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Recognizing d-interval graphs and d-track interval graphs

Minghui Jiang Utah State University FAW August 12, 2010

Geometric intersection graph The intersection graph Ω(F ) of a family of sets F = {S1 , . . . , Sn } is the graph with F as the vertex set and with two different vertices Si and Sj adjacent if and only if Si ∩ Sj 6= ∅. The family F is called a representation of the graph Ω(F ). Examples: interval graph, unit disk graph. . .

d-interval and d-track interval A d-interval is the union of d disjoint intervals on the real line. A d-track interval is the union of d disjoint intervals on d disjoint parallel lines called tracks, one interval on each track. A d-interval graph is the intersection graph of a family of dintervals. A d-track interval graph is the intersection graph of a family of d-track intervals. d-track interval graphs ⊂ d-interval graphs The d disjoint tracks for a d-track interval graph can be viewed as d disjoint “host” intervals on the real line for a d-interval graph. Thus the class of d-track interval graphs is contained in the class of d-interval graphs. Throughout, the intervals are all open.

Restrictions A d-interval is balanced if the d disjoint intervals have the same length, A d-interval is unit if the d disjoint intervals all have unit length. A d-interval is (x1 , . . . , xd ) if the d disjoint intervals have integer endpoints and have lengths x1 , . . . , xd , respectively, in sequential order. (x, . . . , x) ⊆ (x + 1, . . . , x + 1) ⊆ unit ⊆ balanced ⊆ unrestricted

For a balanced d-interval graph, the d disjoint intervals that constitute the same d-interval must have the same length, although the intervals from different d-intervals may have different lengths; on the other hand, all intervals of a unit d-interval graph must have the same unit length.

Representation depth The depth of a family of d-intervals is the maximum number of intervals that share a common point. The representation depth of a d-interval graph is the minimum depth of any d-interval representation of the graph. We use the term depth-two as an abbreviation for representation depth ≤ 2. Any multiple-interval representation of a triangle-free graph must have depth at most 2, because three intervals sharing a common point would induce a triangle in the intersection graph. triangle-free ⊂ depth-two

Applications As generalizations of the ubiquitous interval graphs, d-interval graphs and d-track interval graphs have wide applications, traditionally to scheduling and resource allocation, and more recently to bioinformatics. In bioinformatics, 2-interval graphs and 2-track interval graphs are natural models for • similar regions of DNA sequences, • helices of RNA secondary structures, • syntenic blocks in genomic maps.

Recognizing d-interval graphs West and Shmoys (1984): for any constant d ≥ 2, recognizing d-interval graphs is NP-complete (moreover, for any constants d ≥ 2 and r ≥ 3, recognizing d-interval graphs of representation depth at most r is also NP-complete). Gambette and Vialette (2007) observed that interval lengths in West and Shmoys’s construction for 2-interval graphs can be adjusted to meet the balanced restriction. Thus recognizing balanced 2-interval graphs is NP-hard too. Question 1 (Gambette and Vialette, 2007). What is the complexity of recognizing unit 2-interval graphs and (x, x) 2-interval graphs?

Recognizing d-track interval graphs Gy´arf´as and West (1995): recognizing 2-track interval graphs is NP-complete (moreover, for any constant r ≥ 3, recognizing 2-track interval graphs of representation depth at most r is NPcomplete). Conjecture 1 (Gy´arf´as and West, 1995). For any constant d ≥ 2, recognizing d-track interval graphs is NP-hard. It is easy to check that Gy´arf´as and West’s construction for 2track interval graphs can also be adapted to show that recognizing balanced 2-track interval graphs is NP-hard. But whether the proof can be adapted further to prove the NPhardness of recognizing unit and (x, x) 2-track interval graphs, or the NP-hardness of recognizing d-track interval graphs for d > 2, is not at all obvious.

Our main result The following theorem partially answers Question 1 and confirms Conjecture 1: Theorem 1. For any constant d ≥ 2, recognizing d-track interval graphs, balanced d-track interval graphs, unit d-track interval graphs, and (2, . . . , 2) d-track interval graphs are all NPcomplete. Moreover, for any constants d ≥ 2 and r ≥ 3, recognizing d-track interval graphs, balanced d-track interval graphs, unit dtrack interval graphs, and (2, . . . , 2) d-track interval graphs of representation depth at most r are all NP-complete.

