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   Maximal Knightriders
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   Author  Topic: Maximal Knightriders  (Read 414 times)
Patashu
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    Patashu0
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Maximal Knightriders  
« on: Sep 3rd, 2004, 4:26am »
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Let's start off simple. Knights leap to the opposite end of a 3x2 rectangle, moving 2 squares then turning 90 degress and moving 1 square.
Code:

-X-X-
X---X
--N--
X---X
-X-X-

 
How many Knights can you fit on a chessboard without any being able to move onto each other?
 
The answer, is of course 32, since Knights can only move to the opposite colour of the board, thus you can fill one colour with Knights.
 
Now, replace the Knights with Knightriders, which can make as many knight leaps as they want as long as they're in the same direction as the first.
Code:

--X---X--
---------
X--X-X--X
--X---X--
----N----
--X---X--
X--X-X--X
---------
--X---X--
And so on...

 
i) How many Knightriders can you fit on an 8x8 chessboard without any being able to move onto each other?
ii) What about with a larger board? 9x9? 10x10? 12x12? 16x16?
iii) Is it generalizable? If it is, do so.
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Grimbal
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Re: Maximal Knightriders  
« Reply #1 on: Sep 3rd, 2004, 5:12am »
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I don't know if it is optimal, but here is what I can do
 
::
1x1 - 1
2x2 - 4
3x3 - 5
For n>3, I can put 2n knights.
::
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Patashu
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Re: Maximal Knightriders  
« Reply #2 on: Sep 3rd, 2004, 5:18am »
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Not much point doing it with a 4x4 board or smaller, since the board needs to be 5 squares high or wide for the knightrider to make a second leap. Thus, it'd be like doing it with normal knights.
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Grimbal
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Re: Maximal Knightriders  
« Reply #3 on: Sep 3rd, 2004, 6:32am »
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I know.  2x2 is even more silly.  But a generalization means for every nxn.  And the solution for 2x2 is not the one suggested by the example which would be 2 knights on one color only.
« Last Edit: Sep 3rd, 2004, 6:34am by Grimbal » IP Logged
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