UNIVERSITY OF NORTH CAROLINA
Department of Statistics
Chapel Hill, N. C.
Mathematical Sciences Directorate
Air Force Office of Scientific Research
Washington 25, D. C.
AFOSR Report No.
,
AN ALGORITHM FOR A MIN]MUM COVER
OF
AN ABSTRACT COMPLEX
*
by
D. K. Ray-Chaudhuri
February, 1961
Contract No. AF 49(638)-213
Algorithms for a minimum cover and maximum matching of an abstract complex are
obtained. These algorithms are useful
for many combinatorial problems.
*This
research was supported in part by the United States Air Force
Office of Scientific Research of the Air Research and Development
Command, under Contract No. AF(638)-213. Reproduction in whole or
in part is permitted for any purpose of the United States Government.
Qualified requestors may obtain copies of this report from the ASTIA Docment Service Center, Arlington Hall Station, Arlington 12, Virginia. Department of Defense contractors must be established for ASTIA services,
or have their "need-to-know" certified by the cognizant military agency
of their projec~or-~ntfact.
""
Mimeo Series 275
Institute of Statistics
AN
•
ALGORITHM
FOR
A
MINIMUM
COVER
oF
AN
ABSTRACT
COMPLEX
Do K. RAY .. CHAUDHURI
University of Novth Carolina
Introduction.
and
A
co.,
Let X. {Xl' x 2 '
~ t~, A:!' .",
An1
m } be a finite set of
X
be a class o£
n
system of points and sets is called a complex (X,
the class
Xl' x 2 '
A
u~,
A'A
edges Al' A2 ,
1 (x) denotes the
tA 1 and contain the point
c~ass
OOO,~!)
A)
If every set of
0
A complex (X,
For any subclass
For a subset A,
X"
A )~
Such a
m points
Al
of
consisting of the sets which belong to
A
l
~l and
consisting of those sets which belong to
point of Ao
subsets o£ X.
contains two points" the complex is a graph with
x m and n
m points
(A) denotes the class
contain at least one
which every set has the same
\A'
points is called a regular complex"
and
,
~l \
number of
denote respectively
the number of points in A and the number of sets in the class ~l. Let £
c ( x]J , c (x2 ) .•. c (Xu!) l\
be a m- vector of positive integers] A subclass ~ 1 is called a 2. .. cover
if for every point x,
\~ 1 (x) \ ~ c(x)o
matching i f for every point
x
IA
number of sets and matchings with
J. (x)
m~urn
I
A subclass
~ c(x)
0
~l is
a
2. -
Covers with minimum
number of sets are respectively
called minimum covers and maximum matchings 0
For a minimum c .. cover
let Xl be the set of those point x for which
Iv4 1 (xl
Following Fulkerson and Ryser ffiJ
I Xi l
l'
\ : c(xl.
is called the 2. ... height of
~ J. "
Minimum possible c .. height of a minimum c - cover is called the c - height
of the complex
(X~ vA.
)!)
Similarly for :
note the set of those points
x
for which
2. .. matching
,A
~ l ' l:t X2
1 (x) \ < c(x)g
de-
I X2 ,
2
~
is called the
~
~-
- depth of a
~
depth of the
~-
matching
- matching is called the
~
~ 10
Maximum possible
- depth of the complex.
To a complex (X" ~
LJ ) we can associate an incidence matrix A "" «a~J
.. »
with m rows and n columns where a
R 1
i f xi ~ Aj3 and a
ij ~ 0, otherij
j
1, 2, .0 no Fulkerson and Ryser [2] considers
g
the minimum cover problem in terms of the matrix Ao
A set of & oolumns
of the matrix A is oalled an a-set of representatives if in the submatrix
of A consisting of the
than ao
Let
e(a)
be the minimum number of colurnns of A that form an
a-set of representativeso
matrix Ao
Obviously
the complex (X,
'~ery
m rows and the & columns, each row sum is not less
A)
The number
sea) is called the a-width of the
e(a) is the cardinality of the minimum c - cover of
where c
co
(a"
4,
0
u,
a) 0
Fulkerson and Ryser cOIlll.lents,
little is known concerning good computational methods for determining
widths and heights of (0, 1) -matrices.
Efficient algorithms in this domain
would be of great interest tt o
Petersen
(41
introduced the concept of alternating path which led
to algorithms for minimum cover and maximum matching of graphs o
Berge
C1J
first gave an inductive proof of the maximum matching theorem for graphs.
Norman and Rabin
[.3]
ggve an ingenious proof of the maximum matching and
minimum cover theorems for graphs.
In this paper the idea of alternating
path is extended to general complexes and certain theorems are proved about
minimum covers and maximum matchings of general complexes.
this theorem follow the line of proof of Norman and Rabin
The proofs of
[3]
Section 1
proves the main theorem about the minimum cover of a complex and gives the
corresponding algorithm. A llevel transformation 1 is defined.By applying this
tranflformation
I
- on a particular minimum cover all minimmn covers can be obtained. In
section 3 we prove a theorem about the minimum cover of a regular complex.
