$P=QR$ decomposition of given list

Question

For given arbitrary lists, I want to decompose the list into two parts and assign individual lists.

i.e., I want to decompose, $P=QR$.

And I can think of its decomposition in two cases: ordered and non-ordered cases.

For example, Let $P=\{a1,a2,a3\}$,

For ordered case, I have $2$ cases.

$(Q,R)=\{ (\{a1\}, \{a2,a3\}), (\{a1,a2\}, \{a3\}) \}$

where the ordering of a1,a2,a3 are strictly followed.

For non-ordered case, I have $3+3=6$ possible decomposition. $(Q,R) = \{ (\{a1\}, \{a2, a3\}), (\{a2\}, \{a1,a3\}), (\{a3\}, \{a1,a2\}), (\{a1,a2\}, \{a3\}), (\{a1,a3\},\{a2\}), (\{a2,a3\},\{a1\}) \}$.

And in this case I want to assign $Q[[1]]=\{a1\}, R[[1]]=\{a2,a3\}, $ and so on.

For $P=\{a1, a2,a3,a4\}$ case I have $4+6+4=14$ unordered partitions.

explicitly, neglecting $a$'s I have

$1234= (123,4) + (124,3) + (134,2) + (234,1) + (12,34) + (13,24) + (14,23) + (23,14) + (24,13) + (34,12) + (1,234) + (2,134) + (3,124)+ (4,123)$

For given lists, I want to make ordered and non-ordered partitions via Mathematica.

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My little trial, is for ordered case it seems mathematica command table might be good. Because naively, my pseudo code

 Q=Table[{P[1], \cdots, P[[i]]\}, ]
 R=Table[P[[i+1]], ]

seems fine.

And for non-ordered case, subset command might be useful. Because,

Subsets[{a, b, c}]
{{}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

except for the empty set and the whole set, I want them as $Q$, but I have trouble identifying $R$.

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Answer 1

Your intuition for what to try seems good. See if you can study the help pages to understand the following and how it tries to implement your pseudocode.

P={a1,a2,a3,a4};
(*ordered*)
Q=Table[Take[P,i],{i,Length[P]-1}]
R=Table[Drop[P,i],{i,Length[P]-1}]
(*unordered*)
Q=Subsets[P,{1,Length[P]-1}]
R=Map[Complement[P,#]&,Q]