Posts for Randil

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Posted: 4/9/2011 1:05 PM
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Randil
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Location: Norrköping, Sweden
Ok, I got as far as the integral being equal to (is this right?) but I'm having trouble finding an antiderivate to (cothu)^(1/3).
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Posted: 4/8/2011 3:17 PM
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Randil
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p4wn3r wrote:
Prove that Wolfram Alpha is complete crap and determine
I have tried to solve this today, but without much luck. I've tried several variable substitutions, considered integration by parts, and generally searching for a method that could work. Can you provide a hint how to solve this? I think I'm drawing blank.
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Posted: 4/3/2011 7:49 AM
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Randil
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Great job, here's another yes vote!
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Posted: 4/1/2011 3:43 PM
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Randil
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Location: Norrköping, Sweden
I vote yes because of the high rerecord count. And since a high rerecord count is always strongly correlated with the movie being mind blowingly entertaining, I think it's safe to give this a yes, even without watching it. Just looking through the input file in notepad, I can see that this is 100% concentrated awesomeness. I like the goal. A "8955 frames" run is in many senses the holy grail of SMB-TASing, and you definitely did a good job pulling it off. We can now all bathe in the joyful knowledge that this has be done, and with this good a result. I watched this run and was mind blown. Those were 8955 well spent frames of my time! Maybe we will see more runs like this in the future!
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Posted: 4/1/2011 3:32 PM
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Randil
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What just happened? I'll give this a yes, but I'm still confused. :)
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Posted: 3/31/2011 4:08 PM
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Randil
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p4wn3r wrote:
http://en.wikipedia.org/wiki/Stewart%27s_theorem
Wow, that theorem works perfectly! I'm impressed that you could pull up that theorem just 10 minutes after I posted the problem. :) Using that theorem, I found the radius to be r=6/7. Thanks for the help.
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Posted: 3/31/2011 3:35 PM
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Randil
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A friend told me about this problem, and it's been driving me crazy - I feel like there is a simple solution to it, but I simply can't find it. I thought maybe you guys would find it a fun (and maybe challenging) problem! Here's the problem: *Draw a circle C1 with radius 3 with center at (0,0). *Draw a circle C2 with radius 2 with center at (1,0). *Draw a circle C3 with radius 1 with center at (-2,0). Your task is now to find the biggest circle (in terms of radius) that is completely inside the green zone shown in the picture below. Bonus points if you can give the coordinates of the centroid. I realize that it wouldn't be too hard to find a numerical solution to this, but I'm looking for an exact analytical solution (if it exists).
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Posted: 3/29/2011 4:52 AM
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Randil
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Could be interesting - in particular, I think it would be interesting to make a full analysis of the physics of this game - basically making a mathematical model of how the game works, and find the perfect game by this model. Maybe it's only me who finds those ideas interesting. :) I tried doing this some months ago, with some luck, but it's a big project to understand how everything works in this game. Anyway, I have a bunch of RAM addresses if they can be of help: 
00B7 Direction
CE Swing|speed
CD Club
12E Screw
| |
00B0 Ball|Y|sub
00B1 Ball|Y|pos
00AD Ball|X|sub
00AE Ball|X|pos
| |
00B6 Ball|height|pix
00B5 Ball|height|sub
D6 Swing|traject
x96 Wind|dir,|20=R
97 Wind|speed
8F Dist?
| |
E1 Ball|height|sp
E2 Ball|height|sp
db Ball|X|speedpix
da Ball|X|speedsub
| |
| |
sub=sp=subpixel, tracect=tracectory, Dist?=distance to hole (probably) Some more facts: Shot directions: 256 Swing speeds: 3 Clubs: 14 Screws: 5 Possible shots: 53760 This does not count in wind speed, which is a random factor I do not fully understand yet. But ~50k possibilites per shot isn't that bad - it would be quite possible to brute force each possibility, and you can of course cut this down more by eliminating shot directions and swing speeds that are obviously bad.
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Posted: 3/27/2011 8:04 AM
Post subject: Re: Hey i have a question about the input.
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Randil
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sirstarmagic wrote:
how do you get an input?[/b]
I'm not sure I understand the question, but if you're wondering how you go ahead and record input for these 2 games 1 input TASes, see what sgrunt said:
sgrunt wrote:
Many emulators support "background input", i.e. recognising button presses even when the emulator window isn't the main focus. It's my understanding that it is this feature which was used to record multi-run movies.
So basically, you have two emulator windows open, where one of them has "Run in background". Now this emulator window will recognize button presses even if the other emulator window is highlighted. Now, start recording a movie with both emulator windows, and your input will be recognized by both emulators and recorded into both movies. And as adelikat said, make sure that the emulators are on the same frame (to make sure that they synch, always have frame display on) and only use frame advance when recording this way.
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Posted: 3/26/2011 1:35 PM
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Randil
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I really enjoyed that WIP. Keep up the good work!
