Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So, the puzzle is what is the maximum number of knights one can place on a $2 \times n$ ($n\ge 2$) board such that no two knights can attack each other.

Apparently there is a formula for this $2\cdot\left(2\cdot(n/4) + \min(n\,\%\,4, 2)\right)$

How can one explain this formula or how does one derive this?

share|improve this question
    
You can get within one or two of this with the simple checkerboard pattern. You can express it as $\begin {cases} n&n\%4=0\\n+1&n\%4=1,3\\n+2&n\%4=2\end {cases}$ –  Ross Millikan Dec 23 '13 at 18:41

1 Answer 1

up vote 3 down vote accepted

Fill the strip with $2\times 2$ with knights, followed by $2\times 2$ empty squares, rinse and repeat, and you fulfill the formula.

As to why this is the maximum, counting how many fields each knight voids (guards, covers, or whatever term you prefer) should get you at least quite close.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.