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Nice work, looks much better. One thing to notice, the U version might be faster as far as I know, or am I wrong? Have you checked that carefully enough?
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HappyLee wrote:
Mister wrote:
Oh fuck, I couldn't finish TSRP2 before the release of its remake!! Time passes so quick. Thank you for notification, Fortranm.
Edit: seems like the numbers of exits/levels have changed... It might require a different strategy, I guess?
I would suggest you ignore that temporarily and finish your amazing TSRP2 run as quickly as you can. You can do it. :)
Ah, sorry for confusing. I was just wondering what it would be like in TSRP2R. I'm not planning to redo my run at all, and I'll be back to TASing by autumn, hoping it'll be cool soon. :p
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Oh fuck, I couldn't finish TSRP2 before the release of its remake!! Time passes so quick. Thank you for notification, Fortranm.
Edit: seems like the numbers of exits/levels have changed... It might require a different strategy, I guess?
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Zowayix wrote:
Also loving the fact that a TAS of a SMW-based game on this site is finally able to show off the Cloud-Bowser glitch without any ACE-related arguments.
The cloud strategy was originally discovered in SDW and had been meant to be used there. So this is rather a reverse import or something. AND I think this is a good reason for not using this trick in SMW; we have a more appropriate hack to use it, SDW!
EDIT: link added. Could you modify your description on the cloud strat, Panga?
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The two tops belong to the same topological class, namely a compact orientable surface with five boundary components. Thus they are homeomorphic, but there's no transformation (in 3-space) which deforms one to the other. In other words, they are different realizations of that topological space in 3-space. It's tough to explain that, so I'll try making an easier example here.
First, the question reduces to the the following problem. Create two paper strips, and give a full twist (i.e. 360-degree twist ) to one of the two. (If you give it a half twist, it is a famous möbius strip.) Then can we deform the normal strip into the twisted strip? If you make three other holes on them, you see that this turns into the initial problem. Now the answer is no, though it's rather non-trivial. Let me explain further.
Maybe you know what is obtained by cutting the möbius strip along the center line. We do the same thing to the above two strips. The normal strip turns into a pair of strips, while the twisted strip turns into a pair of strips entangled with one another. If we were to turn the normal strip into the twisted one (in 3-space), then the two pairs of strips would also turn into one another, which is obviously impossible.
Untwisting in higher dimensions is also tough to describe. Here is a quick example (in lower dimensions) related to the problem. Consider a circle with a point inside on a plane. Obviously the point cannot escape out of the circle as long as they are forced on the plane, but via 3-space, it can easily escape. I'm not a topologist so I'm not sure how accurately this example accounts for the untwisting problem. (I think it's close enough, though.) Higher-dimensional topology is really confusing. :p
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I wonder if I can say congrats on this umm... It was 11 May that I asked Masterjun if he knew how to reserve a lakitu cloud, showing one video (niconico link). It is known a reserved cloud saves about one minute in bowser, and I just wondered if it would be available in SMW11. The result is this submission. :( I remember that antaasas pointed out that eating some chuck would give rise to some ACE years ago, which is finally proved by this run.
Anyway good work, Masterjun! It's worth while to prove SMW can be beaten within one minute, especially considering the SMW ACE history. Yes vote.
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I'm afraid there're two persons working on SMA2 100% (i.e. all dragon coins) run, ISM and gbreeze. I don't know if you collected dragon coins but ISM's previous run (niconico, or youtube MUGG gave above) beats the game in 1:24:34.05. Still waiting for an encode but voting no maybe.
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FinalFighter has already shown the authors' standpoint above. Well, I'm sorry if the submission description (or in particular, the joke in P.S.) is confusing to you. That's my bad translation, and I think there may be some information lost in translation. When you have any questions, it would be so helpful if you ask them in rather easy/plain English. It's hard for us to follow such sensitive discussion. Thanks. :)
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Congratulations on finishing this great ACE project finally! I saw you guys working so hard every day and I'm really happy to see the final product. But it's unfortunate that this is not by Masterjun! NO VOTE!
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Awesome work as always. I watched this run, and watched other playing movies, and then watched this again. Well, I don't think those can be the same game's videos. Voted yes.
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I watched it twice, and found it highly unoptimized. Super Mario World is a well-studied game and possesses a lot of tricks/glitches, which you may have to learn. For example, you can optimize your speed by releasing the direction key at a correct timing while flying.
Mothrayas has already pointed out that this is slower than a part of any%. Even as 100%, it's slower than the 6.5-year old TAS.
No vote, for poor strategies and suboptimality.
@Spikestuff: the cancelled one is for any%, not 100%. With the latest strategies, the 100% run would be done within 17 minutes or so.
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Bobo the King wrote:
This problem has been on my mind a little lately. I haven't given it a whole lot of thought, but to the extent that I have, I'm stumped.
1) Construct a function of two real numbers that, using only addition/subtraction, multiplication/division, and absolute values that takes the value of the larger of the two numbers. (So construct the max function using only absolute values. This should be very easy.)
