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DIGGINGIN
in BASIC
Programming
Recursive
.'
"
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ON THE LIGHTDRSIDE
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COMING
WHAT'S
ADINDEX...........111
hy L.D. StMder
Co
o Copyright 1919,Benvill PublislingCo+, a Molgan_GBnpian
aPRrL1s7e PersonalComPuiing
I
Recursive
Prograrnmin€
inBASIC
BY HERBERT
L. DERSHEM
Recursive
algoithms
areanathemrto the personal
compul
ingprogrammer.
Theirfeputationsuffersfrom the belief
that fecursion
requires
a large,powerfulcomputeranda
fancylanguage.
programming
Peopleconsiderrecursive
so
complicated
that you needa degree
in computerscience
to
understand
it. But no onewith thst muchexperience
would
bolherwith recursion.
lt is,so the myth go€s,highlyinefficient,wastefulofr€sources
andonly a loy for acad€mics
to
playwith.
But the mythsarcfalse.Recunioncanbe a valuabl€
tool
on a personal
computerusingBASICandshouldbe in every
programmels
reperioire.
Anythingis recursive
if it is definedin termsofitself. In
programming,
a recursive
subroutine
is onethat callson it,
sell Peoplecommonlyconfuserecursion
wilh iterution;
both canbe usedto solvethe sameproblem,
But recucionanditerationarecleaflydistinct.A proc€.
dureis it€rativeifthe sameprocess
is performed
many
times.A procedure
is recursive
ifin themiddleof its execution ii cals uponitseli A fecursive
procedure
beginsanothei ex€cutionofitselfbeforeihe originalis finished.Such
a procedure
hasmorethanoneexecutionin pfogress
ai a
giventime;aniteraiiveprocedure
neverhasmorcthanon€.
NoreverylanguaBe
implemenration
pern'irs
recuision.
In BASIC,recunionis only possible
ifit is permissible
to
calla subroutine
frornitselfandstil retainthe abilityof
the originalexecutionto returnto the properpoint.The
folloMngprogram
canle\l yourBAslc fo- recursion
capability.
10 Rttt TEst !0R Iu! tBrlrrY o! TItl
aasrc t0 plt!oR{ tEcotslolr.
20 [=1
l0 T=0
40 GOSUB
100
50 Ir I<>2*[-1 Tgtll 1?0
8 0 G O T O3 0
16
PenonalComputing aPRrLreTe
100 I=I+1
I1O IF I<[
r ?0 iElltr!
THlN 'I30
1 1 0 e os u B 1 0 0
140 Ir I>2*n- 1 TflElr 't70
1 5 0 I = r +1
1 60 tEl0Rll
1't0 t!IllT tr[ct t!c0RsIv!(it{-1itrTIttEs. tr
180 !!rD
Everycall ofsubroutine100increments
I by one.The
Nth callwill resultin I beingsetto N anda returnwithout
a recursive
call.Thiswill retun to eachof theprevious
tevels,incrementing
I by oneon eachreturn,On the final
return,I shouldequal2*N-I if thepfocess
wascarriedout
correctly.Thus,ifthe aboveprograrn
runsfor a givenvalue
ol N,lhe\,elsion
of BASIC
usedwill allowN.I recursrve
calls.ManyvelsioasofBASIChavevirtuallyno limit on th€
numberof suchcallspossible.
Ifthe BASICis not recursive
to the appropriate
levelfor somevalueofN, ir will usually
respondwith som€iype of diagnostic
message
ratherthan
arriveal statement170.
Ifyour computer's
BASICallowsrecursive
subroutines,
you now facethe problemoflvritingthem.Here\ a general
outlineofa recunivesubroutine
in BASIC.
L Ifthe firsi call,initializethestackpointer2. If teminationcondition,computeresultidecremeni
stackpointerireturn.
3. Do computation.
4. Saven€cessary
valuesin stack.
5. Incrementstackpoinier.
6. Recursively
callthissub.ouiine.
7, Restorcsaved
valuesfrom the stack,
8. Do any remaining
computation.
9. Decrement
stackpointer.
10.Reiurn.
