What this really is (and honestly isn’t)
This toy runs the actual oral-messages algorithm OM(m) from Lamport, Shostak & Pease,
The Byzantine Generals Problem (1982). Each round: every general announces a plan, then the council
does m = ⌊(N−1)/3⌋ rounds of relayed gossip (“he swears she told him…”), then everyone takes recursive
majority votes, with ties defaulting to retreat. Every loyal general assembles the same slate of
settled plans and votes on it. The picture shows only the first round of letters — the gossip after it
grows combinatorially, so it’s drawn as ripples rather than arrows.
Above the line (N ≥ 3f + 1) the 1982 theorem guarantees that no traitor strategy whatsoever
can split the loyal generals. What you watch here is that guarantee holding against every strategy this page
knows how to try. Below the line the theorem says agreement is impossible to guarantee for any
oral-message protocol; this page demonstrates that by letting the traitors play a concrete
camp-splitting, two-faced strategy (it quietly searches a family of such strategies and plays the first that
works). A demonstration is not the proof — the real impossibility argument is a beautiful
indistinguishability construction, and it is worth reading in the original paper.
Two more honest wrinkles. The loyal generals start with divided intel because guarantees are about worst
cases; if you hand them unanimous intel, traitors below the line may still fail — that’s luck, not safety.
And the whole ⅓ bound is specific to unsigned messages: the same paper shows that with signed,
unforgeable messages the bound falls away entirely. Different game.