# ETOOBUSY đźš€ minimal blogging for the impatient

# All partitions of a set - preliminary considerations

**TL;DR**

We go on with finding all partitions of a set, following the track started with PWC108 - Bell Numbers.

In previous post All positive integer sums we laid out a possible strategy for finding all (distinct) partitions of a set.

The first half was to find out all possible ways to express a positive integer as the sum of other (lower or equal) positive integers; this has been addressed and led us to the iterator-based solution described in All positive integer sums, as iterator.

Nowâ€¦ weâ€™re *only* left with generating the sets starting from these
partial sums. Letâ€™s take a first look at the case for $N = 3$:

There is an obvious factor that has to be taken into considerations: we
have *three* distinct expansions for $2 + 1$, but only one for $1 + 1 + 1$.

In general, any subset of equal addends in the sum have to be taken with
care in order to avoid duplicates; this does not happen, of course,
across different values. For this reason, the $2 + 2$ decomposition for
$4$ has to be taken with care too, or we would have duplicates. In other
words, the following are the *only* distinct partitions of the type $2 +
2$:

Any partition with the $a$ in the *second* subset would lead to a
partition that is equivalent to one of the above, i.e. a duplicate.

Summing up, when generating all partitions starting from our
decomposition of the integer input into possible sums, we will have to
address the subsets of equal addends as a kind of *unit* with a specific
algorithm.

For this reason, itâ€™s useful to express the generic sum decomposition like this:

\[N = \sum_{j = 1}^{J}{k_j \cdot n_j}\]where $n_j$ represents the addendum value and $k_j$ represents how many that addendum appears in the decomposition. This would mean the following:

\[3 = 3 = 1 \cdot 3 \\ 3 = 2 + 1 = 1 \cdot 2 + 1 \cdot 1 \\ 3 = 1 + 1 + 1 = 3 \cdot 1\]Enough for the preliminary considerationsâ€¦ stay safe!