如果你觉得太孤单,与I相关的函数有Φ函数。
Φ(0,α)=Ω_α
Φ(0,Φ(0,Φ(0,……)))=Φ(1,0)=Ω_Ω_Ω_……=I(φ函数的味)
ψ(Φ(1,1))=ψ(Φ(0,Φ(0,Φ(0,……Φ(1,0)))))=ψ(ψ_I(1))
ψ(Φ(2,0))=ψ(Φ(1,Φ(1,Φ(1,……Φ(1,0)))))=ψ(ψ_I(I))
ψ(Φ(2,1))=ψ(ψ_I(I2))
ψ(Φ(3,0))=ψ(ψ_I(I2))
ψ(Φ(4,0))=ψ(ψ_I(I3))
ψ(Φ(1,0,0))=ψ(ψ_I(I^I))
ψ(ψ_I(I^I^ω))=ψ(Φ(1,0,0,0,……)),I版的SVO。
ψ(ψ_I(I^I^I)),I版的LVO。
ψ(ψ_I(I^I^I^……))=ψ(ψ_I(ψ_(Ω_I+1)(0)))=ψ(ψ_(Ω_I+1)(0)),I版的BHO。
ψ(ψ_(Ω_I+1)(I))=ψ(ψ_(Ω_I+1)(ψ_I(ψ_(Ω_I+1)(……ψ_I(ψ_(Ω_I+1)(0)))))),I被加强了,琢磨一下。
ψ(ψ_(Ω_I+1)(Ω_I+1))=ψ(ψ_(Ω_I+1)(ψ_(Ω_I+1)ψ_(Ω_I+1)……(0)))
α→ψ(ψ_(Ω_I+α))的序数不动点为ψ_I(2)(0),不要认为它是ψ(ψ_Ω_I(Ω_(I2)))。
ψ(I(3))
ψ(I(I(I……(I(I(I))))))=I(1,0)
ψ(I(1,0,0))多元φ函数:又学我
ψ(I(1,0,0,0))
ψ(I(1@4))
再往下走下去又有ψ(I(1@ψ(I(1@1))))等等。
如果觉得太麻烦,还有χ函数,里面还有M(马洛序数),它与I序列的关系是:
I(α?@(1+β?),……α?@(1+β?),α?@(1+β?),α?
)=χ(M^α?β?+……+M^α?β?+M^α?β?,α?)
ψ(I(ω@ω))=ψ(χ(M^M^ω))
ψ(I(ω@ω@ω))=ψ(χ(M^M^M^ω))
就这样一直下去,你也可以自己造一个新的函数,继续下去,就像有限世界一样……
有限的尽头是无限。那么,按理说,一直这样下去也要有个头啊。
没错,这个头就是非递归序数。
非递归序数,字面意思就是指不能用任何运算和递归关系来达到的数。因此,非递归序数就像一个升级的无穷序数。
非递归序数也分多个。第一个非递归序数是ω???(其中,下标1表示第1个,上标cK表示非递归的教堂克林序数),即教堂克林序数。然而它依然是有限序数:(
第二个非递归序数是ω???,它则是所有由ω???通过各种递归运算得出的序数的集合。
运可以增强下标弄出更多的序数。(ω???)
顺便说一下,教堂克林序数可以表示忙碌海狸函数,Σ(n)的增长率为ω???,Σ?(2)=ω???+1,高阶图灵机的增长率Σ?(n)为ω???
————————
只要图灵机选的恰当,f_w1ck可以被f_w+3超越
LetObeKleene’ssystemofordinalnotations.EveryaPOautomaticallyproduces
afundamentalsequenceforthecorrespondingordinal.Therefore,thenotationaPO
automaticallygeneratesthea-thfastgrowingfunctionfainastraightforwardmanner.
Notethatfadependsonthenotationa,butnotontheordinal|a|O.
Putα0asthenotationfortheordinal0,and
αn`1“mintmPO|m?αnand|m|O?|αn|Ou.
Thesequencep|αn|Oqn?ωisclearlyafundamentalsequenceforωCK
1.Then,wedefine
fωCK
1pnq“fαnpnq.
Evenifweusetheabovenaturalfundamentalsequencepαnqn(orpγnqn),weshowthat
itispossiblethatfωCK
1
maybeveryslowgrowing.Thisisduetothebasicobservation
thatthedefinitionofO(hencefωCK
1)heavilydependsonagivennumberingofpartial
computablefunctions.
