Lintao Zhang ( 2 3 1 4 5) ( 1 2 -3) ( 1 –2) (-1 4) (-1) BCP Algorithm (2/8) Let’s illustrate this with an example:
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Lin
tao
Zh
an
g
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
(-1)
BC
P A
lgo
rith
m (
2/8
)
Le
t’s illu
str
ate
th
is w
ith
an
exa
mp
le:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
2.1
/8)
Le
t’s illu
str
ate
th
is w
ith
an
exa
mp
le:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
(-1)
watc
hed
litera
ls
Conce
ptu
ally
, w
e ide
ntify
the f
irst
two litera
ls in e
ach c
lause
as
the w
atc
hed o
nes
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
2.2
/8)
Le
t’s illu
str
ate
th
is w
ith
an
exa
mp
le:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
(-1)
watc
hed
litera
ls
Conce
ptu
ally
, w
e ide
ntify
the f
irst
two litera
ls in e
ach c
lause
as t
he w
atc
he
d
ones
Chan
gin
g w
hic
h litera
ls a
re w
atc
hed is r
epre
sente
d b
y r
eord
erin
gth
e litera
ls
in t
he c
lause (
wh
ich c
om
es into
pla
y late
r)
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
2.3
/8)
Le
t’s illu
str
ate
th
is w
ith
an
exa
mp
le:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
(-1)
watc
hed
litera
ls
One litera
l cla
use b
reaks invari
ants
: hand
led
as a
specia
l ca
se (
ignore
d h
ere
after)
Conce
ptu
ally
, w
e ide
ntify
the f
irst
two litera
ls in e
ach c
lause
as t
he w
atc
he
d
ones
Chan
gin
g w
hic
h litera
ls a
re w
atc
hed is r
epre
sente
d b
y r
eord
erin
gth
e litera
ls
in t
he c
lause (
wh
ich c
om
es into
pla
y late
r)
Cla
uses o
f siz
e o
ne a
re a
spe
cia
l case
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
3/8
)
We b
egin
by p
rocessin
g the a
ssig
nm
ent v1 =
F (
whic
h is im
plie
d b
y
the
siz
e o
ne
cla
use) ( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
3.1
/8)
We b
egin
by p
rocessin
g the a
ssig
nm
ent v1 =
F (
whic
h is im
plie
d b
y
the
siz
e o
ne
cla
use) ( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
To m
ain
tain
ou
r in
varia
nts
, w
e m
ust
exam
ine e
ach c
lause w
here
the
assig
nm
ent
bein
g p
rocessed h
as s
et
a w
atc
hed litera
l to
F.
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
3.2
/8)
We b
egin
by p
rocessin
g the a
ssig
nm
ent v1 =
F (
whic
h is im
plie
d b
y
the
siz
e o
ne
cla
use) ( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
To m
ain
tain
ou
r in
varia
nts
, w
e m
ust
exam
ine e
ach c
lause w
here
the
assig
nm
ent
bein
g p
rocessed h
as s
et
a w
atc
hed litera
l to
F.
We n
eed n
ot
pro
cess c
lauses w
here
a w
atc
hed litera
l has b
een s
et
to T
,
because t
he c
lause is n
ow
sa
tisfied a
nd s
o c
an n
ot
becom
e im
plie
d.
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
3.3
/8)
We b
egin
by p
rocessin
g the a
ssig
nm
ent v1 =
F (
whic
h is im
plie
d b
y
the
siz
e o
ne
cla
use) ( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
To m
ain
tain
ou
r in
varia
nts
, w
e m
ust
exam
ine e
ach c
lause w
here
the
assig
nm
ent
bein
g p
rocessed h
as s
et
a w
atc
hed litera
l to
F.
We n
eed n
ot
pro
cess c
lauses w
here
a w
atc
hed litera
l has b
een s
et
to T
,
because t
he c
lause is n
ow
sa
tisfied a
nd s
o c
an n
ot
becom
e im
plie
d.
We certainly
need n
ot
pro
cess a
ny c
lauses w
here
neither
watc
hed litera
l
chang
es s
tate
(in
this
exam
ple
, w
here
v1 is n
ot
watc
hed).
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
4/8
)
No
w le
t’s a
ctu
ally
pro
ce
ss th
e s
eco
nd a
nd
th
ird
cla
use
s:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
4.1
/8)
No
w le
t’s a
ctu
ally
pro
ce
ss th
e s
eco
nd a
nd
th
ird
cla
use
s:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
For
the s
econd
cla
use,
we r
ep
lace v
1 w
ith ¬
v3 a
s a
ne
w w
atc
hed litera
l.
Sin
ce ¬
v3 is n
ot
assig
ned t
o F
, th
is m
ain
tain
s o
ur
invari
ants
.
