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Abstract
In this work we present the specification and the implementation of a new abstract data type (ADT) called STORE. This new ADT allows the storage of a given collection of elements offering an abstract mechanism that supplies a direct access to them, alternative to the method defined by the standard operations of usual ADTs. The interest of the new mechanism stems from the efficiency of pointers, while avoiding the loss of modularity that usually occurs when pointers are used. The implementation of the operations offered by the new ADT is done by derivation from the equational specification. The representation chosen for the implementation of the new ADT makes the efficiency previously mentioned possible, even when the representation strategy requires the movement of the elements.
Full text
Sho cu s: Abs ac Poin e s"
J. Ma co
[email p o ec ed] c.es
X. F anch
anc[email p o ec ed] c.es
Dep . Llengua ges i Sis emes In o ma ics (LSI)
Uni e si a Poli ecnica de Ca alunya (UPC)
Campus No d
Jo di Gi ona Salgado, 1-3
08034 BARCELONA
Abs ac
In his wo k we p esen he sp ecica ion and he implemen a ion o a new abs ac da a yp e
(ADT) called
STORE
. This new ADT allows he s o age o a gi en collec ion o elemen s oe ing
an abs ac mechanism ha supplies a di ec access o hem, al e na i e o he me ho d dened
by he s anda d op e a ions o usual ADTs. The in e es o he new mechanism s ems om he
eciency o p oin e s, while a oiding he loss o mo dula i y ha usually o ccu s when p oin e s a e
used. The implemen a ion o he op e a ions oe ed by he new ADT is done by de i a ion om
he equa ional sp ecica ion. The ep esen a ion chosen o he implemen a ion o he new ADT
makes he eciency p e iously men ioned p ossible, e en when he ep esen a ion s a egy equi es
he mo emen o he elemen s.
1
Con en s
1In o duc ion 3
2 The ADT
STORE
4
2.1 Op e a ions . ... ... .. ... ... ... ... ... .. ... ... ... ... ... . 5
2.1.1 Cons uc o Op e a ions . ... ... ... ... .. ... ... ... ... ... . 5
2.1.2 Obse e Op e a ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.1.3 Mo die Op e a ions ... ... ... ... ... .. ... ... ... ... ... . 6
2.2 Equa ional sp ecica ion .. ... ... ... ... ... .. ... ... ... ... ... . 6
2.2.1 Equa ions b e ween Cons uc o Op e a ions .. .. ... ... ... ... ... . 6
2.2.2 Equa ions o he Obse e Op e a ions .. ... .. ... ... ... ... ... . 8
2.2.3 Equa ions o Mo die Op e a ions . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3 Implemen a ion in Main Memo y 14
3.1 The Typ e Rep esen a ion . ... ... ... ... ... .. ... ... ... ... ... . 15
3.2 De i ing co de o he op e a ions ... ... ... ... .. ... ... ... ... ... . 19
3.2.1 De i a ion o
pu
.. ... ... ... ... ... .. ... ... ... ... ... . 19
3.3 Chosing implemen a ions o he comp onen s o he ep esen a ion . ... ... ... . 25
4 An Example: The Tennis Ladde 26
4.1 Aho, Hop c o & Ullman's Solu ions o he P oblem . .. ... ... ... ... ... . 26
4.2 The solu ion wi h he ADT
STORE
. ... ... ... .. ... ... ... ... ... . 28
5 Co dica ions 29
6 Conclusions 30
2
1 In o duc ion
The p esen wo k deals wi h he me ho dology o mo dula p og am de elopmen by means o abs ac
da a yp es (ADT). The mo dula me ho dology oe s many imp o an p op e ies such as unde s anding,
abs ac ion, eusabili y, e c., bu , o en, i implies a loss o eciency due o he imp ossibili yo
accessing implemen a ion o ADTs om o he mo dules.
O en, while designing a da a s uc u e, we ace he p oblem o educing he space cos (i.e.,
minimising edundan da a) o he emp o al cos (i.e., accessing mo e as e o da a). Too e come
hese p oblems wemay need an implemen a ion which p e mi s he access o he da a by means
o p oin e s (see, e.g, 1, pag. 149]). Recall ha he no ion o
poin e
co esp onds o he memo y
add ess o a gi en ob jec (wi h some high-le el acili ies, which a y among he die en p og amming
languages). Un o una ely, he use o p oin e s causes in gene al a comple e loss o mo dula i y b ecause:
I is necessa y o know he unde lying da a s uc u e used in he implemen a ion and he e o e,
in o ma ion hiding is no accomplished.
Co ec ness p o o s and so wa e main enance and eusabili y a e, in gene al, mo e complica ed.
Also, p og ams a e mo e dicul o unde s and.
An implemen a ion using p oin e s needs o gua an ee ha da a will no change hei p osi ion in
he s uc u e o he wise wewould no access he desi ed in o ma ion (unless all clien mo dules
keep acko changes and his may b e imp ossible).
When he implemen a ion is done by means o p oin e s no only he ADT op e a ions can b e
used bu also we can access di ec ly o i s ep esen a ion and manipula e i . The e o e, i is
p ossible o manipula e ADTs wi hou esp ec ing he p olicy dened by hei equa ions.
