Geometric Puzzle (followed by Chess Problem)

[DonM435]

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!

- - - - - -

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.

- - - - - -

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.

[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?

[Sarawak]

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.

[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.

[grapes]

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

[Sarawak]

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?)

[Sarawak]

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.

[Solfe]

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.

[Sarawak]

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

But, I had nothing better :)

[grapes]

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.

[DonM435]

You're all doing fine.

The "official" solution does ignore the board markings ("squares"). 64 is not an issue.

[Sarawak]

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.

[DonM435]

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."

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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?

[DonM435]

16419

[Sarawak]

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)

[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?

[Sarawak]

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?

[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.

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.

Headache!

[Sarawak]

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.

[grapes]

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. :)

[Sarawak]

Their solution is the same as the one I described in post#5

Agreed.


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!

[DonM435]

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.

[DonM435]

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

http://www.theproblemist.org/what-are-chess-problems/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.

[Sarawak]

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

[DonM435]

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

;)

[Sarawak]

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

[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...)

[Sarawak]

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.

[DonM435]

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!)

[Sarawak]

Have I missed something big here?

I don't know if you have missed anything big here, but I know I have!

[DonM435]

I suppose that I should ask a moderator to rename this "Geometric Puzzle (followed by Chess Problem)."

[Solfe]

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.

What I lack in math and geometry, I make up for with good humor!

[DonM435]

I'm prepared to nominate Coelacanth for Grandmaster of Pawn Placement Chess, a subtle sidestreet of the Royal Game.

[Sarawak]

I'm prepared to nominate Coelacanth for Grandmaster of Pawn Placement Chess, a subtle sidestreet of the Royal Game.

I am flattered, but I couldn't possibly accept the honour, as it was grapes who solved the puzzle.

[DonM435]

Well, grapes deserves a title on practical application, but you're the leading therorist.

[DonM435]

I finally got a hold of some graph paper (it's tough to do this kind of thing without an indefinitely-sized chessboard), and it looks as if the white queen has to move back and over more and more each time to extend the soluition, so the 69-move solution could be much as described in the quotation. :o

Once I verify it, I'll note this here, if anybody still cares.

[Sarawak]

I just clicked on the link to the chess problem. That is one weird configuration to be in.

[DonM435]

It's a composed problem. The rules are that the position has to be possible by conventional play, but it doesn't have to be a probable one. The black pieces and pawns would have to work hard to get into this kind of a fix.

In Dawson's problem, Black has material superiority, but almost no mobility (those black pawns are moving down the board: by convention, White is at the bottom of diagrams). Black’s pieces and pawns are all blocked and frozen, except that his king can move back and forth between two squares. Black does have a knight that could move. However, after making a waiting move to set things up, White’s queen can continually pin the black knight who always ends up shielding his king, vertically and diagonally in alternation, whilst sneaking in closer each time until delivering a checkmate.

And, as they note, the key zig-zag manuever by the white queen could have begun earlier and from farther back, if only the board were bigger in that direction. It's fun.

[Sarawak]

(those black pawns are moving down the board: by convention, White is at the bottom of diagrams)

The configuration is so weird, I just assumed they must be going the other way, and wondered why black didn't move them, for a while . . .

[DonM435]

Here's H and R's presentation of the Dawson Problem.

16436

Solution:
1. Bb1 Kd1
2. Qd6 Kc1
3. Qf4 Kd1
4. Qd4 Kc1
5. Qe3 Kd1
6. Qd3 Kc1
7. Qc2 mate

Move the white queen back to l10 (file L, rank 10) and we have a mate-in-9, by adding two moves in a bigger zigzag
1. Bb1 Kd1
2. Qd10 Kc1
3. Qh6 Kd1
... and continue as above ...
4. Qd6 Kc1
... and so on.

[Swift]

I suppose that I should ask a moderator to rename this "Geometric Puzzle (followed by Chess Problem)."
So it is written, so it is done http://www.wetcanvas.com/Community/images/29-Apr-2007/102905-egyptian_smilie.gif

[DonM435]

There's a good fellow!

[Sarawak]

I finally got a hold of some graph paper (it's tough to do this kind of thing without an indefinitely-sized chessboard), and it looks as if the white queen has to move back and over more and more each time to extend the soluition, so the 69-move solution could be much as described in the quotation. :o

Once I verify it, I'll note this here, if anybody still cares.

It does look like a geometric solution to me, so the 34th pair of moves would be a little under 10 billion squares long.

