✬
Algìrijmoi kai Poluplokìthta
✩
Diˆlexh 12
1
Μέγιστη κοινή ακολουθία
Orismìc
X =< x1 , x2 , · · · , xm >, mÐa ˆllh akoloujÐa
Z =< z1 , z2 , · · · , zk > eÐnai upì-akoloujÐa thc X eˆn upˆrqei mÐa gnhsÐwc aÔxousa
akoloujÐa < i1 , i2 , · · · , ik > apì deÐktec thc X tètoia ¸ste gia ìla ta j = 1 · · · k
isqÔei xi = zj .
j
Me dedomènh mÐa akoloujÐa
Parˆdeigma
X =< A, B, C, A, D, E, A, F, B >
Upì-akoloujÐa thc
•
X
AkoloujÐa deikt¸n thc
Z =< A, C, D, E, A >
< 1,
3, 5, 6, 7
>
Z =< A, B >
< 1,
2
>, < 1,
9
X
>, < 4,
9
>
Mia kalÔterh anaparˆstash:
X: <A B C A D E A F B>
✫
✪
Z: <A C D E A>
✬
Algìrijmoi kai
Poluplokìthta
Diˆlexh
12
To prìblhma thc mègisthc koin c upì-akoloujÐac (MKU)
(
✩
2
The longest common subsequence (LCS) problem)
EÐsodoc:
Duo
'Exodoc:
H
akoloujÐec
X
kai
Y
.
mègisth koin upì-akoloujÐa.
ParadeÐgmata
X: A B C B D A B
X: A B C B D A B
Y: B D C A B A
Y: B D C A B A
X: A B C B D A B
Y: B D C A B A
✫
✪
✬
Algìrijmoi kai
Poluplokìthta
Diˆlexh
✩
12
3
GiatÐ meletˆme to prìblhma thc MKU?
•
LÔnetai me
•
'Eqei arketèc
qr sh
dunamikoÔ
programmatismoÔ.
efarmogèc.
Efarmog Me
dedomènec
2
diaforetikèc
prosdioristoÔn oi allagèc
ekdìseic
pou
(versions)
èginan apì
thn
enìc
pr¸th
progrˆmmatoc, na
èkdosh
èwc
th
deÔterh.
LÔsh
BreÐte thn
MKU
apì
MKU
twn dÔo
proèrqetai apì
thn
mÐa
èkdosh
programmˆtwn.
prosjèseic proc
thn
To
keÐmeno
afairèseic pou
pou
den
eÐnai mèloc
èginan katˆ thn
metˆbash
ˆllh.
✫
✪
✬
Algìrijmoi kai Poluplokìthta
✩
Diˆlexh 12
4
Mia aploðk prosèggish
•
'Estw
•
'Elegxe kˆje upì-akoloujÐa tou
Er¸thsh
Apˆnthsh
thc
S
to sÔnolo ìlwn twn dunat¸n upì-akolouji¸n twn
S
kai
Y.
kai prosdiìrise thn MKU.
Poio eÐnai to mègejoc tou sunìlou
'Estw
X
S?
k = min(legth(X), length(Y )).
Tìte,
|S| = 2k .
Autì kajistˆ thn aploðk prosèggish anedafik !
✫
✪
✬
Algìrijmoi kai
Poluplokìthta
Diˆlexh
✩
12
5
MÐa anadromik lÔsh
'Estw
X =< x1 , x2 , · · · , xm >
Xi
Sumbìlise me
kai me
'Estw
Mac
Yi
c[i, j]
hn
upì-akoloujÐa
to mègejoc
na



c[i, j] =


.
Xi =< x1 , x2 , · · · , xi >
Yi =< y1 , y2 , · · · , yi >
thn upì-akoloujÐa
endiafèrei na
MporoÔme
Y =< y1 , y2 , · · · , yn >
kai
thc
.
MKU
twn
upologÐsoume to
grˆyoume
thn
Xi
kai
Yj
.
c[m, n]
.
anadromik sqèsh:
0
eˆn
i=0
c[i − 1, j − 1] + 1
eˆn
i, j > 0
kai
xi = y j
max(c[i, j − 1], c[i − 1, j])
eˆn
i, j > 0
kai
xi 6= yj
Er¸thsh
i) EÐnai h
ii) Pwc
Apˆnthsh
parapˆnw
katal xame
sqèsh
se
j=0
swst ?
aut ?
i) NAI
✫
✬
Algìrijmoi kai
ii)
•
eˆn
✪
Poluplokìthta
c[i, j] =
i=0





ken •
eˆn
6
eˆn
i=0
c[i − 1, j − 1] + 1
eˆn
i, j > 0
kai
xi = y j
max(c[i, j − 1], c[i − 1, j])
eˆn
i, j > 0
kai
xi 6= yj
kai
j=0
:
apì
upì-akoloujÐa
i, j > 0
✩
12
0
j=0
Toulˆqiston mÐa
Diˆlexh
tic akoloujÐec eÐnai ken .
(sumbolÐzetai me
xi = y j
to
ǫ
'Ara, h
mình upì-akoloujÐa eÐnai h
).
:
< x1 , x2 , · · · , xi−1 , xi >
=
< y1 , y2 , · · · , yj−1 , yj >
LCS(Xi , Yj ) = LCS(Xi−1 , Yj−1 ) ⊕ xi
=⇒ c[i, j] = c[i − 1, j − 1] + 1
(
⊕
sumbolÐzei
th
sunènwsh
sumboloseir¸n
/
akolouji¸n)
•
eˆn
i, j > 0
kai
xi 6= yj
:
< x1 , x2 , · · · , xi−1 , xi >
6=
< y1 , y2 , · · · , yj−1 , yj >
✫
LCS[Xi , Yj ] =





