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Yes. It works more like the pijack movie than bortreb's item-turned-program movie. It requires some DV manipulation and box renaming, but no items forming program code.
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I'm currently looking into Pokemon Gold (Coin Case glitch). After a bit of research, I am almost 100% certain that arbitrary execution of code from the Coin Case glitch can be triggered right away when entering Goldenrod City and obtaining the Coin Case as soon as possible. Messing around with items is not necessary.
Edit: It requires a few manipulations, but I believe it would not take more than a minute of extra time to pull off.
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http://forums.glitchcity.info/index.php/topic,6716.msg191372.html#msg191372
MrWint indicates in his post that Bellsprout's cry serves the same purpose as Machop's cry (indeed, they leave the same string in memory). Bellsprout is also far quicker to encounter, since it is a random encounter on Route 31 (the route near the beginning just prior to Violet City).
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I watched this run now. Personally, I feel that the TAS is only rather OK. As in, this run is good when compared to a non-Sonic TAS, but there are better Sonic TASes out there.
As for screenshots, I like 5.
Edit: Speaking in terms of entertainment, not tech quality.
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Radiant wrote:
Since the site is now steadily getting more runs in some of the newer categories, on this index page, I think it would be helpful if the line for "computer" was split into three lines for DOS, MSX, and Windows separately (they are three separate categories/tags in the database, after all); and several more categories now have enough movies that it makes sense to link to the stars and moons only, and have a second button for 'all' movies (the way is done for NES/SNES movies as well), e.g. the MSX, Arcade, and DOS categories.
Thank you for your time!
I have now split the computer category into DOS, MSX, and Windows separately.
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Please note that the May 19 thing (which actually requires setting your operating system's clock) was a glitch in VBA-rr 17 and an error on my part. Stuff like this does not happen with more recent versions of VBA-rr.
There are other things that may influence the TAS, such as DS Firmware or emulated RTC. As CoolKirby stated, savestates handle these issues so that there isn't supposed to be a problem.
The input files should contain any information that is important for sync.
As for the second question, the input files for VBA (vbm) do not allow changing the RTC in the middle of the movie. It may be possible to change the RTC using Lua scripting, but no one has tried yet, and it breaks the rules anyway.
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I tend to find modern fighting games in general to be confusing, as things simply fly everywhere at high speed and it is extremely hard for a casual watcher to follow.
That being said, I like this TAS. It is clearly a playaround TAS on an insanely popular game aiming for entertainment above all else. For that, I throw my vote into the ring like everyone else has done.
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Ingredients:
- 1 Shanoa
- 10 cups Metroidvania
- 1 cup vandalism
- 8 cups boss destruction
- 16 cups floor shuffling
- 1 giant enemy crab
- 1 cup switching
- 2 cups superfast mode
- 1 Dracula
- 3 tbsp ドゥエ
- 1 tbsp ホァイ
Directions:
Mix all the ingredients together using 57103 rerecords. Keep mixing until NKT. Let it sit until IGAAAAAAAAA.
Result:
Yet another awesome Castlevania TAS.
I watched it twice (the Nicovideo encode and cwizkim's encode), and I was impressed. I would vote Very Yes if there was such an option.
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Any time mtbRc makes a submission here, that always means that he uploaded an encode to Nicovideo previously.
Link to videoLink to videoLink to video
Not sure why the time given in the Nicovideo description is different from here, but oh well. It's close enough.
Edit: Strangely the Nicovideo encode uses Japanese voices, even though it is still the English version and the TAS is the same. Anyway, encode with English voices is in the post below this one.
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feos, how long would it take for the encodes to be ready? I have a temp encode ready now, but I will hold off if your encodes are ready in a short time.
By the way, Youtube encodes require TASBlend/ng_deblink (also, ng_deblink fails with the spotlight effects in this TAS, I tested it).
私はKPMさんの言うことが意味分からなくてすみません。
どのウェブサイトや文章が分かりにくいですか？ accept(受理)されたエミュレータ表を頼みますか？
I'm sorry. I don't understand what you are saying.
