Posts for flagitious

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Posted: 7/20/2006 6:47 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
FractalFusion, you should submit that 8 second win and have the shortest TAS video ever ;)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/19/2006 6:59 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hmm I am having a hard time even coming up with a way to make 7 work.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/18/2006 9:10 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
AzHP wrote:
Is there a value in memory that changes when a boost is ready? That would make things real easy. =)
boost counter 7e10ca 1u (need 128)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/15/2006 2:50 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Tub, the generic strategy would be nice to hear, the optimal is one less than that though. Omni, do the cylinders all have to be the same height and diameter?
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/14/2006 2:46 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
OmnipotentEntity wrote:
Old puzzle, new twist: You work for the Secret Service, but are not particularly well respected. You were given sixteen coins, and told that two of them are counterfeit. The counterfeit coins either weigh less than or more than the other fourteen coins, and they both weigh the same amount. You also have a scale. Except this scale is triangle shaped and has three pans. (No one ever gives you the good equipment.) Unfortunately, the Secret Service is beholden to the Scale Industry Corporate Interests, and are forcing the Secret Service to pay them every time a scale is used by anyone. How many weighings are required to determine which two coins are false?
Revival an ancient topic time. I just read another variant of this type of problem, you have a regular scale (2 sides). How and how many weighings does it take to find the 1 odd ball from a group of 12 balls? How about 120 balls? Note the odd ball may be heavier or lighter than the others. I haven't worked on them yet but the 120 sounds very challenging. (Both have been solved already, without the use of a computer as aide.) When I read the problem it gave the answer of how many weighings and just asked for how it is done, but I think the problem is better if that part is unknown too.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/14/2006 2:35 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
nah, i guess i would suggest something as short as possible like, the pink panther (82 mins). only if a scene is unescapable and impossible to be shorter should they really focus on making it entertaining
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/14/2006 1:11 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
personally i think speed should be first priority
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/12/2006 6:24 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
It is not really like wall jumping, because wall jumping is an accident, a glitch. Whereas boosts are an intential feature.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/12/2006 12:52 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
About the bumping, my explanation explains that too. The 2 versions of it correspond to the values -1 and 1. Both are glitched but they are offset it different directions only 1 of which happens to work. However this would mean that each has an equal likely chance of happening unless for whatever reason the amount of time allowed to set one of the values is longer during the sword spin than the other. That being said, I do not really totally understand how it works either, I have never messed with it on emulator. However I am certain it can be done in 1 bounce, and I suspect it does not need to be done bouncing between 2 bumpers and could be done using just one, such as in the 3rd pendant dungeon.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/12/2006 12:44 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Nitsuja, I'm just curious, how does the save state contain the movie file without being nearly as large as the movie file? Huffers: even if its not as fast as the WR I'd still be curious what your time was, to compare it to the non-NBT record. And again even if its not as fast as the non-NBT record its still quite an accomplishment considering you automated the whole thing and these records are close to perfect with 12+ years of people working on them.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/11/2006 2:40 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Ok I just whatched your demo run. And here's my take. The business with disallowing pressing A 30 times a second is not arbitrary. Your goals are clearly defined. Beat the game as fast as humanly possible on original hardware. This is a fine goal, but this site is just not about what is humanly possible, so I don't think it could ever be published as a normal run, but as a concept video or something, sure. Since your goal is to do it as fast as humanly possible, maybe once you master and understand all the techniques you could actually do it on console and record it and send it in to SDA or something, that seems more fitting. Several years ago I have beaten the game in pretty much this same way on console, it took me more like 40 minutes though. Opinion stuff aside here are some game play related things: You mentioned skipping the lantern, but I don't think this can be done without UP+DOWN, it won't let you push that thing with zelda if you skip it. When you get in the 3rd crystal dungeon, it is possible to pull the glitch off much faster. I think you have read some faulty information on how it works (I remember reading something like you have to bounce until the hand comes 7 times or something). But it is really only nessecary to bounce back and forth once. I can't find a link to how it really works, but here is the jist: If at a certain point in your sword spin you hit the left bumper it caues an offset of 1. If at that point in your sword spin you hit the right bumper it causes an offset of -1. If neither of those happens the offset remains unchanged. If you get 1 then -1 or -1 then 1 it goes back to 0. The glitch works when you leave the room with of an offset of -1 or 1 (can't remember which). If you bump a wall before exiting the room the offset goes back to 0. Since the glitch can be done faster you won't have to kill any of the mummies. Anyways best of luck with your run.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/11/2006 2:10 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
SXL wrote:
in bullet-proof rerecording, aren't movies recorded within the savestates ? if snes9x isn't bullet-proof, maybe it would be time to add it :)
I think so, but working with a very long movie (1.7MB), the save states while recording are typically around 150KB. So I'm not really sure any more... maybe it uses compression?
