Hemibrain v1.2 1,745 Kenyon cells 27 facts per cell Readout: mushroom body

Can a Fly Solve Sudoku?

A mushroom body is a classifier, not a solver — so give it the one part of sudoku that is classification. The fly lands on a square, reads 27 raw facts about it, and answers with a digit or with don't know. Everything before the readout is the measured connectome. Watch it go, then take away the layer it doesn't have.

Projection neurons — 27 facts
Kenyon cells — top 20% firing
Output neuron votes
Run
Readout The fly has the first. No fly has the second.
Ready.
Filled0
Remaining0
Wrong0
Confidence—

Why it stalls

The task itself is easy: a small network reading the same 27 facts gets it perfectly right. The sparse Kenyon cell code is not the problem either — the information survives the expansion almost intact. What fails is the last step. A mushroom body has exactly one learned layer, Kenyon cells onto output neurons, and one linear layer cannot pull the answer back out of the code.

Same input, same Kenyon cells, different readout

Accuracy on squares that have exactly one possible digit. Every row uses the identical 27 facts; rows two and three additionally share the same measured connectome and the same feedback inhibition.
The Kenyon cell code carries 99.7 % of what is needed. The fly cannot reach it, because reaching it takes a second learned layer and the mushroom body has only one.

What that costs on a whole puzzle

Forty puzzles, each solvable entirely by single-square deduction. Both readouts answer only above a confidence threshold calibrated to 99.9 % precision, which is why neither ever corrupts a board — the fly stops rather than guesses.

What is honest here

The fly supplies the inference; flying square to square supplies the iteration. That division is not a dodge — it is how an insect actually operates, making local decisions as it moves rather than solving its world in one shot. What the fly is never given is help: it sees only which digits already occupy its row, column and box, and has to learn the deduction from that. No candidate lists, no pre-computed possibilities.

And the honest answer to the question in the title is no. Two puzzles out of forty is not solving sudoku. The interesting part is that the failure is architectural and locatable: not the wiring, not the sparse code, but the single learned layer at the end.