Pebbling a Chessboard - Numberphile
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as the standard explanation of the weight-invariant proof for pebbling or cloning problems on an infinite chessboard.
Recommended by 1 people across 1 different months, 2022 to 2025.
What it is
A Numberphile video in which Zvezdelina Stankova of Mills College and the Berkeley Math Circle explains the 'pebbling a chessboard' (clone escaping prison) puzzle on an infinite quadrant board. It walks through why the clones cannot all escape, using a weighting invariant. Filmed by Brady Haran, it is the first of a short series with continuations.
What people said
Recommends awkward · 2024-11 · Mathematics Stack Exchange · on “Invariants on changing locations in coordinate plane”
You are off to a promising start. To conclude, you might view Zvezdelina Stankova's explication of Pebbling a chessboard [this video] on YouTube. The problems are not identical, but close enough that essentially the same proof should work in both cases.
Cites Vepir · 2022-10 · Mathematics Stack Exchange · on “Checking if a configuartion is possible in an infinite chessboard”
This is also known as Pebbling a Chessboard [this video] . It can't be done. Their proof is as follows: Give the bottom left piece a weight (value) of $1$ . Every time we remove it and place two new ones, give them half the weight each. The total weight will remain invariant (will not change), and will remain $1+\frac12+\frac12=2$ . The sum of all possible weights (whole board) is $4$ . This means that in order to empty out the first three pieces, we need to occupy every other square on the board. But that would require an infinite number of moves, hence it is impossible.
Cites hdecristo · 2025-10 · Mathematics Stack Exchange · on “Adapted Kontsevich "Pebbling a Chessboard" game: can you win it?”
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? [LINK] 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