Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is inspired by the settings of this MO thread.
Let $ (a_{ij})$ be an $ n\times n$ matrix with entries $ 1,...,n^2$ and increasing rows and columns. We'll call such a matrix of constant widths or, more precisely, $ (k,n^2-1-k)$ -wide iff $ a_{in}-a_{i1}=k\ $ and $ a_{ni}-a_{1i}=n^2-1-k $ for all $ i=1,...,n$ .
Call $ k$ feasible if a $ (k,n^2-1-k)$-wide matrix exists.
Call a $ p\times q$ board filled with $ pq$ consecutive integers in order, either row by row or column by column, trivial.

Obviously the feasible values are in $F_n:=\{n-1,...,n^2-n\}$, and the two extrema of this set are trivial in the above sense. More generally, if $ p,q|n$ then we obtain a $ (qn+p-pq-1, n^2-qn+pq-p)\ $ -wide and a $ (pn+q-pq-1, n^2-pn+pq-q)\ $ -wide matrix by putting together trivial $ p\times q$ matrices. We may consider all those cases also as trivial.

Question:

(1) For given $ n$ , which values $ k$ can yield non-trivial matrices of constant widths?

Playing around, you may have at first glance the impression that if $ n$ is a prime, no non-trivial $ k$ exist. This may be true for $ n=5$ , but for $ n=7$ , we have e.g. the $ (18,30)$ -wide matrix $$ {\left|\begin{array}{ccccccccc}\hline 1 &2&|&7&10&13&|&18&19\\ 3&4&|&8&11&14&|&20&21\\ 5&6&|&9&12&15&|&22&23\\ \hline 16&17&&24&25&26&&33&34\\ \hline 27&28&|&35&38&41&|&44&45\\ 29&30&|&36&39&42&|&46&47\\ 31&32&|&37&40&43&|&48&49\\ \hline \end{array}\right|}$$ composed essentially of trivial $ 3\times2$ and $ 3\times3$ matrices as shown.

Note that as the number of conditions grows only linearly with $n$, finding solutions will get easier as $n$ grows. So I'd expect only relatively few $k\in F_n$ not to be feasible. (I have not written a program to check that!)

(2) More interesting may thus be the question of finding the smallest non-trivial feasible $ k$ , and also the ones closest to $\frac{n^2-1}2$.

Such a matrix doesn't need to be 'antisymmetric' with respect to its center, meaning $a_{ij}+a_{n+1-i,n+1-j}=n^2+1$. But the boundary conditions seem to impose some sort of poised-ness. So the next question is:

(3) For odd $n=2m-1\ $, are there matrices of constant widths with central entry $a_{mm}\ne\frac{n^2+1}2$ ?

There may be an easy argument to show that there aren't, but I don't see it.

Edit: for the last question, the answer is yes. Take $n=21$ and start with nine trivial $ 7\times7$ blocks filled row by row. Now modify the first four lines of the central block by filling them column by column instead of row by row. This decreases the central entry of the block (and thus of the whole matrix) by $9$.

Now, such a construction won't work if $n$ is prime. So I suggest to replace question (3) by

(3') Is there a prime $n=2m-1\ $ such that there exists a matrix of constant widths with central entry $a_{mm}\ne\frac{n^2+1}2$ ?

share|improve this question
add comment

1 Answer

The answer to (3') is yes. Here's a matrix of constant widths with $n=7$ and central entry $27\ne25$:

$$\pmatrix{1&2&5&6&8&11&12\cr3&4&7&9&10&13&14\cr15&16&17&18&19&20&26\cr21&22&23&27&28&29&32\cr24&25&30&31&33&34&35\cr36&37&40&41&44&46&47\cr38&39&42&43&45&48&49\cr}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.