As of now, claws can hold 8-9 stars max, give or take. Suppose you could guarantee n stars would be in the near zone on the other side. How many stars would you need to guarantee a win?
Assuming every other object far zones, and the opponents win auto and hang, your opponents get 2(24-n)+4+16+12.
48-2n+32 = n > 3n = 80 > n = 27 stars.
So that’s not possible. But what if your partner could push all the other object into the near zone? Opponents get 24-n+8+12+4 points.
48-n = n > n = 24 stars to tie. It’s possible, at least. What if, in addition to pushing all the objects into the near zone, your partner does a single dump at the last second?
A claw could hold 2 cubes, and 2 stars. Assuming these near zone, the score is 24-n-2+8-4+12+4 = n+4+2 > 42-n=n+6 > n = 18 stars to tie.
A pushbot with a claw on the front can guarantee the stars in the claw will be scored in the near zone. Paired with a competitive clawbot with a fairly large capacity, a pushbot can guarantee victory if it has a claw that can hold 18 stars.
Ways to make this feasible? Possibly extend the claw into the far zone once it has gathered enough stars. Also, winning auto makes the requirement to tie 10 stars, which is certainly possible even with a current claw. And, this assumes that your opponents can high hang, which many have seen as unnecessary this season. Without a hang, the requirement goes down to 6 stars, which is very easily possible. Moreover, we assume that all the objects from the final dump near zone, which is unlikely. If it all far zones (which is again unlikely), the pushbot only needs to hold 8 stars even with losing auto.
Game-breaking strategy? Possibly. Have I made some egregious error in my train of thought? I certainly hope so.