When I took the right ladder earlier in the level (which is faster in theory)...
Memory address 05B7
At the left turnaround point, the platform reaches X-position 088.
It moves 2.5 pixels per 3 frames.
So its X-position goes like 088-091-093-096-098... when it moves rightward.
The right turnaround point of its journey is 010 after going beyond 255.
So the rightward pattern finishes as ...253-000-002-005-007-010.
Moving leftward is just the reverse order. 010-007-005... until 088.
So 178 pixels from one side to the other at 2.5 pixels per 3 frames.
The whole one-way journey takes 213 frames.
So if the platform is traveling away from you when you reach the cliff, you'll have to wait almost double that amount (426 frames = 7.1 seconds).
So you have to manipulate the position of the platform with sub-optimal play earlier in the level.
It first appears at the memory address when you reach this point:
When I tried the left ladder instead, the platform didn't show up at 05B7. I don't know its memory address. But on either path, going through optimally is going to lead to a wait of about 6 seconds for the platform. It has to be manipulated by wasting some frames, like the big man on stage 2. It's trial and error.
It's very complicated.
Fortunately it's the last super-difficult thing to do in the game.
I'm almost ready to move on from this part.
It should be a few frames faster than v9.
But it's disappointing because a perfect spawn could save about 3 seconds.
It always spawns at X-position 093, which is almost the left edge. It moves left to the edge (088) and turns around toward the right edge (010).
While making your way to it:
0. Mini-right to left (093-088)
1. Left to right 1 (088-010)
2. Right to left 1 (010-088)
3. Left to right 2 (088-010)
4. Right to left 2 (010-088)
5. Left to right 3 (088-010)
6. Right to left 3 (010-088)
So we want the third full right-to-left cycle to sync perfectly with when Arthur reaches the edge of the cliff.
Optimally, I can get there around 22825.Language: luafunction stuff() Xcam = memory.readbyte(0x64)+memory.readbyte(0x65)*0x100 for i = 0, 18 do id = memory.readbyte(0x4f0+i) y = memory.readbyte(0x509+i) x = memory.readbyte(0x554+i) if x< 10 then x = 10 elseif x>250 then x = 250 end if y< 0 then y = 0 elseif y>210 then y = 210 end if id > 0 then gui.text(x,y,string.format("%X",id),"#00ffffff","#000000ff") end gui.text(30, 10, Xcam) end end emu.registerafter(stuff)
The casual viewer of this movie won't even think about how perfect this timing is.
But for me it's satisfying to finally see this platform perfected after first TASing this game 12 years ago.
I get to the boss at 24418.
In v9, I get to the boss at 24652.
So it is now 234 frames (3.9 seconds) faster overall than v9.
The platform improvement saved 168 frames (2.8 seconds).
If you perform a crouching attack on the first skeleton monkey on the exact right frame, the skeleton monkey on the floor above doesn't activate.
If it does activate, it ends up standing on top of the ladder when you climb it (so you have to take damage or wait).
Taking damage there isn't an option because of the damage boost earlier.
I'm going to try to finish it this weekend.At 5 and 6 levels of the game it's possible to save huge chunk of time by damage boosting into the walls. At the 6 level by boosting of the red arremer you can start clipping upwards by jumping as shown in the movie. With good optimization this trick can save ~5 second over current TAS. At the beggining of the level 5 it's possible to perform the same trick and skip first part of the level (as shown in the movie). Also by doing this you prevent enemies from spawning. But I can't met platform cycle. So I end up saving not that much time. But if it possible to manipulate that platform... as i think it can save ~10 second.
