Adapted Kontsevich "Pebbling a Chessboard" game: can you win it?

Question

In 1981, Kontsevich proposed the following problem.

Imagine a chessboard that extends infinitely in only two directions: upwards and to the right. Put one pebble in each colored cell of the figure (the "L" shaped region formed by three cells in the corner). Only one move is allowed, called cloning: if for a given pebble the cells directly above and to its right are empty, then add a new pebble to each of these cells and remove the original pebble. Using cloning, is it possible to clear the pink region of pebbles?

In this standard form, the problem can be solved by considering an invariant. Assign a weight to each cell: the corner cell has weight $1$, and given a cell wigh weight $k$, assign weight $k/2$ to the cell directly above it and to the cell directly to its right. Then the sum $S$ of weights of all the cells which contain a pebble is invariant under cloning. A full solution is on Numberphile. The answer is that such a task is impossible, because the initial sum is $S = 2$ and the sum of all weights in the board is $4$.

My question is: suppose we started with only two pebbles (instead of three), one in the corner cell and one next to it. In this case, can we still clear the L region in the figure? The invariant argument does not work ($S = 3/2$, while the sum outside could be up to $2$). I worked out some cases, but couldn't solve it; I wonder if it is possible.

Answer 1

Lovely question! I think the answer is that the game cannot be won.

In Pebbling a Chessboard by Chung, Graham, Morrison, and Odlyzko, the variant where we start with only 1 pebble on the board (in the lower-left corner) is considered.

They give an algorithm (Theorem 1) to determine the following:

Given a set $X$ of squares, is it possible to achieve a state where no square in $X$ contains a pebble?

You execute the algorithm by playing a modified version of the pebbling game (starting with one pebble in the lower-left corner), where you are allowed to place multiple pebbles on each square. The algorithm tells you which moves to make and when to stop, and at the end, you check to see if any square contains at least 3 pebbles. If so, it is not possible to achieve a state where $X$ does not contain a pebble.

I believe the same algorithm works (simply by modifying the initial state of the board in the algorithm) to answer the same question for arbitrary starting configurations of pebbles. Warning: I have not checked this carefully.

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When you run this algorithm for your game, you end up with the following states:

$$\begin{array}{|c|c|c|c|} \hline \hphantom{0}&\hphantom{0}&\hphantom{0}&\hphantom{0} \\ \hline &&& \\ \hline 1&&& \\ \hline 1&&& \\ \hline\end{array} \rightarrow \begin{array}{|c|c|c|c|} \hline \hphantom{0}&\hphantom{0}&\hphantom{0}&\hphantom{0} \\ \hline &&& \\ \hline 2&&& \\ \hline &1&& \\ \hline\end{array} \rightarrow \begin{array}{|c|c|c|c|} \hline \hphantom{0}&\hphantom{0}&\hphantom{0}&\hphantom{0} \\ \hline 2&&& \\ \hline &3&& \\ \hline &&1& \\ \hline\end{array}$$

(with 0's in all unmarked cells). This configuration has a $3$ in it! So your game cannot be won.

Answer 2

EDIT: With the additional precision that the forbidden zone is the pink L, including the cell without pebble at the beginning, the task is impossible.

To see this, note that the first row will always contain two pebbles. In order to clear the pink zone, you must clone once in the first cell (that will set a pebble in the second cell), twice in the second cell (to get rid of the initial pebble and the one from the first) and at least once in the third cell (to isolate pebbles coming from the second cell), giving the pattern $(1,2,1)$.

This pattern shows pebble arrivals on the second row, and in order to clear the first cell and separate pebbles, you must clone at least along the pattern $(1,2,2,1)$.

But as soon as you see the previous pattern (or any pattern with larger values, and possibly translated along the horizontal axis), you must shift the central overflowing pebbles by two cells on the right, using the pattern $(0,1,2,2,1)$. So the pattern appears on the row above.

Therefore you cannot clear the pink zone.