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I heard about this game early on (as being possibly the best Uzebox game) when Uzebox was first released in BizHawk.
Obviously it obviously looked like a nightmare to optimize. It's great that you got so many people interested in collaborating to make a TAS for this game. Also TASes on co-op modes are way more than twice as good as a TAS on a single-player mode.
I like this page.
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EZGames69 wrote:
When watching the single screen mode (as is with the temp encode I made), it's watchable for the most part. However when I watched this in emulator with 3D mode turned on, I quickly realized it ISNT watchable because you can't make out what you see in the hallways. I know that youtube has a feature that lets you turn off 3D mode, but unfortunately that isnt a feature in youtube mobile last time I checked. So I have to give this a meh in entertainment because of this.
Last time I checked, Youtube's 3D mode is trash. I don't have 3D glasses, so the only way I can try to see it is with cross-eye view, but I couldn't figure out how to enable it in Youtube or if it was even possible.
At least Bizhawk has cross-eye view there (you can set it in VirtualBoy -> Preferences... -> ThreeDeeMode -> SideBySide) but unless you're trained in cross-eye view it's kind of difficult to do sometimes, and even then who knows about the actual art quality of the game itself.
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I think there should be a program these days that is capable of stitching them together (and keeping them at 60fps). Maybe Avidemux? I normally just use AviSynth because I tend to do a lot of post processing.
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Watched it just now. The TAS looks pretty good (I'm not familiar with ZX unfortunately) but the cutscenes are long. Models exist in this game essentially so you can cosplay as Mega Man characters? That's my guess. Looks cool anyway.
Also, nice "Movie finished playing." at 1:08:08. (I don't think DeSmuME AVI recording is supposed to put that there, but what do I know.)
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Encode (720p60 is available, always use 60fps for Mega Man):
Link to video
Looks interesting, seeing how the 2p controller exploits were used in the levels.
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It says in the description that it is a PC game.
Didn't expect otherwise; this is too good to be a NES/FC ROM.
Edit: Also ファミコン風 means Famicom-style which is a big indicator that the game is a PC game; if it were a NES/FC ROM, "Famicom-style" would not be the selling point.
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I haven't been following Pokemon save corruption recently (totally glitched runs just aren't my thing anymore), but I have a question. Did using glitch items (at least one which I believe allows you to execute RAM) ever factor into your TAS? I managed to find one on the JP version, but I can't remember if it was the same for the US version.
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Thanks! At least it's nice to get something submitted after a long time of not doing anything, even if it gets rejected.
Spikestuff wrote:
You'd think someone who has submitted on April Fools in the past would submit this also on April Fools (2021)... especially when the user has been part of the site for so long...
April Fools man. April Fools is the time you submit things like these.
Whatever, I was 6 months off. April Fools is overrated anyway. :)
Twisted Eye wrote:
I kinda wish some more time was wasted before starting the loop. Time it so that the counter wraparound gets exactly to 1 when the time runs out, so that the one death in the "Maximum Lives" run actually leads to a game over
Yeah, I actually hoped it would have ended in the negatives for the elusive one-death game over. Oh well, can't have everything I suppose.
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OmnipotentEntity wrote:
If e^z = ln(z), because e^z and ln(z) are inverse functions, then it is the case that also e^z = z. Then solving using the Lambert W function is rather trivial:
ze-z = 1
-ze-z = -1
-z = W(-1)
z = -W(-1)
While -W(-1) is one solution (I'm not going to talk more about this; go see blackpenredpen if you're interested), this reasoning doesn't explain why there aren't any other solutions, since any z such that e^(e^z) = z (not necessarily e^z=z) could be a candidate solution for e^z = ln(z).
OmnipotentEntity wrote:
The answer to the problem posed in the question is known, and is a large irreducible fraction
Technically, the problem isn't asking for a number. It's asking "Is the teacher correct?". Which is far easier to answer and justify.
This site answers the question, and also gives probabilities (up to n=17), for which the student coming forward would have heads; I see you also posted here going up to n=20.
As far as calculating the exact probability, I don't have a nice way to figure that out. But to answer the original question, "Is the teacher correct?":
No, the student is more likely to have flipped tails. Given that at least one student has neighbors that both flip heads, and a random candidate stepped forward, that student is more likely to have stepped forward the fewer candidates there are, which is more likely when that student has tails. It's related to similar Bayesian paradoxes like the Monty Hall problem and the Boy or Girl paradox.
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Forgot to reply despite watching it a week ago. The game still makes an entertaining TAS even up to now (that's probably why it has a star).
I can't remember what the previous run was like though; it would be nice if there was
a comparison encode.
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Warp wrote:
That got me thinking: Are there ellipses (other than the circle) where the major radius, the minor radius and the perimeter are all expressible with closed-form expressions?
(In this context "closed-form expression" excludes integrals.)
I assume by "closed-form expression" you are also excluding infinite sums.
Other than the circle and the degenerate ellipse (minor radius = 0), I don't think it is possible. However there are ways to approximate perimeter of an ellipse using infinite sums.
