grantsanderson

2026-10-25 · open the original ↗

• Puzzle: find the expected volume where two random intersecting spheres overlap. Both spheres have radius 1. Equivalently, fix one at the origin and pick the second's centre uniformly at random within distance 2 of it, which guarantees an intersection whose shape depends on where that centre lands. He poses it as this month's puzzle without solving it. Solutions to several monthly puzzles will be bundled into fuller videos near the end of the year, with previews for Patreon supporters meanwhile.

2026-10-16 · open the original ↗

• Proof should now be treated as raw data, not the goal. For most of math's history proof stood in for understanding: a problem counted as solved the moment it was proved. He argues machine proofs turn proof into something like numerical calculation or more digits of pi. The prize belongs to whoever turns a proof's ideas into coherent theory, such as a Navier–Stokes blow-up result, and he calls proof "overvalued for decades".
• Machine theorem-proving barely touches math's value or its funding. He lists three sources of value: applications, transferable problem-solving skill, and aesthetic joy. Applications almost always come from theory building, as with Galois's group theory rather than the quintic result, or the theory grown around Fermat's Last Theorem. So public and private funding stays rational, and theory builders gain leverage. What changes is the culture and how credit is assigned.
• Skill and joy need human engagement, like lifting weights. Machines lifting more than you doesn't make your own gym work pointless, and the same holds for doing proofs yourself. Teaching is math's biggest funding source, and it pays for exactly these individual goods. He argues education is limited by motivation, not by how clear the explanation is, and motivation is social. That makes teaching one of the jobs safest from AI displacement.
• AI's current position is overrated, but its velocity is not. This is his conclusion from interviews with mathematicians that he has recorded since mid-2025. In 2023 DeepMind's Olympiad geometry system looked bespoke and brute-force. Within a couple of years general language models could solve IMO problems by any reasonable standard. Machine proofs of Erdős problems went from dismissed as low-hanging fruit to serious results, though the frontier is "spiky at a fractal level".
• It's a good time to enter because the field must change. His reason isn't new AI tools. Newcomers can shape the field at a pivotal moment while it looks for guidance. As a 2026 PhD student he would try to get visual explanation like 3Blue1Brown's accepted as a thesis. His other advice is to give lots of talks, since oral presentation will increasingly be how understanding is measured. He also says to walk with a fellow student and talk through what each loves about math.

2026-09-25 · open the original ↗

• A phone number has a multiple made of ones and zeros. Multiplying any whole number by some positive integer can yield a larger result whose digits are entirely 1s and 0s, and if the original number ends in 1, 3, 7, or 9, a multiple whose digits are entirely 1s can be found instead. This is posed as the month's puzzle, left for the viewer to prove rather than solved on screen.
• A naive brute-force search for this multiple is computationally infeasible. He notes that understanding why the result holds points toward a smarter, more targeted brute-force search that could actually find it within reach.

2026-09-18 · open the original ↗

• A once-impossible AI benchmark fell within two years. Google DeepMind's model answered four of six questions on the 2024 IMO; by 2025 several organizations' models answered every question but one; by 2026, publicly available reasoning models could answer all six just from prompting.
• Problem six of the 2025 IMO stumped AI entirely. Only six of the 603 competitors — each already among the top problem-solvers in their own country — earned full marks on it, and it was the sole question none of the AI systems solved that year.
• The puzzle asks for the fewest possible tiles. On the grid described in the actual IMO problem — 2025 by 2025 unit squares, shown smaller here for illustration — non-overlapping tiles whose edges align with the grid must cover the board so that every row and every column has exactly one uncovered unit square; the task is to find the minimum tile count achieving this.
• The real challenge is proving the minimum, not just finding it. The task isn't merely a numerical answer or even a construction that achieves it — it requires a rigorous proof that no other tiling arrangement can use fewer tiles, which the video calls a path into "intuition and discovery."

