Here are the values of some Nx2 Domineering positions. Here "O" represents an empty square, and "X" represents a played square. 1/2 | -2 = -3/4 +- 5/4 ; XX XX XX XO OO OO XO OO XX OO OO OO XX XX XX 5/2 | 1 = 7/4 +- 3/4 ; XX XX XX XO XO XO XO XO XO XO OO OO XO OO XX XO XO XO XO XO XO OO OO OO XX XX XX 2 ||| 3/2 | 0 || -1/2 | -5/2 XX XX XX XX XX XX XX XX XX OO OO OO OO OO OO OO OO OO XO OO XX XX XX XX XX XX XX XO OO OO OO OO OO OO OO OO yy OO XO XO XO OO XX XX XX OO OO XO XO XO OO OO OO OO yO OO XO OO XX OO XO OO XX yO OO XO OO OO OO XO OO OO XO XO XO XO XO XO XO XO XO XO XO XO XO XO XO XO XO XO XX XX XX XX XX XX XX XX XX 2 ||| 3/2 | 0 || -1/2 | -5/2 cools by 3/4 to 5/4 ||| 3/4* || 1/4 | -1/4 = 3/4 + 1/2 ||| * || -1/2 | -1 = 3/4 * + 1/2*||| 0 || -1/2* | -1* = 3/4* + int 1/2 to 1/2* of 0|||0||0|1 A CANONICAL STRATEGY FOR SUMS/DIFFERENCES OF 2xN DOMINEERING We'll play Left on a sum of 2xN and Nx2 Domineering boards, for assorted odd values of N. To begin, we partition each of the Nx2 boards into 2x2 "parcels". We act as though horizontal (Right) Dominoes have already been played just above the top and just below the bottom of the board, and we include the top right Domino within our topmost parcel. So, if N = 2n+1, we have one topmost parcel in which Right has already played, and then n initially empty parcels beneath it. Just below the bottom of the board would be another Right parcel, which we will view as out of our range. Similarly, in every other available column of each 2xN board, we declare a 2x1 parcel. If N = 2n+1, there are n such initially empty parcels, bounded on the west by a boundary parcel which we view as already having been played by Left, and on the east by another Left domino which we will view as out of our range. We then play Blockbusting on the parcels. Each move by the opponent affects only one parcel, and each of our moves will be entirely within whichever parcel provides a best Blockbusting move. We continue to follow this strategy until no empty parcels remain. We then stop, and bound the score according to the following accounting system: SCORE FORMULA On vertical (Nx2) boards, score >= #Left parcels - #Right parcels/2 - #Right BB pts/2, and on horizontal (2xN) boards, score >= #Left parcels/2 + #Left BBpts/2 - #Right parcels ### Verification of scoring formulas, assuming no doubly-occupied Right parcels: On vertical boards, define the "closed" edge as the top or bottom parcel edge which touches the domino Right played within this parcel. The other bottom or top edge of the parcel is called "open". Each closed edge cuts the board into disjoint pieces. If an open edge is adjacent to a Left parcel, then together these two parcels have value +1/2, which agrees with the formula, +1 for the Left parcel and -1/2 for the Right parcel. If the open edge is adjacent to the closed edge of an adjacent Right parcel, then it has value -1, agreeing with -1/2 for the Right parcel -1/2 for the Right BB point. If the open edge is adjacent to an open edge of an adjacent Right parcel, then the value of this pair of parcels is 1|-1, which is greater than the formula value of -2/2 (for two Right parcels) -1/2 for the Right BB point. For each Left parcel edge not adjacent to an open edge of a Right parcel, we (as Left) may arbitrarily agree henceforth never to play across that edge, thereby cutting the board. Each Left parcel adjacent to no Right parcels then has value +1, matching the formula. If a Left parcel adjacent to two open edges of Right parcels, we find the combined value of all three parcels together as 0, matching the formula's +1 for the Left parcel, -2/2 for the two Right parcels. On horizontal boards, each Left domino cuts the board, and we may divide the board into regions, each region containing its western Left domino (or western edge of the original board) and all consecutive Right parcels going eastward until, but not including, the next Left domino or end of the original board. A typical region contains 2 x (2n+2) squares, the Leftmost one occupied by