Shading a honeycomb board The following figure shows a honeycomb  board that is bounded by an equilateral triangle. 

The $i^{\mathrm{th}}$ row contains $9 - (i - 1)$ cells for each integer $1 \leq i \leq 9$. A game is played on the board. Every time a token is placed on the board,
all the cells on the same row and on the same diagonal as it are shaded. (An instance of the shading from a turn is shown.) Determine the minimal number of turns required to shade the figure.
Comments
By induction, for every positive integer $n$, the maximum number of turns to shade a similar honeycomb figure with either $2n - 1$ or with $2n$ cells along an edge is $n$. As there are 9 cells along the base of the given figure, and as $9 = 2(5) - 1$, the maximum number of turns required to shade the given figure is 5.
Requests
What is the minimum number of turns to shade the board? Is there a convenient rule for computing the minimum number of turns needed on a similar board with $n$ cells along the bottom row?
 A: Let $f(n)$ be the minimal number of tokens required to cover a triangle of side length $n$.
Theorem. $f(n)=\lceil\frac n2\rceil$.
Proof.
This is certainly true for $n=1$ and $n=2$.
Also, by placing a token on a vertex of the triangle, we reduce the side length by $2$ and so obtain
$$f(n)\le 1+f(n-2) $$
for $n>2$. Thus by induction, 
$ f(n)\le \lceil\frac n2\rceil$ for all $n$.
Assume place $m$ tokens on an $n$-triangle, where $m<\lceil \frac n2\rceil$, i.e., $2m<n$.
Each token by itself covers exactly $2n-1$ of the $\frac{n(n+1)}{2}$ places (as this is true at a vertex, and moving by one step changes one line by $+1$, one by $-1$, and one not at all).
Hence, ignoring any overlaps for different tokens,  $m$ tokens cover $(2n-1)m$ places.
Question: How much overlap is between two tokens?
There are three ways to designate an "up" direction and view one of the edges of the triangle board as the "bottom" edge.


*

*If the two tokens are on different "heights" (for all three interpretations of "up"), one of the tokens is the "lower" token for at least two choices of "up". For both these choices, the "horizontal" row of this "lower" token intersects the two "skew" rows of the "upper" token in a place that is not the lower token itself. Thus we count at least $4$ places of intersection.

*If the tokens are on the same "height" for one of the three interpretations of "up", the overlap is a full row of at least two places plus at least one additional intersection of the skew directions "above" that row. This immediately gives us at least four places of intersection, except when the common row has length $2$, i.e., both tokens are next to a vertex; but  in this special case, one directly verifies (using $n>2$) that there are four intersections as well.
Thus for each pair of tokens, we have to subtract (at least) $4$ from the total count of covered places. Hence $m$ tokens cover at most
$$m\cdot(2n-1)-4{m\choose 2} =2\cdot m\cdot(n+\tfrac12-m)$$
places. The function $x\mapsto x\cdot(a-x)$ is strictly increasing for $x\le \frac a2$. Hence for $m<\frac n2$, the above expression is strictly smaller than 
$$2\cdot \frac n2\cdot\left(n+\tfrac12-\frac n2\right)=\frac{n(n+1)}{2}, $$
the total number of places on the board.
In other words, $m<\frac n2$ tokens  cannot cover the full board.
We conclude that  $f(N)\ge \frac N2$, and as $f(n)$ is an integer, we have $f(n)\ge \lceil \frac n2\rceil$, as desired. $\square$