This theorem gives a shorter algorithm for the minimmn cover of a regular
3
•
complex"
In section
3 is proved a theorem about the height of a minimum
cover which gives an algorithm for the height of a complex o A 'parallel
this tr~sformation
transformation' is defined.By applying -,).1' ~ on a minimum cover with minimum
can.be obtained.
height::" the family of all minimum covers with minimum heights J Section 4
gives a simpler algorithm for the height of a regular complsxe
5
Section
establishes a close relationship between a matching problem and a cover problem..
It is shown that
minimum cover"
~
a maximum
-
0 ..,
matching can be obtained from a certain
Also the .£ .., depth~.:l of a complex.. can be determined from the
- height of the complex for a certain vector
d.
E4tending the ideas of the present paper, it is possible to get an
algorithm for any integer
subsequent paper"
progra~ning
problemo
This will be developed in a
There are many important practical applications of the
minimum cover and maximum matching algorithm..
For some special applications
it is possibJa to obtain an algorithm sharper than the general algorithm.
These applications also will be discussed in the subsequent paper.
1. Minimum.£ - cover.
Let
C of distinct sets AI' Bl ,
sider a finite sequence
Ai e
A 1 . ~d
sets Al , B2,
B
i
oeo_
c.A 1 and !JJ 1 be two classes of sets
$ l'
6
Ai' Bi ,
i '" 1, 2,
i '" 1, 2,
000
Ul 0
•
()
~
Let
Di =
~i ~d ~
and
Fi
=
Ix
I
000
Ai and
~-
XS
cover and $J. =
Bl~
B2,.oo Bi ,
Hi'
Gi ,
lob
i(x) ) <
I
gfi(X)
We shall use these notations throughout sections 1 and 2 0
a
respec~~~elY
i
denotes the null class.
{x \x.
< C(X)!.
where
000
Let Hi be the union of the
denote the classes consisting of the sets AI' A2,
"
A2, B2,
Con-
o
vA - ~
J.'
A. sequence C
=
i
I} , =
i
Let
A:J,.BJ.'
cAl
J.,2, •••
be
~, B2 • ••• j1.
is an alternating chain i f
~6
(1 1)
0
o
••
0
$
Bi P (
~':-l
and
f, (
A1
!f;.-.J.) (Di )
1'"
A1 -
J) i)
0.. .
i .... 1" 2"
An alternating chain 0
t:s {
~ Bl l
chain if Dp+l is a null set.
is a null set" i
set
~
Q
80
(Fi )"
f
.... Ap ' BpI Ap +l
A single set
i f for every point
~
x of
is called a reducible
is a reducible chain if
\~
A:L,
lex) \ > c(x)o
of the reducible chain is called the leader of the chain.
D
l
The
A.2. - cover
vii 1 is an irreducible £ ... cover if there exists no reducible cha~ with
respect to (w.r.t)
cAl~
\
It is eaay to see that i1' .AI is a.!<
Ap +l }
is a reducible chain w.r.t
~ cover
and C·
.AI' then A 2 • ..A 1 ~
131
g
Let
~ ... cover
~, .Bl1 ~,
C of sets
i
be a
B2 ,
ou
and
1),'
A A .. ~
Bp
1 ...
where
~
Bp
J!) pol U '!lip
A 1-
is a .2. ... cover with cardinality less than that of
Let
i A..!.' B:!., .... '),
P
10
~A 1
Consider a sequence
and
Bi
Q
gal'
1" 2" ••• p.
Di '
i
g
= 1,
{
xJ x e2 , •••
Hi"l~ i_l(x)/ <'~:i(x) r}
1
•
Such a sequence is called a level chain if
Bi e
•
Ai +l e
(
$1'"
~-l)
(v'{1 ... J)
(F i )
>
and
for every x
i)
(Di)"
II:
in Hp and i
takes the c - cover
~
a
to
1,,2, •••
•
t]B2 where
c(x)
Oonsider the transformation which
6]2
qj 2 is
Obviously
fBI - gfp U ~
=
also a c - cover o
Since the cardinality of $2 is same
l
~1~ the above transformation is called a 'level transformationio
as that of
by applying the level transformation a finite number of times.
tAl and
A
fi 1" A
d(
0'
~
tJ31
\
JI. r£
c.Il1 cA
Let
cover in the family
cA
1"
L(
and
~ 0)
0
J1
L( c.A
nearest to AI.
where
eA
Ai +1
Proof.
1 -
L(
cA
10
If there exists a set
Let
c!fB 1 be
reducible chain with
o
•
•
Since
o
•
•
0
~
(
as leader
~1
•
0
-
and Bi
••
10
•
1 -
f,
$1 -
1Jf 0)
(D1 )
C
Al
Al
w.r.t.
cAl'
i
= 1,
D
J?
1 -
2,
u.
p.
~ 1) with
Since there is no
)0
If possible, suppose
J) l '
it follows from (1.1) that for every x
/1)1 I
This contradicts the definition of DID
can choose B where
1
(
l
D1 is not empty.
g;0)
~1 is a ~ - cover~
in D1
•
13
a
~, Bp ' ~+l }
•••
£1
Consider (
0) •
covers.
