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{{Short description|Theorem in computability theory}}
{{distinguish|text=[[Kleene's theorem]] for regular languages}}
{{Use shortened footnotes|date=May 2021}}
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== Rogers's fixed-point theorem ==
Given a function <math>F</math>, a '''fixed point''' of <math>F</math> is an index <math>e</math> such that <math>\varphi_e \simeq \varphi_{F(e)}</math>. Note that the comparison of in- and outputs here is not in terms of numerical values, but in terms of their associated functions.
Rogers describes the following result as "a simpler version" of Kleene's (second) recursion theorem.{{sfn|Rogers|1967|loc=§11.2}} {{math theorem | name = Rogers's fixed-point theorem | math_statement = If <math>F</math> is a total computable function, it has a fixed point in the above sense.}}
This essentially means that if we apply an [[Effectiveness|effective]] transformation to programs (say, replace instructions such as successor, jump, remove lines), there will always be a program whose behaviour is not altered by the transformation. This theorem can therefore be interpreted in the following manner: “given any effective procedure to transform programs, there is always a program that, when modified by the procedure, does exactly what it did before”, or: “it’s impossible to write a program that changes the extensional behaviour of all programs”.
=== Proof of the fixed-point theorem ===
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:'''The second recursion theorem'''. For any partial recursive function <math>Q(x,y)</math> there is an index <math>p</math> such that <math>\varphi_p \simeq \lambda y.Q(p,y)</math>.
The theorem can be proved from Rogers's theorem by letting <math>F(p)</math> be a function such that <math>\varphi_{F(p)}(y) = Q(p,y)</math> (a construction described by the [[Smn theorem|S-m-n theorem]]). One can then verify that a fixed-point of this <math>F</math> is an index <math>p</math> as required. The theorem is constructive in the sense that a fixed computable function maps an index for
=== Comparison to Rogers's theorem ===
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== The first recursion theorem ==
While the second recursion theorem is about fixed points of computable functions, the first recursion theorem is related to fixed points determined by enumeration operators, which are a computable analogue of inductive definitions. An '''enumeration operator''' is a set of pairs (''A'',''n'') where ''A'' is a ([[Gödel number|code]] for a) finite set of numbers and ''n'' is a single [[natural number]]. Often, ''n'' will be viewed as a code for an ordered pair of natural numbers, particularly when functions are defined via enumeration operators. Enumeration operators are of central importance in the study of [[enumeration reducibility]].
Each enumeration operator Φ determines a function from sets of naturals to sets of naturals given by
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:# For any computable enumeration operator Φ there is a recursively enumerable set ''F'' such that Φ(''F'') = ''F'' and ''F'' is the smallest set with this property.
:# For any recursive operator Ψ there is a partial computable function φ such that Ψ(φ) = φ and φ is the smallest partial computable function with this property.
The first recursion theorem is also called Fixed point theorem (of recursion theory).<ref>{{Cite book |last=Cutland |first=Nigel |title=Computability: an introduction to recursive function theory}}</ref> There is also a definition which can be applied to [[Primitive recursive functional|recursive functionals]] as follows:
Let <math>\Phi: \mathbb{F}(\mathbb{N}^k) \rightarrow (\mathbb{N}^k)</math> be a recursive functional. Then <math>\Phi</math> has a least fixed point <math>f_{\Phi}: \mathbb{N}^k \rightarrow \mathbb{N}</math> which is computable i.e.
1) <math>\Phi(f_{\phi})=f_{\Phi}</math>
2) <math>\forall g \in \mathbb{F}(\mathbb{N}^k)</math> such that <math>\Phi(g)=g</math> it holds that <math>f_{\Phi}\subseteq g</math>
3) <math>f_{\Phi}</math> is computable
=== Example ===
Like the second recursion theorem, the first recursion theorem can be used to obtain functions satisfying systems of recursion equations. To apply the first recursion theorem, the recursion equations must first be recast as a recursive operator.
