An exact thirteen-point fractional covering obstruction

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The following argument and finite computation are author-recorded by Plasma AI. Indices in this proof start at zero. Take the thirteen rows

$$\begin{array}{c|ccccc|r} i&v_{i,0}&v_{i,1}&v_{i,2}&v_{i,3}&v_{i,4}&a_i\\\hline 0&1&0&0&1&1&69633\\ 1&1&1&1&0&1&65809\\ 2&1&0&1&0&0&257\\ 3&1&1&1&0&0&273\\ 4&0&0&1&1&1&69888\\ 5&1&1&0&0&1&65553\\ 6&0&0&1&1&0&4352\\ 7&1&0&1&1&0&4353\\ 8&0&0&0&1&1&69632\\ 9&0&0&1&0&0&256\\ 10&0&1&0&1&1&69648\\ 11&0&1&1&0&1&65808\\ 12&1&1&0&1&1&69649 \end{array}$$

Here $a_i=\sum_{j=0}^4 16^jv_{i,j}$. A signed sum of the vectors has integer coordinates of absolute value at most thirteen. If its highest nonzero coordinate is $s$, its contribution to the scalar encoding has absolute value at least $16^s$, whereas all smaller coordinates contribute at most $13(16^s-1)/15<16^s$. Thus a scalar signed sum vanishes exactly when the vector signed sum vanishes. All thirteen scalars are distinct and positive.

Exhaustive dissociation and extraction

The checker generates every subset sum of the vectors. Two equal subset sums give a ternary relation supported on their symmetric difference. Conversely every nonzero ternary relation arises from two disjoint subsets with equal sums. A subset is therefore dissociated exactly when it contains none of these collision supports. The checker also obtains the supports by a separate split enumeration of all $3^{13}$ ternary coefficient vectors and compares the resulting families.

The numbers of dissociated subsets of sizes zero through thirteen are

$$(1,13,78,284,673,982,652,16,0,0,0,0,0,0).$$

For every one of the $8191$ nonempty subsets $F$, dynamic programming over its submasks computes $r(F)$. The exact integer inequalities $13r(F)\ge7|F|$ hold throughout. The full set has $r(A)=7$ and thirteen elements, so the hereditary ratio is exactly $7/13$.

For every subset $D\subseteq A$, at least one of $D$ and $A\setminus D$ is dependent. This exhausts all two-color partitions; allowing overlapping covers does not help, since assigning each element to one covering class gives a partition into subsets of the same dissociated classes. A three-color partition, specified by the indices above, is

$$\{0\},\qquad \{6,8,9,10,11,12\},\qquad \{1,2,3,4,5,7\}.$$

Each class passes the same exact independence check. Hence $\chi_1(A)=3$.

Matching fractional certificates

Give the thirteen elements dual weights with numerator vector

$$(2,4,2,3,3,4,2,2,4,4,2,2,3)$$

and common denominator nineteen. The numerators sum to 37, and every dissociated subset has numerator sum at most nineteen. Thus any fractional cover has total weight at least $37/19$, by summing its covering inequalities against these dual weights.

For the reverse inequality, encode a subset $D$ by the integer bit mask $\sum_{i\in D}2^i$. The following dissociated subsets, with their weight numerators over nineteen, form a fractional cover:

\begin{array}{r|r@{\qquad}r|r@{\qquad}r|r} 1207&6&1431&1&1687&2\\ 1960&1&3659&4&3689&4\\ 4454&2&4578&1&4792&2\\ 4995&2&6018&1&6492&8\\ 7072&3&&&& \end{array}

The numerators total 37, and each point is covered with numerator exactly nineteen. The checker verifies dissociation and all thirteen coverage equalities. Hence $\chi_f(A)=37/19$.

For any nonnegative element weights $w$, these same sets $D_j$ and cover weights $\lambda_j$ satisfy $\sum_j\lambda_jw(D_j)=w(A)$. Since $\sum_j\lambda_j=37/19$, one set has $w(D_j)\ge(19/37)w(A)$. This proves the stated weighted extraction consequence without any external duality theorem.

The claim is finite: no extrapolation from these checks to a growing family is part of the argument.