Which website or part of text is hard to understand? Are you asking for a list of which systems are allowed?
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I agree with Mothrayas.
I think [725] SNES Mega Man X by FractalFusion in 29:57.88 should be default (no branch name) and [2059] SNES Mega Man X "password glitch" by FractalFusion in 16:56.88 should have something like "password glitch".
I also changed branch name of [980] NES Snow Bros. "2 players" by Randil in 13:56.85 to "2 players" and moved it to Moon tier.
Edit: IIRC, MrWint's first TAS used item list glitch and overflow (buffer overflow?), like p4wn3r's TAS before that. His second TAS used Transform glitch to execute arbitrary data. I don't know if it is accurately described as buffer overflow. Possibly, since the description of the "zero move" in the Transform glitch involves using data from unintended areas of memory.
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arflech wrote:
The challenge is to describe the set of cycles of this LFSR as the number of bits increases; based on reading that Wikipedia article, I suspect it has to do with the factorization of xn+xn-1+1, and I've seen from Wolfram Alpha that this polynomial is reducible over the Reals (and therefore over GF(2)) when n is 5, 8, 11, or 14, and additionally is reducible over GF(2) (but not the Reals) when n is 10, 12, 13 or 16.
It has to do partially with the factorization of xn+xn-1+1 (mod 2) (or equivalently, factorization of the reverse xn+x+1 (mod 2)), but it is more complex than just factorization. I can't explain everything in complete detail, but this is how it goes (everything below is in mod 2 land):
- Every LFSR is equivalent to a Galois LFSR. Example for x16+x14+x13+x11+1 (credit Wikipedia):
- The internal state of the Galois LFSR above (lower image) can be represented by the polynomial given by sum(ith bit is 1) xi, where the leftmost bit is bit 0 and bit position counts left to right. Example above: 1+x2+x4+x5+x8+x9+x10+x15.
- Cycling the Galois LFSR is as simple as multiplying the polynomial internal state by x and taking the remainder upon division by the reverse of the LFSR's polynomial 1+x2+x3+x5+x16.
- Let f(x) be the reverse of the LFSR's polynomial and p(x) be the initial Galois LFSR state. Then the question is: what is the smallest positive n such that f(x) divides p(x)(xn-1)?
- Factor f(x) into irreducible factors f1(x)f2(x)...fk(x). Then for each irreducible factor that does not divide p(x), the question is: what is the smallest positive ni such that fi(x) divides xni-1?. If f(x) itself is irreducible, then it cannot divide p(x) (since deg f > deg p) unless p(x) is the zero polynomial.
- Even if f(x) is irreducible, that doesn't necessarily mean that the LFSR has period 2deg f-1 (the maximum) for nonzero initial values. In order for that to happen, the smallest positive n for which f(x) divides xn-1 must be 2deg f-1. In this case, f(x) is a primitive polynomial (field theory definition of primitive, not gcd definition).
- Note that every polynomial with nonzero constant term is a divisor of xn-1 for some n. In particular, an irreducible polynomial g(x), other than x, is a divisor of x^(2deg g-1)-1. This follows from field theory but I won't go into it here.
At this point, it is easier just to do small examples. The polynomials of interest are the reversals of xn+xn-1+1, that is, xn+x+1.
n=5:
x5+x+1=(x2+x+1)(x3+x2+1).
Note that both x2+x+1 and x3+x2+1 are primitive polynomials.
Let p(x) be the polynomial for the initial Galois LFSR state. Let g(p(x))=(u(x),v(x)) where u(x) is remainder of p(x) divided by x2+x+1 and v(x) is remainder of p(x) divided by x3+x2+1 (this is unique by the Chinese Remainder Theorem).
If g(p(x))=(0,0) (p(x) is the zero polynomial), then the cycle has length 1.
If g(p(x))=(u(x),0) for nonzero u(x), then the cycle has length 22-1=3 (since x3+x2+1 divides p(x), its only contribution is from x2+x+1)
If g(p(x))=(0,v(x)) for nonzero v(x), then the cycle has length 23-1=7.
