Paths in interval graphs

Transcription

1 1 / 47 Paths in interval graphs Peng Li (o+), Yaokun Wu (Ljn) Shanghai Jiao Tong University Halifax, June 19, 2012

2 Outline 2 / 47 Background A linear time algorithm to solve 1PC Some equivalent connectivity properties

3 Interval graph 3 / 47 An interval representation I of a graph G sends each v V(G) to an nonempty interval [l I (v), r I (v)] such that vw E(G) if and only if v w and I(v) I(w). A graph possessing an interval representation is an interval graph.

4 Interval graph 3 / 47 An interval representation I of a graph G sends each v V(G) to an nonempty interval [l I (v), r I (v)] such that vw E(G) if and only if v w and I(v) I(w). A graph possessing an interval representation is an interval graph.

5 Interval graph 4 / 47 Figure:

6 Interval graph 5 / 47 There are several linear time algorithms to generate a right-endpoint ordering, a left-endpoint ordering and an interval representation of an interval graph [6, 8, 21]. Interval graphs look to be very simple. It is thus reasonable to expect easy algorithms on them for various kinds of combinatorial problems.

7 Interval graph 5 / 47 There are several linear time algorithms to generate a right-endpoint ordering, a left-endpoint ordering and an interval representation of an interval graph [6, 8, 21]. Interval graphs look to be very simple. It is thus reasonable to expect easy algorithms on them for various kinds of combinatorial problems.

8 Interval graph 5 / 47 There are several linear time algorithms to generate a right-endpoint ordering, a left-endpoint ordering and an interval representation of an interval graph [6, 8, 21]. Interval graphs look to be very simple. It is thus reasonable to expect easy algorithms on them for various kinds of combinatorial problems.

9 6 / 47 Path problems 1-HP: Given a fixed vertex, find a Hamiltonian path with that vertex as an endpoint. 2-HP: Given two vertices, find a Hamiltonian path with those two vertices as endpoints. k-pc: Given a set of k vertices, find a minimum number of vertex disjoint paths that cover all vertices of the graph and contain those given k vertices as endpoints.

10 6 / 47 Path problems 1-HP: Given a fixed vertex, find a Hamiltonian path with that vertex as an endpoint. 2-HP: Given two vertices, find a Hamiltonian path with those two vertices as endpoints. k-pc: Given a set of k vertices, find a minimum number of vertex disjoint paths that cover all vertices of the graph and contain those given k vertices as endpoints.

11 6 / 47 Path problems 1-HP: Given a fixed vertex, find a Hamiltonian path with that vertex as an endpoint. 2-HP: Given two vertices, find a Hamiltonian path with those two vertices as endpoints. k-pc: Given a set of k vertices, find a minimum number of vertex disjoint paths that cover all vertices of the graph and contain those given k vertices as endpoints.

12 Is 1HP polynomial time solvable? 7 / 47 Till now there are no "natural" graph problems known to be NP-complete for interval graphs, except ACHROMATIC NUMBER [5]. May 1HP and 2HP be further candidates? Only by very superficial consideration this seems to be unlikely. Peter Damaschke (1993) [9]

13 1PC/1HP is polynomial 8 / 47 Theorem 1 (Asdre-Nikolopoulos (2010) [3]) There is an O(n 3 )-time algorithm to solve the 1PC/1HP problem on interval graphs. 1 1 Indeed, their algorithm runs in O(nm)-time.

14 Outline 9 / 47 Background A linear time algorithm to solve 1PC Some equivalent connectivity properties

15 Main result 10 / 47 Theorem 2 (Li-W. [16]) There is an O(n + m)-time algorithm to solve the 1PC/1HP problem on interval graphs.

16 11 / 47 The design and the analysis of our algorithm rely heavily on a careful analysis of a greedy algorithm, the so-called Straight-Path Algorithm (Arikati-Rangan 1990 [1]). The basic idea of the Straight-Path Algorithm also appears in Keil (1985) [15], Manacher-Mankus-Smith (1990) [18] and Damaschke (1993) [9]. Note that the solution of the longest path problem for interval graphs by Ioannidou, Mertzios and Nikolopoulos (2011) [14] also makes essential use of similar ideas.

