SOME S I M P L I F I E D N P - C O M P L E T E PROBLEMS
by
M. R. G a r e y and D. S. Johnson
Bell Laboratories
M u r r a y Hill, New Jersey
and
L. Stockmeyer t
M a s s a c h u s e t t s Institute of T e c h n o l o g y
Cambridge, M a s s a c h u s e t t s
kBSTRACT
c o m b i n a t o r i a l problems, form a class, the
N P - c o m p l e t e * problems, no m e m b e r of w h i c h
is k n o w n to have a p o l y n o m i a l time
algorithm, but such that if any one of
the problems does h a v e such an algorithm,
then they all do.
It is w i d e l y b e l i e v e d that showing a
p r o b l e m to be N P - c o m p l e t e is tantamount
to p r o v i n g its c o m p u t a t i o n a l intractability.
In this paper we show that a
number of N P - c o m p l e t e problems remain
N P - c o m p l e t e even w h e n their domains are
s u b s t a n t i a l l y restricted.
First we show
the completeness of SIMPLE MAX CUT (MAX
CUT w i t h edge weights r e s t r i c t e d to
value i), and, as a corollary, the comp l e t e n e s s of the OPTIMAL LINEAR
A R R A N G E M E N T problem.
We then show that
even if the domains of the NODE COVER
and D I R E C T E D H A M I L T O N I A N PATH problems
are r e s t r i c t e d to p l a n a r graphs, the two
problems remain NP-complete, and that
these and other graph problems remain
N P - c o m p l e t e even w h e n their domains are
restricted to graphs w i t h low node degrees.
For GRAPH 3-COLORABILITY, NODE
COVER, and U N D I R E C T E D H A M I L T O N I A N
CIRCUIT, we determine e s s e n t i a l l y the
lowest possible upper bounds on node
degree for w h i c h the problems remain
NP-complete.
These results have stimulated many
researchers to examine other c o m b i n a t o r i a l
problems for w h i c h no p o l y n o m i a l time
a l g o r i t h m s are known, to determine w h e t h e r
they too are NP-complete, and their
efforts have resulted in the d i s c o v e r y of
a d d i t i o n a l members of this class [Sa72,
Se73,U73].
Such results have considerable
p r a c t i c a l significance.
If one knows that
t h e p r o b l e m he wishes to solve is NPcomplete, and thus is u n l i k e l y to have
any p o l y n o m i a l time algorithm, he may
feel justified in c o n c e n t r a t i n g on more
hopeful a l t e r n a t i v e approaches.
He can look for algorithms which, a l t h o u g h
a d m i t t e d l y e x p o n e n t i a l in the worst case,
seem to w o r k q u i c k l y on most p r a c t i c a l
p r o b l e m s [K70,L73], or even w h i c h are
just "less exponential" than previous
algorfthms, and hence extend somewhat the
m a x i m u m size p r o b l e m w h i c h can be solved
w i t h i n p r a c t i c a l time limits [Sa73].
A l t e r n a t i v e l y , he can look for fast
algorithms which, a l t h o u g h they do not
a c t u a l l y find optimal solutions for the
problem, are guaranteed to yield solutions
w h i c h are "close" to optimal [Ga72,Gr72,
Jo72, Jo73 ] .
INTRODUCTION
Certain c o m b i n a t o r i a l problems, such as
the traveling salesman p r o b l e m and
theorem p r o v i n g in the p r o p o s i t i o n a l
calculus, have long b e e n notorious for
their c o m p u t a t i o n a l intractability, in
that, despite the effort of m a n y clever
people, no algorithms have been found
for them w h i c h can be g u a r a n t e e d to
require time b o u n d e d by a p o l y n o m i a l
in the length of the input.
The belief
in the inherent d i f f i c u l t y of these
problems has b e e n strengthened by results
of Cook and Karp [C71,K72].
These show
that simple forms of the above problems,
together w i t h a wide variety of other
A third approach, w h i c h m o t i v a t e s this
paper, relates to the fact that the practical s i t u a t i o n of concern often imposes
a d d i t i o n a l restrictions on the domain of
the problem, w h i c h may or may not make
the p r o b l e m easier to solve.
We examine
"polynomial complete" problems,
t e r m i n o l o g y of Karp [K72].
The w o r k of this author was done, in
part, at Project MAC, and supported by
The N a t i o n a l Science F o u n d a t i o n u n d e r
research grant GJ-34671.
47
in the
integers into strings of O's and l's
omitted).
Our proofs can then consist
of showing that k n o w n N P - c o m p l e t e
languages reduce to the ones we are considering.
(A list of the k n o w n N P - c o m plete languages we shall use, together
w i t h their definitions, is given in the
Appendix).
In general, we leave to the
reader the s t r a i g h t f o r w a r d v e r i f i c a t i o n
that (a) the language is in NP and (b)
the d e s c r i b e d m a p p i n g can be p e r f o r m e d in
p o l y n o m i a l time.
I.
SIMPLE MAX CUT A N D R E L A T E D P R O B L E M S
certain n a t u r a l restrictions on the
domains of a number of known N P - c o m p l e t e
problems, to d e t e r m i n e w h e t h e r the resultant subproblems are still NP-complete,
or if they do have p o l y n o m i a l time
algorithms.
Our results indicate that many k n o w n NPcomplete p r o b l e m s remain NP-complete,
even w h e n their domains are s u b s t a n t i a l l y
restricted.
In a d d i t i o n to the immediate
s i g n i f i c a n c e of k n o w i n g that these restricted p r o b l e m s are NP-complete, the
nature of the r e s t r i c t i o n s makes the comp l e t e n e s s results useful in two other
ways.
First, they increase our knowledge of the e s s e n t i a l elements w h i c h
made the o r i g i n a l p r o b l e m s NP-complete.
Second, they give us valuable tools for
p r o v i n g other c o m p l e t e n e s s results.
For
instance, by o b s e r v i n g that SATISFIA B I L I T Y W I T H AT M O S T 3 L I T E R A L S FER
CLAUSE, a r e s t r i c t e d form of SATISFIABILITY, is still NP-complete, Karp
[K72] was able to derive the N P - c o m p l e t e ness of C H R O M A T I C NUMBER, EXACT COVER,
MAX CUT, and a n u m b e r of other problems.
In [K72], the f o l l o w i n g p r o b l e m was
to be NP-complete:
shown
MAX CUT
~ :
G r a p h t G = (N,A), w e i g h t i n g
f u n c t i o n w:
A ~ Z, p o s i t i v e
integer W.
PROPERTY:
There is a set S C N such that
w(lu,vl) ~ w.
[u,v]~A
ucS,veN-S
In the first half of this paper, we show
that an important r e s t r i c t e d version of
MAX CUT is still NP-complete, and from
that derive the completeness of the
(until now) i n t r a n s i g e n t O P T I M A L LINEAR
A R R A N G E M E N T problem, as well as a number
of more closely related problems.
The
second hail of the p a p e r considers the
effect of r e s t r i c t i n g the allowable types
of graphs for N P - c o m p l e t e graph p r o b l e m s
such as NODE COVER, C H R O M A T I C NUMBER, and
H A M I L T O N I A N CIRCUIT, by either restricting the m a x i m u m degree of the nodes, or
a l l o w i n g only p l a n a r graphs, or both.
Karp p r o v e d the N P - c o m p l e t e n e s s of this
p r o b l e m by a r e d u c t i o n from the P A R T I T I O N
problem.
Thus, his p r o o f relies on the
fact that the edge w e i g h t s can be represented in space p r o p o r t i o n a l to the loga r i t h m of their magnitudes, since there
is a dynamic p r o g r a m m i n g a l g o r i t h m for
P A R T I T I O N w h i c h runs in time p o l y n o m i a l
in the input length, if those inputs are
e x p r e s s e d in u n a r y (i.e., in length prop o r t i o n a l to their magnitudes).
One might conjecture, therefore, that if
we r e s t r i c t e d MAX CUT by r e q u i r i n g each
edge w e i g h t to be exactly i, the new
problem, w h i c h we call SIMPLE MAX CUT,
might become easy.
As added support for
this view, notice that if W = IAI, then
this p r o b l e m simply asks w h e t h e r G is
bipartite, w h i c h can be d e t e r m i n e d quite
easily.
We summarize here the basic definitions,
r e f e r r i n g the reader to [K72] for a more
complete discussion.
Let B = [0,i] and
let B* denote the set of all finite
strings of elements from B.
Any subset
L of B* is called a language.
Let ~ be
the class of functions f:B* ~ B * w h i c h
are computable in p o l y n o m i a l time by onetape d e t e r m i n i s t i c T u r i n g machines.
If
L and M are languages, we say that L is
p o l y n o m i a l reducible to M, w r i t t e n L ~ M,
w h e n there is a f u n c t i o n f¢~ such that
f(x)~M if and only if xeL.
M is NPcomplete if M e N P (the class of languages
r e c o g n i z a b l e in p o l y n o m i a l time by onetape n o n d e t e r m i n i s t i c T u r i n g machines)
and every language in NP is p o l y n o m i a l
reducible to M.
In fact, if L is NPcomplete and L ~ M, then M e N P implies
that M is NP-complete.
In fact, however, SIMPLE MAX CUT is
NP-complete, as we show u s i n g a two-step
r e d u c t i o n from S A T I S F I A B I L ! T Y W I T H AT M O S T
3 L I T E R A L S PER CLAUSE (SAT3-- for formal
definition, see the Appendix).
