last battle against Brendan, Pelipper should be manipulated to use something other than Protect.
- I'm not sure clearing the Kecleon outside Fortree Gym before doing the corruption is faster. Biking is 4 times faster than walking, so it should be possible to lose HP quickly by biking, faster than the time it takes to walk back and forth on the Fortree steps.
- I noticed wall bumping going on as manipulation to avoid encounters. This is OK when running/walking, but it loses a lot of time when biking. Going back to the last frame when starting the biking motion, and starting one frame later is probably best. Even taking a couple extra steps to bike (going back and forth) is better although it costs 8 frames.
In any case, I think the run could also be improved by using a different main Pokemon for battle and perhaps some X Attacks, but Emerald is somewhat tricky, so some study would have to be done regarding this.
Mr. Pwnage wrote:
For some inexplicable reason, the game only checks against those two non-essential badges instead of all 8
I think gym 2's badge is essential to get to that point. I think you need it to activate Norman's gym (gym 5). I know though that gym 6's badge is non-essential, and that the badge check will reject you if you try to enter without it.
grassini wrote:
you mentioned u can skip a lot of badges in the endgame check at ever grande and the japanese versions can walk through water with a glitch and possibly reach lilycove faster(thru slateport)?
I don't know about this glitch. Do you mean the walk-on water glitch using Dive? Because that exists in English versions as well, and you still can't walk through boundaries.
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Lary, I watched your run.
The proper stopping point of this TAS is frame 18959 (when you touch the last axe). This puts the run at about 5:15.
Regarding backwards Mario: The goal choice "Mario Backwards almost the entire time" isn't clear, since it doesn't precisely say when Mario can face forwards and when he cannot. Watching this TAS, I can't say what it is either, since there are more than a few places where Mario walks facing forward. In addition, plenty of alternative goals for this game in the past have been rejected in the past; I think "Mario facing backwards" (including backwalking Mario) is not significant enough to warrant a new category and would be rejected.
Regarding post editing: There is an edit button on all your posts, in the top right corner of each post. It is customary to edit your post instead of posting again (double posting), if no one else has posted in the meantime and not much time has passed (such as, not more than 6 hours or so).
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Kurabupengin wrote:
How do you exactly win points in a video?
The Nico ranking says (at about 1:10):
Total points = Number of views + Number of comments + (Number of Mylist × 20).
For monthly ranking, all videos labelled or tagged TAS or such, uploaded in that month (Japan time), qualify for the ranking. The calculation for points occurs at the end of the fifth day of the next month (Japan time).
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arandomgameTASer wrote:
Wasn't expecting Shantae to rank so high o.o
Apparently it was a general month of disinterest. I didn't watch the previous five rankings, so I don't know if it is just this month, or if viewership has steadily declined.
Also, 11th place has 27K points. Then the number of points goes up rapidly to 110K for 5th place.
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I feel that there is a consistent lack of respect going on within the TASVideos community at so many levels, not that this is a new or surprising thing. In particular, I think that judges should be supporting each other, not tearing each other down.
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dunnius wrote:
But I would like to know what people think of how I handled the input in this run (and not what they think of the game or its music).
I don't think a lot of us are experienced with this game, but, from your submission comments, it sounds like you did a lot of work to optimize this run. About how many rerecords were used (by bot, and manually)?
The stopping and going on grass tiles threw me off initially, but I got used to it later. The alternative is waiting for a long time, I guess.
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Well, this is a lot faster than the previous TAS, and definitely more entertaining. It cuts down on both grinding and final boss wait.
The best part of the previous run was fishing monsters off the bottom of the screen in the grave. Unfortunately, this run doesn't have it, but it uses the four-corner area of the lake to fish eels instead, which is good in its own right.
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Scepheo wrote:
If I, however, take it as the length of the protruding tetrahedra, it gives me (2√2)/3, not √2 ...
That's just the volume of the 8 protruding tetrahedra. The volume of the octahedron (whose edge length is also 1) is √2/3; add it to the volume of the 8 tetrahedra to get √2.
Warp wrote:
What's the volume if the unit length is the length of the side of the cube that contains the stellated octahedron (iow. the side length of the square I posted, ie. the distance between the tips of two adjacent "spikes", ie. the height of the polyhedron when it's on the floor like on the rendering I posted)?
