Posts for petrie911

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Posted: 11/2/2009 7:49 AM
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petrie911
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Posts: 635
The comments mention this aims for real time. Out of curiosity, how does it compare to the previous run in terms of ingame time?
Posted: 10/18/2009 9:00 PM
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petrie911
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So, the question is, would this be useful in a run? http://www.youtube.com/watch?v=eBZ1a6WP1Bk I estimate the time loss for getting the Red/damage tunic to be in the 2:00-2:30 range, so the question is how much time would having that 2x damage on non-spin attack hits save?
Posted: 8/25/2009 9:58 PM
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petrie911
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MrGrunz wrote:
it is definately not possible to do everything on the first frame possible, if you play with 5% speed.
Since OoT runs at 20fps, 5% speed is 1fps. I think it probably would be possible to hit things on the first frame possible consistently when you have 1 second intervals to do it in. Frame advance is still better, though.
Posted: 8/25/2009 6:54 PM
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petrie911
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Quick question: Does the emulator currently support starting movies from savestates?
Posted: 8/14/2009 12:24 AM
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petrie911
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Sir VG wrote:
What they want to do is basically hit the reset button, but the emulator is not working that way - it's working as if you quick turn the system off then back on. This causes the playback file to desync.
Ah, so the sole problem with Mupen's recorded reset is that it causes desyncs, correct? There's not some rule against using it?
Posted: 8/12/2009 10:18 PM
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petrie911
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Comicalflop wrote:
IIRC the reset command does 2 things. 1) In Windows, it resets the emulator, not just the ROM. Which is not how resetting is supposed to work. 2) in Linux, the reset command causes Mupen to close. (I think)
Ah, which brings us to question 2...how is resetting the emulator any different from resetting the ROM? Is it merely the distinction between switching the N64 on and off vs pressing the "reset" button on the N64, or is there more to it than that?
Posted: 8/12/2009 10:10 PM
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petrie911
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Since apparently this recorded reset thing is holding up the whole run, could someone please remind me what exactly is so wrong with Mupen's current recorded reset?
Posted: 7/17/2009 6:59 AM
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petrie911
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Introverted (I) 72.73% Extroverted (E) 27.27% Intuitive (N) 65.79% Sensing (S) 34.21% Thinking (T) 73.53% Feeling (F) 26.47% Perceiving (P) 56.41% Judging (J) 43.59% Your type is: INTP INTP - "Architect". Greatest precision in thought and language. Can readily discern contradictions and inconsistencies. The world exists primarily to be understood. 3.3% of total population.
ooh, big surprise, INTP again. With about the same ratios, too, though I appear to be less introverted here. At least it's consistent, I suppose. However, their description of INTP is...interesting.
loner, more interested in intellectual pursuits than relationships or family, wrestles with the meaninglessness of existence, likes esoteric things, disorganized, messy, likes science fiction, can be lonely, observer, private, can't describe feelings easily, detached, likes solitude, not revealing, unemotional, rule breaker, avoidant, familiar with the darkside, skeptical, acts without consulting others, does not think they are weird but others do, socially uncomfortable, abrupt, fantasy prone, does not like happy people, appreciates strangeness, frequently loses things, acts without planning, guarded, not punctual, more likely to support marijuana legalization, not prone to compromise, hard to persuade, relies on mind more than on others, calm
OK, while I admit many of those things do apply to me, a lot of them sound rather negative when they put it that way. The three bolded ones definitely do not apply to me, though. That third one really came out of the blue.
favored careers: philosopher, game designer, scientist, software engineer, freelance artist, research scientist, assassin, freelance writer, physicist, software developer, mathmetician, geologist, computer scientist, philosophy professor, webmaster, slacker, medical researcher, painter, mortician, systems analyst, comic book artist, computer technician, website designer, scholar, archeologist, computer repair, forensic anthropologist, astronaut, researcher, historian, systems engineer, genetics researcher, astronomer, enviromental scientist, egyptologist
OK, I must say I've considered a lot of those, and actually am one of them, but assassin? Really? I didn't know that was considered a valid career. The same goes for slacker, though for a different reason. Mortician is an odd one, but now that I think about it, I could see myself in it. On a related note, I really, really hate philosophy, so it was interesting to see that as a first choice. Considering I temporarily worked on a OoT hack, perhaps game designer isn't too far off, though.
disfavored careers: human resources, public relations, social worker, guidance counselor, health care worker, trainer, school teacher, wedding planner, movie star, hospitality worker, supervisor, child care worker, fundraiser, customer service, stay at home parent, office administrator
Well, they're got me there. I wouldn't be able to stand any of those.
