AbstractWe show that the problem of deciding whether a given planar graph (complete with planar embedding) of degree at most 7 has a cubic subgraph is NP-complete

Stewart, Iain A.

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Deciding whether a planar graph has a cubic subgraph is NP-complete 

Discrete Mathematics 126 (1994) 349-357 
North-Holland 
349 
Deciding whether 
graph has a cubic 
is NP-complete 
a planar 
subgraph 
Iain A. Stewart 
Computing Laboratory, University of Newcastle upon Tyne. Claremont Tower, Claremont Road, 
Newcastle upon Tyne, NE1 7RU, UK 
Received 26 September 1991 
Revised 30 April 1992 
Abstract 
We show that the problem of deciding whether a given planar graph (complete with planar 
embedding) of degree at most 7 has a cubic subgraph is NP-complete. 
In Garey and Johnson’s compendium of NP-complete problems [l], many of the 
references are personal communications between themselves and others, and the 
proofs of some NP-completeness results have not appeared before in print. One such 
(to our knowledge) is the proof that CUBIC SUBGRAPH is NP-complete, attributed 
to Chvatal in Cl], where CUBIC SUBGRAPH is the problem of deciding whether 
a given graph has a nontrivial subgraph with each vertex of degree 3, i.e. a cubic 
subgraph: the hint given in [l] is to reduce GRAPH 3-COLOURABILITY to 
CUBIC SUBGRAPH. The nonexistence of such proofs in the literature has probably 
restricted further interest in these and related problems. 
In this paper we exhibit a logspace reduction from ONE-IN-THREE 3SAT 
to CUBIC SUBGRAPH and then show how, from this reduction, a logspace reduc- 
tion from ONE-IN-THREE 3SAT to CUBIC SUBGRAPH PLANAR can be 
constructed: ONE-IN-THREE 3SAT is the version of 3-SATISFIABILITY where 
exactly one literal in each clause of exactly 3 literals is satisfied and it was shown to be 
complete for NP via logspace reductions in [3], and CUBIC SUBGRAPH PLANAR 
is the version of CUBIC SUBGRAPH where all instances are planar graphs. In fact, 
we show that the problem obtained from CUBIC SUBGRAPH PLANAR where all 
Correspondence to: Iain A. Stewart, Department of Computer Science, University College of Swansea, 
Singleton Park, Swansea SA2 8PP, UK. 
0012-365X/94/$07.00 0 199GElsevier Science B.V. All rights reserved 
SSDI 0012-365X(92)00435-2 
350 I.A. Stewarl 
Fig. 1. A tag on a vertex u and its pictoral abbreviation. 
instances are of degree at most 7 and come with a planar embedding is complete for 
NP via logspace reductions. Our results provide more information on those proper- 
ties which are hard to verify computationally for both general and planar graphs. 
Our technique is similar to that used in [2], although the word ‘technique’ is 
probably a misnomer. What we do is to convert the logspace reduction from 
ONE-IN-THREE 3SAT to CUBIC SUBGRAPH into one from ONE-IN-THREE 
3SAT to CUBIC SUBGRAPH PLANAR by replacing any crossing edges with 
a suitable planar graph. As remarked in [2], it is often feasible to use this approach 
when specializing a given reduction to planar graphs, but finding appropriate graphs 
with which to replace crossing edges is usually hard. 
Before we proceed with the proof of Theorem 1, we need to introduce some 
nonstandard notation. We say that there is a tag on a vertex u of a graph G if there is 
a subgraph of G as in Fig. 1 such that no other edges of G involve the vertices of the 
tag (except for perhaps u): we abbreviate a tag pictorially as in Fig. 1. A vertex can 
have any number of tags on it in some graph. Notice that one edge of a tag occurs in 
a cubic subgraph of G if and only if all of the edges of the tag do. 
Theorem 1. There exists a logspace reduction from ONE-IN-THREE 3SAT to CUBIC 
SUBGRAPH. 
Proof. Let (X,C) be an instance of ONE-IN-THREE 3SAT, where 
X=(X1,X*, . ..) X,} is a set of Boolean variables and C = { Ci, CZ, . . . , C,} is a set of 
clauses, each containing exactly 3 literals. We build the corresponding instance G of 
CUBIC SUBGRAPH in stages. 
Stage A: For each iE{1,2, . . . . p>, we build the graph Gi as follows. 
(i) Gi has vertices: 
{xi~lXi~xi,j~lxi,j~ K::, 1 Yi, Yi,/c,l Yi,k, Ui,/c,l ui.1~3 K,k,l V,k: j=1,2,3; 
k=l,2 )...) q}, 
amongst others. 
