2014-national
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Closing his eyes Towsif begins to place knights on a Chess board of $ {19} \times {21}$. After placing how many knights
Towsif will be sure that on the next move at least one knight will attack another one. (In one move knight goes
straight for ${2}$ steps and the 3rd step should be at right angle to the previous path.)
Towsif will be sure that on the next move at least one knight will attack another one. (In one move knight goes
straight for ${2}$ steps and the 3rd step should be at right angle to the previous path.)
Re: 2014-national
It is a nice problem.Have you solved this?If not,try to solve it.I haved enjoyed it.I think you will also enjoy.
"Questions we can't answer are far better than answers we can't question"
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Re: 2014-national
no bhai. i did not solve it. i need help
- seemanta001
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Re: 2014-national
You can place the knights in this order.This in an $8 \times 7$ chessboard.Hope this helps you.
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Last edited by seemanta001 on Fri Jul 10, 2015 8:10 pm, edited 1 time in total.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"
- seemanta001
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Re: 2014-national
The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"
- asif e elahi
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Re: 2014-national
Can you please explain your solution??seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
Re: 2014-national
Imagine such a chessboard. A knight placed at a black square will go to a white square on its next move and vice versa. So the maximum number of knights placed in the board will be the number of black or white squares, assuring that none attacks anyone. Assuming that the board starts with a black square, the number of black squares will be= (ceiling of (19.21)/2) = 250. So after placing at least 251 knights, it can be made sure that on the next move at least one knight will attack another.
- Kazi_Zareer
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Re: 2014-national
How? Can you please tell it?seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
We cannot solve our problems with the same thinking we used when we create them.
- asif e elahi
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Re: 2014-national
This isn't obvious. You have to prove it. (This isn't even true for a $2\times 2$ chessboard.)RJHridi wrote: So the maximum number of knights placed in the board will be the number of black or white squares, assuring that none attacks anyone.