Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Adding to what Spikestuff said:
As far as I know, VBA Lua savestates have never worked properly (tends to result in desyncs). So if you were thinking of botting Pokemon, it won't work in VBA.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Bigbass wrote:
It was nice that the volume equation canceled out the square root.
Yes, that's what I had in mind (disc integration). It's interesting because if you tried to find the area of the 2D shape (not the volume of the 3D egg formed as a surface of revolution), you would get an elliptic integral which is not elementary.
p4wn3r wrote:
Prove that, given the exact sequence (*), if one can "reverse the arrows", that is, find an exact sequence of the form
0 -> C -> B -> A -> 0,
then the sequence (*) splits.
This is just the splitting lemma, right? Anyway I trust that what Wikipedia says is true since I'm not going to analyze it any further.
I ended up learning that "abstract nonsense" is an actual term, used by actual mathematicians. And here I thought it was a term p4wn3r made up. Now I'm even more confused.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
I was thinking about egg-like shapes and their volumes, when I created this.
The graph below is the part of the elliptic curve y^3 - y = -3x^2 above the x-axis:
Rotate this graph around the y-axis to get a 3D shape that looks like an egg standing on end. What is the volume of this egg?
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
p4wn3r wrote:
Saw this one on the internet and it was fun.
What are the last two digits of 7^7^7^7^7^7^7?
Note: of course if you do lots of computations the problem is pretty easy. Try to solve it in the style of math contests, finding some property that makes the solution come out easily.
Do you mean 7^(7^(7^(7^(7^(7^7)))))?
(It would be easier to see this if it was typeset in proper math notation.)
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
I didn't know there was a puzzle game based entirely on pentomino tilings!
Considering pentominoes are the only n-ominoes (other than n=1 or 2 which are trivial) for which all of them can be tiled into a rectangle, I'll post screenshots of some beautiful rectangle tilings:
Not sure what the "dancing links algorithm" is (I'm not familiar with algorithms for coverings), but it's interesting how you use mathematical algorithms to solve a problem related to a TAS.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
CasualPokePlayer wrote:
MrWint wrote:
I'm not even convinced it is faster with animations on to begin with.
gifvex actually measured that out (for another RTA category, unrelated to this submission), and Stomp is 11 frames faster with animations on instead of off. Of course, I doubt this changes anything, given factors you laid out along with 10 frame timeloss for each Tauros send out with animations on.
Don't know if it is related to what you said, but I first found out because of this video:
Link to video
where it was discussed that Stomp is the fastest move in the game with animations on, and in fact much of this run was done with animations on. So I was wondering how much faster it was with animations on instead of off. Thanks for the information.
I did eventually (like three weeks ago) come to the feeling that it was slower anyway to implement in the any% (since very few battles are actually all-Stomp battles) and the 2 or so seconds it takes to switch options twice. But I didn't bother updating my post because it appeared (until two days ago) everyone lost interest in posting here.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
I've just verified that it syncs on emulator (both for the normal Mega Man X6, as well as Mega Man X6 with the Zero Tweak).
BizHawk 2.4 appears to be required (doesn't sync on 2.5.2).
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Yes, that's right.
I got the same equations: Solve xyz= 22*3*5*72*11 and x+y+z=100.
To see how to get the solution, note that x,y,z are at most 48. This is because if any of them are 49, then the most that xyz can be is 49*25*26= 49*5*2*65 (less than 22*3*5*72*11 = 49*5*2*66), and higher numbers would result in even lower xyz (the numbers would be further apart which decreases the product).
So two of x,y,z are multiples of 7 (let's say x,y). Then the factor of 11 must go with z. Since x+y+z=2 mod 7, z must be a multiple of 11 that is 2 mod 7; the only possibility is z=44=11*22. The factor of 5 goes with one of x,y (let's say y), and so y=7*5=35 (can't be higher), and x=7*3=21. So the distribution of colors must be 44, 35, 21.
This problem was from yesterday's Riddler Express.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Saw this somewhere. Combinatorics and number theory:
A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent.
How many marbles of each color are in the bag?
(If it helps, you may assume that the colors are red/green/blue, with #red>=#green>=#blue. Or not.)
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
brunovalads wrote:
My suggestion is adding I have no name as a co-author too, since his proposed solution had the greatest impact in this improvement.
Sounds good. I will add I have no name as a co-author.
I was surprised that you apparently never saw I have no name's post. I thought you saw it already and chose to ignore it.
By the way, there is another weird thing that got me looking into this game: Prior to your submission, I didn't know anything about The Witness (I may have heard about it 5 years ago but never paid attention.) So when I saw your TAS and then I have no name's post, I started seeing "improvements" in the lower section's puzzles (the ones where you separate differently-colored dots) that didn't actually exist, because I thought I knew the rules but actually didn't. These false improvements were actually what got me to download TheWit.nes and try them out myself. So I guess ignorance actually does pay off sometimes.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
This reminds me of the first "bigame" TAS, although this submission technically isn't two different games, just Mega Man X6 and a mod of Mega Man X6 (to enable Zero).
Did you use any special scripts to do this TAS? Or just turn on background input and do it slowly? It seems it would take quite a bit of effort to do this type of TAS, though certainly not on the scale of what Masterjun did (I can't believe someone was insane enough to actually do that).
And I also didn't expect HappyLee to show any interest in any Mega Man X games. :D (I thought normally that Mario TASers stick with Mario.)
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
I was wondering about something, since I saw something that was discussing move animations (with the battle animation on) and I remembered that Tauros uses Stomp.
How much faster is Stomp with battle animation on compared to when battle animation is off?
