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Math-1015 · mathematics · Combinatorics

Wide partition conjecture

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Statement

Definitions and mathematical statement

An integer partition is a finite nonincreasing sequence $\lambda=(\lambda_1,\ldots,\lambda_s)$ of positive integers. A subpartition $\mu$ is obtained by selecting any submultiset of its parts and putting the selected parts in nonincreasing order. Thus entire rows are removed; individual parts are not shortened. Define the conjugate partition by $\mu'_j=|\{i:\mu_i\ge j\}|$.

For partitions of the same integer, the dominance relation $\alpha\succeq\beta$ means $\sum_{i=1}^t\alpha_i\ge\sum_{i=1}^t\beta_i$ for every $t\ge1$, with missing parts treated as zero. Say that $\lambda$ is wide if $\mu\succeq\mu'$ for every subpartition $\mu$, including the empty one.

The diagram of $\lambda$ has cells $(i,j)$ with $1\le i\le s$ and $1\le j\le\lambda_i$, with all rows left justified. Call $\lambda$ Latin if its cells admit integer entries $T_{ij}$ such that row $i$ is a permutation of $1,\ldots,\lambda_i$, and entries in each column are pairwise distinct. No condition requires distinctness across different columns beyond the row condition.

The wide partition conjecture states that for every partition $\lambda$, \[ \lambda\text{ is wide}\quad\Longleftrightarrow\quad \lambda\text{ is Latin}. \] The dominance tests are imposed on every subpartition, not only on $\lambda$ itself. The constructive direction asks for one simultaneous filling of the entire diagram from those inequalities; separate fillings of individual rows are not sufficient.

Short English statement

Are exactly the partitions passing every subpartition dominance test those whose rows can be filled as permutations without repeated symbols in a column?

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Math-1015-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1015

Wide partition conjecture

Definitions and mathematical statement

An integer partition is a finite nonincreasing sequence $\lambda=(\lambda_1,\ldots,\lambda_s)$ of positive integers. A subpartition $\mu$ is obtained by selecting any submultiset of its parts and putting the selected parts in nonincreasing order. Thus entire rows are removed; individual parts are not shortened. Define the conjugate partition by $\mu'_j=|\{i:\mu_i\ge j\}|$.

For partitions of the same integer, the dominance relation $\alpha\succeq\beta$ means $\sum_{i=1}^t\alpha_i\ge\sum_{i=1}^t\beta_i$ for every $t\ge1$, with missing parts treated as zero. Say that $\lambda$ is wide if $\mu\succeq\mu'$ for every subpartition $\mu$, including the empty one.

The diagram of $\lambda$ has cells $(i,j)$ with $1\le i\le s$ and $1\le j\le\lambda_i$, with all rows left justified. Call $\lambda$ Latin if its cells admit integer entries $T_{ij}$ such that row $i$ is a permutation of $1,\ldots,\lambda_i$, and entries in each column are pairwise distinct. No condition requires distinctness across different columns beyond the row condition.

The wide partition conjecture states that for every partition $\lambda$, \[ \lambda\text{ is wide}\quad\Longleftrightarrow\quad \lambda\text{ is Latin}. \] The dominance tests are imposed on every subpartition, not only on $\lambda$ itself. The constructive direction asks for one simultaneous filling of the entire diagram from those inequalities; separate fillings of individual rows are not sufficient.

Short English statement

Are exactly the partitions passing every subpartition dominance test those whose rows can be filled as permutations without repeated symbols in a column?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and plan.
Chow, Fan, Goemans and Vondrák [R1] formulate the wide-partition problem and prove partial results. Aharoni, Berger, Guo and Kotlar [R2] prove a dual version; this is not automatically the desired simultaneous filling theorem. We prove the necessary dominance conditions directly and construct fillings for rectangles and for every admissible partition with at most two rows.
Latin implies wide: a capacity proof.
Let $\mu$ be any selected subpartition of a Latin partition, keeping the selected rows in nonincreasing length order. The inherited filling is still Latin. For any $t\ge1$, the first $t$ columns contain \[ \sum_i\min(t,\mu_i)=\sum_{j=1}^t\mu'_j \] cells. Symbol $s$ occurs exactly $\mu'_s$ times in the entire filling, because it occurs once in each row of length at least $s$. Among the first $t$ columns it can occur at most $t$ times, by column distinctness. Hence the number of cells there is at most \[ \sum_{s\ge1}\min(t,\mu'_s)=\sum_{i=1}^t\mu_i. \] The last identity counts the boxes of the first $t$ rows by columns. Comparing the two displays proves $\mu\succeq\mu'$ for every $t$. Since the selected rows were arbitrary, the original partition is wide. This proves one full direction of the conjecture without any unproved matching or integrality assumption.
Rectangles.
For $\lambda=(a^b)$, the first dominance inequality requires $a\ge b$. If this holds, fill \[ T_{ij}=1+((i+j-2)\bmod a),\qquad 1\le i\le b,\quad1\le j\le a. \] Each row is a cyclic permutation of $[a]$. Within a column, distinct row indices have distinct residues modulo $a$, since there are at most $a$ rows. Thus the rectangle is Latin, and the preceding necessity proof shows it is wide. If $a<b$ it is neither, because the first column cannot have $b$ distinct entries drawn from $[a]$. This settles the equivalence for every rectangular shape.
All shapes with two rows.
Write $\lambda=(a,b)$ with $a\ge b\ge1$. The shape $(1,1)$ is neither wide nor Latin. For $a\ge b\ge2$, put $1,2,\ldots,a$ in the first row and $2,3,\ldots,b,1$ in the second. The lower row is a derangement of $[b]$ relative to the first $b$ entries, so no column repeats. For $b=1$ and $a\ge2$, put $2,1,3,\ldots,a$ in the first row and $1$ in the second. This also works. Hence every other two-row shape is Latin and therefore wide. A one-row shape is filled in its natural order. These checks include the row-length-one boundary that a single cyclic-shift formula would otherwise miss.
Why the capacity argument does not construct a general filling.
For each symbol $s$, one needs to place it once in every row of length at least $s$ and at most once in every column. Such placements for different $s$ must occupy disjoint cells. Solving these matching problems separately does not enforce that last requirement. In the square shape $(2,2)$, for example, each symbol separately can be placed in the two diagonal cells, giving two individually valid matchings which conflict at both cells. The shape is Latin, but this demonstrates the logical failure of choosing arbitrary separate matchings and superimposing them. The dominance inequalities above are necessary capacities; an additional coordinated selection theorem is needed to make them sufficient in all shapes.
Verification and final status.
Subpartitions delete whole rows, exactly as required; no shortened rows or arbitrary skew diagrams are introduced. The capacity count uses the prescribed row alphabet $[\mu_i]$, not just rowwise distinctness with arbitrary symbols. UNRESOLVED: necessity and the stated infinite families are proved, but sufficiency for an arbitrary wide partition remains the first missing step.

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