( 2 3 1 4 5)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F)
Pending:
State:(v1=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
4.2
/8)
No
w le
t’s a
ctu
ally
pro
ce
ss th
e s
eco
nd a
nd
th
ird
cla
use
s:
( 2 3 1 4 5)
( 1 2 -3)
( 1 –2)
(-1 4)
For
the s
econd
cla
use,
we r
ep
lace v
1 w
ith ¬
v3 a
s a
ne
w w
atc
hed litera
l.
Sin
ce ¬
v3 is n
ot
assig
ned t
o F
, th
is m
ain
tain
s o
ur
invari
ants
.
The t
hird c
lause is im
plie
d.
We r
ecord
the n
ew
im
plic
ation o
f ¬
v2,
and a
dd it
to t
he q
ueue o
f assig
nm
ents
to p
rocess.
Sin
ce t
he c
lause c
annot
ag
ain
becom
e n
ew
ly im
plie
d,
our
invariants
are
ma
inta
ine
d.
( 2 3 1 4 5)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F)
Pending:
State:(v1=F)
Pending:(v2=F)
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
5/8
)
Next, w
e p
rocess ¬
v2. W
e o
nly
exam
ine the first 2
cla
uses.
( 2 3 1 4 5)
(-3 2 1)
( 1 –2)
(-1 4)
For
the f
irst
cla
use,
we r
epla
ce v
2 w
ith v
4 a
s a
ne
w w
atc
he
d litera
l. S
ince v
4
is n
ot
assig
ned
to F
, th
is m
ain
tain
s o
ur
invari
ants
.
The s
econd c
lause is im
plie
d.
We r
ecord
the n
ew
im
plic
atio
n o
f v3,
and a
dd
it t
o t
he q
ueue o
f assig
nm
ents
to p
rocess.
Sin
ce t
he c
lause c
annot
ag
ain
becom
e n
ew
ly im
plie
d,
our
invariants
are
ma
inta
ine
d.
( 4 3 1 2 5)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F, v2=F)
Pending:
State:(v1=F, v2=F)
Pending:(v3=F)
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
6/8
)
Next, w
e p
rocess ¬
v3. W
e o
nly
exam
ine the first cla
use
.
( 4 3 1 2 5)
(-3 2 1)
( 1 –2)
(-1 4)
Fo
r th
e f
irst
cla
use
, w
e r
ep
lace
v3
with
v5
as a
ne
w w
atc
he
d lite
ral. S
ince
v5
is n
ot
assig
ne
d t
o F
, th
is m
ain
tain
s o
ur
inva
ria
nts
.
Sin
ce
th
ere
are
no
pe
nd
ing
assig
nm
en
ts,
an
d n
o c
on
flic
t, B
CP
te
rmin
ate
s a
nd
we
ma
ke
a d
ecis
ion
. B
oth
v4
an
d v
5 a
re u
na
ssig
ne
d.
Le
t’s s
ay w
e d
ecid
e t
o a
ssig
n v
4=
T
an
d p
roce
ed
.
( 4 5 1 2 3)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F, v2=F, v3=F)
Pending:
State:(v1=F, v2=F, v3=F)
Pending:
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
7/8
)
Next, w
e p
rocess v
4. W
e d
o n
oth
ing a
t all.
( 4 5 1 2 3)
(-3 2 1)
( 1 –2)
(-1 4)
Sin
ce t
here
are
no p
end
ing a
ssig
nm
en
ts,
and
no c
onflic
t, B
CP
term
inate
s
and w
e m
ake a
decis
ion.
Only
v5 is u
nassig
ned.
Let’s s
ay w
e d
ecid
e t
o
assig
n v
5=
F a
nd p
rocee
d.
( 4 5 1 2 3)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F, v2=F, v3=F,
v4=T)
State:(v1=F, v2=F, v3=F,
v4=T)
Lin
tao
Zh
an
g
BC
P A
lgo
rith
m (
8/8
)
Next, w
e p
rocess v
5=
F. W
e e
xam
ine the first cla
use.
( 4 5 1 2 3)
(-3 2 1)
( 1 –2)
(-1 4)
Th
e f
irst
cla
use
is im
plie
d.
Ho
we
ve
r, t
he
im
plic
ation
is v
4=
T,
wh
ich
is a
du
plic
ate
(sin
ce
v4
=T
alr
ea
dy)
so
we
ig
no
re it.
Sin
ce
th
ere
are
no
pe
nd
ing
assig
nm
en
ts,
an
d n
o c
on
flic
t, B
CP
te
rmin
ate
s a
nd
we
ma
ke
a d
ecis
ion
. N
o v
ari
ab
les a
re u
na
ssig
ne
d,
so
th
e p
rob
lem
is s
at,
an
d w
e a
re
do
ne
.
( 4 5 1 2 3)
(-3 2 1)
( 1 –2)
(-1 4)
State:(v1=F, v2=F, v3=F,
v4=T, v5=F)
State:(v1=F, v2=F, v3=F,
v4=T, v5=F)
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