2) Construct a function of three numbers with the same requirements as above (using only a/s, m/d, and absolute values) that takes the value of the middle number. (This is only a bit harder.)
3) Construct a function of four numbers with the same requirements as above that takes the value of the second-largest number. (I don't know off the top of my head how one might do it.)
So for example, without loss of generality, for any a<b<c<d, we seek functions f, g, and h (corresponding to cases 1, 2, and 3 above) that only involve addition, subtraction, multiplication, division, and the absolute value, such that:
f(a,b) = f(b,a) = b
g(a,b,c) = g(b,c,a) = g(c,a,b) = g(c,b,a) = g(b,a,c) = g(a,c,b) = b
h(a,b,c,d) = {23 other permutations of the arguments} = c
Can this be done?
Edit: I found one way to do it. It's a little tricky and I'm sure there are many equivalent ways. I also suspect the process can be generalized to yet more variables.
For f, the function f(a,b) := (a+b+|a-b|)/2 will do, and the min function can be also constructed as f'(a,b) := -f(-a,-b) = (a+b-|a-b|)/2.
For more variables, I'm just going to explain the process, not showing the functions explicitly (you can write down functions with ease, though). Now assume we have three real numbers a, b and c, and let A, B and C be the max of the pairs (a,b), (a,c) and (b,c), respectively. Then the set {A,B,C} consists exactly of the largest and the middle among {a,b,c}. So we have only to pick up the smallest of {A,B,C}, which can be done by iterating the min function of two numbers.
For the second-largest of four numbers, pick up the largest of each triple among {a,b,c,d} to form the set {A,B,C,D}, and the smallest of this set is the second-largest of {a,b,c,d}.
This is somewhat redundant but has the advantage of being generalized easily to the case of the m-th largest of n numbers.
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gbreeze wrote:
Mister wrote:
I think it's also useful in VoB2b of SMW.
Nice find! It looks like it's definitely useful in that level. What about Forest ghost house? It may be possible to clip through some blocks in the beginning and avoid having to corner clip later.
I'm afraid it's impossible there. Unfortunately, Mario falls too quickly while in a wall, so the wall must be long enough to pass through (12 blocks long or so?). Although I looked through SMW, VIP1 and VIP3 (from the viewpoint of capeless for VIPs), I couldn't find any level where it's strikingly useful.
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I happened to know, one and a half months ago, that we can zip through walls downwards from top-left-corner-clipping, even if the corner is filled with solid blocks. It had been thought (at least by me) that Mario should die in that way.
Here is a quick demo tested in SDW. It improves PSWa by like 9 seconds, so it's available in any% (or warps in recent terms). It's a pity that it looks much worse than the current strategy with two keys.
Link to video
Or from my TSRP2 WIP:
Link to video
As shown in the latter video, we can pass through a one-block-thick wall. We will be able to see its power in some puzzle hacks maybe.
I think it's also useful in VoB2b of SMW.
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You may have to change a RAM address, not ROM one. Change 7E0F30 (counter for timer) to 01, or change 7E0F31 (hundreds of timer) to 09, for example. Both worked for me.
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It's been a long while since the ladder glitch was found, and I'd been wondering if Masterjun would move on to this game. We really need the branch "Masterjun". Great work, yes vote. :D
PS, I'm also looking forward to seeing a normal run finished.
念のために書いておくと日本語タイトルは星のカービィスーパーデラックスですよ。
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RGamma wrote:
As x --> 0, f(x) --> e (because cos x --> 1). We get (g is > 0 and continuous on (0,infinity) so we can replace f(x) with e):
1/x^2 * abs(e - (a +b*x + c*x^2))
You missed something here. For instance, suppose that f(x):=x and g(x):=(f(x)-0)/x, and consider their limits at x --> 0. You see that you can't replace f(x) by 0 though f(x) --> 0, since g(x)=1 whenever x =/= 0.
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Warp wrote:
I know this isn't in any way a new idea, but just for the fun of it I made a small program that draws the so-called ulam spiral, but with composite numbers colored as well.
Each pixel is colored with a shade of gray depending on how many terms there are in its prime factorization. The shade of gray is 255/n, where n is the amount of prime factors. (Ie. it's 255 for primes, 127 for numbers with 2 factors, 85 for numbers with 3 factors and so on.)
In other words, the darker the pixel, the more prime factors it has.
(image)
Besides the traditional prime (ie. white) diagonal patterns, I find it interesting that there are some relatively wide diagonal dark areas. These seem to be diagonals where there are many numbers with a significant amount of prime factors.
Another interesting detail is that this coloring also emphasizes many horizontal and vertical lines with lots of primes in them (and which are not so evident in an image where only the primes are drawn.)
