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Thanks for giving everyone unique names this time.
The new skip actually makes it better by reducing the number of bosses where you use Slots (still a lot of Slots usage though). In particular, the fight against Materia Keeper where you manipulate its HP to 7777 and poison it so it would take 7777 damage because of the All Lucky 7s status is ingenious.
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Here is an encode (don't forget to enable 720p60, Mega Man must be watched in 60fps):
Link to video
Rockman: Spirits of Hackers is a Mega Man 6 hack by Puresabe (the ROM hacker who did Rockman 4: Minus Infinity) and a number of others. The ROM hack can be downloaded here and you'll need to patch the JP version of Mega Man 6. (Note the checksum of the ROM I used doesn't match the information given at romhacking.net, but it matches the one in the movie file).
As posted here, Version 0.00 is the only version of this hack, and it is definitively final (no further updates will be made) despite its "INCOMPLETE" designation. So there should be no problem in submitting TASes for this hack with regards to which version should be used, will there be a newer version, etc.
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I mentioned the endings because it is possible to save time by killing one of the players in the last screen so they don't have to get to the rope to finish the game, but you would end with the 1-player ending. This is not a strat that I would use myself (because getting the 1-player ending in a 2-player run really does look like a bad ending) but it has been used before in this 2-player speedrun.
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p4wn3r wrote:
The game works like this. The dealer deals an amount N>=2 of cards face down in a straight line (the content of the cards doesn't matter). The player and the dealer take alternating turns, with the player starting. At any point, the person who's playing can either:
(a) Remove exactly two consecutive cards from any point in the interval.
(b) Remove a single card, provided that it's isolated from the rest of the pack.
The game goes on like this until it's someone's turn and there are no cards left. In that case, the person who has the turn loses.
I'll just cut straight to the answer since it would otherwise take a giant post to explain every little thing here:
This game is the octal game ".17". This OEIS sequence gives the nim values (or "Sprague-Grundy values") for this game; starting N=0:
0,1,1,0,2,1,3,0,1,1,3,2,2,3,4,1,5,3,2,2,3,1,1,0,3,1,2,0,1,1,4,4,2,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
...
The starting position of N cards is a win for the player (who has the first move) if and only if its nim value is nonzero. Since it is periodic with period 34, and only 4 values in each period are zero, assuming that there is no limit to how large N can be, this gives a winning probability of 30/34 = 15/17. So a fair payoff would be 2/15 a dollar for every dollar bet.
----
PS: Every position in an impartial game under normal play convention (last to play wins) can be assigned a nim value; if the nim value is n, there exist moves to positions with nim value 0, 1, 2, ..., n-1, but no moves to positions with nim value n. Hence positions can be thought of as heaps in a slightly-modified Nim game. Positions that can be divided into independent subpositions have a nim value equal to the nim sum (binary XOR sum) of the nim values of all its subpositions.
The nim value of a position can be calculated as: a position where no move is possible has nim value 0, and for all other positions, the nim value is the smallest non-negative integer not appearing as the nim value of any position it can move to. Using this, it is possible to calculate the nim values for the octal game ".17", and it turns out to be periodic with period 34 and just 4 exceptional values; this can be verified computationally.
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Some context:
Cowlitz Gamers 2nd Adventure is a free NES homebrew game created in 2017 for Cowlitz Gamers for Kids Expo. It is a platformer where you collect coins to open up the path upward; there are 48 screens (not counting bonus screens) in this game. You can download the ROM from the game's main site.
The only difference between difficulty levels is the number of lives you start out with and how many coins you need to get more lives.
It is faster to complete the game with 2 players since they can both collect the coins necessary to open the path to the next screen. Deathwarps are abused a lot in this TAS. There are a couple endings depending on whether you finished with both players alive or with just one player.
Note that there are collectible gems in each screen which allow you to access hidden bonus screens. Collecting all the gems (and coins) is a goal I would like to see as well.
I uploaded an encode (will take some time before it is ready):
Link to video
Note that the ending music is a few minutes long before it ends. A button press is necessary to get to the "THE END" screen with all the stats; I included it in the encode.
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p4wn3r wrote:
Prove or disprove:
Let a, b, c be real numbers. Suppose that l is a real number satisfying l > max(|a-b|,|b-c|,|c-a|) and l < min(a+b,b+c,c+a), so that it's possible to form a triangle having one of the sides measuring l, and any two numbers from a, b and c. Then, in these conditions, it's always possible to construct a triangular pyramid having an equilateral triangle of side l as basis, and having a, b and c as the lengths of the remaining edges.
