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McBobX wrote:
If we look at BCAS, the most enjoyable thing is the high speed and the air sliding.
I just watched BCAS now, and actually that was the least enjoyable part of the run for me. The most enjoyable part of the run was the autoscroller.
Did it combine two gimmicks or something? (Just from its name I thought it was a hack like that.) It just doesn't feel like Mega Man.
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p4wn3r wrote:
It seems to me that it only makes sense to define the ring of polynomials mod p, where it is, in fact, Euclidean. If we define them over a composite number n we get the problem that the group of units mod n is not closed under addition, so in this case, the ring wouldn't even be an integral domain. That's why polynomial mod a composite number can have more roots than their degree, but not modulo a prime number.
For example, take x^2-1 over Z/8Z. Not only does it not have unique factorization (x^2-1 = (x-1)(x+1) = (x-3)(x+3), for example), you could even factor x^2-1 into two polynomials of higher degree, like x^2-1 = (4x^4+x-1)(4x^4+x+1). Certainly there is no Euclidean property here.
p4wn3r wrote:
What I thought when he presented it was: Every permutation can be broken up into cycles, the sign of the permutation is found by multiplying (-1)^(length+1) for each cycle.
Oh, I didn't even think about it in terms of cycles. Well in that case rearranging and relabelling is just equivalent to relabelling the numbers in the cycle representation, and whether a permutation is even or odd depends only on cycle structure and not on the labels.
Note that conjugation also preserves cycle structure.
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p4wn3r wrote:
Maybe it's just me, because I know most of the math concepts, but I also thought his final observation, where he uses a relabeling to prove that the sign of the permutation is the Legendre symbol very difficult to follow. The argument in the blog post, which uses permutation cycles, although more mathy, is more natural in my opinion.
I assume you are referring to the proof of Zolotarev's Lemma. I'll try to summarize what I got from it in the video.
Every prime p>2 has a primitive root. There is a proof which uses field theory: Basically, if the max order of elements in (Z/pZ)* (multiplicative group) is some m<p-1, then every element satisfies x^m-1=0 over the field Z/pZ, and so x^m-1 has p-1 factors, contradicting that x^m-1 can only have at most m factors. It is nonconstructive (no algorithm is known in general that gives a primitive root).
If g is a primitive root of Z/pZ, then g^2, g^4, ... , g^(p-1) are all squares, and so those are all the quadratic residues and the rest (g^1, g^3, ... , g^(p-2)) are all quadratic nonresidues.
The rearranging and relabelling trick is better described as conjugation. So for a permutation x, rearranging and relabelling turns it into zxz-1: left-multiplying by a permutation z corresponds to rearranging, and right-multiplying by z-1 corresponds to relabelling. Because parity of a permutation when multiplying is like even/odd numbers under addition, x and zxz-1 have the same parity. This is why rearranging and relabelling does not change parity. (So you don't have to wait for Mathologer to prove it next time.)
So because of the rearranging and relabelling trick, the permutation based on x→bx (mod p) is turned into a permutation based on x→x+c (mod p-1), by taking discrete logarithms, where g is a primitive root and b=g^c. The parity of x→x+c (mod p-1) is even if c is even, and odd if c is odd. Based on the property of primitive roots above, x→bx is even if and only if b is a quadratic residue mod p.
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p4wn3r wrote:
Determine all values of a0 for which there is a number A such that a0 = A for infinitely many values of n.
I think you mean an = A for infinitely many values of n. Otherwise the answer would be all natural numbers.
I've seen this problem before, and yes, we finally get to use mathematical induction!
I'll comment a bit later on how to do this; I would normally be all over these problems but I've been way too busy with real life for the last 8 months or so.
Ferret Warlord wrote:
That's the jeep problem, related to the harmonic series
Referring to this:
https://en.wikipedia.org/wiki/Jeep_problem
There was actually a related problem we discussed before about camels and bananas. It's somewhere in this thread but I don't have time to go back through everything to find it.
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ThunderAxe31 wrote:
The mandatory choice is Piplup, because it starts with the move Leer. Which it doesn't lower the adversary defense, but instead it raises our own Attack. By two stages. Like every other stat-affecting move in the game.
I'm pretty sure 叫声 is Growl. Not that it makes it any less ridiculous.
Also I'm not sure what Garchomp's 暴走 is. From your description, it seems like you identified it as Take Down. I thought at first it was Thrash. Whatever, I guess. It's not like any of these moves work like they are supposed to anyway.
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Forgot to reply until now.
I like that this ROM hack has original music (how many hacks of any kind even have that?) and there were some neat things during the stages. Certainly left more of an impression on me than the last Mega Man hack Baddap1 did.
Still it's difficult to beat Rockman 4 Minus Infinity which was pretty much the best Mega Man hack I've ever seen. Also, with all the Mega Man level hacks available right now, it's somewhat hard to stand out nowadays.
