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CoolHandMike wrote:
Huh. So I take it CUI's inputs using the actual previous publication rom do not sync then?
CUI is using the same ROM as woabclf (previous submission).
However, the ROM used differs from the one posted at romhacking.net, by a single byte.
More details at https://tasvideos.org/Forum/Posts/235094
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I made a semi-comparison encode with a left sidebar showing parts of the previous best submission for comparison purposes. This TAS video and audio remain intact without any mixing or pausing.
Link to video (available in 1080p60)
Also, this is an interesting TAS, with the new strategy of using Ice Mario and Penguin Mario.
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I only watched around 30% of this TAS. All it did was remind me of this submission.
I think I might have found it a bit more interesting, if there was some use for Super Knuckles after Emerald Hill Zone. Unfortunately, it was not meant to be.
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OmnipotentEntity wrote:
So to find critical points of this function we just need to take the derivative. This is (x^2 - x - 6)e^x which factors into (x-3)(x+2)e^x. Set equal to zero and we have critical values of m at y = 0 (from e^x at negative infinity), y=7e^-2 (from the (x+2)), and y = -3e^3 (from the (x-3)). So this function has zero real roots for m < -3e^3, exactly one at m = -3e^3, two on the range -3e^3 < m <= 0, three on the range 0 < m < 7e^-2, two at m = 7e^-2, and finally one for m > 7e^-2.
Just to add:
From the derivative, f(x) is continuous and strictly increasing on (-inf,-2], strictly decreasing on [-2,3] and strictly increasing on [3,inf).
Furthermore, lim[x->-inf] f(x) = 0 and lim[x->inf] f(x) = inf.
So f(x) is increasing from (but not including) 0 to 7/e2, then decreasing from 7/e2 to -3e3, then increasing from -3e3 to infinity. A graph of f(x) is shown below:
So the equation (x2-3x-3)ex - m = 0 has:
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Nice to see you back, p4wn3r.
I'm guessing this is just calculus?
Edit: Deleted wrong solution. I really should have looked up what the curve looked like first.
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(very large; recommended to view image in new tab)
Here is a PNG that I labelled showing what I believe are all the colorable regions (let me know if I missed any!). The only purpose of this PNG is to identify the colorable regions by labels 1..9a..zA..V (excluding l and I). Generally I let 1 be the background, 2 the ground if separately colorable, and 2/3 the largest region of the dino. Beyond that, there is no rule to the labelling. I tried to keep labels relatively close together, and the sclera usually has an earlier label than the eye itself (sorry for the mediocre positioning).
#2, #6 and #14 were the only ones I found that have phantom bounded regions that are forced to be the same color as the background "1" (they are indicated in the PNG).
Using 2x3 or 3x2 overlapping rectangles as the basis for when regions touch each other, I made a list of adjacencies. Each pair of regions that touches each other by this rule is specified, with the earlier label (that comes earlier in the ordering 1..9a..zA..V) on the left of ":" and the later label on the right.
I have also included a number of possibly debatable "touches" that are not covered by the above rule. All of them concern the eyes. The debatable ones are in brackets () and also has a question mark somewhere in there. #3, #9, #10, #11, #12, #13 and #16 are affected.
From discussion so far, it seems that the eyes should be deemed touching whichever regions make up the sclera? Or anything nearby? In that case, the rule would need a bit of updating. (The goal is to have a rule that aligns with common sense.)
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Is it this hack: https://www.romhacking.net/hacks/7839/ ?
(It would help if the game title was correct.)
Anyway, I didn't watch this TAS (since the author hasn't exactly posted their submissions in an accessible manner), but I saw a longplay, and this hack doesn't deserve to be named after Rosenkreuzstilette. Sure, the backgrounds in this hack are sometimes cool. But not justified in having "Kreuzstilette" in its name.
(BTW the enemies and bosses are completely unchanged from MM3, and the weapons too. So it's pretty much a level modification hack of MM3 with some changed foreground/background graphics. Also the music is taken almost entirely from other NES Mega Man games.)
