Posts for rhebus


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Joined: 2/19/2010
Posts: 248
what happened to the Zelda tricks page? If I'm not mistaken it used to be at http://tasvideos.org/GameResources/NES/LegendOfZelda.html (and there are still links to it, eg from http://tasvideos.org/GameResources/NES/LegendOfZeldaOutlands.html) but it's just a 404 now. In particular, is the item generation chart still available somewhere? EDIT: I found the table on this submission
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This might not be the right topic to report this, but the previous movie publication text (1685M) copies text wholesale from the one before that (868M):
This is the fifth Legend of Zelda TAS at this site (not counting the second quest movie), and is ~85 seconds faster than the previous movie,
The text is correct on 868M but wrong on 1685M.
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Posts: 248
It's not actually a big enough boost to be useful, but I was surprised when I disassembled the Castlevania: Aria of Sorrow code to discover that the Heart Pendant makes hearts more likely to drop, and the Gold Ring makes gold more likely to drop. These are analogous to the Soul Eater and Rare Rings which are more explicitly documented.
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As well as assuming there are no children, you also need to assume there are no same-sex marriages in the remaining 30% of men or 10% of women.
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Joined: 2/19/2010
Posts: 248
I think I have a solution that gives 833 bananas. split the 3000 bananas into 3 piles of 1000: take each pile 334 miles and drop them now have 1998 bananas and 666 miles to go split into 2 piles of 999: take each 499 miles and drop them now have 1000 bananas and 167 miles to go Take them to market. You have 833 bananas left. The idea is, at each stage to split the bananas into as few piles as possible to minimise consumption. The more piles you transport bananas in, the more the camel eats. So you stop and drop as soon as you're able to reconfigure in to a smaller number of piles.
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Excellent improvement! Nice to see Claimh Solais, Blocking Mail, and Black Belt being used again, great to see the earlier kicker/malphas. One question: previous runs have got Erinys right before Balore, in order to benefit from a significant EXP (and hence STR) boost for killing Balore. Are you too weak to make this worthwhile at that stage? (Also, where the submission text says "Skeleton Blaze" it really means "Kicker Skeleton")
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Posts: 248
Yes, I think this is a bug in the general drop script. Kriole reported a similar problem to me in a PM once upon a time:
Kriole wrote:
Not quite sure what's wrong, but it shows me too many heart drops, most of which are bogus.
I'm afraid I've lost the original source code and all my notes when switching laptops a few years ago, so I'm not in a position to fix the bug.
Hetfield90 wrote:
Don't you need to be below a certain amount of mana for hearts to drop? Or does that only apply to candles and not enemies?
I believe this is just candles, so far as I can remember.
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Another way to show this is false, without the distinguished player being in a tied place: one player beats every other player except our distinguished player. All other matches are ties. The player who wins gets 19.5 points, taking first place. Our hero gets 10 points, taking second place. The 19 other players gets 9.5 points each, taking third equal.
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I tried tackling this by simplifying the problem and solving that. Imagine flipping a biased coin with 2/3 chance heads and 1/3 chance tails, and you keep flipping until you have seen a head and a tail. How many heads and tails do you expect to see? My answer: 7/3 heads and 7/6 tails. My reasoning: I built a state machine diagram. You have two goals to satisfy: a head or a tail. You can therefore be in one of four states: neither seen (initial state), head seen, tail seen, both seen (the terminal state). In the initial state, you have a 1/3 chance of entering the "tails seen" state, and 2/3 chance of entering the "heads seen" state. In the "heads seen" state, you have a 2/3 chance of getting an extra (unneeded) heads, and a 1/3 chance of entering the terminal state. You can expect to see 2 heads in this process. (I solved the infinite series E(h) = 2/3 + 2/3*2/3 + 2/3*2/3*2/3 ... to obtain this figure.). A similar argument shows that in the "tails seen" state, you will continue flipping tails until you flip your first heads, and that you will expect to see 1/2 heads. Therefore, in total, you can expect to see 1 + 2/3 * 2 = 7/3 heads, and 1 + 1/3 * 1/2 = 7/6 tails. (The 1 term is the heads/tails you get on the state transition to satisfy the "heads/tails seen" goal). This technique can be generalized to the monsters and gems problem, but there are two niggles: 1. there are twice as many states (8 in total). 2. the expected number of gems calculation is more complex in the "seen two types" states I'm going to have a go at solving it without resorting to drawing out the whole state diagram.
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The answer cannot be 3:2:1. You will always get at least one of each gem, but you have a one in 36 chance of having 2 rare gems from the first two monsters. This puts a lower bound on the expected number of rare gems at (1/36 * 2 + 35/36 * 1) = 37/36. The exact number will be higher than this, but it's enough to disprove the 3:2:1 answer.
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Patashu wrote:
1) Think about the central limit theorem - the normal distribution you get from summing infinitely many uniform distributions. Now think about what kind of distribution you get from summing finitely many uniform distributions. For n = 2 you get a triangular shape, for n = 3 you get a shape that looks like parabolic sections stuck together and so on. How do you generate the formulas that create these curves for each value of n? (For normalization purposes, let's say that the domain is x >= 0, x <= 1 and the integral of the curve should be equal to 1.)
