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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.)

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.

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

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

The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.

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??

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.

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?

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

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.)