I wish I could say the same about Mega Man X series though; the number of hacks of Mega Man X games can be counted on one hand.
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Huh. I thought you forgot about this. Oh well, cool that you finished the all bombs TAS finally.
BTW, if anyone is wondering what this game is about, I made a few detailed posts years ago starting with:
http://tasvideos.org/forum/viewtopic.php?p=429211#429211
Seeing this TAS reminds me again of how far the game goes to troll the player on Hard mode:
- Hard mode throws emergency timers everywhere.
- The various devices that threaten to set off the bomb get more difficult; a really good example is the pendulum in both Rail Crossing and Pinball Machine; on Easy and Normal, the pendulums don't shake a whole lot. On Hard, they nearly set off the bomb switch just by turning a screw once (and you get timers on them too).
- Hint messages are placed in hard-to-find areas (like dead ends). Sometimes they are false and you have to find another message to confirm that.
That being said, I do like how a lot of the game works like a logic puzzle (except Gunder which is just rock-paper-scissors, and the last bomb on Hard which is a 50/50 guess). Some of the logic is a little difficult to figure out though (it took me a long time to fully understand the one with the Doppler effect essay in it).
Ice Coffee is still an amusing little bomb (EXPLODE, NOT EXPLODE, EXPLODE, NOT EXPLODE, ...). Also teaches you that triangles are important, not just in real life, but video games as well.
Finally, for those with an interest in seeing things get destroyed, here is a video showing all the explosion cutscenes. Don't question why things like the Moon, the Pacific Ocean, your own shadow, and other people can all be bombs. Or why an orange explodes like a nuke. Or how an engine bomb causes the Earth to blow up. This is a weird game after all.
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TASVideoAgent wrote:
Unfortunately, this game is meant to played with the DS rotated 270 degrees to the right, and Bizhawk doesn't seem to do avi recording in that way at the moment.
You can rotate it using Avisynth. For example:
v = avisource("video.avi")
TurnLeft(v)
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Just watched this. Co-op modes TASes are usually great, and this is no exception! In fact, a lot of these tricks wouldn't be possible in single-player.
I noticed you didn't enter a couple of stages right away; I suppose this is for RNG reasons.
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merrp wrote:
I found a way to crash the game. If you do the tile glitch on Route 110 and then walk out through the tiles towards the left 4 tiles, the game will freeze. It works in a couple other places as well.
Did it freeze as in completely crash, or was the music still playing?
The music fades out, as if going to a new area. Otherwise, the display freezes.
Oh, I know about that! I found about it because of epicdudeguy's speedrun using this glitch on JP Any% to dupe Rare Candies:
https://www.speedrun.com/pkmnrubysapphire/run/yjk2ldnm
Some time ago, I tried some number of metatiles to see if it would do anything strange. So far, I haven't encountered any tiles that do something completely weird. I did find a tile (spawned by either of the two mail words ちからもち or ちくでん, under the category for abilities) that warps you to your secret base if you enter them from below. Other than that, I didn't really find anything else. But you probably know better how to manipulate these metatiles.
I found a way to crash the game. If you do the tile glitch on Route 110 and then walk out through the tiles towards the left 4 tiles, the game will freeze. It works in a couple other places as well. I didn't find it to be useful though.
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merrp wrote:
Was that mentioned in the last list you helped make of what moves to use when? I thought I copied it over and followed it pretty exactly but maybe I made a mistake.
Oh, I assumed that you were still using Torrent strategy. But maybe I wrote something down wrong. Or I assumed that Water Gun and Mud Shot were interchangeable in a lot of places, so I left it as a puzzle for you to work out. :)
merrp wrote:
Isn't the super effective message longer than the critical hit one?
Only in English. In JP version, SE message is shorter than crit message.
merrp wrote:
I've been working on trying to crack open JP Ruby lately, as well as whether sprite ACE could be faster on JP. But we'll see!
How are you planning to crack open JP Ruby? Is there a glitch that allows potential ACE?
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merrp wrote:
キ is broken.
Did something break?
Yeah, it's the adventures of fake BAD EGG red キ. This is how broken キ is:
I don't even know what move キ is trying to use here.
On a more serious note, I have not forgotten you; I just have nearly no time these days to do anything TAS related, in case you were wondering whether I disappeared. I did have time to watch your TAS, and you did manage to get it below 58 minutes.
I should mention that you could have used Mud Shot on Wally's Ralts to avoid the critical hit there, and switch out one of the Mud Shots on either of the Numel you would eventually use it on with Water Gun. (Also, there was no need to use Water Gun on Magnemite before then; you could have just used Mud-Slap). Either that or you could have taken the Tackle damage from Wingull to set up Torrent and do as you did before. However, I don't think it would be too appropriate right now to nitpick this anymore, especially since you already submitted this, and you are testing out a new strategy that may even be faster.