We first
c o n s i d e r the f o l l o w i n g r e s t r i c t e d version
of the M A X I M U M S A T I S F I A B I L I T Y p r o b l e m of
[JOY3]:
M A X I M U M S A T I S F I A B I L I T Y W I T H AT MOST 2
L I T E R A L S PER CLAUSE
INPUT:
D i s j u n c t i v e clauses C~,C~, .... Cp,
each c o n t a i n i n g at mo~t ~wo
literals, p o s i t i v e integer k.
PROPERTY:
There is a truth a s s i g n m e n t to
the variables w h i c h satisfies
k or more clauses.
In accord w i t h the above definitions,
the "problems" we shall consider in this
paper, a l t h o u g h m a n y are more n a t u r a l l y
thought of as o p t i m i z a t i o n problems,
shall be p r e s e n t e d as r e c o g n i t i o n p r o b l e m s
(with the s t r a i g h t f o r w a r d details of the
e n c o d i n g of entities such as graphs and
We use the ordered pair (N,A) to denote
a graph G h a v i n g node set N and edge set
A.
48
We use the a b b r e v i a t i o n MAX SAT2 to
denote this problem.
Observe that,
when k:p, this p r o b l e m can be solved in
polynomial time [Co71].
However, we now
show that SAT3 can be reduced to MAX SAT2,
proving that MAX SAT2 is NP-complete.
Theorem I.i:
SAT 3 ~ MAX SAT2
N = [Ti:O<i<3p] <9 [Fi:O!i£3p]
%) [tij:l!i<_n,O<j<3P]
<* [fij:lii<_n,OiJi3P]
k) [xi:l!i<_n] k~ [xi:l!i<_n].
Proof.
Suppose we are given an input for
SAT~, that is, a set S of disjunctive
clauses, each containing at most 3 literals.
If any clause has fewer than 3 literals, we
may replace it by an equivalent clause
which has exactly 3 literals merely by
repeating one of the literals w h i c h it
contains.
Hence, we may assume that each
clause in S contains exactly 3 literals,
and we label them (alVblVCl) through ( a v b ~ c ~
. . . .a varmao~e
r~bi~
where e~ch ai,bi, andci represents either
or its negation.
The corresponding set S r
of clauses and value k forMAX SAT2 are given by :
S' :
~
[(ai), (bi), (°i), (di),
i=i
(a,ivbi), (aivci),
(bi,,~i) ,
(air'i), (bivdi), (ci~d i) ],
k : 7m.
7m or more of the clauses in S' can be
satisfied s i m u l t a n e o u s l y if and only if
the original set S is satisfiable.
For
note that, if we have any satisfying
assignment for S, then either one, two,
or three of ai,bi,ci must be set
true
for each i. The reader may verify that,
in all three cases, there is a truth
setting for d i causing p r e c i s e l y seven
of the clauses in S' arising from clause
i to be satisfied.
Furthermore, no setting of d. will permit more than seven
m
of the ten clauses to be satisfied,
~nd at most six of the clauses can be
satisfied if all of a i, bi, and c i are
"false. " •
The basic
framework A I of edges is
A 1 : [[Ti,Fj]:Oiii3p,0<Ji3P]
[[tij,fij]:l<i<_n,O<j<3P]
k9 [(xi,fij]:l!i<_n,O!j!3P]
[[xi,tij]:l~i<_n,O<j<3P]"
For any given partition N = S I ~ S 2,
S I ~ S~ = ~, we will say that edge [u,v]
is "ha~" if both u and v belong t~ the,,
same set in the partition and is go~a
otherwise.
Notice that all edges in A 1
will be good for any partition
N : S I ~ S 2 which obeys (a) all T~ belong
to the same set in the partition ~nd all
F i belong to the other set, and (b) for
each i, x i and all ti~ belong to the same
set in the
partition
~nd x~i and all f-.
•
'
i O
belong to the other set.
Furthermore,
if any pair F$,F. belong to different sets
in the partitmon~ then at least 3p+l edges
from A I will be bad, since each such pair
of nodes are mutually adjacent to 3p+l
other nodes.
Similarly, if any pair
xi,~ i belong to the same set in the partitlon, then at least 3p+l edges from A~
will be bad, since there are_3p+l disjoint
3-edge paths between x i and x i.
The following additional
included in G:
A 2 = ([ai,bi]:l~i<p
We now prove the completeness of SIMPLE
MAX CUT by reducing MAX SAT2 to it.
Theorem 1.2:
MAX SAT2 ~ SIMPLE MAX CUT.
edges are
and a i ~ b i]
[[ai,F2i_l]:l<i<P]
tO [[bi,F2i]:l<i<P].
Proof:
Let clauses CI,C2,...,C
and
~-~er
k be given as input forPMAX
SAT2.
In analogy with the proof of
Theorem i, we may assume that each clause
contains exactly two literals, not
necessarily distinct, and label them as
(a l~bl) , !a2~b2) ..... ( ~ n bp )r a We shall
cons~ruc~ a correspon
g g p , as input to SIMPLE MAX CUT, in two steps,
first giving the nodes and a basic framework of edges, and then adding in some
additional p r o b l e m - s p e c i f i c edges.
Let
Xl,Xg, ....,x n be the variables occurring
(either complemented or uncomplemented)
in the p clauses.
The set N of nodes
for the graph G is
The input for SIMPLE MAX CUT is the graph
G = ( N , A I k O A 2 ) and W =IA I] + 2k.
Given a truth assignment for the n
variables which satisfies k or more
clauses, construct the p a r t i t i o n
N = S l < 9 S 2 as follows:
S I : [Fi:O!i£3p] <3 [xi:x i is false,l<i<n]
[tij:x i is false,
l<i<n, OiJi3P]
[xi:xi is true,l<i<n]
S2 = N - SI .
49
(fij:xi
is true,
l!i<n,
O!j!3P],
Since, for each satisfied clause, one or
b o t h of a. and b~ b e l o n g to S~, exactly
two edges in A~ a r i s z n g from ~hat clause
must be good. -Furthermore, by our previous comments
every edge in A I is good.
Thus we have a~ least W = I A I I + 2k good
edges.
is positive, b o t h S I and S 2 are nonempty.
Let j = n -ISll.
Form
,
,
S~ = S ~ < ) [ul,u o ..... u.] and S^ = N' - S I.
r
n
T~en N = S~<-) ~ is a J p a r t i t i ~ n for G'
w i t h 14j = I~21 : ~, u I c S~, Un ~ S~, and
I
I
I[[u,v]¢A':u~SI,VCS2]I
Now, suppose we have a p a r t i t i o n
N = S I ~ S 2 for w h i c h W or more edges are
good.
Since k > 0 and IAoI < 3P, the
n u m b e r of bad edges cannoZ e~ceed 3P.
By our previous discussion, this implies
that all the F. must belong to the same
set, say S~.
~or the same reason,
e x a c t l y on~ of each pair xi,~ i must belong to S I.
Thus, a consistent truth
a s s i g n m e n ~ is obtained by setting x i
"true" if and only if x~ belongs to S~
For this truth assignment, clause i is
satisfied w h e n e v e r a~ or b i or b o t h
b e l o n g to S 2.
HowevSr, it is not difficult to see that, of the edges in A 2
a r i s i n g from clause i, e x a c t l y two are
good if one or both of a. and b i b e l o n g
to S 2 and none are good ~f a~ and b~
b o t h b e l o n g to S I.
Therefor@, sinc@ at
least 2k edges from A 2 must be good,
this truth a s s i g n m e n t must satisfy at
least k c l a u s e s . ~
=
n 2
-l[[u,v]6A':ucS~,v¢So]l
: n 2 - l[[u,v]~A:UCSl,V¢S2]l
< n 2 -W
:W'.
Ngw, suppose there i9 a p a r t i t i o n N' =
$7 k9 S~,, w i t h u~ e S~, u n ~ $2, and
IS~II = I~-I = n,sdch, t~at
I[[u,v]c~':ucST,vcSo]I < n 2 - W = W'.
Then N : S~ <9~S~, ~ h e r ~ S~ : S~ (-l N and
S o = S~(-~ ~, isma p a r t i t i S n fo~ G satisfying
l[[u,v]eA:UeSl,VeS2]l
=l Cu,vlZA,:uS;,vS;l,
I
I
= n 2 - l[[u,v]~A' :UCSl,VCS2] I
n 2 - (n2-W) = W.
Thus G has a cut of w e i g h t greater than
or equal to W if and only if G' has a cut
w i t h w e i g h t not e x c e e d i n g W', w h i c h
separates u I and u and divides the nodes
of the grap~ into ~wo equal sized subsets.
The r e d u c t i o n is p r o v e d . |
An easy c o r o l l a r y to the completeness of
SIMPLE MAX CUT concerns the f o l l o w i n g
problem:
M I N I M U M CUT INTO E Q U A L - S I Z E D SUBSETS
INPUT:
Graph G : (N,A), two distinguished nodes s and t, positive integer W.
PROPERTY:
There is a p a r t i t i o n
N = $ 7 < 2 S 2 w i t h Slgh S 2 =
I Sll = ~ S I, scaT, tgSo, and
A u s e f u l r e s t a t e m e n t of SIMPLE MAX CUT is:
MINIMUM EDGE-DELETION BIPARTITE SUBGRAPH
Observe that this p r o b l e m can be solved
in p o l y n o m i a l time if no r e s t r i c t i o n is
made as to the sizes of the subsets
[K72].
However, as defined, the p r o b l e m
is NP-complete, as we can conclude from
the completeness of SIMPLE MAX CUT and
the following:
That the f o l l o w i n g n o d e - d e l e t i o n version
of this p r o b l e m is also N P - c o m p l e t e
follows from T h e o r e m 1.4 below.
