2014-national

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Mahfuz Sobhan
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2014-national

Unread post by Mahfuz Sobhan » Wed Jul 08, 2015 6:14 pm

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.)

tanmoy
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Location: Rangpur,Bangladesh

Re: 2014-national

Unread post by tanmoy » Wed Jul 08, 2015 9:47 pm

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"

Mahfuz Sobhan
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Joined: Sat Feb 07, 2015 5:40 pm

Re: 2014-national

Unread post by Mahfuz Sobhan » Fri Jul 10, 2015 2:04 am

no bhai. i did not solve it. i need help

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seemanta001
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Re: 2014-national

Unread post by seemanta001 » Fri Jul 10, 2015 3:15 pm

You can place the knights in this order.This in an $8 \times 7$ chessboard.Hope this helps you. :)
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8 X 7 Chessboard
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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"

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seemanta001
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Re: 2014-national

Unread post by seemanta001 » Fri Jul 10, 2015 7:48 pm

The solution is:
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"

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asif e elahi
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Location: Sylhet,Bangladesh

Re: 2014-national

Unread post by asif e elahi » Mon Jan 25, 2016 10:46 pm

seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
Can you please explain your solution??

RJHridi
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Re: 2014-national

Unread post by RJHridi » Wed Mar 30, 2016 1:09 pm

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.

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Kazi_Zareer
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Re: 2014-national

Unread post by Kazi_Zareer » Sat Apr 23, 2016 1:40 am

seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
How? Can you please tell it?
We cannot solve our problems with the same thinking we used when we create them.

abmonim
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Re: 2014-national

Unread post by abmonim » Fri Jun 17, 2016 12:54 pm

19.21/2=250 ??? Confused!!!

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asif e elahi
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Location: Sylhet,Bangladesh

Re: 2014-national

Unread post by asif e elahi » Thu Jul 28, 2016 11:18 pm

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.
This isn't obvious. You have to prove it. (This isn't even true for a $2\times 2$ chessboard.)
Try to find a mapping from the white squares to the black squares.

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