Depth-two and triangle-free West and Shmoys (1984) also posed the following natural question after proving the NP-completeness of recognizing dinterval graphs of representation depth at most r for any constants d ≥ 2 and r ≥ 3: Question 2 (West and Shmoys, 1984). What is the complexity of recognizing d-interval graphs of representation depth at most 2? In particular, what is the complexity of recognizing d-interval graphs that are triangle-free? Gy´arf´as and West (1995) later posed the following question on recognizing 2-track interval graphs: Question 3 (Gy´arf´as and West, 1995). What is the complexity of recognizing 2-track interval graphs that are triangle-free?

Our following theorem complements Theorem 1 and partially answers Question 3: Theorem 2. Recognizing (i) depth-two 2-track interval graphs and (ii) depth-two unit 2-track interval graphs are both NP-complete. Our following theorem partially answers Question 2: Theorem 3. Let G be a graph of n vertices and m edges. (i) There is an O(poly(m + n)) time algorithm that determines, for any d ≥ 2, either that G is not a depth-two d-interval graph, or that G is a depth-two (d + 1)-interval graph. (ii) There is an O(2m poly(m + n)) time algorithm that determines, for any d ≥ 2, whether G is a depth-two d-interval graph. (iii) There is an O(m + n) time algorithm that determines the smallest number d such that G is a depth-two unit d-interval graph.

The O(poly(m + n)) time algorithm in Theorem 3(i) implies an approximation algorithm with additive error 1 for finding the smallest number d such that G is a depth-two d-interval graph. This approximation would be best possible if it turned out that recognizing depth-two d-interval graphs is NP-hard for any constant d ≥ 2. The O(2m poly(m + n)) time algorithm in Theorem 3(ii) improves a previous O(3m (m + n)) time algorithm by Maas (1983) for recognizing depth-two d-interval graphs. Also note the sharp contrast between Theorem 2(ii) and Theorem 3(iii): it is interesting that while recognizing unrestricted 2interval graphs and 2-track interval graphs are both NP-complete, recognizing their restricted (depth-two, unit) variants are so drastically different in complexity.

Summary of complexities

unrestricted balanced unit (2, . . . , 2) depth-two depth-two, unit

d-interval NP-C [WS84] NP-C (d = 2) [GV07] ? [GV07] ? [GV07] ? T3(i-ii) P T3(iii)

d-track interval NP-C T1 NP-C T1 NP-C T1 NP-C T1 NP-C (d = 2) T2 NP-C (d = 2) T2

Interval number and track number For a graph G, the interval number i(G) is the smallest number d such that G is a d-interval graph, and the track number t(G) is the smallest number d such that G is a d-track interval graph. A graph is a d-interval graph (resp. d-track interval graph) if and only if its interval number (resp. track number) is at most d. Similarly define the unit interval number iu (G) as the smallest number d such that G is a unit d-interval graph, and the unit track number tu (G) as the smallest number d such that G is a unit d-track interval graph. i(G) ≤ iu (G)

t(G) ≤ tu (G)

i(G) ≤ t(G)

iu (G) ≤ tu (G)

Caterpillar and linear arboricity A caterpillar is a tree containing a dominating path such that every vertex not on the path is adjacent to some vertex on the path. A caterpillar forest is a graph in which every connected component is a caterpillar. The caterpillar arboricity ca(G) of a graph G is the minimum number of caterpillar forests into which its edges can be decomposed. A linear forest is a graph in which every connected component is a path. The linear arboricity la(G) of a graph G is the minimum number of linear forests into which its edges can be decomposed. ca(G) ≤ la(G)

Bounds Let G be a graph of n vertices, m edges, and maximum degree ∆. It is known that i(G) ≤ ⌈(n + 1)/4⌉, iu (G) ≤ ⌈(n − 1)/2⌉, √ i(G) ≤ ⌈ m/2⌉ + 1, and i(G) ≤ iu (G) ≤ ⌈(∆ + 1)/2⌉. Akiyama, Exoo, and Harary (1980) conjectured that every graph G of maximum degree ∆ satisfies la(G) ≤ ⌈(∆ + 1)/2⌉. Alon (1988) showed that the conjecture is correct asymptotically as ∆ → ∞. Theorem 4. (i) t(G) ≤ ⌊n/2⌋; (ii) tu (G) ≤ ⌈m/2⌉, and this bound is best possible; (iii) t(G) ≤ ca(G) and tu (G) ≤ la(G). In particular, if la(G) ≤ ⌈(∆+1)/2⌉ as conjectured, then t(G) ≤ √ tu (G) ≤ ⌈(∆ + 1)/2⌉ ≤ ⌈n/2⌉ and t(G) ≤ ⌈ m ⌉.