MINIMUM NODE-DELETION BIPARTITE SUBGRAPH
INPUT:
Graph G = (N,A), p o s i t i v e integer
k.
PROPERTY:
G has a b i p a r t i t e s u b g r a p h
formed by d e l e t i n g k or fewer
edges.
l![u,v]~:UCSl,~s2]l ~ w.
INPUT:
G r a p h G = (N,A), p o s i t i v e integer
k.
PROPERTY:
G has a b i p a r t i t e s u b g r a p h
formed by d e l e t i n g k or fewer
nodes.
T h e o r e m 1.3:
SIMPLE MAX CUT a M I N I M U M
~ A L - Z I Z E D
SUBSETS
Proof.
Given a graph G = (N,A) and posimZi-V~--integer W, as input for SIMPLE MAX
CUT, let n =IN'I and U = [Ul,U 2 ..... Un]
satisfy U(-h N = ~.
The c o r r e s p o n d i n g input for M I N I M U M CUT INTO E Q U A L - S I Z E D SUBSETS is the graph G r = (Nt,A'), nodes u l
and u_, and p o s i t i v e integer W ~, define~
as fo~lows:
N t
=
N QDU;
At
=
[ [u, v] :u, vcN ~ and
W t
=
n 2
-
T h e o r e m 1.4:
CLIQUE ~ M I N I M U M NODED E L E T I O N B I P A R T I T E SUBGRAPH.
Proof:
Given a graph G = (N,A) and
p o s - ~ i v e integer j as input to CLIQUE
(for a formal d e f i n i t i o n of CLIQUE See
the Appendix), let n = INI and
U = [u 1 ,u2,.'''.,u n ] where U(~ N = ~
The
c o r r e s p o n d m n g input for M I N I M U M NODED E L E T I O N B I P A R T I T E S U B G R A P H is the g r a p h
G, = (N',A') and integer k defined as
follows:
N' = N tJU;
A' : ( [ u , v ] : u , v ~ N ' , f u , v ] 6 A , and
[u,v] ~ A];
W.
Ifu,v]gh ul < i];
Suppose there is a p a r t i t i o n N = S L-) S
that l[[u,v]cA:U~Sl,V~S2]l ~ W.
Since W 2
k = n - j.
50
The reader may verify that G contains a
clique of j nodes if and only if G' has a
bipartite subgraph formed by deleting
n-j or fewer n o d e s . |
The final result of this section concerns
the OPTIMAL LINEAR ARRANGEMENT problem
[A73], defined as follows:
OPTIMAL
If(u)-f(~)l
:
{U, v}[A'
If (u)-f(~)l
LINEAR ARRANGEMENT
INPUT:
Graph G = (N,A), weighting
function w:A ~ Z , positive
integer W.
PROPERTY:
There is a i-i function
f:N ~ Z such that
w({u,v]). If(u)-f(v)l i
=
3
W.
< < nS+n+lh
-
This problem is a special case of the
well-known quadratic assignment problem
and a number of related facility location
and component placement problems.
We use
a reduction from SIMPLE MAX CUT to show
that this problem is NP-complete, even
in the restricted case where all edge
weights are required to be i (which we
call SIMPLE OPTIMAL LINEAR ARRANGEMENT).
Theorem 1.5:
~ A R
/
_ V If(u)-f(~)l
{u,~}~A
(u,v]~A
3
Now s u p p o s e t h e r e e x i s t s
f:N'---~Z such that ~
y
_kn ~ =
w.
a 1-1 function
Jf(u)-f(v)l < W.
Then there exists such an f having range
[i 2
n~+n].
Let F denote the set of
±-{ ~ d ~ i o n s
f:~,' ~ [1,2 ..... n'+nl.
Observe that for any f[F
SIMPLE MAX CUT ~ SIMPLE
ARRANGEMENT
Proof:
Given a graph G = (N,A) and
pos--~ive integer k as inpu~ for SIMPLE
MAX CUT, let n = INI, r = n-, and
U = fuq,u o .... ,Ur] where U ~ N = ~. The
corresDonding input for SIMPLE OPTIMAL
LINEAR ARRANGEMENT is the graph
G' = (N',A ' ) and positive integer W
defined as follows:
N'
h -~
: ( n~+n+l
3
Z
If(u)-f(v)l +
)~
If(u)-f(v)q
[u,vI~A/
[u,v]~A'
l<_i<j~<__~n~+n'J-i1: ~
\
3 /"
= N <J U;
there
There fore,
A' : [[u,v]:u,v~N'
and
exists an fcF such that
[u,v]6A];
~'
If(u)-f(v)
I > kn 4
(u,,,T]~A
Define
(Notice
th;t-(t~:
>-]
(v-u) which
l<u<v<t
the minimum W achievable--for
graph on t nodes.)
is
W* = max
f~F
a complete
(u)-f(v) I
'qf
/,
[U,V]cA
and
Suppose we have a partition N : SItN
which satisfies l[[u,v]~A:u~Sl,v~S~]1%2k.
Let S~ = [al,a 2 ..... at] and ~
~
-S o = ~bq,b o ..... b t].
Define f as
fbllows?
n-_
F* = IfcF:
£
If(u)-f(v)l = W* I "
[u,v]~A
Clearly W* > kn 4 and F* is nonempty.
each f~F*, ~efine the set
f(ai)
= i,
1 < i < t
f(ui)
= t + i, 1 <_ i <_ n
f(bi)
: n ~4 + t + i, 1 <_ i <_ n - t.
For
4
S(f) : [vcN:zui,uj~U
with
f(ui) < f(v) < f(uj)]
and let m(f) : I S (f)l.
Then there exists
a function g~F* such that m(g) < m(f) for
all fcF*.
We first show that m~g) : O.
51
Suppose m(g) > O.
Let v eS(g) be such
that g(v~) > g(v) for alY yeS(g).
For
each yen w, ~efine
L(v) = l[ueN':[u,v]¢A
+
2,
I g(u)-g(v) 1
(u,v]cA
ue S I , ve S 2
and g(u)<g(v)] I
and
<
R(v) = l~ueN':[u,v]eA
~ W* and m ( g ) < m ( g )
veS2]l
n3
n3
n3
< -~- + -6- + - ~ + n 4.1 [[u,v]eA:neSl,VeS2]l
which,
This
which
~(Vl) : g(v 2) and ~(v2) : g(vl),
satisfies
[u,v[cA
:W @ ,
of W*.
= O.
Since m(g) = O, the elements of U are
mapped by g to a set of consecutive
integers.
Define a p a r t i t i o n N = SIk3S 2
by
S I = [vcN:g(v)<g(u)
for all ueU],
S 2 : [veN:g(v)>g(u)
for all ueU].
v~
~
Graph Problems
~
Ig (u)-g(v) I
[u,v]~A
L,,,
the proof of Theorem 1.5.
More interesting are the cases where the
answer is not readily apparent.
For
instance, it is implicit in the l i t e r a t u ~
that MAX CUT, when restricted to planar
raphs, can be solved in polynomial time.
072] presents a polynomial time procedure for reducing the p r o b l e m of finding
the m a x i m u m cut in a w e i g h t e d planar
graph to that of finding a m i n i m u m
w e i g h t e d m a t c h i n g in a complete graph
derived from the dual of the original
graph.
A l t h o u g h [072] then resorts to a
n o n - p o l y n o m i a l branch and bound technique,
the w e i g h t e d matching can be found in
p o l y n o m i a l time using a method of Edmonds
[E70].
We now have
kn 4 <
I ~ k.
In certain eases, we can observe trivially
that it is.
For example, consider the
p r o b l e m CLIQUE [K72].
Since the largest
clique possible in a planar graph has
size 4, and the largest clique possible
in a graph w i t h m a x i m u m node degree k has
size k + i, we can find the largest
clique in either case in p o l y n o m i a l time
by examining all subsets of 5 or fewer
(k+l or fewer) nodes~ in time p r o p o r t i o n a l
to at most n ~ or n K+±, respectively.
g(u)-g(v)l
(u,v]~A
c o n t r a d i c t i n g the definition
Therefore, we must have m(g)
implies
M a n y of the reductions w h i c h were first
used to show certain graph theoretic
problems to be NP-complete involved the
c o n s t r u c t i o n of rather complicated graphs,
highly non-planar and w i t h nodes having
a r b i t r a r i l y high degree.
Since in many
p r a c t i c a l problems node degree may be
bounded (e.g., fan-in, fan-out restrictions on circuit elements), or graphs may
be planar, it is w o r t h w h i l e to determine
w h e t h e r the complexity of the graphs
involved in these reductions was necessary.
contradicts either the definition of W*
or the choice of g.
Thus, L(v ) < R(v ~
Let t = max[g(v):veN',g(v) < g~v )~ :d
L(v) > R(v)].
The value of t is~we~! defined since there exists a ueU w i t h
g(u) < g(vo) and L(u) : R(u) = O.
Thus,
if g(v]) = t and g(v2) = t + i, we must
have L~v2) < R(v2).
The function ~eF,
w h i c h is identical to g except that
>
completes
II, Restricted
5]
>
since k is an integer,
l[[u,v]eA:UeSl,VeS2]
{u, hA
l~(u)-~(v)I
+ (nh+n)l[ [u, v]eA :UeSl,
and g(u)>g(v)]l .
~ote that veU implies L(v) = R(v) = O.
Suppose L(v ) > R(Vo).
Let UoeU be such
that g(u ) J g~u) for all ueU.
Then by
definition ~f v , g(v ) < g(v) < g(u )
implies that veU.
CoNsider the--function
gcF which is identical to g except that
g(Vo) = g(u ) and g(u ) : g(Vo).