If g(p(x))=(u(x),v(x)) for nonzero u(x) and v(x), then the cycle has length gcd(3,7)=21; this covers all possible values as well.
n=8:
x8+x+1=(x2+x+1)(x6+x5+x3+x2+1).
Note that both x2+x+1 and x6+x5+x3+x2+1 are primitive polynomials.
Let p(x) be the polynomial for the initial Galois LFSR state. Let g(p(x))=(u(x),v(x)) where u(x) is remainder of p(x) divided by x2+x+1 and v(x) is remainder of p(x) divided by x6+x5+x3+x2+1.
If g(p(x))=(0,0) (p(x) is the zero polynomial), then the cycle has length 1.
If g(p(x))=(u(x),0) for nonzero u(x), then the cycle has length 22-1=3.
If g(p(x))=(0,v(x)) for nonzero v(x), then the cycle has length 26-1=63.
If g(p(x))=(u(x),v(x)) for nonzero u(x) and v(x), then the cycle has length gcd(3,63)=63. For a given v(x), there are 3 possible values for u(x) in order to cover all possible values; thus, this gives 3 classes of length 63, for a total of 4.
n=9:
x9+x+1 is irreducible.
Now since the smallest positive n such that x9+x+1 divides xn-1 is n=73 (instead of n=29-1=511), x9+x+1 is not primitive. In fact, if p(x) is the initial Galois LFSR state, and p(x) is not the zero polynomial, then the cycle has length 73. There are 7 such cycles. Of course, if p(x)=0, then the cycle has length 1.
Now, it's pretty hard to characterize the factorizations of xn+x+1, even if polynomial factorization can be done quickly (faster than integer factorization for sure). The only other thing I can say is that x2+x+1 is always a divisor of xn+x+1 when n=2 (mod 3), and xn+x+1 is always square-free, since the gcd of xn+x+1 and its derivative is always 1. That's about it.
Here are full factorizations of xn+x+1 (mod 2) from n=2 to 20. The symbol * after a factor means that it is not a primitive polynomial; in this case the smallest m for which it divides xm-1 is given.
x2+x+1
x3+x+1
x4+x+1
x5+x+1=(x2+x+1)(x3+x2+1)
x6+x+1
x7+x+1
x8+x+1=(x2+x+1)(x6+x5+x3+x2+1)
x9+x+1*
x10+x+1=(x3+x+1)(x7+x5+x4+x3+1)
x11+x+1=(x2+x+1)(x9+x8+x6+x5+x3+x2+1)
x12+x+1=(x3+x2+1)(x4+x3+1)(x5+x3+x2+x+1)
x13+x+1=(x5+x4+x3+x+1)(x8+x7+x5+x3+1)
x14+x+1=(x2+x+1)(x5+x3+1)(x7+x6+x5+x2+1)
x15+x+1
x16+x+1=(x8+x6+x5+x3+1)(x8+x6+x5+x4+x3+x+1)*
x17+x+1=(x2+x+1)(x3+x+1)(x12+x11+x10+x9+x8+x6+x4+x+1)*
x18+x+1=(x5+x2+1)(x13+x10+x8+x7+x4+x3+x2+x+1)
x19+x+1=(x3+x2+1)(x4+x+1)(x5+x4+x2+x+1)(x7+x5+x4+x3+x2+x+1)
x20+x+1=(x2+x+1)(x5+x4+x3+x2+1)(x13+x11+x10+x9+x7+x4+1)
x9+x+1: 73
x8+x6+x5+x4+x3+x+1: 85
x12+x11+x10+x9+x8+x6+x4+x+1: 273
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thatguy wrote:
are attention spans really so short nowadays that people can get bored in 25 seconds?
Assuming that one must "get bored" in order to find it not entertaining. Perhaps these people were never entertained to begin with.
Anyway, I voted no, because I did not find this movie entertaining. That being said, I like that someone improved this silly game and welcome its publication according to the rules.