Posted: 7/16/2009 2:39 AM
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petrie911
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I see I'm still INTP, with a very strong I (89), strong N and T (both 63), and a moderate P (22).
Posted: 7/15/2009 5:31 AM
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petrie911
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Nice use of Sesquipedalian Loquaciousness, Warp. hoo, boy, let's see what I can do... 1.) Felicity - Joy or happiness 2.) Lexicon - essentially, the vocabulary of a particular area 3.) Jostle - bump into or shake 4.) Pious - faithful, in a religious sense 5.) Vivacious - Lively 6.) Beguile - Charm (ie, tricking someone with it) 7.) Ubiquitous - everywhere; all-pervasive 8.) Superfluous - extra and unneeded 9.) Obstinate - refuses to yield 10.) Cacophony - din (ie, a whole lot of loud noise) 11.) Deprecate - to speak ill of or belittle 12.) Facetious - not serious; joking 13.) Veritable - means something similar to "real" (as in "That was a real big one.") 14.) Cavort - dance around, usually in merriment (or perhaps felicity) 15.) Inundate - flood (the verb, not the noun) 16.) Inane - dull, pointless 17.) Nimiety - wow, got me there. can't even parse this one out 18.) Convivial - friendly, nice to be around 19.) Staid - fixed, unmoving 20.) Copse - a small group of trees 21.) Incipient - beginning 22.) Glower - a very angry look. 23.) Badinage - no idea. moving along 24.) Perfunctory - cursory; without much care 25.) Obfuscate - to obscure. These words do an excellent job of obfuscating language, though I'm surprised how many I actually use in everyday speech. 26.) Hermitage - presumably going off and living alone, ie, becoming a hermit. 27.) Abscond - Escape, often from a crime 28.) Bombinate - I can think of some joke guesses 29.) Vagary - a statement meant to be vague and devoid of information 30.) Aberrant - not normal, in a negative way. 31.) Temerarious - hmm...from the root, I'd guess it has something to do with trembling in fear. 32.) Mephitic - root seems to indicate being like the devil or hell, so I'll go with that 33.) Egress - eh, I'll skip this one 34.) Frission - Definitly never heard of this. you sure it's spelled right? 35.) Fervid - vehement, highly-charged, that sort of thing. 36.) Bedlam - chaos, pandemonium, that sort of thing 37.) Expectorate - to spit 38.) Carom - well, it's a type of plant, for one. Carom seeds are used in Indian cooking. 39.) Somniferous - sleep-inducint 40.) Venial - minor. Nearly always used in reference to sin, to distinguish from mortal sin. Well, assuming my definitions that weren't guesses are right (and the ones that were are wrong), that's 32/40.
Posted: 7/15/2009 4:24 AM
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petrie911
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I feel like weighing in on this...
OmnipotentEntity wrote:
Does a 100% run include catching a 21-pound fish? What about catching the Hylian Loach?
The loach isn't kept track of, IIRC, so it can't really count. As for the fish, see below
Does a 100% run include the Cow in Link's House?
I say yes. Others will disagree.
What about filling up the high score list in Link's House?
I suppose you could. I mean, we get 100 skulltulas anyways, and no reason not to go perfect on Gerudo Archery. If we do the cow, there's the horse race, so all that leaves is the fish (which wouldn't exactly be hard for a TAS), and the marathon man. Though, IIRC, you can get a 00:00 time against him with timer manipulation, so I'm not sure how good of an idea that one is. There's certainly a case for this.
Will you be obtaining partial upgrades as part of the 100%? What about optional partial upgrades (like the Giant's Knife?)
Since they get replaced, it's hard to make a case for them counting.
Will you be obtaining all of the masks?
Considering you need the Mask of Truth to get a Deku Nut upgrade, yes.
Opening all of the chests?
I suppose you could, as the game does keep track of it, but I think we can all agree to forgo this one. Besides, there's no guarantee we even know where all the chests are, though UoT can help with that.
etc.