Huffers wrote:
Currently, changing input every 4 frames, the bot takes overnight to do 5 laps of Mario Circuit 1 on time trial. As soon as I work out how to get my bot to record to smv files I'll post one here :)
Oh nice, what was the time? (I'm too excited to wait for the smv version!)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/7/2006 2:28 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Good ones. Here's one I stole from another forum: A girl asks her dad, "Why am I called Rose?" The father replies, "Because a rose petal fell on your head when you were born" Then her sister asks, "Why am I called Lilly?" The father replies, "Because a lilly petal fell on your head when you were born" Her brother says, "ERTGTHREGERG£$%£$^£EGRD" The father replies, "Shut up, Cinderblock."
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/6/2006 2:19 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Also it could be in pepsi's best interestest to decline paying for secrets if they suspect it is a trap by Coke to try and get them to do something illegal.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/4/2006 4:17 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
You could make like a pseduo mail chess match. Where every frame you send your opponent your 'move' (inputs). This would be cool but due to so many frames would take an extremely long time to make. The idea of 1 person plays 2 players against each other is a cool idea too, and a bit more practical. I'd certainly watch both ideas if they were done.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/4/2006 3:29 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
i think this was even written before their latest version which is by far the worse than anything they have ever produced before. (note i dont use aol, this is what i hear from aol users...)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/4/2006 4:15 AM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
well dont waste your time on the delux version, after 1 hour it boots you out of the game and 1 hour aint enough time to get anywhere you couldnt get on the web version. on a sad note i tried to crack it and got a virus like the noob i am...
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/3/2006 7:14 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
oh ive been playing the web version, i assumed the delux cost money, but ill give that a shot
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/3/2006 4:03 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
you guys are talking for GROW right? i havent noticed limit on play for popcap yet.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 7/3/2006 2:31 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
call me a simpleton but i really like this game, its called insaneaquarium at popcap.com. my best score for tank 1 is 26266.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 6/27/2006 12:45 AM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hmm good point, some games only check inputs on odd numbered frames (solstice for example). So this would be really easy to do, but would have 0 entertainment factor added for the viewer. Lets say the movie inputs were the first row, then you could turn it into a palindrome and it would play identical using the second row. 
a b c d e f g h i j k l m n
anbmcldkejfighhgifjekdlcmbna
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 6/27/2006 12:35 AM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511 
12345678
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Here's a way that only uses 4 extra (gravity independent too), not that this of any use as you already proved the point, I was just bored.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 6/26/2006 9:12 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Aw that makes it harder :( It's an interesting challenge but I think it would be hard for a viewer to appreciate the palindromocity of it. I bet certain games (short and simple ones) would be doable without new tools being created. You could play the game half way and then paste it in reverse. Then watch it, make a few adjustments repeat. This is easier said then done ofcourse.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 6/26/2006 8:23 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
It's easy so long as you dont worry about getting a great time. Just double the movie length and paste all the inputs in backwards, sure they will do nothing, but you never said they had to :)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Posted: 6/23/2006 2:20 PM
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flagitious
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Haha good idea. Has there been problem with spam bots here?
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
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