# LibraryReExportIsNPComplete

### From APIDesign

(→Converstion of 3SAT to Module Dependencies Problem) |
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== Module Dependencies Problem == | == Module Dependencies Problem == | ||

- | Let <math>A, B, C, ...</math> denote | + | Let <math>A, B, C, ...</math> denote various [[API]]s. |

Let <math>A_1, A_{1.1}, A_{1.7}, A_{1.11}</math> denote compatible versions of API <math>A</math>. | Let <math>A_1, A_{1.1}, A_{1.7}, A_{1.11}</math> denote compatible versions of API <math>A</math>. |

## Revision as of 11:26, 4 April 2009

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 |

## 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 various APIs.

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 it re-exports *B'*s 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?

## Conversion of 3SAT to Module Dependencies Problem

Let there be 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) of repository *R* and there is a solution to the 3SAT formula.

## Proof

"": Let's have an evaluation of each variable to either true or false that evaluates the whole 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 *3-or* would evaluate to false and as a result also the 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? From all the three versions, there is just one , the one its *x*_{iq} evaluates to true. However *x*_{iq} can either be without negation, and as such depends on which is included as *v*_{j} is true. Or *x*_{iq} contains negation, and as such depends on which is included as *v*_{j} is false. **qed**.