17 11 / 47 The design and the analysis of our algorithm rely heavily on a careful analysis of a greedy algorithm, the so-called Straight-Path Algorithm (Arikati-Rangan 1990 [1]). The basic idea of the Straight-Path Algorithm also appears in Keil (1985) [15], Manacher-Mankus-Smith (1990) [18] and Damaschke (1993) [9]. Note that the solution of the longest path problem for interval graphs by Ioannidou, Mertzios and Nikolopoulos (2011) [14] also makes essential use of similar ideas.

18 Straight-Path Algorithm 12 / 47 Procedure SP(G, I, a); { Input: an interval graph G, an interval representation I of G and a vertex a of V(G) which belongs to the leftmost cliques of G with respect to I; Output: an ordering π of G; } begin Let π 1 = a. For i = 1 to V(G) 1 do If there is some unvisited vertex of N G (π i ), then visit the unvisited vertex of N G (π i ) with minimum r I value, say x, that means π i+1 = x; Else if there is some unvisited vertex of V(G), then visit the unvisited vertex with minimum r I value; Else, exit. end;

19 13 / 47 An ordering of the vertex set of an interval graph generated by the Straight-Path Algorithm is called an almost straight ordering. If we further require r I (a) = min{r I (v) : v V(G)} in the procedure SP(G, I, a), then the output ordering of SP(G, I, a) is called a straight ordering of G.

20 13 / 47 An ordering of the vertex set of an interval graph generated by the Straight-Path Algorithm is called an almost straight ordering. If we further require r I (a) = min{r I (v) : v V(G)} in the procedure SP(G, I, a), then the output ordering of SP(G, I, a) is called a straight ordering of G.

21 14 / 47 Figure: A straight ordering If you start from interval 2, you will get an almost straight ordering.

22 Lemma 3 (Arikati-Rangan [1], Manacher-Mankus-Smith [18]) If an interval graph G has a Hamiltonian path, then each of its straight orderings gives a Hamiltonian path. 15 / 47

23 Let I be an interval representation of a graph G. Assume that r I (a) = min{r I (v) : v V(G)} and l I (b) = max{l I (v) : v V(G)}. We say that a is the leftmost interval and b is the rightmost interval in the given interval representation I. Lemma 4 (Li-W.) The interval graph G has a Hamiltonian path if and only if it has a Hamiltonian path connecting a and b. Proof. It follows easily from Lemma 3. Theorem 5 (Li-W.) For any v V(G), G has a Hamiltonian path starting from v if and only if it has a Hamiltonian path starting from v and ending at either a or b. Proof. A direct consequence of Lemma / 47

24 17 / 47 If N G (π i ) {π 1,..., π i 1 }, we say that i is a breaking time for π. A slice of π is a maximal set of the form π[i, j] such that j is the only breaking time for π among [i, j].

25 Reduce 1PC to 1HP for interval graphs 18 / 47 Let x be the fixed vertex of the interval graph G. Let S 1,..., S k be the k slices in that order for a straight ordering π of G. Assume x S i. Let v be the rightmost vertex of S i+1 with respect to the original interval representation. Do straight ordering in S i+1 from right to left starting from v and let the last visited vertex be z. Take W = i j=1 S j \ N G (z) and consider a straight path τ of G[W]. Suppose S is the slice of τ containing x. Then the answer to 1PC for G and x is k if there is an HP in G[S] with x as an endpoint and is k + 1 else. We can explicitly construct the minimum path cover after knowing the required HP in G[S] or knowing its nonexistence. We can solve 1PC on interval graphs in linear time if and only if we can solve 1HP on interval graphs in linear time.