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Randomno wrote:
like I'm the intended demographic for Color a Dinosaur?
lol
Though I realized shortly after my post that I could have just used overlapping rectangles. So like (words in [square brackets] are optional):
Two [colorable] regions are deemed to be touching if there is a 2x3 pixel rectangle [in any position and orientation] that overlaps both of them.
In case of certain edge (no pun intended) cases, we could include/replace with a slightly larger rectangle with an additional restriction:
Two [colorable] regions are deemed to be touching if there is a 2x3[ and/or 3x3 and/or 3x4] pixel rectangle [in any position and orientation] that overlaps both of them, and no other [colorable] regions.
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Maybe: Two separately colorable regions A and B are deemed to be touching if there is a colorable pixel in A and a colorable pixel in B where the distance between their centers is less than 2.5 pixel lengths.
The goal is to consider minimum(-distance) offsets (±1,±1), (0,±2), (±2,0), (±1,±2) and (±2,±1) as touching/adjacent, but (±2,±2), (0,±3) or (±3,0) as NOT touching/adjacent (so as to avoid the problem of "corner adjacency" which is not a valid interpretation of adjacency in the four-color theorem). The former offsets all have distance sqrt(5)≈2.236 or less, and the latter ones have distance sqrt(8)≈2.828 or greater.
Possible issues: Colorable regions could have minimum offsets of (0,±1) or (±1,0), which may lead to "crowding" (but I'm not aware of any cases so far). Colorable regions could also have minimum offsets of (±2,±2), (0,±3) or (±3,0) and should be considered touching by common sense (but I'm also not aware of any cases so far).
Using this metric, the pink segments have a minimum offset of (±2,±1) and so are deemed touching. Around the 2x2 black square in the same image near the top center, the outside and inside blue segments are not touching, and the left yellow and right pink segments are also not touching, because the minimum offsets are (0,±3) and (±3,0), respectively.
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Since when did Roll become Zero?
This homebrew is quite impressive. There are a lot of features, even ones that come from more recent games like Mega Man 11. And of course there are multiple characters you can play as. Even an option to play as "Man" (one has to wonder if it had anything to do with this April Fools joke). At least with "Man" it is possible to do the fabled "slideless" category now. It was impossible before then.
Anyway, I played around a little bit and identified some ideas that may improve this TAS (I didn't test completely so I may be wrong, and this is by no means complete):
* Use Dive Missile on Ring Man stage first miniboss (and the similar miniboss later).
* Use Dust Crusher on Dive Man stage minibosses.
* In Dust Man stage, destroy the junk blocks with jumping slashes, rather than standing slashes. Ring Boomerang not required.
* In Dust Man stage, after the junk block section, destroy the enemy blocking the way with Pharaoh Shot (one charged, three uncharged) instead of taking a hit.
* Possible route change: Pharaoh -> Dust -> Dive -> Ring -> Bright -> Toad -> Drill -> Skull
* Cossack 3 boss: Try taking damage so you can slash the bottom spider bot? Not sure about this one (on Hard it is extremely easy to die).
One other thing to note is that there is an easter egg based on a MM4 glitch, which is triggered by getting pushed by the auto-scrolling screen in Cossack 3 while hanging with the Wire item. This allows skipping the first part of Cossack 3. However, it's not faster to do this because of the time required to get the Wire item. Only if you had a goal which required you to get the Wire item (or you used the setting that gives you all the items right away) would it be worth doing.
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Based on the information I am given so far, I can only presume that maximum likelihood estimation was intended as the solution method. (After all, we are not allowed to assume anything about the prior distribution ... what else can we possibly do?)
So, with regard to MLE, let n be the total number of tiles and k be the number of blue tiles. The number of ways of choosing 4 blue and 8 yellow tiles is f(k)=C(k,4)C(n-k,8), and we need to find the k that maximizes this. (Here, C(a,b) denotes a choose b.)
