Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Dear children,
I am sorry for not having had the opportunity to respond before, but I am here with you right now.
At D03A50 is the chest table. There is one entry for each of the 84 enemy types. Each entry consists of five bytes. The first byte indicates the chance that a chest appears, in 64ths. The second byte has two nibbles. The lowest nibble is the necessary level of the Analyze spell that will disarm the trap. The highest nibble tells if you can escape the trap. The player's agility must be more than eight times that value, otherwise the trap will be triggered half the time. Bit 3 through 5 of the third byte indicates the rare item rate, in 64ths. If the top two bits equal 0, then the chest contains money. If they equal 1, then the chest contains an item. If they equal 2, the common item is money and the rare item is an item. If they equal 3, the common item is an item and the rare item is money. The fourth byte indicates the common item or money amount, and the fifth byte indicates the rare item or money amount.
Item types are:
1 - 20: Hats
22 - 41: Sweaters
43 - 62: Mittens
64 - 75: Sweets
128 - 135: Balls
The upper three bits of each byte at 7ECC48 has the number of sweets. If they are 0, you have eight sweets. You can raise the limit by changing the byte at C0655A.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
It's simple to compute Fibonacci and Lucas numbers using only integer math, by taking advantage of the following facts:
L(A+B)=(L(A)L(B)+5F(A)F(B))/2
F(A+B)=(L(A)F(B)+L(B)F(A))/2
L(2A)=(L(A)^2+5F(A)^2)/2
F(2A)=L(A)F(A)
I have an old program which computes the Nth Fibonacci number and the Nth Lucas number at once using the above relations, for N up to 46999, which takes about 60ms to compute.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Yagamoth wrote:
What does this mean?
7e002b Easter egg counter
This just counts the number of times you've pressed R while holding A and L. After pressing R 39 times, the word NAS appears on the screen. I am guessing that Nasir implemented this to prove his involvement, in case Squaresoft should forget to pay him.
And further about something else. Someone posted a video of a "let's play SoM" with an interesting glitch. I'd like to add, I had something similar several times. However, the NPC-Character wasn't pushed back through a wall, it rather simply went backwards into the "nothing", using his AI to position itself. This happened Particularly often after a whip-jump in the mana-fortress. Does this have to do something with the so called "float"-status? I think the NPC backed of into the nothingness before I had control over my characters again.
It could be. This is the part of the game script that is related to whip-jumping:
06F8:
CA924E hold
CA924F ifbetween 03 1 1 jump 06FA ; If the "on jump tile" flag is set, proceed to 06FA
CA9254 setchrvar talker.19B 00
CA9258 action talker 80
CA925B complete
...
06FA:
CA927A ifpeq talker.1E4 04 jump 06FB ; If the acting character's weapon is the whip, proceed to 06FB
CA9280 call 0007
06FB:
CA9283 loopaction talker 95 ; Whip fastening animation
CA9286 complete
CA9287 call 0410
CA9289 gather
CA928A inc 01 ; Set float status
CA928C iftalker man call 06FC ; Move man, woman, then sprite
CA9290 iftalker woman call 06FD ; Move woman, sprite, then man
CA9294 iftalker sprite call 06FE ; Move sprite, man, then woman
CA9298 center
CA929A dec 01 ; Clear float status
So, if the script is somehow interrupted between CA928C and CA929A, the float status remains set. I don't know how that can happen, though. 7E00D0 holds the current script state.
About the Raffi in the Mountain: Is this actually a bug, that he doesn't always appear? I can remember seeing a screenshot, where there was a chatbox but no Raffi - so I guess it's a bug.
This happens when there are already 3 enemies on the screen. The game can't handle more than 3 NPCs in the fighting areas.
I guess there is somewhere a flag set to 1 while you normally can't move that allows your characters to travel through non-walkable-obstacles (as with my occurance, I believe the NPCs backed off while I still was unable to control the character). I wonder whether this also would allow us to go out of the screen with a character...
In case there really is such a flag, is it reset to non-walkable when a screen transition happens? I mean, it's worth a try to 'activate' the 'floating' while in the same frame changing area (though, that may be impossible due to the 5 frame(?) rule described earlier)
Bit 0 of 7ECF01 enables walking through walls. It is not reset during screen transitions.
Using this technique we may be able to use some pole-skips without even equipping the rope. Only problem being, that either an exit/entrance or another pole is required to regroup.
The characters only gather at jump tiles if the player has the whip equipped, but they also gather when talking to Neko.
Which "attack" is being executed when you do the rope-jump? May it be possible to force this via charge-glitch or is it bound to the poles?
Action 9E, but it's useless, and there is no 9E in the attack table either. The aforementioned bit must be on to pass through walls.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
That's easy.
Assume that is false. Then, exists. Because , the difference does not exists, and there is a contradiction. I have proved that the expression is not false, therefore it must be true.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
By the way, if there is a way to zip into Knuckles' room from the left in Carnival Night 2, you can also do the same glitch there, if you passed him earlier in the level.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Here is what needs to be done:
- You must be playing with Sonic as your main character.