Proposition1.Thereisanadmissiblenumberingofpartialcomputablefunctionssuch
thatfωCK
1
isdominatedbythepω`3qrdfastgrowingfunctionfω`3.
Proof.LetΦbeacanonicaladmissiblenumberingofpartialcomputablefunctions.One
canassumethatforalmostallethereisdwithe?d?2esuchthatΦdp0q“eand
Φdpn`1q“2Φdpnq.Inotherwords,ifeisacodeofanordinalα,thenΦdpnqisacodeof
α`n;hence35disacodeforα`ω.Forexample,inausualprogramminglanguage,
thereisaconstantc(independentofe)suchthatthelength|d|?log2pdqofsucha
programdisboundedby|e|`c.
WeconstructanewnumberingΨbydefining
Ψ2epnq“
#
2òòn,
ifn?fp3q
ω`3p352e`3q,
Φe,
otherwise.
andΨ2e`1“Φe.ItiseasytocheckthatΨisadmissible.Wenowassumethatthenew
numberingΨisusedtodefineO(hencepαnqn).
Wefirstclaimthatαn-“352e`1foranyn,e.If352e`1RO,theclaimtrivially
holds,soweassume352e`1PO.Then,thefunctionΨ2e`1isincreasingw.r.t.?O;
hence2òòn?OΨ2e`1pnqforanyn.ThisimpliesthatΨ2eisalsoincreasingw.r.t.?O,
whichmeans352ePO.SinceΨ2epnq“Ψ2e`1pnqholdsforalmostalln,wehave
ta:a?O352eu“ta:a?O352e`1u.Thus,352e`1cannotbeequaltotheleastm
suchthatαn′1?mand|αn′1|O?|m|O(orαn′1?Om).
Assumethatαn“2bforsomebandn?0.
Weclaimthatαn′1“b.
First,
|αn′1|O?|αn|O“|b|O`1implies|αn′1|?|b|O.Ifαn′1?bthen|αn′2|O?|αn′1|O?
|b|Oischosenasαn′1,acontradiction.Thus,wecanassumeαn′1?b.Notethat
|αn′1|O?|b|Oisimpossible;otherwisebmustbechosenasαnbyourassumption
αn′1?b.Hence,|αn′1|O“|b|O.Ifαn′1?b,wehave2αn′1?2b;hence2αn′1ischosen
asαn.Asaconsequence,weobtainαn′1“b.
Hence,ifαncodesanordinalλ`pforsomelimitλandfinitep,thenwehavea
sequenceαn′p?Oαn′p`1?O?Oαnwhichcodesλ?λ`1??λ`p.Put
m“n′p.Thenαmisoftheform352e.Ifmissufficientlylarge,asmentioned
above,thereisdwithe?d?2esuchthat352dcodestheordinalλ`ω.Then
1
——————————
352d?352e?expp4q
2peq:“2222e
;hence,thenumber352dcodingλ`ωhashigher
prioritythanthenumberexpp4q
2peqcodingλ`4.Inotherwords,wemusthavep?3.
WenowcalculatefωCK
1pnq.Asobservedabove,foralmostalln,αnisoftheform
expppq
2p352eqforsomep?3.Ifp“0,asαnis?-increasing,weclearlyhaven?352e.
Hence,foranyp?3,wehaven?352e`3.Ifp“0,ourdefinitionofΨ2eensuresthat
fαnpmq“fmpmq“fωpmqforanym?fp3q
ω`3pa`3q,wherea“352e.Ifm?fp2q
ω`3pa`3q,
then
fpmq
ω
′
fp2q
ω`3pa`3q
ˉ
?fω`1
′
fp2q
ω`3pa`3q
ˉ
?fp2q
ω`3pa`3q
andthus,ifp“1,thenfαnpmq“fpmq
a
pmq“fpmq
ω
pmq“fω`1pmqforanysuchm.
Similarly,onecanobservethat,ifp“2,thenfαnpmq“fω`2pmqforanym?fω`3pa`3q,
andifp“3,thenfαnpmq“fω`3pmqforanym?a`3.Inanycase,wehavefαnpnq?
fω`3pnqsincen?a`3asobservedabove.Consequently,fωCK
1pnq“fαnpnq?fω`3pnq
foralmostalln,thatis,fωCK
1
isdominatedbyfω`3.
□
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