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Treebeard's Stumper Answer
17 January 2003

Peaceful Knights

We're having our Dunn Middle School Waffles 'N Jammies chess tournament today, so here's an ancient chess stumper to play between games. Start with any 3 by 3 part of a chessboard. Place two white knights in the upper corner squares and two black knights in the lower corners. The stumper is to swap the white and black knights in the fewest possible moves. Of course they must make the usual knight moves, and only one peaceful knight can be on a square at a time. If that's too easy, then what is the smallest size board you need to exchange three (or more) knights on each side?

I call this stumper "Peaceful Knights" because only one knight can be on a square at a time, and there are no captures or kills allowed in this peaceful solitaire game. It's also appropriate because Monday, January 20, 2003 is Dr. Martin Luther King, Jr. Day. It is a good time to make a personal stand for effective non-violence when it's needed. It's a basic rule of computer programming (and life) that there is always another way!

A move is a move, and there is a minimum number required. It's more challenging if you count one play as moving a single piece through succesive moves until stopped by the rules. What is the smallest number of plays that can solve this stumper? Exchanging three or more pieces on each side (and recording the moves) is a real stumper!

It takes at least sixteen moves to exchange black and white knights on a 3 by 3 chessboard, and it's easy to get lost. It can be done in just seven plays, if successive moves by one piece count as one play. If you number the squares in rows from left to right, the winning moves are: [1-6], [3-8-1], [9-4-3-8], [7-2-9-4-3], [6-7-2-9], [1-6-7], [8-1]. There are many solutions, and they all share some interesting geometry as steps around the virtual circle {1-6-7-2-9-4-3-8-1}. It's much easier to exchange 6 knights on a 3 by 3 board. The details are below.


Another solution that's easier to remember is to work your way around the board in either direction, moving each knight in turn the same direction. After four turns around the board, you have solved the stumper in 16 moves and 16 plays. Max was the first Dunn Middle School student to find this strategy when I gave the stumper to my kids at school last week. With ten minutes to work at it, 28 kids got it and 34 didn't, with about the same ratio in each grade. Even if you get it, it's hard (and tedious) to figure out what you did!

Graybear found another elegent solution in nine plays:

The best solution I have found is also simple and elegant. Each knight ends up diagonally opposite his starting point. It takes 16 moves in 9 plays. By symmetry you can start with any knight and go either clockwise or counterclockwise, so there are 8 nearly-identical solutions; here's one of them:

[1-6], [3-8-1], [9-4-3], [7-2-9], [6-7-2], [1-6-7], [3-8-1], [9-4-3], [2-9].

This is an ancient chess puzzle. It is attributed to Paulo Guarini di Forli in 1512, and maybe even further back to al-Adli ar-Rumi - "The Great Player" - who wrote a comprehensive book on chess (with chess problems) in Arabic in about 840. The book is lost, but there are references.

I found this stumper and that elegant seven play solution as problem 341 - "The Four Frogs" - in Henry Ernest Dudeney's Amusements in Mathematics (1917; Dover 1950). Dudeney gives an interesting topological solution using his "buttons and string" method. Put a "button" on each square, and use a string to show the legal knight moves. Now disentangle the string to make the circle on the right.

Look at the circle on the right. Put the white knights on the red circles 1 and 3, and put the black knights on the black circles 7 and 9. As Dudeney says,

You have simply to move the knights round the circle in one direction or the other. Play over the moves given above, and you will see that every little difficulty is removed.

I'm sure it's possible to figure exactly how many solutions are possible. Start with any knight, go clockwise or counterclockwise, and multiply the solutions by 8. But it's such a beautiful day here in California for a walk... I'll leave that as a combinatorial math stumper for the hardcore. See the "Enumeration" links on the Knight's Tour Notes webpage for inspiration.

I thought the stumper with three knights on each side would be harder, but in fact it's much easier. The board is so congested that it's hard to make a wrong move. Use the same circle above with extra knights on the 2 and 8 circles, and move them around the circle. You will do it in 8 moves. Four knights on a side with a 4 by 4 board is a piece of cake in 16 moves. I figure most setups are possible.

Here are some Web links for your own research on this stumper.

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