2026-09-18 · open the original ↗

• A cube-slicing puzzle supplies the key proof strategy: one edge, one tile. Proving a 3×3×3 cube needs at least six planar slices to cut into 27 unit cubes — even allowing pieces to be rearranged between cuts — rests on the fact that each of the inner unit cube's six faces requires its own unique slice, a one-to-one correspondence between faces and cuts. That same "count what needs its own tile" logic becomes the template for the harder tiling proof.
• The optimal tiling is a windmill of squares, using k²+2k−3 tiles. Stacking k×k square tiles so each gap sits at the shared corner of four tiles, in a windmill pattern, builds a grid of side k² using (k−1)² interior tiles plus 4(k−1) boundary tiles. Since 2025=45², k=45 gives the answer — and, Grant notes, happens to match the floor tiling at the airport where the 2025 contestants landed.
• A two-path trick reduces the proof to a classic permutation theorem. Highlighting only one edge per gap proves at best k²−1 tiles, too weak, because the tiles it misses cluster on one side; drawing a longest increasing and a longest decreasing path through the gaps instead splits the grid into four regions, each with its own highlighted edge, closing the gap to the tight bound via the Erdős–Szekeres theorem.
• Human understanding, not the proof itself, is what math should credit. DeepMind's Tang Luong told Grant models struggle because they aren't trained to be patient enough to understand a problem before attacking it; since a correct proof of this result already existed on YouTube, the video's own contribution is what Grant calls a "motivated explanation" — a narrative that makes the discovery feel inevitable rather than a "bolt from the blue," which he argues deserves the same academic credit as the proof itself.

2026-08-18 · open the original ↗

• The setup starts three pegs at three corners of a square. The legal move is a jump-over: a peg leaps over an adjacent peg to land an equal distance beyond it on the opposite side, and this can be repeated indefinitely.
• Jumps aren't limited to straight lines — diagonal jumps are also legal. Two pegs can hop over each other repeatedly along a straight line, but a peg can also jump diagonally over another, opening up different angles of movement across the board.
• The question: can a peg ever reach the square's fourth, empty corner? If it's possible, the puzzle asks for the smallest number of moves required, with a proof that no faster route exists.
• If it's impossible, that too has to be proven, not just asserted. The puzzle offers no hint as to which outcome holds — reaching the corner or ruling it out both demand a rigorous argument, in keeping with the video's puzzle-of-the-month format.

2026-07-24 · open the original ↗

• The setup: 64 sugar cubes fill a 4×4×4 block, each colored. Every cube is white, blue, or brown, and the arrangement of colors across the 64 positions can be anything at all.
• The task: prove six same-colored pairs always exist, no matter the coloring. That means finding 12 individual cubes total — grouped into six pairs, each pair a single matching color — regardless of how the 64 cubes happen to be colored.
• A further constraint: the six pairs' distances between centers must all match. For each pair, measure the distance between the centers of its two cubes; the puzzle requires that all six of these distances come out equal to each other.
• The puzzle comes from the 2024 British Math Olympiad. It's presented as part of Grant's recurring "puzzle of the month" series, posed without a solution in this short.

2026-07-16 · open the original ↗

• Cross entropy measures the cost of using the wrong code. It's the average bits per symbol when a code optimized for one distribution (Q) gets used on symbols that actually follow another (P). In the video's robot-instruction toy example, a code optimized for a 50/25/12.5/12.5% split costs 2.625 bits per instruction once the true distribution shifts to 12.5/12.5/25/50%.
• Cross entropy hits its minimum only when the two distributions match. At that point its value equals the entropy of the true distribution P — the reason it makes a sound loss function. A Lagrange-multiplier argument on a "my name is ___" example shows this property forces the per-token loss to be a logarithm specifically, rather than any other decreasing function of probability.
• A 2002 gzip trick reconstructs language-family trees from compression alone. Appending a snippet of one document to another and comparing the compressed file sizes approximates cross entropy — how well one text's patterns compress another's — letting researchers cluster and even rank the relatedness of languages purely from resulting file sizes, without any built-in linguistic knowledge.