NP-hardness We show that recognizing d-track interval graphs (RDT) is NP-hard by a reduction from the following NP-hard problem: Hamiltonian path in triangle-free cubic graph (HP3): Given a triangle-free cubic graph G = (V, E) and an edge uv ∈ E, decide whether there is a Hamiltonian path in G starting at u and ending at v. In the following, we first study the simple case of d = 2 and prove that recognizing 2-track interval graphs (R2T) is NP-hard, then turn to the general case and prove that recognizing d-track interval graphs for any constant d ≥ 2 (RDT) is NP-hard. Our proof, although presented in a progressive way, will reduce HP3 to both R2T and RDT directly.

Edge realization In any multiple-interval representation of a graph, each edge uv must be represented by two overlapping intervals, one interval of u and one interval of v. In a d-track interval representation, each edge is realized on at least one of the d tracks, that is, each edge uv is represented by two overlapping intervals, one interval of u and one interval of v, on at least one of the d tracks. If the intervals of u and v overlap on exactly one of the d tracks, then the edge uv is uniquely realized on that track.

Continuous representation If the union of a group of intervals is a single interval (with no holes), then we say that these intervals are continuous. Lemma 1. In any multiple-interval representation of a graph, if the depth of the representation is at most 2, then at most t − 1 edges can be represented by t intervals. Moreover, if exactly t − 1 edges are represented by t intervals, then these t intervals must be continuous. When we read the representation from left to right we obtain at most one new edge at the left endpoint of each interval. If exactly t − 1 edges are represented by t intervals, then these t intervals must be continuous because the left endpoint of every interval except the left-most one is covered by the preceding interval.

K4,3 A d-track representation of a graph is continuous if the union of the intervals on each track is a single interval, is continuous on some track if the union of the intervals on that particular track is a single interval. Track 1

Track 2

a2

a3

a4

a4

b3

b2

b1 a3

b3

b2

b1 a1

a1

a2

Bipartite =⇒ triangle-free =⇒ depth-two. 12 edges in total; 6 edges represented by 7 intervals on each track.

K2d,2d−1 Track 1

b2

b1 a1

b2d−1 a2d−1

a2

a2d

.. .

Track k

b2

b1 a2k−1

a2k

b2d−1 a2k−3

a2k−2

.. .

Track d

b2

b1 a2d−1

a2d

b2d−1 a2d−3

a2d−2

Lemma 2. K2d,2d−1 has a continuous (2, . . . , 2) d-track interval representation. Every d-track interval representation of K2d,2d−1 must be continuous on each track.

HP3 ≤ R2T Given a triangle-free cubic graph G = (V, E) with the vertex set V = {v1 , . . . , vn } and an edge v1 vn ∈ E, we will construct an extended graph G2 such that G has a Hamiltonian path between v1 and vn if and only if G2 has a 2-track interval representation. G has exactly 3n/2 edges, so the number n of vertices must be even.

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

¨ is the graph G with the edge v1 vn deleted. G Q = {q1 , . . . , qn } is an independent set of n vertices. P = hp0 , p1 , . . . , p2n+2 i is a path of 2n + 3 vertices.

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

K4,3 is a complete bipartite graph. X4 is a gadget consisting of a vertex x and four copies of K4,3 , where x is connected to a vertex of degree 3 in each copy of K4,3 .

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

The extended graph G2 is composed by connecting the graph ¨ the independent set Q, the path P , n copies of the complete G, bipartite graph K4,3 , and 2n + 3 copies of the gadget X4 .

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

¨ to Q by n edges vi qi , 1 ≤ i ≤ n. Connect G

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

Connect Q to P by 3n edges qi p2i−1 , qi p2i , and qi p2i+1 , 1 ≤ i ≤ n.