26 Reduce 1PC to 1HP for interval graphs 18 / 47 Let x be the fixed vertex of the interval graph G. Let S 1,..., S k be the k slices in that order for a straight ordering π of G. Assume x S i. Let v be the rightmost vertex of S i+1 with respect to the original interval representation. Do straight ordering in S i+1 from right to left starting from v and let the last visited vertex be z. Take W = i j=1 S j \ N G (z) and consider a straight path τ of G[W]. Suppose S is the slice of τ containing x. Then the answer to 1PC for G and x is k if there is an HP in G[S] with x as an endpoint and is k + 1 else. We can explicitly construct the minimum path cover after knowing the required HP in G[S] or knowing its nonexistence. We can solve 1PC on interval graphs in linear time if and only if we can solve 1HP on interval graphs in linear time.

27 Reduce 1PC to 1HP for interval graphs 18 / 47 Let x be the fixed vertex of the interval graph G. Let S 1,..., S k be the k slices in that order for a straight ordering π of G. Assume x S i. Let v be the rightmost vertex of S i+1 with respect to the original interval representation. Do straight ordering in S i+1 from right to left starting from v and let the last visited vertex be z. Take W = i j=1 S j \ N G (z) and consider a straight path τ of G[W]. Suppose S is the slice of τ containing x. Then the answer to 1PC for G and x is k if there is an HP in G[S] with x as an endpoint and is k + 1 else. We can explicitly construct the minimum path cover after knowing the required HP in G[S] or knowing its nonexistence. We can solve 1PC on interval graphs in linear time if and only if we can solve 1HP on interval graphs in linear time.

28 Reduce 1PC to 1HP for interval graphs 18 / 47 Let x be the fixed vertex of the interval graph G. Let S 1,..., S k be the k slices in that order for a straight ordering π of G. Assume x S i. Let v be the rightmost vertex of S i+1 with respect to the original interval representation. Do straight ordering in S i+1 from right to left starting from v and let the last visited vertex be z. Take W = i j=1 S j \ N G (z) and consider a straight path τ of G[W]. Suppose S is the slice of τ containing x. Then the answer to 1PC for G and x is k if there is an HP in G[S] with x as an endpoint and is k + 1 else. We can explicitly construct the minimum path cover after knowing the required HP in G[S] or knowing its nonexistence. We can solve 1PC on interval graphs in linear time if and only if we can solve 1HP on interval graphs in linear time.

29 Reduce 1PC to 1HP for interval graphs 18 / 47 Let x be the fixed vertex of the interval graph G. Let S 1,..., S k be the k slices in that order for a straight ordering π of G. Assume x S i. Let v be the rightmost vertex of S i+1 with respect to the original interval representation. Do straight ordering in S i+1 from right to left starting from v and let the last visited vertex be z. Take W = i j=1 S j \ N G (z) and consider a straight path τ of G[W]. Suppose S is the slice of τ containing x. Then the answer to 1PC for G and x is k if there is an HP in G[S] with x as an endpoint and is k + 1 else. We can explicitly construct the minimum path cover after knowing the required HP in G[S] or knowing its nonexistence. We can solve 1PC on interval graphs in linear time if and only if we can solve 1HP on interval graphs in linear time.

30 Perfect ordering 19 / 47 Another key concept in our work is perfect ordering. Let π be an ordering of the vertex set of an n-vertex graph G. The right-neighbor sequence of π, denoted by π, is a sequence from Z n 0 such that π v is the number of neighbors of π i among π i+1, π i+2,..., π n. We say that π is k-perfect provided π (i) min(k, n i) for any i = 1,..., n.

31 Perfect ordering 19 / 47 Another key concept in our work is perfect ordering. Let π be an ordering of the vertex set of an n-vertex graph G. The right-neighbor sequence of π, denoted by π, is a sequence from Z n 0 such that π v is the number of neighbors of π i among π i+1, π i+2,..., π n. We say that π is k-perfect provided π (i) min(k, n i) for any i = 1,..., n.

32 Perfect ordering 19 / 47 Another key concept in our work is perfect ordering. Let π be an ordering of the vertex set of an n-vertex graph G. The right-neighbor sequence of π, denoted by π, is a sequence from Z n 0 such that π v is the number of neighbors of π i among π i+1, π i+2,..., π n. We say that π is k-perfect provided π (i) min(k, n i) for any i = 1,..., n.

