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I don't know how this game works, but in SMW, at least, getting 99 lives would be certainly slower, even if you don't have to slow down inside levels. That's because the game has to convert from hexadecimal to decimal your score, lives and coins every frame, and that causes lag if the digits are high (in the lives and coins counter, only the first digit maters).
I don't think it is that 'entertaining' to see 99 as the number of lives, as the entertainment in a TAS can go much beyond that. Despite I do understand why there is this request of a run with 99 lives, this is not a significant change in the entertainment of a TAS, and thus, it doesn't worth to perform a speed/entertainment tradeoff just for the sake of that.
Yes vote, by the way.
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dwangoAC wrote:
I regret to say... that while this is a fantastic movie that should absolutely be published as a star run and I voted yes and everything is right in the world, I was unable to console verify this fully. And by that I mean I was able to console verify this up until what appears to be the very last frame, at which point Mario's final throw appears to miss and instead Mario meets his demise between two of Bowser's mechanical minions. I do not know why. I may try altering the file to delay the final action in some way such as delaying the last frame but it's annoying that it doesn't "just work".
All that to say, congratulations on a fantastic run - keep doing more of this! :)
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coldsun0630 wrote:
BrunoVisnadi wrote:
Yoshi's Island 3 (104)
Here happens the biggest improvement...
...If you keep doing this, then slot 2 will stop spinning, then 1, then 0, then 3 again. So, it is impossible to 4 blocks to spin at the same time.
Seems typo. It should be 5.
Also, I think saying like be taken instead of stop spinning would be better, because it not only stops spinning but also generates another spinning one.
Anyway, another awesome TAS! Voted Yes.
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Ah, the (8/9) was actually because of the B not being a multiple of 9 condition, but now I realize that since 27 is multiple of 9, 27N + 23 will never be a multiple of 9 for every N.
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Thanks for answering, again. I'll point out something about the other problem you solved:
I think there is a way to avoid using integrate do determinate Σ (2/(2x+1)^2), k from 3 to infinity. See if you think that's correct:
We know that Σ 1/x^2 is pi²/6. Thus, Σ 1/(2x)^2 = Σ 1/4x^2 = pi²/24.
It's easy to tell that Σ 1/(2x)}2 + Σ 1/(2x-1)^2 = Σ 1/x^2.
That means Σ 1/(2x-1)^2 = pi²/6 - pi²/24 = pi²/8
Since we only wanted values from 3 to infinity for Σ 1/(2x+1)^2, we will only want values from 4 to infinity for Σ 1/(2x-1)^2.
So we got R(N)/2 = pi²/8 - (1 + 1/9 + 1/25) , which is less than 0.085 .
So R(N) is less than 0.17.
Here I'm considering N as the whole natural numbers set. Since (1/2)(2/3)(12/3)(23/25)(8/9) = ~0.25, and 0.25-0.17 = 0.08, there must be 0.08N, and thus, infinite solutions for f(n) = f(n-1) + 1 in N.
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I'll put here another problem from that olympiad. I think this one is a bit easier:
Is it true that there is a polynomial f(x) with rational coefficients, but at least one non integer, which degree is n>0, a polynomial g(x) with only integer coefficients, and a set S with n+1 integer elements so that g(t) = f(t) to every t that belongs to S?
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Warp wrote:
This may sound a bit insensitive, but I really can't understand the relevance of this topic at a website like this one.
It's like I were to go to a forum about... I don't know... chess, and declare there "hey, did you know that I'm a man?" Seems a bit superfluous. The expected answer would be that of puzzlement, and "yeah, so?"
If someone comes to a forum about making runs of computer games and says "did you know that such and such person is trans?", I'm like, well, "yeah, so?" I don't really care what somebody is, especially not in a forum that's rather unrelated to such issues. Isn't this like a personal thing? You are what you are; I am what I am. So?
In fact, I find it even more bothering when it's "did you know person X is Y", rather than "I am Y". It's a bit like gossiping.
I do understand that some people struggle with their identities and it's important to them, but there's the right place for everything.
I don't understand if you mean that this kind of 'news' should be as casual, and consequently not worth sharing, as something like 'OoT speedrunner hair color is blond' (in which case I would agree with you, but that's not happening because society will never accept transsexualism that way); or if you mean that this new is so unimportant that it doesn't have place even in the off-topic forum of this site. If that's the case, your post is something to worry about.
In a world where manytranssexualandtransgenderpeople are murdered or attacked due to intolerance, have a extremely high rate of suicide, and often never get to reveal to others their gender identity, it is absolutely not significant that a person related to this community has come out as transgender?
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What is 'O'?
Supposing it is the number of frames, it would be 2^(n*O), right?
For instance, in SNES there are 16 inputs. If you use them all, there would be 2^16 = 65536 possibilities.
If you try for 2 frames there would be 2^32 = 4294967296 possibilities, which might be too much.
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Ah, so this got cancelled? Sorry for pointing out a improvement and then not collaborating in making the new movie, this notebook problem was very unexpected.
Maybe we can collab on this game in a few months, when I manage to get a way to TAS.
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ars4326 wrote:
Bruno, are you still ironing out improvements on this one?
I was, until 1 hour ago. Certainly every fight can be improved a lot.
Now, however, the cooler of my notebook just stopped working, and it overheats and turns off in few seconds after turned on. I don't really know what to do - I'll try to open and clean it, and if it doesn't work I'll send it to be fixed. Until then, I have almost no way to TAS anything. The computer I'm using to write this message has a terrible keyboard which makes it impossible to TAS, and with the few hours I will have per week in a decent computer, I will rather focusing on SMW any%.