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

Connect each vertex qi in Q to a vertex of degree 3 in a distinct copy of K4,3 .

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

Connect each vertex pj in P to the vertex x and two neighbors of x in a distinct copy of X4 .

X4 (y) Denote by X4 (y) the graph consisting of a vertex y and a copy of the gadget X4 , where y is connected to the vertex x and two neighbors of x in X4 . Observe that each vertex pj and the corresponding copy of the gadget X4 induce a subgraph X4 (pj ) in G2 . Lemma 3. The graph X4 (y) has a (2, 2) 2-track interval representation. Moreover, in any 2-track interval representation of X4 (y), there is a track on which the interval of y is completely covered by the intervals of X4 . The implication: each vertex pj has at most one free interval on one of the d tracks that is available to overlap with the intervals of other vertices of P and the vertices of Q.

κ1 Track 1

κ2

κ3

κ2

κ3

κ4

y

x

κ1

κ4

y

Track 2

x

Each of the four copies κ1 , κ2 , κ3 , and κ4 of K4,3 in X4 , shown schematically as an oval, is represented continuously: Track 1

Track 2

b1 a1

b2

b1 a3

b3 a3

a2

b2 a4

a4

b3 a1

a2

=⇒ G has a Hamiltonian path between v1 and vn if and only if G2 has a 2-track interval representation. We first prove the “only if” direction. Suppose that the input graph G has a Hamiltonian path hv1 , . . . , vn i. We will construct a 2-track interval representation of the extended graph G2 .

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

Realize the Hamiltonian path hv1 , . . . , vn i by consecutively overlapping intervals on track 1. ¨ that Note that the edges of G are not on the Hamiltonian path form a perfect matching.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

¨ Realize each edge vi vj of G that is not on the Hamiltonian path, together with the two edges vi qi and vj qj , by disjoint groups of four intervals on track 2.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

Also realize the edge between each vertex qi in Q and the corresponding copy of K4,3 on track 2, such that the free interval of qi is on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

Realize the subgraph X4 (pj ) for each vertex pj in P following the pattern for X4 (y), such that the free interval of pj is also on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

Realize the path P = hp0 , p1 , . . . , p2n+2 i by consecutively overlapping intervals on track 1. Then realize the edges between Q and P by choosing the same interval for qi and p2i on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

This representation of G2 can be easily adapted to a (2, 2) 2track interval representation.

⇐= G has a Hamiltonian path between v1 and vn if and only if G2 has a 2-track interval representation. We next prove the “if” direction. Suppose that the extended graph G2 has a 2-track interval representation. We will find a Hamiltonian path between v1 and vn in the input graph G.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

Each vertex pj has at most one free interval. Thus all edges of the path hp0 , p1 , . . . , p2n+2 i must be uniquely realized on the same track, say track 1, by consecutively overlapping intervals.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

Then all edges between Q and P must be uniquely realized on track 1 too.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

The interval of each vertex qi is completely covered by the three consecutively overlapping intervals of p2i−1 , p2i , and p2i+1 on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

Then the other two edges incident to qi (the edge vi qi and the edge between qi and the corresponding copy of K4,3 ) must be uniquely realized on track 2.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

The representation of K4,3 is continuous on each track. Since qi is adjacent to only one vertex of K4,3 , at least one endpoint of the interval of qi must be covered by the union of the intervals of K4,3 on track 2.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

Similarly, since vi is adjacent to qi but not to K4,3 , at least one endpoint of the interval of vi must be covered by the interval of qi on track 2.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

For each edge vi vj , if the two intervals of vi and vj overlap on track 2, then they must overlap at their endpoints that are not covered by the intervals of qi and qj .

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

Since no three vertices of a triangle-free graph can have three overlapping intervals sharing a common point, neither the interval of vi nor the interval vj can overlap with the interval of a third vertex vk on track 2.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

Thus at most n/2 edges of G can be realized on track 2. X4

vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

On the other hand, by Lemma 1, at most n − 1 edges of G can be realized on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

The cubic graph G has 3n/2 ¨ is obtained from G edges. G by deleting one edge v1 vn , so it has 3n/2−1 = n/2+(n−1) edges, with two edges incident to each of v1 and vn , and three edges incident to each other vertex vi .

