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The Hamiltonian circuit problem for circle graphs is NP-complete
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THE
Letters
3 July 1989
32 (1989) l-2
HAMILTONIAN
CIRCUIT
PROBLEM
FOR CIRCLE GRAPHS
IS NP-COMPLETE
Peter DAMASCHKE Sektion Mathematik,
Friedrich-Schiller-Uniuersitiit
fena,
UniuersitZtshochhaus,
6900 Jew
German Dem. Rep.
Communicated by T. Lengauer Received 8 November 1988 Revised 6 March 1989
We show that the problem
Keywords:
Hamiltonian
of finding
circuit,
Hamiltonian
circle graphs,
circuits
in intersection
graphs
of chords
in a circle is NP-complete.
NP-completeness
1. Introduction
2. The NP-completeness
The intersection graph of a finite family of sets is defied as follows: There is a one-to-one correspondence between the vertices of the graph and the sets of the family so that two vertices are adjacent iff the corresponding sets have a nonempty intersection. Intersection graphs of chords in a circle are called circle graphs. The class of circle graphs exactly coincides with the class of interval overlap graphs [3]. A Hamiltonian circuit (HC) in a graph is a simple circuit including all vertices. The problem of finding an HC is NP-complete even when restricted to undirected path graphs [l], double interval graphs [4], chordal bipartite graphs, strongly chordal split graphs [2], and some other classes; cf. also the surveys in [2,5]. Now we establish a new NP-completeness result. This corrects a mistake in [5], where it has been asserted that the HC-problem is polynomialtime solvable for circle graphs. Let us remark that permutation graphs are a subclass of circle graphs [3], and that the complexity of our problem in this class is still an open and, apparently, very difficult question. For definitions not mentioned here we refer to
It is trivial that our problem belongs to NP. We reduce HC for bipartite graphs with maximum degree 3, which is known to be NP-complete [4], to our problem. (The degree of a vertex is the number of vertices adjacent to it.) Let G be such a bipartite graph and let X, Y be the colour classes of G, i.e. the partition of the vertex set of G into independent sets. First suppose that X and Y have the same cardinality, otherwise no HC can exist. Further, we can introduce additional edges in one of the colour classes, say Y. A HC cannot pass through these edges, since the vertices of X and Y must occur altematively in each HC. Now we describe the construction of a circle graph H. We arrange the chords representing the X-vertices in such a way that for each chord holds that one of the arcs marked by the endpoints of the chord contains no endpoint of another chord. These chords are called X-chords. Each Y-vertex of degree 2 is represented by a chord which intersects the chords of the two adjacent vertices and no other X-chord. Note that this is possible due to the arrangement of the X-chords. Let y be a Y-vertex of degree 3. xi, x1, x3 denote the vertices adjacent to y. We represent y
131. 0020-0190/89/$3.50
0 1989, Elsevier Science
Publishers
B.V. (North-Holland)
proof
1
Volume
32, Number
1
INFORMATION
PROCESSING
by a component consisting of three chords U, u, w. The chord u connects the chords of xi and x2 as described in the degree-2 case. Analogously, u connects the chords of x2 and x3. The so-called bridge w intersects only the chords u and u. We must be careful about bridges of distinct Y-vertices not intersecting further vertices. Note by geometrical perception that this can be really done. We have to show that G has an HC iff H has an HC. It suffices to show that each component of a Y-vertex y of degree 3 plays the same role as the vertex y in G. (For degree 2 it is clear, and intersections of non-X-chords do not perturb our reduction, due to the remarks above.) First note that in an arbitrary HC in H the vertices U, w, u must occur consecutively, with w in the middle. The following sequences of vertices in G and H are equivalent: . x,, y, x2 in G - x,, u, w, u, x2 in H, . x,, y, x3 in G - xi, u, w, u, xj in H, . x2, y, x3 in G - x2, u, w, u, x3 in H. This already completes the proof.
LETTERS
3 July 1989
A slight modification of our reduction yields that the Hamiltonian path problem is also NPcomplete.
References [l] A.A. Bertossi and M.A. Bonuccelli, Hamiltonian circuits in interval graph generalizations, Inform. Process. Left. 23 (1986) 195-200. [2] P. Damaschke and H. Miiller, Hamiltonian circuits in convex and chordal bipartite graphs. Discr. Ma&., to appear. (31 M.C. Golumbic, Algori’lhm Graph Theory and Perfect Graphs (Academic Press, New York, 1980). [4] A. Itai, C.H. Papadimitriou and J.L. Szwarcfiter, Hamiltoman paths in grid graphs. SIAM /. Comput. 11 (1982) 676-686. [S] D.S. Johnson, The NP-completeness column, an ongoing guide, J. Algorithms 6 (1985) 434-451.
