# Bacteria Grid Puzzle Solution

## Метаданные

- **Канал:** 3Blue1Brown
- **YouTube:** https://www.youtube.com/watch?v=Qoes9bTJtn0
- **Дата:** 21.03.2026
- **Длительность:** 3:00
- **Просмотры:** 188,459
- **Источник:** https://ekstraktznaniy.ru/video/15079

## Описание

Part of a monthly series of puzzlers, in collaboration with MoMath and Peter Winkler

## Транскрипт

### Segment 1 (00:00 - 03:00) []

Last month, I shared a puzzle about bacteria replicating on a grid. We start with just one cell at the origin. And at any point, if the spots, one above and one to the right of that bacterium are empty, then it's allowed to replicate, populating both of those spots, leaving the previous spot it was on unoccupied. The puzzle is to find the minimum number of moves required to clear all 16 lattice points inside this box here. A natural place you might start is with a smaller box. For example, at the smallest case, clearing that one first lattice point obviously just takes one move. If you move up to a 2 by two box with four lattice points, after a bit of playing around, you'll find that it takes eight moves to clear that box. So, you might be thinking, you know, what's this pattern? Could it be something related to powers of two? But as soon as you jump up to a 3x3 box with nine lattice points, it becomes clear this is going to take an astronomical number of moves to clear it out. Just to make space for the cell at 20 to replicate once, we need all these other cells to replicate first. In fact, it starts to seem unlikely that it's even possible to clear this box of nine lattice points, let alone one with 16 lattice points. Often in problem solving, if you're trying to reason about a situation that can unfold in some huge or infinite number of ways, it's often helpful to find some quantity that always must remain the same no matter what choices are made. In this case, if you think about how each replication moves you from one of these diagonal lines to its neighbor on the right, then because each replication turns one cell into two, you might think to count the total number of cells with a weight where each line has half the weight of the preceding one. So, for example, the point at the origin has a weight of one, all of the points on this line have a weight of 1/2. So, after the first replication, the total weighted sum of the cells stays fixed at one. And more generally, with the weights, as we have them labeled here, any move that you make doesn't change the total weight. So no matter what you do, the weighted sum of all the bacteria on the grid has to stay fixed at where it started, namely one. So in order to clear a box, you need the sum of all of the weights outside of that box to be at least one. Without this, even given infinite time, the bacteria could never possibly escape. So a natural next step is to add together all of the weights on this infinite grid. Now you might know that 1 plus a half plus a/4 and so on converges to two. This gives us the sum of the first row of lattice points. Repeating this for each of the rows above and adding up the results, you get a very similar sum that converges to four. Meaning the total weight on the entire grid is only four. But the sum of the weights inside that box with 16 lattice points adds up to be a little above 3. 5. So the sum of the weights outside that box is notably less than one, meaning there is simply no way to cram all of the descendants of that first bacterium outside of that box. So the puzzle is a little bit cruel. The answer is that it's impossible. And this is the proof. In fact, there's no way for the bacteria to even escape this smaller 3x3 box of nine lattice points since their weights add up to 3. 0625. And you could even make the argument that this shape with eight lattice points could only be cleared just barely if you had infinite time since the sum of its weights is exactly