MATH.9:5 - Archetypal Grounding

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.

MATH.9:5.2 - Restrict a unique optimum by exchange symmetry

Let x1,x2 be nonnegative real numbers with x1+x2=10, and minimize J=x1²+x2². Exchanging the two allocations preserves feasibility and cost.

If the minimizer is known to exist uniquely, the exchange must fix it. Hence x1=x2, and the constraint gives the candidate (5,5). For this example, existence, uniqueness and its optimality can also be established directly: every feasible point has the form (5+h,5-h), and

J(5+h,5-h)=50+2h².

The minimum is attained only at h=0. The direct calculation can finish this small problem; the symmetry deduction is reusable when uniqueness has another available justification.

If the same J is maximized on the segment, its maxima are (10,0) and (0,10). The exchange moves one to the other. Its fixed midpoint is the minimum, so the fixed-point equation alone would solve the wrong optimization question.

Now minimize x1²+2x2² with the same resource constraint. Exchange changes the criterion. Substituting x2=10-x1 gives

J=3*(x1-20/3)²+200/3,

so the unique optimum is (20/3,10/3). The old equality x1=x2 no longer follows. MATH.10 supplies the more general admissible-variation construction when the remaining optimization is the difficulty.

MATH.9:5.3 - Return a direction after geometric normalization

The input is an unordered pair P={-v,v} of opposite nonzero vectors in the plane. The requested answer is one of the two unit directions along its axis. Rotating the input should rotate the selected direction.

For P0={(-1,0),(1,0)}, a half-turn leaves the pair unchanged and negates every permitted unit direction. No permitted output is fixed, so a rotation-equivariant deterministic direction cannot be selected from this input.

The same failure appears in normalization. Both the identity and a half-turn map P0 to the same standard pair. Selecting (1,0) there and undoing those transformations returns opposite directions. The standard pair did not determine a direction of return.

Returning both directions, {-v/|v|,v/|v|}, is compatible with rotation when that set is the required output. Averaging them returns zero, whose length is zero rather than one.

Alternatively, mark one endpoint w in P and return w/|w|. Every rotation R preserves length, so R*w/|R*w|=R*(w/|w|). The marked-input rule is therefore equivariant. It answers a question with additional supplied information, which the unmarked pair lacked.

MATH.9:5.4 - Choose a position despite an ambiguous normalization

Let the six permutations of positions 1, 2 and 3 act on the triples formed by permuting d0=(a,a,b), where a and b differ. An answer may be any of the three positions, and relabeling the triple must relabel the answer.

The transformations fixing d0 are the identity and the swap of positions 1 and 2. All three positions are permitted answers, but only position 3 is fixed by both transformations. Thus B(d0)={3}; choose y0=3.

For the input d=(a,b,a), two transformations normalize it to d0. The first, h1, swaps positions 2 and 3. The second, h2, sends positions 1 to 2, 2 to 3 and 3 to 1. Both carry the b entry to position 3 while placing the two a entries in the remaining positions.

Returning y0 uses their inverses:

h1^-1(3)=2=h2^-1(3).

For every permutation of d0, the transported answer is the position carrying b. This gives the rule on the entire three-input orbit. The stabilizer calculation makes the return independent of the normalizer even though the input initially permits three different answers.