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Warp wrote:
At some point I realized that the Cantor's diagonal argument technically speaking is a demonstration that the set of all possible infinite strings of digits is uncountable. (If we assumed that we can enumerate all possible infinite strings of digits, the argument shows that the assumption leads to a contradiction.)
Technically speaking that in itself doesn't yet prove that the set of reals is uncountable. We would still need to prove that the set of reals has the same size as the set of all possible infinite strings of digits.
I'm not completely sure that's trivial, given that some different strings of digits represent the same real number.
Decimal representation of reals is sometimes not unique: for example, 1.00... = 0.99999..., but Cantor's argument can avoid this ambiguity by choosing digits 1 or 2 when constructing a new real number from a countably-indexed list of reals.
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Flip wrote:
Just from the fact that the integers are a unique factorisation domain
We don't require the uniqueness of factorization. In fact, we only have to require a slightly weaker condition that any pair of integers a and b has a greatest common divisor (and hence a least common multiple), which is deduced from Euclid's algorithm in the case of integers.
Now recall the two conditions on an integer p (or an element p of an integral domain).
(I) p cannot be written as a product of two non-invertible (non-zero) elements. (Such p is said to be irreducible. This is a usual definition of a prime number.)
(P) If p divides ab, then p divides a, or p divides b. (Such p is said to be prime.)
(P) always implies (I), but in general, the converse is not true. One of the sufficient conditions for the converse to be true is the GCD condition above. Suppose that an irreducible element p divides ab, and assume p doesn't divide a, so that d := gcd(a,p) ≠ p. Since d divides p and p is irreducible, d = 1. Since both p and a divides ab, l := lcm(a,p) divides ab, hence lc = ab for some c. On the other hand, we have ap = dl = l. Altogether, we have ab = lc = apc. Thus b = pc, or equivalently p divides b. Furthermore, it follows by induction that if an irreducible element p divides a^n, then p divides a, which was a key property in proving the n-th root of a prime is irrational.
Amaraticando wrote:
In general, if n^p is not integer (n integer and p rational), then it's irrational.
I shall prove this in a slightly different setting: if p is a prime, then p^r is irrational for any rational r with 0 < r < 1. (The case r > 1 follows from the observation that if r = s + r' with s an integer and 0 < r' < 1, then p^r = p^s p^r', and p^s is an integer.)
Proof. Let p be a prime, and r = a/b rational with gcd(a,b) = 1 and 0 < a < b. Assume p^r is rational: p^r = c/d, where gcd(c,d) = 1 and d >0. Then we have dp^(a/b) = c, hence d^b p^a = c^b. Since p divides LHS, p divides c^b and hence p divides c. We can write c = pc'. Then we have d^b p^a = p^b c'^b. Dividing it by p^a gives d^b = p^(b-a) c'^b. Again, RHS is a multiple of p, hence so is d. Thus p is a common divisor of c and d, which is a contradiction. The proof is complete.
The case of a composite number is harder to deal with, but essentially the above proof applies, with some appropriate case reduction.
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Warp wrote:
I think the problem can be generalized: If you have two curves on a plane, if you evenly scale the whole thing larger or smaller (I don't know how to express this mathematically) the ratio of the lengths of those two curves remains the same. The ratio between a circumference and diameter is just a special case of this.
Of course proving this general case might not be trivial.
First we have to specify what we mean by curve and by length. As long as a curve is defined as a continuous mapping [a,b] → R^2 and its length is defined to be the "limit" of the length of line segments which approximate the curve, the assertion is a consequence of the elementary facts:
d(rx,ry) = r*d(x,y) for any points x and y and any scalar r;
d(x-z,y-z) = d(x,y) for any points x, y and z,
where d(x,y) denotes the distance of x and y. If you want to go with another set of definition, let us know what it is.
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Bobo the King wrote:
Bobmario511 wrote:
Saw this abstract linear algebra question on a test today and wanted to share it (posting it off of memory so somethings may be off):
Let A,B,M be nxn matricies which have real entries where n is a natural number, where AM=MB [corrected] and A and B share the same characteristic polynomial.
