To formalize this, we can assume there are two three operations we can do: If you want the white ones in the corners, you have just 2 of them in place (2,11). How do you decide that upfront? Should you take into account how many balls are already in place? In my example, if you want the grey ones in the corners, you already have 3 in place (balls 1,10,14). There's also the decision of which types of balls we want in the corners to be made. 5 is not, we swap it with 2, then we swap 4 with 3 (or with 8), but this would already be inefficient because we've either moved the 4 to the center or the 8 in 4's position - i.e. So for example, we'd assume 1 is at the correct position. How do you proceed?Ī simple approach would be to start in order, top -> bottom and left -> right. The two remaining balls (a stripe and a solid) don't matter.Īssume you just finished a game, gather the balls, put them in the rack and proceed to arrange them to start a new one. the 8-ball must be in the center, and along the sides the stripes and the solids must alternate. Since racking of billiard balls for the 8-ball game can be done under multiple rules, here's the racking I refer to:

Proffesional billiards rack triangle