THE
Letters
3 July 1989
32 (1989) l-2
HAMILTONIAN
CIRCUIT
PROBLEM
FOR CIRCLE GRAPHS
IS NP-COMPLETE
Peter DAMASCHKE Sektion Mathematik,
Friedrich-Schiller-Uniuersitiit
fena,
UniuersitZtshochhaus,
6900 Jew
German Dem. Rep.
Communicated by T. Lengauer Received 8 November 1988 Revised 6 March 1989
We show that the problem
Keywords:
Hamiltonian
of finding
circuit,
Hamiltonian
circle graphs,
circuits
in intersection
graphs
of chords
in a circle is NP-complete.
NP-completeness
1. Introduction
2. The NP-completeness
The intersection graph of a finite family of sets is defied as follows: There is a one-to-one correspondence between the vertices of the graph and the sets of the family so that two vertices are adjacent iff the corresponding sets have a nonempty intersection. Intersection graphs of chords in a circle are called circle graphs. The class of circle graphs exactly coincides with the class of interval overlap graphs [3]. A Hamiltonian circuit (HC) in a graph is a simple circuit including all vertices. The problem of finding an HC is NP-complete even when restricted to undirected path graphs [l], double interval graphs [4], chordal bipartite graphs, strongly chordal split graphs [2], and some other classes; cf. also the surveys in [2,5]. Now we establish a new NP-completeness result. This corrects a mistake in [5], where it has been asserted that the HC-problem is polynomialtime solvable for circle graphs. Let us remark that permutation graphs are a subclass of circle graphs [3], and that the complexity of our problem in this class is still an open and, apparently, very difficult question. For definitions not mentioned here we refer to
It is trivial that our problem belongs to NP. We reduce HC for bipartite graphs with maximum degree 3, which is known to be NP-complete [4], to our problem. (The degree of a vertex is the number of vertices adjacent to it.) Let G be such a bipartite graph and let X, Y be the colour classes of G, i.e. the partition of the vertex set of G into independent sets. First suppose that X and Y have the same cardinality, otherwise no HC can exist. Further, we can introduce additional edges in one of the colour classes, say Y. A HC cannot pass through these edges, since the vertices of X and Y must occur altematively in each HC. Now we describe the construction of a circle graph H. We arrange the chords representing the X-vertices in such a way that for each chord holds that one of the arcs marked by the endpoints of the chord contains no endpoint of another chord. These chords are called X-chords. Each Y-vertex of degree 2 is represented by a chord which intersects the chords of the two adjacent vertices and no other X-chord. Note that this is possible due to the arrangement of the X-chords. Let y be a Y-vertex of degree 3. xi, x1, x3 denote the vertices adjacent to y. We represent y
131. 0020-0190/89/$3.50
0 1989, Elsevier Science
Publishers
B.V. (North-Holland)
proof
1
Volume
32, Number
1
INFORMATION
PROCESSING
by a component consisting of three chords U, u, w. The chord u connects the chords of xi and x2 as described in the degree-2 case. Analogously, u connects the chords of x2 and x3. The so-called bridge w intersects only the chords u and u. We must be careful about bridges of distinct Y-vertices not intersecting further vertices. Note by geometrical perception that this can be really done. We have to show that G has an HC iff H has an HC. It suffices to show that each component of a Y-vertex y of degree 3 plays the same role as the vertex y in G. (For degree 2 it is clear, and intersections of non-X-chords do not perturb our reduction, due to the remarks above.) First note that in an arbitrary HC in H the vertices U, w, u must occur consecutively, with w in the middle. The following sequences of vertices in G and H are equivalent: . x,, y, x2 in G - x,, u, w, u, x2 in H, . x,, y, x3 in G - xi, u, w, u, xj in H, . x2, y, x3 in G - x2, u, w, u, x3 in H. This already completes the proof.
LETTERS
3 July 1989
A slight modification of our reduction yields that the Hamiltonian path problem is also NPcomplete.
References [l] A.A. Bertossi and M.A. Bonuccelli, Hamiltonian circuits in interval graph generalizations, Inform. Process. Left. 23 (1986) 195-200. [2] P. Damaschke and H. Miiller, Hamiltonian circuits in convex and chordal bipartite graphs. Discr. Ma&., to appear. (31 M.C. Golumbic, Algori’lhm Graph Theory and Perfect Graphs (Academic Press, New York, 1980). [4] A. Itai, C.H. Papadimitriou and J.L. Szwarcfiter, Hamiltoman paths in grid graphs. SIAM /. Comput. 11 (1982) 676-686. [S] D.S. Johnson, The NP-completeness column, an ongoing guide, J. Algorithms 6 (1985) 434-451.