Is the answer ceiling of $n^2/2$.

That's my answer too, but I can't prove that more than that is not attainable

well in my opinion the idea is to partition the board in a way like so that each region has exactly 2 kappas and cannot hold more. I tried this idea for all day long. But couldn't find the right partition. I thought this was very similar to imo 1999 p3. You can have a look there.Someone out there with complete solution.

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@kayak For every positive integer n. For construction start from the 1 column and bottom row put kappas in odd colums,do the same for 2nd last row,but in the 3,4th row do it for even columns, then repeat the 1st process and we would be done. we can achieve ceiling of $n^2/2$ this way. For even n maybe we can partition into 2x2 boards apply pigeon hole principle prove the lemma that there cannot exist 2x2 square with 3 kappas for $n>4$.

Is there any construction for $\frac{5n^2}{9}$ kappa ? I could only show that $d \le \frac{5}{9}$ I guess the main idea is to partition the board into $3$x$3$ blocks!

From what I make out of the statement, we're looking for reals $d$ so all $n$'s work for that each particular $d$; i.e., $d$ is constant when $n$ varies. So an answer like $\left\lceil\tfrac{n^2}{2}\right\rceil$ doesn't make much sense to me. On the other hand, I suspect $d\le \tfrac 49$ is the answer, but I can't prove that $d$ can't be bigger. Edit: actually, ignore that, I don't think that's the answer. $d=\tfrac12$ is possible as pointed out below.

you are right i am very sorry.hello but $n^2/2$ is always reached.

@sauditya you can post your partial progress it may help us to find the proof.

Ok! So basically we can put atmost $6$ kappas in a $3$ x $3$ board But there are only $2$ ways of putting $6$ kappa's in a $3$ x $3$ board ( you cannot place a kappa in the center sq. if $\ge 6)$ Now Consider a $6$ x $6$ board and divide it into four $3$ x $3$ board if we could place more than $20$ kappa's (say 21) atleast 3 squares must have $\ge 5$ kappa's and atleast one has $\ge 6$ Suppose the top most sq has $6$ kappa's then one can see that the two adjacent sq has less than $5$ kappa's So that gives $d \le \frac{5}{9}$

Construction for $\frac{n^2}{2}$ for $n=8$ and $9$

yes that is what i wrote.I have finally finished the proof but it is difficult to express without diagram.The proof goes like partition the board into two into two squares for n=even as that would suffice. Lemma-there can be no 3 kappas in a two into two board for $n>4$ and $n=2^m$ and other boards containing 2 kappas exactly. there are 4 types of pieces of squares which have 3 kappas name them T,D,L,R and 6 types of pieces with 2 kappas name them N,O,P,Q,R,S You need to consider 4 cases.

Official Solution by Yevhenii Diomidov, Kada Williams and Mihir Singhal: The answer is $d \le \frac{1}{2}$. The construction consists of placing kappas in all squares of the forms $(2k, 4\ell)$, $(2k, 4\ell+1)$, $(2k+1, 4\ell+2)$, and $(2k+1, 4\ell+3)$. To prove that this is minimal, consider all connected components consisting of squares that contain kappas that are connected via edges. It is easy to see that there are only five different kinds of connected components. Extend each connected component into a larger figure as shown below: [asy][asy] size(9cm); int N = 12; for (int i=0; i<=N; ++i) { draw((2*i,0)--(2*i,2*N), grey); draw((0,2*i)--(2*N,2*i), grey); } pen efill = opacity(0.2)+lightcyan; pen eborder = heavycyan; pen ifill = opacity(0.8)+grey+blue; pen iborder = black; picture pic1; filldraw(pic1, (1,-1)--(3,1)--(1,3)--(-1,1)--cycle, efill, eborder); filldraw(pic1, scale(2)*unitsquare, ifill, iborder); add(shift(4,18)*pic1); picture pic2; filldraw(pic2, (1,-1)--(3,1)--(3,3)--(1,5)--(-1,3)--(-1,1)--cycle, efill, eborder); filldraw(pic2, box((0,0),(2,4)), ifill, iborder); add(shift(12,18)*pic2); picture pic3; filldraw(pic3, box((-1,-1),(5,5)), efill, eborder); filldraw(pic3, scale(4)*unitsquare, ifill, iborder); add(shift(16,12)*pic3); picture pic4; filldraw(pic4, (-1,-1)--(3,-1)--(7,3)--(7,5)--(3,5)--(-1,1)--cycle, efill, eborder); filldraw(pic4, (0,0)--(4,0)--(4,2)--(6,2)--(6,4)--(2,4)--(2,2)--(0,2)--cycle, ifill, iborder); add(shift(10,4)*pic4); picture pic5; filldraw(pic5, (1,-1)--(5,-1)--(5,1)--(1,5)--(-1,5)--(-1,1)--cycle, efill, eborder); filldraw(pic5, (0,0)--(4,0)--(4,2)--(2,2)--(2,4)--(0,4)--cycle, ifill, iborder); add(shift(4,10)*pic5); [/asy][/asy] Due to the fact that there are no three kappas in a line and due to the nature of the extensions, one can see that after extension, the interiors of the figures remain disjoint. However, note that the extended area of each figure is at least twice its original area (it is exactly twice except for the 2 by 2 square, for which it is $\frac{9}{4}$ times the original area). Thus, Nicky can cover at most $\frac{1}{2}$ of the squares with kappas.