Assume that there exists no such reducible chain (w.rot
as the Ie ader g
(1.1)
{~, 1).,
&
~-
1 -
=
i
0)
be two
nearest to
Jl r), A.
eA BBl' then there exists a reducible chain
C
~
1) ;; d(
l'
For two covers
q]31 be a cover in the family
is said to be a cover of the family
Lenuna ,+.
in
Let
0
0) such that
d(
Then
0
1
denotes the number of sets of the class
0
vii. 1 which do not belong to cA
L(
~
L( $1) be the family of all classes which can be obtained from
Let
(x)
+
' ¥ o (x)
I;;
o(x) •
Hence (1.1) is not true and we
1\
eo
(
0
ll.
Since
0
•
~l)
3
•
•
•
•
Q
\qBl (x) \
•
From (1 0 4)
181"
~(x)
)
we can easily see that
~l and
worot o
If possibie, suppose
(A
1 is a c - cover, from (1.3) it follows that for every x
F
l
(104)
(Fl)o
(cAl" cf)l) (Fl)C
••
A
-
A 1 .. J) 1)·
gfO) (Dl ) .. (
1 -
(~l
Next we consider
(103)
1J
$2
$1 . . f/i l
=
=
C
I I~ I;;
+
(x)
i~'B:L}
e-
U ))1
in
c(x).
is a level chain
L(
u'1
0
)0
Also we have
••
C:
"
•
Cl
d(c.A l'
•
which contradicts the assumption that
(1.3) is not true and we can choose
e
A2
~1
is nearest to
cA 1.
Hence
A where
2
(CA 1 ... J) 1)
(Fl )
(eJa 1" tf(1 )
..
Using the same arguments by induction we can show that there exists an infinite
sequence of distinct sets which is a contradiotiono
reducible chain with
Theorem
2. ..
c~ver
Proof.
A 2.
as the leader
- cover
if and only i f
fi
Necessity is obvious.
irreducible.
family
;I,~
-\
L(
Let
vA
0)
~0
nearest to
.A:t
contained in
reducible chain wor.t o
is empty.
the required property"
1 of a complex (X"
fi)
is a minimum
1 is irreducible.
To prove sufficiency, assume that
be a minimum
it is sufficient to show that
exists a set
cA
i~-ith
Hence there must exist a
~ ..
cover"
Let
Ar
rJd1
v'l1 C !dlo
Since
fB1 be a
"~1 is
co~~r in the
is also a minimum .2. - cover"
If possible" suppose tnere
cA 1 - 13 1:>
Then by lemma 1, there is a
cfi1 which is a contradiction.
This completes the proof of the theorem..
Henca
vii1.. 131
7
Theorem 2.
c~ver A
c -
Proof
""It 1
is a IlUm.mum .2. - cover, then any other minimum
2 belongs to the family
L(
~ 1).
1st $ 1 be the cover in the family
Q
Since
If
A
A 1) nearest to A 2~
L(
2 is a minimum .2. - cover, it is irreducible.
Now by the same _argu-
vA
ments as used in the proof of theorem 1, we can show that t(/3l ...
2.
Theorem 1 gives the following algorithm for the minimum cover of a
c~~lex..
woro t.
We start with a .£ - cover
A 10' A
1 is a minimum
looking for reducible chains..
v4 l
.2. -
If there is no reducible chain
o
So the algorithm consists in
cover.
Whenever a reducible chain is obtained, we get
a new c - cover whose cardinality is one less than that of the original.£ cover.
In this way, finally we get a .£ - cover which is irreducible.
.2. -
i f there is a reducible chain worq.t a
cA
cover
A
1 as the leader, we can proceed as follows. :Let
We start with
A:t
i f possible we choose
~
B
1
A2
and test whether
~
and test i f
~2 .i~
reducible chain wor.t o
6
(
vii
1 -
a null set..
A
~ of
.. ~ l' •
I with a given set
djBI ... A
is a reducible chain wor"t.
VJd 1 (Dl )
To test
A
I"
If not,
Next i f possible we find
0
J) 1)
If
(Fl )
o!J 2 TJf.·Jf.iS a
D2 is a null set,
1 and the test is completed.
If not, if possible
we find
In this manner, we proceed to build a chain until we get a reducible chain or
until no further addition of sets to the chain satisfying the conditions (1.1)
is possible.
~,
Bl'
uu
For instance the chain will terminate after the selection of
Ap ' Bp
if
(A 01).
~,
c.A (fS
1 -
terminate after the selection of
is a reducible chain w.rot.
class.
p)
(Fp ) is a null class.
Bl , .. eo
1 or
Ap,
1 ..
Bp ' Ap+l
(If
p)
BIt A
2
41 U
o!) P+'l U g;P
(Dp +1 ) is a null
At the terminating stage, we have a maximal chain.
choices of the sets
if
The chain will
By varying the
we can get all possible maximal chains and
B
wor~t~ vA 1
find out i f there is a reducible chain
Theorem 2 gives an algorithm for
from a given minimum
level chains.
A
A 1.
