The same bit-flip channel moves a pure input further than an already-mixed one, and leaves a fifty-fifty input unchanged. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Every earlier bit-flip row started from a pure state

Chapter six's bit-flip scans always started with 100 zero shots and zero one shots before the channel. At flip fraction one fifth, that pure input maps to 80 zero and 20 one shots.

(100,0)(80,20)(100,0)\to(80,20)
Pure input rowThe familiar pure-zero input opens this scan.1zero0one4/5zero1/5onechannel map

A channel applied to an already-mixed input moves less

The same channel is applied to inputs that already carry some one-shots before the channel runs. The output moves less far from the input as the input becomes less pure.

inputNout,zNout,o(70,30)6238(50,50)5050\begin{array}{c|c|c}\text{input}&N_{\text{out},z}&N_{\text{out},o}\\(70,30)&62&38\\(50,50)&50&50\\\end{array}
Partially mixed input rowThe seventy-thirty input still moves toward balance.7/10zero3/10one31/50zero19/50onechannel map

An even split is a fixed point of this channel

The fifty-fifty row is unchanged by the channel: 50 zero and 50 one shots go in, and the same counts come out. This symmetric bit-flip channel moves any input toward this balanced point, and an input that starts there simply stays.

(50,50)(50,50)(50,50)\to(50,50)
Fixed-point input rowThe final row is rendered as the checked fixed-point case.1/2zero1/2one1/2zero1/2onechannel map