A linear-time algorithm for the strong chromatic index of Halin graphs

A L I N E A R - T I M E A L G O R I T H M F O R T H E S T R O N G C H R O M A T I C I N D E X O F H A L I N G R A P H S T on Kloks 1 ⋆ and Y ue-Li W ang 2 1 Department of Computer Science National T sing Hu a University , No. 101, S ec. 2, Kuang Fu Rd., Hsinchu, T aiwan kloks@cs.n thu.edu.tw 2 Department of Information Management National T aiwan University of Science and T echnology No. 43, Se c . 4, Keelung Rd., T aipei, 106, T aiwan ylwang@cs. ntust.edu.tw Abstract. W e show that there exists a linear-time algorithm that computes the strong chromatic index of Halin gr aphs. 1 Introduction Definition 1. Let G = ( V , E ) be a graph. A strong e dg e coloring o f G is a proper edge coloring such that no edge is adjacent to two edges of the same color . Equivalently , a str ong edge coloring of G is a vert ex coloring of L ( G ) 2 , t he square o f t h e linegraph o f G . T he strong chromatic index of G is the minimal integer k such th at G has a strong edge co l oring with k colors. W e denote the strong ch romatic index o f G by sχ ′ ( G ) . R ecently it was shown that the str ong chromatic index is b ounded by ( 2 − ǫ ) ∆ 2 for some ǫ > 0, where ∆ is the maximal de g ree of the graph [12]. 3 Earlier , Andersen showed that the strong chr o matic index of a cubic graph is at m o st ten [1]. Let G b e the class of chordal graphs or t he class of cocomparability graphs. If G ∈ G t h en also L ( G ) 2 ∈ G and it follows that the strong chromat ic index can be compute d in polynomial time f o r these classes. Also for graphs of bounded treewidth there ex is ts a polynomial t ime algorithm that computes th e str ong chromatic index [13]. 4 ⋆ This au tho r is supported by the National Science Council of T aiwa n , under grant NSC 99–2218 –E– 00 7–016. 3 In the i r paper Molloy and Reed state that ǫ > 0.0 02 whe n ∆ is sufficiently large. 4 This algorithm checks in O ( n ( s + 1 ) t ) time whe the r a partial k -tr ee has a str on g edge coloring that u ses at most s colors. Her e, the exponent t = 2 4 ( k + 1 )+ 1 . Definition 2. Let T be a tree wit hout vertices of degree t wo. Consider a plane em- bedding o f T and connect the l e av es of T by a cycle t hat c rosses no edges of T . A graph that is construct ed in this way is call ed a Halin graph . Halin graphs have treewidth at most three. Furthermore, if G is a Halin graph of bounded degree, the n also L ( G ) 2 has b ounded treewidth and thus the strong chromatic index of G can be computed in linear t ime. Recently , K o- W ei Lih, et al. , proved that a cubic Halin graph othe r than one of the two ‘ neckla ces’ N e 2 (the compleme nt of C 6 ) and Ne 4 , has strong chromatic index at most 7. The two exceptions have strong chromatic index 9 and 8, respectively . If T is the underlying tree of the Halin graph, and if G 6 = Ne 2 and G is not a wh eel W n with n 6 = 0 mod 3, then Ping- Y ing T sai, et al. , show that the stro ng chromatic index is bo un ded by sχ ′ ( T ) + 3. (See [14, 15] for earlier r e su lts that appeared in regular pap e rs. 5 ) If G is a Halin graph then L ( G ) 2 has bo unded r ankwidth . In [5] it is shown that there exists a polynomial algorithm th at computes the chromatic number of graphs with bounded rankwidth, thus th e strong chromatic index o f Halin graphs can be computed in po lynomial time. In passing, let us mention the following result. A class of graphs G is χ -bounded if there ex ists a function f such that χ ( G ) 6 f ( ω ( G )) for G ∈ G . Here χ ( G ) is the chr omatic number of G and ω ( G ) is the clique number of G . Recently , Dvo ˇ r´ ak and Kr´ al showed t hat for every k , the class of graphs with rankwidth at most k is χ -bounded [3]. O b vious ly , the graphs L ( G ) 2 have a uniform χ -bound for graphs G in the class of Halin graphs. In t his note we show that ther e exists a linear-time algorithm that compute s the stro ng chromatic index of Halin graphs. 