# Thread: Geometric Puzzle (followed by Chess Problem)

1. ## Geometric Puzzle (followed by Chess Problem)

You may enjoy this. It was published in I.A. Horowitz and P.L. Rothenberg, The Complete Book of Chess, Collier Books, 1963. Chapter 14, "Chessboard Recreations." However, its chess trappings are just for show: no knowledge of the game is required. Warning: it is tedious reading!

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What, a blank Chessboard? That is right. What is more, it is the only prop you need for what is the greatest puzzle ever conceived by man. You need no knowledge of Chess, checkers or any other game or stunt on the Chessboard to appreciate the wholesomeness and beauty of this poser. But you do have to muster that little bit of common sense and elementary reasoning which is often the key to ostensible mysteries. In short, finding the answer is most gratifying; failing to find it exacts a gasp at the fiendish simplicity of this gem. Observe.

Two men sit at a perfectly constructed Chessboard, each square of which is, to the nth fraction of an inch, exactly like any other. Each man has an unlimited number of Pawns at his disposal, each of which is also perfectly constructed, so that the diameter of the base of any one Pawn does not vary, by so much as the nth fraction of an inch, from any other. Each of the two men, in turn, places a Pawn – always in upright position – anywhere on the surface of the Chessboard. In the center of a chosen square, or at the edge, or at the line of intersection between one square and the one contiguous to it, or, actually, at the very edge of the board with but part of the base of the Pawn resting on the board.

Now then, it is stipulated that he who succeeds in placing the last Pawn on the board, with no space whatever left for his opponent to place an additional Pawn, is declared the winner. Who wins? The man who is first to place a Pawn on the board, or his opponent?

In the interest of guarding against captious stratagems, we add that the base of the Pawn is not larger than the surface of the entire chessboard. This would be a bit too inanely simple. It may be reasonably assumed that the base of the Pawn is much smaller than the are of each square.

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Can either side force a win? Okay, does the first player always win, or the second?

I'll invite speculation before revealing their answer, which is interesting, though I'm not sure that I agree with it.

2. Well, I'd say if the second player mirrors the placement of the first player's pawns, he should win. Or is that too simple an answer?

3. Originally Posted by Sp1ke
Well, I'd say if the second player mirrors the placement of the first player's pawns, he should win. Or is that too simple an answer?
The first player can prevent this by placing a pawn covering the centre.

I've got some special cases worked out, but I don't see any general solution yet.

4. I would have to say that the winner would have to be the first player to place a piece for the reason that there are 64 squares on the board and 64 is not evenly divisible by 6.

5. Originally Posted by Coelacanth
The first player can prevent this by placing a pawn covering the centre.

I've got some special cases worked out, but I don't see any general solution yet.
The first person can prevent it, but if the first person puts it at the exact center, then the first person can mirror subsequent moves of the second person.

So first person wins

6. Originally Posted by grapes
The first person can prevent it, but if the first person puts it at the exact center, then the first person can mirror subsequent moves of the second person.

So first person wins
Huh, it's as simple as that, is it.

I had it in my hands, but failed to reel it in

Nicely done. (Assuming it's correct, and I think it is, no?)

7. Originally Posted by Solfe
I would have to say that the winner would have to be the first player to place a piece for the reason that there are 64 squares on the board and 64 is not evenly divisible by 6.
Not sure what the argument is, but it seems to me the 64 is irrelevant - I would say the puzzle is the same even without dividing the board into squares.

8. 64/6 is the answer I give when answering at 1 am. Well, trust me... it seemed very clear to me last night despite the fact that it is wrong by the logic I used to come by it!

But I thought my logic was kind of cool so I will share it.

White goes first and places a pawn in the center a square. Both players proceed by this method until all 64 squares are full. White attained 32 pawns on the board first, but so did black and black placed his last. So the game continues.

Assume* for a moment that there is JUST enough space between the rows of pawn (not squares) for another pawn to fit. Now you have one of two situations playing out. Either you can construct rings of six pawns around the previously placed pawns or you can create groups of six pawns by adding 5 more to the first placed pawn to make a hex shape. This proceeds until you reach the edges of the board.

The edges will remove the two pawns from your rings, either two forming the point of the hex shaped groups of pawns or one flat side of two pawns. This continues until the corners are reached. The corners force 4 pawns to be left of the packing of pawns.

