This puzzle is inspired by this one: Board with all 2020s
Zeroes are written in all cells of a 3×3 board. Pressing a cell increases by 1 the number in this cell and all cells having a common side with it. Is it possible to obtain different numbers in each cell? Bonus question: what is the least number of presses needed to achieve this? Good luck!
Make these presses:
\begin{matrix} 0 &4 &2\\ 1 &2 &0\\ 0 &0 &0 \end{matrix}
To get these values:
\begin{matrix} 5 &8 &6\\ 3 &7 &4\\ 1 &2 &0 \end{matrix}
You can solve the problem via integer linear programming as follows. Let nonnegative integer decision variable $x_{i,j}$ be the number of times that cell $(i,j)$ is pressed. Let binary decision variable $y_{i,j,v}$ indicate whether cell $(i,j)$ contains value $v$. Let $N_{i,j}$ be the neighborhood of cell $(i,j)$, including $(i,j)$ itself. The problem is to minimize $$\sum_{i,j} x_{i,j} \tag1$$ subject to: \begin{align} \sum_{v\in V} y_{i,j,v} &= 1 &&\text{for all $i$ and $j$} \tag2 \\ \sum_{i,j} y_{i,j,v} &\le 1 &&\text{for all $v$} \tag3 \\ \sum_{(\bar{i},\bar{j})\in N_{i,j}} x_{\bar{i},\bar{j}} &= \sum_{v\in V} v\ y_{i,j,v} &&\text{for all $i$ and $j$} \tag4 \end{align} The objective function $(1)$ is the total number of presses. Constraint $(2)$ enforces one value per cell. Constraint $(3)$ enforces at most one cell per value. Constraint $(4)$ links the number of presses in the neighborhood to the value in the cell.
Yes, it is possible:
Press the corners 1, 3, 6, and 2 times. This gives the resulting grid of [143/705/682].
