Dr. Ken Housley,
Most of this page is in ASCII, because I sent it to myself from home as an email message and some diagrams are less complicated to write in ASCII than in HTML. However, I will try to emphasize some stuff in the text with HTML formatting.
I hope everything is clear to you. If you want clarification on
something, send a message to
Shlomi Fish
Recent Note: I managed to prove that the two-way algorithm works for 3k and 3k+1 square boards. It was added as theorems 4 and 5.
More Recent Note: Added section about the dimension of solutions space in 3k+2 boards and a note about the efficiency of the algorithms.
First, I’ll prove some analogous statements about a one-dimensional toggle-squares board, i.e: conjectures that apply to the set of rows of size 1*n and in which the presses are: 110000 , 111000 , 011100 , 001110 , 000111 , 000011 (n=6 was given as an example). Mathematically, it can be treated as vectors that belong to the linear space (or vector field) {0,1}^n where the operations in the set {0,1} are XOR for ‘+’ and AND for ‘*’.
A 1-D toggle field is solvable for every state if (and only if) the vector 100...000 can be formed as combination of the presses.
Proof: If 1000000 can be solved then along with the press 110000...00 than 010000000 can be solved too. Furthermore: 100000...00 + (xor) 010000...00 + 111000...00 = ------------- 001000...00 and from here it can be proved inductively that the vectors with all 0s and one 1, are solvable. Since this is a basis of {1,0}^N then every state is solvable. (i.e contained in Sp{Presses} )
A 1-D toggle field of size N is solvable for every state if N = 3k or N = 3k+1 and isn’t if N = 3k+2 for some non-negative integer k.
Proof: One way to show it is by inspecting the matrixes of the presses set: 1 , 11 11 , 110 111 011 , 1100 1110 0111 0011 and calculate their determinants (according to the boolean operations). After calculating the determinants of the first three by hand: 1, 0 and 1 respectively, we can show that the determinant of every matrix where N > 3 is: based on the first row: det (Presses Matrix (N-1)) XOR - det(Matrix that contains 1 at the top-left cell and below a Presses Matrix of size N-2) = det (PM(N-1)) XOR det(1 * PM(N-2)) = det(PM(N-1)) XOR det(PM(N-2)). Hence, it’s a XOR of the determinants of the two previous matrixes. (A boolean Fibonacci series :-) ) Recursively, we get a repeating sequence of N | 1 2 3 4 5 6 7 8 9 ... det | 1 0 1 1 0 1 1 0 1 ... Matrixes with a zero determinant cannot form every vector, and vice versa. It can also be proved (at least for the solvable cases) using Lemma 1. For N=3k 1000....000 can be formed by: 1110000...0000 XOR 0001110...0000 XOR . . 0000000...0111 = --------------- 1111111...11111111 XOR 0000000...00000011 XOR 0000000...00011100 XOR 0000000...11100000 XOR 011100000000000000 = (3k = 2 + (k-1)*3 + 1) ------------------ 100000000000000000 (Q.E.D) and likewise for 3k+1. For 3k+2, we can show that the vector 11111...111 can be formed in two different ways: 11000000...000000 XOR 00111000...000000 XOR 00000111...000000 XOR . . 00000000...000111 = ----------------- 11111111111111111 and 11100000...000000 XOR 00011100...000000 XOR . . 00000000...011100 XOR 00000000...000011 = ----------------- 11111111111111111 and thus the dimension of the span is less than n, and so not every state is included in it, and is solvable.
Now for the actual proofs:
2-D (3k+2) * (3k+2) boards are not solvable for every state.
Proof: If we take a look at the topmost row - it is affected only by the presses on the two topmost rows. The effective portions of all the presses on that row is the set: 11000..00 11100..00 . . 00000.111 00000.011 and Lemma 2 proved that for N=3k+2 it has unsolvable states. If some formations of the first row cannot be solved by any combination of the presses on the block, much less the entire board can be solved for every state.
2-D boards of size 3k * 3k or (3k+1) * (3k+1) are solvable for every state.
Proof: If we restrict ourselves to the presses on cells of one column, then we get an array of rectangular vectors with a fixed horizontal position and width and some y-dimensions that resemble the 1-D toggle-squares vector set. e.g: 110000 110000 000000 000000 000000 000000 , 110000 110000 110000 000000 000000 000000 , 000000 110000 110000 110000 000000 000000 , . . According to Lemma 2, with 1-D presses sets with size 3k and 3k+1 every vector can be formed. Thus we can achieve the following rectangles, with the presses of one column: 110000 000000 000000 000000 000000 000000 000000 110000 000000 000000 000000 000000 000000 000000 110000 000000 000000 000000 . . 000000 000000 000000 000000 000000 110000 and the same for every other column. Now, if we look on any row of cells, we will see that the “sticks” that were generated from the various columns in that row are also the 1-D TogSqrs sequence. Because N=3k,3k+1 again, every row is solvable for all states, and therefore, the entire board is solvable for all states. Diagram: 110000 , 111000 , 011100 , 001110 , 000111 , 000011 000000 . . . . . . . . . . .
Solvable states of 3k+2 boards can be solved by the one-way algorithm.
The one way algorithm is described as follows: Scan every row from top to bottom. For every row, scan every cell from right to left, and for every filled cell, click the cell to the bottom-left of it. Proof: Every row can be cleared by pressing a cell to the left or the bottom-left of every filled cell, because that way one consistently regress the right-most filled cell. Moreover, it’s impossible that only the left-most cell will be filled in a solved state (Lemma 1). By going down, we ensure that no new cells will be filled instead of the new one, and we can clear that way the first N-1 rows. After the iteration when the first N-1 rows are cleared, the last row cannot be partially filled because it’s a column-wise contradiction of Lemma 1. Therefore, it will also be cleared.