33 2-perfectness implies the existence of Hamiltonian cycles 20 / 47 Procedure HC(G, π); { Input: an interval graph G and a 2-perfect almost straight path π of G; Output: an HC ρ of G such that π 1 and π 2 are consecutive on ρ and π n 1 and π n are consecutive on ρ; } begin If V(G) 3, Let ρ = π 1,..., π n. Else if V(G) > 3. Take j > 2 such that π 1 π j E(G). Do Procedure HC(G[π [j 1, n] ], π [j 1, n] ) to get an HC of G[π [j 1, n] ] say ρ = π j 1, π j, x 1,..., x h. Let ρ = π 1, π 2,..., π j 1, x h, x h 1,..., x 1, π j. end;

34 21 / 47 Solving 1HP { Input: a graph G, an interval representation I of G, x V(G); Output: an HP of G which starts at x or "there is no HP of G which starts at x"; } Let a = arg min v V(G) r I (v) and b = arg max v V(G) l I (v). Run SP(G, I, a) and SP(G, I, b) to get two straight orderings π and π. Compute the right neighbor sequences π and π. Suppose x = π p = π p. If π is not 1-perfect, then return "there is no required HP". If π is 2-perfect, do Procedure HC(G, π) to get an HC ρ of G and then return an HP starting at x. Else, set i 0 = min{j : π j = 1}, ĩ 0 = min{j : π j = 1}.

35 21 / 47 Solving 1HP { Input: a graph G, an interval representation I of G, x V(G); Output: an HP of G which starts at x or "there is no HP of G which starts at x"; } Let a = arg min v V(G) r I (v) and b = arg max v V(G) l I (v). Run SP(G, I, a) and SP(G, I, b) to get two straight orderings π and π. Compute the right neighbor sequences π and π. Suppose x = π p = π p. If π is not 1-perfect, then return "there is no required HP". If π is 2-perfect, do Procedure HC(G, π) to get an HC ρ of G and then return an HP starting at x. Else, set i 0 = min{j : π j = 1}, ĩ 0 = min{j : π j = 1}.

36 21 / 47 Solving 1HP { Input: a graph G, an interval representation I of G, x V(G); Output: an HP of G which starts at x or "there is no HP of G which starts at x"; } Let a = arg min v V(G) r I (v) and b = arg max v V(G) l I (v). Run SP(G, I, a) and SP(G, I, b) to get two straight orderings π and π. Compute the right neighbor sequences π and π. Suppose x = π p = π p. If π is not 1-perfect, then return "there is no required HP". If π is 2-perfect, do Procedure HC(G, π) to get an HC ρ of G and then return an HP starting at x. Else, set i 0 = min{j : π j = 1}, ĩ 0 = min{j : π j = 1}.

37 Solving 1HP, Contd. 22 / 47 By symmetry and according to Theorem 5, we reduce 1HP to the following subroutine: 1HP(G, I, π p, b, i 0, π, π, π, π ). { Input: an interval graph G, an interval representation I of G. Let a be a vertex of V(G) with minimum r I and let b be a vertex of V(G) with maximum l I. Run SP(G, I, a) and SP(G, I, b) to get two orderings π and π. Compute the right neighbor sequences π and π. Let x = π p. Suppose π is 1-perfect but not 2-perfect and input i 0 = min{j : π j = 1}. Output: an HP of G which starts at x and ends at b; or "there is no HP of G which starts at x and ends at b"; }

38 Solving 1HP, Contd. 22 / 47 By symmetry and according to Theorem 5, we reduce 1HP to the following subroutine: 1HP(G, I, π p, b, i 0, π, π, π, π ). { Input: an interval graph G, an interval representation I of G. Let a be a vertex of V(G) with minimum r I and let b be a vertex of V(G) with maximum l I. Run SP(G, I, a) and SP(G, I, b) to get two orderings π and π. Compute the right neighbor sequences π and π. Let x = π p. Suppose π is 1-perfect but not 2-perfect and input i 0 = min{j : π j = 1}. Output: an HP of G which starts at x and ends at b; or "there is no HP of G which starts at x and ends at b"; }

