# Partitioning set including empty set; P=QRST

Suppose I have $$l$$ length of words denoted as $$P$$, I want to divide this set into four $$P=QRST$$ including empty set, i.e., $$Q=R=S=\phi, T=P$$.

For non-empty set decomposition, I can do

 p = Permutations[P];
n = Length[P];
pn = IntegerPartitions[n, {4}];
pnn[i_] := Permutations[pn[[i]]];

Tablen = Length[pn];
in = Flatten[Table[pnn[i], {i, 1, Tablen}], 1];
tn = Flatten[Outer[TakeList, p, in, 1], 1];
newn = DeleteDuplicatesBy[tn, Sort /@ # &];

Q = Map[#[[1]] &, tn];
R = Map[#[[2]] &, tn];
S = Map[#[[3]] &, tn];
T = Map[#[[4]] &, tn];


Actually, I can do more general decomposition by replacing {4} -> {m}. But here the problem is in the process of making in and tn, the empty set was not included.

I want to make Q,R,S,T including an empty set. Any ideas?

[This post contains some answer in my previous post, but here I want to include empty set]

Example

$$P=\{1,2,3,4\}$$

then

$$Q= \{ \{ \}, \{1\}, \{2\},\{3\},\{4\},\{1,2\}, ...\}$$

$$R=\{ \{\}, ....\}$$

$$S=\{ \{\}, .....\}$$

$$T=\{\{1,2,3,4\}, ...\}$$

where $$P=QRST$$

• $P=QRST$ and $T=P$ : Is this a recursive definition?
– Syed
Dec 13, 2022 at 6:57
• @Syed what I mean $P=QRST$ is dividing $P$ into $Q,R,S,T$. If $Q,R,S$ becomes an empty set then $P=T$. The example case above is the case $Q,R,S$ becomes empty and so $T$ becomes the whole set $P$ Dec 13, 2022 at 7:01
• Please load a minimal example including a definition for P such that it evaluates.
– Syed
Dec 13, 2022 at 7:04