Wi h he aim o a oiding hese d awbacks, we p op ose he design o a new ADT, which no only
p e mi s o s o e da a and o access hem bymeans o akey, bu also allows o di ec access o da a
wi hou knowing how hey a e s o ed. We should men ion ha he addi ion o his mechanism will
b e done by dening i s o mal seman ics. Ce ainly, he idea o s udy his new ADT comes om he
necessi y o a comp omise b e ween mo dula i y and eciency.
The goal o he p esen wo k is o p esen he design o his new ADT which oe s, in he one hand,
unc ions o da a s o age and access bykey o da a s uc u es, indep enden ly o hei o ganiza ion
3
and o he key yp e and, on he o he hand, he p ossibili y o a di ec access o da a by means o
an add ess we call he
sho cu
. Essen ially,we ob ain an abs ac mechanism inside he ADT which
co esp onds o he p oin e s in b o h concep and eciency.
The es o he pap e is o ganized as ollows. In sec ion 2, we p esen he algeb aic sp ecica ion
o he new ADT showing he so s and he op e a ions oe ed as well as he equa ions ha dene
he b eha iou o each o hem. In sec ion 3, wecho ose s he ep esen a ion o he new da a yp e,
hen we show he comple e de i a ion o one o he op e a ions and nally we indica e he cos o
each op e a ion, dep ending on he de ailed implemen a ion o he ep esen a ion used. In sec ion 4, we
conside a classic example ha equi es he use o p oin e s o eciency easons, and whe e we can see
ha he use o he new ADT assu es he same eciency wi hou losing mo dula i y a all. In sec ion
5, we include se e al conclusions ega ding he co dica ion o he new ADT in ADA, Mo dula-2 and
C++. Finally, in sec ion 6, we summa ise he ad an ages o he new ADT and we p esen some op en
ques ions o u he esea ch. This pap e is based on 11], whe e all he emaining discussions can b e
ound, ega ding he de i a ion o all he op e a ions and de ails o he h ee implemen a ions.
2 The ADT
STORE
Since wewan o ob ain an ADT which p e mi s o s o e any kind o in o ma ion, weha e designed an
ADT ha is gene ic" ( pa ame e ized"). The new ADT, called
STORE
, has he da a o b e s o ed
as a pa ame e , decomp osed in o
key
and
in o ma ion
. The pa ame e
key
is used o access he da a.
The e a e no op e a ions equi ed on
in o ma ion
he so
key
mus ha e compa ison op e a ions
1
.
The ADT
STORE
oe s wo so s:
s o e:
whe e he da a is s o ed.
sho cu :
which supplies he di ec access o he s o ed da a.
1
F om now on, we use he op e a ion symbol
cmp
o e e o he equali y op e a ion.
4
2.1 Op e a ions
2.1.1 Cons uc o Op e a ions
c ea e:
;!
s o e
c ea es an emp y s o e.
pu : s o e, key, in o ma ion
;!h
s o e, sho cu
i
e u ns a pai o med by he esul ing s o e a e adding he pai o key and in o ma ion,
and he sho cu whichgi es di ec access o his pai .
Al e na i ely,we could decomp ose
pu
in o wo op e a ions, one o each comp onen o he esul .
Howe e , weha e p e e ed he ab o e men ioned s uc u e o he sake o clea ness o algeb aic
sp ecica ion p esen ed in Subsec ion 2.2.
2.1.2 Obse e Op e a ions
ge In Key: s o e, key
;!
in o ma ion
e u ns he in o ma ion asso cia ed o he key. P econdi ion: he key is in he s o e.
ge In Sho: s o e, sho cu
;!
in o ma ion
e u ns he in o ma ion asso cia ed o he sho cu . P econdi ion: he sho cu has da a
asso cia ed inside he s o e.
ge Key: s o e, sho cu
;!
key
e u ns he key asso cia ed o he sho cu . P econdi ion: he sho cu has da a asso cia ed
inside he s o e.
isIn?: s o e, key
;!
bool
e u ns a b o olean alue indica ing whe he he key is in he s o e.
ge Sho cu : s o e, key
;!
sho cu
e u ns he sho cu asso cia ed o he key. P econdi ion: he key is in he s o e.
isEmp y?: s o e
;!
bool
e u ns a b o olean alue indica ing whe he he s o e is emp y.
5
2.1.3 Mo die Op e a ions
emo e: s o e, sho cu
;!
s o e
e u ns he esul ing s o e a e emo ing he da a asso cia ed o he sho cu . P econdi-
ion: he sho cu has da a asso cia ed inside he s o e.
modi y: s o e, sho cu , in o ma ion
;!
s o e
e u ns he esul ing s o e a e subs i u ing he in o ma ion asso cia ed o he sho cu ,
by he new in o ma ion. P econdi ion: he sho cu has da a asso cia ed inside he s o e.
2.2 Equa ional sp ecica ion
Following he gene al me ho d o algeb aic sp ecica ion wi h ini ial seman ics, wegi e s he equa-
ions b e ween cons uc o op e a ions 4, 5, 6].