39 Solving 1HP, Contd. 22 / 47 By symmetry and according to Theorem 5, we reduce 1HP to the following subroutine: 1HP(G, I, π p, b, i 0, π, π, π, π ). { Input: an interval graph G, an interval representation I of G. Let a be a vertex of V(G) with minimum r I and let b be a vertex of V(G) with maximum l I. Run SP(G, I, a) and SP(G, I, b) to get two orderings π and π. Compute the right neighbor sequences π and π. Let x = π p. Suppose π is 1-perfect but not 2-perfect and input i 0 = min{j : π j = 1}. Output: an HP of G which starts at x and ends at b; or "there is no HP of G which starts at x and ends at b"; }

40 Solving 1HP, Contd. 23 / 47 Let R = π[i 0 + 2, n] and L = π[i 0 + 1]. Suppose N G ( π n 1 i0 ) \ R = {π j1, π j2,..., π jh } and j 1 < j 2 < < j h = i Assume that l I (π js ) = max{l I (π jt ) : t [h]}. Figure: L = [1, 9], R = [10, 11]

41 Solving 1HP: A sketch of 1HP(G, I, π p, b, i 0, π, π, π, π ) If x R {π j1,..., π jh }, output "there is no HP starting at x and ending at b"; For the remaining case of x L \ {π j1,..., π jh }, taking advantage of the existing path (π js, π n 1 i0, π n 2 i0,..., π 1 = b) we will either find an HP from x to π js in G[L] (and assert that there is an HP of G connecting x and b) or find that no such path exists and then conclude that there is no HP connecting x and b. 24 / 47

42 Solving 1HP: A sketch of 1HP(G, I, π p, b, i 0, π, π, π, π ) If x R {π j1,..., π jh }, output "there is no HP starting at x and ending at b"; For the remaining case of x L \ {π j1,..., π jh }, taking advantage of the existing path (π js, π n 1 i0, π n 2 i0,..., π 1 = b) we will either find an HP from x to π js in G[L] (and assert that there is an HP of G connecting x and b) or find that no such path exists and then conclude that there is no HP connecting x and b. 24 / 47

43 Straight orderings of G in both directions derived from that of G 25 / 47 Let G = G[L π js ]. Note that i 0 = V(G ). τ = 1, 2, 3, 4, 9, 6, 7, 8; (directly read from π = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) τ = 8, 7, 6, 9, 4, 3, 2, 1; (directly read from π = 11, 10, 5, 8, 7, 6, 9, 4, 3, 2, 1)

44 Straight orderings of G 26 / 47 Let τ be the straight path of G[L \ {π js }] obtained by putting τ i = π i, i / {j s, j s+1,..., j h } and τ js = π js +1 for any s [s, h 1].

45 Reduction to G : Find an HP of G from x to N G (π js ) V(G ) 27 / 47 τ(1) is the leftmost interval and τ(1) is the rightmost interval.

46 Case 1: τ is 2-perfect 28 / 47 Find HC in G in which τ i0 and τ i0 1 are consecutive.

47 Find an HP in G[L] from x to π js : τ is 2-perfect 29 / 47 If τ is 2-perfect, then h = 1, namely N G ( π n 1 i0 ) \ R = {π j1 } and s = 1. We do Procedure HP(G, I G, τ, x) to get an HC C in G. It is known that π js τ i0, π js τ i0 1 E(G). Based on C we can construct an HP in G from x to τ i0 or τ i0 1 and then by moving one step further to π js we get a required HP in G[L] from x to π js.