First we find out all
find out ITli.z:imum
chains
~ •. cover
fin~ing
~
- covers
wor.t o each of
k+:1.' •••
A
2'
minimum~
as the leader o
- covers starting
The algorithm consists in looking for
possib~e ~evel ch~ins worcto~ 1
A 2' A 3' 00.. A
~ 3'
and so one
kt
out all
~
with
QUO
fi
ke
and
Next we find level
k and find new minimum.£ -covers
Finally we shall reach a stage when we do
not find any more minimum.£ - coverSll
ao
complexes~
Minimum covers of regular
of a regular complex
(X,
fer at least one point
A)o
x of
A,
A set
J
A of
A
1st cAl be a.£ - cover
1 is called a heavy set i f
L.A 1 (x) I > a(xL
A reducible chain
whose leader is a heavy chain is called a heavy reducible chain.
section we consider only regular complexeso
Lemma 2 0
A reducible chain
worot~ a.2. -
cover
A
In this
1 contains at
least one heavy set o
Proof.
w.r~t.
Suppose
fi l~
~
c
{ \. l3:L' :...
Ap ' BpJ Ap +l } is a reducible chain
Let H 1 be the union of the sets in the chain Co Let HI
p+
be the union of the sets
.
~,A2'
suppose there is no point x
:In
000
A +1
HI
r:r which. \
q be the number of points in every set of
oo.oe.o •
Hence
•
for
Q
x
in Ht
and H*
A
0
=
A
Hp+l - HI.
1 (x) \ >
If possible,
c(x).
Let
Then it follows easily that
9
(2.2)
•
+
For
x
•
•
•
0
•
•
0
I~
p(x)
I
-
c(x)
I
...
o.
p+l(X)
-
l:
Ifif(x) ,
(
I~(x)1
+
I 0 1(x)
I
;>
c(x).
Let cAlba any
L(
~I are
Proof.
It is sufficient to show that
C =
B
i
~
contained in
.A..t
t!IdA:t.. B:L.vi?... '\.
1 -
x
-
H,*
e
c(x)
;;
I~
I(x)
IoZt p+1(x) I
1"
0'
...
::I
~l'
~d
then
Ji
I
a minimum.£ - cover.
If a
nearest to
~I is
C
A
10
If all heavy
a minimum - covero
OjjlO
If not, suppose
Al -
Then by lemma 1, there exists a reducible chain
i '" 1,2,9
~-
w~r~t~ ~ l '
in
Ult
CI
u
po
As.+l
e
A
1 - <J81
and
By lemma 2 there is a heavy set
This contradicts our assumptiono
A
Hence the lemma
follows.
ducible chain,
-3-
c(x»
~oba
Bp ' '\+1 } whara
in the reducible chain Co
Theorem 30
I
<
... c(x»
Hence there must exist a point x
2. - cove;'and
sets of
contains a set
for
In other words there is a heavy set in
etb. be a ;:, - cover in the family
qJ3 1
Hence
+
the reducible chaino
Le""'" 3.
•
Hp+1
\J1i(X) I
for which
IA .(x)1
( 2 03 )" we get
The inequality (2 0 4) contradicts (2 0 2)0
HI
:i
c(x)
c(x)
and
•
-
p+1(x) \
(08 p+1(x) I
A'(x) f +
X e,
(
l:
X .~ ~+1
+
I
0
(2 0 2)
From
o •
•
~(x)'
(
(2.4)
•
J1
J
Ij}
H,*
in
I
I A,(x)
• • •••
0
cover
then
~l is
vII
such that there is no heavy re-
1 is a minimum ;:, -
cover~
10
Proof.
Let..A 0 be a minimum
family
L(
fi 0) nearest
~.
contains a heavy set
A1~
fi 10
By
A 1are contained in fll 1
all heavy sets of
w.rQt.
to
~ 1 be a ~ - cover
lemma ~, it is sufficient
.£ - .cover and
c=
to show that
fi 1 - $1
Then by lemma 1 there exists a reducible chain
t'\, ~, ...
Ap ' Bp ' Ap+l} •
This contradicts our hypothesis o
C is a heavy reducible chaino
By definition
If not, sy.ppose
IJ
in the
Hence the theorem is trueo
Theorem 3
gives a shorter algorithm for minimum covers of regular complexes.
It shows that in the case of rGgular complexes it is necessary to look for
only heavy reducible chains.
3 •
comPlex..~
Height of a
sider a sequence
Pi
1
='.... ~l Ej ,
J=
Let Aland
E of sets
E .l' E ,
1
2
i=1,2,.oo2po
Let
i = 1, 2,
t
Let
=
Ri
(3,,1)
"
..
and
x \
• •
8
i
i
000
o
"
<..tJ
x
2,
000
~1
and
E belonging to
2p
i
A sequence
E
~1 which
e.
x
Ij( I
Pi'
f,
i(X}
Pi'
If(
i(X} \
>
If
E2k
<
I
Yi(X}
3 and 4c-
A
1 is a.£ - cover
~d
131
=
k
fl:} 1 and (~1 - !/22k) (S2k)
(A 1 -f(2k) (82k )
E2k+1
e-
I]
2po
f,
and E2k+2
~I
occur among
is called a properly connected sequence
~
Let
r Yi(X}g
if tho following four conditions are satisfied for
(i)
A.
Con-
respectively denote the
We shall use the above notations throughout' sections
Now we assume that
fi.~
2po
x
fI
=
= 1,
E _ '
2p 1
tJD and e/
v""L i /J
classes consisting of those sets of
E2 , •• 0 E ,
1
oou
$1 be two subclasses of
(f!81 -
cr
2k+l)
(R2k+1 ) •
~..,~ 1.