2.2.1 Equa ions b e ween Cons uc o Op e a ions
I seems ha we could assume ha he
s o es
b eha e as se s, in ha he o de ing in which he da a
a e s o ed is no imp o an . Ac ually his happ ens in able" like ADT's 6]. In o de o assu e his
p op e ywe should ha e o gua an ee ha , o a gi en
key
,we ob ain always he same
sho cu
,
ega dless o he momen in which he
key
and i s
in o ma ion
a e s o ed. This would imply ha
die en keys could no ha e he same
sho cu
o he wise, s o ing an
in o ma ion
and a
key
implies a
isk o mo di ying he
in o ma ion
o ano he key (o he same
sho cu
) whichwas s o ed p e iously.
The e o e, he e should b e a unique
sho cu
co esp onding o a
key
, and his, in u n, equi es he
numbe o keys and he keys hemsel es o b e known b e o e hand. Al e na i ely,we can do i byan
injec i e unc ion om he se o keys o he se o sho cu s. The las equi emen is e y es ic i e,
i wewan he
s o e
o b e indep enden om he
key
.
Weo e come his es ic ion, by le ing he
sho cu
o dep end also on he
s o e
, and in his way
die en keys can ha e he same
sho cu
, i hey a e s o ed in die en s o es. The e o e, he o de
in which da a a e s o ed is imp o an , and hus
s o es
will no b eha e as se s.
Now, le us conside he ela ions b e ween he wo consecu i e
pu
op e a ions, when we s o e
some
in o ma ion
wi h a
key
whichwas p e iously in he
s o e
. Since we can ha e only a unique
in o ma ion
asso cia ed o each
key
, he esul will b e ha o eplacing he
in o ma ion
asso cia ed o
he
key
by he new
in o ma ion
. The ques ion ha a ises he e is: whichis he
sho cu
ob ained?
6
Since he e was a
sho cu
asso cia ed o he
key
( i.e., he
sho cu
ob ained when s o ed he p e ious
in o ma ion
) and since wewan a unique
sho cu
co esp onding o he
key
, hen i is p ecisely he
o me
sho cu
he one we will ob ain.
Le us see he equa ions which exp ess hese ela ions. Le us s a by hose ela ed o s o es we
will use he op e a ion
isIn?
which ell us whe he he key is in he s o e. This op e a ion is used only
when he s o e is no emp y, b ecause o he wise he s o e would no ha e his ela ion.
In he case when he
key
is he ein, we conside he ollowing wo cases:
1. The key is he las one added o he s o e.
2. The key exis s bu i is no he las one.
In he s case, he equa ion is simple i jus exp esses he ac ha he ob ained s o e will b e
he same as i he las ime we had s o ed he new in o ma ion.
h
isIn?(pu (s ,k
1
,i
1
).s o e,k
2
)
^
cmp(k
1
,k
2
)
i
)
pu (pu (s ,k
1
,i
1
).s o e,k
2
,i
2
).s o e = pu (s ,k
2
,i
2
).s o e
In he second case, he equa ion exp esses he swap b e ween he wo las keys ob aining his:
h
isIn?(pu (s ,k
1
,i
1
).s o e,k
2
)
^:
cmp(k
1
,k
2
)
i
)
pu (pu (s ,k
1
,i
1
).s o e,k
2
,i
2
).s o e = pu (pu (s ,k
2
,i
2
).s o e,k
1
,i
1
).s o e
No ice ha in his case he e is no a subs i u ion o he in o ma ion asso cia ed o he key, bu
applying his equa ion successi ely,we will a i e o he s case since he key is he ein. The e o e,
we can apply he s equa ion and in his way o subs i u e he in o ma ion asso cia ed o he key.
Le us see now he equa ions o sho cu s. As b e o e, we dis inguish wo cases:
1. The key is he las one added o he s o e.
2. The key exis s, bu i is no he las one.
In he s case, we only ha e o exp ess ha he ob ained
sho cu
will b e he one ob ained
p e iously.
h
isIn?(pu (s ,k
1
,i
1
).s o e,k
2
)
^
cmp(k
1
,k
2
)
i
)
pu (pu (s ,k
1
,i
1
).s o e,k
2
,i
2
).sho cu = pu (s ,k
1
,i
1
).sho cu
7
In he second case weha e o jus exp ess ha he ob ained
sho cu
will b e he same as he one
ha wewould ha e ob ained, i he new key and he new in o ma ion had b een s o ed in he p e ious
s o e.
h
isIn?(pu (s ,k
1
,i
1
).s o e,k
2
)
^:
cmp(k
1
,k
2
)
i
)
pu (pu (s ,k
1
,i
1
).s o e,k
2
,i
2
).sho cu = pu (s ,k
2
,i
2
).sho cu
This do es no sp eci y ye comple ely he b eha iou o sho cu s, b ecause weha e no conside ed
he case in which he key is no he ein in ac , in his case,
pu
may e u n any alue as sho cu
bu he access o he sho cu should supply he app op ia e esul s. The equa ions o he op e a ion
ge In Sho
will desc ib e his b eha iou , as gi en nex .