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

qi

p2i x

vn

qn

qj

v1

···

P

X4 vi

p2n+2

vj vi

Q

vn

p2n+1

p0

p2n p2n+1 p2n+2

¨ has exactly n/2 edges G uniquely realized on track 2, and has exactly n − 1 edges uniquely realized on track 1.

p1 Track 1

Track 2

¨ G

···

v1

p2

p2n

q1

qn

qi

···

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

p2n+2

vj vi

Q

vn

p2n+1

p0

qj

v1

···

The n/2 edges on track 2 are represented by n/2 pairs of intervals, and form a perfect matching. The n − 1 edges on track 1 are represented by n continuous intervals, and form a Hamiltonian path between v1 and vn .

R2T → RDT Recognizing 2-track interval graphs is NP-hard. How about d-track interval graphs? Track 1

b2

b1 a1

b2d−1 a2d−1

a2

a2d

.. .

Track 1

b3

b2

b1

a4

a3

a2

a1

Track k

b2

b1 a2k−1

b2d−1 a2k−3

a2k

a2k−2

.. .

Track 2

b3

b2

b1 a3

a4

a1

a2

Track d

b2

b1 a2d−1

K4,3 → K2d,2d−1

a2d

b2d−1 a2d−3

a2d−2

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

¨ is the graph G with the edge v1 vn deleted. G Q = {q1 , . . . , qn } is an independent set of n vertices. P = hp0 , p1 , . . . , p2n+2 i is a path of 2n + 3 vertices.

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

K2d,2d−1 is a complete bipartite graph. X2d is a gadget consisting of a vertex x and 2d copies of K2d,2d−1 , where x is connected to a vertex of degree 2d − 1 in each copy of K2d,2d−1.

¨ G

Q v1

q1

p0 p1 p2

P

X4 vi

qi

p2i x

vn

qn

p2n p2n+1 p2n+2

¨ the indeThe extended graph Gd is composed by connecting the graph G, pendent set Q, d−1 copies of the path P , n copies of the complete bipartite graph K2d,2d−1 , and (d − 1)2 (2n + 3) + (d − 2)n copies of the gadget X2d .

k X4 (y) → X2d (y) k Denote by X2d (y) the graph consisting of a vertex y and k copies of the gadget X2d , where y is connected to the vertex x and two neighbors of x in each copy of X2d . k Lemma 4. Let 1 ≤ k ≤ d. The graph X2d (y) has a (2, . . . , 2) d-track interval representation. Moreover, in any d-track interval k representation of X2d (y), there are k tracks such that on each of the k tracks the interval of y is completely covered by the intervals of X2d . d−2 • X2d (vi ): each vi has at most 2 free intervals. d−1 • X2d (pj ): each copy of pj has at most 1 free interval.

p2n+1

p1 Track 1

···

Track 2

···

Track 1

···

p0

p2

p2n

q1

qn

vi

p1

.. .

···

.. .

Track d

···

p0

···

qj

p2n+1 p2

p2n

q1

qn

p2

p2n

q1

qn

p1 Track k

v1

vj

qi

p0

vn

p2n+2

vn

p2n+2

v1

p2n+1 p2n+2

vj

qi vi

qj

···

Restricted variants Recognizing d-track interval graphs is NP-hard. How about those restricted variants? • For the “only if” direction, the d-track interval representation in our construction has depth exactly 3, and can be easily adapted to a (2, . . . , 2) d-track interval representation. • For the “if” direction, our arguments do not depend on any special assumptions on the depth or the interval lengths of the given d-track representation.

All your base are belong to us! Thus our proof implies the stronger result that, for any constant d ≥ 2, recognizing balanced, unit, and even (2, . . . , 2) dtrack interval graphs are all NP-hard, and moreover, for any constant r ≥ 3, recognizing such graphs of representation depth at most r are also NP-hard.

Open questions

unrestricted balanced unit (2, . . . , 2) depth-two depth-two, unit

d-interval NP-C [WS84] NP-C (d = 2) [GV07] ? [GV07] ? [GV07] ? T3(i-ii) P T3(iii)

d-track interval NP-C T1 NP-C T1 NP-C T1 NP-C T1 NP-C (d = 2) T2 NP-C (d = 2) T2