[DonM435]

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 hadn't actually explored this particular problem until tossing it in here, but I'm really fascinated by it, as you may have guessed.

Just as we expect the laws of physics to prevail in distant galaxies, exercises such as the T. R. Dawson (1889-1951) problem apply the laws of chess to actions at fantastic distances. The rules state that a rook can move any number of squares along a rank or a file. As there's no limit set on his range, we must assume that a rook on a board of 10,000 squares can still go from one edge to the opposite edge. A check from 100,000 squares away is still a threat. A queen who can travel a million squares on a single move can't do anything about an attacking knight because he isn't on one of her lines. Knights are still limited to their standard short hops, and pawns and kings alike lumber on in their limited fashion no matter how big their domain. Them's the rules.

Now, Dawson's white queen has to start on the diagonal that extends out from the c1 square, so as to pin the black knight who stands on d2 shielding the king at c1. Then she has to scoot over to the d-file (file #4) to again pin the knight after the king has moved just behind it. (The knight cannot legally move if doing so would expose the king to capture.) Geometry demands that the queen back up in increasing steps to maintain these relationships.

The queen cannot let up on the knight pins for even one move, as once Black gets a chance to move the knight, it frees up his trapped bishop and in turn the other pieces stuck on the first rank, and there will be no checkmate by White. Indeed, Black would eventually win the game via superior forces. (Black would gladly lose the knight, as doing so would free its pieces.)

But, happily for White, due to the logjam of the pieces, all Black can do is move the king back and forth, allowing the knight to be pinned again and again by the rapidly approaching queen. (Poor knight! Were this a rook or bishop, the white queen couldn't safely pin it both vertically and diagonally.)

The conventional, mate-in-7 version of the problem requires the queen to start on square h6 (i.e., file 8; rank 6). The unconventional mate-in-9 requires a start at square l10 (file 12; rank 10), i.e., the queen has to back up four ranks and move over four files. For this, we need to extend the chessboard to the north and the east, though all of the other rules still apply.

The mate-in-11 version requires the queen to go back eight more squares and over eight more, to start at a square we'd call t18 (file 20; rank 18), so the board now must be 20 x 20 at a minimum. The mate-in-13 requires backing up 16 more and moving over 16 more, and starting at file 36, rank 34. We could designate this as square (z+10)34, but letters will soon become inadequate.

(I worked this out by assigning the starting conditions, then having a computer program move the starting position back again and again in ever-increasing distances and convert these to squares, inches and miles for print out. I'm sure there's an elegant way to compute the starting rank and file from the number of moves, though I haven't tried to discern it. Not seriously, I mean.)

You can see where this is heading. Each new two-moves-longer checkmate requires moving the queen back and over twice as far as the last time, and the two added moves cover more and more territory. By the time we set up the mate-in-33, we need a board with 32,772 squares along a side, and assuming 2-inch squares, the queen starts about a mile and a half away from the doomed king. In the 61-mover, the board would be so large as to cover most of the surface of the Earth.

In the mate-in-69 version of the problem, the white queen would have to begin on the 8,589,934,594th rank, at the 8,589,934,596th file. This means that you'd require a board with at least 8,589,934,596 squares on a side. Corner to corner, the hypotenuse of the board would measure 8,589,934,596 times the square root of two, or 12,148,002,005.56 square-equivalents. As we have two-inch squares (as specified in the discussion of the original problem), you double this to get 24,296,004,011.12 inches. At 12 inches to the foot and 5,280 feet to the mile, this is 383,459.66 miles.

Now, this figure represents the diagonal measurement of the chess board. I'd say that the fact that the white queen starts out two squares away from her corner, and the targeted king meets his fate a similar distance from his corner doesn't matter very much, but you can chop a few inches off that 383,000+ miles if this bothers you. In any event, this board spans the Earth-Moon distance and then some. Try to picture a chessboard with one corner somewhere in a lunar crater and the opposite corner somewhere in a New York City park. (That's the Dawson chessboard. The one that I calculated is about 60% bigger.) It's not just an oversized board; it's an intricate one with literally trillions of 2-inch squares on it! But we need one this big and this detailed for the 69-move version of the problem.
(A regulation board would have 2-1/4 inch squares rather than 2 inch, and that would increase our mileage by 11.1 per cent, but I don't think that we need that additional complication.)

So, the 69-move solution begins thus. (We'll have to use numeric coordinates, as we have only 26 letters and over eight billion files to designate.)