2.2.2 Equa ions o he Obse e Op e a ions
Le us s udy now he b eha iou o he obse e op e a ions.
The equa ions o
ge In Sho
:
{
The s one will b e an e o equa ion, indica ing ha he op e a ion is no dened o an
emp y
s o e
. Recall ha his op e a ion is pa ial.
ge In Sho(c ea e,sc) = e o
{
We also include an equa ion o exp ess, he b eha iou when he
sho cu
whose
in o ma ion
wewan o consul is he same as he one e u ned in he las op e a ion o
s o e
.The
in o ma ion
ha we will ob ain will b e he one o his las op e a ion. In o de o compa e
he
sho cu s
we will use a p i a e op e a ion, deno ed
cmp
.
h
cmp(sc,pu (s ,k,i).sho cu )
i
)
ge In Sho(pu (s ,k,i).s o e,sc) = i
{
The nex equa ion indica es ha , i he
sho cu
is no he same as he one ob ained in
he las op e a ion o
pu
, hen he esul o consul ing he in o ma ion asso cia ed o he
sho cu
will b e he same as consul ing he p e ious
s o e
.
h
:
cmp(sc,pu (s ,k,i).sho cu )
i
)
ge In Sho(pu (s ,k,i).s o e,sc) = ge In Sho(s ,sc)
8
No ice ha , i he e was no
in o ma ion
asso cia ed o he
sho cu
,by applying successi ely
his equa ion, wewould ob ain
s =c ea e
, and he e o e om he s equa ion his is an
e o .
Equa ions o
ge Key
, simila o hose o
ge In Sho
:
ge Key(c ea e,sc) = e o
h
cmp(sc,pu (s ,k,i).sho cu )
i
)
ge Key(pu (s ,k,i).s o e,sc) = k
h
:
cmp(sc,pu (s ,k,i).sho cu )
i
)
ge Key(pu (s ,k,i).s o e,sc) = ge Key(s ,sc)
The Equa ions o
ge In Key
:
{
The s equa ion s a es ha , i he
key
o b e consul ed is he las one s o ed hen we will
ob ain he las he
in o ma ion
s o ed.
h
cmp(k
1
,k
2
)
i
)
ge In Key(pu (s ,k
1
,i).s o e,k
2
)= i
{
I he condi ion o he s equa ion do es no hold hen weha e a second equa ion, and
he co esp onding esul will b e ha o applying
ge In Key
on he p e ious
s o e
.
h
:
cmp(k
1
,k
2
)
i
)
ge In Key(pu (s ,k
1
,i).s o e,k
2
)=ge In Key(s ,k
2
)
{
Since he op e a ion in ques ion is a pa ial one (b ecause i is no dened when he
key
is no in he
s o e
)wemus also add a las equa ion which indica es ha consul ing he
in o ma ion
asso cia ed o a
key
in an emp y
s o e
will b e an e o .
ge In Key(c ea e,k) = e o
Equa ions o
isIn?
:
{
The s equa ion is in ui i ely clea . I we apply
isIn?
o a
key
and o an emp y
s o e
,
hen he esul will b e alse.
isIn?(c ea e,k) = alse
{
In he same way, he second equa ion comes na u ally. I he
key
o whichwe apply he
op e a ion
isIn?
is he las one added o he
s o e
, hen he esul will b e ue i he
key
is no he las one added, hen he esul will b e ha o applying
isIn?
o he p e ious
s o e
.
9
Using his ep esen a ion o he yp e
s o e
we s ill ha e an op e a ion o high cos , namely he
pu
op e a ion. When we s o e he da a by means o a
key
which is al eady in he
s o e
,weha e only o
ob ain he
sho cu
asso cia ed o he
key
and hen using i o access o he a ay, assigning he alue
o he new da a. In his case, he cos is ha o consul ing he able. Howe e , when he
key
is no
in he
s o e
,weha e o sea ch o ee p osi ion in he a ay, and his, in he wo s case, has linea
cos . Toa oid his cos we add ano he s uc u e o nd quickly a ee p osi ion. This s uc u e is a
queue whe e he ee p osi ions o he a ay a e s o ed (see Fig. 3). Wi h his mo dica ion, he cos
o nding a ee p osi ion is cons an (p o ided ha he implemen a ion o he queue is go o d enough,
i co esp onds jus o ob aining he s elemen o he queue). The cos o pu ing a p osi ion as a
ee one, while emo ing, could b e made also cons an i i jus co esp onds o he cos o pu ing an
elemen o he queue.
k
1
k
6
k
5
k
3
k
2
k
4
1 2 3 4 5 6 7 8 9 10 11 12 13
k
5
k
1
k
3
k
6
k
4
k
2
in
5
in
1
in
3
in
6
in
4
in
2
j
q
q)
-
-
666I 66
Figu e 3: a ay, able and queue
16
The ep esen a ion o he new so s o da a is:
s o e
is implemen ed by a eco d consis ing o h ee elds:
1.
a
: a ayo
n
comp onen s, whe e he alid indices a e he na u al numbe s be ween 1 and
n
, o a ce ain na u al
n
he comp onen s o he a ay a e pai s consis ing o wo elds:
in
(o yp e
in o ma ion
)and
key
(o yp e
key
).