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I did a very quick (not very optimized) TAS of Crystal Clear 2.0 BABY, using the speedrun route (but not doing the 25 Rare Candies):
Link to video
Possible that Ursaring or some other Pokemon is faster.
Actually I did a TAS of 2.0 Beta a year ago. BABY is faster because you can beat 8 Gym Leaders + E4 to fight the Mt. Silver trainer, whereas in Beta you need 16 Gym Leaders + E4.
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Arcorann wrote:
Wikipedia's page on parabolas gives the expression for the vertex and focus of the parabola obtained by transforming the unit parabola using a given affine transformation. The vertex and focus of the unit parabola are (0,0) and (0,1/4) respectively, so all that is required now is to solve for the similarity transformation for those two points, which is relatively straightforward.
Nice! I was trying to figure out how to get either of the two (there are two of them; one is a reflection of the other) similarity transformations from the affine transformation but couldn't quite figure it out.
By the way, the reasoning I gave for a parabola being a parabola under affine transformation is wrong; I assumed that it would hold for compositions of the transformations if it held for each transformation, which is wrong reasoning. After all, a rectangle remains a rectangle under the transformations I gave, but a rectangle is no longer a rectangle under a skew transformation.
Looks like you need some room to air-dash onto the top of the door (an overhang or such), then slide down the door and switch to Zero at the last moment. There isn't enough room to do this on the final door leading to a Maverick boss battle. If you can find a glitch that allows this on a final door, then that would definitely save time.
Otherwise, it doesn't seem to save time, not even against Byte.
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OmnipotentEntity wrote:
For instance y=x^2 and y = 1/2 x^2 are not technically similar, but they are related by an uneven scaling.
y=x^2 and y=x^2/2 are similar; scaling the graph of y=x^2 uniformly by a factor of 2 gives the graph of y=x^2/2.
(It is also true that they are related by an uneven scaling, but that is beside the point.)
OmnipotentEntity wrote:
When I had written my last post I thought that It may be the case that "parabolaness" is preserved under all (non-degenerate) affine transformations, but when I attempted to prove it ...
Yes, "parabolaness" is preserved under all (bijective) affine transformations; a parabola remains a parabola when you apply uneven x/y scaling or any isometry (rotation, reflection, translation), and since those operations generate the affine transformations, a parabola remains a parabola under any affine transformation.
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OmnipotentEntity wrote:
A similarity transform is a little bit too restrictive. It doesn't allow for stretching in either the x or y direction.
I assumed that Warp was referring to uniform scaling, based on what was said:
Warp wrote:
All parabolas are similar. This means ...
Warp wrote:
... scaling it equally in all directions ...
Warp wrote:
(They are not affine transformations because those allow uneven scaling.)
"scaling it equally in all directions" I assumed meant uniform scaling.
"uneven scaling" I assumed meant stretching x and y by unequal factors.
Edit: Also, I'm pretty sure that if you have uneven scaling together with all the isometries (rotation, reflection, translation), that is enough to give you all the affine transformations. So that probably explains why there isn't a special name for "similarity transformations plus uneven scaling", because that just gives you the affine transformations.
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Warp wrote:
All parabolas are similar. This means that you can take any parabola and by simply rotating it, flipping it, translating it and scaling it equally in all directions, you can map it exactly on any other parabola.
I was wondering if there's a name for those allowed transformations. (They are not affine transformations because those allow uneven scaling.)
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p4wn3r wrote:
Evaluate:
We know that the sum of the first n terms of the harmonic series is given by:
Hn = ln(n)+γ+O(1/n)
where γ is the Euler-Mascheroni constant, and since the O(1/n) terms go to 0, we have:
lim[n→∞] sum[ i=n-2020 to 2n+2020 ] (1/i)
= lim[n→∞] ln(2n+2020)+γ) - (ln(n-2020)+γ)
= lim[n→∞] ln(2n+2020) - ln(n-2020)
= lim[n→∞] ln( (2n+2020)/(n-2020) )
= ln( lim[n→∞] (2n+2020)/(n-2020) )
= ln(2)
Do you like the harmonic series?
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Warp wrote:
So my thinking is: If we assume that the list is countable, then it shouldn't make a difference if we replace every element with a natural number (eg. its position in the list). It shouldn't make any difference what we replace every digit of every element in the list with, as long as the resulting remapping is unique. From the point of view of the proof, it shouldn't make a difference.
So if we do that, we are replacing the original list of infinite strings with a list of finite strings, and the only thing that the diagonalization does is to tell us "there are no infinite strings in the list", which is self-evident to the point of being tautological. Of course there aren't, because we replaced them with finite lists in the first place. The only thing that the "proof" is now saying is, in essence, "if you replace all infinite strings with finite strings, there will be no infinite strings."
Cantor's diagonal argument is not that.
Cantor's diagonal argument says only: Given any countable list of real numbers that I so choose, you can always find a real number not in this list, using the diagonal argument. This number you choose is dependent on the list I choose.
Cantor's diagonal argument is not about proving that a specific listing lacks some real number; it is not reducible to replacing every element with a natural number and then showing that there are real numbers that are not in the list. So replacing numbers does make a difference.
