Math-3819 · mathematics · Combinatorics

Math-3819

Dittert–Hajek conjecture

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For $n\ge2$, put \[ \mathcal K_n=\left\{A=(a_{ij})\in[0,\infty)^{n\times n}:\ \sum_{i,j=1}^{n}a_{ij}=n\right\}. \] For $A\in\mathcal K_n$ define row sums $r_i=\sum_j a_{ij}$, column sums $c_j=\sum_i a_{ij}$, and \[ \operatorname{per}(A)=\sum_{\sigma\in S_n}\prod_{i=1}^{n}a_{i,\sigma(i)},\qquad \phi(A)=\prod_{i=1}^{n}r_i+\prod_{j=1}^{n}c_j-\operatorname{per}(A), \] where $S_n$ is the set of permutations of $\{1,\ldots,n\}$. Let $J_n$ be the all-ones matrix and $U_n=J_n/n$. Thus $\phi(U_n)=2-n!/n^n$. The case $n=1$ has just one admissible matrix.

Statement recorded in the supplied source.
Is it true for every $n\ge2$ and $A\in\mathcal K_n$ that \[ \phi(A)\le 2-\frac{n!}{n^n},\qquad \phi(A)=2-\frac{n!}{n^n}\ \Longleftrightarrow\ A=U_n? \] [S2]

Short English statement

Among nonnegative square matrices whose entries sum to the size of the matrix, prove that the Dittert functional has its unique maximum at the constant matrix.

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