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I'm using them in TASTudio. You can select for each frame a value to Ax, and a value to Ay, which are kinda of 'coordinates' to the analog control. You can select a value between -128 and 127, for both Ax and Ay.
Edit: still without analog control, I've already saved 2 more frames. Let's try with analog control now.
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Have you seen the movie?
I think it is still possible to save a few frames by using the analogical control. It's indeed really hard to use it in this game (especially for me, since I don't have a controller and I have to use TASTudio), but it allows you to move directly to the enemy if you press the right combination.
Edit: still without analogical control, 224 frames saved (analogical seems to potentially save more 5 frames): http://tasvideos.org/userfiles/info/26622595336975652
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Bizhawk 1.11.3 is not saving the buttons I chose for hotkeys and controllers. When I close the emulator and open it again, it goes back to deafult. What am I doing wrong?
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Thanks for your answer!
Yeah, this Olympiad does require pretty specific knowledge sometimes, unfortunately.
I could understand most of your solution, and I'm happy because what I wrote in the Olympiad was pretty similar with the beginning of your solution, so I might gain some partial points. I got to f(27) = 27, f(169) = 26, f(a*27) = 27 and f(b*169) = 26, being a a square free not divisible per 3 and b a square free not divisible per 13, so that 27a = 169b + 1. But at that moment I couldn't get any beyond that.
I can kinda understand the principle of your solution, but I have some questions about it:
1) Why didn't you include 5^2 as one of the possibilities of odd k^2? Is there a reason to that?
2) I'm not used at all with math terms in English. I didn't even know what gcd means. Could you say what 'ceiling' means in ''2*ceiling(N/k^2)?
3) I know this is the part that requires some further knowledge, so I might not understand it, but could you explain a bit deeper how you got to ''R(N) <= N*( sum{i=3 to infinity} 2/(2i+1)^2 ) + 2*sqrt(169N+144)/2''?
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Kurabupengin wrote:
BrunoVisnadi wrote:
I managed to save 9 7 frames in the intro by selecting some stuff faster! Now I'll try to improve the fighting as well. Dekutony, let's to co-op on this?
Um... sure, why not. I really have nothing better to do so I guess that's allright.
One question though, what did you do different on the first fight which saved that many frames?
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Spikestuff wrote:
It's 7 frames, not 9, I have no idea where you got the extra 2 from.
Has spliced the input into the older one, which is why he wrote 7, causes a desync so will not be providing it.
I wished your input did not look like that, it doesn't not look great, and it's just random mashing, when you can just have a single press for each time.
Thankfully (sarcasm) BizHawk crashes on this game when using TAStudio.
I suppose you mean the random mashing in the beginning. Sorry about that, but anyway that's not a problem since input can't affect anything there. If you are referring to the animation skipping, I do have to press the buttons that way, otherwise frames would be lost.
Well, here it seems I saved 9 frames, I didn't understand what exactly the problem is.
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I managed to save 9 frames in the intro by selecting some stuff faster! Now I'll try to improve the fighting as well. Dekutony, let's to co-op on this?
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FractalFusion wrote:
BrunoVisnadi wrote:
Being n's prime factorization = (p1)^q1 + (p2)^q2 + (p3)^q3 (...) + (pk)^(qk).
And being f(x) = q1*p1^(q1-1) + q2*p2^(q2-1) + q3*p3^(q3-1) (...) + qk*pk^(qk-1).
Prove that there are infinite natural numbers n so that:
f(n) = f(n-1) + 1
Just to be clear.
Do you mean n = (p1)^q1 * (p2)^q2 * (p3)^q3 * ... * (pk)^qk ? That is normally how a prime factorization is written out.
If so, is it correct that f(n) = q1*p1^(q1-1) + q2*p2^(q2-1) + q3*p3^(q3-1) + ... + qk*pk^(qk-1) ?
Of course! Sorry, I wrote that completely wrong! The '+' symbols in both n prime factorization and f(x) should replaced by '*' symbols. Thanks for pointing that out, I'll edit my post.
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This problem appeared in a Brazilian math olympiad, so it should be solvable without much math knowledge required:
Being n's prime factorization = (p1)^q1 * (p2)^q2 * (p3)^q3 (...) * (pk)^(qk).
And being f(x) = q1*p1^(q1-1) * q2*p2^(q2-1) * q3*p3^(q3-1) (...) * qk*pk^(qk-1).
Prove that there are infinite natural numbers n so that:
f(n) = f(n-1) + 1
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Firstly, being h=OE the height of the OBC triangle, and n=EC, we see that:
h² = 23² - (30-n)² = 16² - n²
Thus, (30-n)² - n² = 23² - 16² and n = 209/20
And h² + (209/20)² = 256. So h = (23/20)*sqrt(111)
Now, being g=AF the height of the triangle ABC, and m = FC.
Again, g² = 27² - (30-m)² = 22² - m².
(30-m)² - m² = 27² - 22², and m = 131/12.
Thus, g² + (131/12)² = 22², and g = sqrt(52535)/12.
Let's put this image in a Cartesian plane, being b = (0,0).
O = (30-n,h) and A = (30-m,g).
So, if d = OA, d² = [(30-n) - (30-m)]² + (g-h)².
d² = [(30-209/20) - (30-131/12)]² + (sqrt(52535)/12-(23/20)*sqrt[111])²
Solving this equation, d is *almost* 7:
d = sqrt([61421/30] - 23*sqrt(388759/15)/2)/2
which is approximately 7.0000000857...