I do think maps and compasses should be part of 100%, though obviously many disagree. Still, they even show up on the start menu, so why do we neglect them? And before you say "they don't increase Link's abilities", may I remind you that neither do the Stone of Agony or any of the Gold Skulltula Tokens. And yet we require them anyways. Also, about the 37th heart piece. Surely there is at least 1 Heart Piece that is slower than it. So why not just throw that one out and get the 37th instead? You'd still have 20 Hearts at the end, so no need to worry about Lastly, I do suggest RBA be banned for 100%, considering you can very well RBA yourself songs and equipment.
Posted: 7/3/2009 9:17 PM
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petrie911
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Warp wrote:
I have the need for this algorithm in practice. Suppose you have the rotation notation <angle1>. That means: "First rotate around the X axis by angle1 degrees, then around the Y axis by angle2 degrees, and finally around the Z axis by angle3 degrees." Also assume that you can apply any amount of such rotations in succession. For example, you could first apply the rotation <0>, then <15> and then <5>. It is, in fact, possible to create from any amount of such successive rotations one single <angle1> triplet which produces the exact same transformation as all those successive rotations. In other words, any amount of successive rotation triplets can be collapsed into one. However, I don't know how this is done.
Well, each rotation can be expressed in the form of a 3x3 matrix. Although the exact form depends on the convention you're using for the axes of rotation, an explicit formula for changing angles into matrices can be found. In order to combine rotations, you multiply the matrices together. Then just apply the formula from before in reverse to the resulting matrix.
Posted: 6/27/2009 10:01 PM
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petrie911
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Heh, I actually wrote my code for my TI-89 to generate my answers. It was rather slow, probably due to the 89's bad processor and inefficiencies in the code, but it worked well for generating the series up to 100. This is what I used. 
tasg(n,m)
Func
Local s
Local i
Local j
Local k
{1,2,...,n}->s
For i, n+1, m
    For j, 1, i/n
        For k, 1, n
            If product(i-k*j-s) =/= 0
            goto Y
        EndFor
        Goto N
        Lbl Y
    EndFor
    {s,i}->s
    Lbl N
EndFor
s
EndFunc
{s,i} and {1,2,...n} are not the exact TI-89 commands, due to the fact that the TI-89 inexplicably has no list-joining command. But it is their effect. The actual commands are mat>list(augment(list>mat(s),[])) and cumSum(mat>list(diag(identity(n)))). The TI-89 language is pretty easy to understand, so you should be able to work out what it does. I'd try running your code, but I'm not sure what language it's in, nor what I'd need to run it. EDIT: Not exactly 3, eh? Well, that's strange. Are we certain it converges? It seems like it should, but still.
Posted: 6/26/2009 12:01 AM
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petrie911
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To OmnipotentEntity: 1 and 2) I haven't proven that this is maximal, but it seems like it is. I shall ignore arithmetic progressions with common difference 0, as they aren't really important to the problem. If we wish to find a sequence that contains no arithmetic sequence longer than n, then the algorithm is as follows. Start with a sequence S={1,2,...,n}. Now, starting with N=n+1, see if N makes an arithmetic sequence of length n+1 when combined with the sequence S. If no, add N to S. Then move on to N+1 and repeat. Using this algorithm, I came up with this sequence for part 2 (terms less than 100 shown). 1 2 3 5 6 8 9 10 15 16 17 19 26 27 29 30 31 34 37 49 50 51 53 54 56 57 58 63 65 66 67 80 87 88 89 91 94 99... I'll get to part 3 later. However, an interesting thing is that the sum in part 1 seems to sum to 3 (again, ignoring common differences of 0). Does it really sum to 3? Is there a simple relation between the sum of the series and n?
