Sustainability of ST/ZT stacking

[Question_Mark]

The wiki states that ST/ZT stacking is more or less sustainable under the Random Generator. After careful analysis I rigorously prove that it is not sustainable without an occasional wasted T. I also prove that LST/JZT stacking is not sustainable at all, and finally outline an efficient (no wasted T's) method of stacking for infinite TSD's and Tetrises--which I also rigorously prove is perfectly sustainable.
I attached my analysis. Enjoy reading--it's more than four pages long and has only a couple of pictures.

QM

[vipjun]

Nice analysis ,  

[Question_Mark]

Thanks, I'm glad you like it

[caffeine]

I'm having a little trouble "seeing" the "4-bag cycle." Four bags, or 28 pieces, will clear 11.2 lines. However, 4 T-Spins and 1 Tetris = 12 lines.

For this reason, in the past when I analyzed ST-stacking, I used a "5-bag cycle." That's 35 pieces and 14 lines. That means 5 T-Spins and 1 Tetris (14 lines). It's all nice and neat this way.

Left partition (yellow) = 3 I's, 5 O's, 5 J's

Middle partition (green) = 5 S's, 5 L's, (1/4)*5 T's

Right partition (red) = 5 Z's, 2 I's, (3/4)*5 T's

Here's what it looks like after it's all stacked up:

[fumen]110@pb3lmzSpLcushbssjbYiwsLwaiusLwaiusLwaiusLw?aits8eLwai+e3lai+e3llzYiRp8e3lmzSp3lmzSp3lmzSp/?emzSp/emzSp/emzSppbA4G[/fumen]

So, it appears as though you're right in that it's not stable. Nice work.

[Paul676]

nice analysis - any chance you could give us a fumen of the zt/jzt stacking idea - doesn't need to be all 24 bags used, only as many as give us a picture of what it would be like - thanks

[Question_Mark]

I'm having a little trouble "seeing" the "4-bag cycle." Four bags, or 28 pieces, will clear 11.2 lines. However, 4 T-Spins and 1 Tetris = 12 lines.

For this reason, in the past when I analyzed ST-stacking, I used a "5-bag cycle." That's 35 pieces and 14 lines. That means 5 T-Spins and 1 Tetris (14 lines). It's all nice and neat this way.

Left partition (yellow) = 3 I's, 5 O's, 5 J's

Middle partition (green) = 5 S's, 5 L's, (1/4)*5 T's

Right partition (red) = 5 Z's, 2 I's, (3/4)*5 T's

So, it appears as though you're right in that it's not stable. Nice work.

I understand if it's hard to visualize. I was trying to make things as "loopy" as possible and a 4-bag cycle seemed to work best because of the way the right partition became empty after the fourth bag of that cycle (think ZT for first bag, ZT for second bag, ZTI for third bag with the I going in column 10 after the TSD, ZTI for fourth bag with the I going in column 8 and doing the tetris--this clears the right substack completely and the loop can start over).
With your 5-bag cycle, although I openly admit it is more easy to analyze in terms of lines cleared/lines stacked, I just couldn't get the loop to work out on the right side. And I actually wasn't thinking about it in terms of lines stacked based purely on the counts of the cells--I was trying to analyze it substack-by-substack and that probably led to me over-thinking some parts of the proofs.

nice analysis - any chance you could give us a fumen of the zt/jzt stacking idea - doesn't need to be all 24 bags used, only as many as give us a picture of what it would be like - thanks

As soon as I figure out how to use the Fumen tool--yes, I know about the tutorials and everything, no need for a link, I'm just a bit slow --I will build one and try my best to explain it. In the meantime, this should provide what is basically a mirror image of ZT stacking and JZT stacking. The most difficult part, by far, is properly switching between ZT and JZT. The height of the center stack is crucial to getting the switch right. I'm working on a NullpoMino play-through of the technique, but it's not quite done.
Like I said, I will try to fumen it later. The JZT stacking is actually harder than ZT because of the five-wide space on the right, but I'll figure it out eventually. My technique for doing JZT>ZT is to stack my excess L's and S's in the center field until I've stacked four of each there. Then I stack up my left field in columns 1-5 until column 5 in the left field is the same height as column 6 is in the center. Then I switch to ST stacking.
In the right field I just do JT, ZT, JT, ZT, JT, ZT, JT, ZT, I (Tetris), then the field is empty and I do ZT, ZT, ZT, ZT, I.

I'm working out the kinks for the ZT>JZT switch...like I said earlier, the 4-bag cycle for ZT is crucial for this method.
I'll see how much I can do.
~QM

[Alexsweden]

Yay!

I will play around with this once I come back home, and once I figure out exactly how it should work

[XaeL]

its sustainable in the sense that u can do it forever.
There is no assumption about wasting T's or not.

There is a simple proof that is shorter than your .doc...

4 T-spins and 1 Tetris clears 12 lines.

However, you need 4 T pieces, i.e. 4 bags. 4 bags produces 28/2.5 lines of stack = 11.2 lines.

Since 12 > 11.2, you must dump a T piece every 20(?) bags to keep it sustained.




SIMPLE ALGORITHM:

CRITICAL PIECE clears (x) lines.
getting Critical Piece creates (y) lines of "stack"

i.e. X = rate of clearing. Y = rate of growth.


if (x = y) PERFECTLY sustainable, all CRITICAL pieces will clear lines
if (x > y) some CRITICAL piece must be dumped on "stack"
if (x < y) not enough CRITICAL pieces. "stack" will grow faster than clearing lines.


e.g. Tspin TRIPLES only:
T clears (3) lines
getting critical piece creates (7 /2.5) 2.8 lines of "stack"

since (X > Y) some T's must be dumped onto stack.