It's much easier to explain why composite numbers form vertical/horizontal (half) lines. For example, if we put numbers in an Ulam spiral as follows:
3 2 9
4 1 8
5 6 7
then we can find a half line starting from 18, 39, 68,.... This half line can be expressed in n as 4n^2+9n+5, which factorizes as (4n+5)(n+1). Hence every number x on this half line must have at least two factors. Moreover, n+1 is even if n is odd, so x often has at least three factors.
In general, the numbers on some half line can be expressed by a quadratic polynomial, and polynomials corresponding to dark lines tend to be decomposable.
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Thank you, BrunoVisnadi. So we spent a few days trying to optimize the yi4 stun bug, and haven't got faster yet than in the usual way. At the first visit to yi4, there are some crucial differences from BrunoVisnadi's video. The main problems are as follows.
We don't have a cape at yi4. This means that the setup of double-tongue glitch can't be so easy; we have to take a hit or so. In fact, we can get one in yi3 via the null-spit glitch on rotating platforms, but the loss would be too large to save time in total, I think.
In our situation, we have a red shell instead of a blue one, and it's very troublesome to deal with. For example, it is impossible to stun the shell to make a damage source.
To perform the stun bug there, we have to take yoshi to yi4, so the possible overworld route is yi2 -> yi3 -> yi4 -> yi1 -> c1; then we lose about 135 extra frames from backtracking.
The next run will be made on lsnes, which is more laggy (or laggier?) than snes9x. Therefore, it can happen that it is faster in snes9x while slower in lsnes.
It is challenging to optimize such a cool trick. I hope that it will get faster.
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Tub wrote:
@Scepheo: so your proof says that you can remove any number F(n) after you've already removed a number F(k) with k<n? We know that's not possible, so there must be a flaw in there.
I think this is the part where your proof goes poof:
A) we will never have used both of them, as F(n - 1) + F(n - 2) = F(n) and b - F(n) < F(n)
B) we can safely remove the other one, as per our initial assumption.
A) Yes, after removing F(k), k < n, you will still have a sum, and that sum has only used either of F(n-1) or F(n-2)
B) ...but now you're retroactively changing k to either n-1 or n-2! By doing so, you get a different sum for b, invalidating your assumption that the other number wasn't used!
If you remove F(n-1), your sum may contain F(n-2), and if you remove F(n-2), your sum may contain F(n-1). At no point are both used (so A still stands), but there's no proof that you can remove both at the same time and still represent any b you need.
Scepheo's statement in B) is that you can remove F(n-1) instead of F(n-2) (not both) if F(n-2) is removable, isn't it?
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FractalFusion wrote:
Mister wrote:
The existence of such a representation is not so difficult to imagine once you know any integer can be written as the sum of distinct Fibonacci numbers, as you can see that the densest sum F(2)+F(3)+F(4)+F(5)+F(6)+F(7)+... can change into F(4)+F(6)+F(8)+... as desired. The proof of the uniqueness, however, is a little formal since it often requires a non-constructive argument.
Proof of uniqueness can be done constructively.
Let's say n is an integer with F(m)≤n<F(m+1). Then F(m) must be a summand of n. Otherwise, n is at most:
● F(m-1)+F(m-3)+...+F(2), but n≥F(m)=F(m-1)+F(m-3)+...+F(2)+1, or
● F(m-1)+F(m-3)+...+F(3), but n≥F(m)=F(m-1)+F(m-3)+...+F(3)+1,
contradiction. (A summand of 1 is always considered to be F(2) for the purpose of consecutive Fibonacci number.)
Then n-F(m)<F(m-1). We can then repeat the argument on n-F(m) to write it uniquely as the sum of distinct Fibonacci numbers. Adding F(m) then gives a unique representation of n as the sum of distinct Fibonacci numbers.
(Using induction, of course.)
In fact, this representation is the same as those of the greedy algorithm posted above. ("Greedy algorithm" here means "take the largest Fibonacci number F(m) as summand and repeat for n-F(m).)
So you have proven two identities, which are summarized as a more general statement that the sum of a set of non-consecutive distinct Fibonacci numbers with F(n) largest must be strictly bounded by F(n+1). These only ensure that your "existence proof" process, which is constructive, gives a unique representation, but it doesn't mean that the proof of the uniqueness itself is constructive. This is what I wanted to mean in the above post; I might have confused you...
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The existence of such a representation is not so difficult to imagine once you know any integer can be written as the sum of distinct Fibonacci numbers, as you can see that the densest sum F(2)+F(3)+F(4)+F(5)+F(6)+F(7)+... can change into F(4)+F(6)+F(8)+... as desired. The proof of the uniqueness, however, is a little formal since it often requires a non-constructive argument. That's the way of mathematics. :) I'm looking forward to seeing how people prove it.
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It's been one and a half month since it got rejected. I asked maziari if he would work together then, and he answered no if I remember correctly. However, I don't know his final decision nor whether he's still working on it. He also said that he was going to post his WIP (and he hasn't yet). I guess he suffers from the language barrier. In any case, SMA2 doesn't have its own topic, so does anyone want to make one? It would be a more suitable place to discuss further the future run than here.
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