The Knights Tour By Colleen Raimondi The Knights

The Knight’s Tour By Colleen Raimondi

The Knight’s Tour n n n The Knight’s Tour is a geometrical problem played on a chessboard. It consists of moving a knight on a chessboard in such a way that will hit every cell once and only once. There are many, many different solutions for this problem.

Knight’s Tour Terms n Re-entrant: the last cell that the knight lands on to finish the problem is also a cell that the first cell, that the knight started from, commands. n Cell: each square on a chessboard.

Re-entrance n n n On a board containing an even number of cells the solution may or may not be re-entrant. You cannot have a solution be re-entrant on a board that contains an odd number of cells. Reason: If a knight begins on a white cell, its first move will end up on a black cell, and its next move to a white cell, and so on. That is why if the solution passes through all the cells, on a board with an odd number of cells, the last move will leave it on a cell of the same color of the cell which it started from. Therefore these cells cannot be connected by one move.

Different Solutions n n De Moivre Solution Euler Solution Vandermonde Solution Warndorff Solution

De Moivre n n n n n These solutions pertain to an ordinary eight by eight chessboard (64 cells). Divide the board into an inner 4 x 4 square that is bordered by an outer ring of two cells deep. Either start the path in the outer circle or by starting in the inner square. If you start from the outer circle, the knight needs to move around the outer circle in one direction to fill it up completely. Only go into the inner square when it is absolutely necessary When the outer ring is filled up, there are only a few cells left in the inner square. You can pass through these pretty easily with a little ingenuity. If you start from the inner square, you have to reverse the process. You can apply this method to different variations of square and rectangle boards.

An Example of De Moivre

Euler 1759 n Euler would start by randomly moving a knight over a board until there were no more moves available to it. Doing this carefully left a few cells on the board that were not hit by the knight and he would label these cells a, b, …. His method consisted of establishing certain rules by which these labeled cells could be inserted into the knight’s path, and also rules to make the solution re-entrant.

Euler’s Method Let us take the example of a path formed by the knight of 1, 2, 3, …, 59, 60; and with four cells left empty to be labeled a, b, c, d. 55 58 29 40 27 44 19 22 60 39 56 43 30 21 26 45 57 54 59 28 41 18 23 20 38 51 42 31 8 25 53 32 37 a 47 16 46 17 9 24 50 3 1 4 52 36 7 12 15 34 5 33 48 b 14 c 10 49 35 6 11 d 13 2

Euler’s Method We need to begin by making the path from 1 to 60 re-entrant. Cell 1 commands a cell p, where p is 32, 52, or 2. Cell 60 commands a cell q, where q is 29, 59, or 51. If any of the values of p and q differ by 1 then we can make the path re-entrant. For this case we have p=52 and q=51. So the cells 1, 2, 3, …, 51; 60, 59, …, 52 form a re-entrant path of 60 moves. Therefore, if we replace the numbers 60, 59, …, 52 by 52, 53, …, 60, the steps will numbered successively.

Euler’s Method 57 54 29 40 27 44 19 22 52 39 56 43 30 21 26 45 55 58 53 28 41 18 23 20 38 51 42 31 8 25 46 17 59 32 37 a 47 16 9 24 50 3 60 33 36 7 12 15 1 34 5 48 b 14 c 10 4 49 2 35 6 11 d 13

Euler’s Method Next, we need to add the cells a, b, d to this path. In the new path of 60 cells a commands the cells 51, 53, 41, 25, 7, 5, and 3. It does not matter which of these we pick, for the example we will pick 51. We want to make 51 the last cell of the path of 60 cells so that we can continue with a, b, d. To do this we want to add 9 to every number in path of 60 cells. Then replace 61, 62, …, 69 with 1, 2, …, 9, then we will have a path that starts from the cell initially occupied by 60, the 60 th move is on the cell initially occupied by 51, and the 61 st, 62 nd, 63 rd moves will be on the cells a, b, d in that order.

