# what's the disposition of the pieces?

Lahaina
Posers and Puzzles 26 Apr '06 05:49
1. Lahaina
Cruel sun
26 Apr '06 05:49
I just had an idea of puzzle by colouring all the field my pieces could reach in order to capture a piece of my enemy if he would came there.

For example, if white can capture on a4 and c4, white got a pawn on b3.

The question of the position of all the pieces is now to be solved with the following informations:

White: with 1 king, 1 queen, 2 knights, 1 bishop and 5 pawns
can reach, in order to capture, fields:
a7 - c7 - g7 - d6 - f6 - h6 - a5 - e5 - g5 - b4 - d4 - h4 - a3 - b3 - c3 - d3 - e3 - g3 - b2 - c2 - d2 - e2 - f2 - c1 - d1- e1 - f1 - g1 - h1

Black: with 1 king, 1 queen, 1 knight, 2 bishops, 4 pawns
can reach, in order to capture, fields:
a8 - b8 - d8 - h8 - a7 - b7 - c7 - d7 - f7 - h7 - b6 - c6 - d6 - e6 - f6 - h6 - c5 - d5 - f5 - h5 - d4 - e4 - f4 - g4 - h4 - b3 - d3 - e3 - f3 - h3 - d2 - e2 - f2 - g2 - h2 - c1 - d1 - e1 - f1 - g1

Of course, an own piece could lay in a protected field, and the same fields could be protected by more pieces.

This position is taken from a finished game played on this chess-site, with ID number somewhere between 1976546 and 1979545, and the winner was black (checkmate).

The answer should come in such form:
White: Kd1 / Qb2 / ...etc

Good luck!

Another question: how could we call this crazy game? chesso-ku?
2. XanthosNZ
Cancerous Bus Crash
26 Apr '06 09:481 edit
Originally posted by Lahaina
I just had an idea of puzzle by colouring all the field my pieces could reach in order to capture a piece of my enemy if he would came there.

For example, if white can capture on a4 and c4, white got a pawn on b3.

The question of the position of all the pieces is now to be solved with the following informations:

White: with 1 king, 1 queen, 2 knight ...[text shortened]... / Qb2 / ...etc

Good luck!

Another question: how could we call this crazy game? chesso-ku?
Say we have a queen on a1 and a pawn on a4. All of row 1, all of the a-h diagonal and a2,a3 are captureable. What about a1 and a4? My understanding is that a4 would be but a1 wouldn't. Is that right?

Assuming this is true and working on the first one.

Firstly, finding possible locations for the knights (as nothing can block their attacks). We can see that the only possible locations are e8, b5, a2, a1, b1 and h1. Any other position would attack a square not on the list.
Also the king must be on d2, d1 or e1 as again any other location would attack a non-attacked square.
The queen must also occupy d2, d1 or e1 by the same reasoning (It must attack at least every square adjacent to it). But the queen cannot occupy d2 (no piece can be placed on e3 which will not attack f4), d1 (no piece can be placed on e2 that does not attack f3) or e1 (no piece can be placed on e2 or e3 that will not attack f3 or f4).

Therefore my assumption was incorrect and Lahaina isn't counting occupied squares that are attacked as attacked.
3. Lahaina
Cruel sun
26 Apr '06 11:184 edits
Originally posted by XanthosNZ
Say we have a queen on a1 and a pawn on a4. All of row 1, all of the a-h diagonal and a2,a3 are captureable. What about a1 and a4? My understanding is that a4 would be but a1 wouldn't. Is that right?
Right. The Queen in a1 protects the pawn in a4, and the way to this pawn (a2, a3). The queen protects also the diagonal b2, c3, d4 ,... and the row 1. The pawn protects b5 only (in the attack, the pawn takes one step diagonaly).

Edit: XanthosNZ, it's not so bad. You gave several positions for knights and you got two of them correct. Anyway, you call a field as unattacked even if it was on the given list, that causes you a false position for the third knight.
For the Kings, you're on a good way too.

Edit: don't forget that white plays "up" and black "down". This gives an influence on the fields protected by pawns. * The same field may be protected by more that one piece and can have a piece on it (in your example, a4 is containing a pawn and is protected by the a1-queen).
4. XanthosNZ
Cancerous Bus Crash
26 Apr '06 13:172 edits
Originally posted by Lahaina
Right. The Queen in a1 protects the pawn in a4, and the way to this pawn (a2, a3). The queen protects also the diagonal b2, c3, d4 ,... and the row 1. The pawn protects b5 only (in the attack, the pawn takes one step diagonaly).

Edit: XanthosNZ, it's not so bad. You gave several positions for knights and you got two of them correct. Anyway, you call a fie piece on it (in your example, a4 is containing a pawn and is protected by the a1-queen).
Let's go over this again. Here's where the contradiction lies.

Both the king and queen must attack all square adjacent to the square they are on. This is clear.
Now, there are only three squares where this is possible given the arrangement of attacked squares. They are d1, e1 and e2.

If I place the queen on d1 then I must have a piece on e2 for f3 to be unattacked. It can't be a bishop or a pawn as they would both attack f3. So would the king. And a knight on e2 would attack f4. Therefore as it not possible for f3 to be unattacked if the queen is on d1 it cannot be so.

