LibraryReExportIsNPComplete

From APIDesign

This page describes a way to convert any 3SAT problem to a solution of finding the right configuration from conflicting libraries in a system that can re-export APIs. Thus proving that the later problem is wikipedia::NP-complete.

Contents 1 3SAT 2 Module Dependencies Problem 2.1 Module Dependency Problem 3 Conversion of 3SAT to Module Dependencies Problem 4 Proof

3SAT

The problem of satisfying a logic formula remains NP-complete even if all expressions are written in wikipedia::conjunctive normal form with 3 variables per clause (3-CNF), yielding the 3SAT problem. This means the expression has the form:

Failed to parse (Can't write to or create math temp directory): (x_{11} \vee x_{12} \vee x_{13}) \wedge

Failed to parse (Can't write to or create math temp directory): (x_{21} \vee x_{22} \vee x_{23}) \wedge

Failed to parse (Can't write to or create math temp directory): (x_{31} \vee x_{32} \vee x_{33}) \wedge

Failed to parse (Can't write to or create math temp directory): ...

Failed to parse (Can't write to or create math temp directory): (x_{n1} \vee x_{n2} \vee x_{n3})

where each Failed to parse (Can't write to or create math temp directory): x_{ab}

is a variable Failed to parse (Can't write to or create math temp directory): v_i
or a negation of a variable Failed to parse (Can't write to or create math temp directory): \neg v_i

. Each variable Failed to parse (Can't write to or create math temp directory): v_i

can appear multiple times in the expression.

Module Dependencies Problem

Let Failed to parse (Can't write to or create math temp directory): A, B, C, ...

denote various APIs.

Let Failed to parse (Can't write to or create math temp directory): A_1, A_{1.1}, A_{1.7}, A_{1.11}

denote compatible versions of API Failed to parse (Can't write to or create math temp directory): A

.

Let Failed to parse (Can't write to or create math temp directory): A_1, A_{2.0}, A_{3.1}

denote incompatible versions of API Failed to parse (Can't write to or create math temp directory): A

.

Let Failed to parse (Can't write to or create math temp directory): A_{x.y} > B_{u.v}

denote the fact that version x.y of API A depends on version u.v of API B.

Let Failed to parse (Can't write to or create math temp directory): A_{x.y} \gg B_{u.v}

denote the fact that version x.y of API A depends on version u.v of API B and that it re-exports B's elements.

Let Repository Failed to parse (Can't write to or create math temp directory): R=(M,D)

be any set of modules with their various versions and their dependencies on other modules with or without re-export.

Let C be a Configuration in a repository Failed to parse (Can't write to or create math temp directory): R=(M,D) , if Failed to parse (Can't write to or create math temp directory): C \subseteq M , where following is satisfied:

Failed to parse (Can't write to or create math temp directory): \forall A_{x.y} \in C, \forall A_{x.y} \gg B_{u.v} \in D \Rightarrow \exists w >= v \wedge B_{u.w} \in C

- each re-exported dependency is satisfied with some compatible version

Failed to parse (Can't write to or create math temp directory): \forall A_{x.y} \in C, \forall A_{x.y} > B_{u.v} \in D \Rightarrow \exists w >= v B_{u.w} \in C

- each dependency is satisfied with some compatible version

Let there be two chains of re-exported dependencies Failed to parse (Can't write to or create math temp directory): A_{p.q} \gg ... \gg B_{x.y}

and Failed to parse (Can't write to or create math temp directory): A_{p.q} \gg ... \gg B_{u.v}
then Failed to parse (Can't write to or create math temp directory): x = u \wedge y = v
- this guarantees that each class has just one, exact meaning for each importer

Module Dependency Problem

Let there be a repository Failed to parse (Can't write to or create math temp directory): R=(M,D)

and a module Failed to parse (Can't write to or create math temp directory): A \in M

. Does there exist a configuration Failed to parse (Can't write to or create math temp directory): C

in the repository Failed to parse (Can't write to or create math temp directory): R

, such that the module Failed to parse (Can't write to or create math temp directory): A \in C , e.g. the module can be enabled?

Conversion of 3SAT to Module Dependencies Problem

Let there be 3SAT formula with with variables Failed to parse (Can't write to or create math temp directory): v_1, ..., v_m

as defined above.

Let's create a repository of modules Failed to parse (Can't write to or create math temp directory): R . For each variable Failed to parse (Can't write to or create math temp directory): v_i

let's create two modules Failed to parse (Can't write to or create math temp directory): M^i_{1.0}
and Failed to parse (Can't write to or create math temp directory): M^i_{2.0}

, which are mutually incompatible and put them into repository Failed to parse (Can't write to or create math temp directory): R .

For each formula Failed to parse (Can't write to or create math temp directory): (x_{i1} \vee x_{i2} \vee x_{i3})

let's create a module Failed to parse (Can't write to or create math temp directory): F^i

that will have three compatible versions. Each of them will depend on one variable's module. In case the variable is used with negation, it will depend on version 2.0, otherwise on version 1.0. So for the formula 

Failed to parse (Can't write to or create math temp directory): v_a \vee \neg v_b \vee \neg v_c

we will get:

Failed to parse (Can't write to or create math temp directory): F^i_{1.1} \gg M^a_{1.0}

Failed to parse (Can't write to or create math temp directory): F^i_{1.2} \gg M^b_{2.0}

Failed to parse (Can't write to or create math temp directory): F^i_{1.3} \gg M^c_{2.0}

All these modules and dependencies add into repository Failed to parse (Can't write to or create math temp directory): R

Now we will create a module Failed to parse (Can't write to or create math temp directory): T_{1.0}

that depends all formulas: 

Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^1_{1.0}

Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^2_{1.0}

...

Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^n_{1.0}

and add this module as well as its dependencies into repository Failed to parse (Can't write to or create math temp directory): R .

Claim: There Failed to parse (Can't write to or create math temp directory): \exists C

(a configuration) of repository Failed to parse (Can't write to or create math temp directory): R
and Failed to parse (Can't write to or create math temp directory): T_{1.0} \in C
Failed to parse (Can't write to or create math temp directory): \Longleftrightarrow
there is a solution to the 3SAT formula.

Proof

"Failed to parse (Can't write to or create math temp directory): \Leftarrow ": Let's have an evaluation of each variable to either true or false that evaluates the whole 3SAT formula to true. Then

Failed to parse (Can't write to or create math temp directory): C = \{ T_{1.0} \} \bigcup

Failed to parse (Can't write to or create math temp directory): \{ M^i_{1.0} : v_i \} \bigcup \{M^i_{2.0} : \neg v_i \} \bigcup

Failed to parse (Can't write to or create math temp directory): \{ F^i_{1.1} : x_{i1} \} \bigcup \{ F^i_{1.2} : \neg x_{i1} \wedge x_{i2} \} \bigcup \{ F^i_{1.3} : \neg x_{i1} \wedge \neg x_{i2} \wedge x_{i3} \}

It is clear from the definition that each Failed to parse (Can't write to or create math temp directory): M^i

and Failed to parse (Can't write to or create math temp directory): F^i
can be in the Failed to parse (Can't write to or create math temp directory): C
just in one version. Now it is important to ensure that each module is present always at least in one version. This is easy for Failed to parse (Can't write to or create math temp directory): M^i
as its Failed to parse (Can't write to or create math temp directory): v_i
needs to be true or false, and that means one of Failed to parse (Can't write to or create math temp directory): M^i_{1.0}
or  Failed to parse (Can't write to or create math temp directory): M^i_{2.0}
will be included. Can there be a Failed to parse (Can't write to or create math temp directory): F^i
which is not included? Only if Failed to parse (Can't write to or create math temp directory): \neg x_{i1} \wedge \neg x_{i2} \wedge \neg x_{i3}
but that would mean the whole 3-or would evaluate to false and as a result also the 3SAT formula would evaluate to false. This means that dependencies of Failed to parse (Can't write to or create math temp directory): T_{1.0}
on Failed to parse (Can't write to or create math temp directory): F^i
modules are satisfied. Are also dependencies of every Failed to parse (Can't write to or create math temp directory): F^i_{1.q}
satisfied? From all the three versions, there is just one Failed to parse (Can't write to or create math temp directory): F^i_{1.q}

, the one its Failed to parse (Can't write to or create math temp directory): x_{iq}

evaluates to true. However Failed to parse (Can't write to or create math temp directory): x_{iq}
can either be without negation, and as such Failed to parse (Can't write to or create math temp directory): F^i_{1.q}
depends on Failed to parse (Can't write to or create math temp directory): M^j_{1.0}
which is included as Failed to parse (Can't write to or create math temp directory): v_j
is true. Or Failed to parse (Can't write to or create math temp directory): x_{iq}
contains negation, and as such Failed to parse (Can't write to or create math temp directory): F^i_{1.q}
depends on Failed to parse (Can't write to or create math temp directory): M^j_{2.0}
which is included as Failed to parse (Can't write to or create math temp directory): v_j
is false. 

"Failed to parse (Can't write to or create math temp directory): \Rightarrow ": Let's have a Failed to parse (Can't write to or create math temp directory): C

configuration satisfies all dependencies of Failed to parse (Can't write to or create math temp directory): T_{1.0}

. Can we also find positive valuation of 3SAT formula?

For Failed to parse (Can't write to or create math temp directory): i -th 3-or there is Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^i_{1.0}

dependency which is satisfied. At least by one from Failed to parse (Can't write to or create math temp directory): F^i_{1.0}
or Failed to parse (Can't write to or create math temp directory): F^i_{1.1}
or Failed to parse (Can't write to or create math temp directory): F^i_{1.2}

. The one Failed to parse (Can't write to or create math temp directory): F^i

that has the satisfied dependency reexports Failed to parse (Can't write to or create math temp directory): M^j_{1.0}
(which means Failed to parse (Can't write to or create math temp directory): v_j = true

) or Failed to parse (Can't write to or create math temp directory): M^j_{2.0}

(which means Failed to parse (Can't write to or create math temp directory): v_j = false

). Anyway each Failed to parse (Can't write to or create math temp directory): i

3-or evaluates to Failed to parse (Can't write to or create math temp directory): true

.

The only remaining question is whether a Failed to parse (Can't write to or create math temp directory): C

configuration can force truth variable Failed to parse (Can't write to or create math temp directory): v_j
to be true in one 3-or and false in another. However that would mean that one Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^i_{1.x} \gg M^j_{1.0}
and also Failed to parse (Can't write to or create math temp directory): T_{1.0} \gg F^p_{1.u} \gg M^j_{2.0}

. However those two chain of dependencies cannot be in one Failed to parse (Can't write to or create math temp directory): C

as that would break the last condition of configuration definition.

qed.