22
$\begingroup$

A wazir is a fairy chess piece that moves like a rook, but can go only one square.

                                                              enter image description here

We wish to place a number of wazirs on a 9 $\times$ 9 chessboard so the following conditions are satisfied

  1. Each wazir is being attacked by at least one other wazir.
  2. Each empty square is being attacked by at least one wazir.

What is the minimum number of wazirs we need to place to satisfy the conditions above?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ @PotatoLatte Try this: chess.com/analysis $\endgroup$
    – Jafe
    Commented Jul 5, 2019 at 11:46
  • 1
    $\begingroup$ lichess.org/editor allows to set up a position, and, if you click on "Analysis Board" you can also insert moves and link to the game. $\endgroup$
    – shoopi
    Commented Jul 5, 2019 at 14:14
  • 1
    $\begingroup$ (+1) Any reason why you chose the name wazir? $\endgroup$
    – TheSimpliFire
    Commented Jul 5, 2019 at 19:53
  • 3
    $\begingroup$ @TheSimpliFire en.wikipedia.org/wiki/Wazir_%28chess%29 :) $\endgroup$
    – Jafe
    Commented Jul 5, 2019 at 20:08
  • 1
    $\begingroup$ Hah, wazir. That's a funny name :) $\endgroup$
    – Mr Pie
    Commented Jul 6, 2019 at 3:07

3 Answers 3

Reset to default
24
$\begingroup$

I have a solution with

25 wazirs:
enter image description here

Step by step:

We can cover all border squares with 8 pairs of wazirs whose attacking squares do not overlap. Note that each wazir covers two border squares, which is the maximum because it's not possible for a piece to be orthogonally adjacent to 3 border squares on a 9x9 board. So there is no way to cover the entire border with fewer wazirs.
enter image description here

Also note that this arrangement covers the maximum amount of non-border squares as well. Every wazir covering two border squares covers exactly one non-border square, unless it is located one step diagonally from a corner (in which it covers 2). Every square one step diagonally from a corner is in use, so the wazirs cover the maximum possible amount of non-border squares.

There are 17 dark squares left, so we need a minimum of 5 wazirs to cover them all.
enter image description here

Then we need four more wazirs to cover the remaining 12 light squares.
enter image description here

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ What a pity you have to break the symmetry with that one piece, as you can't put it in the centre. Lovely solution though. $\endgroup$
    – Jaap Scherphuis
    Commented Jul 5, 2019 at 12:27
  • 1
    $\begingroup$ this is minimal :) $\endgroup$
    – Oray
    Commented Jul 5, 2019 at 12:40
  • 2
    $\begingroup$ Very nice argument, I like the diagrams. I was hoping somebody would be able to present something like this. $\endgroup$
    – hexomino
    Commented Jul 5, 2019 at 15:11
7
$\begingroup$

Here is an attempt which needs

27

wazirs:

enter image description here

Improve this answer
$\endgroup$
6
$\begingroup$

A perhaps more elegant way to do

27 wazirs

than @Glorfindel's solution is:

Put wazirs on every square on the b, e, and h files. Alternatively, the 2nd, 5th, and 8th rank. You get the idea.

Improve this answer
$\endgroup$
0

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