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Tub wrote:
12: not a clue, though my observations above about level 7 may apply.
Nope, still not necessary. Though now I'm curious as to your solution for level 7.
EDIT: For an indication of it's difficulty, I have 0-7 and 12. I'm still trying to do 8 without using external files.
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Warp wrote:
I can at least give an upper bound. If you use weights of powers of 2, then with 6 weights you can represent any value between 1 and 63 as a "binary number" using the weights.
I don't know, however, if when using 1-40 in particular, some other scheme would allow 5 weights to be used.
You can easily lower the upper bound to 5 by ommiting the weight of value 1. You could only compare to even values, but if unknown weigth X is heavier than known weights Y and Y+2, X is obviously Y + 1.
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The problem is that your mm1 movie does not sync, but the mm4 one does. This means they do not contain the same input. As such, wether your fault or not, means the idea behind the 9-game run is lost.
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According to the thread on F-Zero Central, the game checks the code by decrypting it (generating a player name) and checking that against the actual name. If you could find the corresponding part of the game's code, it should be too hard to make a password -> name script. This would allow you to check the relatively short list of very fast passwords (the best would probably be some 0's, followed by some 1's, followed by some 3's, ending with the mandatory 30), seeing which one gives the best player name. Still, this one is really fast already.
On the other hand, if you were to do master and use dirty SRAM, you can enter the password in the verification movie, making the speed irrelevant.
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1) F-Zero Central
2) I think it's Jet Vermillion. Although you ussually unlock it by (yeah, really) beating every cup with every vehicle on master (again, yeah, really), Nintendo left a password in. Although dependent on save game name, someone decrypted it, so it could be used in a TAS.
3) Ah, right. Again, F-Zero Central should be able to help you with useful articles.
Sorry if I sounded harsh, I didn't mean to. I honestly just want this to be good and I respect the simple fact you're willing to take this up.
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Personman wrote:
Scepheo:
Adelikat's infinite mario TAS is definitely playing around in the same space as this, and may have subconsciously influenced my thinking about it. Not quite the same, though, since it has a complete playthrough in each loop, designed to account for the small variations, and never terminates. But it is very cool!
I was talking more about the emulator hack. It allows you to record some input, and then loop the rest of the input. If you instantly start looping, this emulator hack could actually be used to record and play back such movies as you suggest.
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Didn't adelikat hack FCEUX to do sort-of-this with Super Mario Bros.? If that version still exists, it should be possible to use for this. At the very least, it's an interesting concept.
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Tub wrote:
It's interesting to note that the function name allows []' as characters, but neither . nor (). So if you want to call document.getElementById you could write it as document['getElementById']
Good observation that it allows brackets, but it's not necessary, actually.
I don't think the above is a really big spoiler, but I made it one either way.
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Bobo the King wrote:
There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless.
2n3 + 3n2 + n =
(2n2 + n)(n + 1) =
n(2n + 1)(n + 1)
I think it's obvious that either n or n + 1 will be divisable by 2. Then remains to see that at least one term is divisable by 3. There are three cases:
1) n = 3k
2) n = 3k + 1
3) n = 3k + 2
In the first case the first term (n) will be divisable by three, in the last case the third term (n + 1) will be divisable by three. In the second case, we can show that
2n + 1 = 6k + 2 + 1 = 3(2k + 1)
which is again, divisable by 3. As at least one term is divisable by 2, and another by 3, the result will be divisable by 6. The proof is rather long, but I find the result to be at least somewhat intuitive, looking at the numerator.
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Sir VG wrote:
Tell Google to implement a "STOP PROTECTING ME" function? Or a "I trust this domain/site" function?
Cause as much as I love Chrome, sometimes it really is farking annoying.
Google should really just learn that if trusted site A has a teensy bit of content from non-trusted site B, it suffices to just block that contect. I mean, I can live with one avatar not being shown.
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Warp wrote:
Could someone give an intuitive explanation of how there can be two points on opposite sides of the Earth with the same two parameters (such as temperature and height)?
I think I have the general idea:
You already said that it worked for every circle you can take, so let's take all the circles on which the north and south pole lie. Each of these circles will have at least one point (for the sake of intuitiveness, let's say its around the equator) at which the parameters are equal on both sides.
Due to the fact that the function is continuous, you can draw a line through all of these points and get a new "circle"; it doesn't have to be round per se, but we are working with points on opposite sides of a sphere, so this line will divide the sphere into two equal halves.
Now it goes that for any other parameter, this circle-of-equal-points can also be made. As this one divides the sphere into two equal halves as well, the two circles must either lie on eachother or cross eachother. Either case, there's at least one point were both parameters are equal to those at the other side.
I haven't looked at the link FractalFusion posted, but I have a feeling this works for N parameters on any N+1 (or greater) dimensional sphere.
