2n i?1 ? 1} of integers, for every i ? 2 ,
Furthermore, every S i has the following property regarding I i : 597 Observation 4.8. Let i ? 1 be fixed. For every ? ? I i , the graph S i has no subset 598 ,
,
, Note that every non-leaf 601 vertex of S i different from r is a cut-vertex. Under all assumptions, this yields that, by 602 the value of ?, necessarily the two neighbours r and r of r belong to V ? . Since r r is a 603 cut-edge of S i , this means that V ? has to cover all vertices different from r
, We are now ready to prove the NP-hardness of Realization for claw-free graphs
, Realization is NP-hard when restricted to connected claw-free graphs
, Let <A, B, s> be an 608 instance of 3-Partition, where we use the same terminology as in these proofs. We may 609 assume that s(a 1 ) ? · · · ? s(a 3m ). Free to modify this instance following Observation 2.1, 610 we can assume that there is an ? such that s
, We add m disjoint copies 612 of K 4 to the graph, where the vertices of the ith copy are denoted by a i , b i , c i , d i, p.613
m ? 1, we then identify the vertices d i and a i+1 , so that the K 4 's form a kind of 614 path connected via cut-vertices. For every i = 1, . . . , m ? 1, we then add a copy of S ? to 615 the graph, and we identify its root with b i . Finally, we consider every i = 1, . . . , m, and: 616 ? for i = 1 or i = m, we add a complete graph K B?2 to the graph ,
m ? 1}, we add a complete graph K B?1 to the graph, and we identify 619 one of its vertices and c i ,
, Note that G is claw-free (it is actually a line graph). The |V (G)|-sequence ? we consider 621 for the reduction is ? =
s(a 3m ) ? I ? . For this 623 reason, by Observation 4.8, in any realization of ? in G, the part V 1 with size mn ? + m ? 1 624 has to contain the vertices of all S ? 's added to G, and, because G[V 1 ] must be connected, 625 also all vertices d 1 , . . . , d m?1 . Then G ? V 1 is a disjoint union of traceable B-graphs, and 626 we have to find a realization of (s(a 1 ), . . . , s(a 3m )) in it, and that ? was chosen so that s(a 1 ) ,
, On the 632 other hand, we have provided, in Section 3, new kernels of sequences showing that the AP 633 problem is in NP for a few more classes of graphs. However, we are still far from a proof 634 that 1) every graph has a polynomial kernel of sequences (which would establish the full 635 NPness of AP), and that 2) the AP problem is complete for some complexity class
, which was 638 mentioned in [10] by Broersma, Kratsch and Woeginger. It can easily be noted that the 639 reduction in our proof of Theorem 3.3 yields cographs, so Realization is NP-hard for 640 these graphs. It is still open, though, whether there is a polynomial kernel of sequences for 641 cographs. Note that Theorem 3.1 makes a step in that direction, as 1-sequential graphs 642
, The second line of research we have considered in this work is the weakening, p.644
, It would be 645 interesting if there were such a weakening for every condition for Hamiltonicity, as it would 646 emphasize the relationship between Hamiltonicity and APness. However, previous investi-647 gations and some of our results seem to, ness, of well-known sufficient conditions for Hamiltonicity (or traceability)
, We believe, however, that it would be nice dedicating more attention to this direction; 650 let us thus raise an open question which might be interesting. As mentioned in the intro-651 ductory section, Ore's well-known condition for Hamiltonicity can be weakened to APness
, In particular, all n-graphs G with ? 2 (G) ? n ? 2 having a (quasi-) perfect matching are 653
, This result implies one direction of upcoming Question 5.1, which, if true, would stand 654 as a result à la Bondy-Chvátal, AP
, the k-closure of G is the (unique) graph obtained by repeatedly 656 adding an edge between two non-adjacent vertices with degree sum at least k. A celebrated 657 result of Bondy and Chvátal states that an n-graph is Hamiltonian if and only, p.658
, Analogously, an n-graph is traceable if and only if its (n ? 1)-closure is 659 traceable. However, it is not true that every n-graph is AP if and only if its (n ? 2)-closure 660 is AP: In the complete bipartite graph K n/2?1,n/2+1 , every two non-adjacent vertices have 661 degree sum at least n?2, p.662
, This is 663 actually not the only exception. Indeed, let G be a graph of order n = 4k + 2 consisting 664 of two complete graphs K n/2 with a common vertex and a pendant edge attached to this 665 vertex. Then the (n ? 2)-closure of this graph is complete, but G has not realization of 666 (n/2, n/2). We wonder whether there are many such exceptions, in the following sense: 667 Question 5.1. Is there an "easy" class of graphs G, such that if G ? G is an n-graph
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