MATH.11:5 - Archetypal Grounding
MATH.11:5.1 - Derive weights for an exchange construction
States are triples (a,b,c) of nonnegative integer counts. Two rules are allowed:
- Replace two A units with three B units when a≥2:
(a,b,c) → (a-2,b+3,c). - Replace one B unit with one C unit when b≥1:
(a,b,c) → (a,b-1,c+1).
Starting with four A units, can the construction end with exactly five C units and nothing else?
The unweighted count changes under the first rule. Try I(a,b,c)=α*a+β*b+γ*c. The two differences are -2*α+3*β and -β+γ. Their joint equations have the solution (α,β,γ)=(3,2,2), giving:
I(a,b,c)=3*a+2*b+2*c.
Every allowed step preserves this value. The start (4,0,0) has value 12; the target (0,0,5) has value 10. The target is impossible under the stated rules, however many steps are attempted.
The same invariant helps formulate a useful alternative: six C units have value 12. They are attainable. Apply the first rule twice to obtain (0,6,0), then the second rule six times to obtain (0,0,6). This sequence supplies the contribution that equality of the invariant alone left open.
Now start at (0,3,0) and ask for (2,0,0). Both values are 6, but the first rule cannot create A and the second only consumes B to create C. Thus the requested target is unreachable. Adding the reverse exchange (a,b,c) → (a+2,b-3,c) when b≥3 makes that target reachable in one step. Its invariant change is 2*3-3*2=0, so the old invariant survives while the reachability answer changes.
For a material or operational application, interpret the counted kinds and allowed exchanges before using this mathematical result. The calculated weights express preservation by these rules; identifying them with physical mass or monetary value requires the corresponding subject account.
MATH.11:5.2 - Derive an accumulated sum from an update
A construction starts at (n,s)=(0,0) and repeatedly applies:
(n,s) → (n+1,s+n+1).
The question is what s will be when n has reached a chosen nonnegative integer. The new term added to s depends on n, so try I(n,s)=a*s+b*n^2+c*n+d.
Substitution gives:
I(n+1,s+n+1)-I(n,s)=(a+2*b)*n+(a+b+c).
Set a+2*b=0 and a+b+c=0. Choose a=2, b=-1, c=-1 and d=0. The resulting invariant is I(n,s)=2*s-n^2-n. It starts at zero, so every state reached by the update satisfies:
2*s=n^2+n, hence s=n*(n+1)/2.
After three steps, (3,6) satisfies it. The proposed state (3,7) has invariant value 2 and cannot result from these updates.
The same relation can be used with a different start. Starting at (2,10) gives invariant value 14, so later states satisfy 2*s-n^2-n=14. One step produces (3,13), which satisfies that changed equation. The preservation proof is unchanged.
Now change the update to (n,s) → (n+1,s+2*n+1). Substitution into the old invariant gives change 2*n, so the old formula fails in general. Reusing the same candidate family gives equations 2*a+2*b=0 and a+b+c=0. Choose a=1, b=-1, c=0: the new invariant is s-n^2. From (0,0), it gives s=n^2.
These constructions use integer arithmetic without overflow. They also use the update as a simultaneous substitution: the expression for the new s contains the old n. An implementation that increments n before evaluating that expression would need a different calculation.
MATH.11:5.3 - Change the value arithmetic to find a parity obstruction
The state is an integer n. Allowed steps add 2 or subtract 2. Starting at zero, can the construction reach 1?
A rational-valued linear candidate I(n)=a*n+b changes by 2*a under addition of 2. Requiring zero forces a=0, leaving only constants in this family.
Instead take I(n)=n mod 2, with values 0 and 1. Both +2 and -2 preserve the remainder. The start has remainder 0 and the target remainder 1, proving impossibility.
Here the necessary condition also leads to a construction. For any even target n=2k, use k additions of 2 when k≥0, or -k subtractions of 2 when k<0. Thus the reachable states are exactly the even integers. The invariant and the explicit sequence establish the two directions of that statement.
Adding the steps +1 and -1 destroys this parity invariant: 0 can now move to 1. Every integer becomes reachable by repeated unit steps. The earlier two-step rules and their parity calculation remain correct, but no longer cover all allowed steps.