The height difference will always be between -20 and +20, since we play on a 20 rows high matrix. This means the height difference (average height of Z lanes minus average height of S lanes) increases by 2 every 12 pieces. In average, that is 2 pieces per S lane and 3 pieces per Z lane. If we do not want to violate the condition, we have to place the 6 S pieces on the S lanes and the 6 Z pieces on the Z lanes. Let us say the initial surface is shaped in the way, that there are 3 lanes for S pieces and 2 lane for Z pieces. The condition above means that all pieces must be placed vertically in 5 lanes, each lane 2 columns wide with odd-numbered left-most column. Not more than 240 pieces can be played before either a special placement is made or a hole is created.ДЄ hole denotes an enclosed empty cell. Which corresponds to the statement that not more than 120 special S and Z placements can be made. V 2 + H 2 + H 3 ) в‰¤ 20 + 20 = 40 (2)Д«y using (в†в†) and beginning on the right side we conclude.This can lead to a recursive estimate of the terms on the left side.Д«y using (в†) and beginning on the left side we conclude What have we won? Since our matrix is 20 rows high, we know that B i+1 - B i в‰¤ 20 and B i - B i+1 в‰¤ 20. Now let us compare the number of blocks in two neighboured columns. Let V i be the number of dropped S and Z pieces that contributed blocks to the i th and (i+1) st columns.Д« i = 2 Let H i be the number of dropped S and Z pieces that contributed blocks to the (i-1) st, i th and (i+1) st columns. Let B i be the total number of blocks (Minos, filled cells) located in the i th column. Columns are numerated from 1 to 10 (from left to right). Here a placement is called special, if the piece is placed with its leftmost blocks in an even-numbered column. Not more than 120 special S and Z placements can be made. This also means that certain ('special') placements can only be made a finite number of times - or the height difference between the first and the second column becomes too big. As a result, at each point in time there will be never more blocks in the first column than in the second column (starting with an empty field). What have all those placements in common? There is no placement where more blocks (filled cells, Minos) are added to the first column than to the second column. 2 blocks in 1st column 2 blocks in 2nd columnД blocks in 1st column 2 blocks in 2nd columnДЎ block in 1st column 2 blocks in 2nd columnД blocks in 1st column 1 block in 2nd column

Tetris pieces