Euler’s Method Replaced 61, 62, …, 69 with 1, 2, …, 9 and a, b, d with 61, 62, 63 Added 9 to all cells 66 63 38 49 36 53 28 31 6 3 61 48 65 52 39 30 35 54 1 48 5 52 39 30 35 54 64 67 62 37 50 27 32 29 4 7 37 50 27 32 29 47 60 51 40 17 34 55 26 68 41 46 a 8 41 59 12 9 10 59 56 25 18 33 12 69 42 45 16 21 24 23 c 19 13 58 11 44 15 20 d 22 10 43 14 57 b 38 49 2 53 28 31 61 56 25 18 33 42 45 16 21 24 43 14 57 62 23 c 19 15 20 63 22 13 58 46 36 11 44

Euler’s Method We still have to introduce the cell c to the path. Since c now commands the new cell number 25, and 63 commands the cell now numbered 24, we can use the method we used earlier to make the path reentrant. In fact the cells numbered 1, 2, . . , 24; 63, 62, . . , 25, c form a knight’s tour. Therefore we need to replace 63, 62, . . , 25 by the numbers 25, 26, … 63, and then we can label c with 64.

Euler’s Method 6 3 50 39 1 40 5 36 49 58 53 34 4 7 51 38 61 56 59 48 17 54 33 62 2 41 28 37 42 52 35 60 57 27 32 63 18 55 46 43 16 21 24 45 14 31 26 23 64 19 15 20 25 8 47 29 12 9 10 13 30 11 44 22

Euler’s Method The last thing we have to do is make the solution re-entrant. First we need to get cell 1 and cell 64 near each other. Do this by taking one of the cells commanded by 1, like 28, so 28 commands 1 and 27. Therefore the cells 64, 63, …, 28; 1, 2, …, 27 form a path. We represent this by replacing the cells numbered 1, 2, …, 27 by 27, 26, …, 1. The cell now occupied by 1, commands the cells 26, 38, 54, 12, 2, 14, 16, 28, and the cells commanded by 64 are 13, 43, 63, 55. The cells 13 and 14 are consecutive, and therefore the cells 64, 63, …, 14; 1, 2, …, 13 form a path. We need to replace the numbers 1, 2, …, 13 with 13, 12, … 1. By doing this we achieve a re-entrant solution that covers the entire board.

Euler’s Method Final Board: Replaced 1, 2, …, 27 by 27, 26, …, 1. 22 25 50 39 52 35 60 2, …, 13 with 13, 12, … 1 57 22 25 50 39 52 35 60 57 27 40 23 36 49 58 53 34 24 21 26 51 38 61 56 59 41 28 37 48 11 54 33 62 41 28 37 48 3 54 33 62 1 32 63 13 32 63 4 55 46 43 12 7 4 29 16 19 46 43 2 7 2 64 9 18 45 14 31 12 9 64 5 1 11 20 47 42 29 16 19 18 45 14 31 15 30 17 44 5 13 8 10 55 3 6 20 47 15 30 42 17 44 6 10 8

Vandermonde’s Method n n Cover the board by two or more independent paths taken at random, and then to connect the paths. Define the position of a cell was by a fraction x/y, where the numerator x is the number of the cell from one side of the board, and the denominator y is its number from the adjacent side of the board. In a series of fractions representing a knight’s path, the difference between the numerators of two consecutive fractions can be only one or two, while the corresponding difference between their denominators must be two or one respectively. Also x and y cannot be less than 1 or greater than 8.

Vandermonde’s Method 1 1 2 3 4 5 6 7 8 5/5, 8/2, 7/1, 6/6, 2/6, 3/5, 3/1, 4/2, 4/4, 8/8, 8/4, 4/3, 6/1, 5/2, 5/4, 1/8, 1/4, 2/3, 3/4, 5/6, 6/7, 7/2, 2/4, 7/3, 6/4, 4/6, 3/7, 2/2, 1/1, 1/5, 7/5, 8/6, 5/1, 4/5, 8/1, 8/5, 2/5, 1/6, 4/1, 3/2, 2/7, 8/7, 7/8, 6/3 5/3, 6/2, 7/7, 1/7, 2/8, 3/3, 1/3, 4/8, 6/8, 5/7, 7/4, 8/3, 5/8, 3/8, 4/7, 1/2, 2/1, 3/6, 7/6, 6/5,

Warnsdorff’s Method n n n His rule is that the knight must be moved always to one of the cells from which it will command the fewest number of cells not already touched upon. The solution is not symmetrical or reentrant In most cases a single false step, except in the last three of four moves will not affect the result. http: //w 1. 859. telia. com/~u 85905224/knight/e. Warnsd. htm