A 2-D board of size (3k) * (3k) can be solved by using the two-way - two-way algorithm.
The two way algorithm is described as follows: Scan the rows from top to bottom. For every row, scan the cells from right to left, and for every filled cell click the cell to the bottom-left of it. Then scan the cells in that row, from right to left and for every filled cell click the cell to the bottom-right of it. After you reached the bottom, do the same thing from bottom to top, but with pressing the cell to the top-right or top-left instead. Proof: One can show that the two-way algorithm works for N=3k in a one-dimensional board. Even if there is a state of 100...000000000 , by pressing to the right of the filled squares one eventually gets to a 000...000000011 state which is soon cleared to 000...000000000. This provides us with a method to clear rows by pressing cells in the row beneath them, and we can use it to clear all rows except the bottom one. Now, let’s take a look at a situation in which there is a partially filled row, and above it, at least two empty ones. I will prove that clearing it by pressing the cells in the row above it, will make the two rows above it a duplicate of its original state. For example if the status was: 000000 000000 <- I am pressing this row 011010 then it will turn into 011010 011010 000000 The proof is quite simple. If we take a look at one of a cells above an initially filled cell, then the cells that affect it and the bottom cell were pressed an uneven number of times, during the clearing process. (or else the state of the bottom cell would have remained filled). Thus, it will also change its state and become filled. Likewise, a cell above an unfilled cell in the bottom row, was switched an even number of times and will be unchanged at the end. With the same deduction it can be shown that two duplicate rows once cleared will fill the row above them in the same formation (provided it was initially blank). E.g: 000000 110011 110011 will become 110011 000000 000000 Since N is equal to 3k, then after the bottom-to-top iteration we will end up with the two topmost rows filled in the same formation. Then, the topmost row will be cleared along with the lower one. Now, for the remaining modules: In a 1-D board where N=3k+1 the two-way method as is, will ping the possible remaining 1 back and forth between the two edges. Therefore if there is a state of 1000...0 than one should use the leftmost press to change it into 0100...000. Then, by repetitive pressing to the right of the leftmost filled square, one gets to the 0...11 state, which is afterwards cleared.
A 2-D board of size (3k+1)*(3k+1) can be cleared by an down-(edge)-up iteration of left-(edge)-right row clearing scheme.
This algorithm is similar to the two-way algorithm, except that if a at the end of the left iteration, there is a single filled cell, one has to press on its lower or upper cell, before moving on. Likewise, if at the end of the down scanning, you have filled cells in bottom row, you have to clear them by pressing cells of the bottom row. Proof: Very similar to Theorem 4 except those notes: 1. In N=3k+1 the two-way method as is, will ping the possible remaining 1 back and forth between the two edges. Therefore if there is a state of 1000...0 than one should use the leftmost press to change it into 0100...000. Then, by repetitive pressing to the right of the leftmost filled square, one gets to the 0...11 state, which is afterwards cleared. To clear the rows in a 3k+1 board one should use this method. 2. The bottom row (if still partially filled) should be moved to the row above it, by clearing it by presses of cells of the bottom row. Then it can be transposed in the same manner described in Theorem 4 to the the two topmost rows and then cleared. For example, if the status is 0000000 0000000 0110110 | | press those cells and it will be be moved to 0000000 0110110 0000000
(3k+2)*(3k+2) boards are not solvable for every state. The solvable states form a vector field which is formed by the spanning all the presses. Since, in this case it does not include all the states, I wondered if we can calculate its dimension or characterize it somehow. I think I know the answer. First in 1-D: the presses of {0,1}*(3k+2) span a field of size 3k+1 because it takes only one extra vector - 100...000, so they will form every possible state (Lemma 1). Since the 1-D clearing mechanism for 3k+2 uses only the leftmost 3k+1 cells to clear every solvable state (part of Theorem 3), then their presses are a valid basis for this linear space. The one-way - one-way clearing algorithm which was proved in Theorem 3, uses only a (3k+1)*(3k+1) square of presses to solve the entire board. Ergo, the dimension is (3k+1)^2 or less. I believe it _is_ (3k+1)^2 and I think I can prove it by showing that the presses of a corner (3k+1)*(3k+1) are linearly independent. In the boolean field the meaning is that any number of them other than zero XORed together will not generate the clear state. I’ll demonstrate on a 5*5 board: | | - 00000 00000 00000 - 00000 00000 The ‘-’s and ‘|’s mark the relevant presses. Let’s assume that a certain number of presses in the square can form the clear state. If so, then Pr(4,4) cannot be one of them because it is the only press that can affect square (5,5). Moreover, Pr(3,4) and Pr(4,3) cannot be included either, because, once Pr(4,4) is eliminated they are the only ones that can affect (4,5) and (5,4) respectively. And so forth, proving that that all the presses of cells (1,4)-(4,4) and (4,1)-(4,4) cannot be part of the zero sum. The same can be deduced for the next layer starting from Pr(3,3), which is the only vector left that can affect square (4,4). And so-on for the other layers, proving that they are a linearly independent set of vectors. In conclusion, there cannot be less than (3k+1)^2 vectors that span the space of the solvable states of (3k+2) boards.
I believe the previous section showed that the 1w-1w algorithm for clearing (3k+2) boards is as efficient as it could be, i.e. requires the minimal number of presses. As for boards of other sizes: the 2w-2w algorithm is not the most efficient one in relevance to the number of presses it takes to solve the board. I noticed that when I solved a couple of boards and realized that the number of moves it took me with it was greater than the number which was returned from my matrix-canonization program. However, I believe it is agreeable that it is the simplest algorithm (yet) regarding “CPU” requirement and growing complexity, and has the advantage that it can be utilized without the aid of a computer.