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Decidability results for primitive recursive algorithms
René David
To cite this version:
René David. Decidability results for primitive recursive algorithms. Theoretical Computer Science, Elsevier, 2003, 300 (1-3), pp.477-504. �hal-00384668�
algorithms
R.David
January 30,2002
Abstract
InthispaperIusethenotionoftracedenedin[9]toextendT.Coquand's
constructive proof[6]oftheultimateobstinationtheoremofL.Colsontothe
case whenmutualrecursionisallowed. As aby productI get analgorithm
thatcomputes thevalue ofaprimitiverecursive combinatorappliedto lazy
integers(inniteorpartiallyundenedargumentsmayappear). Ialsoget,as
T.Coquandgotfromhisproof,that,evenwhenmutualrecursionisallowed,
thereisnoprimitiverecursivedenitionfsuchthatf(S n
(?))=S n
2
(?).
1 Introduction
In [3], Colsonproved theso-called"ultimate obstination theorem". This theorem
asserts that a primitiverecursive algorithm always reads,and so locks, ona par-
ticularinputargumenttocompleteitscomputation.Thisbehaviourdoesnotallow
the computation toshift from oneargumentto another onein orderto eÆciently
computeafunction,asforexampletheinf function studiedoriginallybyColson.
In[9],byusingthesyntacticnotionoftracewhichisasimpliedversionofthese-
quentialalgorithmsofBerryandCurien(see[2]or[1],chapter14)areformulation
of thistheorem isgivenandproved.
Theproofofthistheorem,asgivenin[3]or[9],isnotconstructive: itdoesnot
giveawaytodetermineonwhichparticularinputargumentthealgorithmlocks.
In this paper, I show that this can be done in a constructive way. This was
proved by T.Coquand in [6] for primitive recursivealgorithms. This is extended
hereinthecasewheremutualoralternaterecursionisallowed. Alternaterecursion
(seedenition2)hasbeenintroducedbyP.Valarcher[17]togiveagoodalgorithm
to computetheinf function.
TheproofofthemaintheoremneedsadiÆcultcombinatorialresult(proposition
18)whichhassomeinterestbyitself.
Decidability results : I show that, even when mutual or alternate recursion is
allowed, various problems are decidable. For example, it is possible to compute
the intentional behaviour of aprimitiverecursive combinatorf, i.e. the valueof
f when applied to lazy integers (innite or partially undened arguments may
appear). Note that, when alternate recursion is allowed, theultimate obstination
theoremobviouslyfails.Thisshowsthatthedecidabilityresultshavenothingtodo
withultimateobstination! Ialsoshowthat whenlistsorchangeofparametersin
therecursionschemeareallowed,theseproblemsbecomeundecidable.
Asaby-productoftheproofIalsoget:
LaboratoiredeMathematiques. CampusScientique. 73376LeBourgetduLacCedex. email
david@univ-savoie.fr
example, that, even when mutual or alternate recursion is allowed, there is no
primitive recursive denition f such that f(S n
(?)) = S n
2
(?). This was proved
by T.Coquandin the ordinarycase.I believethesametechnic could implyother
similar results such as, for example, if mutual recursion is restricted to at most
twofunctions,thereisnof suchthatf(S n
(?))=S [n=3]
(?)wherewhere[x]isthe
integer partof x but this seemsto need a combinatorial lemma that I have not
beenableto prove(nordisprove).
Irecallhereafterthemainintuitionsconcerningthenotionoftrace. LetN be
the domain of lazy integers.An element e of N canbeseenas apartial function
that lls someaccessiblecells (in thesense of [2])with theconstructors S,0and
?. Since?correspondstoalackofinformation,acelllledwith?isoftensaidto
beunlled. Forexample(seegure1)in e
0
=S(0)theaccessiblecellsaretheones
denoted bytheiraddress0and1.TherstoneislledwithSandthesecondwith
0:Ine
1
=S 2
(?)theaccessiblecellsaretheonesdenoted 0;1;2. Thecells0and1
arelledwithS andthethirdoneis unlled.
e
0
=
cellnumber 0 1
constructor S 0
and e
1
=
cellnumber 0 1 2
constructor S S ?
g. 1
Theset of traces isdened asfollows. LetW betheset of(nite orinnite)
wordson thealphabet fx
n
=n 0; x is aletterg. A traceis apair(e;) where
e2 N and isalabellingi.e. afunction from theaccessible cellsof e to W (see
examplesingure2).
