Diagonally cyclic equitable rectangles

Abstract

An equitable $(r,c;v)$-rectangle is an $r \times c$ matrix $L=(l_{ij})$ with symbols from $\mathbb {Z}_v$ in which each symbol appears in every row either $\left\lceil c/v \right\rceil$ or $\left\lfloor c/v \right\rfloor$ times and in every column either $\left\lceil r/v \right\rceil$ or $\left\lfloor r/v \right\rfloor$ times. We call $L$ diagonally cyclic if $l_{(i+1) (j+1)}=l_{ij}+1$, where the rows are indexed by $\mathbb {Z}_r$ and columns indexed by $\mathbb {Z}_c$. We give a constructive proof of necessary and sufficient conditions for the existence of a diagonally cyclic equitable $(r,c;v)$-rectangle.

Appendix: Technical lemmata

In the following lemmata, we say $p^k$ exactly divides $n$ if $p^k$ divides $n$ and $p^{k+1}$ does not divide $n$. For these results, we assume $r, c$ and $v$ are arbitrary positive integers, and, as in the rest of the paper,

$$\begin{aligned}N_{\mathrm {row}}=\frac{c\,\gcd (r,v)}{v\,\gcd (r,c)} \quad \text { and } \quad N_{\mathrm {col}}=\frac{r\,\gcd (c,v)}{v\,\gcd (r,c)}.\end{aligned}$$

Lemma 8

Suppose $v$ divides ${{\mathrm{lcm}}}(r,c)$. Then $N_{\mathrm {row}}$ and $N_{\mathrm {col}}$ are positive integers.

Proof

Let $p$ be a prime. Suppose $p^a$ exactly divides $v$, and $p^b$ exactly divides $r$ and $p^x$ exactly divides $c$. If $N_{\mathrm {row}}$ is a positive integer, then it would be exactly divisible by $p^{x+\min (a,b)-a-\min (b,x)}$. Since $p$ is arbitrary, it is sufficient to show that

$$\begin{aligned} x+\min (a,b)-a-\min (b,x) \geqslant 0 \end{aligned}$$

(8)

as, if $N_{\mathrm {row}}$ were not a positive integer, then (8) would be false for some prime $p$. Note that $a \leqslant \max (b,x)$ since $v$ divides ${{\mathrm{lcm}}}(r,c)$.

Case I $\min (a,b)=a$. Then (8) follows immediately.

Case II $\min (a,b)=b$ and $\min (b,x)=b$. The left hand side of (8) becomes $x-a$, which is non-negative, since $a \leqslant \max (b,x)=x$.

Case III $\min (a,b)=b$ and $\min (b,x)=x$. The left hand side of (8) becomes $b-a$, which is non-negative, since $a \leqslant \max (b,x)=b$.

We can show that $N_{\mathrm {col}}$ is a positive integer by switching $r$ and $c$.

Lemma 9

Let $\chi =c\, \gcd (r,v)/\gcd (r,c)$. Then $\chi$ divides $c$ if and only if $\gcd (r,v)$ divides $c$.

Proof

If $\gcd (r,v)$ does not divide $c$, then $\chi$, which is a multiple of $\gcd (r,v)$, also does not divide $c$.

Conversely, assume $\gcd (r,v)$ divides $c$. Let $p$ be a prime. Suppose $p^a$ exactly divides $\gcd (r,v)$, and $p^b$ exactly divides $\gcd (r,c)$ and $p^x$ exactly divides $c$. Hence $p^{x+a-b}$ exactly divides $\chi$. Since $p$ is arbitrary, it is sufficient to show that $b \geqslant a$. Since $p^a$ divides $\gcd (r,v)$, we know $p^a$ divides $r$, and since $\gcd (r,v)$ divides $c$, we know $p^a$ also divides $c$, so $p^a$ divides $\gcd (r,c)$. Hence $b \geqslant a$.$\square$

Lemma 10

We have:

• $\gcd (r,v)$ divides $\gcd (r,c)$ if and only if $vN_{\mathrm {row}}$ divides $c$ and

• $\gcd (c,v)$ divides $\gcd (r,c)$ if and only if $vN_{\mathrm {col}}$ divides $r$.

Proof

$$\begin{aligned} \gcd (r,v) \text { divides } \gcd (r,c)&\iff \gcd (r,v) \text { divides } c \\&\iff c\, \frac{\gcd (r,v)}{\gcd (r,c)} \text { divides } c&\text {by Lemma 9} \\&\iff v N_{\mathrm {row}}\text { divides } c. \end{aligned}$$

The second dot-point is the same as the first with $r$ and $c$ switched.$\square$

Lemma 11

Suppose $v$ divides ${{\mathrm{lcm}}}(r,c)$. Suppose also that $\gcd (r,v)$ divides $\gcd (r,c)$. Then $v$ divides $c$.

Proof

Let $d$ be a prime power divisor of $v$. Since $v$ divides ${{\mathrm{lcm}}}(r,c)$ and $d$ is a prime power, we know that $d$ divides $r$ or $c$ (or both). Since we want to prove that $d$ divides $c$, assume $d$ divides $r$. Since $d$ divides both $r$ and $v$, we know that $d$ divides $\gcd (r,v)$ and hence $d$ divides $\gcd (r,c)$ by assumption. Therefore $d$ divides $c$. Since $d$ is an arbitrary prime power divisor of $v$, we conclude that $v$ divides $c$.$\square$

Lemma 12

If $v$ divides $\gcd (r,c)$, then $vN_{\mathrm {row}}$ divides $c$ and $vN_{\mathrm {col}}$ divides $r$.

Proof

If $v$ divides $\gcd (r,c)$, then $v$ divides $r$ and hence $v$ divides $\gcd (r,v)$. But since $\gcd (r,v)$ divides $v$, we must have that $v=\gcd (r,v)$. Hence $N_{\mathrm {row}}=c/\gcd (r,c)$ and

$$\begin{aligned} vN_{\mathrm {row}}= \frac{c}{\left( \frac{\gcd (r,c)}{v}\right) }, \end{aligned}$$

which divides $c$ (since $v$ divides $\gcd (r,c)$, and $\gcd (r,c)$ divides $c$).

The second claim, that $vN_{\mathrm {col}}$ divides $r$, follows from the first claim with $r$ and $c$ switched.   $\square$