# LibraryReExportIsNPComplete

### From APIDesign

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:<math>\{ M^i_{1.0} : v_i \} \bigcup \{M^i_{2.0} : \neg v_i \} \bigcup</math> | :<math>\{ M^i_{1.0} : v_i \} \bigcup \{M^i_{2.0} : \neg v_i \} \bigcup</math> | ||

:<math>\{ 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} \}</math> | :<math>\{ 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} \}</math> | ||

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+ | It is clear from the definition that each <math>M^i</math> and <math>F^i</math> can be in the <math>C</math> 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 <math>M^i</math> as its <math>v_i</math> needs to be true or false, and that means one of <math>M^i_{1.0}</math> or <math>M^i_{2.0}</math> will be included. Can there be a <math>F^i</math> which is not included? Only if <math>\neg x_{i1} \wedge \neg x_{i2} \wedge \neg x_{i3}</math> but that would mean the whole [[wikipedia::3SAT]] formula would evaluate to false. This means that dependencies of <math>T_{1.0}</math> on <math>F^i</math> modules are satisfied. Are also dependencies of every <math>F^i_{x.y}</math> satisfied. Obviously yes... |

## Revision as of 11:02, 25 May 2008

This page describes a way to convert any wikipedia::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 |

## wikipedia::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:

- ...

where each *x*_{ab} is a variable *v*_{i} or a negation of a variable . Each variable *v*_{i} can appear multiple times in the expression.

## Module Dependencies Problem

Let *A*,*B*,*C*,... denote an API.

Let *A*_{1},*A*_{1.1},*A*_{1.7},*A*_{1.11} denote compatible versions of API *A*.

Let *A*_{1},*A*_{2.0},*A*_{3.1} denote incompatible versions of API *A*.

Let *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 denote the fact that version *x.y* of API A depends on version *u.v* of API *B* and that *B* re-exports its elements.

Let *Repository* *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 *R* = (*M*,*D*), if
, where following is satisfied:

- - each re-exported dependency is satisfied with some compatible version
- - each dependency is satisfied with some compatible version
- Let there be two chains of re-exported dependencies and then - this guarantees that each class has just one, exact meaning for each importer

**Module Dependency Problem**: Let there be a repository *R* = (*M*,*D*) and a module . Does there exist a configuration *C* in the repository *R*, such that the module , e.g. the module can be enabled?

## Converstion of wikipedia::3SAT to Module Dependencies Problem

Let there be wikipedia::3SAT formula with with variables *v*_{1},...,*v*_{m} as defined above.

Let's create a repository of modules *R*. For each variable *v*_{i} let's create two modules and , which are mutually incompatible and put them into repository *R*.

For each formula
let's create a module *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

we will get:

All these modules and dependencies add into repository *R*

Now we will create a module *T*_{1.0} that depends all formulas:

- ...

and add this module as well as its dependencies into repository *R*.

**Claim**: There a configuration *C* of repository *R* and there is a solution to the wikipedia::3SAT formula.

## Proof

"": Let's have an evaluation of each variable to either true or false that evaluates the whole wikipedia::3SAT formula to true. Then

It is clear from the definition that each *M*^{i} and *F*^{i} can be in the *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 *M*^{i} as its *v*_{i} needs to be true or false, and that means one of or will be included. Can there be a *F*^{i} which is not included? Only if but that would mean the whole wikipedia::3SAT formula would evaluate to false. This means that dependencies of *T*_{1.0} on *F*^{i} modules are satisfied. Are also dependencies of every satisfied. Obviously yes...