48 Find an HP in G[L] from x to π js : τ is 2-perfect 30 / 47 An HP: 4, 3, 2, 1, 6, 5, 7,8, 9

49 Case 2: τ is not 2-perfect 31 / 47 If τ is not 2-perfect, then it is 1-perfect and we can find j 0 = min{j : τ j = 1}. Let L = τ[1, j 0 + 1], R = τ[j 0 + 2, i 0 ], R = R {τ j0, τ j0 +1}. Figure: τ j0 = 4, τ j0+1 = 9, L = {1, 2, 3, 4, 9}, R = {6, 7, 8}, R = {4, 9, 6, 7, 8}, τ j0 = 4, τ j0+1 = 9

50 L and R 32 / 47 Suppose x = τ q = τ q. x L if and only if q j 0 + 1; x R if and only if q > j τ j0 is the leftmost interval of R.

51 Case 2.1: x L. Find an HP in G from x to the rightmost vertex in G Do 1HP(G[L π js ], I L\{πj s }, x = τ q, τ 1, j 0, τ, τ, τ, τ ) to either get an HP of G from x to τ 1 or conclude that no such path exist. In the former case, the obtained path can be extended one step further to reach π js and thus produces a HP in G[L] from x to π js. In the latter case, we can conclude that there is no HP in G[L] from x to π js. 33 / 47

52 34 / 47 Figure: τ j0 = 4, τ j0+1 = 9, L = {1, 2, 3, 4, 9}, R = {6, 7, 8}, R = {4, 9, 6, 7, 8}, τ j0 = 4, τ j0+1 = 9 = π jh x = 2 L, try to find HP in G = G[L \ {π js }] from 2 to τ 1 = 8, get 2, 1, 3, 4, 9, 6, 7, 8, 5,10,11.

53 Case 2.2: x R. Find HP in R from x to the leftmost vertex in R ; Find HC in L. Figure: green HP from x to the leftmost interval τ j0 in R ; red HC in L with τ j0 and τ j0+1 consecutive; brown edge connecting π js and τ j0, blue HP in G[L] from x to π js 35 / 47

54 Case 2.2: x R. Do Procedure HC(L, τ) to get an HC ρ of L where τ j0, τ j0 +1 are consecutive in ρ. Suppose ρ L = [τ j0 +1, ρ 1, ρ 2,..., ρ j0 1, τ j0 ]. Define two straight paths σ, σ of R : σ i = τ i, i [i 0 j 0 1], σ i0 j 0 = τ j0 +1, and σ i0 j 0 +1 = τ j0 ; σ i = τ i+j0 1, i [i 0 j 0 + 1]. Compute σ, σ and i 2 = min{j : σ j = 1}. Let x = σ q. Do 1HP(R, I R, x = σ q, τ j0, i 2, σ, σ, σ, σ ) to either get an HP of G[R ] which starts at x and ends at τ j0, which must be of the form ρ R = (x, ϱ 2 2,..., ϱ2 i 0 j 0 1, τ j 0 +1, τ j0 ) or output that there is no HP of G[R ] from x to τ j0. In the former case, we walk along ρ R from x to τ j0 +1 (skip the last vertex τ j0 on ρ R ) and then continue with walking along ρ L from τ j0 +1 to τ j0 and then walk one step from τ j0 to π js and in this way we get a HP of G[L] from x to π js. In the latter case, we assert that there is no HP in G[L] from x to π js. 36 / 47

55 37 / 47 Figure: τ j0 = 4, τ j0+1 = 9, L = {1, 2, 3, 4, 9}, R = {6, 7, 8}, R = {4, 9, 6, 7, 8}, τ j0 = 4, τ j0+1 = 9 = π jh x = 7 R, try to find HP from 7 to τ j0 = 4 in R, there is no such path! x = 6 R, try to find HP from 6 to τ j0 = 4 in R, get 6, 7, 8, 9, 4; find HC of L, get 3, 1, 2, 4; putting together we have the required HP 6, 7, 8, 9, 3, 1, 2, 4, 5, 10, 11.