It is
not difficult to see Zhat
Ig(u)-g(v)l
+ \3/
Ig(u)-g(v) I
[u,vlzA
u,veS 1
On the other hand, one of the authors has
shown [St73] that the question of w h e t h e r
a graph is 3-colorable (GRAPH 3-COLORABILITY, itself a restriction of CHROMATIC
NUMBER) is NP-complete, even when the
domain is restricted to include only
planar graphs.
In this section, we shall
present further completeness results,
Ig(u)-g(v)l
[u,v]cA
U, V6 S 2
52
c o n c e r n i n g GRAPH 3-COLORABILITY, NODE
COVER, and H A M I L T O N I A N CIRCUIT.
The
formal d e f i n i t i o n s of these problems
appear in the appendix.
know w h e t h e r PLANAR D I R E C T E D I ~ M I L T O N I A N
CIRCUIT, PLANAR U N D I R E C T E D H A M I L T O N I A N
CIRCUIT, or PLANAR U N D I R E C T E D ~ M I L T O N I A N
PATH are NP-complete, and these remain
s i g n i f i c a n t open problems.
The f o l l o w i n g table gives the p r i n c i p a l
r e s t r i c t e d versions of these problems
w h i c h we prove to be p o l y n o m i a l complete:
The proofs of the results given in the
table follow.
For each problem, we show
that there is a known N P - c o m p l e t e p r o b l e m
w h i c h reduces to it.
PROBLEM
i. PLANAR GRAPH
2. U N D I R E C T E D
3. PLANAR
4. NODE
3-COLORABILITY
HAMILTONIAN
DIRECTED
T h e o r e m 2.1.
PLANAR GRAPH 3 - C O L O R A B I L I T Y
PLANAR GRAPH 3 - C O L O R A B I L I T Y W I T H NODE
DEGREE AT MOST ~.
CIRCUIT
HAMILTONIAN
PATH
COVER
5. PLANAR NODE
NODE
Proof: The key to our c o n s t r u c t i o n will
~ e
use of "node substitutes."
Figure
I(A) shows the 3-outlet node substitute
H3, w i t h its first, second, and third
output nodes labelled.
For k > 4, the koutlet node substitute H k is formed by
a d j o i n i n g to Hk_ I a copy of H~ having its
first outlet c o i n c i d i n g w i t h 5utlet k-i
of H~ 7.
The outlet nodes of H k are the
n o d e ~ - ~ a v i n g degree 2.
The outlets w h i c h
o r i g i n a l l y b e l o n g e d to H k I retain the
same labels, w i t h the second outlet of th~
a d j o i n e d HR b e c o m i n g outlet k-i and its
third outl~t b e c o m i n g outlet k.
Figure
I(B) shows H 5.
COVER
DEGREE AT M O S T
i.
4
2.
3
3.
}~-OUT,
4.
5.
3
6
3-1N
For results 3 and 5, it was not p r e v i o u s l y
known if the p l a n a r problems were complete, even if no r e s t r i c t i o n s were
p l a c e d on node d-~grees.
The degree constraints in 1,2, and ;~ are all best
possible, in that each of the problems
becomes easy if the r e s t r i c t i o n on node
degree is reduced by I.
NODE COVER and
U N D I R E C T E D } ~ M I L T O N I A N CIRCUIT are
clearly trivial for graphs w i t h m a x i m u m
degree 2, and a w e l l - k n o w n result of
Brooks [B!$1] implies that a c o n n e c t e d
graph w i t h m a x i m u m degree 3 is 3-colorable if and only if it differs from K~4,
the complete graph on four nodes, w h i c h
is easy to determine.
It is easy to prove by i n d u c t i o n that,
for all k > 3, the f o l l o w i n g facts hold:
(2.1A)
H~ has 7(k-2) + i nodes,
i ~ c l u d i n g k outlets.
(2.1B) No node of H k has degree
e x c e e d i n g ~.
(2.1C) H k is planar.
(2.1D) Hb is 3-colorable, but not
2Ucolorable, and every valid
3 - c o l o r i n g of H k assigns the
same color to every outlet
node.
Given any p l a n a r graph G, we show how to
c o n s t r u c t a p l a n a r graph G', using node
substitutes, w h i c h has m a x i m u m degree I~
and w h i c h is 3-colorable if and only if G
is 3-colorable.
In a d d i t i o n to the above results, there
are a n u m b e r of more or less immediate
corollaries.
Result 2 implies that
D I R E C T E D ! ~ H I L T O N I A N CIRCUIT w i t h node
degree b o u n d e d by 3-OUT, 3-1N is NPcomplete; however, the largest degree
bounds for w h i c h we know this p r o b l e m to
be easy are 2-OUT, I-IN or 2-1N, 1-OUT.
Also, we may substitute PATH for CIRCUIT in result 2.
However, we do not
2
Fix a p l a n a r e m b e d d i n g of G and a r b i t r a r i ly index the nodes of G w h i c h have degree
e x c e e d i n g 4 as V l , V o , . . . , v r.
We construct
a sequence of graph~ G = G~,G~,
,G = G'
i
"'"
r
as follows: G i is c o n s t r u c t e d from Gi_ I.
2
(A)
3
(B)
Figure
I. Hode
substitutes
53
H 3 and H 5
4
It is easy to prove by induction that the
f o l l o w i n g facts hold for all k > i:
Let d be the degree of v~ in Gi_ I and
let {u_,v.],[u2,v.],
%
... ,~Ud,Vi] be the
edges ineldent w i t h vi, taken in clockwise order.
To form Gi, delete node v i
from Gi_l, replacing it w i t h a copy of
H~, and replace each edge [ui,vi] by an
ecge joining ui to outlet j ~f the node
substitute.
(2.2A) F k contains k 2 + k - i nodes,
none w i t h degree e x c e e d i n g 3.
(2.2B) Fb has one inlet node of degree
1-and k outlet nodes, none w i t h
degree e x c e e d i n g 2.
(2.2C) For any outlet node of Fw, there
exists a path from the inlet to
that outlet w h i c h includes each
node of F k exactly once.
It follows from the c o n s t r u c t i o n and
p r e v i o u s l y stated facts that, for
0 < k < r, G k is planar, G k has r-k
no~es ~ i t h degree e x c e e d i n g )~, and G k
is 3-colorable if and only if G is 3colorable.
Thus, G' : G r satisfies all
the required properties, c o m p l e t i n g the
proof.
One more p r o p e r t y of F~ will be required
and, since its p r o o f iZ not quite as
straightforward, we present it as a lemma.
Lemma 2.2.1.
Suppose a graph G contains
a s u b g r a p h H isomorphic to Fk• k ~ i, in
such a way that
(i) no two nodes of H are adjacent in
G unless the c o r r e s p o n d i n g nodes
of F~ are adjacent• and
(ii) any ~ode of H w h i c h is adjacent to
a node of G not b e l o n g i n g to H
c o r r e s p o n d s to either an inlet or
outlet node of F k.
T h e o r e m 2.2.
UNDIRECTED HAMILTONIAN
CIRCUIT ~ U N D I R E C T E D HAMILTONIAN CIRCUIT
W I T H VERTEX DEGREE AT M O S T 3
Proof. This c o n s t r u c t i o n will also use a
"node substitute," w h i c h is formed from
a special graph, called a fan.
The oneoutlet fan F I consists simply of a sin-gle node.
The single node, labelled
Uq I, is both the inlet and the outlet
T
.
.
.
or • PI.
_nductively,
assume
we. have
defiffed the k - o u t l e t fan Fk, k > i, w i t h
inlet U I i and outlets U~ I through
U~ ~.
T~e (k+l)-outlet ~&n Fk+ I is
fb£Red from F k by adding the f o l l o w i n g
nodes and edges:
Nodes: Uk+l,i,
Proof of Lemma.
The Lemma holds
t r i v i a l l y for k = i.
Suppose it holds
for F k i, k > i, and c o n s i d e r a graph G
w h i c h ~ontains a s u b g r a p h H isomorphic to
Fk in the s p e c i f i e d manner, and w h i c h
contains a H a m i l t o n i a n circuit C. We consider the nodes of H as b e i n g l a b e l l e d
i d e n t i c a l l y w i t h the c o r r e s p o n d i n g nodes
of F~.
Observe that H, and hence G, contain~ a s u b g r a p h H' isomorphic to Fk_ I
w h i c h has inlet node U I i and w h i c h
satisfies the two conditions of the
Lemma.
By the i n d u c t i o n hypothesis, C
contains a path from U1 1 to some U k i "
w h i c h includes p r e c i s e i # the nodes
- 'J
b e l o n g i n g to H'.
The node Uk_ I j and the
remaining nodes of H are shown ~n Fig. 3.
i <_ i <_ k + i; Sk+l,i,
i<
Edges:
Then, any H a m i l t o n i a n circuit of G contains a p a t h from the "inlet" of H to
some "outlet" of H, c o n s i s t i n g p r e c i s e l y
of all the nodes of H.
i<k+l.
[Uk, i,Sk+l,i] ,
[Uk+l,i,Sk+l,i],
i <_ i < k;
1 <_ i <_ k
[Uk+l,i+l,Sk+l,i],
i
+
i;
<_ i <__ k;
[Sk+l,k+l'Uk+l, i ] "
The inlet of Fk+ I is UI, I and its o u t l e ~
are Uw~ l
Figure 2
_, I through U k + l , k + l "
shows Pi,92, and F 3.
