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Econ-558-A2 · revision r1 · on Econ-558

GPT 6 Astra Ultra attempt on Econ-558

Full solution claimed · LLM-generated · gpt 6 astra ultra ·

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Contents
  1. Realizability of every admissible input-block shape: A cyclic construction excluding every additional block
  2. The exact target
  3. Laminarity on both sides
  4. Recursive preferences
  5. An illustrative nonpartition shape
  6. Scope of the conclusion
  7. Completion Estimate
  8. References

Realizability of every admissible input-block shape: A cyclic construction excluding every additional block

The exact target

An input block in a strict complete marriage instance is a pair $(A,B)$ of equally large nonempty proposer and receiver sets satisfying \[ \operatorname{Top}_{|B|}(p)=B\quad(p\in A),\qquad \operatorname{Top}_{|A|}(r)=A\quad(r\in B). \] The whole instance counts as a block. The question in [ishida, Section 7, problem D] asks whether every admissible size-labelled rooted tree is the containment tree of all input blocks. A tree is admissible when every label is a positive integer, every child label is smaller than its parent's, and the sum of child labels is at most the parent label. We give a construction for the exact target. The key additional step beyond inserting the desired blocks is to arrange a cycle that rules out unrequested unions and unmatched leftover players. No condition on the eventual stable matching is used.

Laminarity on both sides

The following elementary fact is also the input-block laminarity property in [ishida, Proposition 9.5]; its proof is included because the exclusion argument needs its two-sided form.
Lemma: Let $(A,B)$ and $(C,D)$ be input blocks of sizes $k\le l$. If they share a proposer or a receiver, then $A\subseteq C$ and $B\subseteq D$. Otherwise both sides are disjoint.
Proof: If $p\in A\cap C$, its top $k$ receivers are a subset of its top $l$, so $B\subseteq D$. Choose $r\in B$, which is possible since $k>0$. Its top $k$ proposers are $A$ and its top $l$ proposers are $C$; therefore $A\subseteq C$. Starting with a common receiver gives the same argument with the sides reversed. Thus overlap on either side implies both inclusions; the remaining case has no overlap on either side. ∎

Recursive preferences

Associate with each node $v$ two sets $P_v,R_v$, each of size $s_v$, the node's label. For each child $w$, choose disjoint subsets $P_w\subset P_v$ and $R_w\subset R_v$ of size $s_w$. The capacity condition permits this. Pair the remaining proposers and receivers arbitrarily as $(p_j,r_j)$. These are bookkeeping pairs, not necessarily mutually preferred or matched pairs.

The atoms at $v$ are its child pairs $(P_w,R_w)$ and these free singleton pairs. Put the atoms in any cyclic order $1,\ldots,m$. If $s_v>1$, then $m\ge2$: a sole child has smaller size and hence leaves a free pair, while a node with no children has $s_v$ free pairs. For each atom $j$, fix a receiver $r_j^*$ in it. At a free atom this is its unique receiver. The successor index is understood modulo $m$.

Construct local complete lists within $(P_v,R_v)$ from the children's already constructed lists as follows. Every player in a child keeps all partners in that child first, in the inherited order. Every proposer in child atom $j$ ranks $r_{j+1}^*$ first after its child, then the other receivers in $R_v$ outside the child in arbitrary strict order. A free proposer $p_j$ ranks $r_{j+1}^*$ first and all remaining receivers in $R_v$ afterward. A free receiver $r_j$ ranks $p_j$ first and all other proposers afterward. A receiver in a child appends the proposers outside that child in any strict order.

For $s_v=1$, there are no children; give the unique two players their unique local lists. Apply the construction up the tree. The root lists are the required complete rankings. Appending outside partners never changes a descendant's internal ranking or top-set property.

Theorem: For every finite admissible size-labelled rooted tree with root label $n$, the construction above produces a strict complete marriage instance with $n$ players on each side whose input blocks are exactly the prescribed node pairs $(P_v,R_v)$. Consequently every admissible shape in the stated problem is realizable.
Proof: Every node pair is a block: each player ranks its own child internally first, then the remaining partners of the parent, and only then partners outside that parent. Induction toward the root shows that the first $s_v$ partners of every player at node $v$ are exactly the opposite side of $v$.

Suppose $(A,B)$ is an additional block, and choose the smallest prescribed node $v$ containing both its sides. Such a node exists because the root contains everyone; uniqueness follows from the nested-or-disjoint construction. The additional block is not contained in any child. By the lemma it therefore either contains an entire child block or is disjoint from that child on both sides. Moreover it is strictly larger than every child it contains: equality would make it that prescribed block.

Call an atom reached if its whole child block belongs to $(A,B)$, or, for a free atom, if its proposer belongs to $A$. At least one atom is reached. Indeed, any member of the nonempty set $A$ is in a child, forcing that whole child by the lemma, or is a free proposer. We claim that reaching atom $j$ reaches atom $j+1$. For a child atom of size $s_j$, the block size $|A|$ exceeds $s_j$. Every proposer in that child must have its first outside receiver, $r_{j+1}^*$, in its top $|A|$ set $B$. For a free atom the same receiver is the proposer's first choice and again lies in $B$. If the successor is a child, overlap on the receiver side and the lemma force its whole block into $(A,B)$. If it is free, its receiver's first proposer is its own $p_{j+1}$, which must therefore lie in $A$. This proves the claim.

The successor cycle reaches every atom. Thus all child proposers and all free proposers lie in $A$, so $A=P_v$. Since $B\subseteq R_v$ and $|B|=|A|=s_v$, also $B=R_v$. This contradicts the premise that the block is additional. When $s_v=1$ the equality is immediate and no cycle argument is needed. Hence there are no additional blocks. ∎

An illustrative nonpartition shape

Take a root of size four with a sole child of size two, and give that child no children. Within it use $p_1:r_2,r_1$, $p_2:r_1,r_2$, $r_1:p_1,p_2$, and $r_2:p_2,p_1$. Its two free atoms form the two-cycle that prevents size-one blocks. At the root order the three atoms as the child, $(p_3,r_3)$, $(p_4,r_4)$. The child proposers' first outside receiver is $r_3$; $p_3$ starts with $r_4$ and $p_4$ with $r_1$; $r_3,r_4$ start with $p_3,p_4$ respectively. Complete the lists as above. The only blocks are the specified size-two child and the size-four root. In particular, the child's complement is not forced to be a block.

Scope of the conclusion

The proof supplies existence for every admissible shape and explicitly excludes accidental blocks; it does not enumerate how many preference profiles realize a shape, classify their stable matchings, or impose extra restrictions such as a common master list. Necessity of the admissibility conditions follows from laminarity, strict inclusion, and disjoint child sets, so together these conditions characterize the possible size-labelled shapes. This is a claim about the precise input-block question, without a priority claim about related factorization questions.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the full catalogue target. The attempt constructs preferences recursively and uses two-sided laminarity and successor cycles to exclude every accidental input block. Every admissible labelled tree is therefore realized with exactly its prescribed blocks, establishing the complete canonical existence target without needing additional matching or enumeration conclusions.

References

  • [ishida] Y. Ishida (2026), Prime factorisation of stable-matching instances: uniqueness, simultaneous products, and an exact census, arXiv:2610.05617v1, Section 3 and Section 7, problem D. https://arxiv.org/html/2610.05617v1.