6
$\begingroup$

This is a sequel to Gentrification in Chessshire.

Due to the febrile state of the Chesster housing market you and your flat mates are forced to rent out half your place to another group of sharers.

Keeping every one (including punters BTL) happy was not easy before and will only become more tricky in the crammed new conditions.

You quickly find that the new lot are not at all compatible with you and your friends and to prevent open warfare arrangements have to be made that minimise contact between you and them while still maintaining the delicate dynamics within each group.

Task

Place a complete (black and white) set of proper pieces (no pawns) on a standard chess board such that

  1. the bishops are pairs, i.e. light- and dark-squared
  2. no black piece threatens any white piece and vice versa
  3. each white piece attacks exactly one other white piece and every white piece is attacked by exactly one other white piece. This results in a single daisy chain running through all white pieces. The same holds for the black pieces

Bonus / Tie breaker

Find

  1. a symmetric solution
  2. a non symmetric solution
  3. a pair of solutions that differs only by the placement of a single piece
  4. a solution that permits drawing a straight line such that all squares occupied by white pieces are entirely on one side and all squares occupied by black pieces are entirely on the other side
  5. a solution with alternating corner occupancy, i.e. a white piece on at least one of a1,a2,b1 and of g8,h7,h8 and a black piece on at least one of a7,a8,b8 and of g1,h1,h2.
  6. a solution with a queen in the centre
  7. a solution with a king in the centre
  8. a solution with another piece in the centre
  9. a solution with the centre empty
  10. a solution with four squares that can be connected by a straight line that is entirely covered by the squares such that the first and third square are occupied by white and the second and fourth by black pieces
  11. a solution where every square is attacked

Note: solutions (not necessarily distinct) for each of the above properties exist.

Hints:

1/(4)

There are a total of 9 solutions up to rotation, mirroring and swapping colours. As a hint I'm giving rough territorial maps for all of them:

      BBBBB.WW    BBBB.BBB    BBBBBBBB
      BBBBB.WW    BBBB.BBB    BBBBBBBB
      BBBBB.WW    BBBB.BBB    BBBWWWWW
      BBBBB.WW    .....WWW    BBBWWWWW
      BBBBB.WW    BBB.....    BBBWWWWW
      ......WW    WWW.WWWW    BBBWWWWW
      ......WW    WWW.WWWW    BBBWWWWW
      WWWWWWWW    WWW.WWWW    BBBWWWWW

      WBBBBBBB    BBBBBBWW    BBBBBBWW
      WWBBBBBB    BBBBBBWW    BBBBBBWW
      WWWBBBBB    BBBBBBWW    BBBBBBWW
      WWWWBBBB    BBBBBBWW    BBBBBBWW
      WWWWBBBB    WWWWBBBB    WWWWBBBB
      WWWWWBBB    WWWWBBBB    WWWWBBBB
      WWWWWWBB    WWWWBBBB    WWWWBBBB
      WWWWWWWB    WWWWBBBB    WWWWBBBB

      XXXXXXXX    BBBBBBBB    BBBBBWWW
      XXXXXXXX    BBBBBBBB    BBBBBWWW
      WWWWWWW.    WBBBBBBB    BBBBBWWW
      WWWWWW.B    WWWWWBBB    BBB..WWW
      WWWWW.BB    WWWWWBBB    BBB..WWW
      WWWW.BBB    WWWWWBBB    BBBWWWWW
      WWW.BBBB    WWWWWBBB    BBBWWWWW
      WW.BBBBB    WWWWWWBB    BBBWWWWW

     W: white  B: black  X: both  .: neither
 

Please note that rectanglese of B,W,X are not necessarily minimal. They may contain empty ranks or files, possibly at the boundary.

2/(4)

The first three maps at better resolution:

      BBBBB...    BBBBBBB.    BBBBBBB.
      BBBB....    BBBBBBB.    BBBBBBB.
      BBBB..WW    BBBBBB..    BBBWWWWW
      BBBB..WW    B......W    BBBWWWWW
      BB....WW    B......W    B..WWWWW
      ......WW    ..WWWWWW    B..WWWWW
      ......WW    .WWWWWWW    B..WWWWW
      .WWWWWWW    .WWWWWWW    ...WWWWW
 

3/(4)

And maps 4-6:

      ....BBBB    BBBBBB..    BBBBB...
      WW..BBBB    BBBBBBWW    BBBBB.WW
      WW...BBB    BBBBB.WW    BBBBB.WW
      WWWW.BBB    ......WW    ......WW
      WWWW.BBB    WW......    WW......
      WWWW.BBB    WW...BBB    WW...BBB
      WWWWWW.B    WWWW.BBB    WWWW.BBB
      WWWWWW..    WWWW....    WWWW....
 

4/(4)

And the last lot:

      ..XXXX..    .BBBBBB.    .....WWW
      ..XXXX..    .BBBBBB.    B.....WW
      WW......    W.......    BBBB..WW
      WW.....B    W......B    BB.....W
      WWWW...B    WWWW.BBB    BB.....W
      WW.....B    WWWWWBBB    BB.WWWWW
      WW....BB    WWWWWBBB    B...WWWW
      WW.BBBBB    WWWWWW.B    B.......
 