~C]iS3, 3
{'ho~$2,2
U1,1
tl
U1,1
•
~oS2,
2
I ~ U2, 2
U2,2
U1,1
U
3, 3
- S3, 2
I ~ $2' t
U2, 1
e
I ] S3' 1
U3, 1
F~
F3
F2
Figure 2. Fans FI, F2,
54
and F 3
Uk-1, j
--0--
Uk,1
Uk, j
Sk, 1
Figure
Consider
Uv i and
s~guction
there are
n o d e s of
•
'
.
.
.
--0
.
Sk, j Uk, j+l
3. R e m a i n i n g
the set T of n o d e s of the f o r m
S~ ~, i < i < k.
By the con~ f ~ F k aid t~e a s s u m p t i o n s on H,
o n l y two w a y s by w h i c h the
T can be a c c e s s e d by c i r c u i t C:
•D
Sk,k
of H
v l ( i ) , v 2 ( i ) ..... Vn(i ) and its o u t l e t s by
ul(i),u2(i),
,un(i ).
The s p e c i f i c a t i o n
of G' is c o m p l e t e d by i n c l u d i n g in G',
for e a c h edge [vl,vi]eA
the two edges
[ u j ( i ) , v i ( J ) ] an~ [ ~ i ( J i , v j ( i ) ] .
(a) F r o m n o d e s of G n o t in H via an
outlet U- •
(b) F r o m an Ko d ~ l e t U -~ - ~i . , l• of H' • via
the c o r r e s p o n d i n g ~k,i"
to
Since C c o n t a i n s a p a t h f r o m U I
U ~ i . w h i c h u s e s all the n o d e ~ ' ~ f H',
t~e 6 ~ l y w a y that (b) can o c c u r is via
the edge [Uk_ 1 4,S k ~].
U s i n g this edge,
the p a t h from ~
~ £~ U, ~ . can be
i
~-i,J
.
e x t e n d e d to a p a t ~ f r o m U~ ~ to e i t h e r I%
•
.
.
I
~
~,j
U k ~+i c o n s i s t i n g p r e c l s e ~
of th_ n o d e s
b e i 6 n g i n g to H.
If this is not w h a t
occurs in C, t h e n e i t h e r that edge is not
used, or it is u s e d and C exits f r o m the
set of n o d e s T b e f o r e all n o d e s of T have
b e e n covered.
In e i t h e r case, a n o n - z e r o
and e q u a l n u m b e r of U - t y p e and S - t y p e
n o d e s f r o m T w i l l r e m a i n to be c o v e r e d by
s u b s e q u e n t v i s i t s of the H a m i l t o n i a n circuit to the set T.
H o w e v e r e a c h such
visit of T b y C m u s t b o t h e n t e r a n d leave
via U - t y p e node, and hence m u s t use one
m o r e U - t y p e n o d e t h a n S - t y p e node.
Thus,
not all the S - t y p e n o d e s c o u l d be c o v e r e d
b y C, c o n t r a d i c t i n g the fact that C is a
H a m i l t o n i a n circuit.
Therefore, C must
c o n t a i n a p a t h f r o m U 1 I to some o u t l e t
Uk i c o n s i s t i n g p r e c i ~ @ l y of all the
n o ~ e s of H, as claimed.
The L e m m a follows by i n d u c t i o n . ~
nodes
Uk,k
The f o l l o w i n g u s e f u l p r o p e r t i e s of
d o u b l e - f a n s D n are i m m e d i a t e c o n s e q u e n c e s
of the c o r r e s p o n d i n g p r o p e r t i e s for Fn:
(2.2a)
(2.2b)
or
2.2c)
2.2d)
D n c o n t a i n s 2 ( n 2 + n - l ) nodes,
n o n e w i t h d e g r e e e x c e e d i n g 3.
The n inlet n o d e s and n o u t l e t
n o d e s of D
e a c h have d e g r e e not
e x c e e d i n g ~.
For e a c h o u t l e t and e a c h inlet
of D_, there is a p a t h b e t w e e n
them w h i c h i n c l u d e s e v e r y node
of D n e x a c t l y once.
S u p p o s e an u n d i r e c t e d g r a p h G
contains a subgraph H isomorphic
to D n in the m a n n e r s p e c i f i e d in
L e m m a 2.2.1.
Then every
H a m i l t o n i a n c i r c u i t of G contains a p a t h f r o m an "inlet" of
H to an " o u t l e t " of H, c o n t a i n ing p r e c i s e l y all the n o d e s of H.
O b s e r v e that G' has m a x i m u m d e g r e e 3,
since e a c h inlet or o u t l e t of a Dn(i ) has
at m o s t one edge j o i n i n g it to a node not
b e l o n g i n g to Dn(i), u s i n g p r o p e r t i e s
(2.2a) and ( 2 . 2 b ) .
S u p p o s e G' has a H a m i l t o n i a n c i r c u i t C,
i.e., an o r d e r i n g of the n o d e s as
YT,Y2, .... Ym, w h e r e m = INrl, such that
f6r all j, i < j < m, [y~,y._~]~A' and
s u c h that [y_~y~]c--A'.
W$ m ~ m a s s u m e
withm
±
.
out loss of g e n e r a l i t y that Y7 and Ym do
not b e l o n g to the same d o u b l e ' f a n Dn~i ).
Thus, by c o n s t r u c t i o n of G', one of Y l , Y m
must be an inlet of some fan Dn(i ) an~ t~e
o t h e r m u s t be an o u t l e t of a n o Z h e r fan
Dn(j).
Since G' is an u n d i r e c t e d graph,
we m a y a s s u m e that Yl is an inlet of some
fan D_(i).
Then, b y - ( 2 . 2 d ) , C d e t e r m i n e s
an o r d e r i n g D n ( i l ) , D n l i 2 ) ..... Dn(in) of
the d o u b l e - f a n s s u c h that all the n o d e s
of Dn(ik) p r e c e d e all the n o d e s of Dn(i. )
in C w h e n e v e r i < k < j < n.
But the
J
c o n s t r u c t i o n of G' i m p l i [ s that
v i l , v i o , . . . , V i n m u s t then be a H a m i l t o n i a n
c i r c u i t of G.
(Notice that this a r g u m e n t
G i v e n a g r a p h G = (N,A), we n o w show how
to c o n s t r u c t a g r a p h G' = (N',A~), h a v i n g
m a x i m u m node d e g r e e 3, such that G' contains a H a m i l t o n i a n c i r c u i t if and o n l y
if G does.
Let N = [Vl,V 2 ..... Vn] w h e r e n = INf.
Let D n d e n o t e the d o u b l e - f a n f o r m e d b y
j o i n i n g two c o p i e s of F n w i t h an edge
c o n n e c t i n g t h e i r inlet nodes.
The outlet n o d e s of one of the copies of F n are
the inlet nodes for Dn, and the o u t l e t
n o d e s of the o t h e r are the o u t l e t n o d e s
of D_
For e a c h node v. in N, the g r a p h
Sr
c ~o°n t a i n s a c o p y D n ( i ~ of D n.
The
i n l e t s of Dn(i ) w i l l be d e n o t e d b y
55
fails when n = 2, however,
we may let G' = G.)
The set of edges is made up of two parts,
A : A] L_)Ao, the first of w h i c h depends
only Dn n Znd t and forms a skeleton for
G (see Figure 4):
in this case
Conversely, suppose vi~,vio,._.,Vin is a
H a m i l t o n i a n circuit C 6f G? By construction of G some outlet of Dn(i~ ) must be
connected to some inlet of Dn@i~+~,, in
G', for i < j < n, and similarl9 ~or
Dn(in) and--Dn(il).
It is then a simple
matter to construct a H a m i l t o n i a n circuit in G' using (2.2c).
A I = <ao,j,Y0,j>,
<Yn, j,an+l,j>:
l<j<t]
L3 [<Y0,j~a0, j+l> ' < a n + l , j , Y n + l , j + l > :
l<j<t-l]
Therefore, we have shown that G' contains a H a m i l t o n i a n circuit if and only
if G does, completing the proof of
Theorem 2.2.I
U
< Y o , t , f l ~' < f n + l ' y n , l ~]
< fi' di>' < fi' el>' < di' fi+l>"
<ei,fi+l>, <di,ai,l>,
<ei,ci,t) , <ci,t,ei>:
Theorem 2.3:
EXACT COVER ~ PLANAR
D-J~MILTONIAN
PATH.
Proof: Given any collection S of sets,
we must construct a planar directed
graph G w h i c h has a H a m i l t o n i a n path if
and only if S contains an exact cover.
We first introduce some terminology.
<ai,l,di>,
l<i<n]
< ci, j,ai,j+l>, <ai,j+l,Ci,j> :
l<i<n,l<j<t-l}.
Let G = (N,A) be a directed graph with
nodes Pl,Po,-..Pm.
The edges in A are
ordered pa~rs < P i , P j > "
We will call Pi
and p j a d j a c e n t w h e n e v e r either
<Pi,p~eA
or <,pi,Pi>eA.
Any H a m i l t o n i a n
path ~H-path) of G can be identified
with a s - - ~ g
pj (l~Pi(2],. ~..pi(-)
~
_°f nodes
such that n = m~<' tP4filJ and \PI(ii,
Pi(j+l)>CA forJ~l i ! ~ < m.
The remaining edges in A = A ~ A
connect, for each i and j, ±i\ ~ ~ n,
I < j < t, certain nodes from
[a~ ",~i -,c- .y- - .,y- .], depending on
whe~er
6~ n ~ t 0 u ~ - @ $ 1 o n ~ J t o
S i.
If
u. ~ S~, then A~Ucontains the eight edges
s~own in Fig. 5~.
