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| {{About|algorithmic word problems in mathematics and computer science|other uses|Word problem (disambiguation){{!}}Word problem}}
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| In [[mathematics]] and [[computer science]], a '''word problem''' for a set S with respect to a system of finite encodings of its elements is the [[decision problem|algorithmic problem of deciding]] whether two given representatives represent the same element of the set. The problem is commonly encountered in [[abstract algebra]], where given a presentation of an algebraic structure by [[generating set|generators]] and [[relator]]s, the problem is to determine if two expressions represent the same element; a prototypical example is the [[word problem for groups]]. Less formally, the word problem in an algebra is: given a set of identities ''E'', and two expressions ''x'' and ''y'', is it possible to transform ''x'' into ''y'' using the identities in ''E'' as [[rewriting]] rules in both directions? While answering this question may not seem hard, the remarkable (and [[deep result|deep]]) result that emerges, in many important cases, is that the [[undecidable problem|problem is undecidable]].
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| Many, if not most all, undecidable problems in mathematics can be posed as word problems; see the [[list of undecidable problems]] for many examples.
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| == Background and motivation ==
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| Many occasions arise in mathematics where one wishes to use a finite amount of information to describe an element of a (typically infinite) set. This issue is particularly apparent in computational mathematics. Traditional models of computation (such as the [[Turing machine]]) have storage capacity which is unbounded, so it is in principle possible to perform computations with the elements of infinite sets. On the other hand, since the amount of storage space in use at any one time is finite, we need each element to have a finite representation.
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| For various reasons, it is not always possible or desirable to use a system of ''unique'' encodings, that is, one in which every element has a single encoding. When using an encoding system without uniqueness, the question naturally arises of whether there is an algorithm which, given as input two encodings, decides whether they represent the same element. Such an algorithm is called a ''solution to the word problem'' for the encoding system.
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| == The word problem in combinatorial calculus ==
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| {{main | Combinatory logic#Undecidability of combinatorial calculus}}
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| The simplest example of an undecidable word problem occurs in [[combinatory logic]]: when are two strings of combinators equivalent? Because combinators encode all possible [[Turing machine]]s, and the equivalence of two Turing machines is undecidable, it follows that the equivalence of two strings of combinators is undecidable.
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| Likewise, one has essentially the same problem in [[lambda calculus]]: given two distinct lambda expressions, there is no algorithm which can discern whether they are equivalent or not; [[Lambda calculus#Undecidability of equivalence|equivalence is undecidable]].
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| == The word problem in universal algebra ==
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| In [[algebra]], one often studies infinite algebras which are generated (under the [[finitary]] operations of the algebra) by finitely many elements. In this case, the elements of the algebra have a natural system of finite encoding as expressions in terms of the generators and operations. The word problem here is thus to determine, given two such expressions, whether they represent the same element of the algebra.
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| Roughly speaking, the word problem in an algebra is: given a set ''E'' of identities (an [[equational theory]]), and two [[Term (logic)|terms]] ''s'' and ''t'', is it possible to transform ''s'' into ''t'' using the identities in ''E'' as [[rewriting]] rules in both directions?.<ref>Franz Baader and Tobias Nipkow, ''Term Rewriting and All That'', Cambridge University Press, 1998, p. 5</ref>
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| <!--- wrong: The act of discovering such equivalences is known as [[unification (computer science)|unification]]. --->
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| A proper extension of the ''word problem'' is known as the ''[[unification (computer science)|unification problem]]'' (a.k.a. as ''equation solving problem'').
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| While the former asks whether two terms ''are'' equal, the latter asks whether they have ''instances'' that are equal.
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| As a common example, "<math>2 + 3 \stackrel{?}{=} 8 + (-3)</math>" is a word problem in the [[Integer#Algebraic properties|integer group ℤ]],
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| while "<math>2 + x \stackrel{?}{=} 8 + (-x)</math>" is a unification problem in the same group; since the former terms happen to be equal in ℤ, the latter problem has the [[substitution (logic)|substitution]] <math>\{x \mapsto 3\}</math> as a solution.
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| Substitutions may be ordered into a [[partial order]], thus, unification is the act of finding a [[join]] on a [[lattice (order)|lattice]]. {{clarify|reason=In which lattice precisely? Unification maps two terms to a substitution, while term lattice meet maps two terms to a term, and substitution lattice meet maps two substitutions to a substitution.|date=July 2013}}
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| In this sense, the word problem on a lattice has a solution, namely, the set of all equivalent words is the [[free lattice]].{{clarify|reason=The solution of a word problem was explained above to be an algorithm, not a set.|date=June 2013}}
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| One of the most deeply studied cases of the word problem is in the theory of [[semigroup]]s and [[group (mathematics)|group]]s.
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| There are [[word problem for groups|many groups for which the word problem]] is not [[Decidability (logic)|decidable]], in that there is no Turing machine that can determine the equivalence of two ''arbitrary'' words in a finite time.
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| The word problem on [[ground term]]s is not decidable.<ref>Yuri Matijasevich, (1967) "Simple examples of undecidable associative calculi", ''Soviet Mathematics Doklady'' '''8'''(2) p 555-557.</ref> {{clarify|reason=In which algebra? It is certainly decidable in ℤ|date=June 2013}}
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| The word problem on free [[Heyting algebra]]s is difficult.<ref>Peter T. Johnstone, ''Stone Spaces'', (1982) Cambridge University Press, Cambridge, ISBN 0-521-23893-5. ''(See chapter 1, paragraph 4.11)''</ref>
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| The only known results are that the free Heyting algebra on one generator is infinite, and that the free [[complete Heyting algebra]] on one generator exists (and has one more element than the free Heyting algebra).
