A detuning tolerance source scan holds the candidate distance fixed and varies only the tolerance, so the accepted bit flips when the tolerance reaches the distance. With candidate square 2 and target 1, the distance is 1; tolerances 1/2, 1, and 2 give accepted bits 0, 1, and 1 exactly for the stated toy detuning ledger; real devices depend on microwave calibration, resonator drift, flux noise, fabrication spread, and finite-shot uncertainty.

The same detuning distance is compared with three exact tolerance sources. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Tolerance row 1 gives detuning bit 0

The candidate square stays 2 and the distance from target stays 1. The tolerance source is 1/2.

∣2−1∣=1,t=12,a=0|2\mathbin{-}1|=1,\quad t=\frac{1}{2},\quad a=0
Tolerance source row 1The detuning check compares the same distance with each tolerance.distance=1, tolerance=1/2, accepted=0

Tolerance row 2 gives detuning bit 1

The candidate square stays 2 and the distance from target stays 1. The tolerance source is 1.

∣2−1∣=1,t=1,a=1|2\mathbin{-}1|=1,\quad t=1,\quad a=1
Tolerance source row 2The detuning check compares the same distance with each tolerance.distance=1, tolerance=1, accepted=1

Tolerance row 3 gives detuning bit 1

The candidate square stays 2 and the distance from target stays 1. The tolerance source is 2.

∣2−1∣=1,t=2,a=1|2\mathbin{-}1|=1,\quad t=2,\quad a=1
Tolerance source row 3The detuning check compares the same distance with each tolerance.distance=1, tolerance=2, accepted=1

Tolerance is the boundary source

The distance is fixed. Below the distance rejects; equality and larger tolerance accept.

dta1120111121\begin{array}{c|c|c}d&t&a\\1&\frac{1}{2}&0\\1&1&1\\1&2&1\\\end{array}
Tolerance boundary scanThe equality row is displayed.tolerance=distance accepts