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1555M wrote:
The (J) version of FFVI also has an equipment glitch which is not present in the (U) version. Otherwise, this run of FFVI is quite similar to our single-game FFVI movie.
Whichever one is more technically well-done can only be determined by watching both movies. I can't say because I didn't watch either of them.
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It seems like the consensus is to remove the "glitched" branch from a TAS which is the fastest/only movie (which I agree with). I will remove those labels now.
Edit: Couldn't figure out a good branch name for http://tasvideos.org/950M.html , so I just called it "less glitched" for now.
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maul91 wrote:
"all you have to do is make a savestate where you leave off. Then when you wake up in the morning and open up VBA and the game, just press Shift+8 (or whatever combination you set), load your savestate, and resume working."
I've done this - but it does not start recording when this happens it says (no movie).
After opening VBA and the game, go to Tools -> Movie -> Play Movie... and select the VBM that you have recorded from the previous session. You must do this or else it will not record.
After you have done this, then you can press Shift+8 so it says "Read+Write" (if it says "Read-Only", press Shift+8 again) and then load the savestate and resume working.
maul91 wrote:
I know I have to record with fraps for example
Please don't use fraps. VBA and most other emulators have built-in tools to dump AVI, and it doesn't skip frames.
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Mega Kick and Mega Punch for power, Pound to clean up weak Pokemon, Water Pulse for Rock and Ghost types, and maybe one other strong non-Normal-typed attack (Fire Blast, Blizzard) or Metronome.
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This TAS is funny, even if the game is rather simplistic. The voices progressively go crazy as the TAS speeds through the courses. This game is also weird.
That, and because of Japanese interests, is probably why this video has 175K views on Nicovideo. (The average uploaded encode of a completed TAS of a game, I think, gets about 15K views on Nicovideo.)
Glitcher wrote:
racing game with a ball
No, it is a racing game with a rod. (Even though it might as well be a ball from looking at the gameplay.)
- The internal state of the Galois LFSR above (lower image) can be represented by the polynomial given by sum(ith bit is 1) xi, where the leftmost bit is bit 0 and bit position counts left to right. Example above: 1+x2+x4+x5+x8+x9+x10+x15.
- Cycling the Galois LFSR is as simple as multiplying the polynomial internal state by x and taking the remainder upon division by the reverse of the LFSR's polynomial 1+x2+x3+x5+x16.
- Let f(x) be the reverse of the LFSR's polynomial and p(x) be the initial Galois LFSR state. Then the question is: what is the smallest positive n such that f(x) divides p(x)(xn-1)?
- Factor f(x) into irreducible factors f1(x)f2(x)...fk(x). Then for each irreducible factor that does not divide p(x), the question is: what is the smallest positive ni such that fi(x) divides xni-1?. If f(x) itself is irreducible, then it cannot divide p(x) (since deg f > deg p) unless p(x) is the zero polynomial.
- Even if f(x) is irreducible, that doesn't necessarily mean that the LFSR has period 2deg f-1 (the maximum) for nonzero initial values. In order for that to happen, the smallest positive n for which f(x) divides xn-1 must be 2deg f-1. In this case, f(x) is a primitive polynomial (field theory definition of primitive, not gcd definition).
- Note that every polynomial with nonzero constant term is a divisor of xn-1 for some n. In particular, an irreducible polynomial g(x), other than x, is a divisor of x^(2deg g-1)-1. This follows from field theory but I won't go into it here.
At this point, it is easier just to do small examples. The polynomials of interest are the reversals of xn+xn-1+1, that is, xn+x+1.
n=5:
x5+x+1=(x2+x+1)(x3+x2+1).
Note that both x2+x+1 and x3+x2+1 are primitive polynomials.
Let p(x) be the polynomial for the initial Galois LFSR state. Let g(p(x))=(u(x),v(x)) where u(x) is remainder of p(x) divided by x2+x+1 and v(x) is remainder of p(x) divided by x3+x2+1 (this is unique by the Chinese Remainder Theorem).