If false, try to come up with a condition that allows the pyramid with equilateral basis to be constructed.
Ptolemy's inequality on the tetrahedron (=triangular pyramid) with equilateral triangle base of side l and other edges a,b,c gives al+bl>cl, which gives a+b>c. Note that it is equality only if the four points lie on a circle in a plane, which is not the case here. Likewise b+c>a and c+a>b. If any one of these inequalities does not hold, then there exists no such tetrahedron. E.g. For a=1, b=1, c=2.5, l=1.75, it is not possible, even though the triangle inequality on all faces is satisfied.
Note that a+b>c, b+c>a, c+a>b is still not sufficient; for example, as explained in the posts above, l=1, a=b=c=0.55 still does not give a tetrahedron even though all necessary conditions given so far are satisfied.
Searching online reveals a paper giving a necessary and sufficient condition: Six given edge lengths determine a tetrahedron iff the triangle inequality holds for all faces and its Cayley-Menger determinant is greater than 0. Using the last formula on the Wikipedia page and substituting A=a2, B=b2, C=c2, L=l2 gives:
4ABC+(A+B-L)(B+C-L)(A+C-L)-A(B+C-L)2-B(A+C-L)2-C(A+B-L)2>0
or, after expanding and writing as a polynomial in L:
-L3+(A+B+C)L2+(AB+AC+BC-A2-B2-C2)L>0
Since L must be positive, divide by -L to get:
L2-(A+B+C)L+(A2+B2+C2-AB-AC-BC)<0
Now A2+B2+C2-AB-AC-BC=(1/2)((A-B)2+(B-C)2+(A-C)2)>0, and A+B+C is positive, so solving x2-(A+B+C)x+(AB+AC+BC-A2-B2-C2)=0 gives us positive solutions for x. We get:
x=(1/2) (A+B+C ± sqrt(3)sqrt(2AB+2AC+2BC-A2-B2-C2))
But note that 2AB+2AC+2BC-A2-B2-C2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c). And so this gives us:
x=(1/2) (a2+b2+c2 ± sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
Thus:
(1/2) (a2+b2+c2 - sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c))) < L < (1/2) (a2+b2+c2 + sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
and since this interval is in the positive reals, this gives for l:
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
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In case anyone wants an encode of my 1:46 TAS, I made one just now:
Link to video
(I didn't make one before because I never expected the sudden interest in Pitfall.)
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p4wn3r wrote:
Prove that:
First, note the triple angle identity
which we can rewrite as
Using the identity on itself multiple times, we get
As k goes to infinity, we get
The left hand side is of the form theta*sin(x)/x (where x=theta/3k) as x goes to 0, for which the limit is theta. Then plugging in theta=pi gives
and dividing by 4 gives you the identity you posted.
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PJ wrote:
The biggest change is the cutscene after fighting Mash in the Robot Museum. You can hold start to advance the text during that cutscene; you let it play out in the WIP.
Thanks! That explains why I wasn't gaining any time over the 100% TAS. Now I feel better about it.
Note that there are some things I did in the WIP which are actually suboptimal (I think one or two Rush Coil instances are like that), so it isn't completely well done. I used my WIP mainly to try out different strategies.
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I did it incrementally. And similar to what p4wn3r did, you don't ever need to leave the world of symmetric polynomials.
● (a+b+c)2 = a2+b2+c2 + 2(ab+ac+bc) → ab+ac+bc = (12-2)/2 = -1/2
● (a2+b2+c2)(a+b+c) = a3+b3+c3 + a2b+ab2+a2c+ac2+b2c+bc2 → a2b+ab2+a2c+ac2+b2c+bc2 = 2(1) - 3 = -1
● (ab+ac+bc)(a+b+c) = a2b+ab2+a2c+ac2+b2c+bc2 + 3abc → abc = ((-1/2)(1)-(-1))/3 = 1/6
● (ab+ac+bc)2 = a2b2+a2c2+b2c2 + 2a2bc+2ab2c+2abc2 → a2b2+a2c2+b2c2 = (-1/2)2 - 2abc(a+b+c) = 1/4 - 2(1/6)(1) = -1/12
● (a2+b2+c2)2 = a4+b4+c4 + 2(a2b2+a2c2+b2c2) → a4+b4+c4 = 22 - 2(-1/12) = 25/6.
I imagine you would get higher symmetric polynomials using this method.
