Disjunctive normal form

In boolean logic, a disjunctive normal form (DNF) is a standardization (or normalization) of a logical formula which is a disjunction of conjunctive clauses; it can also be described as an OR of ANDs, a sum of products, or (in philosophical logic) a cluster concept. As a normal form, it is useful in automated theorem proving.

Definition

A logical formula is considered to be in DNF if and only if it is a disjunction of one or more conjunctions of one or more literals.[1]:153 A DNF formula is in full disjunctive normal form if each of its variables appears exactly once in every clause. As in conjunctive normal form (CNF), the only propositional operators in DNF are and, or, and not. 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 formal grammar for DNF:

  1. disjunction → (conjunctiondisjunction)
  2. disjunctionconjunction
  3. conjunction → (literalconjunction)
  4. conjunctionliteral
  5. literal → ¬variable
  6. literalvariable

Where variable is any variable.

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

However, the following formulas are not in DNF:

Conversion to DNF

Converting a formula to DNF involves using logical equivalences, such as the double negative elimination, De Morgan's laws, and the distributive law.

All logical formulas can be converted into an equivalent disjunctive normal form.[1]:152-153 However, in some cases conversion to DNF can lead to an exponential explosion of the formula. For example, in DNF, logical formulas of the following form have 2n terms:

(X_1 \or Y_1) \and (X_2 \or Y_2) \and \dots \and (X_n \or Y_n)

Any particular Boolean function can be represented by one and only one[note 1] full disjunctive normal form, one of the canonical forms.

Complexity issues

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 clause contains at most k literals. Dually to CNFs, the problem of deciding whether a given DNF is true for every variable assignment is NP-complete, the same holds if only k-DNFs are considered.

See also

Notes

  1. Ignoring variations based on associativity and commutativity of AND and OR.

References

  1. 1 2 B.A. Davey and H.A. Priestley (1990). Introduction to Lattices and Order. Cambridge Mathematical Textbooks. Cambridge University Press.

External links

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