\vhenyou re.uajvelycaila sJbrouline,rom itself.Ihe
variablesin the calledexecutiondestroythe variablesof the
samenamein rhe callingexecurion.To preserveihe original
valuesof thesevariables,savein a dirnensionedvanable
you needlo recalllater.Sup(caleda stack)thosevanables
three
variables(xJ andz) it
has
subroutine
;ose vour
whenil retumslroma re.ursive
wirhesto savetor recall
call. The format of the stackdimensioned riable A lbr
wouldbe:
thissubroutine
A(l) X from execuiionI
A (2) Y from execution 1
A (3) Z from executionI
A (4) X from execution2
A (s) Y irom execution2
A(6) Z from execution2
To keeptrackof tlle positionin the stackwhelethe
its values,a pointeris used'This
currenlexecuiionsaves
by
oointeris incremeflted the requirednunber,in the
cal is made,and
atoveexanple3, everytime a.ecursive
by lhesameamounlon ea!hIelurn'
emenleo
de('the
shoss
above
procedure
ourlined
recursive
seneral
genetal
sleps3
In
lhe
subroudne.
call
rn
onlyon; tecursive
timesbefolethe returnat
several
to 8 maybe repeated
9 and 10.
steDs
andseehow lhesesleps
sr.rbrouline
Lt\ takea recursive
is
the
standaldfitst example
The
factorial
areimplemented.
a
it alsorepresents
because
unfortunately,
ofrec;nion
Droblemwherercculsionrsnot thebestwayro oblaina
ir i( thernosrfamiliarandsimpleslof
;olulion.HoweveI,
sowhy bleaktradiiion?
all examples.
Thecommondefiniiionof N facrorial(Nl) is iterative:
0 ! =I
N:=N- (N-l) (N-2}" *2alforN=I.2
definitionofN faclorial:
But thereis alsoa Iecursive
0 ! =I
N ' = N * O { - 1 )fro rN = l , 2 , . . .
ln thiscas;the factoriaiis definedin termsofitself,but
clausewhichoccursat 0!. Therecuaive
with oneescape
form of a subloutineto computeN! isl
960 iE[ tnrs st!Ro0Tr[! coltPoTEsN
FACIORITI, RECI'!SI!!I,Y
AND
970 rlt Srolls lll! RlsolT rI !.
tltE lrRsT cAL!, rs To 990'
980 !!tt tErs lRocRltt rs rt{ RlDro
sBlcK l,!v!L r BASrC990 s:l
1 O 0 OI F n = 0 l D B l l l = 1 : s - s - 1 : R I T S R N
l 0 t 0 c ( S )= n
t 020 s=s+t
1 O 3 O{ = I - ' ! l c c s ! ! 1 0 0 0
1 0 4 0N = A{ s )
1050 l=N*r
1050 s=s-1
1070!rruiN
to stepI in
990 colresponds
In thisprogram,statement
l000corre_
given
Slarement
eal[er.
gened
algorilhm
rhe
ro step2, -here rhe terninalioncondilionis N=0
soon-ds
bv I in thisproThestackis increnentedanddecremenied
whena lccursive
N.
is
raved
va
rable.
onlyone
srambecaus€
correspond
plogram
this
in
No
compulations
;aI is nade.
1010to
and
slatements
general
atgorithm,
of
the
step
3
to
lo(Iepr4ro l0 reipecrivelv
l07Ocorespond
of lhefactorial
ileraiveversion
of cou$;,lhesimpra'
subroutineexecutesmuch faster:
Sl]ElOoTIllE
910 8eI IIIRATII'
I O C O I T P O I E BN I A C T O i I I L
axD sro!! r1 rll P.
980 tlr
990 REr{lllls PRocrat rs tn
r Blsrc.
RADrO SqlC!( tt{El
1000r=1
1010 IF !<=1 IllEll llToRl{
1020 ror I=2 10 N
?=FI
1030
't040 lrxT r
1050 llTolll
isurecomputa'
ofrecursion
exampie
Anothercommon
TheNthnum_
ofnumberssequence
tionofihe Fibonacci
F(N),is definedin termsof
berin the Fibonaccisequence,
its two predecessors.
F(0) = 0
F(1) = 1
F ( N ) =F ( N - l ) +F ( N - 2f)o r N = 2 , 3 , . . .
Your BASICrcrsionof thisalgorithmis:
9-70!!r 50BRO0lrN! 990 cALCoLlllS
1llE Ntl FrBOnrccr NotlBlR
o 8 0 ! ! i ! ! c D l s r v E l Y a { D D E T 0 nS
IT Tr{F. n.S. LIVFL T BrSlC.
990 s=1
1 0 0 0 I F ( F = 0 )+ ( ! = 1 )
Tg!I !=I: S=5-2: REtlt !N
1 0 1 0 I ( s )= !
020 s=s+2
0 3 0 n = I - 1 : G O S ! B1 0 0 0
0tr0 lr=a(sl
0 5 0 A ( s + 1 )= !