----
Actually, there is a method to solve this using a recursive formula that gives you all higher powers an+bn+cn for n≥4. If we define f(n)=an+bn+cn, then f(n) is the solution to a recursive formula with f(1)=1, f(2)=2, f(3)=3 and whose characteristic polynomial is:
(x-a)(x-b)(x-c) = x3 - (a+b+c)x2 + (ab+bc+ac)x - abc = x3 - x2 - (1/2)x - (1/6),
that is, f(n) = f(n-1) + (1/2)f(n-2) + (1/6)f(n-3). Plugging in gives f(4) = f(3)+(1/2)f(2)+(1/6)f(1) = 25/6, and you can use this formula to get the higher powers: f(5)=6, f(6)=103/12≈8.5833, and so on.
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Riddler Express this week sounds interesting:
You purchase a new clock but are dismayed to realize that both of its hands are identical. At first, it seems it’s going to be impossible to tell the time because you don’t know which hand is for the minutes and which is for the hours.
However, you realize you don’t need to know which is which for every time — for example, when it’s 12:30, the minute hand will be exactly on the 6 and the hour hand will be halfway between the 12 and the 1. It can’t be the other way around because if the hour hand were exactly on 6, the minute hand would have to exactly on 12, which it’s not. So you know what time it is.
How many times during the day will you not be able to tell the time?
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Mar. 2019 ranking in detail.
Runs from this site that made it on there: Magical Quest 3 starring Mickey & Donald, Tetris TGM2, Super Mario All Stars: Super Mario Bros. 3, and Indiana Jones and the Last Crusade (UBI Soft).
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Oh, you actually did SoftLock%. Cool. I looked at it briefly after seeing Shenanagans do it but then I eventually forgot about it.
While testing, I made a TAS on the Japanese Red version (Pokemon Red JP V1.0) that apparently completes it 1:20. It seems as though mine is faster (just from comparing the gameplay portion of it) but I don't run my stuff off GBC firmware. I don't have time to figure out which one is legitimately faster, but I'll note that it is possible to start the Prof. Oak scene before the girl gets there, because the girl can still move to that square before your character bonks her. If you would like an encode, just ask me for it.
The other thing is that I wish you had chosen a day of the year other than this one to submit this TAS. I hate this day.
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Hi PJ. Haven't seen you around for a while.
I already found out that glitch two years ago (although I never made it public). I'll tell you what I know so far.
- I came to the same conclusion as you, that the Wily Machine can be quick killed with this method (but not Wily Capsule, since it doesn't have a hitbox).
- I disassembled how Rush Search works. Rush Search has three random components. One is the item to be dug up, one is the first Rush delay (wag-tail animation) and the last one is the second Rush delay (digging animation). Note the last one does not apply when digging up important items. There is a 3/32, 3/32, 1/32, and 1/32 chance of digging up small life energy, small weapon energy, large life energy, and large weapon energy, respectively. So the chances of pulling this trick off for the Wily Machine is only 1/4.
- Using it on the Cloud Proto visit saves time, if it works. It is also possible to use on the Turbo Proto visit but is difficult to do and unnecessary because there is a full Turbo Proto visit skip which is much faster and isn't random (you have to enter the room by jumping and then exit the room while touching the ground at the same time).
- As you already found, you can use it to get the Proto Shield while Proto Man is talking. With Shade Man already done you can then leave the level. According to my calculations, it is now faster in a TAS to do Shade Man first, quick kill Turbo Man then revisit Shade Man's stage for the Proto Shield.
- For the Bass scene in Shade Man's stage, if Rush Search has very small delay and digs up a small life/weapon energy, you can save a small amount of time here. In 100% you get a shot at this for free because you need Rush Search three rooms before that for the Energy Balancer.
- I couldn't find other places where Rush Search worked.
The Rush Search before the Proto door in Shade Man is something I never found before though, so thanks for that (by the way, you need to hold left while going through the door for it to work). It definitely saves time in a TAS. However this trick doesn't seem to work for any other door.
I've looked at 100% for a while because there are a lot of possible improvements (and not just because of Rush Search glitches). Here is the test run I made two years ago (requires snes9x 1.51 v7); don't know whether it helps or not.
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I watched both this and the Ultimortal 1HP version. While I don't particularly admire this game, I do appreciate that you took the time to do these modes.
That being said, I prefer this TAS for having more variety and tricks (such as damage abuse / flying across the screen) as well as having shorter boss fights.