If the length between the tips of two adjacent spikes is 1, then the length of an edge of a protruding tetrahedron is √2/2 (think of two adjacent spikes at (1/2,1/2,1/2) and (1/2,1/2,-1/2), and a corner where the respective tetrahedra meet which is at (1/2,0,0)). Since the volume of a stellated octahedron where the unit length is the edge of a protruding tetrahedron is √2, then the volume of this stellated octahedron is √2 (√2/2)3=4/8=1/2. So, the stellated octahedron takes up half of the volume of the cube it is contained in.
Now that Warp suggested it, there is actually another way to compute the volume. The stellated octahedron is a cube with (non-regular) tetrahedra cut out from each of its edges. If the cube has unit side length, then each of the tetrahedra cut out clearly has base area 1/4 and height 1/2, and so the volume cut out by each one is (1/3)(1/4)(1/2)=1/24. Since 12 of them are cut out, a total volume of 12(1/24)=1/2 is cut out of the cube, leaving a remaining volume of 1-1/2=1/2.
Wolfram Alpha is wrong. The volume is not 1/(3√2), regardless if unit edge length means the side length of the large tetrahedron that is intersected with its dual (volume is √2/8), or unit edge length means the side length of the tetrahedra extending out from the octahedron (volume is √2).
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I can't say "what it is in 3D" (more on that later), but I can say what the 2D envelope (the curve that is the upper bound) is.
The envelope is one section of a parabola. Note that it is related to the first example on the Wikipedia page for envelope, and it even explains how to find it. It is also a quadratic Bézier curve with control points (1,0), (0,0), (0,1).
As for "what it looks like in 3D", it looks similar to the standard hyperbolic paraboloid diagram (also the standard saddle point diagram). Conveniently, the envelope itself is a section of a parabola, just like the cross section of the hyperbolic paraboloid z=y2-x2 in the plane x=0.
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The Lucas numbers are defined as: L0=2, L1=1, Ln=Ln-1+Ln-2 for n≥2.
We claim that Ln=φn+(1-φ)n. Proof is as follows:
The generating function L(x) satisfies
(1-x-x2)L(x) = L0+(L1-L0)x = 2-x,
and so L(x)= (2-x) / (1-x-x2).
Note that φ is a root of x2-x-1, so φ2-φ-1=0. We can write this as φ(φ-1)=1.
Then 1-x-x2 is factored as (1-φx)(1+(φ-1)x) and so we write
L(x)= (2-x) / (1-x-x2) = 1/(1-φx) + 1/(1+(φ-1)x).
Therefore, Ln=[xn]L(x)=φn+(1-φ)n.
(The above can also be proved using induction instead of generating functions.)
Since -0.7<1-φ<0, then |(1-φ)n|<0.5 for n≥2, and so Ln=round(φn) for n≥2.
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My method of calculation for the 4A/RF situation is as follows. Same assumptions as in Flip's opening paragraph.
We wish to find all possible permutations of 4A/RF cards/dummy in the form
xx yyyyy zz
where xx is Player A's hand, yyyyy is the board (the middle), and zz is Player B's hand, and Player A cards + board form a royal flush, and Player B cards + board form a 4-aces hand. The order of cards is important for this calculation.
Assume for the moment that spades is the RF suit. Then the ace of spades must be on the board (since it is shared by both 4A and RF. Remove this card for the moment, so we count
xx yyyy zz
where the 8 cards are the RF cards except the ace (KS, QS, JS, TS), the other aces (AC, AD, AH), and the dummy card (DU). KS, QS, JS, TS must occur on x or y positions, and AC, AD, AH on y or z positions.
Suppose we are given the order from left to right of {KS, QS, JS, TS} as well as of {AC, AD, AH}. There are two cases. If DU is in A's hand, then one RF card is in A's hand, and the other three are on the board, and the aces fill out the rest. Then there are 2C1*4C3=8 arrangements. On the other hand, if DU is not in A's hand, then two RF cards are in A's hand, the other two are on the board, and the aces and DU fill out the rest. Then there are 4C2*4C3=24 arrangements.
So, given the order as above, there are 32 arrangements. Multiply this by:
- all possible permutations of {KS, QS, JS, TS} (4!)