I came up with the official approach as well, but my main movitation was to do some kind of generalized Voronoi diagram (with some suitable metric), which would split up the non-Kappa squares as above.

My idea for proving minimality was to consider 4 contiguous rows and prove that if the $i$-th of those rows has $k_i$ kappas, $k_1 + k_2 + k_3 + k_4 = 2n \pm c$. Here is $c$ is a small error due to the edges, and as $n$ grows it is negligible. For example, if $k_2=k_3=\frac{2}{3} n$ then we must have $k_1=\frac{1}{3} n$ by counting the squares where a kappa may not appear in row 1 due to two-in-a-row in rows 2 and 3, and noting that a kappa is blocked from a cell by at most one or two two-in-a-rows, never three. The argument doesn't generalize to other distributions very easily though, and might require some more ingenious counting. I'm afraid I haven't found a way to resolve this I wonder if there is a solution aside from the original one. Still, very cool idea. It looks like it was inspired by origami

v_Enhance wrote:

I am sorry but I have a silly doubt. For connected components at the edges of the board, it may happen that the extended region transcends the boundaries of the board which can create a problem in proving the required bound as then the extended regions won't cover double the area belonging to the board . EDIT : I understand @below that the error I am mentioning is of order $\mathcal O(n) $ and the bound is of the order $\mathcal O(n^2)$ so if makes no difference. But why wasn't it mentioned in the official solution. I think it's a non-trivial observation.

It's fine. It proves an upper bound of $\frac{(n+1)^2}{2}$. Take $n\to\infty$ and we are done.

babu2001 wrote: EDIT : I understand @below that the error I am mentioning is of order $\mathcal O(n) $ and the bound is of the order $\mathcal O(n^2)$ so if makes no difference. But why wasn't it mentioned in the official solution. I think it's a non-trivial observation. I didn't include it because I thought it was obvious, but I'll make a note to this effect when I upload the packet to my website. (You're right that I should have.)

My solution is the same as the official.

whatshisbucket wrote: Proposed by Mihir Singhal and Michael Kural And Inspired by ARMO 2013, 9th grade, P8

I thought immediately of the idea of in the official sol but I tried to use $\frac13$ for bounding instead of $\frac{3}{8}$ -- this did not end well. Another larger issue was the lack of seeing the area argument ;-; Anyways, here's a new sol? We claim that all $d \le \frac12$ can be done. Construction The construction is coloring a stretched checkerboard where $x + \left\lfloor\frac{y}{2}\right\rfloor \equiv 0 \pmod{2}$ (and the bottom right square is $(0, 0)$). Bounding Now, define the ring of a square as the $8$ or less squares surrounding it and let $G$ be the grid for a $n \times n$ config. Define $R(s)$ for a square $s$ to be the number of kappas in the surrounding ring. Let $\kappa(s)$ be the indicator function for the presence of a Kappa. Then note that \[ \sum_{s \in G} R(s) = \sum_{s \in G} 8 \kappa(s) + O(n) \]so it follows that \[ \frac{\sum_{s \in G} R(s)}{8n^2} = \sum{s \in G}{n^2} + O\left(\frac1n\right) \]It thus remains to show that $\sum_{s \in G} R(s) \le 4n^2$. Now, note that if $\kappa(s) = 1$, it follows that $R(s) \le 4$. Claim: If $\kappa(s) = 0$, then we can pair $R(s)$ with some unique surrounding $R(s')$ such that their average is at most $4$. Proof. First suppose that $\kappa(s) = 6$. Then the structure of the ring ends up being two $L$-pieces. We can manually check that for each $L$ piece with the two "leg" squares as $L_1, L_2$, we have that $\min\{R(s_1), R(s_2)\} \le 3$, or else we would get a diagonal run. As such, it follows that we can consume these two $L$-pieces which have both legs touching $s$ to get an average of $4$. Similarly, if $\kappa(s) = 5$ and there's an $L$ piece we finish similarly. Note that a leg in an $L$-piece can only be assigned to one other square, or else that would give a run. Finally, consider the $\kappa(s) = 5$ case where the surrounding squares in the ring form a U. Then pair up the bottom of the U with $s$, which can be checked to have an average of at most $3$. We can check that this square then can't be the leg of an L piece and also can't be shared between two Us (or else that would cause a diagonal run). $\blacksquare$