56 Outline 38 / 47 Background A linear time algorithm to solve 1PC Some equivalent connectivity properties

57 39 / 47 A spanning k-trail of a graph G between u and v is a set of k pairwise-internally-disjoint (u, v)-paths such that the interior points of these paths are exactly V(G) \ {u, v}. Theorem 6 (Li-W.) If a graph G has a k-perfect Hamiltonian path π 1,..., π n, then G has a spanning k-trail between π 1 and π n.

58 Theorem 7 (Li-W.) For any positive number k, the following statements for an interval graph G are equivalent: All straight orderings are k-perfect. There is a k-perfect straight ordering. For every interval representation of G, there exists a spanning k-trail of G between the two vertices corresponding to the leftmost interval and the rightmost interval in the representation. There exists an interval representation of G such that there is a spanning k-trail of G between u and v where u and v correspond to the leftmost interval and the rightmost interval in the representation. The deletion of any vertex subset of size at most k 1 results in a graph having a Hamiltonian path. Let Ϝ k stand for the class of interval graphs satisfying any/all of the above conditions. 40 / 47

59 Theorem 7 (Li-W.) For any positive number k, the following statements for an interval graph G are equivalent: All straight orderings are k-perfect. There is a k-perfect straight ordering. For every interval representation of G, there exists a spanning k-trail of G between the two vertices corresponding to the leftmost interval and the rightmost interval in the representation. There exists an interval representation of G such that there is a spanning k-trail of G between u and v where u and v correspond to the leftmost interval and the rightmost interval in the representation. The deletion of any vertex subset of size at most k 1 results in a graph having a Hamiltonian path. Let Ϝ k stand for the class of interval graphs satisfying any/all of the above conditions. 40 / 47

60 41 / 47 Theorem 8 (Li-W.) For k 2, Ϝ k is exactly the set of interval graphs from which the deletion of at most k 2 vertices results in a Hamiltonian graph and is also the set of interval graphs G such that there exists a spanning k-trail of G between u and v for every two different vertices u and v of G. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices. 2 Theorem 9 (Li-W.) An interval graph is Hamiltonian-connected if and only if it lies in Ϝ 3. In general, for k 3, Ϝ k is the set of interval graphs which keeps to be Hamiltonian-connected after deleting at most k 3 vertices. 2 Every 8-connected claw-free graph is Hamiltonian-connected [12].

61 41 / 47 Theorem 8 (Li-W.) For k 2, Ϝ k is exactly the set of interval graphs from which the deletion of at most k 2 vertices results in a Hamiltonian graph and is also the set of interval graphs G such that there exists a spanning k-trail of G between u and v for every two different vertices u and v of G. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices. 2 Theorem 9 (Li-W.) An interval graph is Hamiltonian-connected if and only if it lies in Ϝ 3. In general, for k 3, Ϝ k is the set of interval graphs which keeps to be Hamiltonian-connected after deleting at most k 3 vertices. 2 Every 8-connected claw-free graph is Hamiltonian-connected [12].

62 42 / 47 The next example says that a direct generalization of Theorem 9 from interval graphs to general graphs is not correct. Example 10 (Xuding Zhu) Kneser graph KG(5, 2) (i.e., the Petersen graph) has the property that deleting two vertices, the resulting graph has an H-path. Consider the vertices of KG(5, 2) as 2-subsets of {1, 2, 3, 4, 5} where disjoint subsets are adjacent. There are two types of pairs of vertices to check: {1, 2} and {3, 4} (disjoint), or {1, 2}, {1, 5} (intersect). However, no H-path of KG(5, 2) ends with {1, 2} and {3, 4}.

63 k-trail between extreme intervals to general k-trails 43 / 47 Figure: Use Lemma 4 to rearrange the red parts into the missing path

64 k-trail between extreme intervals to general k-trails 44 / 47 Figure: Use Lemma 4 to rearrange the red parts into the missing path

65 From 3-trail to Hamiltonian-connectedness 45 / 47 Figure: a: the leftmost interval; b: the rightmost interval

66 From 3-trail to Hamiltonian-connectedness 46 / 47 Figure: a: the leftmost interval; b: the rightmost interval

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72 47 / 47!Thank You! ykwu@sjtu.edu.cn