Consider the recursion equations for the [[factorial]] function ''f'':<math display="block">\begin{align}
&f(0) = 1 \\
&f(n+1) = (n + 1) \cdot f(n)
\end{align}</math>The corresponding recursive operator Φ will have information that tells how to get to the next value of ''f'' from the previous value. However, the recursive operator will actually define the graph of ''f''. First, Φ will contain the pair <math>( \varnothing, (0, 1))</math>. This indicates that ''f''(0) is unequivocally 1, and thus the pair (0,1) is in the graph of ''f''.▼
▲The corresponding recursive operator Φ will have information that tells how to get to the next value of ''f'' from the previous value. However, the recursive operator will actually define the graph of ''f''. First, Φ will contain the pair <math>( \varnothing, (0, 1))</math>. This indicates that ''f''(0) is unequivocally 1, and thus the pair (0,1) is in the graph of ''f''.
Next, for each ''n'' and ''m'', Φ will contain the pair <math>( \{ (n, m) \}, (n+1, (n+1)\cdot m))</math>. This indicates that, if ''f''(''n'') is ''m'', then {{nowrap|''f''(''n'' + 1)}} is {{nowrap|(''n'' + 1)''m''}}, so that the pair {{nowrap|(''n'' + 1, (''n'' + 1)''m'')}} is in the graph of ''f''. Unlike the base case {{nowrap|1=''f''(0) = 1}}, the recursive operator requires some information about ''f''(''n'') before it defines a value of {{nowrap|''f''(''n'' + 1)}}.
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The first recursion theorem (in particular, part 1) states that there is a set ''F'' such that {{nowrap|1=Φ(''F'') = F}}. The set ''F'' will consist entirely of ordered pairs of natural numbers, and will be the graph of the factorial function ''f'', as desired.
The restriction to recursion equations that can be recast as recursive operators ensures that the recursion equations actually define a [[least fixed point]]. For example, consider the set of recursion equations:<math display="block">\begin{align}
&g(0) = 1\\
&g(n + 1) = 1\\
&g(2n) = 0
\end{align}</math>There is no function ''g'' satisfying these equations, because they imply ''g''(2) = 1 and also imply ''g''(2) = 0. Thus there is no fixed point ''g'' satisfying these recursion equations. It is possible to make an enumeration operator corresponding to these equations, but it will not be a recursive operator.▼
▲There is no function ''g'' satisfying these equations, because they imply ''g''(2) = 1 and also imply ''g''(2) = 0. Thus there is no fixed point ''g'' satisfying these recursion equations. It is possible to make an enumeration operator corresponding to these equations, but it will not be a recursive operator.
=== Proof sketch for the first recursion theorem ===
The proof of part 1 of the first recursion theorem is obtained by iterating the enumeration operator Φ beginning with the [[empty set]]. First, a sequence ''F''<sub>''k''</sub> is constructed, for <math>k = 0, 1, \ldots</math>. Let ''F''<sub>0</sub> be the empty set. Proceeding inductively, for each ''k'', let ''F''<sub>''k'' + 1</sub> be <math>F_k \cup \Phi(F_k)</math>. Finally, ''F'' is taken to be <math display="inline">\bigcup F_k</math>. The remainder of the proof consists of a verification that ''F'' is recursively enumerable and is the least fixed point of Φ. The sequence ''F''<sub>''k''</sub> used in this proof corresponds to the Kleene chain in the proof of the [[Kleene fixed-point theorem]].
The second part of the first recursion theorem follows from the first part. The assumption that Φ is a recursive operator is used to show that the fixed point of Φ is the graph of a partial function. The key point is that if the fixed point ''F'' is not the graph of a function, then there is some ''k'' such that ''F''<sub>''k''</sub> is not the graph of a function.
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* [[Denotational semantics]], where another least fixed point theorem is used for the same purpose as the first recursion theorem.
* [[Fixed-point combinator]]s, which are used in [[lambda calculus]] for the same purpose as the first recursion theorem.
* [[Diagonal lemma]] a closely related result in mathematical logic.
== References ==
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