This may be a question with a well known answer, but I've been thinking on it for two days, and can't quite come up with a satisfactory answer.

Consider the problem of dividing $p n$ bins numbered $1$ through $pn$ into $p m + 1$ segments by placing $pm$ balls. If we let $k = pn - pm \bmod pm+1$, then we can show that we may attain a placement of the $pm$ balls such that there are exactly $(pm+1) - k$ segments of empty bins of length $\lfloor \frac{pn-pm}{pm+1} \rfloor$ and $k$ segments of empty bins of length $\lceil \frac{pn-pm}{pm + 1} \rceil$.

Here's the tricky question: can we accomplish this task while ensuring that there are exactly $m$ balls in each interval $[(j-1)n + 1, jn]$, for $j \in \{1, 2, \dots, p\}$?

Every concrete example I work through answers in the affirmative, but I can't seem to get an algorithmic way of doing it, or mathematical proof that one such arrangement exists.

For particular examples, it always seems to work. See an example below with $p = 2,$ $n = 7,$ $m = 2.$

1| |X| | |X|1|X| | |X| | |

where 1 denotes the beginning of a new 'period', and | denotes the 'wall' of a bin, and X denotes a ball. Note that 1 and | both denote 'walls' of bins.

  • $\begingroup$ Your example is actually not regular. You want ||X|||X|||X|||X||. $\endgroup$ Mar 25, 2016 at 0:34
  • $\begingroup$ Not sure I agree with that; my example has $pn = 14$ bins, with $7$ in each period. Yours appears to only have $12$ bins? $\endgroup$
    – notwatson
    Mar 31, 2016 at 14:08
  • $\begingroup$ I count 14. Are you counting the extreme ones? $\endgroup$ Mar 31, 2016 at 18:35
  • $\begingroup$ Ah, no I was not. I only counted bins as spaces with lines on either side. So, I missed the two most extreme bins in your example. Thanks! $\endgroup$
    – notwatson
    Apr 5, 2016 at 18:09

1 Answer 1


For $0 \leq i \leq pm+1$, define $$ x_i = \left\lfloor i \frac{pn+1}{pm+1} \right\rfloor. $$ We put ball $i$ in bin $i$ for $1 \leq i \leq pm$. The length of the $i$th space (for $1 \leq i \leq pm+1$) is $$ x_i - x_{i-1} = \left\lfloor i \frac{pn+1}{pm+1} \right\rfloor - \left\lfloor (i-1) \frac{pn+1}{pm+1} \right\rfloor \in \left\{ \left\lfloor \frac{pn+1}{pm+1} \right\rfloor, \left\lceil \frac{pn+1}{pm+1} \right\rceil \right\}, $$ so your first condition is satisfied (the spaces are as equal to each other as possible).

For your second condition, we need to verify that $1 \leq x_1,\ldots,x_m \leq n$, $n+1 \leq x_{m+1},\ldots,x_{2m} \leq 2n$, and so on. It's clearly enough to verify that $x_{Cm} \leq Cn$ for $1 \leq C \leq p$ and that $x_{Cm+1} > Cn$ for $0 \leq C < p$. For the first condition, we have $$ x_{Cm} = \left\lfloor Cm \frac{pn+1}{pm+1} \right\rfloor < Cm \frac{pn+1}{pm+1} + 1 = \frac{Cpmn + Cm + pm + 1}{pm+1} \\ < \frac{Cpmn + Cn + pm + 1}{pm+1} = Cn+1, $$ using your implicit assumption $m < n$. For the second condition, $$ x_{Cm+1} = \left\lfloor (Cm+1) \frac{pn+1}{pm+1} \right\rfloor \geq (Cm+1) \frac{pn+1}{pm+1} = \frac{Cpmn + Cm + pn + 1}{pm+1} \\ \geq \frac{Cpmn + Cn + pm + 1}{pm+1} = Cn+1, $$ using $Cm + pn \geq Cn + pm$, which follows from $(p-C)(n-m) \geq 0$.

  • $\begingroup$ Looks as if you're on track to hit 100K rep before long; see here. Congratulations! Wish I could see it happen, but I'm for bed. $\endgroup$ Mar 25, 2016 at 1:35

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