By John J. Watkins
Around the Board is the definitive paintings on chessboard difficulties. it isn't easily approximately chess however the chessboard itself--that easy grid of squares so universal to video games all over the world. And, extra importantly, the interesting arithmetic at the back of it. From the Knight's travel challenge and Queens Domination to their many adaptations, John Watkins surveys the entire famous difficulties during this unusually fertile quarter of leisure arithmetic. Can a knight keep on with a direction that covers each sq. as soon as, finishing at the beginning sq.? what percentage queens are wanted in order that each sq. is concentrated or occupied via one of many queens?
Each major subject is handled intensive from its old belief via to its prestige at the present time. Many attractive strategies have emerged for simple chessboard difficulties because mathematicians first begun engaged on them in earnest over 3 centuries in the past, yet such difficulties, together with these concerning polyominoes, have now been prolonged to third-dimensional chessboards or even chessboards on strange surfaces equivalent to toruses (the an identical of enjoying chess on a doughnut) and cylinders. utilizing the hugely visible language of graph idea, Watkins lightly courses the reader to the leading edge of present study in arithmetic. by way of fixing a number of the many workouts sprinkled all through, the reader can percentage totally within the pleasure of discovery.
Showing that chess puzzles are the place to begin for very important mathematical principles that experience resonated for hundreds of years, around the Board will captivate scholars and teachers, mathematicians, chess lovers, and puzzle devotees.
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Additional info for Across the Board: The Mathematics of Chessboard Problems
Moreover, since there are an equal number of squares colored A and squares colored C, when a knight moves from a square colored A to a square colored C, it can’t then go on from the square colored C to a square colored D, for the knight would still have to eventually return to another square colored C before being able to get back to a square colored A. In other words, the knight has to avoid squares colored B and D entirely in order to visit all of the squares colored A. Therefore, no knight’s tour is possible.
Now, let’s follow the cycle from X to A and see where it goes. Continue past A from edge to edge until you get to C. From C, the Hamiltonian cycle can’t go back to X yet, so it must go to B next, from where we follow it on to D. At D there again seems to be a choice of which way to go, but remember that there must be an edge attached to vertex Y coming from the left, so our Hamiltonian cycle must now go from D to Y. Since the remaining situation is now once again symmetric, we may as well assume that the Hamiltonian cycle continues next from Y on to E, rather than F, and from there we follow the cycle along until we reach vertex Z and ﬁnd ourselves unable to continue.
Since a knight on a red square can only move to a blue square, every time a knight is at a red square during the knight’s tour it must visit a blue square next on the tour. Moreover, since there are the same number of red squares and blue squares to be visited, a knight can’t dare visit two blue squares in a row anywhere along the tour since it wouldn’t be able to make up for this later by visiting two red squares in a row. Therefore, we conclude that the knight’s tour must strictly alternate between the red and the blue squares.