Toeach primitiverecursivedenition (prc)f weassociateafunction[[f]]from
traces to traceswhich "codes" theway f gets its result: thefact that the token
x
i
occursin (n)intuitivelymeansthatthecellioftheelementnamedxhasbeen
used togete(n):
An exampleis given in gure 2 : dene add as usual by add(0;m) = m and
add(Sn;m)=Sadd(n;m):
-Thetracet
2
meansthat to getS thealgorithmhas usedthecell0of t
0 and
to get0thealgorithmhasusedrstthecell1oft
0
andnextthecell0oft
1 .
-Thetracet
3
meansthatto getS thealgorithmhasused rstthecell0oft
1
andnextthecell0oft
0
andtoget0thealgorithmhasusedthecell1oft
0 :
t
0
=
cellnumber 0 1
constructor S 0
labelling x
0 x
1 t
1
=
cellnumber 0
constructor 0
labelling y
0
t
2
=[[add]](t
0
;t
1 )=
cellnumber 0 1
constructor S 0
labelling x
0 x
1 y
0
t
3
=[[add]](t
1
;t
0 )=
cellnumber 0 1
constructor S 0
labelling y
0 x
0 x
1
g. 2
Since a trace carries the informations on a computation and not only on the
result,this notionallowsto alsocompose the computations.I believeitalsomakes
ofColson. Inparticular,theextensionofCoquand'sconstructiveresulttothecase
wheremutualrecursionisallowedwouldprobablybeimpossiblewithoutthenotion
of trace.
Thisnotionoftraceisrelatedto thesequentialalgorithmsintroducedbyBerry
and Curien ([2] or [1], chapter 14) as follows. In their terminology, a sequential
algorithmisatree. Eachbranchofthistreecorrespondstothecomputationofthe
algorithm on particular arguments, that isexactly (witha slight variationon the
syntax andtheterminology)what Icallatrace. Inparticular [[f]] canbeseenas
thesequentialalgorithmassociatedtof.
Thepaperisorganizedasfollows: I recallthenotion oftrace(section 2)and
itsmain properties(section5). Themain resultisgiveninsection3. Insection4,
I giveacombinatorialresultthatiscrucialfortheproofofthemain theorem. The
sections6,7and8aredevotedto itsproof. Insection9,Igivetheundecidability
results. Finally the appendix gives the proof of the combinatorial proposition of
section4.
2 The trace
InthissectionIrecall,forselfcompleteness,themaindenitionsabouttraces.More
details canbefoundin[9].Since,in thispaper,theonlydatatypeI amconcerned
with,isthedatatypeofintegersmanythingsaresimplerthanin thegeneralcase.
I thusadaptthedenitionsof[9]tothiscase.
Denition 1 1. The scheme for primitive recursionis : f(0;
!
r)=g(
!
r) and
f(Sx;
!
r)=h(f(x;
!
r);x;
!
r).
2. The scheme for mutual recursion is : f
i (0;
!
r) = g
i (
!
r) and f
i (Sx;
!
r) =
h
i (f
1 (x;
!
r);:::;f
k (x;
!
r);x;
!
r).
3. Thescheme for alternate recursionis:
f(0;y
2
;:::;y
k
;
!
r)=g
1 (y
2
;:::;y
k
;
!
r)
f(Sy
1
;0;y
3 :::;y
k
;
!
r)=g
2 (y
1
;y
3
;:::;y
k
;
!
r)
f(Sy
1
;Sy
2
;0;:::;y
k
;
!
r)=g
3 (y
1
;y
2
;:::;y
k
;
!
r)
...
f(Sy
1
;:::;Sy
k 1
;0;
!
r)=g
k (y
1
;:::;y
k 1
;
!
r)
f(Sy
1
;:::;Sy
k
;
!
r)=h(f(y
1
;:::;y
k
;
!
r);y
1
;:::;y
k
;
!
r)
Denition 2 1. Thesetsofprc(primitiverecursivecombinators)aredenedas
the leastsets containing the projections, the constructors S and0and which
areclosedundercompositionandprimitive recursion.
2. Thesetsofprc
mut
aredenedinthesamewayasprcbutdenitionbymutual
recursionisallowed.
3. Thesetsofprc
alt
aredenedinthesamewayasprcbutdenitionbyalternate
recursionisallowed.
