# LibraryReExportIsNPComplete

### From APIDesign

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- | Reusing libraries | + | Reusing libraries produced by others is essential aspect of [[DistributedDevelopment]]. It simplifies [[Time To Market]], it reduces long term cost of ownership and leads to creation of [[Good Technology|good technologies]]. However it does not comes for free. Read details or directly jump to the [[LibraryReExportIsNPComplete#Implications|implications]] that shall improve your every day development habits. |

- | This page starts by describing a way to convert any [[3SAT]] problem to a solution of finding | + | This page starts by describing a way to convert any [[3SAT]] problem to a solution of finding whether there is a way to satisfy all dependencies of a library in a repository of libraries. Thus proving that the later problem is [[wikipedia::NP-complete|NP-Complete]]. Then it describes the importance of such observations on our [[DistributedDevelopment|development practices]]. |

== [[3SAT]] == | == [[3SAT]] == | ||

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where each <math>x_{ab}</math> is a variable <math>v_i</math> or a negation of a variable <math>\neg v_i</math>. Each variable <math>v_i</math> can appear multiple times in the expression. | where each <math>x_{ab}</math> is a variable <math>v_i</math> or a negation of a variable <math>\neg v_i</math>. Each variable <math>v_i</math> can appear multiple times in the expression. | ||

- | == Library Versioning | + | == Library Versioning Terminology == |

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

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:Let there be two chains of re-exported dependencies <math>A_{p.q} \gg ... \gg B_{x.y}</math> and <math>A_{p.q} \gg ... \gg B_{u.v}</math> then <math>x = u \wedge y = v</math> - this guarantees that each class has just one, exact meaning for each importer | :Let there be two chains of re-exported dependencies <math>A_{p.q} \gg ... \gg B_{x.y}</math> and <math>A_{p.q} \gg ... \gg B_{u.v}</math> then <math>x = u \wedge y = v</math> - this guarantees that each class has just one, exact meaning for each importer | ||

- | + | == Module Dependency Problem == | |

Let there be a repository <math>R=(M,D)</math> and a module <math>A \in M</math>. Does there exist a configuration <math>C</math> in the repository <math>R</math>, such that the module <math>A \in C</math>, e.g. the module can be enabled? | Let there be a repository <math>R=(M,D)</math> and a module <math>A \in M</math>. Does there exist a configuration <math>C</math> in the repository <math>R</math>, such that the module <math>A \in C</math>, e.g. the module can be enabled? | ||

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For each formula | For each formula | ||

<math>(x_{i1} \vee x_{i2} \vee x_{i3})</math> | <math>(x_{i1} \vee x_{i2} \vee x_{i3})</math> | ||

- | let's create a module <math>F^i</math> 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 | + | let's create a module <math>F^i</math> 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 formula |

:<math>v_a \vee \neg v_b \vee \neg v_c</math> | :<math>v_a \vee \neg v_b \vee \neg v_c</math> | ||

we will get: | we will get: | ||

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:<math>F^i_{1.2} \gg M^b_{2.0}</math> | :<math>F^i_{1.2} \gg M^b_{2.0}</math> | ||

:<math>F^i_{1.3} \gg M^c_{2.0}</math> | :<math>F^i_{1.3} \gg M^c_{2.0}</math> | ||

- | All these modules and dependencies | + | All these modules and dependencies are added into repository <math>R</math> |

Now we will create a module <math>T_{1.0}</math> that depends all formulas: | Now we will create a module <math>T_{1.0}</math> that depends all formulas: |

## Revision as of 19:38, 26 August 2009

Reusing libraries produced by others is essential aspect of DistributedDevelopment. It simplifies Time To Market, it reduces long term cost of ownership and leads to creation of good technologies. However it does not comes for free. Read details or directly jump to the implications that shall improve your every day development habits.

This page starts by describing a way to convert any 3SAT problem to a solution of finding whether there is a way to satisfy all dependencies of a library in a repository of libraries. Thus proving that the later problem is NP-Complete. Then it describes the importance of such observations on our development practices.

## 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.

## Library Versioning Terminology

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 formula

we will get:

All these modules and dependencies are added 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.

"": Let's have a *C* configuration satisfies all dependencies of *T*_{1.0}. Can we also find positive valuation of 3SAT formula?

For *i*-th *3-or* there is dependency which is satisfied. At least by one from or or . The one *F*^{i} that has the satisfied dependency reexports (which means *v*_{j} = *t**r**u**e*) or (which means *v*_{j} = *f**a**l**s**e*). Anyway each *i* *3-or* evaluates to *t**r**u**e*.

The only remaining question is whether a *C* configuration can force truth variable *v*_{j} to be true in one *3-or* and false in another. However that would mean that there is re-export via and also another one via . However those two *chain of dependencies* ending in different versions of *M*^{j} cannot be in one *C* as that breaks the last condition of configuration definition. Thus each *M*^{j} is represented only by one version and each *v*_{j} is evaluated either to true or false, but never both.

The 3SAT formula's evaluation based on the configuration *C* is consistent and satisfies the formula.

**qed**.

## Implications

If there is a repository of modules in various (incompatible) versions, with mutual dependencies and re-export of their APIs, then deciding whether some of them can be enabled is NP-complete problem.

As NP-complete problems are hard to solve, it is usually our best desire to avoid them in real life situations. What does that mean in case one decides to practise DistributedDevelopment (and it is inevitable that software for 21st century needs this development style)? If you want to avoid headaches with finding the right configuration of various version of the libraries that allows to execute them together, then stick with following simple rules.

#### Be Compatible!

If you develop your own libraries in backward compatible way, you can always select the most recent version of each library. That is the configuration you are looking for. It is easy to find (obviously) and also it is the most desirable, as it delivers the most modern features and bugfixes that users of such libraries want.

#### Reuse with Care!

If you happen to reuse libraries (and you should because reuse lowers Time_To_Market, just like it did for me when I was publishing my first animated movie), then choose such libraries that can be trusted to evolve compatibly.

#### Hide Incompatibilities!

If you happen to reuse a library that cannot be trusted to keep its BackwardCompatibility, then do whatever you can to not re-export its APIs! This has been discussed in Chapter 10, Cooperating with Other APIs, but in short: If you hide such library for internal use and do not export any of its interfaces, you can use whatever version of library you want (even few years older) and nobody shall notice. Moreover in many module systems there can even be multiple versions of the same library in case they are not re-exported.

## Conclusion

Avoid complexities and NP-complete problems. Learn to develop in backward compatible way. Reading TheAPIBook is a perfect entry point into such compatible software design of the 21st century.