I have log output which seems to differentiate failed updates from actual updates. (This is just a tiny snippet of the log but it’s the part that seems relevant).
This one (a NOTE_ON) successfully updated:
32842882 Calling FillFrameWithDigitalData 232
32843078 sendSerialBufferDec set 4 71 0 1 0 230 8 14 255
32844137 Calling SendFeedbackData from SendDataIfReady
32844411 ...waiting for ACK on try 0
32844583 ...received 170 //ACK
The very next call sequence (processing a NOTE_OFF) failed to update, resulting in a stuck pixel:
32844829 Calling FillFrameWithDigitalData 233
32845037 sendSerialBufferDec set 4 70 0 0 0 0 0 0 255
32846076 Calling SendFeedbackData from SendDataIfReady
32846344 ...waiting for ACK on try 0
32846485 ...received 249 // SHOW_IN_PROGRESS
32846760 ...waiting for ACK on try 1
32848008 ...waiting for ACK on try 2
32849229 ...waiting for ACK on try 3
32850495 ...waiting for ACK on try 4
32851733 ...waiting for ACK on try 5
32852957 ...waiting for ACK on try 6
32853417 ...received 170 // ACK
I’ve observed two stuck pixels so far, with exactly the same structure (try 0 gets 0xF9, timeouts until try 6 which gets 0xAA).
Other strangeness is present in the log but I won’t be able to look at it until tonight.
EDIT: Yep, this is quite consistently associated with stuck pixels. All the extra .print() and .println() calls seem to massively increase the incidence (which is independently interesting).
I just got eight stuck pixels (7 of them in the last position, one in the first) in one 16-step march with a 500ms delay. Every one of them got SHOW_IN_PROGRESS (249, 0xF9) back from try 0 instead of ACK. They took a variable number of retries to get 170: 2 more (n=1), 5 more (n=1), 6 more (n=3), 7 more (n=2), and 10 more (n=1, with 248 (CHECKSUM_ERROR) on retry #8 and 250 (SHOW_END) on retry #9). So 5-7 retries before ACK is typical but the distribution extends in both directions.
Additionally, one other update got 250 (SHOW_END) back from try 0, but 249 on try 1, and that one seems to have updated (not 100% sure).