The last IMO problem AI could not solve

Overview

3Blue1Brown explores the 2025 International Mathematical Olympiad (IMO) Problem 6, the last problem that AI models, including Google DeepMind's AlphaProof, could not fully solve. The problem involves tiling a 2025x2025 grid with unit squares, leaving exactly one uncovered square per row and column, and finding the minimum number of tiles. The solution relies on a proof involving longest increasing and decreasing subsequences (Erdős-Szekeres theorem) to establish a lower bound on the number of tiles, demonstrating how human intuition and appreciation for beauty are crucial for complex problem-solving.

Key takeaways

The 2025 IMO Problem 6, a tiling puzzle, was the last problem AI could not fully solve due to its reliance on visual intuition and 'feel' for the problem.
The optimal tiling construction for the 2025 IMO Problem 6 uses k^2 + 2k - 3 tiles, where k^2 is the grid side length (2025 = 45^2).
A key proof technique involves highlighting edges of uncovered squares and relating them to tiles using permutations and the Erdős-Szekeres theorem.
The Erdős-Szekeres theorem guarantees that for a permutation of size n, the product of the lengths of the longest increasing and decreasing subsequences is at least n.
The value of solving complex mathematical problems like this lies in the human understanding and creative process ('motivated explanation'), not just the final proof.
AI's ability to generate proofs without human understanding challenges traditional metrics of academic progress and mathematical discovery.

Chapters

0:00 Introduction to IMO Problem 6 and AI's Challenge

The 2025 IMO Problem 6 was exceptionally difficult, with less than 1% of participants achieving full marks.
AI models, including Google DeepMind's AlphaProof, struggled with this problem, marking it as a significant challenge.
IMO problems require creativity and deep understanding beyond rote memorization or computation.

6:49 The Tiling Problem: Grid, Tiles, and Gaps

The problem involves tiling a 2025x2025 grid of unit squares with rectangular tiles.
Each row and column must have exactly one unit square not covered by any tile (marked as 'X').
The goal is to find the minimum number of tiles and rigorously prove this minimum.

11:50 Cube Slicing Analogy for Proof Strategy

A 3x3x3 cube cutting puzzle is introduced to illustrate proof by focusing on essential elements.
The insight from the cube puzzle is that proving the minimum number of cuts (6) requires focusing on the inner cube's faces.
This highlights the importance of identifying critical components for rigorous proof in complex problems.

17:38 Developing Intuition: Edges, Tiles, and Efficiency

Each 'X' (gap) has four edges that must be covered by tiles.
Tiles are more efficient when they cover multiple 'X' edges.
The ideal tile covers four 'X' edges, suggesting a 'windmill' pattern around the 'X's.

22:26 Constructing an Optimal Tiling with Square Tiles

Hypothesizing square tiles and the windmill pattern leads to a 'puzzle-like' assembly.
This construction yields a pattern of square tiles where 'X's are at the corners.
For a grid of side length k^2, this construction uses k^2 + 2k - 3 tiles.

30:56 Proof Strategy: Highlighting Edges and Lower Bounds

The proof strategy involves highlighting specific edges of the 'X' squares.
Each tile can touch at most one highlighted edge, establishing a lower bound on tiles.
Highlighting only right edges gives a bound of k^2 - 1, which is insufficient.

44:03 Refined Proof: Four Regions and Highlighted Edges

Dividing the grid into four regions (right, top, left, bottom) allows highlighting edges based on region.
This strategy ensures each tile touches at most one highlighted edge.
The number of highlighted edges is calculated based on 'X' positions relative to two paths (increasing/decreasing subsequences).

51:52 Permutations, Subsequences, and the Erdős-Szekeres Theorem

The problem reduces to analyzing permutations and their longest increasing (LIS) and decreasing (LDS) subsequences.
The Erdős-Szekeres theorem states that for a permutation of size n, LIS * LDS >= n.
The sum of LIS and LDS lengths is related to the number of highlighted edges.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, 3Blue1Brown.