If u~ e S~, then Ao
contains the seven edge~ sho~n in
Fig. 5B.
Let N* denote the set of all finite
strings of elements from N. A partial
H-path oJ for G is an element of N* w h i c h
is a ~refix of some H-path for G.
For
any partial H-path ~ for G, ~eN* is a
k - e x t e n s i o n of ~ if the length of • is k
and m~ is a partial H-path for G.
This completes the d e s c r i p t i o n of G.
The
reader should observe from Figs. 4 and 5
that G is planar.
Informally, the relationship b e t w e e n G
and the covering p r o b l e m is as follows.
Suppose that we are b u i l d i n g an H-path
for G, in step-by-step fashion, and, at
the same time, generating an exact cover
for U.
Clearly, the H-path must begin
w i t h a n I,Yo 1,an o,Yn o,...,ao t,Yn ,,
f].
F~eac~k,~l~<
~'< n, at the ~ @ p
t~e H-path reaches Z~, We choose edge
<f,.,ek> if S,. is to ~e included in the
coger, and e~ge < f ~ , d k > if it is not to
be included.
We t~en proceed left (right)
along the a k ,J~,b-K .,c.K " line of the
graph, exiting fr$~ dk($k) , and finally
a r r i v i n g at fk+1"
Helping us in the
--- direction to travel tha t
choice of which
line is the fact that the nodes Y~_]
i < j < t, "remember" which e l e m e ~ t ~ ' ~ f U
ha~e b~en covered by p r e v i o u s l y selected
sets.
Specifically, Yk-i ~ has already
been visited by the parti$~ H-path if and
only if element u. has not been covered
yet.
Note that, ~hen w~ ~re about to
make our first choice, at fl, all nodes
YO . are already in the path and no elemefl~ from U has been covered.
When we
reach fn+l, after having made a decision
for each set Sk, none of the y
. can have
been visited by the partial H-~&@h (or it
could not be extended to an H-path), so
all elements of U have been covered by the
For any string ~ = b l b p . . . b k e N * , we say
that yeN belongs to ~ if and only if
v = b i for some i,i---< i < k, and < u , v > e A
belongs to ~ if and ~nly--if u = b i and
v = bi+ I for some i, I <_ i < k.
Now suppose we are given a collection
S = [SI,S 2 ..... Sn] of sets w i t h ~
Si
5=1
= '1 = ( U l , U 2 , .... u~:.
,The p l a n a r
directed graph G
~,~,A) is specified
as follows:
The set N of nodes,
on n and t, is
which depends
on'ly
N = [bi,j,ci, j: i < i i n, i <__ j < t]
~
[a. .:
l,O
LO [Yi,j:
0 < i < n + I, i < j < t]
.
.
.
.
0 <__ i <_ n, i <_ j <_ t]
k9 [di,ei,fi:
i < i <_ n] t_) [fn+l].
56
C
C
•
.
/ \./\./
- t ~f ' ' u .
.
i
• ~'~.
.~--~.-
•
....
.,#r-'--"
•
J
,, " ~ . ~ .
•
&z &~
•
~.o,
.
~.
•
~
4b~ - - 0
C~.~
I~%~ c~o .~%~ ~
•
•
q~
~
Q (~--- •
. . . .
•
~
a~|a,~,l %,i c,,,i a~,~. b',,,., c.,,.+.
L
•
•
~.v"
8~.~ ~z,+. c ~
•
-.
• •
O(
ez
•
•q~ b~ c~,~
~ j /
~1~,1 ~"t.-I
J
en
"",jl,o
a..,~h,,t
~÷1
Figure 4. The Skeleton of G.
ci.,3
CA)
ai,~ •
\ ,,'~
. ci,~
" Jt~3
Figure
5. Edges
57
in A 2
B
selected sets.
The edges of Ap force the
t r a n s m i s s i o n of i n f o r m a t i o n from one row
of Yk ~ nodes to the next, and also prevent £~t S~ from b e i n g chosen w h e n e v e r
any of its-members has already been
covered.
This latter p r o p e r t y insures
that the sets in the cover are disjoint,
as required by E X A C T COVER.
insure that < x , , c i .> belongs to 7; that
is, c. • does not ~$t b e l o n g to the H-path
and t~@Jonly way that path can reach c. .
is through x'.
This c o n t r a d i c t i o n pro~@~
(iS).
The other eases follow similarly.|
U s i n g Lemma 2.3.1 and our informal
d e s c r i p t i o n of the c o r r e s p o n d e n c e b e t w e e n
H-paths and exact covers, the reader
should have no d i f f i c u l t y in v e r i f y i n g
that G has an H - p a t h w h e n e v e r S contains
an exact cover.
To complete the p r o o f of
T h e o r e m 2.3, we must show the converse.
The f o l l o w i n g Lemma-shows how the edges in
A 2 force the desired paths, by giving the
relevant o r o p e r t i e s of Figs. 5A and 5B.
Figures 6A and 6 B show all edges of G
w h i c h are incident w i t h nodes of interest.
Lemma 2.3.1:
Fix i and j, i < i < n,
1 < j <_ t.
Let x = d i if j =--i a~d
x = c i . i otherwise; x I = e i if J
n and
x I = a ~ - + l otherwise.
Let co be
partiam J-path of G satisfying:
any
Suppose that G has an H-path 7.
Let
T = [kl < f k , e k > belongs to Z].
We shall
show that S' = [Skl keT] forms an exact
cover for U.
Define the partial unions
U o : ~ and U k : l ~ < k Si• 1 ! k ! n.
leT
(i) every yeN - [a. .,b~ ..,c. .]
l,J
,
1 J
adjacent to Yi-l,j ~e~ong~ to co,
and
(ii) None of ai•j,bi,j,ci, j belongs to
co.
Then we have the following:
Lemma 2.3.2:
For each k, 1 < k < n + i,
if co(N* and cofk is a partial--H-p~th of G,
then
Then the f o l l o w i n g hold:
Case i.
uj~S i (Figure 6A)
(IL) if • = coxai ~ is a partial H - p a t h
of G, the o~Yy p o s s i b l e 4-extensions of T are bi,jYi_l, jCi, j.x t and
b. "Yi .c. .x' .
(IR) I~'@ = ' ~ x ~ ' ~
~ is a partial
H - p a t h of G,-tNe only p o s s i b l e
%-extensions of • are
bi,jYi_l, jai,jx and bi,jYi,jai,jx.
Case 2.
2R
(i) None of {ai,~,b- .,c~ ~,d~,eJ :
k<i<n, ~<j<t~ bel~ng t~ co,
(ii) ATl-of 1[~--~,b~ ~,c. .,d.,e. •
.
~
•
i
i
i"
l<l<k-l, i<J<t} 6eloh~ to co,
(iii) For each j, 1 < j < t• Y~Yi{]'J
belongs to co i~ a n ~ on
uj~Uk_ ~
(iv) If there exists j such that
u l c U w _ l f h Sk, then dv is the only
i Z e x Z e n s i ° n - ° f cofk"
uieS i (Figure 6B)
Proof of Lemma.
The proof is by induction.
The basis k=l is immediate since any H-path
must b e g i n w i t h a 0 ~ Y O ~ l a O , 2 ~ . . a o ~ y
n ~ i ~ ef~ y.
The induction step'zoilows almost'
by Lemma 2.3.1.
The only other p o s s i b i l i t y
is that cofv might be e x t e n d e d by d k f~+l
(or by ek ~k.~).
But then an a r g u m e n ~
similar ~o Z~$ proof of Lemma 2.3.1 shows
that e k ( r e s p e c t i v e l y dk) must be the last
node oT the H-path• w h i c h is i m p o s s i b l e . ~
If • = o~x'c, i is a partial
H-path of G,~'the only possible
3 - e x t e n s i o n of • is bi,jai,jx.
Proof of Lemma:
(Case IL) Let 7 be an
H - p a t h w i t h p r e f i x T.
The last node of
7 must be a . . . . . The only possible
iI~"
• O
4 - e x t e n s i o n o~ • which has been excluded
is b i ~y. .vv, w h e r e v ~ c- .. But then
c.
node of 7,
i . @gula0have to be the ~ t
smf~@e p r o p e r t i e s (i) and (ii) of co
ai,]<!/bi'j ci,j
x,<-'Y"
.
It is now clear that
x'
Figure 6. Cases for Lemma 2.3.1
58
[SwIkeT ] is an exact
c o v e r for U!
Since a n y H - p a t h m u s t have
the form ~ f n ly n i a +- -Yn -a +I ....
Yn tan+l t' ~ e m m d 2 . ~ . ~ ' ~ i m i ~ ~ o r ' { = n + l
i n { u r e s t~at <9~ S~ = U
= U, and (iv)
% n s u r e s the d m s j o l n t n e s s of the sets in
~ne cover.
This c o m p l e t e s the p r o o f of
T h e o r e m 2.3.2. I
BI :
and
B2 =
Ti, 1
Fi, 1
Ti, 2
Figure
as,t
= xi is
the
variable
jth
of
x i in
C].
The
graph
N :
A=
G =
(N,A i is d e f i n e d
~
NiQJ
i=l
i=l
AiU
P
h<9=l N h
by:
,
C
BIUB 2.
h=l
O b s e r v e that e v e r y n o d e of G has d e g r e e
at m o s t 3.
W e s h o w that the set C of
c l a u s e s is s a t i s f i a b l e if and only if G
has a n o d e c o v e r of size 5P.
The f o l l o w i n g p r o p e r t i e s
H i are e a s y to verify:
W e d e s c r i b e the g r a p h G : (N,A) in
s e v e r a l steps.