If g(p(x))=(0,0) (p(x) is the zero polynomial), then the cycle has length 1.
If g(p(x))=(u(x),0) for nonzero u(x), then the cycle has length 22-1=3 (since x3+x2+1 divides p(x), its only contribution is from x2+x+1)
If g(p(x))=(0,v(x)) for nonzero v(x), then the cycle has length 23-1=7.
If g(p(x))=(u(x),v(x)) for nonzero u(x) and v(x), then the cycle has length gcd(3,7)=21; this covers all possible values as well.
n=8:
x8+x+1=(x2+x+1)(x6+x5+x3+x2+1).
Note that both x2+x+1 and x6+x5+x3+x2+1 are primitive polynomials.
Let p(x) be the polynomial for the initial Galois LFSR state. Let g(p(x))=(u(x),v(x)) where u(x) is remainder of p(x) divided by x2+x+1 and v(x) is remainder of p(x) divided by x6+x5+x3+x2+1.
If g(p(x))=(0,0) (p(x) is the zero polynomial), then the cycle has length 1.
If g(p(x))=(u(x),0) for nonzero u(x), then the cycle has length 22-1=3.
If g(p(x))=(0,v(x)) for nonzero v(x), then the cycle has length 26-1=63.
If g(p(x))=(u(x),v(x)) for nonzero u(x) and v(x), then the cycle has length gcd(3,63)=63. For a given v(x), there are 3 possible values for u(x) in order to cover all possible values; thus, this gives 3 classes of length 63, for a total of 4.
n=9:
x9+x+1 is irreducible.
Now since the smallest positive n such that x9+x+1 divides xn-1 is n=73 (instead of n=29-1=511), x9+x+1 is not primitive. In fact, if p(x) is the initial Galois LFSR state, and p(x) is not the zero polynomial, then the cycle has length 73. There are 7 such cycles. Of course, if p(x)=0, then the cycle has length 1.
Now, it's pretty hard to characterize the factorizations of xn+x+1, even if polynomial factorization can be done quickly (faster than integer factorization for sure). The only other thing I can say is that x2+x+1 is always a divisor of xn+x+1 when n=2 (mod 3), and xn+x+1 is always square-free, since the gcd of xn+x+1 and its derivative is always 1. That's about it.
Here are full factorizations of xn+x+1 (mod 2) from n=2 to 20. The symbol * after a factor means that it is not a primitive polynomial; in this case the smallest m for which it divides xm-1 is given.
x2+x+1
x3+x+1
x4+x+1
x5+x+1=(x2+x+1)(x3+x2+1)
x6+x+1
x7+x+1
x8+x+1=(x2+x+1)(x6+x5+x3+x2+1)
x9+x+1*
x10+x+1=(x3+x+1)(x7+x5+x4+x3+1)
x11+x+1=(x2+x+1)(x9+x8+x6+x5+x3+x2+1)
x12+x+1=(x3+x2+1)(x4+x3+1)(x5+x3+x2+x+1)
x13+x+1=(x5+x4+x3+x+1)(x8+x7+x5+x3+1)
x14+x+1=(x2+x+1)(x5+x3+1)(x7+x6+x5+x2+1)
x15+x+1
x16+x+1=(x8+x6+x5+x3+1)(x8+x6+x5+x4+x3+x+1)*
x17+x+1=(x2+x+1)(x3+x+1)(x12+x11+x10+x9+x8+x6+x4+x+1)*
x18+x+1=(x5+x2+1)(x13+x10+x8+x7+x4+x3+x2+x+1)
x19+x+1=(x3+x2+1)(x4+x+1)(x5+x4+x2+x+1)(x7+x5+x4+x3+x2+x+1)
x20+x+1=(x2+x+1)(x5+x4+x3+x2+1)(x13+x11+x10+x9+x7+x4+1)
x9+x+1: 73
x8+x6+x5+x4+x3+x+1: 85
x12+x11+x10+x9+x8+x6+x4+x+1: 273