2.
: a queue o
sho cu s
i.e., a queue con aining he indices co esp onding o he ee p osi-
ions o he a ay.
3.
: is a able o pai s
h
k ey sho cu
i
.
The
sho cu
is implemen ed by a na u al in he ange 0
:::n
. The alue 0 is used as an undened
alue. The es o he alues will co esp ond o he indices o he a ay.
To e i y ha he implemen a ion o he new ADT is co ec wi h esp ec o i s sp ecica ion we
ollow he me ho d p esen ed in 8]. I consis s o dening he abs ac ion unc ion, which ans o ms a
alue o he implemen a ion o a alue o i s mo del, deno ed by a e m which b elongs o i s asso cia ed
equi alence class in he quo ien - e m algeb a 4]. Since his unc ion is pa ially dened, wemus
also dene he in a ian o he ep esen a ion which es ablishes he condi ion ha he alues o he
implemen a ion mus sa is y so o ep esen alid alues o he ADT. Also, we need o indica e when
wo die en alues o he implemen a ion co esp ond o he same alue o ADT unde implemen a ion,
in o he wo ds, weha e o sp eci y he edeni ion o he equali y.
He e is a sho desc ip ion o he a ay, queue and able op e a ions, esp ec i ely, ha we a e going
o use.
cons
(
a p
) e u ns he alue in he p osi ion
p
o he a ay
a
ass
(
a p
) assigns he alue
in
he p osi ion
p
o he a ay
a
pu F i s
(
q
) pu s he alue
in he on o he queue
q
assig n
(
k
)
asso cia es he alue
o he key
k
in he able
deassig n
(
k
) asso cia es an undened alue o he
key
k
in he able
and
l ook up
(
k
) e u ns he alue asso cia ed o he key
k
in he able
.
The Abs ac ion Func ion
con e
(
h
a
i
)=
8
>
>
>
>
<
>
>
>
>
:
c ea e
i
=TABLE.c ea e
pu
(
con e
(
h
a pu F i s
(
sc
)
deassig n
(
k
)
i
)
k cons
(
a sc
)
:in
)
:s o e
i
=assign( ',k,sc)
17
The Rep esen a ion In a ian
The ep esen a ion in a ian mus assu e ha e e y p osi ion o he a ay is ee o o ccupied bu no
b o h o hem a he same ime. The e o e, i indica es ha any p osi ion o he a ay, i.e., any na u al
be ween 1 and
n
mus b e ei he in he queue o ee p osi ions (only once) o as a alue asso cia ed o
a
key
in he able. In his las case, he
key
canno b e wha e e : i mus coincide wi h he alue o
he eld o he a ay in his p osi ion. Since a
sho cu
canno b e asso cia ed o mo e han a
key
,a a
ce ain momen , wemus assu e also ha , gi en a p osi ion he e is only one
key
in he able whichi
is asso cia ed o.
Bu he
sho cu
dep ends on he
s o e
, he e o e he ep esen a ion in a ian mus indica e which
is he
sho cu
ha will assign ou ep esen a ion when we s o e an in o ma ion by means o a
key
ha is no ound in he
s o e
. Since he only condi ion ha mus sa is y he
sho cu
, is no o b e
asso cia ed o ano he
key
,any o he queue p osi ions can b e used, o ins ance i s head.
We can now w i e down he in a ian .
Fi s we will dene a p edica e ha indica es whe he a
sho cu
b elongs only once o he queue.
8
2
q ueue
8
sc sc
1
sc
2
2
sho cu
uniq ue
(
c ea e sc
)=
alse sc
2
c ea e
=
alse
uniq ue
(
enq ueue
(
sc
)
sc
)=
:
(
sc
2
)
sc
2
enq ueue
(
sc
)=
ue
:
ig
(
sc
1
sc
2
)]
)
:
ig
(
sc
1
sc
2
)]
)
uniq ue
(
enq ueue
(
sc
1
)
sc
2
)=
uniq ue
(
sc
2
)
sc
2
2
enq ueue
(
sc
1
)=
sc
2
2
In Rep(
h
a
i
)
8
j
:1
j
n
:
uniq ue
(
j
)
^:9
k
:
l ook up
(
k
)=
j
_
:
(
j
2
)
^9
!
k
:
lookup
(
k
)=
j
^
8
k
:
l ook up
(
k
)
6
=
unde ined
:
cons
(
a l ook up
(
k
))
:key
=
k
^
8
k
:
:
isI n
?(
con e
(
h
a
i
)
k
):
pu
(
con e
(
h
a
i
)
ki
)
:sho cu
=
head
(
)
The Equali y Redeni ion
The equali y edeni ion exp esses ha he alues o he a ay in i s ee p osi ions ( hose ound in
he queue) a e no ele an o he abs ac alue. This means ha he same
s o e
is ob ained i any
ee p osi ion o he a aywe e mo died. Also, i should exp ess ha wo
s o es
a e equal when wo
ables, cons uc ed in die en ways, con ain he same se o pai s. Bu his has b een con empla ed in
he equa ions o he yp e able, and he e o e he e is no need o include i in he equali y edeni ion.