- You must start the game from the beginning.
- You must not die or enter a special stage during Angel Island Zone.
If you're playing as Sonic and Tails, perform these steps:
- When you are around X position $41B0, have Tails running behind you and brake with Tails for 5 frames before he falls.
- When Robotnik is leaving the screen and you reach position $4680, brake for 2 frames with Tails and start running again.
- When you reach position $4700, brake with Tails again for 5 frames.
If you're playing as Sonic alone, do this instead:
- When you reach position $41E8, brake for 5 frames.
- When you reach position $4700, brake for 2 frames.
If everything went right, the switch will now mistake itself for Knuckles, and hit itself as soon as it loads.
It's also possible to make the bridge explode before the capsule is hit, but this is useless. To proceed to Hydrocity, the score tally must have ended before the switch loads.
The main thing you want to accomplish with all this is to make sure that the switch is loaded in slot 6 ($FFB1BC). This is where Knuckles was loaded at the beginning of Act 1. If Tails is with you, slot 5 must also be free, because of an additional object that is created that keeps you from controlling Tails.
Check list:
- Before you begin, $FFFAA4 should contain the value $B1BC.
- When the bomb plane is flying above you, $FFB1BC should contain the value $00050390.
- When entering the boss area, $FFB1BC should contain the value $00000000. If playing as Sonic and Tails, $FFB172 should also contain the value $00000000.
I'd be interested in knowing what happened earlier in the stage, because you've somehow managed it without doing anything special at position $4680 and $4700.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
There is a problem with the main page where names of published movies are missing.
Some movie pages also have problems, for example http://tasvideos.org/1400M.html. It says:
error_handler:
errstr=Invalid argument supplied for foreach()
errfile=/home/tasvideos/public_html/modules/moviechangelog.php
errline=113
And:
error_handler:
errstr=array_key_exists() [function.array-key-exists]: The second argument should be either an array or an object
errfile=/home/tasvideos/public_html/inc/wikifun.php
errline=571
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Generating subtitles automatically is easy.
Here is an example, displaying your HP in Mega Man 5.
prevtext="";
prevframe=emu.framecount()
sub=io.open("test.sub","w+b")
srt=io.open("test.srt","w+b")
srtcount=1
while true do
text=string.format("HP = %3d",AND(memory.readbyte(0xb0),127));
if(text~=prevtext) then
frame=emu.framecount()
time=string.format("%02d:%02d:%02d,%03d",(frame/21600)%60,(frame/3600)%60,(frame/60)%60,(frame*50/3)%1000);
if(frame~=prevframe) then
sub:write("{"..prevframe.."}{"..frame.."}"..prevtext.."\r\n");
srt:write(srtcount.."\r\n"..prevtime.." --> "..time.."\r\n"..prevtext.."\r\n\r\n");
prevframe=frame
srtcount=srtcount+1
end
prevtext=text
prevtime=time
end
emu.frameadvance()
end
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
I have found out what causes the glitch.
Many of the game's objects remove themselves when they are far from the screen. One of the objects that do this is the one responsible for handling running on water (function $383BC). This object's X coordinate follows the first player. It is placed at the fixed address $FFB172 when starting any of the Hydrocity stages. At the end of Hydrocity 1, it is again placed at $FFB172. Normally, this just overwrites the walking on water object from Hydrocity 1 with a fresh one.
To figure out whether to remove an object, the game checks the X coordinate against the left edge of the screen, in 128 pixel increments. If the object's X & $FF80 is 640 or more pixels to the right of the left edge & 0xFF80, or 256 or more pixels to the left, the object is removed.
At frame 12276, you have zipped through the wall so that you are at X coordinate $210A, while the left edge of the screen is at $1EFA. Notice that your X coordinate, anded with $FF80, minus the value at $FFF7DA now equals $30A. When it is $300 or more, the water running handler disappears. The slot is free, and the object that replaces it is the boss. When the boss is defeated, the very last thing it does is to set up a smooth movement of the edges to the next level's settings. However, before it gets the chance, the next act loads, and the walking on water object is again loaded to $FFB172. So, the edges aren't set.
If you turn around, dust will appear under Sonic's shoes, taking up a slot so that the loop above the boss room gets the water running handler's slot, and the boss gets another slot, and it won't work.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Warp wrote:
I understood your explanation, and it makes it clear what A is measuring using its own clock. However, I still have hard time deducing the answer to the original problem: Let's assume there's a visible clock on the bow of ship B, which A can see during the entire experiment. What does someone located at the bow of ship A (or at some other point in ship A) see as the value of this clock at different times, and how does this relate to what someone on ship B sees (when looking at his own clock)?
At time t according to A, B's bow clock shows t/2 and B's stern clock shows t/2+1.5. The time difference is equal to the relative speed times the distance (as measured by B) divided by c squared.