Now the ratio of successive terms is f(k+1)/f(k)=[C(k+1,4)C(n-k-1,8)]/[C(k,4)C(n-k,8)] = [(k+1)(n-9-k)]/[(k-4)(n-k)], and so find when f(k+1)/f(k)<1:
[(k+1)(n-9-k)]/[(k-4)(n-k)] < 1
-k^2+(n-10)k+(n-9) < -k^2+(n+4)k-4n
5n-9 < 14k
(5n-9)/14 < k
So f(k+1)/f(k)>1 when k<(5n-9)/14 and f(k+1)/f(k)<1 when k>(5n-9)/14. This implies that the sequence f(0), f(1), ... , f(n-1), f(n) is maximized at the first k value when f(k+1)/f(k)<1 (that is, ceiling((5n-9)/14) ).
For n=100, this value is ceiling(491/14) = 36, so the drawing of 4 blue and 8 yellow is most likely when there are 36 blue and 64 yellow tiles in the bag to begin with. After drawing, this leaves 32 blue and 56 yellow, so the probability the next draw is blue is 32/88 = 4/11.
For n=30, this value is ceiling(141/14) = 11, so 11 blue and 19 yellow. After drawing, 7 blue and 11 yellow are left, so the probability of drawing blue is 7/18.
As the number of tiles goes to infinity, the proportion of blue tiles in the maximum likelihood estimate approaches 5/14, so the limiting probability of drawing blue is 5/14.
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OmnipotentEntity wrote:
FractalFusion wrote:
As for the bag of tiles question, I don't have an answer yet (I'm still trying to be sure of what is assumed. A priori, is each tile blue or yellow with equal probability, independent of the others?
So the initial probability that a blue or yellow tile is drawn is an unknown. Assuming the bag initially has 50% blue and 50% red tiles is not justified.
BrunoVisnadi wrote:
It feels like it's necessary to make some assumptions about the probability distribution of the colours in the bag.
As you may have observed, the only assumption you need, once you condition upon the likelihood of the observation of the initial draw, is that the probability of any tile being drawn (independent of its color) is uniform.
So, if I'm getting this right, are you saying that we make no assumptions about the initial probability distribution (this is before anything is drawn at all)? Not even that the initial probability distribution is a binomial distribution?
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Regarding the triangle problem I posted:
Obviously 3 is the smallest side; otherwise we can scale up the triangle to get one with larger area.
Hero's Formula is the most direct way of finding the other two sides, although it is more convenient to use a=3, b=r, c=2r-3 in the formula A=(1/4)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)):
A^2 = (1/16)(3r)(3r-6)(r)(6-r) = (-9/16)r^2(r-2)(r-6) = (-9/16)(r^4-8r^3+12r^2)
So solve 4r^3-24r^2+24r=0 ----> r=0 (reject) or r^2-6r+6=0 ---> r = (6±sqrt(36-24))/2 = 3±sqrt(3) (reject negative)
So r=3+sqrt(3) and you can check it gives the maximum.
Now actually if you take 3, r, 2r-3 triangle with angle θ in-between the sides 3 and r (see below), using the Cosine Law and rearranging gives the polar equation r=4-2cos(θ) which is the limaçon shown below. The triangle 3,3+sqrt(3),3+2sqrt(3) gives the red triangle below, and you can see it coincides with the point on the curve with the greatest y-value. (The points on this curve represent the third vertex of a 3,r,2r-3 triangle, if the base is on (0,0) to (3,0). So the one with the greatest y-value will have the largest area.)
----
As for the bag of tiles question, I don't have an answer yet (I'm still trying to be sure of what is assumed. A priori, is each tile blue or yellow with equal probability, independent of the others?
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Comparison encode (now in 1080p):
Link to video
One thing to note is that things are different these days compared to 13 years ago, so I am all right nowadays to having multiple branches.
The other thing is that "normal" and "hard" are actually two different hacks. One hack is named "RockmanNo_constancy_normal.ips" and the other "RockmanNo_constancy_hard.ips". They are both given in the same package and very similar, but they are two different hacks.
The other other thing to note is that this is a TAS of the v1.0 version of this hack. According to the Romhacking.net page, there are two (very very late) additional updates: v1.2 in 2018 and v1.3 in 2019.
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DrD2k9 wrote:
Now I'm curious. Are you referencing the roll-over as bad coding?
What additional coding would it require for this not to happen? Something as simple as an if/then statement along the lines of this:
if Score>999900 then ScoreDisplay = 999900 end
I was somewhat joking when I made that post.