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Posted: 3/18/2011 6:27 PM
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Randil
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It appears that the guy who made it did it just for fun (at least that's how I interpret it). Here's the website: http://facerocker.com/2011/03/17/first-person-mario/
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Posted: 3/18/2011 5:23 PM
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Randil
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First person Mario! Looks pretty cool: Link to video
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Posted: 3/17/2011 5:39 AM
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Randil
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There is a TAS on youtube here: http://www.youtube.com/watch?v=e2a-h1bl18o
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Posted: 3/13/2011 6:23 AM
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Randil
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When it comes to non-integer dimensions, you enter the field of Hausdorff dimension and fractal dimensions. An example of a set of non-integer dimension is the cantor set C - in terms of cardinality, C is as large as R, but in terms of length, C is just a single point. Therefor the dimension of C falls between 0 and 1, in this case 0.631. So I guess a pi-dimensional cube would be an object with 3-dimensional length (where in this case "length" is defined by some metric) but with some cardinality-properties of a 4-dimensional object.
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Posted: 3/12/2011 6:36 AM
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Randil
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I definitely think you should TAS using conventional means, i.e. savestates and frame advance. Although it is possible to record input by this table format, it is mostly used to splice movie files together or change a small section of your movie without having to go back and rerecord from that point. As you said, there is a TAS Edit for FCEUX, and FCEUX:s text based movie file format allows you to easily edit the input in a text editor. For other emulators, such as SNES9X, there is TAS Movie Editor (the latest version seems to be in this post). As for what game you should TAS, I'm not familiar with the games you listed. However, there are a few general things that you should consider when choosing a game to TAS. I'll just list a few that I can think of right now: *How much do you like the game in question? Trust me, you will spend more time and energy on the TAS if it's a game that you really like. *How complex is the game? Roughly speaking, the more complex the game, the harder it is to TAS. As your first game, I recommend a simple game, perhaps a platformer or action game. Note that you don't want the game to be too simple, then it won't make an interesting TAS. *Games with lots of variation, in terms of graphics, music, weapons, etc. usually make a good TAS. You want your TAS to have some variety. This also connects to the next point: *How long do you think the TAS will be? Long games (such as RPGs) require much more planning, commitment and knowledge about the game. The risk of having to redo the movie due to early mistakes in the movie is also quite big. It's also easier to keep the viewer's focus if it's a shorter TAS. I hope that helped!
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Posted: 3/11/2011 8:11 PM
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Randil
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Warp wrote:
Is the set of points inside a unit square larger than the set of points on a unit line?
This is not the case if we can fill the unit square by mapping points from the unit line. Let x be a number on the unit line and let [X,Y] be a point inside the unit square. Let r(x) be the rational part of x and i(x) the irrational part of x. Consider the mapping: [X,Y]=[2*r(x),2*i(x)]. This maps the unit line to a right angled triangle with corners at [0,0],[2,0] and [0,2]. This triangle covers the unit square, and we can fill the entire triangle since both the rational and irrational numbers are dense on [0,1]. Only the rational numbers are filled on the x-axis, and only irrational on the y-axis, but since both these sets are dense, we have filled (more than) the unit square mapping only points from the unit line. This means that the unit square does not have more points than the unit line. I'm not sure I'm allowed to do what I just did, but it was worth a try!
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Posted: 3/10/2011 7:01 PM
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Randil
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Nice! Thanks for your help, petrie911 and p4wn3r!
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Posted: 3/10/2011 6:39 PM
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Randil
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juef wrote:
Well, at first glance (with the numerical approximation you've given), I'm very tempted to say it converges to zeta(2) = Pi²/6.
Interesting! You seem to be right, otherwise it is a very weird coincidence. If that is indeed the limit of the series, all that's left now is to show why the series converges to this. This gives the problem more hope!
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Posted: 3/10/2011 6:12 PM
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Randil
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p4wn3r wrote:
Randil, are you sure that the limit of this series can be found by elementary means? (i.e, is this a textbook problem or something similar?)
Hehe, no, this expression spawned from a probability problem I thought up myself, so I don't know if the limit can be expressed in elementary function, maybe it can't. But on the other hand, I can't prove that it can't, so who knows. I have tried using the fact that H_k(x)=log(k)+gamma+e_k, but it didn't help much.
petrie911 wrote:
Through some manipulation of sequences, that expression is equal to... sum(k=1 to n) (H(n) - H(n-k))/k
Yes, I had something like that from start, but I thought that expression was more difficult to work with, hence I divided it into two terms as I did. But on other hand, I haven't had much more luch with the form I presented the expression in, so I don't know which form is best.
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Posted: 3/10/2011 6:40 AM
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Randil
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Here's a problem I need help with - I've tried to solve it but always end up in a dead end. Consider the expression where H_k is the k:th harmonic number. I think this expression converges as n->Inf, but I'm not sure. Some numerical testing show that it seems to converge to something like 1.644... or so. Can you help me find the limit (if it exists) of the expression above?