I still think the strategy of glitching IVs to move (instead of EVs to move) could work. However, it would take a lot of planning to make sure everything works. Also, the results I posted last year were messed up (I made an error that reported false values), so don't use those results.
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Link to video
Alyosha: For some reason, video dumped from GBHawk core is 60.0 fps instead of 59.72... Not a big deal though (I can correct it in Avisynth), just letting you know.
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Pretty fun and short speed game. I tried it out for myself and even got a couple deathless runs, though nowhere near RTA speed.
You managed to work out the desync problems, which is cool (but why are there even desync problems in the first place? I thought libTAS was invented to play back computer game TASes in a deterministic manner). Certainly I wouldn't have the patience to stick around if desyncs were going to be a problem.
EZGames69 wrote:
ngl, when I saw the thumbnail for your temp encode, I thought you were recording your tas with your phone camera.
Same here! I wonder what the average time was so far for the viewer to figure this out.
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r57shell wrote:
This is for the Nightfire TAS. Basically I'm starting at the black dot in and I need to get to the yellow point. Which of the lines would be the best to take?
I need to jump from A (the ledge i'm walking along) to B (the edge of the platform i'm going to land on)
The speed is constant and the jump length is always the same.
basically the line connecting A to B is always the same length
Let's say D1 (starting point's distance from horizontal edge), D2 (ending point's distance from horizontal edge), R (jump length), W (ending point's distance from vertical edge) and x (horizontal distance of the landing point for the jump) are given above, and we want to find x that minimizes total distance.
In short: Using calculus, we get the following equation: 2x2(W-x)2+D2x2=R2(W-x)2.
For the special case D2=0 (ending point lies on the horizontal edge), clearly x=R/sqrt(2) minimizes the distance, and indeed solving the above equation can be done easily and gives x=R/sqrt(2). However, in general, this is a quartic equation, which I would just solve using WolframAlpha (example is given with D2=8, W=10, R=3). In general, if D2 is not too big compared to W, then the optimal x should be around R/sqrt(2).
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Flip wrote:
If her savings are in $10, $50 or $100 bills anyway, then there would be no error occurring.
Masterjun already tried that; InfamousKnight apparently responded that there are only $20 bills involved.
This is clearly a lateral thinking problem. I came up with my solution after reading InfamousKnight's two posts as literally as possible. There might be other valid solutions, but I believe (unless InfamousKnight says otherwise) that if there is a single intended solution, then mine is the one intended.
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I'm pretty sure the answer is 2160 dollars. Explanation:
The problem only says that (according to her) 4 of the 20-dollar bills makes 100. It does NOT say that (according to her) 4 of the 20-dollar bills makes 100, another 4 of the 20-dollar bills makes 100, another 4 makes 100, ...
This interpretation is reinforced in the statement of the question:
- "Jane was counting her savings with the belief that 4 20s make 100": Exactly, she believes that 4 of the 20s makes 100.
- "So every time she finds 4 20 dollar bills, she adds 100 to her calculator.": She finds 4 20s exactly once: the first (and only) time the count reaches 4 20s.
- "However, she realized that 5 20s make 100 and not 4.": Yes, she realized that 5 of the 20s she had makes 100 and not 4.
- "She counts them like this: 4 20s according to her makes 100.": Yes, the first (and only) time she counts 4 20s, to her it makes 100.
- "So she counts 4 20 dollar bills and puts in her calculator 100 instead of 80 which was the mistake made.": Exactly what it says.
- "And when she was done, she ended up with 2180 under the belief that 4 20 dollar bills make 100 dollars.": Yes, she ended up with 2180 because she believed that 4 of the 20s made 100 dollars.
So her initial calculation was 20 dollars too high. So she actually has 2160 dollars.
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I fixed the Lua script for the RNG analyzer and cleaned up some code:
http://tasvideos.org/userfiles/info/58732884454006571
Turns out the previous one depends on "tables.lua", which I didn't provide. I rewrote it so it is in one file.
merrp wrote:
Some of the memory addresses would have to be changed to match JP's memory map--have you had any trouble doing that?
For the major addresses, I've already accounted for every single Gen 3 version (Ruby/Sapphire/Emerald/FireRed/LeafGreen and U/J for each). It's based on a previous script which I used in the VBA days.
For things like Quick Claw or all the stuff you put in emerald-jp.lua, I haven't done it yet, though.
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Here is the script I use to examine the RNG:
http://tasvideos.org/userfiles/info/58692469321661997
During wild Pokemon generation, the game usually inserts a frame break in between PID generation and IV generation. In some cases (usually when the game rolls for PID a lot), this frame break may occur in other places, such as during the rolls for PID.
I'm currently planning out how to write a Lua script using BizHawk forms so I can integrate all my scripts together instead of having a hundred separate files.