What A actually sees depends on the distance between A and B, since the light has to travel from B to A before A can see it. Using the following variables:
u: Observer's time coordinate according to A
v: Observer's space coordinate according to A
t': Observed clock time
t: Corresponding A's time
x: Relative distance in the direction of movement between A and B
y: Relative distance in the perpendicular direction between observer and B
s': Relative speed
λ': Relative gamma
Then we'll have:
This can also be written as:
If we plug in our gamma, we get:
This gives B's clock time as seen by an observer at [v u] in A's frame.
As for B, he will of course at his coordinates [v u] simply see his bow clock time as .
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
marzojr wrote:
And so on. The trickiest thing to calculate are the speeds of A and B according to the external reference frame, as well as the gamma factor for this speed; this is because you need to calculate the hyperbolic angle and use inverse hyperbolic tangents and so on
Actually, no calculation of angles is necessary, just a plain old quadratic equation:
One can easily see, by writing the last two equations in terms of λ' and factorizing, that in general in this situation with an observer positioned in the middle between two ships according to his own inert frame, one can calculate the relative speed and λ between each ship and the observer from the λ' between the two ships in the following way:
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Warp wrote:
If ship A had a measurement device at the half point of the ship which looks what the clock at B's bow is showing (at the moment when B's bow is exactly at this half point of ship A), would it see a different value than the measurement device located at A's bow? If yes, what would the two readings be?
If there is a clock at B's stern synchronized in B's time with the clock at B's bow, it will show 2 seconds when the clock at B's bow is showing 0.5 seconds. It will always be 1.5 seconds ahead of the clock at B's bow as seen from A.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Warp wrote:
Since it took 1 second for the bow of ship A to go from the bow of ship B to its stern, and since A measured B to be exactly half the length of A, it would mean by reciprocity that it would take 2 seconds for the bow of ship B to go from bow to stern of ship A.
No. B is only half the length of A along A's space direction. Along B's space direction, it is twice the length of A.
Warp wrote:
However, by using the clock of ship B, it takes only a half second for the bow of ship A to go from the bow to the stern of the ship B. Since this is from the time reference of ship B, it would seem that ship B is measuring a half second for this
Again, you are using the clock of ship B to measure the time along A's time direction. If you measure the time along B's time direction, you get 2 seconds as expected.
Call the proper length of each spaceship l. Then, it is apparent that (in implied SI units):
The length of each spaceship in the observer's frame is:
Let's define some events:
E0: when the bows of both ships coincide
E1: when the bow of ship A coincides with the stern of ship B
E2: when the bow of ship B coincides with the stern of ship A
We will write them as [x t] where x is the distance along the direction of movement, and t the distance in the time direction, from event E0. Ship A will be going forwards along the x axis, and ship B will be going backwards.
The transformation from A to observer (and from observer to B) is:
From observer to A (and from B to observer):
The transformation from A to B:
And from B to A:
So, the coordinates for each event in each frame is:
To calculate the length l' of ship B along A's X axis, one has to solve the equation:
Multiply by and we get:
It's easy to see that the process is exactly the same for measuring the length of ship A along B's X axis.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
I would ask "if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?"
That doesn't work.
Let A represent whether he would like to have sex with you, and let B represent whether he is a liar. Then let C be the answer to "Would you like to have sex with me" and D be the answer to "if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?" This has the following solutions:
A=true, B=false, C=true, D=true
A=true, B=false, C=true, D=false
A=true, B=true, C=false, D=true
A=true, B=true, C=false, D=false
Therefore the question has no answer if he doesn't want to have sex with you, and is undecided if he does.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Here are some turkhits I did. Check them out!
Själv med hjälmen gör hon ord - Even with the helmet she makes words
Kalven bruker serviett - The calf uses a napkin (in Norwegian)
Mats sin mor Mats' mom (with some Norwegian)
(For those in the rest of the world who don't know what turkhits are, it's foreign songs with misheard Swedish lyrics.)
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Haha, this was an odd one. I dreamed that I was in heaven, with God. He asked me if my name was (reptilian alien related word) and I said no. I felt a bit uneasy, as if I was lying to God. At some point I asked if equipment existed that could be used to communicate with their world, and how the earth side of such equipment would work, but I think they just gave an avoidant answer. They also talked about a supervillain named Zhao something. I also had the pleasure of talking to one of the Goddesses there, although I don't remember what it was about and I didn't catch her name.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Usage of a bot is a legitimate step in making a movie. A TAS movie may be made in any manner possible (except plagiarizing someone else's input without permission) as long as others can verify it in an emulator.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
It means that the metric is preserved in the embedding. If you regard a torus as a rectangle that wraps in both directions, when you embed it in 3-space, as you move across the rectangle you will in general move at a different speed along the 3-space embedding, so the metric is not preserved here.
Experienced Forum User, Published Author, Active player
(282)
Joined: 10/8/2006
Posts: 356
Most games were in English only. Some games were in English, French and German, and some had different English, French and German version. Secret of Evermore and Illusion of Time were in Spanish. Then there was Shadowgate, which was in Swedish.