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Posted: 3/8/2011 6:11 AM
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Randil
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1. Set g(x)=f'(x). Since f(x) was differentiable, g(x) is continuos and we can apply the Intermediate value theorem on g, which proves that such an x exists. 2. I'm gonna try and use the Intermediate value theorem here. Let v be a small positive real number. f(a-v) is inside the intervall ]-Inf,a[. As x->a, the function value will increase (or at least not decrease, since the function was non-decreasing). We also know that f is bounded above, so it's derivate is bounded. Let M the the highest value of the derivate between a-v and a. We know that lim x->a f(x) is between f(a-v) and v*M+f(a-v). Since M is bounded, f(a-v) exists, and we can make v arbitrarily small, this limit exists. 3a) Differentiate both sides with respect to x. This gives: h(x):=2*f(x)*f'(x)+2*g(x)*g'(x)=2*f(x)*g(x)+2*g(x)*(-f(x))=0 <=> h(x)is a constant c. We now know that f(x)^2+g(x)^2=c. This holds for all x. In particular, for x=0 we get f(0)^2 + g(0)^2 = 0^2 + 1^2 = 1 = c <=> c=1. 3b) We know that f(x)=sin(x) and g(x)=cos(x) will satisfy the differential equations presented. They will also satisty that f(0)=0 and g(0)=1. A system of two linear differential equations with two initial conditions, f(0)=0 and g(0)=1 has a unique solution, and since f(x)=sin(x) and g(x)=cos(x) satisfies this set of diff. equations, it is the unique solution. 3c) Since sin(x) and g(x) are continues, bounded, etc. their power series exist. Set f(x)=a0+a1*x+a2*x^2+... and g(x)=b0+b1*x+b2*x^2+... f(0)=0 -> a0=0. g(0)=1 -> b0=1. Let's use that f'(x)=g(x). We get (1) a1+2*a2*x+3*a3*x^2+...=b0+b1*x+b2*x^2+... g'(x)=-f(x) gives (2) b1+2*b2*x+3*b3*x^2+...=-a0-a1*x+a2*x^2+... Combining (1) and (2) will recursively defined all coefficents, starting with a0=0 and b0=1, showing that the power series presented hold. You can also prove this by induction. Now I should really get ready for work.
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Posted: 3/5/2011 5:37 PM
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Randil
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Here are my RAM addresses, if they can be of any use: 
00000	7E0C05	w	u	0	X pixel
00001	7E0C04	b	u	0	X subp
00002	7E00AE	w	u	0	Scr X pos
00003	7E00B4	w	s	0	Scr Y pos
00004	7E0C08	w	u	0	Y pixel
00005	7E17D5	w	u	0	En1 X pos
00006	7E1825	w	u	0	En2 X pos
00007	7E1875	w	u	0	En3 X pos
00008	7E1805	b	u	0	Enemy 1 HP
00009	7E1855	b	u	0	Enemy 2 HP
0000A	7E18A5	b	u	0	Enemy 3 HP
0000B	7E18F5	b	u	0	Enemy 4 HP
0000C	7E1945	b	u	0	Enemy 5 HP
0000D	7E00D3	b	u	0	Counter 1
0000E	7E00D1	b	u	0	Counter1
0000F	7E0B8B	b	u	0	Move counter
00010	7E0C1A	w	s	0	X speed
00011	7E17B5	b	u	0	Boss HP
00012	7E1995	b	u	0	Boss HP?
I think "move counter" somehow controls when the game accepts your input... Those other counters I'm afraid I don't remember what they stand for.
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Posted: 3/4/2011 5:04 PM
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Randil
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I didn't notice this thread until now! I have been working on this game as well, though I'm not certain I will finish it anytime soon (you go ahead). I am several seconds ahead of the published run (though not as much you I think) at Wolverines stage. Here is the WIP. In particular, Beast's stage and Gambit's stage were improved. Maybe you can find something of use in my WIP, and feel free to ask any questions about it.
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Posted: 3/4/2011 6:02 AM
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Randil
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Yeah, this game is great! There's an older thread here, and both me and Aglar made a WIP of the first level.
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Posted: 3/3/2011 9:23 PM
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Randil
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AndyDick wrote:
Randil wrote:
Any plans on doing a story book atlas video (perhaps Deja Vu) in the near future? I personally think it would look cool.
I'd love to do Deja Vu actually... I'm just unsure of how to logically arrange the different locations. Do you know of a map which shows the different paths between screens?
Unfortunately I don't... But one way to do it could be to arrange them in the order they are visited in the TAS, starting from the top left corner. That way the cursor would basically be moving from top left to bottom right corner, except when backtracking. Though that might in a way ruin the "atlas" purpose of the video. Hmm...
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