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Warp wrote:
It's this thing called "humor". You might have heard of it.
I seem to have the same problem now that you sometimes have: seeming far more on the offense than you actually are. I'm not saying you're not allowed to indulge in applying (semi-)serious math to "silly" problems, I'm just curious as to why. FractalFusion's answer was precisely what I was asking for (and thank you for that).
Warp wrote:
A slightly more semi-serious question: Can you generalize the proof for all rational numbers and all real numbers?
(Since rational numbers are countable, my intuition would be that yes. However, real numbers are a much harder beast.)
I think we can fairly trivially generalize the proof to any set containing the natural numbers, by looking for the "smallest natural number that is not interesting".
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Hm... This is going to be complicated. I think this new route means the RTA route is going to differ from the TAS route.
This new route saves time over the previous (double intermission skip) route, but I don't think it saves more time than a TAS can using that route.
Using tool-assistance, you can avoid the "The Bernoulli Principle" checkpoint, making the one in "Square Root" the last unavoidable Laboratory checkpoint. This means that exiting the Laboratory is going to cost a lot less time in a TAS.
The new way to get to Violet (teleporting, rather than doing Space Station 2 again) doesn't seem to be exclusive to this route, so that time will also be saved over the previous (I think 12:33?) run.
EDIT: Silly me, of course you can't do this when skipping both intermissions, as it requires you to use a teleporter.
I think this means a TAS will be faster if it skips both intermissions, but then I again, I might be completely wrong.
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Warp wrote:
However, that fact is in itself interesting
No it's not.
I've honestly never understood the point or merit of these sort of hypotheses. There's also the one about whether the set containing all sets contains itself (IMO: yes, otherwise it doesn't contain all sets. If this is not allowed, then no such set exists).
It always feel like you are making an assumption (there exist boring numbers) and then saying that assumption must be wrong (this makes them interesting), as such your assumption must be wrong. While I get that this is pretty much what proof by contradiction means, you normally at least deduce something from your assumption.
Sorry for ranting, continue your mathses.
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Derakon wrote:
Weatherton wrote:
Perhaps a preferable approach would be to create a save file the replicates the state of having legitimately unlocked these parts.
That's functionally identical to using an AR code to unlock the parts and then resetting the console. There's still no way to make a verification movie that unlocks those parts.
Functionally, yes. Using a save file (with the parts unlocked through AR) instead of AR during the run does however mean that the viewer can be sure AR isn't used for anything during the run, such as increasing top speed or such.
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I still think we need the input of someone who actually gets to decide these sort of things. I'm fully against using AR normally, but I believe an exception should be made for this game. Especially because it's such common practice for real time players.
One of the special parts is needed to make the best flying machine (Buffalo cockpit, IIRC), which is used for almost all the records in the open ladder. This means there's a chance we won't even beat all existing records.
So it's pretty much just up to the mods which rule we violate, really...
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You can get the tracks, machines and cup from AX but only some of the parts during normal play.
Also, if I'm not mistaken, Black Bull is unlocked through beating all story missions on Normal. You get Deathborn for beating them all on Hard and every chapter gives you an AX machine on very hard. Notably Fat Shark on chapter 4. Which we want.
So that leads to the question, can a verification movie (unlocking Fat Shark -> beating story missions on Very Hard) have its own verification movie (unlocking the Very Hard chapters).
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Assuming you can test wether any given position is "occupyable" by the player, you'd do something like this (this comes from an old game maker project of mine, so it works):
sign = sign(xSpeed);
// Test how far you can move
while (place_meeting(x + xSpeed, y, objBlock) && sign * xSpeed> 0)
{
xSpeed -= sign;
}
// Make sure you don't move backwards
if (sign * xSpeed< 0)
xSpeed = 0;
// Apply movement
player.x += xSpeed;
Also, I disagree with moving in both x and y at the same time. Doing that is not only harder, the seperation of the two is often preferred, to ease movements like sliding down a wall. Of course, the moment your collision surfaces are no longer just along the major axes, you need to generalize anyway.[/code]
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As I don't have a U version, I can't watch the movie right now, unfortunately. But why are you starting with time trials? Isn't it a better idea to unlock some faster vehicles first. Black Bull can be unlocked doing story, and it's a fairly good snaking vehicle. I get that speed is irrelevant for the verification movie, but a shorter movie file will make verifying easier and less desync-prone.
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henke37 wrote:
I am not a fan of "physics engines". They don't provide stable systems.
Yes, they do.
henke37 wrote:
And I am worried about performance when dealing with levels large enough to take more than a few seconds to cross.
No need.
Seriously, try using Box2D or any other physics engine. They've become as well-known as they are for a reason, you know. Just make sure you stick to the parts you need and actually use it well and you'll be fine.
IMHO, if you want stability, don't use flash.