How many knights are necessary...
Burkart
knights are standing).
What is your solution? Which position?
Burkart (who would like to know the best solution!)
The Qurqirish Dragon
>knights are standing).
>What is your solution? Which position?
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Using algebraic notation, a solution with 12 knights is given by:
b3,c3,c4,c6,c7,d6,e3,f2,f3,f6,g5,g6
(this is an "L" shaped group of 3, repeated four times). So 12 is a sufficient
number of knights.
Now, three different knights must be used to cover a1,a2, and b2. Futhermore, a
knight covering any 'a' or 'b' square cannot cover any 'g' or 'h'. Similarly, a
knight covering a '1' or '2' square cannot cover a '7' or '8'. This is because
a knight's coverage area fits in a 5x5 square. Thus, none of the three knights
covering a1,a2,b2 can cover the four squares in any other corner. Thus, three
new knights are needed to cover a7,a8, and b7. Similarly, three more knights
are needed for each of the other two corners. Thus 12 knights are necessary.
Since I listed a 12 knight solution, and 12 is the minimum, my answer is
therefore what you wanted.
--
The Qurqirish Dragon, posting from his home somewhere in Ohlam.
--==<<{{ UDIC }}>>=--
Remember- my address is no laughing matter
Patrick Hamlyn
I define 'best' as most symmetric, and secondarily the one that uses the most
knights.
My solution uses 64 knights, and is easily seen to be the most symmetric and to
use the greatest possible number of knights, and is therefor the best.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
Michael Mendelsohn
> Burkar...@web.de (Burkart) wrote:
> >...to control the whole chess board (without the fields where the
> >knights are standing).
> I define 'best' as most symmetric, and secondarily the one that uses the most
> knights.
>
> My solution uses 64 knights, and is easily seen to be the most symmetric and to
> use the greatest possible number of knights, and is therefor the best.
Nice!
However, your knights control fields where other knights are standing;
this is explicitly excluded by the OP. Interpreted that way, the
Dragon's solution is also wrong (d6xc4); is a solution possible?
Hint below spoiler space
Have fun with knights on jumping horses
Michael
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Think of a knight's tour.
--
When the tongue or the pen is let loose in a phrenzy of passion,
it is the man, and not the subject, that becomes exhausted.
-- Thomas Paine, "On Usenet"
Scott Blomquist
> However, your knights control fields where other knights are standing;
> this is explicitly excluded by the OP. Interpreted that way, the
> Dragon's solution is also wrong (d6xc4); is a solution possible?
I didn't read it that way. I took him to mean that you don't have to have
knights _controlling_ spaces on which other knights are standing. I.e.
standing on a space counts as controlling it. Admittedly, if a solution
exists that meets the requirements as you interpret them, I'll have to
consider it a more elegant solution.
Michael Mendelsohn
> "Michael Mendelsohn" <news123...@michael.mendelsohn.de> wrote:
> > However, your knights control fields where other knights are standing;
> > this is explicitly excluded by the OP. Interpreted that way, the
> > Dragon's solution is also wrong (d6xc4); is a solution possible?
>
> I didn't read it that way. I took him to mean that you don't have to have
> knights _controlling_ spaces on which other knights are standing.
I know. I read it "your" way at first, as well. However, you *can*
read it another way (as Patrick added his interpretation) and thus get
more mileage (= additional fun) out of it.
Have fun with ambiguity
Michael
Mark J. Tilford
There is a solution where every square not occupied by a knight is
attacked by at least one night, and no knight attacks another. I've found
the maximum number of nights that can be on the board under such
conditions, and the position is rather symmetric.
(answer below)
Max: 32 knights. Put one on each black square (or one on each white
square). It's easy to prove this optimal if you know that the 8x8 has a
knight's tour, but it's not too difficult to prove that without it.
--
------------------------
Mark Jeffrey Tilford
til...@ugcs.caltech.edu
Michael Mendelsohn
> > "Michael Mendelsohn" <news123...@michael.mendelsohn.de> wrote:
> >> However, your knights control fields where other knights are standing;
> >> this is explicitly excluded by the OP. Interpreted that way, the
> >> Dragon's solution is also wrong (d6xc4); is a solution possible?
> There is a solution where every square not occupied by a knight is
> attacked by at least one night, and no knight attacks another. I've found
> the maximum number of nights that can be on the board under such
> conditions, and the position is rather symmetric.
Indeed.
Now what is the minimum number of knights needed to attack all
chessboard squares (and only those) that are not occupied by a knight?
Have fun with nights on black squares, days on white
Michael
The Qurqirish Dragon
>> > "Michael Mendelsohn" <news123...@michael.mendelsohn.de> wrote:
>> >> However, your knights control fields where other knights are standing;
>> >> this is explicitly excluded by the OP. Interpreted that way, the
>> >> Dragon's solution is also wrong (d6xc4); is a solution possible?
>
>> There is a solution where every square not occupied by a knight is
>> attacked by at least one night, and no knight attacks another. I've found
>> the maximum number of nights that can be on the board under such
>> conditions, and the position is rather symmetric.
>
>Indeed.
>
>Now what is the minimum number of knights needed to attack all
>chessboard squares (and only those) that are not occupied by a knight?
OK. If a knight is not allowed to threaten another, here is a solution with 16
knights:
bcfg-2367
Take all 16 pairs from there. This forms 4 2x2 squares of knights, each
positioned with one rank/file between it and the board's edge. With 2 empty
ranks/files separating any two of these, no knight can capture one from another
square. With all knights in a square being mutually adjacent, no knight can
capture another within its own square. Since no knight can capture another,
this is a mimimal solution (though I have not shown it to be a minimum
solution)
Michael Mendelsohn
The Qurqirish Dragon schrieb:
> "Michael Mendelsohn" <news123...@michael.mendelsohn.de> wrote:
> >However, your knights control fields where other knights are standing;
> >this is explicitly excluded by the OP.
> >
> >Now what is the minimum number of knights needed to attack all
> >chessboard squares (and only those) that are not occupied by a knight?
>
> OK. If a knight is not allowed to threaten another, here is a solution with 16
> knights:
> bcfg-2367
> Take all 16 pairs from there. This forms 4 2x2 squares of knights, each
> positioned with one rank/file between it and the board's edge. With 2 empty
> ranks/files separating any two of these, no knight can capture one from another
> square. With all knights in a square being mutually adjacent, no knight can
> capture another within its own square. Since no knight can capture another,
> this is a mimimal solution (though I have not shown it to be a minimum
> solution).
Nice. By this reasoning, *every* solution is minimal.
Here's another 16:
a1, b2, b4, d2; a8, b7, b5, d7;
mirror these to fill the other half of the board.
Note that in each quadrant, the knights occupy 4 squares of the same
color, the other 4 squares being attacked from "outside".
I am almost convined that 16 is minimal; now who could prove it?
Have fun with finding 16 knights in the house
The Qurqirish Dragon
>> this is a mimimal solution (though I have not shown it to be a minimum
>> solution).
>
>Nice. By this reasoning, *every* solution is minimal.
Well, that is the thing with minimal solutions. In this case, the rules require
the solution to be minimal (in the original interpretation I used for this,
only the minimization part of the problem required that- the base problem
allowed for non-minimal solutions, such as the 64-knight solution someone
posted))