Английская Википедия:Disjunctive normal form

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Шаблон:Short description In boolean logic, a disjunctive normal form (DNF) is a canonical normal form of a logical formula consisting of a disjunction of conjunctions; it can also be described as an OR of ANDs, a sum of products, or Шаблон:Mdash in philosophical logic Шаблон:Mdash a cluster concept.Шаблон:Sfn As a normal form, it is useful in automated theorem proving.

Definition

A logical formula is considered to be in DNF if it is a disjunction of one or more conjunctions of one or more literals.Шаблон:SfnШаблон:SfnШаблон:Sfn A DNF formula is in full disjunctive normal form if each of its variables appears exactly once in every conjunction and each conjunction appears at most once (up to the order of variables). As in conjunctive normal form (CNF), the only propositional operators in DNF are and (<math>\wedge</math>), or (<math>\vee</math>), and not (<math>\neg</math>). The not operator can only be used as part of a literal, which means that it can only precede a propositional variable.

The following is a context-free grammar for DNF:

  1. DNF → (Conjunction) <math>\vee</math> DNF
  2. DNF → (Conjunction)
  3. Conjunction → Literal <math>\wedge</math> Conjunction
  4. Conjunction → Literal
  5. Literal → <math>\neg</math>Variable
  6. Literal → Variable

Where Variable is any variable.

For example, all of the following formulas are in DNF:

  • <math>(A \land \neg B \land \neg C) \lor (\neg D \land E \land F \land D \land F)</math>
  • <math>(A \land B) \lor (C)</math>
  • <math>(A \land B)</math>
  • <math>(A)</math>

The formula <math>A \lor B</math> is in DNF, but not in full DNF; an equivalent full-DNF version is <math>(A \land B) \lor (A \land \lnot B) \lor (\lnot A \land B)</math>.

The following formulas are not in DNF:[1]

  • <math>\neg(A \lor B)</math>, since an OR is nested within a NOT
  • <math>\neg(A \land B) \lor C</math>, since an AND is nested within a NOT
  • <math>A \lor (B \land (C \lor D))</math>, since an OR is nested within an AND

Conversion to DNF

In classical logic each propositional formula can be converted to DNFШаблон:Sfn ...

Файл:Karnaugh map KV 4mal4 18.svg
Karnaugh map of the disjunctive normal form (¬A∧¬B∧¬D) ∨ (¬A∧B∧C) ∨ (A∧B∧D) ∨ (A∧¬B∧¬C)
Файл:Karnaugh map KV 4mal4 19.svg
Karnaugh map of the disjunctive normal form (¬A∧C∧¬D) ∨ (B∧C∧D) ∨ (A∧¬C∧D) ∨ (¬B∧¬C∧¬D). Despite the different grouping, the same fields contain a "1" as in the previous map.

... by syntactic means

The conversion involves using logical equivalences, such as double negation elimination, De Morgan's laws, and the distributive law. Formulas built from the primitive connectives <math>\{\land,\lor,\lnot\}</math>[2] can be converted to DNF by the following canonical term rewriting system:Шаблон:Sfn

<math>\begin{array}{rcl}

(\lnot \lnot x) & \rightsquigarrow & x \\ (\lnot (x \lor y)) & \rightsquigarrow & ((\lnot x) \land (\lnot y)) \\ (\lnot (x \land y)) & \rightsquigarrow & ((\lnot x) \lor (\lnot y)) \\ (x \land (y \lor z)) & \rightsquigarrow & ((x \land y) \lor (x \land z)) \\ ((x \lor y) \land z) & \rightsquigarrow & ((x \land z) \lor (y \land z)) \\ \end{array}</math>

... by semantic means

The full DNF of a formula can be read off its truth table.Шаблон:Sfn For example, consider the formula

<math>\phi = ((\lnot (p \land q)) \leftrightarrow (\lnot r \uparrow (p \oplus q)))</math>.[3]