Posted: 6/23/2009 11:46 PM
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petrie911
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Oh...well, in that case, I'll post how I solved it. There are 3 parts: the swing, the flight, and the return of the sound. Each of these depends on L in a sufficiently simple way to make the problem numerically solvable. 1) when the pendulum has moved T from vertical, the height difference from its initial position is L*sin(T). Since energy should be conserved, m*g*dh=I*w^2/2 =mL^2*w^2/2=> w=sqrt(2*g/L^2*dh)=sqrt(2*g/L*sin(T))=w0*sqrt(2*sin(T)). So the angular velocity at angle T is w0*sqrt(2*sin(T)). Note that w0=sqrt(g/L) is the natural low-amplitude frequency of the pendulum. now, as we know, w=dT/dt, which can be rearranged into dt=dT/w. integrating from T=0 to T=11pi/18 gives t=INT(dT/sqrt(sin(T)),0,11pi/18) / (sqrt(2)*w0). The integral is a constant, though it can only be analytically represented through elliptic functions. If we evaluate it numerically, we find... t1=2.1034628 / w0 2) Now the ball has been released. For this part, S = sin(pi / 9) and C = cos(pi / 9). This will save space. The ball has initial height 50 - LC, initial upward velocity L*w0*sqrt(2C)S, and of course, a downward acceleration of g. We want to know when it hits the ground. Putting these all together yields... -.5g*t^2 + L*w0*sqrt(2C)S*t + 50-LC=0 => (w0*t)^2 - 2sqrt(2C)S*(w0*t)+ 2C + 100/L = 0 This is a simple quadratic equation, and can be solved quite easily. The answer, choosing the positive root, is t2 = (sqrt(2C)S+sqrt(100/L-2C^3))/w0 3) The total horizontal displacement of the ball is L+LS+L*w0*sqrt(2C)*C*t2, while its total vertical displacement is 50. So, if sound travels at speed Cs, the time it takes to reach you is. t3=sqrt((L+LS+L*w0*sqrt(2C)*C*t2)^2+50^2)/Cs Final) The total time is, of course, t1+t2+t3. While this does not depend on L in a particularly simple way, it is simple enough for a numerical solution to t1+t2+t3=5. And the result you get is L=4.526m (I must have made an error in my previous calculation)
Posted: 6/23/2009 3:01 PM
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petrie911
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andymac, you're on the right track here. But you are right: the reciprocals of composite numbers mess up your current argument.
Warp wrote:
The reciprocals of prime numbers produce a divergent series. I wonder how hard it is to prove that... (I'm assuming it's probably very hard, as almost anything involving primes.)
I was actually getting to that...the proof that the harmonic series diverges implies an inifinite number of primes can be continued to not only show that the sum of the reciprocals of primes diverges, but also gives an estimate of its growth rate. This in turn, can be used to estimate the density of primes. While the above method can be done rigorously, there are a few non-rigorous methods that still get the right answer and are still mostly correct. Also, was my answer to your earlier problem (with the pendulum) correct?
Posted: 6/22/2009 7:30 PM
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petrie911
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Your proofs both work, andymac, as does arflech's. arflech's leads to the non-calculus way to show the harmonic series grows no slower than log(n), as the comparison sequence clearly grows as log2(n).
andymac wrote:
Because the harmonic series is made of converging series (all terms where the denominator is even for example. add them up, you will get a finite number), it must be made on an infinite number of converging series in order to diverge. for every converging series, there is a starting number (1/p) where p is prime, and yeh, you get the drift.
Not quite correct. The sum of the reciprocals of the even numbers, or the sum of the reciprocals of any set of multiples of a number, is just as divergent as the harmonic series.
Posted: 6/19/2009 10:27 PM
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petrie911
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Warp, I'm getting an answer of 16.285 meters. I'll explain how I did it later. Suffice to say, the fact that the large-angle pendulum is a nonlinear problem complicates things a lot, but it's still doable numerically.
Randil wrote:
Hi, I have a question. I was thinking about this last night, but I couldn't come up with any answer: Consider the function p(x) satisfying: p(x) = 1 if x=p/q where p and q are both prime, p>=2, q>=2, and q>p. p(x) = 0 else. What I am wondering is, what properties does this function have on the interval D=[0,1]? Can the limit x->a p(x) be calculated at any point a in D? Is p continous in any point in D? What can be said about the set A={x: p(x)=1}? Is that a closed or open set? Because there are infinitely many primes, the set A will have infinitely many points, but I haven't been able to figure out much more than that.
