MATH.9:5.1 - Select a cheapest position on a cycle

Four positions are arranged in a cycle with a given clockwise direction. A rule receives real costs c=(c0,c1,c2,c3) and must select one minimum-cost position. Relabeling by k moves position j to j+k modulo 4 and moves its cost with it. The selected position must move in the same way.

Take c=(1,3,1,3). Its minimizers are {0,2}. A half-turn leaves c unchanged but exchanges both permitted answers. Neither is fixed, so B(c) is empty. A deterministic rotation-equivariant selector cannot answer this input. Choosing the smallest coordinate label returns 0 both before and after the half-turn, whereas equivariance requires the result to move to 2.

Now supply a marked position m as part of the input. Choose, among minimizers, the one with least clockwise distance

dist_m(j)=(j-m) modulo 4,

using the values 0,1,2,3. The distances are distinct, so this gives one answer. Under joint rotation of costs, mark and position, dist_(m+k)(j+k)=dist_m(j). The rule therefore respects rotation for every cost vector and mark.

With c=(1,3,1,3) and m=3, the distances of minimizers 0 and 2 are 1 and 3, so the rule selects 0. Rotating once gives costs (3,1,3,1), mark 0 and selected position 1. The same reasoning applies to every rotation.

For the unmarked input, returning {0,2} is a compatible set answer. If a probability distribution is wanted, assign probability 1/2 to each of those positions. Both alternatives retain the choice left open by the input.

A change of costs can also settle it: (1,3,0.9,3) has the unique minimizer 2; (1,3,1.1,3) has the unique minimizer 0. Their closeness to the tied input does not preserve that input’s fixed-point obstruction. It also exposes a discontinuity for a rule required to return one minimizing index as these costs vary through the tie.