(ii) Gi has edges: 
On cubic subgraphs 351 
S2 = {two tags on Xi, one tag on x}; 
~~~={(1Xit1Xi,l)r(1Xi,l~1Xi,Z)~(1Xi,2~1Xi,3)~(1Xi,3~1 yi)9 
(1 Xi,j, Xi): j= 1,2,3}; 
i Sz = {two tags on i Xi, one tag on 1 &}. 
(iii) If Xi (resp. 1 Xi) appears in the clauses Cil, CL27 . . . , Ci,, for some r > 1, then Gi has 
edges: 
S3={(Yi,ij, Yi,ij+,),(yi,ij, Vi,i,):j=l,2,...,r-2)U((Yi,i~-,, vi,i,),(yi3 yi,il)> 
U{(Ui,ij, Vi,ij): j= 1,2, ...Y r} u { one tag on Ui, ij: j = 1,2, . . . , r> 
(resp. 
1SJ = ((1 Yi,i,,l Yi,i,+,),(l Yi,i,,l Vi,,,): j= 1,2, ... ,r-2) 
U{(l Yi,i,_l,l vi,i,),(l yi,l yi,il)}u((lui,ij~l Vi,ij):j=1,2,...,r) 
u{one tag on 1 Ui,ij: j= 1,2, . . . ,r}). 
Stage B: The graph G,, is defined as follows. 
(i) Go has vertices: 
{Cj,Dj: j=l,2, . . ..q}. 
amongst others. 
(ii) Go has edges: 
{(Cj,Dj): j=l,2, . . . ,q}U{(Dj,Dj+l):j=1,2,...,4-1} 
u {one tag on D1, one tag on Dq (or two tags on D1 if q= l)}. 
Stage C: The graph G is the (disjoint) union of the graphs {Gi: i=O, 1,. . . , p}, 
together with the edges: 
Ti={(Ui,j,Cj),(Vi,j,Cj): j=l,2, ...,q; XieCj}, 
1 K={(l Ui,j,Cj),(l Vi,j,Cj): j= 192, ... 34; lXiECj>, 
for i=l,2 ,..., p. 
The subgraph of G consisting of Go, some Gi, for k{ 1,2, . . . ,p}, and the edges 
between Go and Gi is shown in Fig. 2: we assume in this particular instance that 
Xi (resp. 1 Xi) appears in the clauses Ci, C3 and C4 (resp. Ci, C2 and C,). For each 
i=l,2, . . . ,p, we call the edges of S1US2US3UTi (resp. ~S~U~S~U~S~U~ Ti) the 
edges belonging to Xi (resp. 1 Xi). 
Suppose that (X, C) is a yes-instance of ONE-IN-THREE 3SAT. So, there is a truth 
assignment t on the Boolean variables of X such that exactly one literal in each clause 
of C is set at True under t. Consider the graph G corresponding to (X, C), as defined 
above. In G, mark all edges of Go, and for each iE{ 1,2, . . . , p}, if Xi (resp. 1 Xi) is set at 
352 I.A. Stewart 
Ui,l vi,1 . . . ui,q vi,, 
Fig. 2. The subgraph induced by Go and Gi. 
True under t and appears in some clause of C then mark all those edges of G belonging 
to Xi (resp. 1 Xi). In the illustration of the subgraph of G consisting of Go, Gi, for some 
iE{1,2, . . . , p>, and the edges between Go and Gi, in Fig. 2, the edges belonging to Xi 
and the edges of Go are drawn in bold. Let E be the set of marked edges of G and let 
H be the subgraph of G with edges E. AS the edges belonging to Xi and 1 Xi, for some 
i, cannot be simultaneously marked, the only way that H could fail to be cubic is for 
some vertex Ci to have degree different from 3 in H. This is impossible as every clause 
of C has exactly one literal set at True under t. Hence, G has a cubic subgraph. 
Conversely, suppose that G has a cubic subgraph H, where G is the graph 
corresponding to some instance (X, C) of ONE-IN-THREE 3SAT. It is easy to see 
that all edges of Go must appear in H, and consequently that each vertex Ci must have 
degree 3 in H. For each Ci, k { 1,2, . . . , q}, let Li be the unique literal for which two of 
the edges belonging to Li involve Ci, and set L = {L,, L2,. . . , L4}. It should be clear 
that Xi and lXi, for some i, cannot both be in L (as, for instance, there is no cubic 
subgraph of Gi whose edges involve both the vertices Yi and 1 Yi). Define the truth 
assignment t by setting t (Li) = True, for each i = 1,2, . . . , q. This truth assignment is 
well-defined and is such that exactly one literal in each clause Ci is set at True under 
t (as each vertex Ci has degree 3 in H). Hence, (X, C) is a yes-instance. 