2 The strong chromatic index of Halin graphs The f oll owing lemma is easy to ch e ck. Le mma 1 (Ping- Ying T sai). Let C n be the cy cle with n vertices and let W n be the wheel with n vertices in the cycle. Then sχ ′ ( C n ) =      3 if n = 0 mod 3 5 if n = 5 4 otherwise sχ ′ ( W n ) =      n + 3 if n = 0 mo d 3 n + 5 if n = 5 n + 4 otherwise. A d o ubl e whe e l is a Halin graph in which the tree T has exactly two vert ices that are no t leaves. 5 The results of Ko- W ei Lih and P in g- Y ing T sai, et al. , were presented at the Sixth Cross- Strait Conference on Graph Theory and Combinatorics which was held at the National Chiao T ung University in T aiw an in 20 11. 2 Le mma 2 (Ping- Ying T sai). Let W be a double wheel where x and y are the vertices of T tha t are not leaves. Then sχ ′ ( T ) = d ( x ) + d ( y ) − 1 where d ( x ) and d ( y ) are the degrees of x and y . Furthermore, sχ ′ ( W ) =      sχ ′ ( T ) + 4 = 9 if d ( x ) = d ( y ) = 3 , i.e. , if W = ¯ C 6 sχ ′ ( T ) + 2 = d ( y ) + 4 if d ( y ) > d ( x ) = 3 sχ ′ ( T ) + 1 = d ( x ) + d ( y ) if d ( y ) > d ( x ) > 3 . Let G be a Halin graph with t ree T and cycle C . Th e n obviously , sχ ′ ( G ) 6 sχ ′ ( T ) + sχ ′ ( C ) . (1) The linegraph of a tree is a claw-free blockgraph. Since a sun S r with r > 3 has a claw , L ( T ) has no induced sun S r with r > 3. It follows that L ( T ) 2 is a chordal graph [9] (see also [2]; in this paper Cameron pro ves that L ( G ) 2 is chordal for any cho r dal graph G ). Not ic e th at sχ ′ ( T ) = χ ( L ( T ) 2 ) = ω ( L ( T ) 2 ) 6 2 ∆ ( G ) − 1 ⇒ sχ ′ ( G ) 6 2 ∆ ( G ) + 4. (2) 2.1 Cubic Halin graphs In this subsection we outline a simple linear-time algorithm for the cubic Halin graphs. Theorem 1. There ex ists a linea r -time algorit hm t hat c omputes the strong chro- matic index of cubic Halin graphs. Proof . Let G be a cubic Halin graph with p l ane tree T and cycle C . Let k be a natural number . W e describe a linear-time algorithm that checks if G has a strong edge coloring with at most k colors. By Equation (2) we may assume that k is at most 10. Thus the co r rectness of this algorithm proves the theorem. R oot the tree T at an arb itr ary leaf r of T . Consider a ve rtex x in T . T here is a unique path P in T from r to x in T . Define th e subtree T x at x as the maximal connected subtr e e of T th at does not contain an edge of P . If x = r then T x = T . Let H ( x ) be the subgraph of G induced by the vertices of T x . Notice that, if x 6 = r then t he edges of H ( x ) t hat are not in T form a path Q ( x ) of edges in C . F or x 6 = r define t he boundary B ( x ) of H ( x ) as the following set of edges. (a) T he unique edge of P that is incident with x . (b) The two edges of C that connect the p ath Q ( x ) of C with t he rest of C . (c) Consider the endp oints of t h e edges mentioned in (a) and (b) that are in T x . Add the remaining two edges that are incident with each of th e se endpoints to B ( x ) . 3 Thus the bo un dary B ( x ) consists of at most 9 edges. T he following claim is easy to check. It proves the corre ctness of the algorithm described below . Let e be an edge of H ( x ) . Let f be an edge of G that is not an edge of H ( x ) . If e and f are at distance at mo st 1 in G then e o r f is in B ( x ) . 