(wor.t
=
A
1, 2,
1)
eoo
is not empty"
po
53
and E2k+2
(iii)
E2k
If
e
11
1
(A
f,
A
1
-!C
1
(iv)
If
A
e
E2k
(vA
&
1
and E2k+2
£,
(
and
cf2k)
!Id
l"
(R2k )
~k
(cAl -1<.
:l2lol-1)
•
(~1 -
•
(S2k+l)
)
::R.. 2k+1 ) (S2k+1)'
('1l1 - Y2k )
1 -
E2k+1
2k+l)
(~1 -
and
E2k+1
and E2k+2
is empty,
and
ie not empty,
(R 2k )
(R2k )
2k)
1. empty,
(S2k)
(R2lol-1 )'
A properly connected sequence E is called an exchange sequence if
v4. A
2
...
quence
E
El
£,
A
_1.-
~ 2p
(wor.tgA 1)
1
and E2
£,
~p
U
is a
£ .. cover.
is called a low sequence
~l (Rl )
An exchange se-
(w.r.,t o
1)
if
and
\{x 11J11{X)! - C{xjl
\vfI I ; C{x>} \
(3.2)
vii
>
\{x I
2 {x)
An exchange sequence
(wer"t o
v'i 1)
if
(woro t.1iI~ 1)
E
El
...
~l
and E2
Obviously if ~ 1 is a c - cover and
~
v'1.l(Sl)
IJl I - 1\ ·
2 (x)
l'
£,
i \A 1(x) \ - C(X1 \
1 xI
(3.3) . . . . . "
cA
is called a parallel sequence
2
is a
and
t
- \ x\
c(x)
E is a low sequence w.r. t~
E. - cover with height less than that of
Similarly if E is a parallel
sequence'~2
A10
is a c - cover with
12
height equal to that of
~2
Let
It):
is obtained
(vii 1 )
f( loIn this
from~/1
case for convenience we say that
by applying a 'paralle1 transformation!.
denote the family of all
~-
covers which can be ob-
e..-A 1 by applying parallel transformation a finite
taine~ from
number of
timefJD
wmma 40
minimum
Suppose
~
v4
1
!E
and
1 are two minimum
- cover with minimum height.
properly connected sequence
A
1 - ffdl
<.A 1 - S
E2
and
1
1 -
1
~2P
6
)
then either
sequence w..r.t o
Proof 0
First we show that
JS>
l·
~P
:iJ
is a
~
2(x) \
we get for
x
£,
I~2(x)
obtain
~
and
~1
Ei
1
where
be a
E
1
c(x)"
s
belongs to either
c
w.r.t.
(u'~t -~2P ) (~p)'
tA
1
or
E
is a parallel
E is an exchange s6quence
-
~p
U
J{
w"r • t~
both
cA 1
c.A 2 • A l - J( 2p
2p are c - coverso
Using
(3.4)
and the fact that
~1
U
To show
- cover, it is sufficient to show that for every x
R2p
(
E21 0"''' E2p
1"
= ~l
2
~I
f!Id 1 being a
C
For this we have to show that both
and
v.bl2
{cA
£
1
f
and each
(S2p)
is a low sequence
E
A
1
=
E
covers"
If'
~: • O(1Jl -JP2P) (~p)
(3.4)
that
both
!!J -A
13 1 - A 1
or
(A
and
£,
wcr.t o
Let
~-
s
~p'
is a .£ -cover,
13
i
x \
x •
~p'
_ t
x \ x • R2p'
(3c5) ~ • • • • • •
tI
x
and
x
{x I x
C
I
~
C(X)}
l~l(X) I
=
C(X)]-
a
c(x>}
IA
2(X)
Jg 2(x) I
I v11
•
S2p'\
6
S2p'
(x.) \
If E is not a low sequence w.r.t.
0.6) .•
0
••
I{x
If x
0
•
II~ 1 (x) J
does
~ot
>
C(X)}
=
~1'
c
c(xj
0
we have
I ; I{ IA
belong to R2pU S2p~
x
I~A
2(X)
=
1(x) \
>
C(x)]
IA2(X»).
Hence from 3.6, it follows that
-I\ I
C(X)}
>
It is easy to check that
R2p '
IJ~L(x) I
= C(X)}
IA
c(x)
1 (x)
1>
\J12(X) \ >
S2p'
IA (x) I
IA
J-
~
I
t \x• ~p'
{x \ x •
~,
C(X)}
1
=
C(X)}
-
2(X) \
>
c(x)]
-
I <-Ar (x) I>
x
C(X)}
{x I
X
t \
•
S2p'
x • S2p'
I
Also we have
•
IA
l\1p'
IA
Q
•
•
(x)
IA
S2p'
l A1. (x) I
From (3 0 6), (307), (3 0 8),
sequence
> .. (X)}
>
C(X)}
2(X)
I
2(X)
IAi(x)1
~p'
=
=
~l
is a
...
5.. :u;t
A
A
E·
1.
If
~ E:L'
~
>
=
I
- cover with least height l we
w.r.t.
~2(X) I = C(X)}
(SSJ:(X)/ = c(X)} 1
I
f!d 1
> \ {
x
I
x • R2p '
(3010) and (3 ..11) we have
G
1
with least height.
to
I
O(X)}
is not a level sequence
which contradicts (3e5).