$\endgroup$
8
  • 1
    $\begingroup$ Talking about me and my band of artists and scientists until we suddenly become chess pieces. Lol. $\endgroup$ – WhatsUp May 16 at 1:54
  • 2
    $\begingroup$ @WhatsUp edited. Can't do anything about the chess bit I'm afraid. $\endgroup$ – loopy walt May 16 at 3:23
  • $\begingroup$ The new restrictions here is condition 2? $\endgroup$ – justhalf May 24 at 2:09
  • $\begingroup$ @justhalf Technically, it also applies to the original puzzle. It just has no effect as there is only one colour. ;-) $\endgroup$ – loopy walt May 24 at 2:47
  • $\begingroup$ Oh ok, so the new condition is the "complete (black and white) set" part. $\endgroup$ – justhalf May 24 at 4:41
4
$\begingroup$

With the new hints this has become much simpler. In fact, OP has (unintentionally?) changed the character of the puzzle. At least some solutions can be step-by-step (like a sudoku for want of a better simile) reconstructed from the hints.

For example number 8 (I'll only do one, so people still have a chance to earn the bounty by solving one of the others):

Let us assume that in each non rectangular area every convex corner not touching the boundary is actually occupied and see how far we can get:

   8   . b b b b b b .
   7   . b b b b b b .
   6   W!. . . . . . .
   5   w . . . . . . B!
   4   w w w W!. B!b b
   3   w w w w W!b b b
   2   w w w w w B!b b
   1   w w w w w W!. B!

       a b c d e f g h

   w: white piece or empty   .: empty    W! white piece
   b: black piece or empty               B! black piece
 

Because of the white pieces on f1 and e3 the black piece on f2 can only be a N. As this N attacks h1 the h3 and g4 squares must be empty. Similarly, f4 must also be a black N and g2 empty. On h1 there cannot be a R or Q because of the white piece on f1. Black Ns are used up leaving a K or a B. But if h1 were a K h2 would have to be a B (the K has to attack something, and that something must not attack him back) but that would leave the N on f2 unreachable. Therefore h1 is a B, f3 is empty and there is a black piece on b7.

   8   . b b b b b b .
   7   . B!b b b b b .
   6   W!. . . . . . .
   5   w . . . . . . B!
   4   w w w W!. bN. b
   3   w w w w W!. b .
   2   w w w w w bN. b
   1   w w w w w W!. bB

       a b c d e f g h
 

Of the black pieces still available only a R can sit on b7 without attacking the white piece on a6. The only way of attacking the knight on f2 without also attacking f4 or h5 or h1 is a B on h4. As the only piece that can attack the Q is a N and the f2 N already attacks another piece h5 must be the Q. To close the cycle g7 must be the K and f8 the other R.

With the black chain complete we can remove the remaining black markers and also all the white ones that are attacked by black pieces:

   8   . . . . . bR. .
   7   . bR. . . . bK.
   6   W!. . . . . . .
   5   . . . . . . . bQ
   4   w . w W!. bN. bB
   3   w . w . W!. . .
   2   w . w w . bN. .
   1   w . w . w W!. bB

       a b c d e f g h
 

The piece on a6 cannot attack b7 ruling out Q,B,K nor can it be a N because there is no suitable white target, so it must be a R. This R can only be attacked via the a6-f1 diagonal. As there is piece on f1 a Q or B between a6 and f1 would attack at least two pieces. Therefore f1 must be the B and the rest of the diagonal empty.

   8   . . . . . bR. .
   7   . bR. . . . bK.
   6   wR. . . . . . .
   5   . . . . . . . bQ
   4   w . . W!. bN. bB
   3   w . w . W!. . .
   2   w . w w . bN. .
   1   w . w . w wB. bB

       a b c d e f g h
 

Of the squares left the only one where the Q doesn't attack two or more other pieces is c2. To shield the black Nf2 there must be another white piece on d2 which in turn forces a2,a4,c1,c3 to be empty.

   8   . . . . . bR. .
   7   . bR. . . . bK.
   6   wR. . . . . . .
   5   . . . . . . . bQ
   4   . . . W!. bN. bB
   3   w . . . W!. . .
   2   . . wQW!. bN. .
   1   w . . . w wB. bB

       a b c d e f g h
 

Only one of the a1 and a3 squares can be occupied. Indeed, a3 would have to be a N and a1 a N or B. They cannot both be Ns and B on a1 could only be reached by a R on e1 which would also attack f1 and e3.

With only one more square useable in the a-file e1 must be occupied. This leaves a1 as the only possible square for the dark-squared B, d4 as the only place for the K and e3 as the only place for the remaining R. The Ns go to d2 and e1.

Fully reconstructed position:

   8   . . . . . bR. .
   7   . bR. . . . bK.
   6   wR. . . . . . .
   5   . . . . . . . bQ
   4   . . . wK. bN. bB
   3   . . . . wR. . .
   2   . . wQwN. bN. .
   1   wB. . . wNwB. bB

       a b c d e f g h
 

Nicer picture (thanks @justhalf !)

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ Unintentionally, indeed. I wonder whether this could be made a new class of puzzle. One would have to modify it somehow so there are more than 9 different puzzles altogether. Maybe a larger board would do the trick. $\endgroup$ – loopy walt May 28 at 21:45
  • 1
    $\begingroup$ I'm all for it @loopywalt. Solving this one was certainly fun. $\endgroup$ – Albert.Lang May 28 at 21:48
  • 2
    $\begingroup$ I added the image of the board here: i.stack.imgur.com/4rHNA.png Feel free to include it into your answer. And here is the lichess link $\endgroup$ – justhalf May 29 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.