050
0 7 0 tl=N-2: G0sltB 1000
080 !-r ts+1, +!
0 9 0 S-S-2
1 1 0 0 RtT0t t!
therearetwo stackenlriesfor eachcall
ln thisexample.
level.TheStil entryin A is the valueof N for that calllev€I,
andthe (St1)st is thevalueof F(N-1)
Again.asin the caseof thefactorial.iterationgivesa
mo!€€fficientsolutionto thisproblem
970 !Et! SttBiolTr[E 1000 crlcol^Tls
tli i !Tll !tBoNlccr llltltB!l
IND lET0RliSrr
980 lEt rt!!altrrlY
990 !!tl It rs llRrTTEllr! llDro
I BASIC.
stilcK l!t!!
000 I=1rP=0
0t0 FOi I=t TO n_l
020
0=r
030 !=!+?
0q0
t[
F'
P:Q
050 Nlrt r
0 RlTotI
015
couldbe
of recursion
Althoughyour first two ex3mples
is the desiredlechnique
betterdoneiterativelv.recursion
for many pfobiemsbecauseit greallv simplili€sihe solution
p'ogram
dnorilhm
-Now ,nd ia i.pt".'.nlarronin a BASIC
t.o suchp oblemslhe filsris rhe fower
consrder
of Hanoi,a well-knownproblemwhichis nicelytreatedln
ofthree pegs,
a recu$ivemanner.Thisproblemconsists
all ol differD
disks,
whichwe will calipegs1,2 and3, and
the
pegs.
Initiallv
on
the
stacked
ent radius,whichcanbe
size
with
of
decreasing
peg
1
in
older
on
thsksarestacked
rhelarqestdiskon the bottom.Theproblemis to movethe
disksfrom peg1 to peg2 with the restriclionth't disks
mustbe movedoneat a time from onepegto another,ano
L\at no disk;ay everbe stackedon top ofa sn'.llerdisk'
Computing l7
aPRrL1e7e Personal
Therecuri\esolution
generajizes
lheproblenio move
^
u osis lron pegI- ro pegF by movingrhetop D.l disks
on
pegE ro rhe{jtrd pe8.movingLheoneremaining
djskon
pegF to pe€f. andrher movingalJof Lhediskson tle
thi,d
pegio pegf.J1;s reduces
rheproblemof movingD disksro
lru moves.ol
D.I di.ks.TherecJrsive
aiCoriLhmlor
moving
the top D disksfrom pegE to pegF is:
I If D=l . movetop diskf(omE to F. rcturn
2 l.er c oe rhenlmber oi rhepegwtuchis nor E or F.
r. Kecu.rrlety
ca lrr procedure
lo moverheropD-l
disksfron E ro c.
4. Movethediskon E io F.
5. R€cursively
caltthisprocedure
to moveth€ top I}1
disksfron C to F.
6. Return
In orrrimplenrentation,store
thenumberof dlskson peg
-.
I in A(I), for I=1,2,3. Thevaluesthatneealto be sav€d
whena recursive
callis madearet, F andD. Our BAslc
rersinnof thrsalSorithmis rhen:
rhelisr relariveto theeventua]
positionofX. Th<!jocessrs
repeat€d
unlilI andJ cross.
Al !}lalpoinllhered,raEe.
menta completed.
Figure1 showsan exampieofthi;
I->12
t2
t2
12
12
75 I >15
3
I
3
b0
ool>o0t.\0
44
13
13
13
13 I >13
x >46 X->46 X-N6X->46 JX_>46IJX
44
44
44I >44
60
94
94
94
94
94
49
49
49
49
49
95
95J >95
95
95
J >3J->3
75
75
75
FrgureI
Theproeramto accornplish
thisprocess
is:
960
9?0 !!d TotlR o! Itlltol suBRo0TlNE
TO I'OVBTIE tOP D DISCS
980 RE! rro
!!c ! To PEG
g I I T T E I I ! I R . S . ! E ! E I . r.
I BASIC.
990
r 0 0 0r ! D = 1 I t ! l = l ( 8 , - t : r ( p , - A ( F , + 1 .
