Help make AI better by inventing a new algorithm! When you play this game and achieve a new
high score, you have discovered a new algorithm that will directly improve the performance of
AI. Each puzzle gives you a single board with three sub-grids — A (bottom-left), B (top-right), and C (bottom-right) —
laid out so the shared dimensions of an A·B = C matrix multiplication line up
edge-to-edge. Fill in cells from the puzzle's alphabet — the classical default is {−1, 0, +1}, but the New Game tab lets you pick a wider one. Your goal is to
wipe the bottom residual grid completely clean while filling in as few pages as
possible. Larger boards are harder to solve, but the reward is greater.
Controls
Click a cell to cycle through the puzzle's alphabet. Shift-click reverses; right-click resets to 0.
Click + drag rotates the 3D stack (flick to spin).
Scroll over the board to step through its rank pages.
Click the small corner tab on a page to make it the active page.
Use the tabs above the board to keep several puzzles open at once.
Why it matters
The schoolbook way to multiply two matrices uses M·N·P multiplications. In 1969,
Strassen showed that two 2×2 matrices need only 7 multiplications
instead of the obvious 8 — a small win that compounds into a much faster
algorithm at large scales. DeepMind's AlphaTensor and the Moosbauer-Poole / Kauers-Wood
flip-graph results have recently extended this kind of trick to bigger matrices, mostly with
small integer entries. This game is a hands-on sandbox for that same search. It is called a bilinear algorithm because it is a generalization of matrix multiplication.
Scoring
Every page you fill in counts against you, and any leftover error in the residual grid adds to
that tally. Fewer pages → faster algorithm → an exponentially bigger score. Doing worse than
the schoolbook algorithm gives a negative score, down to −1,000,000.
The +1,000,000 ceiling sits at the conjectured asymptotic limit ω = 2, but no recursive base case can hit it exactly — divide-and-conquer adds a log N factor from the combine step that keeps the effective exponent slightly above 2. So the cap is approached, never reached; bigger ⟨m,n,p⟩ have shallower recursions and approach it more closely.
Concretely, here is the maximum rank R you can land at on a few small cube boards
and still clear each prize tier (assuming a fully-solved residual). Smaller boards are tougher:
the integer rank ladder is short, so each step you save matters more.
Board
US$1,000 (score ≥ 83)
US$10,000 (score ≥ 1,000)
For reference
⟨2,2,2⟩
R ≤ 6
(proven impossible)
R ≤ 5
(proven impossible)
Strassen R = 7 (score 14)
⟨3,3,3⟩
R ≤ 19
(now open to prize!)
R ≤ 15
(proven impossible)
Schoolbook R = 27
⟨4,4,4⟩
R ≤ 41
R ≤ 32
AlphaTensor R = 49 (score 14)
⟨5,5,5⟩
R ≤ 74
R ≤ 55
Schoolbook R = 125
Prize: if you achieve a score you believe is eligible for a prize, save your
game to a file using the Save game… button and send it to the company linked
at the bottom of the page.