The corresponding truth table is

<math>p</math> <math>q</math> <math>r</math> <math>(</math> <math>\lnot</math> <math>(p \land q)</math> <math>)</math> <math>\leftrightarrow</math> <math>(</math> <math>\lnot r</math> <math>\uparrow</math> <math>(p \oplus q)</math> <math>)</math>
T T T F T F F T F
T T F F T F T T F
T F T T F T F T T
T F F T F F T F T
F T T T F T F T T
F T F T F F T F T
F F T T F T F T F
F F F T F T T T F
  • The full DNF equivalent of <math>\phi</math> is
<math>

( p \land \lnot q \land r) \lor (\lnot p \land q \land r) \lor (\lnot p \land \lnot q \land r) \lor (\lnot p \land \lnot q \land \lnot r) </math>

  • The full DNF equivalent of <math>\lnot \phi</math> is
<math>

( p \land q \land r) \lor ( p \land q \land \lnot r) \lor ( p \land \lnot q \land \lnot r) \lor (\lnot p \land q \land \lnot r) </math>

Remark

A propositional formula can be represented by one and only one full DNF.Шаблон:Refn In contrast, several plain DNFs may be possible. For example, by applying the rule <math>((a \land b) \lor (\lnot a \land b)) \rightsquigarrow b</math> three times, the full DNF of the above <math>\phi</math> can be simplified to <math>(\lnot p \land \lnot q) \lor (\lnot p \land r) \lor (\lnot q \land r)</math>. However, there are also equivalent DNF formulas that cannot be transformed one into another by this rule, see the pictures for an example.

Maximum number of conjunctions

Any propositional formula is built from <math>n</math> variables, where <math>n \ge 1</math>.

There are <math>2n</math> possible literals: <math>L = \{ p_1, \lnot p_1, p_2, \lnot p_2, \ldots, p_n, \lnot p_n\}</math>.

<math>L</math> has <math>(2^{2n} -1)</math> non-empty subsets.[4]

This is the maximum number of conjunctions a DNF can have.[5]

A full DNF can have up to <math>2^{n}</math> conjunctions, one for each row of the truth table.

Example 1

Consider a formula with two variables <math>p</math> and <math>q</math>.

The longest possible DNF has <math>2^{(2 \times 2)} -1 = 15</math> conjunctions:[5]

<math>

\begin{array}{lcl} (\lnot p) \lor (p) \lor (\lnot q) \lor (q) \lor \\ (\lnot p \land p) \lor \underline{(\lnot p \land \lnot q)} \lor \underline{(\lnot p \land q)} \lor \underline{( p \land \lnot q)} \lor \underline{( p \land q)} \lor (\lnot q \land q) \lor \\ (\lnot p \land p \land \lnot q) \lor (\lnot p \land p \land q) \lor (\lnot p \land \lnot q \land q) \lor ( p \land \lnot q \land q) \lor \\ (\lnot p \land p \land \lnot q \land q) \end{array}</math>

The longest possible full DNF has 4 conjunctions: they are underlined.

This formula is a tautology.

Example 2

Each DNF of the e.g. formula <math>(X_1 \lor Y_1) \land (X_2 \lor Y_2) \land \dots \land (X_n \lor Y_n)</math> has <math>2^n</math> conjunctions.

Computational complexity

The Boolean satisfiability problem on conjunctive normal form formulas is NP-complete. By the duality principle, so is the falsifiability problem on DNF formulas. Therefore, it is co-NP-hard to decide if a DNF formula is a tautology.

Conversely, a DNF formula is satisfiable if, and only if, one of its conjunctions is satisfiable. This can be decided in polynomial time simply by checking that at least one conjunction does not contain conflicting literals.

Variants

An important variation used in the study of computational complexity is k-DNF. A formula is in k-DNF if it is in DNF and each conjunction contains at most k literals.Шаблон:Sfn

See also

Notes

Шаблон:Reflist

References

Шаблон:Sfn whitelist

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  1. ↑ However, they are in negation normal form.
  2. ↑ Formulas with other connectives can be brought into negation normal form first.
  3. ↑ <math>\phi</math> = ((NOT (p AND q)) IFF ((NOT r) NAND (p XOR q)))
  4. ↑ <math>\left|\mathcal{P}(L)\right| = 2^{2n}</math>
  5. ↑ 5,0 5,1 Ошибка цитирования Неверный тег <ref>; для сносок noreps не указан текст
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