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The following explanation is what I have in my tutorial about the send-acknowledge method (mirroring the OS's). Please comment/correct.

The Send-Acknowledge Method
In the send-acknowledge method, one unit begins the transmission by pulling one of the lines low. The line that has been pulled low first communicates the bit being sent. If the tip was pulled low first, then the bit was a 0. If the ring was pulled low, then the bit was a 1. Remember that the if one line is pulled low, both units see it as low. The receiver should then read out the value from the link port. Below is a segment of code to read and write to the link port.

; Sending a bit, valued 1, out the link port
; Sender executes this
   ld   a, 2
   out   (0),a

; Reading a bit from the link port
; Receiver executes this
   in   a,(0)
   ; the value in the link port is in a.
   bit   1,a
   ; sets the zero flag if the ring was low   
   jr   z, Do_this_if_Ring_Low
   ; jumps if the ring was low
   bit   0,a
   jr   z, Do_this_if_Tip_Low
   ; jumps if the tip was low
   jr   nz, loop
   ; keep waiting if neither was low

When the receiver sees that the line has been pulled low, and successfully records the value of the bit sent, it proceeds to acknowledge the transmission by pulling the other line low. In the case of the above, the receiver will be pulling the tip low, since the ring is already low.

; Receiver executes this
; Pulls the tip line low
   ld   a, 1
   out   (0),a

The sender should at this point be waiting for such a change in the tip line. This is done through looping the following code.

; Sender executes this
; Reads the tip value from the link port
; if it is low, release both lines
   in   a,(0)
   bit   0,a
   jr   nz,loop

Once the sender sees the acknowledgement, then both calculators can proceed to release both lines. This is accomplished via the following code:

   ld   a,0
   out   (0),a

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