MMP.10:5 - Archetypal Grounding

MMP.10:5.1 - Construct an unknown rule from requirements on its repetitions

A device has three labeled modes A, B and C. The required rule changes the mode on every use and returns to the starting mode after three uses. The question is to construct a deterministic rule, with no additional internal state. The rule itself is the unknown object.

Let S={A,B,C}. Choose one output variable p_i in S for each input i. The requirements become p_i != i and p_(p_(p_i)) = i for every i. Function composition gives the meaning of the repeated application. MATH.1 constructs composable paths; MATH.5 extends an interpretation of their elementary steps to the compounds.

To express the rule by selected pairs instead, choose r_ij in {0,1}. Add sum_j r_ij=1 for each row, r_ii=0, and, for all i,j,k, (r_ij=1 AND r_jk=1) implies r_ki=1. The row condition makes a function. The implication expresses the return after three uses: the first two selected transitions determine a required third.

Recover p by taking the unique selected column in each row. Conversely, p creates the table by selecting exactly its output pair in each row. These constructions are inverse. The triple-application requirement is therefore the same in both formulations, with MATH.7 carrying that relation. A rule A→B→C→A and its reverse both satisfy it.

There are precisely two such rules. From p^3=id, p is invertible with inverse p^2. Its cycles have lengths dividing three. Since a one-element cycle is forbidden, the three modes form one three-element cycle, with two possible orientations. This reasoning proves completeness; listing two examples alone would not.

Change the device to four modes, retaining the three-use return and no unchanged mode. A permutation of four elements cannot partition them into cycles all of length three, so no rule exists under these conditions. Change instead to a return after two uses. The conditions become p_(p_i)=i and p_i!=i; the indicator formulation requires symmetry r_ij=r_ji. It admits three pairings of four modes. Remove the former three-use implication: leaving it in the formulation would make the new, feasible requirement appear impossible.

The result is a rule that can be implemented and its stated scope: deterministic changes of the visible mode without hidden state. A proposal with additional state describes a different device and needs a new account of its operation.

MMP.10:5.2 - Preserve existence while repairing a count

An optional assignment gives each of two named requests either no selected option or one of options 0 and 1. Different requests may select the same option. This is a partial function from the two requests to {0,1}. Each request has three possibilities, so there are nine assignments.

Suppose the storage format gives each request two bits: d says whether an option is defined, and q gives its value when defined. When d=0, q is ignored. All sixteen four-bit records denote valid partial assignments. Every assignment has a record, so the representation can support an existence query with translated requirements.

It does not preserve the count. The empty assignment has four records, each of the four assignments defined on exactly one request has two records, and each of the four total assignments has one record. Thus 4 + 4*2 + 4 = 16. Uniform selection among records gives probability 4/16 to the empty assignment and 1/16 to each total assignment, rather than the 1/9 obtained by uniform selection among assignments.

For a count or uniform assignment sample, one repair is the structural condition d=0 implies q=0 for each request. There are now three admissible bit pairs per request and nine records, one per assignment. Another is a single variable with domain {absent,0,1} per request. If the sixteen-record storage representation must remain, group or weight its records using the multiplicities instead. To sample the nine assignments uniformly, give each record of an assignment with m records probability 1/(9*m). The probabilities of all m records then sum to 1/9 for that assignment. Thus each record of the empty assignment receives 1/36, each record of a one-request assignment 1/18, and each total-assignment record 1/9. C.29.1 supplies the required correspondence; MMP.7 supplies a probability law when sampling is the intended operation.

The original existence use can remain sufficient. The new count or sampling question exposes the need for the additional construction. No change in the underlying possible assignments is intended.

MMP.10:5.3 - Keep quantities and domain restrictions together

A preparation requires one litre containing 35 percent solute by volume, using solutions A and B at 20 and 80 percent. Assume solute is conserved and component volumes add in this preparation. Those subject assumptions supply the relations. Let x and y be the respective volumes in litres; choose nonnegative real domains because the amounts can initially be divided freely.

The joint conditions are x+y=1 and 0.2*x+0.8*y=0.35. Substitution gives x=0.75, y=0.25. Both the total and solute requirements hold for those amounts. Separate bounds 0<=x<=1 and 0<=y<=1 would lose their required total.

Now only whole half-litre doses may be used. Change the representation to x=m/2, y=n/2, with nonnegative integers m and n. The volume equation becomes m+n=2. Its possibilities (m,n)=(2,0),(1,1),(0,2) give solute amounts 0.2, 0.5 and 0.8 litre. None supplies 0.35 litre. Rounding the former solution changes the preparation; it does not satisfy its original condition.

If the required concentration changes to 50 percent, one dose of each solution works. The subject relations and unit remain, while the requirement and feasible assignment change. If mixing changes volume or solute, obtain the replacement subject relation before revising its mathematical expression.