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Annals of Discrete Mathematics 1 (1977) 415-419 © North-Holland Publishing Company
''
THE MINIMAL INTEGRAL SEPARATOR OF A THRESHOLD GRAPH
I
'''
···-`-'·';'·
James ORLIN
r "::.·".· ·'*, ''
Department of Operations Research, Stanford University, Stanford, California 94305, U.S.A.
*.. ·. I.. ··.! . r
A graph is called threshold if there exists a real number b and real numbers aj associated with its vertices w, such that ims a, < b holds iff S is a stable (independent) set of vertices. The vector (a ,..., a; b) associated to a threshold graph is called an integral separator if ai + a > b + 1 for every edge (w,, wj). A simple algorithm is presented to determine for a given threshold graph its (unique) integral separator which minimizes b.
i
j
Let G be a loopless finite graph without multiple edges. If w is a vertex of G, let d(w) be the degree of w. The edge joining vertices u and w will be denoted as (u, w). Graph G is said to have property P if for every two vertices u, v such that (u, v) is
r
·· i
an edge, and for every pair of vertices u *, v * with d(u *)
d(u) and d(v *)
d(v),
(u*, v*) is an edge. In this definition it is possible that u* = u or that vu v. It has been shown in [1] that graph G has property P iff it is a threshold graph.
c
Suppose G is a threshold graph with vertices w, w2, w 3,..., w,. For IC
;,
{1, 2,..., n} let S = {w I i E I}. Let A = (a, a2 ,..., a,) be a real vector and let b be a real number. The pair [A; b] is said to separate G integrally if the following
holds: (1) ai
i i i j
'··
I
''
0 for i = 1,..., n;
(2) ,, 1ai b iff S is a stable (independent) set of vertices; (3) X,,ai~ 3 b + 1 iff S is a non-stable set of vertices. It was shown in [1] that a graph G is threshold iff there exists a pair [A; b] which separates G integrally. The following algorithm determines for a threshold graph G a hyperplane
3 a L ·;f ''' '' -a r
[A *; b*] which separates G integrally and such that b* is minimum. It will also be shown that it is the unique hyperplane with minimum b.
?i II C, i q I i
Algorithm A. Step 0: Relabel the vertices as w,..., w, such that d(w)< d(w 2) < < d(w). Step 1: Let t = minimum index such that (w,, w,,l) is an edge of G. [If no such t
exists let a* = 0 for i = 1 to n and let b* = 0. Then exit from algorithm.] Step 2: If d(w1 )=0 let a =0. If d(w)
~1let a = 1.
415
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-
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---
---
------
-ara
~ ~~
s~~
-
--
--u
-P
-~--
·-
·
416
J. Orlin
Step 3: For i = 2 to t if d(w) = d(wi- 1 ) then let a a* 1; if d(wi)> d(wi, ) then let aT= 1+ a* + a* ++ a-. Step 4: Let b* = a* + a + *- + a*. Step 5: For i = t + 1 to n let si be the minimum index such that (wi,, w,,) is an edge. Then let a= b a *+ 1.
Case 1: T with j j < tion. Thus w algorithm A. G has propel that S is stat
Example. Let G be the graph in Fig. 1. Table 1 shows how the algorithm worked.
[A*, b*] we iEI
Thus for all . Case 2: T Then S, co
/7
j < t then i
d(wi_,)
wi, wi,) is an
n worked.
417
Case 1: There exists a stable set S such that XIE at > b*. Let I = {jf,j2, ... ,jk with j' < j "* A2 jk. If jk a t then Gus, a *a t=1 aTY= b * and we have a contradiction. Thus we may assume that jk t + 1. Let q = sji as chosen in step 5 of algorithm A. Thus (wq, wjk) is an edge of G. If q jk- then d(wq) d(wj_,). Since G has property P, this would mean that (wk_,, wk) is an edge of G, contradicting that S is stable. Hence we may assume that q > jk-. But now by construction of [A*, b*] we have: q-1
a - t. Then by the choice of Si in step 5 it follows that i > sj. But then
3
aha
hEI
h
a*
i
a*t+a*= b* +l. . S
-a
Thus the proposition is ture. 8 26 5
½e threshold
hat d(w) s
~a* and that a is thus minimum. Suppose instead that
=1 to n. i <j. Then Vn)') d(wj). t n-1. io edges. It :his case. :h does not
d(Wk
)
1 +
3
Proof. By (2) then either I by (1) and (3)
a, >1 +
a
a*.
i =I
i=1
Acknowledger acknowledged
Corollary. The value for b* is also minimum. Proof. {w1, w 2 ,..., w,j is a stable set. Thus
Reference
t
a
b
a*
a
i=l
=
b*
i=1
Proposition 3. The algorithm constructs the unique [A; b] which separates G integrally with minimum b.