Lemma
C(X)}.:J
C(X)}
Suppose the lemma is not true, then from
•
R2p'
and (309), we get that if E is not a low
\-11 I
l\1p'
can prove that i f E
•
x x•
.> C(X)}
Similarly using the fact that
Cl
>
tI
:J
w~rot~ ~l
if'":..
(J.1.0) •
I
2(X)/
•
and
1.
~1
Hence the lemma is true.
be a minimum .£ - cover and
Let:!i]1 be a
..
531
0
be a minimum c - cover
.£ - cover in the family M(
contains a set
E2 , ••• E2P } w.r.t.
J1
cA
LA 0) nearest
El , there exists a low sequence
1. where any set in the sequence
E
1.5
A 1 belongs to A 1 - .fJ1
belonging to
Proof G
E:J.
If possible, suppose there is no such low sequence w.r.t.
as the leader.
can show that
that
0
A1
A
Consider 1]1 (Rl ).
1 - f1..,1 is a .2. - covero
is a minimum .2. - cover.
E2
Next we consider
Case 1 0
If
Al
we
E where
2
(AJ.-1(J.)
(A J. - J( ~(82)'
- 9?(1'
This contradicts the assumption
So we can choose
1'JJ.(~):
e
~l(Rl ) C
~1 with
(~)
There are two pcssible cases.
In this case
-
(AJ. -J? 2)
So we choose
E where
3
(3.13 )
E:': (A J. -K2)
(.%>J. (3.14) • .. .. .. .. ..
(3]J. - cJ, )
(~) - (rJJ. -~2
J13)(R3 ).
Then consider
e
) (82),
If possible, suppose
C ~)/J. - 123
(R3 )
In this case we can show that
A
wi~h cardinality less than that of
) (R3 )·
1 - ~ 3 u ~ 3 is a c - cover
cA 1
which is a contradictiono
Hence
(3.14) is not true and we can choose
(3.15)
ct
~4
Case 2 0
-~
0
....
•
"
(f) '\
....
(:2J. - c..J
3.)
In this case
0,16)C A1':!J?2)
Now we consider
(8
2) C($1-(f2) 2)
(8
(g;] :12)
1 -
0.J.7) •( 11J.·: (12 )<R:!)
C
(R:!)'
If possible,
(/iJ. -j( 2)
Then it is easily checked that the sequence E
the conditions of lerruna
4..
Hence either
•
suppose
(R:!)
{E:!.' E2 I satisfies
E is a low sequence
we r.t1>
A
1
16
or
E
tion there is .no low sequence w.r. t.
quence
w.r.t o
the family M
no
(A 0)
assumpti~n that
(3.18)0 • •
E3 e
Next
(3~19)
•
0
0
§(J 1
H~nce
toA 1.
•
•
(
2
Then
0
is a
~
Y
-
fi 10
21-
Jt
U
2 belongs to
This contradicts the
(A 0)
nearest
/1
<.,,{fI2
(J=r 1 -.92 2)
(R2 ) -
0
(S3).
LJ
v91
(R2 )
If possible, suppose
c ('B 1 - f
(53)
LJ
c/--( 2 =
3 )<53 ),
Ll
tJ)
(0
1·' -11 3 U J 3
v-...",
.£ - cover with cardinality less that of
assumption that
2
- cover in the family M
From (3019), we can show that
and
E must be a parallel se-
11
= q.a 1
)
(AI - :R 3)
is a
Since by our assump...
Jio
(3.17) is not true and we can ohoose
•
••
A
2
a:d is nearer to
~ons~der (c.fiIl -~ 3J
•
g
is a parallel sequence worot.
fi r
is a minimum c - cover 0
-
This contradicts the
Hence (3 ~19) is not true
can choose
'!rIe
E4
~
(A J(
1 -
3 )
(12 :f3) (53)
(53) -
1 -
Using the arguments given a.bove by induction we can show that i f there is
no low sequence wOrQt.
~l
with
~
as the leader there exists an
infinite sequence of distinct sets which is a contradictiono
This com-
pletes the proof of the lemma o
Theorem
40
A minimum .£ - cover
is no low sequence
Proof.
A
1 has minimum height i f and only if there
worot~ ~lo
Necessity is obvious.
To prove sufficiency, assume that there is no
low sequence w.r"t. the minimum cover
minimum height.
A
10
10
vA. 1
0
I s t A 0 be a minimum cover with
Let!!J1 be the cover in the :family M
It is sufficient to show thatAl
41 - jj
w.rot.
A
1 contains a set
E10
C :JJ 1&
Then by lemma
This contradicts our hypothesis o
5,
(A
0 )
nearest to
If possible, suppose
there is a low sequence
Hence
A
1
C
E
10
17
Theorem 50
let
minimum cover
jc/ a be a minimum cover with minimum height
fi 1 belongs
to the family
Then any other
0
(A 0) .
M
Eroof o Proof follows from lemma 5 and the fact that there is no low sequence
wor~t. AI.
Theorem
4 gives an algorithm for the height of a complex. We start with a
minimum.£ - cover
A
1 and look for low sequences
Ar
wor.t o
Whenever we
get a low sequence, we get a new minimum cover with height less than that of
the original minimum cover.