This means that the Second player (Not the first player as I said in my first post) will be playing the last piece.

Silly logic, but had I actually used it effectively, I would have given a much more reasonable answer.

*I said assume because if there isn't enough room, the game is over with a black win OR if there is a lot more space and the same packing pattern will fit many times over in the leftover space.

9. I guess the other issue is, it would be possible to place some of the pawns away from the points specified.

10. But, from the OP:
Each of the two men, in turn, places a Pawn – always in upright position – anywhere on the surface of the Chessboard.
Italics on "anywhere" in the original.

11. You're all doing fine.

The "official" solution does ignore the board markings ("squares"). 64 is not an issue.
Last edited by DonM435; 2012-Mar-04 at 08:41 PM.

12. So the "official" solution is not what grapes specified?

Assuming that there is not some flaw in grapes's strategy that I do not perceive, that would prove that there exists a strategy by which the first player can force a win. There could possibly still be other methods by which the first player could also force a solution.

13. Okay, I may as well get this out of the way. I'm sure that more debate will follow.
I will note that Horowitz and Rosenthal use the old descriptive chess notation, e.g., KR1 for "king rook 1." I've provided the equivalent in the modern coordinate-based notation, wherein "a1" is the lower left square and "h8" the upper right.

- - - - - - quote - - - - - -

The man who plays first must win!

The solution of this brilliancy relates entirely to an elementary consideration of the symmetrical features in a regular geometric plane, such as a perfectly constructed chessboard. It is, of course, basic that for every point or subdivision of the plane there is necessarily a counter-point or counter-subdivision, each symmetrically complementing the other. Thus, with relation to the very center of the chessboard, white's KR1 [h1]square is counter-symmetrized by white's QR8 [a8] square, KN3 [g3] by QN6 [b6], K1 [e1] by Q8 [d8], etc. Have we then, succeeded in showing that the second man can always find a counter-point of symmetry for every pawn placed on the board by the first man, and that he, the second man must inevitably be able to make the last play and win? Good grief, what has happened to the answer given at the very outset of the solution? We did say that the first man must win. Be assured that that is correct!

There is one point on the chessboard (and one only) for which there is no counter-point of symmetry. That is the center of the board. All the first player need do is occupy the center (the intersection of the squares K4-K5-Q5-Q4 [e4-e5-d5-d4]) and his opponent cannot match it. The first man then proceeds to counter-symmetrize every one of his opponent's moves. See diagram. Move 1, in the very center of the board, cannot be countered; second player's move 2 is countered by II -- always in the precisely corresponding area; 3 is followed by III; IV comes after 4, and so on, until the first player is the last to fill the board.

Dudeney's masterpiece, which has appeared in a variety of garb, has been adapted to the Chessboard as the greatest common-sense challenge in existence. In the simplicity of its eloquence, it is a living monument to Emerson’s "eloquence in simplicity."

- - - - - end quote - - - - -

They presumably credit Henry Dudeney (1857-1930) with the basis of the problem. You'd swear it rates a Nobel Prize after reading that!
I'll attach the diagram in a follow-up (it may take some work to get it right).

Do you accept their solution?

14. Originally Posted by DonM435
Do you accept their solution?
I do not see any defect, other than minor inaccuracies in the description, which are easily fixed.

But I think you had some criticism against it, if I remember correctly? (too lazy to scroll up to look)

15. Well, the logic seems sound. I wondered, though, why White (i.e., Player #1) has to claim the center on his first move? Seems to me that he could do it any time along the way until space gets tight. After ten or twenty mirror White-then-Black moves, White grabs the center, and suddenly White is mirroring Black (Player #2) instead.

E. g., suppose that White opens (+5, +5) and Black cleverly grabs the center with (0, 0) in response. No matter: White can then mirror his own first move with (-5, -5). From here on, Black moves and White mirrors until White wins.

However ...

The only thing I could figure as a strategy for Black is that if White neglects the center on his first move, then Black should play something like (+0.25, +0.25), i.e., establish a Pawn tangent to the center (or just close enough to it), but not centered over it. This will make it impossible for either side to ever occupy the exact center. Now, White can no longer exploit (0,0), and it looks like Black will eventually win.

So, White has to go for the center immediately to win.

Or does he?