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
p4wn3r wrote:
Prove:
If a triangle with integer side lengths has an area which is also an integer, then the area is a multiple of 6.
If a,b,c are the sides of the triangle, then Heron's formula gives the area as sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.
First, we show that s(s-a)(s-b)(s-c), and therefore the area, must be even. Suppose that s(s-a)(s-b)(s-c) is odd, which can only happen if all factors are odd. This can only occur when s=1 or 3 mod 4, and of a,b,c, either all of them are 2 mod 4 or one is 2 mod 4 and the other two are 0 mod 4. We make the following table of possibilities (up to reordering of a,b,c):
MOD 4
a b c | s s-a s-b s-c | Product
0 0 2 | 1 1 1 3 | 3
0 0 2 | 3 3 3 1 | 3
2 2 2 | 1 3 3 3 | 3
2 2 2 | 3 1 1 1 | 3
In these cases, s(s-a)(s-b)(s-c) is 3 mod 4, so s(s-a)(s-b)(s-c) cannot be a perfect square.
Second, we show that s(s-a)(s-b)(s-c), and therefore the area, must be a multiple of 3. Suppose that s(s-a)(s-b)(s-c) is not a multiple of 3. Then s is 1 or 2 mod 3 and a,b,c must not be the same mod 3 as s, but must be consistent with s=(a+b+c)/2 mod 3. The possibilities (up to reordering of a,b,c) are:
MOD 3
s a+b+c | a b c | s s-a s-b s-c | Product
1 2 | 0 0 2 | 1 1 1 2 | 2
2 1 | 0 0 1 | 2 2 2 1 | 2
In these cases, s(s-a)(s-b)(s-c) is 2 mod 3, so s(s-a)(s-b)(s-c) cannot be a perfect square.
So the area is even and divisible by 3, and so the area is divisible by 6.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Acumenium wrote:
The key difference is that Masterjun's SMB3 is not an arbitrary goal---the poll and thread were to determine Moons, which it, for whatever reason, got.
First of all, I disagree. I believe those submissions have yes votes because people see those submissions as worth voting yes for, and I have no reason to think otherwise. (Same with no and meh votes.)
Second, even if people vote for options for whatever reasons, it's not like we (who are not judges) can do anything about it. Why care so much about an online poll?
Third, those submissions are accepted, published and done with. It's time to move on.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Radiant wrote:
"least A button by framecount"
Except that's not what this submission does.
Submission Text wrote:
The point of the TAS is just beating the game as fast as posible like in any%, but with a restriction:
Every jump has to be the smallest posible
What does this means in gameplay fact is that is not allowed to press the A button for 2 frames or more, is only allowed to make jumps of 1 frame A press
The only exceptions with more than 1 A presses jumps are a 9 frames jump needed to get to the top pipe of 2º room and 2 consecutive 6 frame jumps to get to bowser area, both in 8-4, being that the best I managed to do
This submission's goal is to never press A for 2 consecutive frames or more, with two/three exceptions. Nothing about minimizing A presses by framecount.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Acumenium wrote:
"63 A button presses" (the current theorized minimum without ACE)
I know it's easy to mix up games, but I'm not aware of anything in SMB1 that can be termed "ACE" in any way, even if we tried to stretch the definition of ACE beyond all reason.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
I watched the run and enjoyed it. Didn't expect you'd switch mains yet another time (thought at first you were just use Clefable all the way and get Metronome).
Against the Hiker in Rock Tunnel, was it faster to take them out with Bubblebeam as opposed to missing a move and then manipulating a Selfdestruct miss?
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
CasualPokePlayer wrote:
Oh, and this end up being another thing Toto can do and Graveler (probably) can't :P
Not sure what you mean by this. I thought the main reasons for not using Graveler was that Geodude/Graveler is too slow and needs Quick Claw (which the Toto line doesn't need) and the moveset produces too many super effective / not very effective messages.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Games before Gen 3 tend to have slow damaging moves (even early in the game), and text boxes can be cleared much faster too.
Oh, and before I forget, I noticed that deathwarping from Radio Tower did make it into this version of the TAS. Prior to that, we weren't so sure whether it was faster.
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
kaan55 wrote:
Oh shoot, I guess it's been a while since you guys got an update on the ring attack huh.
Well, you'll be happy to know that I'm about halfway through Scrap Brain 3. Unfortunately it's hard to say when I'll be done because apparently our blue boy needs to breathe (in other words I'm having trouble with bubble rng).
Given how close I am to the end I won't make the gmv public. I would recommend watching the whole thing when it does come out. I've improved a lot of levels from their first iteration. As a result I have no idea which updated levels you may or may not have seen.
To answer your question about getting all rings. Yes it is possible.
Sounds good.
I actually didn't know that anyone was working on a ring attack for this game (since I almost never follow this thread). I thought at first it would be cool to do a TAS of this, but the amount of effort required is huge (not to mention the special stages are terrible and give me a headache even thinking about them, even though they might not be as hard to route as the main stages).
Editor, Experienced Forum User, Published Author, Skilled player
(1941)
Joined: 6/15/2005
Posts: 3247
Apparently, after putting Mordack's wand and using the cheese, you can just pick up Crispin's wand after putting it down too. Don't need to leave the screen.
I'm not going to submit anything for this game again any time soon. I probably missed some more glitches; my testing wasn't insane enough.
It was nice that the volume equation canceled out the square root.
Rotate this graph around the y-axis to get a 3D shape that looks like an egg standing on end. What is the volume of this egg?
Not sure what the "dancing links algorithm" is (I'm not familiar with algorithms for coverings), but it's interesting how you use mathematical algorithms to solve a problem related to a TAS.