Examples and comments
1. For thesimplicityofnotations,I assume,withoutlossofgenerality,that the
recursionalwaysisontherstargumentoftheprc.
Thusaddisaprc:
3. The functions odd and even are dened by : even(0) = 1 and odd(0) = 0:
even(Sx)=odd(x)andodd(Sx)=even(x):Thusoddandevenareinprc
mut :
4. The function inf is dened by : inf(0;0) = 0,inf(Sx;0) = 0andinf
(Sx;Sy) = Sinf(x;y). Thus inf is in prc
alt
and it is easy to see that the
computationtimeofinf(S n
(0);S m
(0))isinf(n;m).
5. Itis wellknownthat,asfunctions,thesets prc,prc
mut
and prc
alt
areequal
butthispapershows,in particular,that,asalgorithms,theyare not.
Denition 3 1. N (resp. N
;Z) is the set of non negative (resp. positive,
negativeor non negative)integers.
2. AnelementeofN isapartialfunctionfromaninitialsegmentofN (denoted
byAcc(e))intofS;0;?gsatisfying :
02Acc(e)
If(n+1)2Acc(e), thene(n)=S:
Ife(n)=?, then(n+1) 2=Acc(e):
3. Anelement eisnite iAcc(e)isnite.
4. Let e;e 0
be elements of N . e e 0
means : Acc(e) Acc(e 0
) and for all
n2Acc(e), ife(n)6=?,thene(n)=e 0
(n):
5. Iwilldenote the elementsof N as:
S n
(0) =f(i;S)=0i<ng[f(n;0)g,
S n
(?)=f(i;S)=0i<ng[f(n;?)g
S
!
=f(i;S)=i2Ng:
Comment
Acc(e)representsthesetofintegersthatareaccessibleine. Inthispresentation,
this simplyis the domain of e. In the generalcase (when various data typesare
allowed)it was moreconvenient to denetwodistinct sets : thedomain of eand
Acc(e). I havekeptthe notationAcc(e) to remaincompatible withthe notations
of [9].
Denition 4 1. Let=fx
n
=xisaletterandn2Ng.Theelementsofare
calledtokens.
2. Awordisanite (possibly empty)or innitesequenceof tokens. The empty
wordisdenoted by":The setof wordsisdenotedbyW.
3. Let u;u 0
be words. uu 0
means that u isa prex of u 0
and u"pdenotes,
for plg(u);the prex of uof lengthp:
4. u+u 0
isthe resultofconcatenatingu 0
attheendofu. Moregenerally, if(u
k )
isa(niteorinnite)sequenceofwordsu
0 +u
1
+::: willbedenotedby P
u
k :
5. Let u
n
be a sequence of words. Say that u
n
! u if for each p there is an
n
0
such that for all n n
0 , u
n
" p = u " p. This unique u is denoted by
Lim(u
n ).
labellingfunction :Acc(e)!W suchthat : 8n2Acc(e), ife(n)6=?,then
(n)isnite.
2. Atrace(e;) isnite ifeisnite andalllabelsarenite.
3. Theorderingon tracesisgiven by: (e;)(e 0
; 0
)iee 0
and for eachn
inAcc(e); (n) 0
(n)and,if e(n)6=?, then(n)= 0
(n):
4. Theset oftraces isdenotedby T.
5. Lete beanelementof N andx bealetter. The trace(e; )where (n)=x
n
foralln2Acc(e)will bedenotedase[x]. Atracease[x]iscalledanelement
namedx.
6. Let t=(e; )be atrace. eiscalled thevalue of tandisdenotedby Val(t):
iscalledthe labellingof tandisdenotedbyLab(t):
Proposition6 T with itsordering formsadomain. Inparticular :
1. Everytraceisaleastupperbound(denotedbySup)ofanincreasingsequence
ofnite traces.
2. Everyincreasing sequencehas aSup.
Thefollowingnotationswillbeconvenientatmanyplaces.
Denition 7 Lett =(e;)be atraceandw beaniteword.
1. w+t isthe trace (e;
0
) dened by : 0
(0)=w+ (0) and 0
(n)=(n) for
n1:
2. h(S;w) ti (or simply (S;w)t if no confusion is possible) is the trace (e 0
; 0
)
dened by : e 0
(0) = S;
0
(0) = w and, for n 0, e 0
(n+1) = e(n) and
0
(n+1)= (n).