16. Originally Posted by DonM435
Well, the logic seems sound. I wondered, though, why White (i.e., Player #1) has to claim the center on his first move? Seems to me that he could do it any time along the way until space gets tight. After ten or twenty mirror White-then-Black moves, White grabs the center, and suddenly White is mirroring Black (Player #2) instead.

E. g., suppose that White opens (+5, +5) and Black cleverly grabs the center with (0, 0) in response. No matter: White can then mirror his own first move with (-5, -5). From here on, Black moves and White mirrors until White wins.

However ...

The only thing I could figure as a strategy for Black is that if White neglects the center on his first move, then Black should play something like (+0.25, +0.25), i.e., establish a Pawn tangent to the center (or just close enough to it), but not centered over it. This will make it impossible for either side to ever occupy the exact center. Now, White can no longer exploit (0,0), and it looks like Black will eventually win.

So, White has to go for the center immediately to win.

Or does he?
So the solution (assuming there is no hidden flaw) gives a strategy by which the first player can force a win. The question you are asking, is whether it is necessary, i.e., if the first player fails to follow this strategy, can the second player force a win? The only possible sort of move that might save the second player is the one you describe, essentially. (It doesn't have to be tangent to the centre, just anywhere that interferes with placing a pawn at the centre.) But does it?

The answer given looks correct to me (I am always anxiously searching for the hidden flaw, though) - the question you are asking is whether it is the only strategy that forces a win. Is that correct?

17. No solid question ... I'm just speculating.

A simple case:

If the situation is such that 16 (4 x 4) pawns can fit on a certain board before it’s full, but no more, we have an even number, and Black will inevitably place the last one. If only nine (3 x 3) Pawns can fit, we have an odd number and White will win.

If the 4 x 4 array is a tight one, with the outer Pawns hanging just short of oblivion, then if White grabs the center, he creates a situation where only nine Pawns (3 x 3 array) will fit.

If, instead the 4 x 4 array left a sizeable margin around the edges, then White’s occupy-the-center strategy might force a 5 x 5, 25 Pawn array, but in this case, he still wins. So, establishing a center does seem to create an odd number of positions.

If we accept their argument, it means that, in a finite plane (like our chessboard), there are an odd number of points. Every (x,y), it's (-x,-y) counter point, and (0,0). But we just showed that we can render the number of points even by making the occupation of (0,0) impossible. I guess that, given that the Pawn bases have area greater than zero, we aren't really talking about points at all, but rather PPPs (possible Pawn positions).

Does an infinite plane have an odd number of points -- i.e., all the pairs plus (0, 0)? I'd think that on an infinite plane, the edge conditions don’t exist, so the odd/even business doesn't enter.

18. Originally Posted by DonM435
No solid question ... I'm just speculating.

A simple case:

If the situation is such that 16 (4 x 4) pawns can fit on a certain board before it’s full, but no more, we have an even number, and Black will inevitably place the last one. If only nine (3 x 3) Pawns can fit, we have an odd number and White will win.
I'd say that's the case if there are constraints on where you can place the pawns. (For example, the pawns must be placed in the centre of the square.) In the 4x4 case, placement of a pawn not on one of the 16 designated points, means that 16 will no longer fit.

19. Originally Posted by DonM435
Do you accept their solution?
Their solution is the same as the one I described in post#5

An infinite playing board isn't really relevant, sine the game would never end--no winner.

20. Originally Posted by grapes
Their solution is the same as the one I described in post#5
Agreed.

Originally Posted by grapes
An infinite playing board isn't really relevant, sine the game would never end--no winner.
There's a thread like that in this forum!

21. Actually, not every checkmate takes place on a board edge, so you could have a finite game of chess on a board with no limits. It probably wouldn't be easy, though!

Horowitz and Rothenberg do present one chess problem that has the potential for infinite extension. I'll try to find that one for you.

22. That was easier than I thought. Check Problem #5 on this page.

http://www.theproblemist.org/what-ar...ems/27-fairies

Despite the site title, it is in fact an orthodox chess problem, but with a twist.

