Record cost at the tipping point
When checks start cutting entanglement apart, how many bits does the hidden record of outcomes hold?
Partly known
Take a row of qubits. Scramble neighbouring pairs with random two-qubit gates, layer after layer, like bricks in a wall. Between layers, check each qubit with odds p. With few checks, entanglement (shared quantum correlation) spreads across the whole row. With many, the checks cut it into short pieces. The switch happens at a tipping point near p = 0.16.
Every check also writes down an outcome. This entry prices that hidden record: how many bits of randomness nature produces per check, right at the tipping point.
A brickwork circuit, watched
starting
left · newest layer at the bottom. Bricks are random two-qubit gates; a dot is a check (ember = read 1, slate = read 0) · right · entanglement between the left and right halves of the row (Rényi-2 entropy, in bits; 5 is the most 10 qubits can hold), with its running average in moss. Slide p from 0.05 to 0.3 and watch the average fall. The record counter adds up the Born-rule surprise of every check: that sum is the number of bits you would have to guess to get this exact run again.
EXACT SIM
EXACT SIM
EXACT SIM
PAPER
In plain words
The toolkit found the tipping point at p_c = 0.157 with 20 qubits, close to the paper's 0.168. At that point each check produces 0.947 ± 0.003 bits of fresh randomness, about 0.16 bits per qubit per step.
That number is why these experiments are so hard. To see the transition directly you must repeat one exact run, outcomes and all. For 20 qubits over 80 steps that is about 256 bits, so about 2256 repeats.
The universal part of the result, the effective central charge c_eff = 0.25 ± 0.03, matches Zabalo et al 2022. The record cost itself is real but belongs to this model. A 24-qubit run at p = 0.21 landed. Where the 20 and 24 qubit curves cross is still unresolved at about 1σ.