Prove det(A- MX) = det(B - XM) if X is an nxn matrix with real entries.
It's not specified whether this is true for any X or if there merely exists an X. I'll assume you mean there exists an X because that's the proof I'm coming up with.
My proof relies on M being invertible, which wasn't specified. I suppose you might be able to re-work it in such a way that it doesn't require inverting M. I kind of doubt it, though, since M-1 shows up explicitly in my solution.
A and B are similar because B = M-1AM. So that's neat.
Also, we know det(A-λI) = det(B-λI). That's more useful. Let's start there.
det(A-λI) = det(B-λI)
Replace I on the left with MM-1. Replace I on the right with M-1M:
det(A-λMM-1) = det(B-λM-1M)
Now just commute λ on the left hand side because it is a scalar.
det(A-MλM-1) = det(B-λM-1M)
Finally, identify X as λM-1. Now we have
det(A-MX) = det(B-XM)
QED.
When M is regular, we can prove it for any X. In fact, we have det(A-MX)det(M) = det(AM-MXM) = det(MB-MXM) = det(M)det(B-XM), and since det(M) is nonzero, we conclude det(A-MX) = det(B-XM). This is easy.
When we want to prove the existence of a non-trivial X (with M arbitrary), one of the answers is X=A (or B), for example. In fact, we have det(A-MA) = det(I-M)det(A) and det(B-AM) = det(B-MB) = det(I-M)det(B). By assumption det(A) = det(B). Thus det(A-MA) = det(B-AM). I think even this is easy.
I'm now trying to prove the general case, but only got some special cases, such as M being the direct sum of a regular matrix and a zero matrix. The point is, I guess, the case of M nilpotent.
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BrunoVisnadi wrote:
We know that Σ 1/x^2 is pi²/6.
To prove this is not so easy as bounding its value. Its proof requires some Fourier calculus, or something related to tough computation of trigonometric functions, which is/are more complicated than FractalFusion's estimate. Except for that, your argument looks fine.
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Bobo the King wrote:
Note that this proof doesn't require a base case for induction or rely on the continuity of the sqrt and log functions. As long as we accept that the sqrt and log functions are definable and positive for all real numbers greater than 1, no matter our starting point, the function blows up to infinity for arbitrarily large x.
I don't think the continuity of f (or whatever like that) is superfluous; there seems to be a slight gap. First we see from the above proof that f(100*x)/f(x) = 10*log(x)/(2+log(x)) = 10-20/(2+log(x)) < 10, hence f(100^n*x) < 10^n*f(x) for any n. Assume we could choose infinitely many a(i), such that (1) 1<a(i)<a(i+1)<100 and (2) f(a(i))>10*f(a(i+1)) for all positive integers i. Then we could show that for any M>0, there exists an x>M such that f(x)<f(a(1)), which would prevent f from diverging to infinity. The proof goes as follows: choose a positive integer n such that M<10^n, and put x := 10^n*a(n+2), so that M<x. Then we conclude that f(x) = f(100^n*a(n+2)) < 10^n*f(a(n+2)) < 10^n*10^(-n-1)*f(a(1)) = f(a(1))/10 < f(a(1)), as desired.
Roughly, if there were a sequence a(i) in an interval of finite length such that f(a(i)) tends to zero, we could no longer expect f to diverge to infinity in the strict sense. We can avoid this trouble by assuming the continuity of f, as we have accepted f is strictly positive on the entire interval (1,inf).
In my opinion, it is basically impossible to discuss a transcendental property by using only algebraic properties, which is why it is transcendental! :P
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Zowayix wrote:
I'm guessing this is an item-swap glitch of some sort where the goal tape somehow replaces one of the Spiny Eggs that Lakitu throws, but how does that work? As far as I know an item swap is only possible if a null sprite is generated or if all sprite slots are filled, and that doesn't seem possible given what was shown on screen. What happened?
An enemy-type sprite on the tongue is turned into a "null sprite" when going off-screen enough, because the something-is-on-the-tongue flag is still on while the sprite despawns. I'm not sure if this is the case, though.
