Kornrade: Yep, that was basically how it was proven - no fancy math, just a lot of deductive reasoning.

As you said, step 1 is to show that the queen and two bishops cannot be placed without losing at least four moves off their optimum mobility. Then, ignoring reflections and rotations, list all the positions where exactly four moves are lost (there aren't that many), and try adding the knights to the position. You will find that the knights cannot be added without losing at least one more move off optimum mobility; furthermore, there is only one way to place the knights so that only one move is lost. That means, for maximum mobility, there is only way to place the queen, the two bishops, and the two knights.

From there, you will see there is only one square for the king that both gives him eight moves and does not cause another piece to lose at least one move. So, the king's position is also precisely defined.

That leaves the rooks. There are three squares that create no interference with the pieces already on the board, but two of those squares are on the same line, which means the rooks would interfere with each other.

That completes the analysis: For optimum mobility, the positions of all but one piece is exact (again, discounting symmetries). The final piece, a rook, can be placed on either of two squares. Kornrade gives one of the two positions. For the second solution, move the rook on g4 to a4.

As for the other unsolved puzzles mentioned by Soltis, one of them was to place the king, queen, two rooks, two knights, and two opposite-colored bishops so that all 64 squares are under attack. There are position where 63 squares are under attack, but none were known where all 64 are under attack. Also, the proviso for opposite-colored bishops is important because a position with bishops of the same color is known: R7/8/2B2K2/3N4/4N3/2Q2B2/8/7R. I'd say it is likely this puzzle has been solved by now, only because computers have gotten so fast - it ought to be possible for a modern computer to brute force its way through the few trillion positions that can be set up and either find a position that meets the requirement or show that none of them do.

The easiest way for the proof, in my opinion, is:

step1: prove that Q+B+B must waste 4 moves (very few placements waste only 4 moves, not counting rotations and reflections)

step2: for each such configuration, test the placement of the remaining pieces, starting with knights, until exhaustion (there are not many cases)

Anyway, here is one configuration in case anyone is wondering how to do it:

FEN: 8/3K4/1Q6/3BBN2/6R1/3N3k/2R5/8

Thanks for the article. btw, what are the other puzzles?

thats cool

Actually, I think the total number of positions is 64! / (8 * 56!), since you have three pairs of identical pieces.  For example, swap the knights and you have the same position, and the same goes for the bishops and rooks.  That means, if you write a computer program to do a brute force analysis of the problem, you can skip the positions where knight A is on a higher square than knight B (however you define higher).

You can reduce the analysis further by taking advantage of symmetry, since reflections and rotations don't change the position.  For example, any position can be rotated or reflected to put the king on one of 10 squares: a1, a2, a3, a4, b2, b3, b4, c3, c4, or d4.  That reduces your monster program loop to a mere (10 * 63!) / 8 * 56!), or about 3.5 trillion positions.  Yeh, computers are probably fast enough now to where you can solve the problem that way.  In fact, I'll bet that's how one of the other puzzles in Soltis' article was finally solved: "Place the eight original pieces so that all 64 squares are under attack."  As this problem stood in 1981, an arrangement that covered 63 squares was known, but none for all 64!

Kotomitsuki, my bad... sorry, correct is 64! / 56! as you say, which is nowhere near as monstrous a number as I gave, and well within the scope of the capabilities of modern tech, at the very least with the bishops set in place, or by using a recursive "back-tracking" method.

When I looked at this on a board today, It isn't actually that difficult to get a solution once you have the starting point right, the number of available squares for each given piece shrinks dramatically. I still can't see a logical proof for it, but It's certainly possible (if somewhat long-winded) to prove it by exhaustion without the aid of a computer. Trying different configurations for the king & knights that will leave open ranks and files for the rooks is sufficient.

But I'm guessing the poster had something more succint and impressive.

@ grey_pieces

There are 64! / 56! possible positions for 8 pieces

Heh, I'd say I haven't started too badly for something I first saw at about 3am this morning and thought about for ten minutes... I'll have to get a board out tomorrow and see if I can find the solutions and the key to the proof.

Again, thanks for posting. I take it you didn't crack or invalidate the other three puzzles then? ;)

To the posters asking if this was solved mathematically: No, I'd say it's more of a logic puzzle.  Yes, grey_pieces, I'd say you're on the right track.  Considering only the diagonal movers (i.e. the two bishops and the queen) is a good way to start, since they are the pieces that lose mobility as they move away from the center.  When you start testing arrangements, I think you'll notice fairly quickly that you can't put the diagonal movers on the board without losing at least four moves.  What's more, there aren't that many arrangements which lose only four moves when reflections and rotations are excluded.

That's probably already more of a hint than some of the analytically-minded chessplayers here need to solve this puzzle and find the 100-move positions.  :) By the way, I didn't use a computer when I solved the puzzle in 1981, and I doubt the other two solvers did, either.  Back in 1981, the only computers most of us knew about were those clunky mainframes.

Interesting post. I thought I'd seen all the great chess problems!

There are 64! - 56! possible positions for 8 pieces, which is a mighty large sum (1.78*10^14) so I can only assume that you didn't use an exhaustive method...

...I'm wondering if the solutions can be found programmatically? Certain heuristic approaches spring to mind that should cut down the number of passes exponentially, and maybe even possible to prove it via this method, as long as the skipped positions are guaranteed to be fruitless. Clearly, eg we could disregard any position where a piece loses more than 4 moves simply because of the board edge, therefore the bishops, queen and king actually have less than 64 squares to choose from. It makes most sense to start with the queen and bishops, because they only have the centre squares for optimal placement but will clearly interefere.

hmmmm...Even if the bishops are placed only 1 square from the centre each, we have already lost 4 moves (2 per bishop) and trying to place the queen in the centre means we've failed (sharing lines with a bishop), as does placing her anywhere else (non-optimal placement). Placing a bishop so that it does not impact the centre will also clearly lose.

Placing the bishop pair side by side in the centre leaves the queen with only 23 moves in a next best spot (check with a board), meaning that we fail if we can't find optimal squares for all of the other pieces. The king only has 2 squares to choose from at this point, but I can't see how to prove without exhaustion that there is no good placement for the other pieces. The knights are next (the rooks have 64 optimal squares on an open board) and without trying it out I can be certain that the last move (for a "solution" of a hundred moves) will be lost because one of the knights moves is blocked by a piece (intuition says king), or the king, queen or a bishop will hit an edge-board rook. No other collision will lose less than 2 moves total. I'm curious about wether the two solutions are these two seperate cases or simply involve the two differing king placements, or even if I'm missing something entirely!

However, I think thats enough already to cut down the number of tests so that a computer could conceivably test all the remaining positions and not have to run for centuries to do it, and if not there are other techniques which can further reduce the overhead, although much more the nitty-gritty of computer science than chess logic - I'm sure this would count as proof, but not certain that would have been your approach back in 1981.

Am I on the right track? Was your proof a logical discursion like this, or was it more purely mathematical?

it looks like you are proficient in algebra, is it the way to demonstration?

Thanks for the article and I agree about soltis' column.