Mutilated chessboard problem

Mutilated chessboard problem (Redirected from Gomory's theorem) Jump to navigation Jump to search The mutilated chessboard Unsuccessful solution to the mutilated chessboard problem: as well as the two corners, two center squares remain uncovered The mutilated chessboard problem is a tiling puzzle posed by Max Black in 1946 that asks: Suppose a standard 8×8 chessboard has two diagonally opposite corners removed, leaving 62 squares. Is it possible to place 31 dominoes of size 2×1 so as to cover all of these squares?

It is an impossible puzzle: there is no domino tiling meeting these conditions. One proof of its impossibility uses the fact that, with the corners removed, the chessboard has 32 squares of one color and 30 of the other, but each domino must cover equally many squares of each color. Plus généralement, dominoes can cover all but two squares of the chessboard, if and only if the two uncovered squares are of different colors. This problem has been used as a test case for automated reasoning, creativity, and the philosophy of mathematics.

Contenu 1 Histoire 2 Solution 3 Application to automated reasoning 4 Related problems 5 Références 6 External links History The mutilated chessboard problem is an instance of domino tiling of grids and polyominoes, also known as "dimer models", a general class of problems whose study in statistical mechanics dates to the work of Ralph H. Fowler and George Stanley Rushbrooke in 1937.[1] Domino tilings also have a long history of practical use in pavement design and the arrangement of tatami flooring.[2] The mutilated chessboard problem itself was proposed by philosopher Max Black in his book Critical Thinking (1946), with a hint at the coloring-based solution to its impossibility.[3][4] It was popularized in the 1950s through later discussions by Solomon W. Golomb (1954),[5] Gamow & Stern (1958),[6] Claude Berge (1958),[4] and by Martin Gardner in his Scientific American column "Jeux mathématiques" (1957).[7] The use of the mutilated chessboard problem in automated reasoning stems from a proposal for its use by John McCarthy in 1964.[8] It has also been studied in cognitive science as a test case for creative insight,[9][10][11] Black's original motivation for the problem.[3] In the philosophy of mathematics, it has been examined in studies of the nature of mathematical proof.[12][13][14][15] Solution The puzzle is impossible to complete. A domino placed on the chessboard will always cover one white square and one black square. Par conséquent, a collection of dominoes placed on the board will cover an equal numbers of squares of each color. If the two white corners are removed from the board then 30 white squares and 32 black squares remain to be covered by dominoes, so this is impossible. If the two black corners are removed instead, alors 32 white squares and 30 black squares remain, so it is again impossible.[16] The same impossibility proof shows that no domino tiling exists whenever any two squares of the same colour (not just the corners) are removed from the chessboard.[17] Several other proofs of impossibility have also been found. For instance a proof by Shmuel Winograd observes that, by induction, in any domino tiling, each row of the mutilated chessboard has an odd number of squares that are not covered by vertical dominoes from the previous row, and therefore an odd number of vertical dominoes must extend into the next row. Ainsi, each of the seven consecutive pairs of rows has an odd number of vertical dominoes, producing an odd total number. By the same reasoning, the total number of horizontal dominoes must also be odd. As the sum of two odd numbers, the total number of dominoes must be even. Mais, to cover the mutilated chessboard, 31 dominoes are needed, an odd number.[16][18] Another method counts the edges of each color around the boundary of the mutilated chessboard. Their numbers must be equal in any tileable region of the chessboard, because each domino has three edges of each color, and each internal edge between dominoes pairs off boundaries of opposite colors. Cependant, the mutilated chessboard has more edges of one color than the other.[19] A B Removing one black and one white square on this Hamiltonian cycle (examples shown with ×) yields one (UN) or two (B) paths with even numbers of squares If two squares of opposite colors are removed, then it is always possible to tile the remaining board with dominoes; this result is called Gomory's theorem,[20] and is named after mathematician Ralph E. Gomory, whose proof was published in 1973.[17][18] Gomory's theorem can be proven using a Hamiltonian cycle of the grid graph formed by the chessboard squares. The removal of any two oppositely colored squares splits this cycle into two paths with an even number of squares each. Both of these paths are easy to partition into dominoes.[20] Gomory's theorem is specific to the removal of only one square of each color. Removing larger numbers of squares, with equal numbers of each color, can result in a region that has no domino tiling, but for which coloring-based impossibility proofs do not work.[21] Application to automated reasoning Domino tiling problems on polyominoes, such as the mutilated chessboard problem, can be solved in polynomial time, either by converting them into problems in group theory[19][22] or as instances of bipartite matching. In the bipartite matching formulation, one obtains a bipartite graph with a vertex for each available chessboard square and an edge for every pair of adjacent squares; the problem is to find a system of edges that touches each vertex exactly once. As in the coloring-based proof of the impossibility of the mutilated chessboard problem, the fact that this graph has more vertices of one color than the other implies that it fails the necessary conditions of Hall's marriage theorem and that no matching exists.[21][23][24] The problem can also be solved by formulating it as a constraint satisfaction problem, and applying semidefinite programming to a relaxation.[25] Dans 1964, John McCarthy proposed the mutilated chessboard problem as a hard problem for automated proof systems, formulating it in first-order logic and asking for a system that can automatically determine the unsolvability of this formulation.[8] Most considerations of this problem in the literature provide solutions "in the conceptual sense" that do not apply to McCarthy's logic formulation of the problem.[26] Despite the existence of general methods such as those based on graph matching, finding a solution to a logical formulation of the mutilated chessboard problem using the resolution system of inference is exponentially hard,[27][28][29] highlighting the need for methods in artificial intelligence that can automatically change to a more suitable problem representation.[30] Short proofs are possible using resolution with additional variables,[31] or in stronger proof systems allowing the expression of additional constraints on local patterns of dominoes that can safely be avoided in any tiling and used to prune the search space.[32] Higher-level proof assistants are capable of handling the coloring-based impossibility proof directly; these include, par exemple, Isabelle,[33] the Mizar system,[34] and Nqthm.[35] Related problems Wazir's tour problem A similar problem asks if a wazir starting at a corner square of an unmutilated chessboard can visit every square exactly once, and finish at the opposite corner square. The wazir is a fairy chess piece that can only move one square vertically or horizontally; diagonal moves are not allowed. Using the same reasoning, this wazir's tour does not exist. Par exemple, if the initial square is white, as each move alternates between black and white squares, the final square of any complete tour is black. Cependant, the opposite corner square is white.[36] This sort of tour of a chessboard also forms the basis of a type of puzzle called Numbrix, in which one must find a tour in which the positions of certain squares match given clues. The impossibility of a corner-to-corner tour shows that a Numbrix puzzle with the clues 1 in one corner and 64 in the opposite corner has no solution.[37] De Bruijn's theorem concerns the impossibility of packing certain cuboids into a larger cuboid. Par exemple, it is impossible, according to this theorem, to fill a 6 × 6 × 6 box with 1 × 2 × 4 cuboids. 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