Also, do you happen to know the timeframe for how long the upper conveyor will continue to move outwards, assuming you do not trigger it to change with your vertical position? It seemed to last about a half-cycle
I think it's more like 3/4 of a cycle.
So this would mean that the probability that it spawns on the bottom is 96/256 = 0.375
Hey Jeff, hopefully you're still following these threads. I've been thinking more about this whole thing and I'm still not sure if I'm convinced. One of these days I'll have to mess around with the MAME debugger and try to get some more concrete values for some things to make the math a bit more concrete -- unless you or somebody else already knows some of this stuff, that would be helpful!
First, your figure of 3/4 of a cycle for outward moving upper conveyors sounds like a ballpark estimate (although probably a pretty good one). I wonder if anyone knows this value more accurately? We could probably break it down into frames. We know that a bottom conveyor cycle, and a hammer cycle lasts 9 BONUS seconds on Level 4+, each of which ticks down after a certain number of frames. So, one cycle = some number of frames (it would be nice to know this number). The amount of time that the upper conveyors can move outward is some other number of frames -- if we knew these we could calculate an exact ratio.
[EDIT: Ok, in the middle of writing this up, I decided to watch a pie factory screen. The estimate of 3/4 of a cycle appears to be VERY accurate. In both directions. Meaning, the upper conveyor will switch from moving inwards to outwards after 3/4 of a cycle AND it will switch from moving outwards to inwards after 3/4 of a cycle. If you just stand at the bottom and watch, you can see the first switch happens with Kong is in the middle moving left, the second reversal is when he is positioned at the right edge of the screen, the third reversal is when he is in the middle moving right, and the next is when he is positioned at the left edge of the screen (3 full cycles). ]
[EDIT: I double checked another source -- verified that the upper conveyor cycle is exactly 3/4 of the bottom conveyor cycle. ]
Also, it sounds like outward moving conveyors can be triggered to immediately reverse and move inwards, but the reverse is not true -- you cannot trigger conveyors to move outwards -- it just happens after Jumpman is below the vertical threshold for a certain number of frames (this is an assumption, and may not be accurate) -- so, does this mean that the switch to moving outward is based on relative or absolute timing? For example, if Jumpman never breaks the threshold while the conveyor is moving outwards, then it will likely switch back and forth between inwards and outwards at some default point (which could be predicted by watching the Timer), but if this cycle is broken by Jumpman breaking the threshold ... how is it handled? Does it now reverse and travel inwards for the "full" period of time that it normally would (which would mean that it's next reversal outwards occurs at a non-default time on the Timer), or does it just finish up some fraction of its cycle and then reverses outwards again according to its default time on the Timer? ... If I had to guess, I would think that once it gets triggered to change from outward to inward, it is forced to remain moving inward for a set period of time. Then, it will periodically check Jumpman's vertical position to determine whether or not to switch to moving outwards (but this periodic check isn't particularly often -- NOTE that you don't usually see it switch from inward to outwards immediately after jumpman moves down below the threshold.)
[EDIT: Did some poking around and I'm fairly certain now that this direction change occurs only with absolute timing. Meaning, after another 3/4 of a cycle goes by, it will check to see if the direction can be reversed, and if so, it reverses (in either direction). Note that this means if it's normally going inward for the first 3/4, then outward for the next 3/4 and then inward again for the next 3/4 -- if Jumpman "triggers" a reversal during the second cycle so that it immediately switches from outward to inward, this means that on the 3rd 3/4 it will now switch to outward (despite the fact that with no triggers it would normally be switching from outward to inward). ]
The next fact that would be nice to know now is the rate of pie deployment...
[EDIT: Ok, for what it's worth, I "THINK" that a pie is deployed once every 124 frames and a full cycle is 512 frames (upper conveyor cycle is 384 frames). However, I think there's an offset with the pies where they wait until the initial fireballs are all spawned before pies come out -- so, depending on where you are in this 124 count when the upper conveyor reverses, I'm not sure that you are (always) guaranteed the 3rd pie -- some confirmation on these numbers would be useful! ]
Anyways, getting back to our "normal", or "best case" scenario for pressing the screen by getting the bottom hammer without delay. Let's assume that there's no pie to "chase down" with our hammer as the conveyor moves from right to left. The two obvious choices are:
1) Grab the hammer with just a bit more than 1/4 of a cycle remaining -- this allows the upper conveyor to reverse to an outward direction just after the hammer is grabbed. This lasts for 3/4 of a cycle. HOWEVER, for the first 1/4 of a cycle you are simply waiting around doing nothing while waiting for the bottom conveyor to reverse! Once it reverses, you are guaranteed some pie smashes for the next 1/2 cycle (how many smashes is this, I think it's still open to debate...). Then, during the last 1/4 cycle, you "might" get more smashes depending on whether or not the pie(s) spawn at the bottom, with a probability of 3/8.
2) Grab the hammer with just a bit more than 0/4 of a cycle remaining. The bottom conveyor almost immediately reverses direction, possibly yielding pies immediately. However, the upper conveyor, which already switched from inward to outward for the most recent 1/4 cycle, now immediately reverses back to inward. BUT, here's the thing -- I'm pretty certain that in this case, the upper conveyor remains moving inward for ONLY a 1/2 cycle! Because the reversal times are absolute -- so the 1/4 cycle that was already used up, plus the next half cycle, means that it will reverse back to outward when the hammer is only HALF expired! So, you have your first 1/2 cycle where you are smashing pies with 3/8 probability followed by another 1/2 cycle where you are guaranteed smashes (again, how many this is is still open to debate). In theory, I believe it's possible to smash 5 pies that are moving left to right with this method (this does NOT include chasing down pies moving right to left).
Granted, some of these figures are probably slightly inaccurate since you would typically grab the hammer slightly before reversal -- and, I think the 3/4 cycle of the upper conveyor might have a very slight offset -- the second reversal doesn't seem to occur EXACTLY when Kong is at the rightmost point of the screen, but a small fraction afterwards -- so some of this is thrown off slightly. Lastly, there is the issue of being somewhat behind and out of position after the hammer expires when using option 2 so that it's impossible to maxamize smashes AND make a right side escape for 5700 Bonus -- instead you will often end up getting caught on the retractable ladder and ending up with 5100 - 5200 Bonus. However, I think the point that I'm trying to make is that after Jeff first posted his observations, it seemed like option 1 had a clear advantage -- and now I'm not convinced. I think it might turn out to be a wash ... which would be good -- it would give the player more flexibility to play the situation based on risk and on fireball behavior rather than based on will happen with the pies. If anyone is able to make any of these observations more concrete, feel free to post your thoughts.