As it stands the problem is an orthodox mate in 7. After the key 1.Bb1, the queen continually pins the knight while moving ever closer to the king. The solution runs 1...Kd1 2.Qd6 Kc1 3.Qf4 Kd1 4.Qd4 Kc1 5.Qe3 Kd1 6.Qd3 Kc1 7.Qc2 mate. Dawson realised that if the board was extended upward and outwards beyond the h-file the manoeuvre could be extended indefinitely, and calculated that if a board with 2 inch squares was assumed and the problem reset as a mate in 69 – “the white queen would start 240000 miles away from c1 - ON THE MOON IN FACT - and we have a vision of a lunar queen sweeping astronomically through space in a tremendous zig-zag path, converging remorselessly to strike the black king to his doom - silver Fate swooping down.”
(T.R. Dawson, Caissa's Fairy Tales 1947)
H & R helpfully state that the queen starts at h6 in the 7-move problem, but could start at i10 in a 9-mover, and so on backward in ever-increasing steps (if additional ranks and files are added to the board), which explains why she's so far away in a 69-move solution.
Last edited by DonM435; 2012-Mar-05 at 12:59 AM. Reason: Added last paragraph

23. OK, for some modified version of infinite chess, but for this pawn-placing game, infinite board means infinite game, no?

24. I'll get back to you on that once I've proven it.

25. Unless I'm missing something, the proof is not difficult.

26. I was working on a proof using my 1:1 scale model of the Earth when I realised that unbounded was different than infinite.

(Thank you Steven Wright...)

27. Originally Posted by Solfe
I was working on a proof using my 1:1 scale model of the Earth when I realised that unbounded was different than infinite.

(Thank you Steven Wright...)
Some basic properties of the pawn-placement game don't seem hard to come by in either situation, though. Bounded means finite game, and unbounded in a surface-of-a-sphere kind of sense would also mean a finite game. Infinite in a sufficiently regular sense (e.g., a Euclidean plane) means infinite game. You could also have an infinite area surface, though, with enough holes that the pawns would be too big for most places.

28. As it stands the problem is an orthodox mate in 7. After the key 1.Bb1, the queen continually pins the knight while moving ever closer to the king. The solution runs 1...Kd1 2.Qd6 Kc1 3.Qf4 Kd1 4.Qd4 Kc1 5.Qe3 Kd1 6.Qd3 Kc1 7.Qc2 mate. Dawson realised that if the board was extended upward and outwards beyond the h-file the manoeuvre could be extended indefinitely, and calculated that if a board with 2 inch squares was assumed and the problem reset as a mate in 69 – “the white queen would start 240000 miles away from c1 - ON THE MOON IN FACT - and we have a vision of a lunar queen sweeping astronomically through space in a tremendous zig-zag path, converging remorselessly to strike the black king to his doom - silver Fate swooping down.”
(T.R. Dawson, Caissa's Fairy Tales 1947)
I was just working with those numbers. Figured out that I could use a program to determine and print out the actual moves in the 69-move solution from Moon to Earth, just for fun, mind you.

But working out the algorithm, I figured out that the white queen has to move back three ranks, and over three files for every additional two moves in the checkmate. She has to be on the black knight’s diagonal so as to pin it, then swing over to the d-file and back to the same diagonal, three ranks closer. It's not some kind of exponential thing. (No, now I see that it is: see Post #36 and following. The preceding statement is wrong. -- Don)

So, she’d start at l10 (that’s el-ten, not eye-ten) for a 9-mover, p14 for an 11-mover, t18 for 13 moves, x22 for 15, and here we’ve run out of letters. In an n-move checkmate, she starts on file 2n-6 and rank 2n-8. Ergo, in a 69-mover the queen starts at file 132 and rank 130.

We’d need at least a board with 132 x 132 squares. Even were these 2-1/4 inch squares (the tournament standard), our board would be something like 25 ft square. Now, that’s one hell of a board (with 17,424 squares on it), but it doesn’t quite reach to the Moon.

Seems as if we’re off by a factor of, oh say fifty million or so. Instead of a 69-move problem, we’d have to specify mate in three billion moves give or take a few to require a start on the Moon. Mind you, the problem would work as advertised; it just would take a lot longer for the Lunar Queen.

(Under the official rules, if 50 moves go by without a pawn move or a capture, either player can claim a draw, so there's no point in announcing a mate in 69 in this manner. But let's ignore that for now.)

Have I missed something big here? If someone can confirm or improve my calculations, we may have devalued this hoary old piece of wisdom. (I hope that adjective doesn't get electronically censored.) (Edit: It may be okay!)
Last edited by DonM435; 2012-Mar-05 at 04:12 PM. Reason: Further thought!

29. Originally Posted by DonM435
Have I missed something big here?
I don't know if you have missed anything big here, but I know I have!

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