Chess and Math Puzzle (nothing to do with CMA)

[David Lyall]

Re: Chess and Math Puzzle (nothing to do with CMA)

The "number of positions" with regard to chess usually means with a fixed board position vis a vis number of positions in computer searches, number of positions possible etc.

Given that the board is not normally rotated ( except at end of game to play another game ) or flipped ( except when game comes to an abrupt end ) it necessary to explicity state these conditions.

The problem is interesting even without the rotation/flipping as it provides insight into how many positions are possible in a regular game especially with regard to the legal positions.

[David Lyall]

Re: List (so far) of Illegal and Illogical Positions

Paul,

The problem did start with an 8x8 chess board.

The definition of legal could include any position that may be arrived at in the regular 8x8 game where the pieces are then moved into the center 4x4 squares.

Such a distinction may allow bishops of the same colour to remain legal as well as those positions with the Queen and Rook battery, for example, the last pawn taken - beside the King - was by a Queen from outside the 4x4 center squares.

[Paul Bonham]

Re: List (so far) of Illegal and Illogical Positions

Paul,

The problem did start with an 8x8 chess board.

The definition of legal could include any position that may be arrived at in the regular 8x8 game where the pieces are then moved into the center 4x4 squares.

Such a distinction may allow bishops of the same colour to remain legal as well as those positions with the Queen and Rook battery, for example, the last pawn taken - beside the King - was by a Queen from outside the 4x4 center squares.

Good thinking about the pawn capture, I had not thought of that. So that means that the Queen & Rook battery, the doubled Rook battery, and the Queen & Bishop battery are legal as long as the checking piece that is adjacent to the King is on the outside edge of the board. If it's in the center, then it couldn't happen. But actually, I think I already mentioned that, so it means the battery check isn't anything special that we need to eliminate.

It also makes case (7) possible, the Rook could have captured a pawn as well.

Technically, the Bishops on the same color are legal, yes. If everyone involved in this thread wants to include it, and even include it for both sides having Bishops on the same color (wow!), then fine, we'll include it.

I was going to insist that it's too far-fetched to include, but then again, all these positions are far-fetched!

For one thing, has there ever been a game in which all 16 pawns have been captured and no pieces have been captured (or alternatively, one or two pieces were captured such as each side's Queen, but pawn promotions replaced them)?

Then you have to have both players bringing all their pieces into the central 4 x 4 grid, like stars being sucked into a black hole.

Maybe that's what we could call this puzzle: the Black Hole Puzzle!

[David Lyall]

Re: List (so far) of Illegal and Illogical Positions

Paul,

A revised list of valid checking squares for the Black Hole Puzzle.

Valid checking squares +
Invalid checking squares -

K=King, R=Rook, B=Bishop, N=Knight, S=Square filled in by any piece

King in Corner Square:

K R+S S
R+B-N+S
S N+S S
S S S S

King in Edge Square:

R+K R+S
B+R-B-N+
N+S N+S
S S S S

King in Middle Square:

B+R+B+N+
R+K R-S
B+R-B-N+
N+S N+S

Queen can check in same squares as R+ and B+
Queen cannot be placed in squares R- and B-

Other King cannot be placed any square adjacent to K

Valid positions do not have King in check by more than one piece.

[Paul Bonham]

Re: Chess and Math Puzzle (nothing to do with CMA)

If we WERE to eliminate all those positions in which either side has both Bishops on same-colored squares (with the assumption that all pawn promotions are to Queen), it seems to be quite simple.

Since all 4 Bishops must be in the grid, here is the matrix of possible placements of them with respect to the color of square they are on:


White Bishops on Same Color Squares: Yes Yes No No
Black Bishops on Same Color Squares: Yes No Yes No
-------------------------------------------------------------------
Allowed: No No No Yes


From this, we can see that there are 4 possible placement combinations, and only 1 of them are allowed.