Left, and n Dominoes played by Right, each blocking one of the other parcels in which Left would have wanted to play. Call this board G. G contains 4n+4 squares, 2 occupied by L and 2n are occupied by Right. This board still contains at least one empty column. Suppose Left plays there, to G^L. Then the board G^L contains only 2n empty squares, and so its value is >= -n. Similarly, if Right moves from G to G^R, he attains a position in which there are 2n empty squares, so at least one column is occupied by two Right dominoes. This column cuts the board into two pieces. If Right has played well, he will have eliminated the empty column and each of the two pieces he has created will have an odd number of empty squares. If Right continues to play as many times as he can, each of these pieces must still contain at least one empty square, so there will still be at least two empty squares remaining when all possible moves have been taken. Hence, even though the position G^R has 2n empty squares, at least 2 of them are unusable, so G^R >= 1-n. Hence, G >= -n | (1-n) = 1/2 - n. This coincides with the score for this region given by the formula, 1/2 for the Left parcel, - n for the n Right parcels. If instead after Right's move from G to G^R, each piece has an even number of empty squares, then at least one of the pieces still contains an empty column, and the prior argument can be applied to it, again yielding a score >= 1/2 - n. Finally, on a horizontal board, a pair of adjacent Left parcels has value 1, equal to the formula's 1/2 for western Left domino + 1/2 for the Left BB point. ### We must also verify the cases in which there are doubly-occupied Right parcels. On vertical boards, a doubly occupied Right parcel has value 0, which exceeds the formula's value of -2/2. On horizontal boards, we can remove one Right domino from each doubly-occupied parcel, compute the score as above, and then observe that replaying back each of the redundant Right dominoes costs Right at least 1/2 point, and so leaves him worse off then if there were no redundant Right Dominoes. Hence, on either vertical or horizontal boards, any redundant play by Right (i.e., playing a second Domino into a parcel on which he has already played before) costs him 1/2 point in value, versus at least a 1/2 point benefit to which he is entitled by the formula. REVISED STATEMENT OF THE FORMULAS: On vertical boards, score >= 1/4 # parcels + 3/4 # Left parcels - 3/4 # Right parcels - #Right BB pts/2 On horizontal boards, score >= -1/4 # parcels + 3/4 # Left parcels, -3/4 # Right parcels, + #Left BB pts/2 where on each horizotonal 2x(2n+1) initially empty board is viewed as initially containing 1 Left parcel and n initially-empty parcels, and each vertical (2n+1)x2 board is viewed as initially containing 1 Right parcel and n initially-empty parcels. Define the "putative value" of the empty 2 x (2n+1) Domineering board as 1/2 - n/4 + Heated to 3/4 ( *.n + Overheated from 1/2 to 1/2* (x[n]) ) where x[n] is the chilled value of the horizontal Blockbusting position LnL. (Notice how the putative formula this corresponds to the revised statement of our scoring formulas. The initial 1/2 = -1/4 + 3/4 accounts for the western-edge Left parcel. The other n parcels account for -1/4 each. The heating to 3/4 matches the value of getting a parcel, and the final term, overheated x[n], accounts for the value of the BB points). However, we needn't dwell too much on these correspondences, because we have proved the following THEOREMS: +++ On a sum/difference of 2xodd Domineering boards in which the putative total value is 0, Left going second can win by the strategy stated above. +++ Hence, +++ If the putative total value is 0, so is the real value. +++ And finally, after verifying that the putative values are really correct for the specific cases of 2x1, 2x3, 2x5, 2x7, 2x9, and 2x11, and that these specific values include a basis in terms of which any putative value can be expressed, we conclude that THE PUTATIVE FORMULA IS CORRECT FOR ALL n.