6 Consider all possible colorings of the edges in B ( x ) . Since B ( x ) contains at most 9 e dg es and since the re are at most k differe nt colors for each edge, ther e are at most k 9 6 10 9 different colorings of the ed ges in B ( x ) . The algorithm now fills a table which gives a boolean value for each coloring of the boundary B ( x ) . This boolean value is TR UE if and only if the co l oring of the edges in B ( x ) extends to an edge coloring of the union of t he sets of e dg es in B ( x ) and in H ( x ) with at most k colors, such that any p air of edges in this set that are at distance at most one in G , have different colors. T hese boolean values are comp ut e d as follows. W e prove the corr ectness by induction on the size o f the subtree at x . First consider the case where the subtree at x co ns ists of the single vertex x . Then x 6 = r and x is a leaf of T . In th is case B ( x ) co ns ists of th r ee edges, namely the three e dg es that are incident with x . These are two edges of C and one edge of T . If th e colors of these three edges in B are diffe rent then the boolean value is set to T RUE . Otherwise it is set t o FALS E . Obviously , t h is is a correct assignment. Next consider the case where x is an internal vertex o f T . Then x has two children in the subtre e at x . Let y and z b e the two children and co nsider the two subtrees rooted at y and z . The algorithm that computes the t ab l es for e ach vertex x processes the sub- trees in order of increasing number of vertices. (Thus the roots of the subtr ees are visited in postorder ). W e now assume t hat the tables at y and z are compute d correctly and show how the table for x is compute d correctly and in constant time. T hat is, we prove t hat the algorithm describe d below computes the table at x such that it contains a color ing of B ( x ) with a value TRU E if and only if there exists an ext ensi on of this color ing t o the edges of H ( x ) and B ( x ) such that any two different edges e and f at distance at most o ne in G , each one in H ( x ) or in B ( x ) , have different co l or s. Consider a coloring o f the edges in the boundary B ( x ) . The boolean value in the tab le of x fo r this coloring is co mputed as fo l lows. No t ice that (i) B ( y ) ∩ B ( z ) consists of one edge and this edge is not in B ( x ) , and (ii) B ( x ) ∩ B ( y ) consists of at most four ed ges, namely the edge ( x , y ) and the three edges of B ( y ) t h at are incident with one vertex of C ∩ H ( y ) . Likewise, B ( x ) ∩ B ( z ) consists of at most four edges. 6 T w o edges in G ar e at distance at most one if the subgraph induced by their endpoints is either P 3 , or K 3 or P 4 . W e assume that it can be checked in constant time if tw o edges e and f are at distance at most one. This can be achieved by a suitable data structure. 4 The algorithm varies the possible colorings of the edge in B ( y ) ∩ B ( z ) . Colorings of B ( x ) , B ( y ) and B ( z ) are consistent if th e intersections are the same co l or and the pairs of edges in B ( x ) ∪ B ( y ) ∪ B ( z ) that are at distance at most one in G have differe nt colors. A coloring of B ( x ) is assigned the value TRU E if there exist colorings of B ( y ) and B ( z ) such that the three colorings are consistent and B ( y ) and B ( z ) are assigned the value T RUE in the tables at y and at z respectively . Notice that the table at x is built in constant time. Consider a coloring of B ( x ) that is assigned the value TRUE . Consider color- ings of the edges of B ( y ) and B ( z ) that are consistent with B ( x ) and that are assigned the value TRUE in the t ables at y and z . By induction, ther e exist ext en- sions o f the colorings of B ( y ) and B ( z ) to the edges of H ( y ) and H ( z ) . The union of these extensions provides a k -coloring of the ed ges in H ( x ) . Consider t wo edges e and f in B ( x ) ∪ B ( y ) ∪ B ( z ) . If their distance is at most one then they have different colors since the coloring o f B ( x ) ∪ B ( y ) ∪ B ( z ) is consistent. Let e and f be a pair of edges in H ( x ) . If they are bo t h in H ( y ) or both in H ( z ) then they have diff erent colors. Assume that e is in H ( y ) and assume that f is not in H ( y ) . If e and f ar e at distance at most one, then e o r f is in B ( y ) . If they are both in B ( y ) , then they have diff erent colors, due t o the co nsistency . Otherwise, by the induction