For the sake of argument, if it were possible to reduce an argument of uncountability to replacing elements with natural numbers and then showing that a number does not exist in the list, then I could "prove" that the set of rational numbers is uncountable as follows:
Suppose the rational numbers are countable. Then from a countable listing of rational numbers, I can replace every rational number with a natural number. Then in this new list, there is a rational number that does not exist in this list: namely 1/2. This is a "contradiction" and so I have "proved" that the set of rational numbers is uncountable.
That isn't what Cantor's diagonal argument is about.
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I'd say you're both right and wrong at the same time.
Right because, in spirit, you are using a special form of the chain rule:
dy/dx = (dy/du) / (dx/du)
Wrong because, pedagogically, use of differentials automatically makes any argument "wrong".
P.S. Don't take me seriously.
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Warp wrote:
But I was thinking that, perhaps, it could be possible to construct a set of size aleph-0 or "smaller" (if that's even possible) where it's not possible to index the elements because there's no "least element" that can be chosen or pointed out.
If the axiom of countable choice (a weaker version of the axiom of choice) holds, then it is possible to index a countably infinite subset of any infinite set just by invoking the axiom, so there cannot be any smaller infinite cardinality than aleph-0.
If we are not allowed to assume any choice axioms, then I think it's up in the air.
Warp wrote:
At first it looks like the rational numbers would be this kind of set: You can't eg. decide that 0 is the "first" rational number and then choose the next one to be "the smallest rational number that's larger than 0" because such a number doesn't exist. However, there are other ways to unambiguously assign unique indices to each distinct rational number.
The Calkin-Wilf sequence is one well-known way to order the positive rationals. It is defined by the breath-first traversal (level-order traversal) of an infinite binary tree whose root is 1/1 and the children of each node are defined as:
- left child of a/b is a/(a+b),
- right child of a/b is (a+b)/b.
The order of the positive rationals here is 1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, ...
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FractalFusion wrote:
[A] tortoise and [a] hare are about to begin a 10-mile race along a “stretch” of road. The tortoise is driving a car that travels 60 miles per hour, while the hare is driving a car that travels 75 miles per hour. [Assume both cars instantly attain those speeds during movement.]
The hare[, wanting to end at the same time as the tortoise,] realizes if it [were to wait] until two minutes have passed, they’ll cross the finish line at the exact same moment. And so, when the race begins, the tortoise drives off while the hare patiently waits.
But one minute into the race, after the tortoise has driven 1 mile, something extraordinary happens. The road turns out to be magical and instantaneously stretches by 10 miles! [The road stretches linearly, taking whatever is on the road with it, including the hare, tortoise, and finish line.] As a result of this stretching, the tortoise is now 2 miles ahead of the hare, who remains at the starting line.
At the end of every subsequent minute, the road stretches by [an additional] 10 miles [in a similar fashion]. With this in mind[, how] long after the race has begun should the hare wait so that both the tortoise and the hare will cross the finish line at the same exact moment?
----------------------------------------------------------
Hint: You don't actually need to find how long it takes for the tortoise to finish the race to answer this problem.
I'll provide the answer I came up with. I find it interesting that the harmonic series (1 + 1/2 + 1/3 + ...) is involved in this question.
Notice that, because of the speeds, the hare should always wait until the tortoise is 20% into the race, regardless of how far the road has stretched. (The road stretches uniformly at all times, so the positions will always remain in proportion.) This means that we do not need to know how long it takes for the race to finish, only how long it takes for the tortoise to get to 20%.
To make things simpler, we can consider the road rescaled to 10 miles at all times, in order to rewrite the problem. So instead of the road stretching by 10 miles every minute, we consider the tortoise slowing down every minute, in a harmonic series-like fashion:
- The tortoise drives 1 mile in the first minute, then 1/2 mile in the next, then 1/3 mile in the next, ... How long then will it take the tortoise to get to 2 miles?
Which is a much simpler question:
- After 1 minute, tortoise is at 1 mi
- After 2 minutes, tortoise is at 1 + 1/2 = 3/2
- After 3 minutes, tortoise is at 3/2 + 1/3 = 11/6 = 22/12
- After 4 minutes, tortoise is at 22/12 + 1/4 = 25/12 > 2
So the tortoise reaches the 2-mile mark somewhere between 3 and 4 minutes. Linear interpolation shows that the tortoise reaches 24/12 at two thirds of a minute (40 seconds) after the 3-minute mark. So if the hare wants to finish the race at the exact same time as the tortoise, the hare should begin the race 3m 40s after the tortoise begins.
Note: How do we know the tortoise is able to finish the race, and it won't go on forever? Simple, 1+1/2+1/3+... diverges. So in fact the race could be a million miles long (stretching a million miles each time) and the tortoise will still eventually finish. Though, even for 10 miles it takes a long time (8.588 days, to be precise), and the amount of time required will increase exponentially based on the distance.
An image of the 8.588-day race is shown in the following solver's twitter post (sorry, the "GIF" (if it is even an actual .gif) is embedded so I can't extract it):
https://twitter.com/ali_geeee8/status/1285183324596535298