1. Bb1 Kd1 -- i.e., B from (4;3) to (2;1) then K from (3;1) to (4;1)
That was easy!
2. Q from (8,589,934,594; 8,589,934,596) to (4; 8,589,934,596) Kc1
On Move #2, the white queen swings all the way from her square beyond the Moon, to another square somewhere out there, but one lined up vertically with the fourth file on the board. The black king moves left one lousy square.

3. Q from (4; 8,589,934,596) to (6; 4,294,967,298) Kd1
On Move #3 the white queen zooms to a square two files over and four billion plus squares down. In fact, this move gets her halfway towards Earth. The black king moves right one crummy square.

This reduces to the Mate-in-67 move problem. Lather, rinse, repeat.

The queen swoops on, in many mighty but ever-decreasing arcs through interplanetary space homing in on Earth, covering half the distance in two moves while the black king goes to d1 on every odd move and to c1 on every even one, over and over and over.

By her 36th move, the queen is within 1,000 miles of her quarry, and plans a checkmate in 33. Around move 50, we have all of the pieces within a good-sized room and are 19 moves from ending it all. Eventually, they're close enough to be accommodated by a 36x36 board, then a 20x20. Ultimately, we will have

67. Qf4 Kd1
68. Qd3 Kc1
69. Qc2 mate

I'm not sure how the previous annotator came up with 240,000 miles for the 69-mover. He probably did it with a slide rule or a logarithm table. Maybe he forgot to apply the Pythagorean factor that I did. Or maybe I missed some subtlety amounting to 100,000 miles?


I felt obliged to provide something of a wrap-up, but invite further discussion!

[DonM435]

[Sorry -- I copied that rather badly. I think you can enlarge it, or simply refer to the better copy above.]

In the interests of completeness, I am compelled to strike a sour note: Extending the board symmetrically up and to the right (as we have been doing mentally) does wreck the problem. Note that the black bishop trapped at h1 would be liberated if any additional empty files are tacked on to his right. If any other black piece can move, Black doesn't have to keep shuffling his king and thus endure the alternating knight pins, and White's scheme thus falls apart. Also, moving this bishop would enable the black rook to get free, followed by the black queen. Black also has three potential queens in those three pawns on the second rank, and so would win pretty easily.

So, once the board swells to 12x12 to create an l10 square for the 9-mover, and after White's first move, Black moves that bishop at h1 into the new space and liberates his rook. If White's queen tries to interfere lower right, black's knight moves somewhere and frees up the bishop on the other side. No, White needs that board edge down there.

Which is probably why Horowitz and Rosenthal's diagram only extends the upper right portion of the board. It's unfortunate, but it does serve to illuminate how well-constructed was the original problem, with every piece and the board geometry contributing to the solution. (The long diagonal of the ever-growing board in our thought-experiment would still reach into outer space, even if the bottom ranks are clipped to a width of eight files, so there!)

Not only did White fail to checkmate in 50 moves without a capture or a pawn move, she probably exceeded the time limit. They usually give you something like 2-1/2 hours for 40 moves, and those early moves, covering hundreds of thousands of miles are going to be a challenge to complete on time. Black can move really fast, as his moves are forced.


I'll be on the lookout for other chess problems that involve infinity or eternity … or astronomical distances. I'd encourage others to do likewise.

[Sarawak]

In the interests of completeness, I am compelled to strike a sour note: Extending the board symmetrically up and to the right (as we have been doing mentally) does wreck the problem. Note that the black bishop trapped at h1 would be liberated if any additional empty files are tacked on to his right.

How about if we put a black pawn in the way, blocking its escape, and then put a white piece in front of the pawn, preventing it from moving. Does that solve the problem?

[DonM435]

How about if we put a black pawn in the way, blocking its escape, and then put a white piece in front of the pawn, preventing it from moving. Does that solve the problem?

I believe that would do it! I tried adding a black pawn there, but couldn't figure out any black piece to put in front of it. Using a white piece is brilliant! Only thing is, they'd have to be present in the nonexistant "i" file, only to be revealed by the board expansion.

For an interesting alternative to the conventional problem, replace black's knight with yet another black pawn. I think the solution still works. Whenever black has a space to advance (and queen) the pawn, it's pinned. When it isn't, the king is in the way. Quite sinister!

[Sarawak]

For an interesting alternative to the conventional problem, replace black's knight with yet another black pawn. I think the solution still works. Whenever black has a space to advance (and queen) the pawn, it's pinned. When it isn't, the king is in the way. Quite sinister!

Heh heh, so close, and yet . . .