P F r N t n f i 0 v En , Ei t r T O , i p : c O t O 1 0
i
90
| 0 1 0 G= 6 - { E + t l
1 0 2 0 a ( s )= t : A ( s + l ) = p : t ( s + 2 ) = D
1 0 3 0 s = s + 3 : D = D -1 : F = c
1 0 4 0c 0 s 0 s 1 0 0 0
1 0 5 0 r = a( s l : r = t ( s l 1 ) : D = r ( S + 2 1
1 0 6 0 r ( E -l r ( l t - 1 . r ( r l : r ( r , + 1 :
PRr!T (Il0V!ir. E. trTOri F
1 0 7 0 s = s + 3 : D = ! - 1 : ! ' = 6 -l ! + ! I
1 0 8 0c 0 s 0 ! 1 0 0 0
1 0 9 0s = s - 3
1 1 0 0R l T 0 t l l
a finatusefulapplication
of lecursion.
_ Now consider
Lhequrcksort
atgorirhm.oneof lhemosrefficjent
and
wjd€Iyusedsortingalgoritlms,is veryeasilyprogrammed
Suplosew€havevaluesstorcdin A(L), . . ., AGD and
vr'ewishto placethevaluesin ascending
orderin ne same
storage
locations_
Thebasicquicksortalgorithmchooses
somearbitraryvaluefiom thislisr,sayX=A(K),laenre"
arranges
tlrevaluesso thai ali valuessmallerthanX are
locatedbefor€ii in the lisr andall valueslarserthanX are
locatedafterit. ThenX will be at its correctsorted
position
in rheiisr.sayA{l). Thecamealgorirhmis fien recu;sivety
appliedwrrhL andl-l in placeof L ard H, andrhenaqain
appliedusingI+l and H. whenthe algorithmiqcaltedwirh
L=8, wesrmpryreturn.
Theonly part ofthe algorithmthai needssomeaddi
X is in its properrosiiionandall othervalueslie or Lne
propersideofX. Keepingtwo pointers,IandJ, accomplhlEs ihis process.I startsby pointing at the lirsr position
in thelist, L. J pointsto H. ThenpointerI is moveddown
thelist until a valueis encountered
whichh no smalier
than X. This valueof A(I) shouldthereforeLiebetowX
in the list. Next pointer J is movedup the list until it encounlersa nunberno large,thanX. Tharnurnberis eychanged
wi$ A(I), andboih arerhenin rtreproperparrof
l8
Personal Computing
apRrL 1e7e
12
1.2
J
J
A
l3r_>lJ
>46X >46
60I_X0
94
e4
49
49
95
95
75
75
FE! QOICT(SORT_SOBIOUIIN' TO
REtnnaic! A(!) THF0 A(fl) SO THlt
c- 7 0 ? E [ l l t v r l o B s < = x L I i i E r o n E
x
I N D T L I V A I U E S> = X l T E A I T E R .
980 iE!{ x Is Atrr{T{(L+0,/2}. INITTAL
caLL T0 990 sorTs r(ll _a(N).
990
1 0 0 0I r l > = i c o t o 1 1 0 0
10 1 0 [=INT ( (I{fl) /2) : r=a (|l] : r=L: JEB
1 0 2 0M ( I ' < t r
THEII=I+1: coTo l02o
1 0 3 0Il l(.1) >I Afi!N it=it-t: cOTo 1030
1 0 r r 0r F I < = J r s B r = r { r , : A( r t = A( i r t:
r ( J l = 1 :r = r + 1 : , t = J -I
1 0 5 0 IF I<=,t THIN GoTO1020
1 0 6 0A ( s ) = I : l l s + 1 ) = f l
1 0 7 0E = J r5 = S + 2 : c o s 0 B1 0 0 0
i 080 I=r (Sl: B:t (s+11
1 0 9 0l = r : s = s + 2 rc o s o ! 1 0 0 0
1 1 0 0S : S - 2 : i ! 1 0 ! N
In thisprogmm,star€ment
lOO0teslsfof termination.
Stat€nentsl0l0 to 1050partjtionthe list into lhosevalues
smallerthanX andthoselargerthanX. Therwo recursive
calhJollosstaremenr
t050.only Lwovalues,I
andH, are
savedduringihe recunivecall,
Anrllustration
ofan execuljon
of lhequicksort
algo.
rithm
isgrvenin Fi8ure2. Theo'rginat
tirrotrennumbers
is
roundon Ihe far lefi. ThecirclednumbelsareLhevaiues
usedfor X. andthenumber-parrs
in recranqtes
arctne
vdlues
oi L andH usedto parntionrhelisl.\oie rharwe
beginwith L:l, H=10.X=46.X is tien placedby partitioningin !}lelfrh position.Nexrrheoroce(srsca edlor X=J,
L-1. H=4andX-49,L=6.H=tO.Theproces\
conlrnues
In
thisway unlil all vaiuesarecorrecdyso ed.
tr