hyp othesis, they have different colors. T his p roves the claim o n th e correctness. Finally , consider the table for the ve r tex x which is the unique neighbo r of r in T . By the induction hypothesis, and the f act that every edge in G is either in B ( x ) or in H ( x ) , G has a strong edge coloring with at most k colors if and only if the table at x co ntain s a coloring of B ( x ) with three diffe rent colors for which the bo olean is set to TRUE . This prove s the th eorem. ⊓ ⊔ Re ma rk 1. The involved constants in this algorithm are impr oved considerably by the re cent results of Ko- W e i Lih, Ping- Ying T sai, et al. . 2.2 Halin graphs of general degre e Theorem 2. There ex ists a linea r -time algorit hm t hat c omputes the strong chro- matic index of Halin graphs. Proof . The algorithm is similar t o the algorithm for the cubic case. Let G be a Halin graph , let T b e the underlying plane tree, and let C be the cycle that connects the leaves o f T . Since L ( T ) 2 is chordal t he chromatic number of L ( T ) 2 is equal t o the clique number of L ( T ) 2 , wh ich is sχ ′ ( T ) = max { d ( u ) + d ( v ) − 1 | ( u , v ) ∈ E ( T ) } , 5 where d ( u ) is the degree o f u in the tree T . By F ormula (1) and Lemma 1 the strong ch romatic index o f G is one of the six possible values 7 sχ ′ ( T ) , sχ ′ ( T ) + 1, . . . , sχ ′ ( T ) + 5. R oot th e tree at some leaf r and consider a subtr ee T x at a node x of T . Let H ( x ) be the subgraph of G induced by the vertices of T x . Let y and z be th e t wo boundary vertices of H ( x ) in C . W e distinguish the f oll owing six types of edges corresponding to H ( x ) . 1. The set o f edges in T x that are adjacent t o x . 2. The edge th at co nnect s x to its parent in T . 3. The edge th at co nnect s y to its neighbor in C that is not in T x . 4. The set o f edges in H ( x ) that h ave endp oint y . 5. The edge th at co nnect s z t o its neighbor in C that is not in T x . 6. The set o f edges in H ( x ) that h ave endp oint z . Notice that the set of edges of every type has bounded cardinality , except t h e first typ e. Consider a 0 / 1-matrix M with ro ws indexe d b y the six types of edges and columns indexed by the colors. A matrix entr y M ij is 1 if ther e is an edge of the row-type i that is co lor e d with the color j and otherwise this entry is 0. Since M has o nl y 6 r ows, the r ank over GF [ 2 ] of M is at most 6. T wo colorings are equivalent if there is a permutation o f the colors that maps one co lor ing to the o ther one. Let S ⊆ { 1, . . . , 6 } and let W ( S ) be the set of colors that are used by ed ges of type i for all i ∈ S . A class of equivalent colorings is fixed b y the set of cardinalities { | W ( S ) | | S ⊆ { 1, . . . , 6 } } . W e claim that the number o f equivalence classes is constant. The number o f ones in the row of the first type is the degree of x in H ( x ) . Every other row has at most 3 ones. This proves t he claim. Consider t h e union of t wo subtrees, say at x and x ′ . T he algorithm considers all equivalence classes of colorings of the union, and checks, by table look-up, whether it decomposes into valid colorings of H ( x ) and H ( x ′ ) . An easy way to do this is as follows. First double the number of types, by distinguishing the edges of H ( x ) and H ( x ′ ) . Then enumerate all equivalence classes of colorings. Each equivalence class is fixed by a sequence of 2 12 numbers, as abo ve. By tab l e look- up, check if an equivalence class restricts to a valid coloring for each o f H ( x ) and H ( x ′ ) . Since this takes constant time, the algorithm runs in linear time. This prove s the th eorem. ⊓ ⊔ 7 Actually , according to the recent re sults of Ping- Y ing T sai, et al. , the strong chromatic index of G is at most sχ ′ ( T ) + 3 except when G is a whe el or ¯ C 6 . 6 R eferences 1. 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