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I need to find a bijective proof that the number of independent partitions in a path of $n$ nodes and number of partitions of a set of $n-1$ elements are equal.

By independent partitions in a path we mean the partitions which create subset of nodes which are not adjacent.

For example, if I have the path:

$a\frac{\;\;\;\;\;\;}{}b\frac{\;\;\;\;\;\;}{}c$

the independent partitions are: $|a|b|c|$ and $|a c|b|$

$a$ and $c$ can be in the same block because they are not adjacent. So, for a path of three nodes we got 2 independent partitions. If we consider the set of $n-1$ elements (then 3-1=2) $\{a, b\}$ we get the following partitions: $|a|b|$ and $|a b|$

So we got again 2 partitions. I'm not going to show you other examples, because the equivalence is easily demonstrable empirically. But I cannot find the connection between the two type of partitions.

I thought about the calculation method for the partitions in a set (with the Stirling Numbers of the second kind) but doesn't seem the right way.

I thought also about the fact that $n-1$ is exactly the number of edge in the path, but even this way seems to lead nowhere

1 Answers 1

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Bijection:

Given a set of 'independent partitions' of a totally ordered set of elements (the independent partitions are defined in the question as a partition such that two consecutive elements are never in the same subset) let the numbers $1\ldots N$ label the elements.

The following process maps the independent partitions of $N$ elements to the partitions of $N-1$ elements.

  1. Order the partitions by the smallest element in each set
  2. For each element $n$, if it is in a higher partition (as defined by the ordering of partitions in 1.) than the element $n-1$, then move it down 1 partition.
  3. Relabel the elements $n \rightarrow n-1$.
  4. Remove the highest partition in the ordering if it is empty.
  5. Remove the element $0$.

Inverting this process gives the opposite map.

Example:

For example, given the independent partitions of 4 elements, ordered according to 1., we perform the steps (2.), (3.) and (4.&5.) whilst highlighting when relevant the elements moved according in step (2.).

$$ \begin{matrix} 13|\color{red}{24} \\ 13|\color{red}{2}|\color{red}{4} \\ 14|\color{red}{2}|\color{red}{3} \\ 1|\color{red}{2}4|\color{red}{3} \\ 1|\color{red}{2}|\color{red}{3}|\color{red}{4} \end{matrix} \xrightarrow{2.} \begin{matrix} 1\color{red}{2}3\color{red}{4}|- \\ 1\color{red}{2}3|\color{red}{4}|- \\ 1\color{red}{2}4|\color{red}{3}|- \\ 1\color{red}{2}|\color{red}{3}4|- \\ 1\color{red}{2}|\color{red}{3}|\color{red}{4}|- \end{matrix} \xrightarrow{3.} \begin{matrix} 0123|- \\ 012|3|- \\ 013|2|- \\ 01|23|- \\ 01|2|3|- \end{matrix} \xrightarrow{4.\, \& \, 5.} \begin{matrix} 123 \\ 12|3 \\ 13|2 \\ 1|23 \\ 1|2|3 \end{matrix} $$