18
8
j
:1
j
n
^
(
j
2
):
con e
(
h
a
i
)=
con e
(
h
ass
(
a j al
)
i
)
3.2 De i ing co de o he op e a ions
In his subsec ion, we p esen he de i a ion o he implemen a ion a pa icula op e a ion,
pu
,asa
case s udy o he whole de i a ion p o cess
2
. In o de o simpli y he demons a ions wewo k unde
he hyp o hesis ha
n
is big enough o assu e ha he queue is ne e emp y. La e , a he nal
implemen a ion, weha e conside ed his p ossible case o e o .
3.2.1 De i a ion o
pu
Acco ding o 8] he op e a ion
pu
mus b e implemen ed by a unc ion sa is ying he ollowing p e-p os
sp ecica ion:
unc ion
impl pu (
h
a:a ay :queue : able
i
k:keyi:in o ma ion)
e u n
hh
a
1
:a ay
1
:queue
1
: able
i
sc:sho cu
i
P e: In Rep(
h
a, ,
i
)
g
P
Pos : In Rep(
h
a
1
,
1
,
1
i
)
^
con e (
h
a
1
,
1
,
1
i
)=pu (con e (
h
a, ,
i
),k,i).s o e
^
sc=pu (con e (
h
a, ,
i
),k,i).sho cu
g
e u n
hh
a
1
,
1
,
1
i
,sc
i
The de i a ion o his unc ion is c ucial o he e ica ion p o cess no only b ecause o i s dicul y
bu also due o he ac ha weha e o p o e, p e iously, a se ie o lemmas necessa y also o he
de i a ion o he es o he unc ions.
The Lemmas ega ding he ables
Lemma 1
lookup(assign( ,k,sc),k
1
)=undened
]
):
ig(k,k
1
)
Lemma 2
lookup( ,k)=undened
]
)
deassign( ,k)=
Lemma 3
lookup( ,k)=undened
]
)
deassign(assign( ,k,sc),k)=
Lemma 4
lookup( ,k)
6
=
undened
]
)
6
=
TABLE.c ea e
2
See 11] o he comple e de i a ion.
19
Lemma 5
lookup( ,k)
6
=
undened
]
)
=assign(
1
,k,lookup( ,k))
Lemma 6
lookup(deassign( ,k),k)=undened
Lemma 7
=assign(
1
,k
1
,sc
1
)
]
)
=assign(
2
,k
1
,sc
1
)
^
lookup(
2
,k
1
)=undened
Lemma 8
isIn?(con e (
h
a, ,
i
),k)=(lookup( ,k)
6
=
undened)
The Lemmas ega ding he queues
Lemma 9
pu Fi s (dequeue(q),head(q))=q
Lemma 10
head(pu Fi s (q, ))=
Using hese lemmas le us see he de i a ion o he unc ion
pu
.
The esul o he op e a ion will dep end on whe he he
key
o b e s o ed is ound in he
s o e
o
no . The e o e, we conside an al e na i e o he design o he unc ion. Since, om Lemma 8, he
esul o
isIn?(con e (
h
a, ,
i
),k)
is equi alen o
lookup( ,k)
6
=
undened
we will pu his las one as a
condi ion o he al e na i e.
unc ion
impl pu (
h
a:a ay :queue : able
i
k:keyi:in o ma ion)
e u n
hh
a
1
:a ay
1
:queue
1
: able
i
sc:sho cu
i
P e: In Rep(
h
a, ,
i
)
g
i
lo okup( ,k)=undened
hen
A
1
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
g
P
1
else
A
2
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened
g
P
2
end i
Pos : In Rep(
h
a
1
,
1
,
1
i
)
^
con e (
h
a
1
,
1
,
1
i
)=pu (con e (
h
a, ,
i
),k,i).s o e
^
sc=pu (con e (
h
a, ,
i
),k,i).sho cu
g
e u n
hh
a
1
,
1
,
1
i
,sc
i
20
Le 's see wha asse ion we can deduce om A
1
.
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
)
( Lemma 8 )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^:
isIn?(con e (
h
a, ,
i
),k)
)
( Deni ion o In Rep )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
)
( Equali y Redeni ion and head( )
2
)
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
con e (
h
a, ,
i
)= con e (
h
ass(a,head( ),
h
k,i
i
), ,
i
)
)
( The in o duc ion o he op e a ion pu in b o h sides o he equali y)
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,head( ),
h
k,i
i
), ,
i
),k,i).s o e
)
( Lemma 3 )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,head( ),
h
k,i
i
), ,deassign(assi gn ( ,k,head( )),k)
i
),k,i).s o e
)
( Lemma 9 )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e=
pu (con e (
h
ass(a,head( ),
h
k,i
i
),pu Fi s (d equeu e( ),head( )),deassi gn(ass ign ( ,k,head( )),k)
i
),k,i).s o e
)
( Eq. o a ay cons(ass(A,i, ),i)= )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,head( ),
h
k,i
i
),pu Fi s (deq ueue( ),head( )),
deassign(assign ( ,k,head( )),k)
i
),k,cons(ass(a,hea d( ),
h
k,i
i
), head( )).in ).s o e
)
( Abs ac ion Func ion )
In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e= con e (
h
ass(a,head( ),
h
k,i
i
),dequeue( ), assign( ,k,head( ))
i
)
)
( In Rep(
h
a, ,
i
)
^
lo okup( ,k)=unden ed]
)
In Rep(
h
ass(a,head( ),
h
k,i
i
),dequeue( ), assign( ,k,head( ))
i
))
In Rep(
h
ass(a,head( ),
h
k,i
i
),dequeue( ), assign( ,k,head( ))
i
)
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head( )
^
pu (con e (
h
a, ,
i
),k,i).s o e= con e (
h
ass(a,head( ),
h
k,i
i
),dequeue( ), assign( ,k,head( ))
i
)
The e o e, om A
1
we can deduce his las asse ion, deno ed A
3
.