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Yes, I know for wilds it picks the nature to roll and then rolls the PID until it matches the nature. By "method 2" (I think it's actually called method H2 elsewhere) I'm referring to the pattern where there is a vblank in between the PID and the IVs. The position of the vblank can change at random but the game usually uses this vblank positioning.
So the random numbers look like this:
#1:
228118f0 <- encounter trigger (this is cycle 80106 of seed 0x3F87)
40311ea3 <- species
615215da
b3381a45 <- nature to roll
3d4295d4 <- pid low
14f8c3b7 <- pid high
9017775e <- vblank
b110d779 <- ivs 1st three
e668fcf8 <- ivs 2nd three
#2:
bf9379af <- encounter trigger (this is cycle 80068 of seed 0xF988)
0d3881f6 <- species
d2ecaa31
fd85c550 <- nature to roll
9ff3c383
daae893a
26fc7a25
6455a834
1df0d697
70b7c0be
4beb5559
a30ad558
12ae06eb
fef6ec82
b9beafcd
354fb0bc
682fe87f <- pid low
06771086 <- pid high
d55f3d81 <- vblank
3110de60 <- ivs 1st three
66304f53 <- ivs 2nd three
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I'm searching both for a good Mudkip in addition to the wild encounter. Note that the proper wild encounter has a very low chance of occurring. To get the proper IV requires a 1/65536 chance; the proper PID: 1/3 chance; and the chance of triggering it: 1/9 * 1/2.
To simplify things, I only aim for Mudkip with 30-31 Atk IV, 31 SpAtk IV and Spe IV is at least 5 (or at least 20 if Spe- Nature). Nature is SpAtk+ with any one of Def-, Spe-, SpDef-. I aim only for the third move to be 0x3110, not the fourth (which is possible, but requires double-corruption, and leaves less backup in case the moves don't work out), and I assume that Method 2 is in effect. (Almost all wild encounters use Method 2.)
I ran a search using a C++ program and got a few results. I'm looking at two of them right now (sorry, I couldn't be bothered to clean up this formatting):
Both have a target cycle of around 80000-80200, which places it approximately at Route 104 after using Teleport, when trying to get the ferry to Dewford Town.
Before I proceed, can anyone confirm whether having these glitch moves is permissible?
(all hex)
1st move - 1300 (Route 104 w/o pkrs) or 1100 (Route 102 w/o pkrs)
2nd move - One of A184, A183, 9984, 9983 (double corruption) or E184, E183, D984, D983 (single corruption)
4th move - One of 3334, 3318 (double corruption) or 7334, 7318 (single corruption)
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I'm currently checking to see if there is a starting seed that makes it feasible to generate a wild encounter Pokemon with 0x3110 as its IV (lower word) and also with the correct PID.
merrp, did you consider the strat of getting 0x3110 as experience at any point?
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merrp wrote:
And yeah, do I still need the X Special for Maxie, or can I skip it?
Well, you can skip it. It's just that, without X Special, Mightyena takes two hits to defeat (Tackle + Torrent Water Gun), and Zubat needs a critical hit.
I don't even know what move キ is trying to use here.
On a more serious note, I have not forgotten you; I just have nearly no time these days to do anything TAS related, in case you were wondering whether I disappeared. I did have time to watch your TAS, and you did manage to get it below 58 minutes.
I should mention that you could have used Mud Shot on Wally's Ralts to avoid the critical hit there, and switch out one of the Mud Shots on either of the Numel you would eventually use it on with Water Gun. (Also, there was no need to use Water Gun on Magnemite before then; you could have just used Mud-Slap). Either that or you could have taken the Tackle damage from Wingull to set up Torrent and do as you did before. However, I don't think it would be too appropriate right now to nitpick this anymore, especially since you already submitted this, and you are testing out a new strategy that may even be faster.
I still think the strategy of glitching IVs to move (instead of EVs to move) could work. However, it would take a lot of planning to make sure everything works. Also, the results I posted last year were messed up (I made an error that reported false values), so don't use those results.
Let's say D1 (starting point's distance from horizontal edge), D2 (ending point's distance from horizontal edge), R (jump length), W (ending point's distance from vertical edge) and x (horizontal distance of the landing point for the jump) are given above, and we want to find x that minimizes total distance.
In short: Using calculus, we get the following equation: 2x2(W-x)2+D2x2=R2(W-x)2.
For the special case D2=0 (ending point lies on the horizontal edge), clearly x=R/sqrt(2) minimizes the distance, and indeed solving the above equation can be done easily and gives x=R/sqrt(2). However, in general, this is a quartic equation, which I would just solve using WolframAlpha (example is given with D2=8, W=10, R=3). In general, if D2 is not too big compared to W, then the optimal x should be around R/sqrt(2).