3. Let x be aletter. thx+ki=(e;
0
) where 0
isobtainedfrom by replacing
x
j byx
j+k
for all j:
Example
y
0
+S(0)[x]=(S;y
0 x
0 )(0;x
1 )
S n
(?)[x]=(S;x
0 )(S;x
1 ):::(S;x
n 1 )(?;x
n ):
Lett=S
!
[x],then t=h(S;x
0
)siwheres=thx+1i.
Denition 8 Letf beafunction fromT n
toT:
1. fisincreasing if for allt
j t
0
j ,f(t
1
; :::; t
k )f(t
0
1
; :::; t 0
k ):
2. fiscontinuousifitisincreasingandpreservestheSupofincreasingsequences.
Proposition9 Every n-ary f 2 prc (resp. prc
mut
, resp. prc
alt
) induces (in a
uniqueway) acontinuousfunction (denotedby[[f]]) fromT n
toT suchthat:
[[0]](t
1
; :::; t
n
)=(0;")
[[S]](t)=(S;")t
Iff isthe i-thprojection then [[f]](t
1
; :::; t
n )=t
i
1 k
[[f]](t
1
; :::; t
n
)=[[g]](r
1
; :::;r
k
)wherer
j
=[[h
j ]](t
1
; :::; t
n )
If f is dened by ordinary or mutual recursion and the recursive equations
are f
i (0;
!
s)=g
i (
!
s)andf
i (Sx;
!
s)=h
i (f
1 (x;
!
s);:::;f
k (x;
!
s);x;
!
s). Then
[[f
i ]](t;
!
s)=
{ (?;w)ift=(?;w):
{ w+[[g
i ]](
!
s) ift=(0;w)
{ w+[[h
i ]]([[f
1 ]](r;
!
s);:::;[[f
k ]](r;
!
s);r;
!
s)if t=(S;w)r
Iff isdenedbyalternaterecursionand(forsimplicityofnotationsIassume
k = 2) the recursive equations are : f(0;y;
!
s) = g
1 (y;
!
s), f(Sx;0;
!
s) =
g
2 (x;
!
s)andf(Sx;Sy;
!
s)=h(f(x;y;
!
s);x;y;
!
s):Then [[f]](t
1
;t
2
;
!
s)=
{ (?;w
1 )if t
1
=(?;w
1 ):
{ w
1 +[[g
1 ]](t
2
;
!
s)if t
1
=(0;w
1 )
{ (?;w
1 +w
2 )ift
1
=(S;w
1
)randt
2
=(?;w
2 ):
{ w
1 +w
2 +[[g
2 ]]((r
1
;
!
s)if t
1
=(S;w
1 )r
1 andt
2
=(0;w
2 ):
{ w
1 +w
2
+[[h]](([[f]](r
1
;r
2
;
!
s);r
1
;r
2
;
!
s) if t
1
= (S;w
1 ) r
1 and t
2
=
(S;w
2 )r
2 :
Commentsand examples
1.
[[add]](S(0)[x];S
!
[y]) = (S;x
0 )(S;x
1 y
0 )(S;y
1 )(S;y
2 )
[[add]](S(0)[x];S 2
(?)[y]) = (S;x
0 )(S;x
1 y
0 )(S;y
1 )(?;y
2 )
[[add]](S 2
(?)[y];S(0)[x]) = (S;y
0 )(S;y
1 )(?;y
2 )
2. Sincethispaperisonlyconcernedwithdecidabilityresults(andnotwithcom-
plexityresults),I donotcareonthestrategyofreductionusedto transform
theequations into algorithms. However, thestrategy that is implicit in this
denitioniscallbyname : intuitively,ateachsteptheleftmostoutermostre-
dexisreducedand,inparticular,twocopiesofthesameredexwillbereduced
twiceiftheyareneededtwice.
3 The main result
Inthissection,I givethemain result(theorem14)ofthepaperandthedenitions
that arenecessaryforitsstatement.
Denition 10 Lett=(e;) beatrace.
1. Thebranchoft(denotedbyBr(t))istheworddenedby: Br(t)= P
k 2Acc(e) (k):
2. Aletterx isunbounded(respectivelybounded)intiffj =x
j
occursinBr(t)g
isinnite(respectively nite).
3. tisultimately obstinate ifithas atmost oneunboundedletter.
4. Nb(t;x;n)isthe least ksuchthat x
n
occurs in (k). If x
n
doesnot occurin
Br(t), Nb(t;x;n)isundened.
Commentsand examples
