We can save one piece by rearranging the order of the 4 operations in each loop. We want the first ball to approach register A from the left, so it doesn’t have to turn back on itself. This means the topmost bit must send the first ball left and locks in the left bit on the second row. The other bit must then send one ball towards the 4 digit of B and dispose of the other.
We can’t make use of the red balls because A and B are too close together, they both discard to the same column.
Link: jstumble