A is countable and dense in [0,1]. This means that p(x) is nowhere continuous and its limit exists at no point. Additionally, it is not closed, as its closure is [0,1] =/= A, and it is not open, as clearly it cannot be expressed as the union of open intervals. I now offer proof that A is dense in [0,1]. The fact that it is countable follows immediately from the fact that it is a subset of Q. Let 1 > r1 > r2 > 0. We wish to find primes p and q such that r1*q > p > r2*q. The gap between our two limits, q*(r1-r2), grows linearly with q. However, for all sufficiently large primes, the gap between successive primes grows with O(p^(3/4)). Thus, by choosing q large enough, we can make q*(r1-r2) large enough that there must be a prime p between r1*q and r2*q. Thus, there exist p and q with r1 > p/q > r2 for all r1, r2 in [0,1], and thus A is dense in [0,1]. EDIT: The thought occurs to me that I should propose a problem. The infinitude of primes was brought up earlier. So let's look at another way to prove that fact. a) The harmonic series is the sum of the reciprocals of all positive integers. Prove that this sum diverges. As a bonus point, show that it grows no slower than log(n). b) Show that the divergence of the harmonic series necessarily implies that there is an infinite number of primes. A few of you probably already know the answer to this one offhand. If you do, don't post it. This is one that's a lot more interesting to work out for oneself.
Posted: 6/14/2009 4:05 AM
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petrie911
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We could always just refill at a fairy fountain. We go to those a few times.
Posted: 6/8/2009 3:09 PM
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petrie911
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After testing it, that's not the problem. You actually can pause before Madame Aroma gives you the bottle. The problem is that the game deducts the item from your inventory as soon as you close the dialog box, instead of waiting a frame like at the Curio Shop. I suspect that this is either because the item is replaced and not consumed, or because the Curio Shop can give different rewards depending on the item. A similar thing happens with the title deeds, but you can't pause with those.
Posted: 6/7/2009 12:55 AM
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petrie911
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I'm thinking this duplication thing could be used with the Letter to Mama. You get a bottle of Chateau for it if you talk to Madame Aroma, so you might be able to duplicate it then. This would mean we could cut out a cycle from the 100% route, which would be huge.
Posted: 6/5/2009 3:06 PM
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petrie911
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I've done more testing on the RNG, and it seems the number of RNG calls made for the wandering NPCs depends on the seed used for the RNG, and possibly the time of day as well. Needless to say, this presents a very complicated problem. So, instead of trying to work out the proper seed to get a good Chimchar, I'm just going to cheat and set the RNG to the proper seed to get a 31/31/31/31/31/28 Rash Chimchar. This can be set anytime between entering Verity Lakefront and selecting Chimchar as the starter. It is likely that this will be impossible to get in an actual TAS, but I suspect the only important things are Rash and 31/31 Atk/sAtk, which leaves a lot of room for adjustment. Using Piplup instead of Chimchar is a consideration. Empoleon has good attacking stats, and Surf/Blizzard/Grass Knot/Waterfall is a really nice moveset, being both a HM slave and a powerful attacker. It saves having to catch a water-type later. However, Empoleon is also really slow, meaning we will lose time when opponents get to attack. Additionally, I foresee Prinplup having problems with Gardenia, as I don't think it has any really good attacks to use there. Quick Claw/Guts Machop I think will end up being a bust. Gardenia presents a problem, as Grass/Poison resists fighting, and Crasher Wake's Gyarados will likely be a huge thorn in our side if we go by that route. Fantina only presents further problems.
Posted: 5/28/2009 9:11 PM
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petrie911
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we have to stop at the trainer school to give the rival the package. Otherwise, though, no. Fighting only required trainers, your Monferno is level 23 after Gardenia, which is pretty good. TM10 is kind of useless. Aside from the difficulty of getting correct hidden powers off this, Grass Knot seems like a far better coverage move.
Posted: 5/28/2009 12:02 AM
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petrie911
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Well, I have good news and bad news. The good news is, there are exactly 59 calls to the RNG between the generation of your ID and secret ID and the generation of Chimchar. Since the RNG is an invertible function, I can figure out what number needs to be in the RNG after the said events needs to be. FYI, it needs to be 8E074B90 The bad news is that the ID and Secret ID generation don't use the normal RNG method. This complicates things. Also, there are a limited number of initial seeds available. So, we'll have to see what the best we can do is.
Posted: 5/27/2009 3:38 AM
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petrie911
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The music mods are rather interesting. So far I've noticed Zelda's Lullaby, Milk Bar (MM), the athletic theme from SMB3, and Good Egg Galaxy. Also, unlike many hacks, this one seems to have a fairly reasonable level of difficulty.
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