The pseudocode is similar to what I had in mind (something like if Score>999999 then Score=999999 end). Obviously this is in reference to pseudocode; stuff like assembly can't be done in one line.
And of course, the simplest answer to why a roll-over exists is: "The developers didn't care."
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It's a bit difficult for me to watch FF7 more than once. I figured though that the TAS is well-done, though using mostly strategies from the no slots TAS.
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A short note regarding the previous post: I did similarly to OmnipotentEntity, but it turns out that Riddler has an explanation which doesn't even use recurrences.
----------------------------
I was thinking of something very recently.
On the left is the classic 3-4-5 triangle, which has an area of 6. Notice the sides are in arithmetic progression.
Now, is 3-4-5 the largest (by area) triangle with sides in arithmetic progression having one side of length 3? The answer is no. There is the 3-5-7 triangle on the right whose area is 15sqrt(3)/4 ≈ 6.495.
Question: What is the largest (by area) triangle with sides in arithmetic progression having one side of length 3?
Note that the other two sides can be any real numbers (do not need to be integers).
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I remember this game.
It's cool that you did a TAS without using some of the more broken glitches. There are some impressive platforming exploits in there.
And of course the battles are well done.
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Is the emulator BizHawk?
You might need to go to File -> AVI/WAV -> Config and Record AVI/WAV and use the Resize Video option (though I haven't personally tested it).
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DrD2k9 wrote:
If one only obtains the maximum possible point value for each duck, completing up through Round 24 will yield a score of 975000 before the "Perfect" shooting bonus.
The 30,000 bonus points will then cause the score display to roll-over (back down) to 5,000 points.
More bad programming that could have been prevented with one line of code.
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If there is one word to describe why Tekken TASes are worth watching, it's "Yoshimitsu". The guy who stabs himself to KO the opponents.
I like that, in this version, the credit sequence replays all your KOs. Sometimes it's worth seeing it twice!
So the equation (x2-3x-3)ex - m = 0 has:
(very large; recommended to view image in new tab)
Here is a PNG that I labelled showing what I believe are all the colorable regions (let me know if I missed any!). The only purpose of this PNG is to identify the colorable regions by labels 1..9a..zA..V (excluding l and I). Generally I let 1 be the background, 2 the ground if separately colorable, and 2/3 the largest region of the dino. Beyond that, there is no rule to the labelling. I tried to keep labels relatively close together, and the sclera usually has an earlier label than the eye itself (sorry for the mediocre positioning).
#2, #6 and #14 were the only ones I found that have phantom bounded regions that are forced to be the same color as the background "1" (they are indicated in the PNG).
Using 2x3 or 3x2 overlapping rectangles as the basis for when regions touch each other, I made a list of adjacencies. Each pair of regions that touches each other by this rule is specified, with the earlier label (that comes earlier in the ordering 1..9a..zA..V) on the left of ":" and the later label on the right.
I have also included a number of possibly debatable "touches" that are not covered by the above rule. All of them concern the eyes. The debatable ones are in brackets () and also has a question mark somewhere in there. #3, #9, #10, #11, #12, #13 and #16 are affected.
Using this metric, the pink segments have a minimum offset of (±2,±1) and so are deemed touching. Around the 2x2 black square in the same image near the top center, the outside and inside blue segments are not touching, and the left yellow and right pink segments are also not touching, because the minimum offsets are (0,±3) and (±3,0), respectively.
----
As for the bag of tiles question, I don't have an answer yet (I'm still trying to be sure of what is assumed. A priori, is each tile blue or yellow with equal probability, independent of the others?
On the left is the classic 3-4-5 triangle, which has an area of 6. Notice the sides are in arithmetic progression.
Now, is 3-4-5 the largest (by area) triangle with sides in arithmetic progression having one side of length 3? The answer is no. There is the 3-5-7 triangle on the right whose area is 15sqrt(3)/4 ≈ 6.495.
Question: What is the largest (by area) triangle with sides in arithmetic progression having one side of length 3?
Note that the other two sides can be any real numbers (do not need to be integers).
I like that, in this version, the credit sequence replays all your KOs. Sometimes it's worth seeing it twice!