In this way finally we get a minimum cover w.r.t.
sequences~
which there are no low
A I with a set
E of
I
pass ible, we choose
E2
Al
To test i f there is any low sequence wor.t.
as the leader, we can
belonging to
After choosing E2 , we examine i f E
~2(X) \
for
necessary that for
sequence,
e
x
x
t
= ~
/fl2(X) I
is necessary to compute
(
R2 0
R2 ,
For
IA
E will be a low sequence
~,
!lcf
k,tf
2(X) \
~
proceed~as follows..
.p
I (RI ) where
1
=L/{f -
Ell E2 } is a low sequence"
IA
J(x)1
If
Ai
0
For this
\J?2(X) I ..
-
E to be an exchange sequence.;l it is
2(x)
t
worot"
~
~
c(x).
fi
C(X)}
If
E
is an exchange
I i f andonly if
1
<
IJll(X) I = C(X)} I ·
If E is a low sequence, the test ter'unates o
J?..
2 ) (82 ),
There will be two possib:l.e cases
is not empty and (11)
we l}house
If not, we consider (.
(AI - J?
(i)~ 1
2 ) (82 ) is empty.
If
ble} we choose
E from the
4
case, i f possible we choose
2)
(i) is the case,
(./11 - R 2) (8 2 )" Next,
class (.J] ]. . . Y3 ') (R )o If (ii.)
3
i f possible, we choose
E from the class
3
E from the class
4
(.:7] 1 - Y
-
-:R.. 2)(8
E belonging to the class
3
~I
i f possi-
is the
2) (R2 ).
(A I - R 3 )(83 ).
After
Next
18
4, we examine
choosing E
if
E
{!l:I'
=
4 } is a low sequence 0": not.
E2 , E), E
In this manner we proceed to build a sequence
J.
=
{Ell E2 , •• 0 E2p
The
sequence terminates at an even stage i f we get a low sequence or if at some
stage (say 2P,th stage) both (
(R
2p
) are empty classBso
A
1"
1i
E
2P)
(S2p)
r:JJ 1 -
and (
By varying the choices of the sets
c.J{jl 2P)
E , E , E ,
2
3 4
••
0
we can find out i f there is a low sequence with E as the leader or not.
2
Theorem 5 gives an algorithm for finding out all minimum covers with minim~
height 0
4«> Height of a regular complex.o
lar complex
(x, A):
IA
in E2 ,
point x
E } wor .. t.
2P
seq'lwnce.
J..eroPl/:i ')"
A
A set
1(x)
Let
I is called a loaded set i f for some
01
>
c(x) + 1.
whose leader
l
A
1 be a minimum c - cover of a regu-
E
2
I
A
A low sequence
E
=
i
Ell E2 _, ..
E is a loaded set is called a loaded low
1
In this section we consider only regular complexes"
wore t" a minimum c - cover
A low sequence
A
1 contai:1s at least
one loaded set"
~
Lot E
=
~ El ,
E2 , . . .
P2p be the union of the sets
sets of
q
A
1 which
be the number of
OCC'.ll'
poir~ts
f\;,
Al'
} be a low Bequere.e WoTot.
Lot
2P
E2 , ".0 EZpc Let P! be the union of those
E
in the sequence
E
and
P* = P2p -)" Pi.
in each set of the comp::"ex
(x, vA
Let
" I t follows
easily that
(4
.. '
\. ,,~)
=
If possible, suppose for no ;lOi':rt
Let
Since
~
A
0=
£It
\./9
E is a low sequence
x.
>
ill
l -!l(2P
U
J
(j)
2p
w~r~t~ ~1'
pq
we have
C'(x) + 1.
19
Let
(4.3)
o
0
t
•
x
•
•
•
jA
l(X) \
>
P2p "
IA
2 (x) \
> c(X)}
x e
pr,
I
xe
P*,
Al(x) I fA 1(X.) ..,
~
Pi"
'~2(X)1
•
tX \ x
•
tX \
{X \ x
1I
x
~
P2P'
\ x
te \
and
CP
-
c(x)
q(x)
>
:::0
c(X)
-
Since by our assumption for no point x
I}
c(x)
e,
G1'
... °2,9
. . I}
> I}
=
c(x)
x e P*- 0 4,lf{2(X)1
:0
:;; G
3'
... G ,
4
==
Gr';)
;; l}
Pi,
:;;
G6
'IJ~l(x)l
c(x)+l,
>
it follows that
G is the union of the disjoint sets G ~~d G 0
1
3
4
Also it is easily checked that G i.s the union of the disjoint sets
2
~4.? ~5 and
(4~4)
••
0
•
~6.
eo?
. IG5 I
(4?J)
Using
(4.5) " •
0
•
From (4,,2), we have
•
Q
IG6 1
.;
>
IG3 I
0
and' (404)$ we get
"
1:
-
x e
:::
C(X)}
=
1:
xe,
}
P2p
l.
+
x
c(x}
20
z
>
x
8
z
:;:z
x
8
G
j l)Jl(xl I
c(x)
~
+
\. G3 1
...