First, for e a c h v a r i a b l e
x.,
a' s u b g r a p h H.i = (N.,A.),
a
1 we have
'
i
i
s i m p l e c i r c u i t w i t h l N.I = IAil = 2m(i),
as s h o w n in Fig. 7.
O ~ s e r v e the a l t e r n a t e l a b e l l i n g of the nodes.
For e a c h
c l a u s e Ch, we have a s u b g r a p h
H~ = (Nh,Ah) w h e r e Nh = [Vh, l , V h , 2 , V h , 3 ]
a n d A ~ c o n s i s t s of an edge j o i n i n g e a c h
p a i r of n o d e s b e l o n g i n g to N h.
The
r e m a i n i n g edges, w h i c h e a c h j o i n a node
f r o m some set N. to a node f r o m some set
Nh, are as foll~ws:
•
[[Fi,j'Vs,t]:
occurrence
COVER WITH
•
jth
of v a r i a b l e
x i in C]
Proof.
S u p p o s e we are g i v e n a set
U--/-TC],C O ..... Cn] of d i s j u n c t i v e
clauseZ, ~ a c h c o n t a i n i n g no m o r e than 3
literals.
As in the p r o o f of T h e o r e m
i.i, we m a y a s s u m e that e a c h c l a u s e
contains exactly 3 literals, possibly
with duplication.
Index the l i t e r a l s
o c c u r r i n g in c l a u s e C h as ah,l,a h 2'
a n d a h ~, i < h < p.
Let x l , x o , . ~ . , x ~
d e n o t e ' ~ h e va--ria[les o c c u r r l n g ~ i n the~p
clauses, and for e a c h i, 1 < i < n, let
m(i) d e n o t e the n u m b e r of o ~ c u r T e n c e s of
v a r i a b l e x i (as l i t e r a l x~ or l i t e r a l
xi) in the clauses.
Arbitrarily index
t~e m(i) o c c u r r e n c e s of v a r i a b l e x i as
o c c u r r e n c e i, o c c u r r e n c e 2,..., o c c u r r ence m(i).
W e shall c o n s t r u c t a g r a p h
G, h a v i n g node d e g r e e at m o s t 3, and
give an i n t e g e r k > O, s u c h that C is
s a t i s f i a b l e if and o n l y if G has a n o d e
c o v e r of size k.
C
as, t = x i is the
occurrence
O b s e r v e that the d i r e c t e d g r a p h G cons t r u c t e d in the p r o o f n e e d not have
a r b i t r a r i l y large degree, in fact, no
node has i n - d e g r e e e x c e e d i n g 3 or outd e g r e e e x c e e d i n g 4.
It is not known,
however, w h e t h e r these are the s t r o n g e s t
p o s s i b l e d e g r e e c o n s t r a i n t s for w h i c h
PLANAR DIRECTED HAMILTONIAN PATH remains
NP-complete.
T h e o r e m 2.4:
SAT3 ~ N O D E
~ Y O ~ ~
AT M O S T 3.
[[Ti,j,Vs,t]:
of the
subgraphs
(2.4A)
T h e r e e x i s t s a n o d e c o v e r for
~i c o n t a i n i n g m(i) nodes,
i n c l u d i n g all n o d e s T. • and no
n o d e s F. .
There al~6Jexists
i
"
such a n 6 ~ e c o v e r w h i c h i n c l u d e s
all n o d e s F. . and no n o d e s
T.i,j"
i,j
(2.~B)
No n o d e c o v e r for H i c o n t a i n s
f e w e r than m(i) nodes, a n d
e v e r y node c o v e r for H i w h i c h
i n c l u d e s b o t h a node F i . and a
n o d e T. _ m u s t c o n t a i n ~ @ r e than
m(i) n~d~s.
Now, s u p p o s e we are g i v e n a t r u t h
a s s i g n m e n t to the n v a r i a b l e s w h i c h s a t i s fies the set C of clauses.
The c o r r e s p o n d i n g node c o v e r S c o n t a i n s the f o l l o w ing nodes:
(i] for e a c h v a r i a b l e x~ w h i c h is set
T, the c o v e r of m(1)
. n± o d e s for H i
"true
w h i c h i n c l u d e s all T i ~.
(ii) For e a c h v a r i a 6 ~ e x i w h i c h is set
"false," the c o v e r of m(i) n o d e s for H i
w h i c h i n c l u d e s all F.l,j"
Ti,m(i)
Fi,2
7. A S u b g r a p h
59
Hi
Fi, re(i)
Now for each i,j, 0 < i,j < 2, let c[i,j]
be the m i n i m u m cardi~ality--for all node
covers C of H obeying
(iii) For each clause C h, all the nodes
of Nh except some one of-them V- • such
that literal a h . is true for t~i{ truth
a s s i g n m e n t (at lSast one such literal
exists since the clause is satisfied).
Clearly, these nodes cover all of the
edges b e l o n g i n g to the sets A~, i < i < n,
and AI~, i < h < p.
E a c h edge~[Ti 7, VsTt ]
in B~ is a~so ~ o v e r e d since eitheg~V~ t
belongs to S or a
= x. is true an
T. . belongs to S~'tSimi~arly, each edge
i~'~ o is covered by S.
Thus, S is a node
coveg.
Furthermore, the number of nodes
in S is
I [V1,V{]
>.
j
(m(i)) = 2p + 3P = 5P,
i=l
a.s required.
Conversely, suppose we have a node cover
S for G such that ISt = 5P.
S must contain at least two nodes from each N h in
order to cover the edges in Ah, for a
total of at least 2p such nodes.
Similarly, by (2.4B) S must contain at
least ~
(m(i)) = 3P nodes from the N i.
i=l
•0
i
2
0
13
14
15
1
13
13
14
2
14
14
15
From the table, we observe
f o l l o w i n g p r o p e r t i e s hold:
that the
(2.5A) For 0 ~ j S 2, c[l,j]
< i and c[j~l]
(2.5B)
-
c[j,1]
=
(i) Draw G in the plane, a l l o w i n g
edges to cross each other.
(ii) Replace each c r o s s i n g by a copy of
H, a s - s h o w n in Fig. 9.
Proof.
The key structure used in this
p ~
is the graph H p i c t u r e d in Fig. 8,
w h i c h w i l l b e called a crossover, w i t h
outlets Vl,Vl, V2, and v2, as labelled.
~
The crossovers w h i c h replace crossings on
the edge [x,y] will be called crossovers
on the x-y line.
The edges c o n n e c t i n g
~Tf~ese crossovers to each other and to x
and y will be called edges on the x-y line.
The endpoints of these edges will be
called the nodes on the x-y line.
Such
nodes w h i c h are a T ~ o c r o s s o v e r outlets
w i l l be called the x-y outlets of their
crossover.
The one w h i c h is nearest x
w i l l be the crossover's y outlet, the one
nearest y its x outlet, f o ~
crosso v e r on the x-~ T T ~ .
~i
Va"
v;
8.
- c[l,j]
= 1
Given a graph G = (N,A) we construct a
p l a n a r graph G'
(NJ,A ') u s i n g these
crossovers as follows:
NODE COVER ~ PLANAR NODE
Figure
- c[0,j]
- e[jfO] < O.
For 0 ! J ! 2, c[2,j]
c[j,2]
Remark.
T h e o r e m 2.4 was o b t a i n e d
~ndently
by Peter H e r r m a n n [H73].
T h e o r e m 2.5:
(-'/ C I = j.
TABLE II. Values of c[i,j].
Hence S must contain e x a c t l y 2 nodes from
each N', and e x a c t l y ~ d e s
from each
N i.
T~us by ~ - A ~ - a n d
(2.4B), S must
contain, from each N_., either all the
T. • nodes and none Sf the F~ ~. nodes,
i,
±,d
or ~ii the F. . nodes and none of the
T i ~ nodes, i A 3 c o n s i s t e n t truth assignme~
can be o b t a i n e d by setting xl
"true" if S contains all the T.. ,.-nodes,
and setting x_. "false" otherwi~& d. The
reader may verify that, because of the
edges in B I and B2, this truth setting
satisfies %he set C of c l a u s e s . ~
UDVE~
and I[V2,V2]
Observe that, by symmetry, w h e n i or j
equals i, the value of c[i,j] is independent of w h i c h element of the c o r r e s p o n d i n g
pair is i ~ .
Table II gives the values
of c[i,j].
We leave to the reader the
s t r a i g h t f o r w a r d but tedious v e r i f i c a t i o n
of the entries.
n
2p +
~h C I = i
Crossover
60
H.
Jy
/i \
BEFORE
Figure
9.
(G)
Construction
AFTER
of G"
(only o u t l e t s
of c r o s s o v e r s
shown).
crossover.
T h u s S < 9 S' <9 S,, is a n o d e
c o v e r of G' h a v i n g k + 2d + l i d : k + 13d
nodes.
Let d be the n u m b e r o f c o p i e s of H u s e d
in c o n s t r u c t i n g G'.
O b s e r v e that the
e d g e s of G" c a n be p a r t i t i o n e d into two
sets:
line edges, t h o s e w h i c h a r e on
the x - y line for some [x,y]¢A, a n d c r o s s o v e r edges, t h o s e w h i c h are p a r t o f one
o~' the d c r o s s o v e r s .
A l l the e d g e s o n
the x - y l i n e c a n be c o v e r e d b y t a k i n g
e i t h e r x a n d a l l the x - o u t l e t s o f c r o s s o v e r s o n the line, or y a n d a l l the y
outlets.
The e d g e s in the c r o s s o v e r s
c a n o n l y be c o v e r e d b y c r o s s o v e r n o d e s .