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Can anyone make an edit to the submission page in order to lead exposure to the forums (or directly to yuigenron's post above and to my post), in case he has no look into here? I guess it is possible he has no idea where to see after submitting a movie, for he has not logged in to the forums yet.
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So the warps run has finally moved to an accurate emulator. Great job. It's surprising that it's still beaten within 10 minutes even on a slower emulator.
bahamete wrote:
Ah, it's worth noting the latter part of ds1 (from the pswitch to cape-spinning the key block) is all Mister's input; I asked him if I could use it a long time ago as I was struggling to optimize, and I forgot about it until just now.
According to the chat log, it was nearly three years ago that bahamete asked me that. Judging from what I saw, it seems like ds1 bubble manipulation, which you cared about then, was not useful, right?
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Eszik wrote:
http://www.twitch.tv/authorblues/c/6454659
All right, new RTA "no credit warp" WR in 9:30. It uses a new strat involving getting Big Boo in the inventory and beating him instead of Bowser.
Now that isn't cloud, so no ACE... so new 11-exit TAS hype?
Doesn't that PI setup involve ACE possibility? Remember the first credits warp, which was done on the rotating platform.
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"This video has been removed because of violation of copyright."
About a hundred of mario games's videos, including TAS videos, were deleted this week. I don't know the exact reason why that's been happening, though.
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Good job! Looks a lot better than the previous. I don't like so much to see mario lose speed with such ease, as it might look as if it were not planned very well. It's just an opinion.
Paused wrote:
I personally did not like the need for the back track to a past level for a Yoshi though. Felt like poor planning I guess? Could you not do the same trick in a future level where you already have access to a Yoshi?
A few rooms after you grab a star in the goal cutscene, you'll get star invincibility (I don't remember the correct number; it can depend on how many stars you grab, if I remember correctly). So, I guess the backtrack is put there in order to manipulate the star timing, besides picking yoshi up.
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IgorOliveira666 wrote:
Mister wrote:
The run itself looks made well but rather boring. I guess the main part of this run is the stun glitch part, which looks very nice, but the preceding part is the classic left-to-right type. So I wonder if there's a better game where we can perform stun glitches without a long boring part (I don't think having only stun glitch is so attractive that it can get viewers' interest, though).
What I couldn't read from description is:
Is it really faster to enter the pipe at 1-1?
Do we really need a shell at 8-1 and 8-2? Can we dodge plants as you did in 4-1?
Why did you release the shell in the watery area of 8-4?
Is it impossible to avoid 6/5 at 9-4?
I believe you could go with better entertainment choice. Especially at 9-3. Flying outside of the screen is the worst choice in my opinion.
That's all. I wish they were explained in your description. Comparison with a seven-year-old run which has no documentation is by no means helpful.
Well I will explain:
In 8-1 is yes faster into the pipe. It is 13 frames faster if I'm not mistaken.
In the 8-1 / 8-2 is yes actually faster, because if I did the same in the 4-1 I would have to go the whole stage 6/5, which would lose some frames, for, taking the shell I could go the the rest of the stage 49 velocity.
At 8-4 I let the shell to kill the sprite because it was not going to lag 8 or 9 frames. Furthermore only did not win lag because I was above the stage. If I had gone under, would have won lag, even killing the sprite.
At 9-4 I went with 6/5 because if it was with speed 49, further ahead in step (on the pass between the blocks) the Mario floating and could not pass between the blocks.
In 9-3 if I remember correctly gave some lag frames if underneath.
Well that's it.
Thank you for your explanation! I quick-tested around, and I noticed that the red "balls" in 9-4 can hurt yoshi, so now it seems impossible to 49-hop there. For 8-4, however, I didn't see such much lag there, but only had 2-4 frames of lag (I am willing to post a test movie for your information if you need). Maybe you didn't have to release the shell, I guess.
Umm I'm not complaining about improvability at all. I've learned the hack's physics and I've tested a few parts myself. I realize further that you did really good work. Still, I doubt if it's worth being published, so no vote from me. I'm looking forward to your next project where you can show your full potential.