On cubic subgraphs 353 
u2 
e2 
Fig. 3. The graph K of Proposition 3. 
Given an instance (X, C) of ONE-IN-THREE 3SAT, the corresponding instance 
G of CUBIC SUBGRAPH can clearly be constructed in logspace and so the result 
follows. 0 
Schaefer’s result, mentioned earlier, now implies that of Chvatal, also mentioned 
earlier. 
Corollary 2. CUBIC SUBGRAPH is complete for NP via logspace reductions. 
The following proposition involves the graph we are to use to replace crossing edges 
in a nonplanar graph (see the initial preamble). 
Proposition 3. Consider the graph K in Fig. 3. If K is a subgraph of some graph G such 
that no other edges of G involve the vertices of K\{ul,u2,vl,~2} and H is a cubic 
subgraph of G, then the edge ei appears in H if and only ly the edge1; appears in H, for 
i= 1.2. 
Proof (sketch). In turn, consider the cases when e, and e2 both appear in H, when one 
of them appears in H, and when none of them appears in H. For each case, mark the 
rest of the edges of K remembering that each vertex of K\{u1,u2,vl,v2} must be of 
degree 3 or 0 in H: notice that either g1 and g2 are both in H or neither of them is in 
H. 0 
Theorem 4. There exists a logspace reduction from ONE-IN-THREE 3SAT to CUBIC 
SUBGRAPH PLANAR. 
354 I.A. Stewart 
Proof. Consider the logspace reduction from ONE-IN-THREE 3SAT to CUBIC 
SUBGRAPH as described in the proof of Theorem 1 (we work throughout with the 
terminology introduced in the proof of Theorem 1). It should be clear that the graph 
G corresponding to some instance (X, C) of ONE-IN-THREE 3SAT is planar except, 
possibly, for some crossing edges joining the subgraphs {Gi: i= 1,2, . . . ,p} to the 
subgraph Go (any Gi is planar even though it is not drawn as such in Fig. 2). 
Consequently, if we can ‘remove’ these crossings (by introducing more vertices and 
edges) whilst still retaining the pertinent properties of G (and do this in logspace), then 
we are done. 
Our strategy is as follows. Firstly, we show that an output graph G from the 
logspace reduction from ONE-IN-THREE 3SAT to CUBIC SUBGRAPH, men- 
tioned above, can be transformed, in logspace, to another graph G’ where the crossing 
edges of G are encoded as line segments on a grid, but where G and G’ are, in fact, 
identical. Secondly, we show that a graph G’ can be transformed, in logspace, to 
another graph G” (essentially by adding some extra vertices and edges) such that G 
has a cubic subgraph if and only if G” has. Finally, we show that the crossings of the 
graph G” (which are essentially the crossings of the graph G) can be removed by 
replacing each crossing with a fixed planar graph in the style of [2]. The grid 
embedding of the relevant subraph of G” (which is still available) enables us to 
systematically identify and remove each crossing whilst using only logspace. 
Consider the subgraph M of such a graph G, above, consisting of all edges with one 
end-vertex in { Ui,j, vi,j, 1 Ui,j, 1 vi,j: i = 1,2, . . . , p; j = 1,2, . . . , 4) and the other in { Cj: 
j=l,2 , . . . , q}. For ease of readability, we prefer to rename the vertices of 
{Ui,j, Vi,j,lUi,j,lK,j: i=l,2 ,..., p; j=1,2 ,..., 4) as fOllOWS: 
for i=l,2, . . . ,p and j=1,2,... ,4q = r, wi,j is defined as 
ui,j+ l/2 if j<2q and odd; vi, j/2 if j < 2q and even; 
l ui,j+ l- Zq/Z if j>2q and odd; l vi,j-2q/2 if j>2q and even. 
Also, for each Cj we introduce new vertices {zj,i: i= 1,2, . . . , r}. 