C(X)]
+
z
4
G~ \Al(X) \
J
x
4
8
1"AI
G
(x)
t
I
-
C(X1
3
{I,A I (x) I - C(Xl}
z
:;:z
...
x e P2p
=
Z
.:x:
f,
P
i kAI(x) ,
2p
(4c5) contradicts
Obviously
contain at least one loaded
~
Lemma 7
Let
with minimum
nearest to
is a minimum
~~
v4.
-
1
Ii: ...
~
to
:;q1
I
...
C(X)}
Hence the low sequence
set~
A
0
E must
be a minimum .£ ... cover
be a minimum .£ - cover in the family
If all loaded sets of
fi
1 are contained in.
M(
vii
0)
~1' A l
cover with minimum height.
It is sufficient
E2 . . . .
'!l31
Let
$1 contains a set
l'1.'
(4~1).
1 be a minimum £ - cover and
height~
A
+ \ q{2P(X)
~
E:!p
J
belongs to
~o
.E1. 0
show that A l C
Then by lemma
w.r. t.
A 1 where
A I - $1.
at least one loaded set of
v6l. 10
5
fiJ
1 cr
If not" suppose
A.
there exists a low sequence
:my set of the sequence
By lemma 6" the sequence
This set also belongs to
1 ...
E
<=
E belcnging
E must contain
AI'" t!lJ 1
which contradicts our hypothesiso
Theorem 6.
A minimum " - cover
A
mum height if and only i f there is no loaded low sequence
Proof.
Necessity is
loaded low sequence
obvious~
~wor.to the
A )bas miniwQr. t. A r
1 of: a regular complex ( X,
To prove sufficiency, assume that there is no
minimum c ... cover
mum £ ... cover with minimum height and
-
AI.
Let
vA
0
be a mini-
f1d1 be a minimum c - cover in the
21
fMJily
M
(A0)
A
nearest to
show that all loaded sets of
A 1 are contained in 3310
c.A :13
1 contains a loaded set
sequence E
=
1 ""
set,
l "S.'
E2 , ... E2p
1-
E is a loaded low sequence
thesis"
Theorem 6
E1 ""
By lemma
",.r.t.
vtl l
A
~ 1f)
w.rot o
Hence all loaded sets of
the proof of the
1" By lemma 7 it is sufficient tc
5,
1·
If no1; suppose
there exists a low
"S.
8mc.
is a loaded
This contradicts our hypo-
are contained in
1J 10
This completes
theorem~
gives a simpler algorithm for the height of a regular complex..
For
(X, ~ J ,when searching for low sequences worot a
cover A l' it is necessary to start with loaded sets of A 1
a regular complex
minimum c -
o
only.
5.
Relationship between maximum matching and minimum covero
=
be a complex and ~
Let ~
integers.
d(~)
where
~
=
(c(X2 ), c(X2 ),
(d(~), d(X ),
000
2
(~(Xi) \
=a'
000
-
c(X »
m
d(Xm»
c(xi ),
(x, A)
Let
be a m - vector of positive
be an associated m - vector
i::s 1,2,
me
000
vil
Let
1 be a
- matching of the complexo
Theorem 10
13
1
Proof,
==
A.£ - matching
A .. vA.
A
1 is a maximum Eo - matching i f and only i f
1 is a mininnun
! -
c:ver and
fi
8mce
1 \
A
\g] 1
that
+
is a
!! -
.£ - matching.
1 .,.
IEI3 I
mininnun c ... cover.
0
\
1
1d 1 is a
Let cA be a maximum
mininnun
is a d - cover.
==
1 is not a maximum Eo - matching,
I fi3 1 I is a
fi A-rd
0
$0 v'Q -A 0
I gL I = IA I = \Jt
) must be greater than
cover,
If possible, suppose
1 is not a maximum
.£ - matching. Then
\A
cover o
£
It is easily checked that i f
is a c - matching and vice, versa.
~-
0
A I
1
Also we have
I I fl I
+
<
0
IA I .
0
Hence
which contradicts the assumption
This completes the proof of the
22
theorem.
Similarly it can be proved that i f
height,?
cA 1 .,. A - j73l.
Ackn~wledgemento
.2 1 is a minimwn
d - cover with minimum
is a maximum .£ - matching with maximum depth.
I wish to express my sincere thanks to Professor Me Po
Schutzenberger who taught me graph theory for suggesting the problem to me
and for many stimulating discussions o
REFERENCES
1. C:> Berge, IITwo Theorems in Graph Theoryll, Proc. Nato Acado Sci. UoS.A"
Vol" 43 (1951) pp, 842~844Q
2.
D. R. Eulkerson andH. J. Ryser, "Traces, Term Ranks, Widths and Heights, II
IBM Journal of Research and Development Vol 4 (1960), No.5
PPo 455-4590
3.
R" Z" Norman and M. o. Rabin, UAn Algorithm for a ¥.linimum Cover of a Graph,
Proc. Amer. Math. SOCf) Vol 10 (1959) pp 315-3190
48
J. Petersen, llDie Theorie dar regularen Graphen,ll Acta Math.
(1891) pp. 193-220.
5.
J. p" Roth, ltAlgebric Topological Methods for the Synthesis of Switching
Systems I,ll Trans. Amer. Math. Soc. Vol. 88 (1958) pp" 302-326..
Vol 51
tJ