[1] V. Chvatal an Math. 1 (197;
Proof. Suppose [A;;b*] separates G integrally. Since t t b* > a> a b* it follows that ai = a* for i = 1 to t. For i = t + 1, t + 2,..., n we have that b* + 1
a +a si-1
ai +
3
at
Annals of Discrete Mathematics 1 (1977) 415-419 © North-Holland Publishing Company
''
THE MINIMAL INTEGRAL SEPARATOR OF A THRESHOLD GRAPH
I
'''
···-`-'·';'·
James ORLIN
r "::.·".· ·'*, ''
Department of Operations Research, Stanford University, Stanford, California 94305, U.S.A.
*.. ·. I.. ··.! . r
A graph is called threshold if there exists a real number b and real numbers aj associated with its vertices w, such that ims a, < b holds iff S is a stable (independent) set of vertices. The vector (a ,..., a; b) associated to a threshold graph is called an integral separator if ai + a > b + 1 for every edge (w,, wj). A simple algorithm is presented to determine for a given threshold graph its (unique) integral separator which minimizes b.
i
j
Let G be a loopless finite graph without multiple edges. If w is a vertex of G, let d(w) be the degree of w. The edge joining vertices u and w will be denoted as (u, w). Graph G is said to have property P if for every two vertices u, v such that (u, v) is
r
·· i
an edge, and for every pair of vertices u *, v * with d(u *)
d(u) and d(v *)
d(v),
(u*, v*) is an edge. In this definition it is possible that u* = u or that vu v. It has been shown in [1] that graph G has property P iff it is a threshold graph.
c
Suppose G is a threshold graph with vertices w, w2, w 3,..., w,. For IC
;,
{1, 2,..., n} let S = {w I i E I}. Let A = (a, a2 ,..., a,) be a real vector and let b be a real number. The pair [A; b] is said to separate G integrally if the following
holds: (1) ai
i i i j
'··
I
''
0 for i = 1,..., n;
(2) ,, 1ai b iff S is a stable (independent) set of vertices; (3) X,,ai~ 3 b + 1 iff S is a non-stable set of vertices. It was shown in [1] that a graph G is threshold iff there exists a pair [A; b] which separates G integrally. The following algorithm determines for a threshold graph G a hyperplane
3 a L ·;f ''' '' -a r
[A *; b*] which separates G integrally and such that b* is minimum. It will also be shown that it is the unique hyperplane with minimum b.
?i II C, i q I i
Algorithm A. Step 0: Relabel the vertices as w,..., w, such that d(w)< d(w 2) < < d(w). Step 1: Let t = minimum index such that (w,, w,,l) is an edge of G. [If no such t
exists let a* = 0 for i = 1 to n and let b* = 0. Then exit from algorithm.] Step 2: If d(w1 )=0 let a =0. If d(w)
~1let a = 1.
415
-----
~-
-
"mI 1.r1---
---
---
------
-ara
~ ~~
s~~
-
--
--u
-P
-~--
·-
·
416
J. Orlin
Step 3: For i = 2 to t if d(w) = d(wi- 1 ) then let a a* 1; if d(wi)> d(wi, ) then let aT= 1+ a* + a* ++ a-. Step 4: Let b* = a* + a + *- + a*. Step 5: For i = t + 1 to n let si be the minimum index such that (wi,, w,,) is an edge. Then let a= b a *+ 1.
Case 1: T with j j < tion. Thus w algorithm A. G has propel that S is stat
Example. Let G be the graph in Fig. 1. Table 1 shows how the algorithm worked.
[A*, b*] we iEI
Thus for all . Case 2: T Then S, co
/7
j < t then i
d(wi_,)
wi, wi,) is an
n worked.
417
Case 1: There exists a stable set S such that XIE at > b*. Let I = {jf,j2, ... ,jk with j' < j "* A2 jk. If jk a t then Gus, a *a t=1 aTY= b * and we have a contradiction. Thus we may assume that jk t + 1. Let q = sji as chosen in step 5 of algorithm A. Thus (wq, wjk) is an edge of G. If q jk- then d(wq) d(wj_,). Since G has property P, this would mean that (wk_,, wk) is an edge of G, contradicting that S is stable. Hence we may assume that q > jk-. But now by construction of [A*, b*] we have: q-1
a - t. Then by the choice of Si in step 5 it follows that i > sj. But then
3
aha
hEI
h
a*
i
a*t+a*= b* +l. . S
-a
Thus the proposition is ture. 8 26 5
½e threshold
hat d(w) s
~a* and that a is thus minimum. Suppose instead that
=1 to n. i <j. Then Vn)') d(wj). t n-1. io edges. It :his case. :h does not
d(Wk
)
1 +
3
Proof. By (2) then either I by (1) and (3)
a, >1 +
a
a*.
i =I
i=1
Acknowledger acknowledged
Corollary. The value for b* is also minimum. Proof. {w1, w 2 ,..., w,j is a stable set. Thus
Reference
t
a
b
a*
a
i=l
=
b*
i=1
Proposition 3. The algorithm constructs the unique [A; b] which separates G integrally with minimum b.
[1] V. Chvatal an Math. 1 (197;
Proof. Suppose [A;;b*] separates G integrally. Since t t b* > a> a b* it follows that ai = a* for i = 1 to t. For i = t + 1, t + 2,..., n we have that b* + 1
a +a si-1
ai +
3
at
