B2B Perfect Clears

[CosmicCommunist]

Is it possible to get a true b2b perfect clear? In the sense, get one pc, clear no other lines, then do another pc without clearing any other lines in between. I know it's possible to get b2bpcs clearing lines in between (my max is 4) but in normal gameplay (i.e. not with garbage) is it possible to get b2bpcs without clearing lines in between the pcs? And if so, what's the limit to how many "true" b2b pcs you can score in this sense? This is given normal bag sort method of pieces.

[Shuey]

You're saying that you'd complete PCs B2B with a Tetris every time, right?  If so, based on what I experienced in my last PC study, it's impossible to do PCs nonstop with the normal bag/guideline sequencing.

[CosmicCommunist]

Sort of. I mean also the JLIOO pc is also up for grabs. But say you do one PC, then do a Tetris pc, or the 5 piece pc only. Is it possible to do it at least once?

[Okey_Dokey]

Tetris-PC into 2-lines PC is possible at the start of a game.

[fumen]http://harddrop.com/fumen/?v115@vhRVQYaAFLDmClcJSAVDEHBEooRBUoAVBsvjFDP+AA?AyxB2uBvkB0mBTlBRgBAAPaAFLDmClcJSAVDEHBEooRBaoA?VBU+TPCPNBAAsnBFiB5oBSxB2vBTpBAAPWAFLDmClcJSAVj?rSAVG88A4N88A5GBAARrBTtBAAA[/fumen]

Tetris-PC into Tetris-PC is impossible at the start of a game. I will write down a proof later. In mid-game, you can perform it though.

[Okey_Dokey]

I have to correct myself: You can start with 2 PC-Tetrises. You'll need a lot of luck though. The only way I found to do so is holding first I piece beyond first PC. Then you need luck to get the second I piece as 10th or 11th piece (and the second T piece before it as well as O piece). And the 21st piece must be either I piece or T piece (and some other conditions).

[fumen]http://harddrop.com/fumen/?m115@ieA8HeB8IeA8HeC8MeQ4ywR4AtRpwhR4wwR4BtRpwh?g0Q4ywAtglRpwhi0wwilRpwhNeB8GeA8BeA8HeA8HeA8HeD?8Meili0ilwhglAtRpQ4g0glAtg0whBtRpR4Btg0whAtzhQ4?Ath0whJeAgHyfAAHeBAIeAAHeCAaeBAGeAABeAAHeAAHeAA?HeDAWeywQaAFLDmClcJSAVDEHBEooRBMoAVBzuzPCP+AAAv?hUlqBsmB2uBzsBvkBAAPbAFLDmClcJSAVDVSAVG88A4c88A?Z+CMCqXkBAXhBFgBziBpoBetBMsBiiBRvQbAFLDmClcJSAV?DVSAVG88A4N88A5mjWCz/1BATlBMpBifBvrBGhBpoBAAA[/fumen]
That's been one of the very few cases still missing when I tried to prove that it was not possible. So much wasted time for nothing. I will post a few theoretical stuff though (damn noticing right now I made a mistake in a proof so have to overthink how to prove the theoretical result) which basically proved to me that this was just possible by placing 2 T pieces and 2 L/J pieces in the first PC-Tetris. And I am pretty sure you need to keep the first I piece (maybe 3 J pieces for second PC also work).

[thickBUT]

Do you get b2b bonus for that?

[caffeine]

Do you get b2b bonus for that?

I asked the same thing seven years ago. Still don't know.

[Okey_Dokey]

I asked the same thing seven years ago. Still don't know.

In Puyo Puyo Tetris I tested T-Spin Double (no PC) into PC-Tetris. In this case you'll get bonus points for the PC - more bonus points than you would expect for just a b2b Tetris: b2b Tetris scores 400 more points than a non-b2b Tetris; a PC-Tetris with b2b from T-Spin line-clear scores 1600 more points than a normal PC-Tetris. That's not what you were looking for but still interesting.

----

Here some theoretical results:

1. To fill a rectangle completely with Tetrominoes, we'll have to place an even amount of T pieces.

Proof: Parity. Imagine laying a checkerboard over the playfield. Any even-sized rectangle contains the same amount of white and black squares in it which we will now fill with Tetrominoes. Each piece but the T piece will fill exactly 2 white and black squares. So once we have placed a single T piece in the rectangle, we have filled more of one color and the only way to even out the colors again is to place another T piece (with the center on the opposite color). (instead of comparing the number of black and white squares, we could have also just looked if the amount of filled white squares was even).

[fumen]http://harddrop.com/fumen/?m115@deA8AeA8AeA8AeA8AeA8BeA8AeA8AeA8AeA8AeB8Ae?A8AeA8AeA8AeA8BeA8wwA8AeA8AewwAeB8ywA8AexwA8BeA?8AeA8AeA8AewwAeB8AeA8AeA8AeA8AeA8BeBtA8AeA8AeQ4?AeB8AeBtA8AeR4A8BeA8AeA8AeA8Q4A8AeB8AeA8AeA8AeA?8AeA8Beg0AeA8AeA8glA8AeB8i0A8AeglAeA8BeA8AeA8Ae?A8hlAeB8AeA8AeA8AeA8AeA8BezhA8RpAeB8AeA8AeA8AeR?pA8BeA8AeA8AeA8AeA8AeB8AeA8AeA8AeA8AeA8BeA8AeA8?AeA8AeA8AeA8JeAgH[/fumen]

2. To fill a rectangle completely with all Tetrominoes but T pieces, we'll have to place an even amount of L&J pieces (amount of L pieces PLUS amount of J pieces is even, that means the sum is even, NOT each addend).