In the 212,497,084,800 combinations we have narrowed down to so far (thanks to Bob Gillanders), all 4 combinations are equally distributed. Therefore only one quarter of these combinations would be allowed, the other three quarters would have same-colored Bishops on the same-colored square.

212,497,084,800 / 4 = 53,124,271,200 combinations.

For anyone who wants to check out a sample Black Hole position, here's one I conjured up, and for interest I made it symmetrical:

A SAMPLE POSITION:

White: Kc3 Qd4 Rc4 Rd3 Be3 Bd5 Nc5 Nf3
Black: Kf6 Qe5 Re6 Rf5 Bd6 Be4 Nc6 Nf4



DOES THE SIDE TO MOVE WIN?



Let's say White to move first. Here's one possible line:

1. Nxe4+ this forces a King move
1. ... Ke7
2. Nxe5 Bxe5 now White Queen is pinned
3. Bxe6 Ne2+
4. Kd2 Ncxd4
5. Bxd4 Nxd4
6. Rxd4 Kxe6 Black is a Rook down with no compensation

if instead Black had moved
2. ... Nxd4 immediately getting the Queen
3. Bxe6

( 3. Kxd4? Bxe5+ 4. Kc5 Nxd3+ and Black is ahead)

3. ... Nb5+
4. Kd2 Kxe6
5. Rxd6+ Nxd6
6. Nxd6 and White is still well ahead.


I did not use a computer for this and do not claim any of these are best lines.

[Jordan S. Berson]

Re: Whoa, you head must be really hurting!

I haven't seen 1=2 yet, so please indulge us. However back in high school I did have a proof for 3=7, probably a similiar theme. ;)

Hi Bob,

Sorry for the delay in responding, I completely forgot about this thread. I've shown this puzzle many times on a chalk board, this is the first time using text, so let's see if this works.

Each of the following steps are conisdered true statements. Find the flaw that allows 1 = 2. Suppose that X = 1 in the following series of steps:

1) X = X

2) X² = X² [Square both sides]

3) (X² - X²) = (X² - X²) [Subtract X² from both sides]

4) (X+X)(X-X) = X(X-X) [Difference of squares on left, Factor out X on right]

5) (X+X)(X-X) = X(X-X) [Remove the common factor (X-X) from both sides]

6) X+X = X

7) 2 = 1

[Bob Gillanders]

Re: Whoa, you head must be really hurting!

Hi Bob,

Sorry for the delay in responding, I completely forgot about this thread. I've shown this puzzle many times on a chalk board, this is the first time using text, so let's see if this works.

Each of the following steps are conisdered true statements. Find the flaw that allows 1 = 2. Suppose that X = 1 in the following series of steps:

1) X = X

2) X² = X² [Square both sides]

3) (X² - X²) = (X² - X²) [Subtract X² from both sides]

4) (X+X)(X-X) = X(X-X) [Difference of squares on left, Factor out X on right]

5) (X+X)(X-X) = X(X-X) [Remove the common factor (X-X) from both sides]

6) X+X = X

7) 2 = 1

Nice try! ;) Statement 4 is wrong. s/b (X+X)(X-X) = 2X(X-X) not X(X-X)

of course (X-X) = 0, so when you go from stmt 5 to stmt 6, you are dividing by zero which isn't kosher, and you can get anything to equal anything.

It's the same trick I employed in my 3=7 proof back in high school.

[Jordan S. Berson]

Re: Whoa, you head must be really hurting!

Nice try! ;) Statement 4 is wrong. s/b (X+X)(X-X) = 2X(X-X) not X(X-X)

No, X(X-X) is correct. You're factoring out X, not 2X.

Example: (XY-XZ) => X(Y-Z) not 2X(Y-Z). Replace the Y and Z with X and it's the same problem, since XX = X².

of course (X-X) = 0, so when you go from stmt 5 to stmt 6, you are dividing by zero which isn't kosher, and you can get anything to equal anything.

Exactly :)