English: Example game for *Star, using a 4-ring board. Blue has created a single star, using the five-pointed bridge in the center of the board to connect the two regions. The Blue star owns 7 peries (R43, *40, *41, S41, T41, A40, and A41) and 2 quarks (*40 and A40). Red has created three stars with a total of 11 peries (the larger star also connects using the five-pointed bridge and owns 8 peries by containing 5: *42, S40, A42, R40, and R42 and surrounding 3: *43, A43, and R41; the smaller star owns 3 peries by containing 2: S42 and T40 and surrounding 1: S43; the smallest star contains 2 peries at T42 and T43, but it does not contribute to the score because it fails to meet the minimum value). In addition, the Red stars collectively own 3 quarks (S40, T40, and R40).

Although Red has won more peries (11) than Blue (7), Blue has taken ownership of those 7 peries using 1 star, while Red has created 3 stars. Therefore the Blue "reward" is 2×(3-1) = +4, while the Red "reward" is 2×(1-3) = -4. The final score is Blue 11, Red 7, and Blue wins this match by more efficiently grouping its pieces.

Under the alternative Schmittberger scoring system, Blue has 7-11+(4-(-4)) = 4 points, while Red has 11-7+(-4-4) = -4 points.

Note that placing additional pieces on the board will not affect the scoring of this game, as the perimeter cells / nodes are all owned by containment or surrounding pieces, and the stars cannot be further connected.