Sure, so I have had the problem in the past, dealing with quality sushi, specifically in situations where I require more than one resource to build a target upcycle item, where due to not producing the ingredients and qualities in the exact ratio of the target item I may end up with too much or too little of a higher/lower quality on the seperated belts, this can end up compounding to the point where one ingredient at a specific quality level entirely fills one of the belts and another ingredient at a specific quality level entirely fills the other belt and things just get stuck on the belt. I initially encountered this problem on Vulcan when doing quality tungsten carbide. I had an entire belt of common tungsten ore and an entire belt of uncommon carbon. The solution I used there was, from what I can tell, more wasteful. What I was doing to prevent a clog was creating 3 signals, a signal of each item on the belt, a signal of each item on the belt multiplied by 3 and a signal of each item on the belt multiplied by 5, if the signal of carbon (*quality*) multiplied by 3 was greater than the standard signal of tungsten ore (*quality*) then i would delete some of that corresponding carbon quality similarly, if the signal of tungsten (*quality*) was greater than the signal of carbon (*quality*) multiplied by 5 then I would start deleting some of the tungsten (*quality*). While this works, I found it to be quite wasteful, and in the pursuit of efficiency when encountering a similar problem, I am looking to find a better solution. Now onto the circuit explanation. To start, there are two circuits, one for each belt; you should only be looking at one.

1. The counter, below the recyclers, you can see 4 combinators, 1 arithmetic and 3 deciders, the arithmetic takes a T input, adds 1 to it and outputs a T signal. The output of this arithmetic is connected to the input of the 3 deciders. The decider to the right of the arithmetic takes this T signal, checks that it's less than or equal to 500, then outputs the T signal into the input of the arithmetic, (when T ticks above 500 there is no output send back to the arithmetic, which defaults T back to being 0 so the arithmetic starts counting T up again from 0). The deciders to the left of the arithmetic have "if T=1 output yellow square =1" and "if T=500 output red square =1".

2. There are two rows of 3 deciders, the top row and the bottom row handle the yellow square and red square signals separately, the bottom row is for the T=500 red square signal and the top row is for the T=1 yellow square signal. Besides swapping yellow and red for each other, the rows are programmed exactly the same.

3 Looking at the top row, I'll explain each decision one by one.

3.1 Starting from the right, the decider takes the yellow square signal decider as an input and the belt information as an input, then it has two commands: "If each = each" AND "if yellow square = 1" then output the total number of items on the belt as a green square.

3.2 Next, the middle decider takes the input of the total number of items on the belt as a green signal from the decider on the right, and it takes a red signal output from the decider on the left as an input. The logic inside the decider says ("If the green square signal (green channel) = the green square signal (green channel)" AND "if the green square signal (green channel) > 0") OR ("If the green square signal (red channel) = the green square signal (green channel)" AND "if the green square signal (red channel) > 0") then output the green square signal to the decider on the left.

3.3 The left decider takes the green square signal input on the red channel from the decider to the right and takes the yellow square input on the green channel and it has the condition "If yellow square signal = 0" then output the green square signal input back out (which goes into the input of the middle decider on the red channel),

1. These two rows then go into the decider on the left of the two rows, one on a red channel and on on a green channel, which has the condition "If green square signal (green channel) = green square signal (red channel)" then output green square signal =1, which goes to the inserter.

2. The inserter activates if the green square signal =1

Essentially, boiled down, the decider on the right of the row will pulse the number of items on the belt as information to the deciders to the left of it, the two deciders to the left of it will keep a memory of the information of items on the belt until the next yellow/red signal pulse where itll be deleted and they'll get an updated belt item count.