LCS[Xi , Yj−1 ]
LCS[Xi−1 , Yj ]
=⇒ c[i, j] = max(c[i, j − 1], c[i − 1, j])
✪
✬
Algìrijmoi kai Poluplokìthta
✩
Diˆlexh 12
7
MÐa anadromik prosèggish
•
Sthn perÐptwsh ìpou oi 2 upì-akoloujÐec den èqoun koinèc upì-akoloujÐec (ektìc thc
ǫ),
o algìrijmoc qreiˆzetai ekjetikì qrìno.
Parˆdeigma
Sqediˆste to grˆfhma anadrom c gia to
C[3, 4].
c[0,4]
c[1,4]
c[2,4]
c[3,4]
c[0,3]
c[1,3]
c[2,3]
c[3,3]
c[0,2]
c[1,2]
c[2,2]
c[3,2]
c[0,1]
c[1,1]
c[2,1]
c[3,1]
c[0,0]
c[1,0]
c[2,0]
c[3,0]
✫
✬
✪
Algìrijmoi kai Poluplokìthta
'Askhsh
Diˆlexh 12
Na upologisteÐ o arijmìc twn kl sewn thc diadikasÐac pou upologÐzei to
✩
8
c[i, j] katˆ th diˆrkeia tou upologismoÔ tou c[m, n], ìpou 0 ≤ i ≤ m, 0 ≤ j ≤ n.
• Profan¸c, upˆrqoun epikaluptìmena upì-probl mata ta opoÐa o algìrijmoc ja
lÔsei parapˆnw apì mÐa forèc.
(O anadromikìc algìrijmìc qreiˆzetai toulˆqiston ekjetikì qrìno.)
Parat rhsh
To
c[m, n] mporeÐ na upologisteÐ eÔkola xekin¸ntac apì th bˆsh
(bottom-up).
(Na sumplhrwjeÐ o pÐnakac
✫
c gramm -gramm apì ta aristerˆ proc ta dexiˆ.)
✪
✬
Algìrijmoi kai Poluplokìthta
✩
Diˆlexh 12
9
LCS length(X, Y )
m = length(X)
n = length(Y )
for i = 1 to m do c[i, 0] = 0
for j = 1 to n do c[0, j] = 0
for i = 1 to m do
for j = 1 to n do
if xi = yj
then c[i, j] = c[i − 1, j − 1] + 1
b[i, j] = ւ
else if c[i − 1, j] ≥ c[i, j − 1]
then c[i, j] = c[i − 1, j]
b[i, j] = ↓
else c[i, j] = c[i, j − 1]
b[i, j] = ←
return c, b
ShmeÐwsh
✫
O pÐnakac
✬
b
Algìrijmoi kai Poluplokìthta
Parˆdeigma
✩
Diˆlexh 12
10
X= A B C B D A B
Y= B D C A B A
j
0
1
B
i
2
D
3
C
4
A
5
B
6
A
1
1
2
2
3
4
4
2
3
3
4
1
2
2
3
3
B
0
0
2
2
1
1
3
C
0
1
1
2
2
2
2
3
2
3
2
2
B
1
0
1
0
1
1
2
A
1
0
2
1
0
0
1
1
0
0
0
0
0
0
0
7
6
B
A
0
5
D
4
0
✫
✪
qrhsimopoieÐtai gia thn anˆkthsh thc MKU.
0
✪
✬
Algìrijmoi kai Poluplokìthta
Diˆlexh 12
Er¸thsh 1 Poia eÐnai h poluplokìthta thc LCS()?
Apˆnthsh: O(mn)
✩
11
Er¸thsh 2 Pìsoc qrìnoc qreiˆzetai gia thn anˆkthsh thc MKU apì ton pÐnaka b?
Apˆnthsh: O(m + n)
Er¸thsh 3 Pìsh mn mh apaiteÐtai gia thn ektèlesh thc LCS()?
Apˆnthsh: O(mn)
Er¸thsh 4
EÐnai dunatì na beltiwjeÐ h poluplokìthta q¸rou sthn perÐptwsh
ìpou jèloume na upologÐsoume mìno to m koc thc MKU?
Apˆnthsh:
Nai! Mìno 2 grammèc tou pÐnaka
=⇒
ShmeÐwsh:
✫
O(min(m, n))
c
eÐnai aparaÐthtoi.
mn mh.
Sthn perÐptwsh aut den mporeÐ na anakthjeÐ h MKU.
✪