Le 's see wha asse ion we can deduce om A
2
.
In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened
)
( Lemma 5 )
In Rep(
h
a, ,
i
)
^
=assign(
1
,k,lo okup( ,k)) )
21
)
( Abs ac ion Func ion )
In Rep(
h
a, ,
i
)
^
con e (
h
a, ,
i
)= pu (con e (
h
a,pu Fi s ( ,lo oku p( ,k)),deassi gn( ,k)
i
),k, cons(a,lo okup( ,k)).in ).s o e
)
( Equali y Redeni ion and
8
k:k
2
pu Fi s ( ,k) )
In Rep(
h
a, ,
i
)
^
con e (
h
a, ,
i
)= pu (con e (
h
ass(a,lo okup ( ,k),
h
cons(a,lo o kup ( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassig n( ,k)
i
),k,cons(a,lo okup ( ,k)).in ).s o e
)
( The in o duc ion o he op e a ion pu in b o h sides o he equali y)
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (pu (con e (
h
ass(a,lo oku p( ,k),
h
cons(a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,cons(a,l o oku p( ,k)).in ).s o e,k,i).s o e
^
pu (con e (
h
a, ,
i
),k,i).sho cu = pu (pu (con e (
h
ass(a,lo o kup( ,k),
h
cons (a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,cons(a,lo oku p( ,k)).in ).s o e,k,i).sho cu
)
( Eq. o s o e pu (pu (s ,k,i
1
).s o e,k,i
2
).s o e= pu (s ,k,i
2
).s o e, and
pu (pu (s ,k,i
1
).s o e,k,i
2
).sho cu =pu (s ,k,i
1
).sho cu )
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,lo okup( ,k),
h
cons(a , lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,i).s o e
^
pu (con e (
h
a, ,
i
),k,i).sho cu = pu (con e (
h
ass(a,lo oku p( ,k),
h
cons(a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,cons(a,l o ok up( ,k)).in ).sho cu
)
( Eq. o a ay cons(ass(A,i, ),i)= )
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,lo okup( ,k),
h
cons(a , lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn ( ,k)
i
),k,cons(ass(a,lo okup ( ,k),
h
cons(a,lo okup( ,k)).key,i
i
),lo ok up( ,k)).in ).s o e
^
pu (con e (
h
a, ,
i
),k,i).sho cu = pu (con e (
h
ass(a,lo oku p( ,k),
h
cons(a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,cons(a,l o oku p( ,k)).in ).sho cu
)
( Lemma 6 )
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,lo okup( ,k),
h
cons(a , lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn ( ,k)
i
),k,cons(ass(a,lo okup ( ,k),
h
cons(a,lo okup( ,k)).key,i
i
),lo ok up( ,k)).in ).s o e
^
pu (con e (
h
a, ,
i
),k,i).sho cu = pu (con e (
h
ass(a,lo oku p( ,k),
h
cons(a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassig n( ,k)
i
),k,cons(a,lo okup ( ,k)).in ).sho cu
^
lo okup(deassig n( ,k),k)=unde ned
)
( Lemma 8 )
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,lo okup( ,k),
h
cons(a , lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn ( ,k)
i
),k,cons(ass(a,lo okup ( ,k),
h
cons(a,lo okup( ,k)).key,i
i
),lo ok up( ,k)).in ).s o e
^
22
pu (con e (
h
a, ,
i
),k,i).sho cu = pu (con e (
h
ass(a,lo oku p( ,k),
h
cons(a, lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassig n( ,k)
i
),k,cons(a,lo okup ( ,k)).in ).sho cu
^
:
isIn?(con e (
h
ass(a,lo ok up( ,k),
h
cons(a, lo okup( ,k)).key,i
i
), pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k)
)
(In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened ]
)
In Rep(
h
ass(a,lo okup( ,k),
h
cons(a,l o oku p( ,k)) .key,i
i
), pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
))
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).sho cu =head(pu Fi s ( , lo okup( ,k))
^
pu (con e (
h
a, ,
i
),k,i).s o e= pu (con e (
h
ass(a,lo okup( ,k),
h
cons(a , lo okup( ,k)).key,i
i
),
pu Fi s ( ,lo okup( ,k)),deassi gn( ,k)
i
),k,cons(as s(a,lo o kup ( ,k),
h
cons(a,lo okup( ,k)).key,i
i
),lo okup ( ,k)).in ).s o e
)
( Abs ac ion Func ion and Lemma 10 )
In Rep(
h
a, ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= con e (
h
ass(a,lo okup( ,k),
h
cons(a,lo okup ( ,k)).key,i
i
), ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).sho cu =lo okup( ,k)
)
( In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened ]
)
In Rep(
h
ass(a,lo okup( ,k),
h
cons(a,lo okup( ,k)).key,i
i
), ,
i
))
In Rep(
h
ass(a,lo okup( ,k),
h
cons(a,lo okup ( ,k)).key,i
i
), ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).s o e= con e (
h
ass(a,lo okup( ,k),
h
cons(a,lo okup ( ,k)).key,i
i
), ,
i
)
^
pu (con e (
h
a, ,
i
),k,i).sho cu =lo okup( ,k)
The e o e, om A
2
we can deduce his las asse ion, deno ed A
4
.