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Excellent job! I'm refraining from leaving a comment now. I have to calm myself down a bit. :P
Here's a 60fps nicovideo encode (thanks to asutoro):
Link to video
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The run itself looks made well but rather boring. I guess the main part of this run is the stun glitch part, which looks very nice, but the preceding part is the classic left-to-right type. So I wonder if there's a better game where we can perform stun glitches without a long boring part (I don't think having only stun glitch is so attractive that it can get viewers' interest, though).
What I couldn't read from description is:
Is it really faster to enter the pipe at 1-1?
Do we really need a shell at 8-1 and 8-2? Can we dodge plants as you did in 4-1?
Why did you release the shell in the watery area of 8-4?
Is it impossible to avoid 6/5 at 9-4?
I believe you could go with better entertainment choice. Especially at 9-3. Flying outside of the screen is the worst choice in my opinion.
That's all. I wish they were explained in your description. Comparison with a seven-year-old run which has no documentation is by no means helpful.
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AmaizumiUni wrote:
つたない英語で迷惑をかけております。
数か所ほど数フレーム短縮できそうなミスがあったので試してみて短くなるなら取り下げたいと思いますので
judging Status new のままで進行しないようにしてもらえますか(作成は1日～数日で終わります)
もし全体として大きく見た目が変わらないのなら、キャンセルという形ではなくムービーの差し替えという形でもかまわないと思いますよ。差し替えてもらう場合は、更新後のファイルを http://tasvideos.org/userfiles/ などにアップロードしてそのアドレスをこちらに書き込めばOKです。もし修正が見つかった場合に参考にしてもらえればと思います。
---
He's going to cancel his run if it can be improved, so I'm telling him to have his movie replaced if there's no big difference.
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puppy610 wrote:
Hello! I would like to finished the unfinished PI 106 TAS. If you do not know what that is then go to the Super Mario World section of the forum or google it. I have a vote setup so I can see if I should really do it. Here it is: http://strawpoll.me/3433840
I don't know at all if you should go or not, but you'd better explain what the PI is if you really want some advice. Or do you want only the smw experts' advice? Maybe no. If you can't explain the PI glitch well enough, you'd better stay for a while and improve your TASing skill and TASing knowledge, as we advised you before.
On the second note, we are not you so there's no means of us knowing what you can do and what you cannot. Please show us more information, such as wip and detailed strategy; anything will do.
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ars4326 wrote:
Techokami wrote:
I'd actually like to suggest that TASVideos get a dedicated localizer for the Japanese language. That way, Japanese TASers can use their native language to write the author notes, and they can then get translated into English for the rest of us, thus preventing hostility and confusing machine translations. Just a person fluent in both languages with editor rights.
Admittedly, I'm not fluent in Japanese, but I have a solid foundation in general sentence and grammatical structure, and tense usage (I took two years in college, from 2011-2013). I would be willing to volunteer for this position, in order to bridge the language gap with our Japanese community (this would also motivate me to continue consistent studies).
If TASVideos is looking to fill such a position, I would love to volunteer and help in an increasing manner.
Sounds great. Language barrier is one of the most serious problems for many Japanese TASers, since machine translator doesn't work well between ja-en in most cases. I suggested to AmaizumiUni a few times that he include his original description written in Japanese with complete sentences as possible, so we can fill the gap more easily. I'm also expecting some Japanese TASers can help translators with such translation, which would make entire translation much more satisfactory. Anyway then, in order for us to avoid any confusion/ambiguity, the author notes would be desired to have good sentences (I mean, to have fewer slang words or abbreviations or something). I think I can help the site make such guidelines to some extent, though I'm almost inactive here these years.
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Welcome to the site, and welcome to SMW TASing. I appreciate your efforts, but I'm afraid your approach has been realized in the previous published run or in the former ones, where neither orb nor cloud is used. Actually, neither glitch is so common in the long history of SMW TAS. You can learn a lot from them and ask your questions in the forum if you have any trouble, to skill yourself up.
Voted no, but I'm looking forward to your next work. Try hard, and you'll be getting skilled.
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