If I place the queen on e1 then I must place a piece on e2 or e3. Once again these squares cannot be occupied by any piece without attacking a square designated unattacked.

If I place the queen on e2 then I have the same problem as I now must place a piece on e3 which cannot be done.

Therefore it is not possible to place a queen on this board under the rules. Therefore you've made a mistake.

EDIT: Wait, are you counting black pieces as blocking influence? If so, that makes the problem almost impervious to logic as you don't state what black pieces remain.
EDIT2: Your second one has an inconsistency also. Only a queen or bishop can attack the c1 square. It cannot do so from d2 as it would also attack c3. It cannot do so from f4 as it would have to attack g5. Therefore it must be on e3. And yet, that would place a piece on f4 (to stop the attack on g5). However, no piece can be placed there without attacking a square that is not to be attacked.
5. Lahaina
Cruel sun
26 Apr '06 13:45
Originally posted by XanthosNZ
Let's go over this again. Here's where the contradiction lies.

Both the king and queen must attack all square adjacent to the square they are on. This is clear.
Now, there are only three squares where this is possible given the arrangement of attacked squares. They are d1, e1 and e2.

If I place the queen on d1 then I must have a piece on e2 for f3 t ...[text shortened]... owever, no piece can be placed there without attacking a square that is not to be attacked.
The black pieces can block the way to a field, of course. The whole set is taken from a real party led to its end. You have to consider in the same time the position of the white army AND the position of the black army. I think it was clear...: I wrote above that the final position shows the black winning by checkmate. You have to collide both lists of attacked fields to determinate where the 19 pieces are located (19 pieces = 10 white + 9 black).

OK, anyway, it's very interesting to see how a good player is looking for the correct answer... 😉 It's good to write the way of your thinking. I can learn something about chess, so thank you to have accepted this puzzle!

white king is located on d1
black king is located on c8
6. BigDoggProblem
Just...the Dogg
26 Apr '06 21:261 edit
Here's one solution:

One quibble: There are extraneous pieces, such as Pa3 and Nb1, that could be removed without changing the solution.

I didn't bother with a logical solution to this one; trial and error is a faster method.
7. Lahaina
Cruel sun
27 Apr '06 05:31
Originally posted by BigDoggProblem
Here's one solution:

[fen]2kn4/4p1p1/8/1N4P1/p1p2Pb1/P3b3/2P5/BN1KQ2q[/fen]

One quibble: There are extraneous pieces, such as Pa3 and Nb1, that could be removed without changing the solution.

I didn't bother with a logical solution to this one; trial and error is a faster method.
not so far from truth, but a white pawn is missing... you only have 4 and there are 5!
8. BigDoggProblem
Just...the Dogg
27 Apr '06 06:46
Originally posted by Lahaina
not so far from truth, but a white pawn is missing... you only have 4 and there are 5!
Looks like I can add wPa2 without changing anything.

I think the problem would be a bit better without the 'extra' pieces, such as this one. It's not really necessary to include every single one from the game, is it?

It's like in forced mate problems, I can show an idea this way:

White mates in 5

OR I can cut the fat, without losing any of the value of the idea (see next post...)
9. BigDoggProblem
Just...the Dogg
27 Apr '06 06:472 edits

White mates in 5

In other words, Pa2, Pa3, and Nb1 should all be removed from your original problem. They serve no real purpose.
10. Lahaina
Cruel sun
27 Apr '06 09:221 edit
I didn't look if the game had useless pieces, that's true.

The solution was

It's game

Thank you to have played...
11. BigDoggProblem
Just...the Dogg
27 Apr '06 16:07
Originally posted by Lahaina
I didn't look if the game had useless pieces, that's true.

The solution was
[fen]2kn4/2p3p1/4p3/1N4P1/2p2P2/P1B1bb2/1PP5/1N1KQ2q[/fen]

It's game

Thank you to have played...
Ahh, too bad. The puzzle would be more interesting if it had only one solution.
12. Lahaina
Cruel sun
28 Apr '06 14:19
Originally posted by BigDoggProblem
Ahh, too bad. The puzzle would be more interesting if it had only one solution.
so let's create one. Mine was done as a test, an experiment. I'm not so involved in chess to create such an ideal-one-answer puzzle...

Let's have a real chess-o-ku puzzle soon!
13. BigDoggProblem
Just...the Dogg
28 Apr '06 16:481 edit
Here's an original using your condition.

White controls the squares marked with white pawns:
[fen]8/8/2P1P1P1/6PP/1P4P1/P1P3PP/8/8[/fen]
Black controls the squares marked with black pawns:

What is the minimum number of pieces that must be on the board?
14. BigDoggProblem
Just...the Dogg
05 May '06 06:37
Originally posted by BigDoggProblem
Black controls the squares marked with black pawns:
[fen]8/8/pp6/ppp5/pppppppp/p7/1p6/p7[/fen]

What is the minimum number of pieces that must be on the board?
Nevermind, the problem has a dual (unintended 2nd solution). It doesn't matter much - this genre seems to generate zero solving interest.
15. 07 May '06 00:10
Originally posted by BigDoggProblem
Nevermind, the problem has a dual (unintended 2nd solution). It doesn't matter much - this genre seems to generate zero solving interest.
Is it 3? Q on a4, pawn on b2 and b3