E.g. Tspin Doubles only:
T clears (2) lines
getting critical piece creates 2.8 lines of stack

since (X < Y) Stack grows faster than T's can clear it.

[Ravendarksky]

This thread makes me think of this webcomic:

[!--ImageUrlBegin--][a href=\\\"http://www.smbc-comics.com/comics/20100703.gif\\\" target=\\\"_new\\\"][!--ImageUrlEBegin--][img width=\\\"400\\\" class=\\\"attach\\\" src=\\\"http://www.smbc-comics.com/comics/20100703.gif\\\" border=\\\'0\\\' alt=\\\"IPB Image\\\" /][!--ImageUrlEnd--][/a][!--ImageUrlEEnd--]

[XaeL]

I find ravendarksky's image offensive.

[Question_Mark]

I agree. Raven, as much as you may find that funny, you don't know how much work I put into the math behind Tetris. Simple math, yes, but not obvious most of the time.

its sustainable in the sense that u can do it forever.
There is no assumption about wasting T's or not.

There is a simple proof that is shorter than your .doc...

4 T-spins and 1 Tetris clears 12 lines.

However, you need 4 T pieces, i.e. 4 bags. 4 bags produces 28/2.5 lines of stack = 11.2 lines.

Since 12 > 11.2, you must dump a T piece every 20(?) bags to keep it sustained.
SIMPLE ALGORITHM:

CRITICAL PIECE clears (x) lines.
getting Critical Piece creates (y) lines of "stack"

i.e. X = rate of clearing. Y = rate of growth.
if (x = y) PERFECTLY sustainable, all CRITICAL pieces will clear lines
if (x > y) some CRITICAL piece must be dumped on "stack"
if (x < y) not enough CRITICAL pieces. "stack" will grow faster than clearing lines.
e.g. Tspin TRIPLES only:
T clears (3) lines
getting critical piece creates (7 /2.5) 2.8 lines of "stack"

since (X > Y) some T's must be dumped onto stack.

E.g. Tspin Doubles only:
T clears (2) lines
getting critical piece creates 2.8 lines of stack

since (X < Y) Stack grows faster than T's can clear it.

I don't think you understand. No math is necessary to understand that ST stacking is sustainable if you periodically dump a T in your stack to build it up. That's pretty straightforward. I'm concerned about not wasting any T's. I explained this concern in my second post in the Tetris Goals thread. If one has to dump T's during ST stacking I find it very suboptimal.
You saying your proof is shorter than my .doc is meaningless--my proof takes up less than the first page, on which there is also introductory text. And thereafter is a similar but opposite proof for LST stacking, an analysis of the problem and feasible solutions, a table, a chart, a conclusion and analysis of its yield.
As for the proof itself, I understand it but like I said earlier I wanted my analysis to include the partitions and was too lazy to bother with the height of the stack on the right--I only bothered with how many lines it cleared. While your proof is simpler it doesn't state at any time that the problem is in the left stack. In fact, I didn't really need math at all.
ST stacking with T-spins keeps the height of the left partition in check with the center partition. The left partition has I's, the center doesn't; thus every cycle when the tetris happens, the left side is affected and the center isn't. This amplifies any difference in the heights of the left and center partitions each bag, which propagates and ultimately destroys ST stacking.
Okay, a little math will prove that ST stacking keeps the left and center in check, but I did that in the paper, so w/e.
The simplest solutions are always the most elegant. The simplest proofs, however, aren't always the same.
~QM

[Alexsweden]

u have time to write essays as responses but no time to learn fumen?
Go make a fumen about how u think it should work or PM me and show the basics and i can make it for you!
Im home in an hour and im gonna do some serious testing on this

[Alexsweden]

4 T-spins and 1 Tetris clears 12 lines.

However, you need 4 T pieces, i.e. 4 bags. 4 bags produces 28/2.5 lines of stack = 11.2 lines.

Since 12 > 11.2, you must dump a T piece every 20(?) bags to keep it sustained.

Since you usually need to place down two T pieces for you  to be able to stack neatly that would mean that you have to use two T pieces in within a short time.



But I have a feeling the hold function can delay this  by some factor as you can continue to build and do T spins even though you are in a defecit of lines since you can hold the T piece so that you could get out a maximum of 6 other pieces before you have to place down a T piece.

So everytime you according to the math have to place down a T piece you actually can steal a few lines from the next bag, this does not solve anything but it does prospone the length for which it can be sustained.

Since you can get out a maximum of 6 pieces before you will be forced to put down a T piece (1 in hold from prev. bag and one from you current bag) this means that you can have a defecit of 2.4 lines and still be able to do the ST. However 6 is a maximum and not every piece might fit where you want it.
Also since you usually have to use two T pieces in the right side (for ZT stack) this is quite good anyway

Duno if my math was correct there but i think we have to take into account that we have hold and that that gives some extra breathing time.


here is a replay of me only using the first two T pieces in the beggining to build the main stack and wasn't forced to use two untill the 74th line. However I did not use not only S and L for the middle stack but also Z O and J so that might mess upp the math as I can smoth out the lack of pieces for the furthermost 4 colums by using some of the S and L pices there.
http://www.speedyshare.com/files/28362432/73lines_ST.rep

Edit: didnt realize this became a double post, sry for that

[blazen_azn]

st stacking isnt possible without dumping at least a t piece into the stack

if you dont, you get tspin nothings

are you question_mark on tf?

[mippo]

I don't see wasting a T as such a bad sign if you can keep going.
I remember using ST stacking all the way up to level 15 or so in Survival Mode on TF, and then I had to switch because the pieces started to drop too fast to place them correctly, how much more sustainability would you need or is this a theoretical argument?