- the position of the ace of spades when put back (5)
- permutation of all suits (this permutes all aces as well as the RF suit) (4!)
- the assignment of the dummy card (44)
This results in a total of 4055040.
Divided by all permutations of 9 cards, P(52,9)=1335062881152000, this gives 3.037x10-9, or 1 in 329 million. Multiply by 2 if it doesn't matter who has the RF and who has the 4A.
Note that Texas hold'em poker is generally played at tables with up to 10 players. For a full table, the probability that this situation occurs in each hand (each deal) goes up by a factor of 90 (choose one player as A, another as B, and the rest are irrelevant).
Flip wrote:
1 Dummy card to be in first hand
5 potential RF cards to be paired with it =5C1=5
...
2 RF cards in hand =5C2 ways
The ace of the RF cannot be placed in a hand; it must be in the middle. That leaves 4C1=4 for the first case and 4C2=6 for the second. Doing so eventually results in the same value as I calculated above.
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Warp wrote:
If you suspect that the program might be using a certain RNG, such as a Linear Congruential Generator, then it's not impossible, but it could ostensibly require a lot of time and, moreover, the answer may be ambiguous
Actually, if you assume that LCG is used, and it is well-chosen (it runs over all possible numbers mod whatever), then the LCG's formula can be discovered just by looking at 3 consecutive numbers and using some number theory. It would not be ambiguous, and can easily be checked for correctness using the next few values. LCGs in video games/computing very often use a modulus which is a power of 2.
Note: You need to look at the entire number used in the LCG which is stored in memory, not just the upper few bits that the game spits out.
Other than LCG and arguably LFSR, other types of pseudorandom number generators are generally too difficult to reverse engineer from their output values (most of them being homemade bit-manipulation functions). In this case, there is no better way to figure out what the formula is than disassembly, however difficult doing that is. Or, if the sequence is always the same, skip the formula altogether and just list all the numbers in sequence, or as many as you need for your purposes.
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MrWint wrote:
At 3E9D is the equivalent of what is called the "Predef" function in the disassembly, which is basically a jump table.
I see.
The versions actually have different locations for Predef function in their ROM. J Red is at 3E9D, J Green is at 3E8B, and J Blue is at 3EB1.
The Japanese Blue version is faster than Japanese Red (even though intro takes a bit longer) because of faster save process.
MrWint wrote:
I wonder if it's possible to use the rival's name (2nd and 3rd character) for the script pointer instead of the Trainer ID. The saved swap may or may not be worth the time spent naming him.
Naming the rival takes like 150 frames, doesn't it? One swap is far shorter than that, and you'd still need to get to the 26th Pokemon.
ALAKTORN wrote:
^I’m not sure why Fractal never spoke about that… it looks like it should be the published movie.
It's not exactly amazing or anything. It doesn't visibly do anything that this submission (or the previous one) doesn't do, other than giving the player a name. It is in a language that most people here don't understand.
I was also not interested in generating any further controversy.
Emulator: VBA v22
Basically, it works like this:
- Manipulate the trainer ID first byte (D2D8) to be D1 or F1. Then D2D7 will read 01 D1 or 01 F1, little endian for the jump into RAM where we want to execute.
- Name the player よッヘマぼ . This corresponds to
D6 AC SUB AC
CD 9D 3D CALL 3E9D
at D11D.
- Do the save and reset thing to corrupt memory.
- After reloading the game, open the Pokemon menu.
- Switch the 7th and 8th Pokemon. This gives a HoF end message that does not require input.
- Switch the 9th and 10th Pokemon. This moves 01 D1/F1 at D2D7 to D2AB.
- Switch the 15th and 26th Pokemon. This moves D2AB to D2ED, the address to an execution pointer, when closing the menu.
- Close the menu. The game reads D101 (F101) off the above address and jumps there. There are 00 bytes (NOP) until D11D, which it executes, resulting in the HoF. The instructions place 0x55 into the A register (01 SUB AC -> 0x55), then jumps to 3E9D, which seems to be a warp manager that uses the A register.
I don't know if such action is considered an acceptable ending for TASVideos (a more extreme version that glitches the end credits directly was rejected in the past), especially since the end is a bit glitchy.
Also, I think my movie is improvable.