By in o ducing hese new asse ions weha e:
unc ion
impl pu (
h
a:a ay :queue : able
i
k:keyi:in o ma ion)
e u n
hh
a
1
:a ay
1
:queue
1
: able
i
sc:sho cu
i
P e: In Rep(
h
a, ,
i
)
g
i
lo okup( ,k)=undened
hen
A
1
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
g
A
3
g
P
1
else
A
2
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened
g
A
4
g
P
2
end i
Pos : In Rep(
h
a
1
,
1
,
1
i
)
^
con e (
h
a
1
,
1
,
1
i
)=pu (con e (
h
a, ,
i
),k,i).s o e
^
sc=pu (con e (
h
a, ,
i
),k,i).sho cu
g
e u n
hh
a
1
,
1
,
1
i
,sc
i
23
Ha ing hese wo new asse ions i is easy o iden i y,by means o he assignmen ule, which
exp essions should b e assigned o
a
1
1
1
, and
sc
in each case. By doing hese assignmen s we ob ain
he ollowing implemen a ion o he
pu
unc ion:
unc ion
impl pu (
h
a:a ay :queue : able
i
k:keyi:in o ma ion)
e u n
hh
a
1
:a ay
1
:queue
1
: able
i
sc:sho cu
i
P e: In Rep(
h
a, ,
i
)
g
i
lo okup( ,k)=undened
hen
A
1
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)=undened
g
A
3
g
a
1
:= ass(a,head( ),
h
k,i
i
)
1
:= dequeue( )
1
:= assign( ,k,head( ))
sc:= head( )
else
A
2
:In Rep(
h
a, ,
i
)
^
lo okup( ,k)
6
=undened
g
A
4
g
a
1
:= ass(a,lo okup( ,k),
h
cons(a,lo okup( ,k)).key,i
i
)
1
:=
1
:=
sc:= lo okup( ,k)
end i
Pos : In Rep(
h
a
1
,
1
,
1
i
)
^
con e (
h
a
1
,
1
,
1
i
)=pu (con e (
h
a, ,
i
),k,i).s o e
^
sc=pu (con e (
h
a, ,
i
),k,i).sho cu
g
e u n
hh
a
1
,
1
,
1
i
,sc
i
24
3.3 Chosing implemen a ions o he comp onen s o he ep esen a ion
Once weha e de i ed he co de o he op e a ions, we s udy wo pa icula implemen a ions o he new
ADT
3
. In b o h o hem we use he dynamic memo y zone as an a ay and he queue o he sys em as
ha o he ee sho cu s. These wo implemen a ions die in he implemen a ion o he able. In he
s one, weha e used a hashing able wi h chaining. Using a hashing able equi es he pa ame e
KEY
o include also a hashing unc ion. In he second implemen a ion weha e implemen ed he
able wi h an AVL
4
1].
We gi e nex he cos o he op e a ions o he ADT dep ending on he implemen a ion used o
able (see Table 1).
Table 1: The cos o he unc ions
COST
FUNTIONS AVL HASHING
TABLE
c ea e
O
(1)
O
(
)
pu
O
(log
n
)
O
(1)
ge In Key
O
(log
n
)
O
(1)
ge In Sho
O
(1)
O
(1)
ge Key
O
(1)
O
(1)
isIn?
O
(log
n
)
O
(1)
ge Sho cu
O
(log
n
)
O
(1)
isEmp y?
O
(1)
O
(1)
emo e
O
(log
n
)
O
(1)
mo di y
O
(1)
O
(1)
As i can b e seen, he cos o he op e a ions
ge Key
,
ge In Sho
and
modi y
is cons an in b o h
implemen a ions b ecause he able is no accessed, in his case. The cos o he emaining op e a ions
3
See 11] o mo e de ails.
4
An AVL is a bina y sea ch ee whe e he di e ence b e ween he heigh o i s sub ees is less o equal han 1 and he
sub ees a e AVL in u n.
25
14]
Wi h, N.
P og amminginModula-2.
Sp inge -Ve lag, 3 d. edi ion, 1988.
32