It's too bad Iji 1.7 is not TASable though. Stuff like this is why I gave up on the notion of Windows TASing being a real thing.
Is this the first time these collections have been in 60fps? I dont think I ever saw it as such until now.
Nicovideo has allowed 60fps uploads for a long time. However, if you access the site when there is a lot of traffic, it may play the 30fps 360p "economy" encode instead of the normal one.
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Masterjun wrote:
But what about the major features of SMW already existing in the game? The only times powerups were used, their purpose was to get hit instantly and give Mario a few invincibility frames. Yoshi was also only used for a few seconds as an extra jumping platform.
For me, it didn't really feel like a game to TAS. The levels were very linear, and a lot of them were on some kind of global timer (either an autoscroller, or waiting for a platform to reach you), making all optimizations useless in the end.
This is typical of Kaizo Mario hacks, which are extremely difficult and aimed at the best SMW players. All the things you listed above are characteristic of Kaizo hacks. Not excusing them, just saying how it tends to turn out.
Additionally, I'm pretty sure wall-jumping was disabled because it was a Kaizo hack. In a Mario hack more suitable for casual play, there would be no reason to disable wall-jumping. So that's another thing against it.
Taralyn wrote:
but I worry that GPW2 or whatever comes down the pipe in the near future will obsolete this extremely quickly.
There is already a TAS of Grand Poo World 2 on Youtube (I don't know how optimized it is). That being said, it might not be submitted here.
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p4wn3r wrote:
Set u = df/dx. Apply the chain-rule and get f'' = du/dx = du/df * df/dx = u*du/df
Now plug everything:
uf = u du/df => f df = du => f² + c = 2u
Actually since f*f' is just (f2/2)', you can directly write:
(f2/2)' = f''
f2/2 + C = f'
f2 + c = 2f'
and so on. The case where u=0 just means that f is constant; that's the case when f2 + c = 0 (that gives f=sqrt(-c)).
p4wn3r wrote:
So, the entire set of solutions should be:
y = -2 * ln(cos(bx/2+k)) + a
y = -2 * ln(cosh(bx/2+k)) + a
y = -2*ln |x+k| + a
y = ax + b
I saw the blackpenredpen video, and one of the comments describes a missing solution.
It turns out that -arctanh(x/b)/b is not the only function (up to adding a constant) whose derivative is 1/(x2-b2). -arccoth(x/b)/b is another such function. This is because the integral of 1/(x2-b2) is -arctanh(x/b)/b but only when -b<x<b. When |x|>b, the integral is -arccoth(x/b)/b. This should give an additional solution of y = -2 * ln|sinh(bx/2+k)| + a.
(Technically the ln functions should all have absolute value signs, though it doesn't matter for ln(cos) (since -cos is just a shifted cos) and ln(cosh) (since cosh>=1).)
Edit: Also, we're assuming that y is continuous and three times differentiable on its domain. Otherwise, y can just be various pieces of the above solutions.
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In the past, I would watch MP4/AVI downloadable files, as at the time they had the highest quality. Nowadays, Youtube encodes are of high quality (thanks to the TASVideos encoders), so I watch it there instead.
Sometimes, if I expect to rewatch or jump around a TAS (especially if it is a long one), I would download the Youtube video using youtube-dl. With a batch file, it is easily customizable so I can choose which resolution to download, etc.
I used to watch by movie file in emulator 10 years ago when it was popular to do so, but I don't do that anymore because everything since then has changed towards making it harder to use emulators for watching TAS replays and at the same time making it easier to watch a video. I only download movie files if I actually need to use emulator features to analyze the game/TAS.
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Spikestuff wrote:
but for some reason you never make encodes anymore so I got to watch it through.
Here's a hint: No one says you have to watch it, and no one says you have to encode it if the submitter won't*. If you don't find any fun in doing this, save your time and do something that is actually fun.
*It's 2019 already; anyone who puts in all the effort to create their TAS should be expected to know how to encode and upload videos of their own TASes.
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Is it only non-tetris all clears that lose frames, or does every non-tetris line clear lose frames? You have a number of line clears that aren't tetrises in this one, especially when doing the CO medal.
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
which we can rewrite as
Using the identity on itself multiple times, we get
As k goes to infinity, we get
The left hand side is of the form theta*sin(x)/x (where x=theta/3k) as x goes to 0, for which the limit is theta. Then plugging in theta=pi gives
and dividing by 4 gives you the identity you posted.