Proof: Alternative Parity. Imagine coloring the odd-numbered columns white and the even-numbered columns black (columns are numerated from 1 to 10); Any even-sized rectangle contains an even amount of white squares. An I, O, S and Z piece will fill an even amount of white squares (and black squares). Since we assumed no T pieces, only L and J pieces will create oddness: They will either add 3 or 1 white squares no matter if placed horizontally or vertically. So, the next L or J piece will even out the number again. (Instead of counting all filled squares in odd-numbered columns, we could have also subtracted filled black squares from white squares on a checkerboard in odd-numbered columns; this definition is more similar to parity as we try to even out the number of filled black and white squares then but this time only squares in odd-numbered columns matter)

[fumen]http://harddrop.com/fumen/?v115@deA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8Ae?A8AeA8AeA8AeA8AeA8AeA8AewwAeA8AeA8wwA8AeA8ywA8A?exwA8AeA8AeA8AeA8AeA8wwA8AeA8AeA8AeA8AeA8AeA8Ae?A8BtAeA8AeA8Q4A8AeA8AeBtA8AeR4A8AeA8AeA8AeA8AeQ?4AeA8AeA8AeA8AeA8AeA8AeA8AeA8g0A8AeA8AeglAeA8Ae?A8i0A8AeglAeA8AeA8AeA8AeA8AehlA8AeA8AeA8AeA8AeA?8AeA8AeA8zhAeRpA8AeA8AeA8AeA8AeRpA8AeA8AeA8AeA8?AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA8AeA?8KeAgH[/fumen]

3. To fill a rectangle completely with 2 T pieces, an even number of L&J pieces as well as some other Tetrominoes, we have to place the T pieces a) both horizontally or b) both vertically. Similarly, for an odd number of L&J pieces, we have to place one T piece horizontally and the other vertically.

Proof: Alternative Parity. Again imagine coloring the odd-numbered columns white and the even-numbered columns black. Any even-sized rectangle contains an even amount of white squares. Let's only consider the case where one T piece is placed horizontally and the other vertically; the other cases are similar. The horizontal T placement adds 2 filled white squares. The vertical T placement adds 3 or 1 filled white squares depening on in which column it is placed. With another L or J piece placed, these 3 pieces will add an even sum to the amount of filled white squares, so keeps it even, as well as any further 2 L or J pieces.

Let's now consider a 4-rows-high rectangle filled with Tetrominoes. We call such a rectangle minimal, if it doesn't consist of any smaller 4-rows-high rectangles. On top of the following Fumen we have those minimal rectangles from left to right: a) a vertical I piece, b) 2 O pieces stacked above each other, 3-wide rectangle consisting of 1 Z and 2 J pieces, 4-wide rectangle consisting of 1 S and I piece each as well as 2 L pieces. Vertical L, J, T piece placements always mark the left resp. right borders of a minimal rectangle (because the area covered by the pieces right resp. left of them must contain a multiple of 4 squares).

4. In a minimal rectangle with less than 2 I pieces, there's at least one L, J piece or vertical T piece on each side (left/right border).

Proof: L, J and vertical T pieces are the only pieces that can add more filled cells on the wall, then on the next column. Horizontally S, Z and T placements do the opposite, so require L, J and vertical T placements automatically. Vertical S and Z placements on the wall will require horizontal J resp. L placements to fill the hook. O pieces require L or J pieces as well unless paired with 2 I pieces.

[fumen]http://harddrop.com/fumen/?v115@dewhRpi00hRpBtg0glQ4hlwhRpg0BtglR4glwhRpi0?hlQ4gl1eA8Eeh0CeA8BeR4g0CeA8BeR4Aeg0DeA8HeA8Eew?wDeA8BewwAexwBeA8BezwDeA8HeA8Eeg0DeA8Aeg0Bei0Ae?A8Bei0JeAg08AnnToDFb0sDPM98AQCtuD0oo2A3iMDEFbcR?ASozNBDoo2AiiSTAS4FNEyoo2Ayeh1D0oo2A3iMDE[/fumen]

5. Each minimal rectangle with no I&T pieces, just 2 L&J pieces and no more than 2 pieces of each kind will be not bigger than 5-wide.

Proof: Trial and error. O pieces and vertical S, Z pieces force a border without I pieces. Horizontal S, Z pieces carry over the overhang towards the left/right side; horizontal S and Z pieces don't fit together well without T pieces.

[fumen]http://harddrop.com/fumen/?m115@deBti0Beh0AtglBtQ4g0Beg0BtglBtR4Beg0AtilBt?Q4BeilJeBthlCejlBtglCeglR4glBtglCeR4ilBtCeilJei?lRpBei0glBtRpBeRpg0RpBtglBeRpglRpilBeilUeBtzhR4?BeglBtRpR4g0BeglBtRpR4g0BehlBtR4h0KeAgWsAyY4dEB?mBDEloo2A3u+1DFb8KCFbkOEpYQrDFbcEE1lbTAS4s4Dn7M?XE[/fumen]

Let's use those 5 theorems to kinda proof that you can't start a game with 2 PC-Tetrises. Let's assume the opposite (it's possible! <- so true) and contradict it (<- not true). First of all, 2 PC-Tetrises are 20 pieces. By theorem 1, each PC-Tetris must consist of either 0 or 2 T pieces. This sadly means that a T piece must stay on hold in the end (resp. come next if hold was never used). Since 3 bags are 21 pieces, we can exactly say how many pieces of each kind are used, namely 3 of each kind except for T pieces. 2 I pieces are placed vertically for the 2 Tetrises; only one I piece is freely available. Also, there're 8 L, J and T pieces in total. Let's make a case differentiation of how many L, J & T pieces are used in the first PC-Tetris:

a) #LJT = 2
...