C o n v e r s e l y , s u p p o s e t h e r e is a n o d e
of G
having k + 13d nodes.
Let
K * = min[[ S I : S is a n o d e
cover
M
of G" a n d
cover
of G'],
and
Now, s i n c e in G ~ e a c h e d g e - c r o s s i n g o f
the p l a n a r r e p r e s e n t a t i o n of G has b e e n
replaced by a planar graph which itself
c o n t a i n s n o c r o s s i n g s , G t is p l a n a r .
Moreover, the s i z e of G t is c l e a r l y at m o s t a
p o l y n o m i a l in the size o f G.
The p r o o f
of the T h e o r e m w i l l thus be c o n c l u d e d
b y s h o w i n g that, for a n y k, G has a n o d e
c o v e r of size k if a n d o n l y if G" has a
n o d e c o v e r of size k + 13d.
S u p p o s e t h e r e is a n o d e c o v e r S of G =
(N,A) w i t h lSl = k.
We construct a node
c o v e r of G" f r o m S as f o l l o w s .
For each
e d g e [x,y]eA, let f ( x , y ) be a n e n d p o i n t
of that e d g e w h i c h is in S.
Then define
S" =
(G')
[v:v is the f(x,y) o u t l e t for a
c r o s s o v e r on the x - y line for
s o m e [x,y]eA].
=
IS: S is a n o d e
cover
ISI = k*]
For each
~(S)
m*
S~M,
define
= I [xeS: x is a n o u t l e t
c r o s s o v e r in G']I ,
: min[m(S):
node
for
some
SEMI,
a n d let S * e M b e some n o d e c o v e r w i t h
m ( S * ) = m *.
S i n c e S* m u s t c o n t a i n 13
n o d e s f r o m e a c h o f the c r o s s o v e r s in
o r d e r for it to c o v e r a l l the c r o s s o v e r
e d g e s (see T a b l e II), w e k n o w that
I S * ( ~ N I < k.
we conclude our proof by
s h o w i n g t~at S = S*(-~ N is a n o d e c o v e r
of G.
S u p p o s e it is not.
Then there exists
some [ x , y ] e A s u c h t h a t S(-~[x,y] : ~ a n d
hence S*~[x,y]
= @.
Let the n u m b e r of
c r o s s o v e r s on the x - y line be £.
Then
t h e r e a r e £ + i e d g e s o n the x - y line,
a n d h e n c e at l e a s t £ + i o f the n o d e s on
the l i n e m u s t be in S*, a n d s i n c e n e i t h e r
x or y is, a l l £ + i m u s t be o u t l e t s .
If
w e let n(i) be the n u m b e r of c r o s s o v e r s
o n the x - y l i n e w i t h i of t h e i r x - y o u t lets i n S*, we thus h a v e n ( 2 ) - n ( O ) ~ i.
W e s h a l l s h o w that this l e a d s to a c o n t r a diction.
S i n c e S is a n o d e c o v e r for G, f is
d e f i n e d for a l l e d g e s in A, a n d so S<3S"
c o v e r s a l l the line e d g e s o f G".
Moreover, s i n c e e a c h c r o s s o v e r is o n two
l i n e s in G', S" c o n t a i n s e x a c t l y two
o u t l e t s for e a c h c r o s s o v e r , one f r o m e a c h
o u t l e t pair, a n d so I S'I = 2d.
All that
r e m a i n s is to c o v e r the as y e t u n c o v e r e d
crossover edges.
F o r these, o b s e r v e
f r o m T a b l e II that a n y set w h i c h c o n t a i n s two n o d e s f r o m a c r o s s o v e r , one
f r o m e a c h o u t l e t pair, c a n b e e x t e n d e d ,
b y a d d i n g ii of the c r o s s o v e r ' s i n t e r n a l
n o d e s , to f o r m a n o d e c o v e r of the c r o s s o v e r m a d e up of c [ l , l ] = 13 n o d e s .
Let
S" b e the set c o n t a i n i n g , f o r e a c h
c r o s s o v e r , a n d 11 a d d i t i o n a l n o d e s n e e d e d
to e x t e n d S" to a n o d e c o v e r f o r that
L e t X i be the set o f n o d e s in the i th
c r o s s o v e r o n the x - y line, 1 < i < £, a n d
let S i = X ~ ( ~ S*.
Let T i C
X ~ be--a n o d e
c o v e r of t~e c r o s s o v e r coh-%-~ihing its x
o u t l e t (but n o t the y o u t l e t ) a n d the same
n o n x - y o u t l e t s as d o e s Si, a n d has m i n i mum cardinality for such node eovers.
For
e a c h i, let r(i) be the n u m b e r of x - y
61
in polynomial time for "transitively
orientable" graphs, "chordal" graphs,
"circle" graphs.
outlets of the i th crossover w h i c h are in
S*.
Then we have, by (2.5A) and (2.5B)
r(i) = 0 implies
Til
r(i)
Ti I <__I Si I , and
= I implies
r(i) = 2 implies
<_I S i I + i,
Both types of results should prove useful
to designers of practical combinatorial
algorithms.
Ti I i l Sil - i.
APPENDIX:
k9 Ti, S = 6
S i.
Since
i=l
i=l
n(2)-n(O) > i we have by the above that
Definitions of Polynomial
Complete Problems
Let T =
I TI
<I sl
S A T I S F I A B I L I T Y W I T H AT MOST 3 LITERALS
CLAUSE .(SAT3) [K72]
- 1
INPUT:
Moreover, T contains at least one fewer
x-y outlet than does S, and exactly the
same number of non-x-y outlets.
Furthermore, T ~9[x] will cover all the line
edges that S did, and so T* = (S*-S)
~9 T ~-J[x] is a node cover of G w i t h
[ T~I
=I S*
-I Sl +I T I + 1 < I
= m(S*
PROPERTY:
There is a truth assignment to
the variables which simUltaneously satisfies all the clauses
in C (a clause is satisfied if
any one of its disjuncts is x i
for some "true" variable xi, or
Z. for some "false" xj).
S*I = k*,
- 1 = m* - 1,
c o n t r a d i c t i n g the definition of m*.
S* k_J N is a node cover of G and the
theo rem--fs p r o v e d . ~
Thus
III.
Concluding
CLIQUE
[K721
INPUT:
GRAPH G = (N,A), positive
PROPERTY:
Notice that, if the graph G given as
input for NODE COVER has no node degree
exceeding 3, the graph G" constructed in
the proof as input for PLANAR NODE COVER
will have no node degree exceeding 6.
This implies that PLANAR NODE COVER W I T H
NODE DEGREE AT MOST 6 is NP-complete.
It
is not known whether this degree bound is
best possible.
PER
Set of clauses C = fCI, C2, .... C-} ~n
variables
x~,x-,
,x-,
@ach cl[use
•
l .~ "':
n
belng the d l s o u n c t l o n of 3 or fewer
literals, where a literal is either
a variable x i or its negation xi"
and
m(T*)
and
integer k
G has a clique of size greater
than or equal to k, i.e., a set
N" C N with I N" J > k and such
that for all n l , n
,~_
N"
[nl,n2]¢A.
EXACT COVER [K72]
INPUT:
Collection of sets S = [SI,$2,... ,
Sn ]
PROPERTY: S has an exact cover, i.e., a
subcollection S" ~ S suc h that
U
Si : O
S~, and for all
S]eS"
i=l
Remarks
Si,
We have seen that a number of graphtheoretic NP-complete problems remain
NP-complete w h e n the structure of the
allowed inputs is substantially restricted.
Similar questions can be asked for
restricted versions of other NP-complete
problems.
For example, it is not yet
known w h e t h e r STEINER TREE [K72] for
planar graphs or m u l t i p r o c e s s o r scheduling
with 3 processors, unit time tasks, and
an arbitrary partial order [U73] are NPcomplete.
The open status of U N D I R E C T E D
H A M I L T O N I A N CIRCUIT for planar graphs has
been m e n t i o n e d previously.
SjeS',
GRAPH 3 - C O L O R A B I L I T Y
INPUT:
9.
[St73]
G has a legal 3-coloring of its
nodes i.e., there is a map
f: N ~ [1,2,3] such that
[nl,n2]cA implies f(nl) ~ f(n2).
U N D I R E C T E D (DIRECTED)
(CIRCUIT) [K72 ]
HAMILTONIAN
Graph G = (N,A).
0
PROPERTY:
In examining such problems, it is important to keep in mind that two types of
results are possible.
Not only is it
important to find simple subcases which
are still NP-complete, but it is also
important to find large subdomains for
w h i c h the p r o b l e m can be solved in polynomial time.
We have given one example
in this latter direction, the case of MAX
CUT for planar graphs.
Other recent
papers [E72,Gav72,Gav73] have shown ~hat
CLIQUE and CHROMATIC NUMBER can be solved
Sj :
Graph G = (N,A)
PROPERTY:
INPUT:
Sifh
=
PATH
(Directed graph
(N,A)).
G has a H a m i l t o n i a n path
(circuit), i.e., an ordering of
the nodes
= in ,n,, .... n I N I ]
such that ~or
~ i~< I NI ,
[ni,n~±1]eA ~ n ~ n ~ 7 > e A
), and
{in th@~circuit-ca~@-only)
Ln I N I 'nl]eA ~ n l N I 'nl)eA)"
NODE COVER [K72]
INPUT: Graph G =
PROPERTY: G has a
than-or
set N ' C
62
(N,A), positive integer k.
node cover of size less
equal to k, i.e., a subN w i t h IN'I< k and such
that for all (x,y]cA,
(x,y]gh N" ~
~.
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