If we take the special case of the discrete uniform distribution with exactly two possibilities, you've just described the binomial distribution. EDIT: and in fact for general discrete uniform distributions, you have a multinomial distribution.
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I think for a proper run, the wall of text could be made more manageable with section titles: - getting polymorphitis and teleportitis - getting Castle wand - getting absurd max hp and pw - getting absurd carry capacity - fixing alignment and luck - get book from Rodney - get candelabrum from Vlad - etc There's generally a subgoal at each stage that's being worked towards and it'd be helpful to see each given subgoal at a time.
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It's always fascinating to read updates on this thread. It's clear that this TAS is unlike many others on the site for the sheer number of possible strategies and the corresponding amount of planning that needs to go in to it. Thanks!
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Joined: 2/19/2010
Posts: 248
Thanks for your input! Also thanks for your previous movie, I've already learned a huge amount studying it.
TheRandomPie_IV wrote:
I noticed that sometimes the game would seem to not fully lag but would appear to do some of the subpixel updates a frame late, leaving you a pixel off where you 'should' be temporarily until it catches up a frame later, but I never really looked at the details of what was going on.
This seems to be consistent with what I'm seeing. On frame 858, my X speed just gets ignored completely and X pos stays constant, like the game forgot to update it. I'll try to do RAM search to find when the dragon gets spawned and see if there's any differences. I might even try splicing your input and my input together in various combinations to see which versions cause the pseudolag and which don't.
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I'm trying to learn how to make a TAS with this game. I've been playing around on the first level and found some places where I might temporarily get a pixel or so ahead of TheRandomPie's movie but I'm stuck at a point where I lose a frame and I can't see why. Here's my attempt: http://dehacked.2y.net/microstorage.php/info/651150305/kidc2.gmv On frame 856, I have exactly the same X and Y position and speed as therandompie, and I enter exactly the same input for the next frame, but therandompie immediately bounces off the rubber block while I have to wait an extra frame to bounce. I don't understand why this is happening. Any ideas? I thought it might be camera position but that seems to be the same in each movie too. The only other thing I can think of is if the dragon later in the level is spawning at different times and causing the game to take longer to process other interactions. Another theory I have is RNG, but I can't see why rubber blocks would depend on RNG.
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I remember this game! I thought the star-scoring system was interesting though I can see why it didn't catch on. I'd be interested in what the fastest strategy for a 7-star match would be -- are Specials too slow to justify winning in two rounds (special + knockdown/counter)? This run was pretty meh though. Good for vault, but I didn't feel entertained by basically the same strategy (with some small variation) in each match.
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Posts: 248
zero. the minimal nonmatching permutation swaps two numbers from their homes. You can't displace one number without also displacing another.
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thanks, FractalFusion. I was going a much more laborious route and had computed an expected value of 1 for n up to 4 but was getting bogged down. Your method is much more elegant.
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I played Apogee's PC clone of this called Jumpman Lives! Many great memories. The end of sreddal was surprising and hilarious. It's clear just how much luck manipulation had to be done to get the result you wanted here. Blackout was also a surprise: you were less than halfway through, then suddenly it ended. Yes vote.
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Randil wrote:
would an equivalent (and in my eyes simpler) description of the routine be to start with your string |||*** and then look for the rightmost instance of the string |* and replace it with *|? The routine would stop when the string |* is not found. It seems to me that this will generate the same solution as you presented.
No. From Bobo's first example, the string |**| progresses to: *||* But your algorithm instead progresses it to: *|*| and will miss the *||* configuration.
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Simon Singh in Fermat's Last Theorem gives the example that all of the following are prime: 31, 331, 3331, 33331, 333331, 3333331, 33333331. But 333333331 is divisible by 17.
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Anty-Lemon wrote:
Samsara wrote:
Updated versions of rhebus' drop scripts (if I can get my hands on them)
http://pastebin.com/0xR8kVkH
Wow, this is way better than my original one (which has been lost to history unless Kriole still has a copy). Great work!
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Given we know nothing about this universe which contains the simulator of our universe, what makes you think 10^83 bits is "too much"? Eg if it were an 83-dimensional universe, that would "just" require a hypercube of data storage with a length of 10 1-bit storage units on each side. Also, there's no reason that our time has to have any relationship with the time in the simulator's universe. For the same reason that emulators can record at one frame per hour but play back at 29.97 frames per second. If the emulator ran slowly, we wouldn't notice it, because our perception would slow down at the same rate as the rest of our observable universe.
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The description on the publication is currently "TODO: describe movie here."
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I think I understand what Warp's getting at: If you have Circuit A, and Circuit B, which have a single conducting link between them, can you not consider the entirety of Circuit A, Circuit B, and the non-conducting medium between them, to be a single large capacitor? It will almost certainly have a capacitance so low as to be negligible, but it's not zero. So you end up with a "circuit" from A over the conducting wire, to B, then back through the "capacitor" to A again. I think this wiki section describes what I'm trying to talk about.
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