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| ==Example: A term rewriting system to decide the word problem in the free group==
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| Bläsius and Bürckert
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| <ref>{{cite book| title=Deduktionsssysteme| year=1992| pages=291| publisher=Oldenbourg| editor=K. H. Bläsius and H.-J. Bürckert| accessdate=30 June 2013}}; here: p.126, 134</ref>
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| demonstrate the [[Knuth-Bendix algorithm]] on an axiom set for groups.
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| The algorithm yields a [[Confluence (abstract rewriting)|confluent]] and [[Abstract rewriting system#Termination and convergence|noetherian]] [[rewrite system#Term rewriting systems|term rewrite system]] that transforms every term into a unique [[Normal form (abstract rewriting)|normal form]].<ref>Apply rules in any order to a term, as long as possible; the result doesn't depend on the order; it is the term's normal form.</ref>
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| The rewrite rules are numbered incontiguous since some rules became redundant and were deleted during the algorithm run.
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| The equality of two terms follows from the axioms if and only if both terms are transformed into literally the same normal form term. For example, the terms
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| :<math>((a^{-1} \cdot a) \cdot (b \cdot b^{-1}))^{-1} \stackrel{R2}{\rightsquigarrow} (1 \cdot (b \cdot b^{-1}))^{-1} \stackrel{R13}{\rightsquigarrow} (1 \cdot 1)^{-1} \stackrel{R1}{\rightsquigarrow} 1 ^{-1} \stackrel{R8}{\rightsquigarrow} 1</math>, and | |
| :<math>b \cdot ((a \cdot b)^{-1} \cdot a) \stackrel{R17}{\rightsquigarrow} b \cdot ((b^{-1} \cdot a^{-1}) \cdot a) \stackrel{R3}{\rightsquigarrow} b \cdot (b^{-1} \cdot (a^{-1} \cdot a)) \stackrel{R2}{\rightsquigarrow} b \cdot (b^{-1} \cdot 1) \stackrel{R11}{\rightsquigarrow} b \cdot b^{-1} \stackrel{R13}{\rightsquigarrow} 1</math>
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| share the same normal form, viz. <math>1</math>; therefor both terms are equal in every group.
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| As another example, the term <math>1 \cdot (a \cdot b)</math> and <math>b \cdot (1 \cdot a)</math> has the normal form <math>a \cdot b</math> and <math>b \cdot a</math>, respectively. Since the normal forms are literally different, the original terms cannot be equal in every group. In fact, they are usually different in [[abelian group|non-abelian groups]].
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| {| style="border: 1px solid grey; float: left; margin: 1em 1em;"
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| |+ Group axioms used in Knuth-Bendix completion
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| |-
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| | '''A1''' || <math>1 \cdot x</math> || <math>= x</math>
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| |-
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| | '''A2''' || <math>x^{-1} \cdot x</math> || <math>= 1</math>
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| |-
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| | '''A3''' || <math>(x \cdot y) \cdot z</math> || <math>= x \cdot (y \cdot z)</math>
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| |}
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| {| style="border: 1px solid grey; float: left; margin: 1em 1em;"
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| |+ Term rewrite system obtained from Knuth-Bendix completion
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| |-
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| | '''R1''' || <math>1 \cdot x </math> || <math>\rightsquigarrow x</math>
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| | '''R2''' || <math>x^{-1} \cdot x </math> || <math>\rightsquigarrow 1</math>
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| | '''R3''' || <math>(x \cdot y) \cdot z</math> || <math>\rightsquigarrow x \cdot (y \cdot z)</math>
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| |-
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| | '''R4''' || <math>x^{-1} \cdot (x \cdot y) </math> || <math>\rightsquigarrow y</math>
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| | '''R8''' || <math>1^{-1} </math> || <math>\rightsquigarrow 1</math>
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| | '''R11''' || <math>x \cdot 1 </math> || <math>\rightsquigarrow x</math>
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| | '''R12''' || <math>(x^{-1})^{-1} </math> || <math>\rightsquigarrow x</math>
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| | '''R13''' || <math>x \cdot x^{-1} </math> || <math>\rightsquigarrow 1</math>
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| |-
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| | '''R14''' || <math>x \cdot (x^{-1} \cdot y) </math> || <math>\rightsquigarrow y</math>
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| |-
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| | '''R17''' || <math>(x \cdot y)^{-1} </math> || <math>\rightsquigarrow y^{-1} \cdot x^{-1}</math>
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| |}
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| {{clear}}
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| ==See also==
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| * [[Munn tree]]
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| * [[Word problem for groups]]
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| * [[Knuth-Bendix completion algorithm]]
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| * [[Unification (computer science)]]
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| ==References==
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| {{reflist}}
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| [[Category:Abstract algebra]]
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| [[Category:Combinatorics on words]]
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| [[Category:Rewriting systems]]
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| [[Category:Computational problems]]
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Nice to satisfy you, my name is Refugia. To perform baseball is the hobby he will by no means stop performing. Managing people is what I do and the wage has been truly satisfying. South Dakota is exactly where I've usually been living.
Visit my website - std testing at home (see this site)