We rearrange the vertices of (Wi,k, Zj,k: i = 1,2, . . . , p; k = 1,2, . . . , r; j = 1,2, . . . , q) on 
the perimeter of a grid as shown in Fig. 4. If there is an edge (Wi,k, Cj) in G then we draw 
a horizontal line segment on the grid from Wi,k to the node of the grid vertically above 
zj,k, and then draw a vertical line segment down to zj,k. The grid, complete with line 
segments corresponding to the edges of the subgraph M of the graph G of Fig. 2, is 
shown in Fig. 4 (with the line segments drawn in bold and assuming that the variable 
i in Fig. 2 is actually equal to 1). By removing each vertex zj,k from the grid and 
extending any vertical line segment involving Zj,k to the vertex Cj, we clearly have 
another representation of the subgraph M of G. In particular, the nodes of the grid 
where line segments cross correspond to the crossing points of the edges of M in the 
original graph G. Denoting our new representation of G, where the edges of M are 
represented as line segments on a grid, by G’, it is easy to see how the graph G’ can be 
On cubic subgraphs 355 
l 
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ZqJ 
21,2 z1,2q+2 Z2,l z2,2q f4 Z3.6 Z3,2q f6 z4,a 
Fig. 4. The grid and some line segments 
obtained from G in logspace (G and G’ are actually identical, but more information is 
present in the encoding of G’ regarding the embedding of M on the grid). 
Having completed the first phase of our strategy, we now amend the graph G’ 
slightly. Consider the embedding of M on the grid in G’, as described above. For each 
crossing point of line segments on the grid, we place a tagged vertex to the left of (resp. 
to the right of, above, below) the crossing point on the horizontal (resp. horizontal, 
vertical, vertical) line segment involved in the crossing: these tagged vertices should 
appear before any other introduced vertices or crossing points. A portion of the 
amended grid of Fig. 4 is shown in Fig. 5 (in order to refer to the line segments on the 
grid, we pretend that the vertices of {Zj,k: j= 1,2, . . , q; k= 1,2, . . . , I} are still present 
even though in reality they are not). Denoting the amended graph by G”, it is easy to 
see that G” can be constructed from G’ in logspace (we retain the grid structure in the 
encoding of G”), and that G’ has a cubic subgraph if and only if G” does. This 
completes the second phase of our strategy. 
356 I.A. Stewart 
. . . 
. . . . . . . . . 
z1,2q+1 cy2q +3 
z1,2q+2 z2,2q f4 
z3,2q+5 24.7 
23,6 23,2q+6 24.8 
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w1,2q+1 
w1,2q+2 
w,2q+4 
w,2q+5 
‘-‘J1,2q+6 
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Fig. 5. The amended graph G” 
Finally, we must remove the crossings of G” (in logspace: remember that G” is 
planar except possibly for the grid crossings of the line segments). Notice that the 
addition of the vertices and edges to G’, to get G”, has increased the number of rows 
and columns of our grid: this is of no consequence. Consider a crossing point in the 
grid. Then according to the construction in the second phase, 4 new tagged vertices 
have been introduced: 1, r, a and b, where 1 (resp. r, a, b) is the vertex to the left of (resp. 
to the right of, above, below) the crossing point. Remove the edges (E, r) and (a, b) from 
the graph and replace them with the graph K of Proposition 3, with u1 = r, u2 = a, 
v1 = 1 and v2 = b. By Proposition 3, the amended graph has a cubic subgraph if and 
only if G” does, and one of the ‘obstructions to planarity’ has been removed. If we 
repeat this procedure for every crossing point then we obtain a planar graph G”’ 
which has a cubic subgraph if and only if G does. The removal of all crossing points 
can be achieved in logspace (as there are only polynomially many of them and we can 
identify them using the grid structure). Consequently, as logspace reductions are 
transitive, there is clearly a logspace reduction from ONE-IN-THREE 3SAT to 
CUBIC SUBGRAPH PLANAR. 0 
Again, Schaefer’s result yields the following corollary. 
Corollary 5. Cl 
tions. 
On cubic subgraphs 357 
JBGRAPH PLANAR is complete for NP via logspace reduc- JBIC S1 
In fact, by scrutinizing the graphs involved in the proofs of Theorems 1 and 4 and 
Proposition 3, it is easy to see that more can be said. 
Corollary 6. The version of CUBIC SUBGRAPH PLANAR where all instances have 
degree at most I and come with a planar embedding is complete for NP via logspace 
reductions. 
References 
[l] M.R. Garey and D.S. Johnson, Computers and Intractability: A Guide to the Theory of NP- 
completeness (Freeman, San Francisco, 1979). 
[Z] D. Lichtenstein, Planar formulae and their uses, SIAM J. Comput. (1982) 329-343. 
[3] T.J. Schaefer, The complexity of satisfiability problems, in: Proc. 10th ACM Conf. on Theory of 
Computing (ACM, New York, 1978) 216-226. 


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Deciding whether a planar graph has a cubic subgraph is NP-complete

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