L(3,2,1)-and L(4,3,2,1)-labeling Problems on Circular-ARC Graphs

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1 I J C T, 9(34) 016, pp Iteratioal Sciece Press L(3,,1)-ad L(4,3,,1)-labelig Problems o Circular-RC Graphs S maathulla * ad Madhumagal Pal * BSTRCT For a give graph G ( V, E), the L(3,,1) - ad L(4,3,,1) -labelig problems assig the labels to the vertices of G. Let * Z fuctio f : V be the set of o-egative itegers. L(3,,1) - ad L (4,3,,1) -labelig of a graph G is a * Z such that f ( x) f ( y d( x, y), for 4,5 respectively, where d( x, y) represets the distace (miimum umber of edges of a shortest path) betwee the vertices x ad y, ad 1 d( x, y) 1. The L(3,,1) - ad L(4,3,,1) -labelig umbers of a graph G, are deoted by 3,,1 ( G) ad 4,3,,1 ( G) ad they are the differece betwee highest ad lowest labels used i L(3,,1) - ad L(4,3,,1) - labelig respectively. I [4], Calamoeri et al. have bee studied L( h, ) -labelig of co-comparability graphs ad circular-arc graphs. Motivated from this paper, we have studied L(3,,1) - ad L(4, 3,,1) -labelig problems o circular-arc graphs. I this paper, for circular-arc graph G, it is show that ( G) 9 6 3,,1 ad 4,3,,1 ( G ) 16 1, where represets the maximum degree of the vertices. These bouds we obtai are the first bouds for the problems o circular-arc graphs. lso two algorithms are desiged to label a circular-arc graph by maitaiig L(3,,1) -ad L(4,3,,1) -labelig coditios. The time complexities of both the algorithms are O( ), where represet the umber of vertices of G. Keywords: L(3,,1) -labelig, L(4, 3,,1) -labelig, frequecy assigmet, circular-arc graph, etwor. Mathematics subect classificatio: 05C85, 68R10 1. INTRODUCTION The frequecy assigmet problem is a problem where the tas is to assig a frequecy (o-egative iteger) to a give group of televisios or radio trasmitters so that iterferig trasmitters are assiged frequecy with at least a miimum allowed separatio. Frequecy assigmet problem is motivated from the distace labelig problem of graphs. It is to fid a proper assigmet of chaels to trasmitters i a wireless etwor. The level of iterferece betwee ay two radio statios correlates with the geographic locatios of the statios. Closer statios have a stroger iterferece ad thus there must be a greater differece betwee their assiged chaels. Hale [14] itroduced a graph theory model of chael assigmet problem which is ow as vertex colorig problem. I, 1988 Roberts proposed a variatio of the frequecy assigmet problem i which * Departmet of pplied Mathematics with Oceaology ad Computer Programmig, Vidyasagar Uiversity, Midapore 7110, INDI, amaat87math@gmail.com, mmpalvu@gmail.com

2 870 S maathulla ad Madhumagal Pal closed trasmitter must receive differet frequecy ad very closed trasmitter must receive a frequecy at least two apart. Two vertices x ad y are said to be very closed ad closed if the distace betwee x ad y is1 ad respectively. Griggs ad Yeh [13] defied the L(,1) -labelig of a graph G ( V, E) as a fuctio f which assigs every x, y i V a label from the set of positive itegers such that f ( x) f ( y 3 d ( x, y), where d( x, y ) represet the distace betwee the vertices x ad y, ad 1 d( x, y). The miimum spa over all possible labelig fuctios of L( h, ) -labelig is deoted by h, ( G ) ad is called h, - umber of G. L(3,,1) -labelig of a graph G ( V, E) is a fuctio f from its vertex setv to the set of oegative it egers such that f ( x) f ( y 3 if d( x, y) 1, f ( x) f ( y if d( x, y) ad f ( x) f ( y 1 if d( x, y) 3. The L(3,,1) -labelig umber, ( G), of 3,,1 G is the smallest oegative iteger such that G has a L(3,,1) -labelig of spa. lso, a L(4,3,,1) -labelig of a graph G ( V, E) is a fuctio f from its vertex setv to the set of o-egative itegers such that f ( x) f ( y 4 if d( x, y) 1, f ( x) f ( y 3 if d( x, y), f ( x) f ( y if d( x, y) 3 ad f ( x) f ( y 1 if d( x, y) 4. The L(4,3,,1) -labelig umber, ( G), of 4,3,,1 G is the smallest oegative iteger such that G has a L(4, 3,,1) -labelig of spa. Frequecy assigmet problem has bee widely studied i the past [,3,7,8,13,14,15,16,17,18,19,5,8,9,30]. Later Calamoeri studied L( 1,,1) -labelig of eight grids [6] ad also tta et al. studied L(4,3,,1) -labelig for Simple Graphs [1]. We focus our attetio o L(3,,1) -labelig ad L(4, 3,,1) -labelig of circular-arc graphs. Differet bouds for ( G) ad ( G) were obtaied for various type of graphs. The upper boud of 3,,1 4,3,,1 ( ),1 G of ay graphg is ( p 1) [5], where is the degree of the graph. I [10], Clipperto et al. showed 3 that ( G) 3 3,,1 for ay graph. Later Chai et al. [9] improved this upper boud ad showed that ( G) 3 3,,1 for ay graph. I [0], Lui ad Shao studied the L (3,,1) -labelig of plaer graph ad showed that ( G ) 15( 1). I [9], Chia et al. also showed that ( G) 5 if 3,,1 T is a complete 3,,1 -ary tree of height h 3 ad for ay tree 1 3,,1( G) 3. I [1,,3], Pal et al. studied some problems o iterval graphs. I [11], Jea studied about L( d,,1) -labelig of simple graph ad showed that ( ) ( 1) 1 d,,1 K d where K is complete graph with vertices ad also sho w that ( K ) d ( m ) 3 d,,1 m,. Kim et al. show that 3,,1 3 ( K C ) 15 whe 8 ad 0( mod5), where K3 C is the Cartesia product of complete graphs K 3 ad the cycle C. gai, for ay graph G [1]. I [6], Paul et al. showed that ( G) w,1 for iterval ( G) 3 6 4,3,,1 graph ad they also show that ( G) 3w,1 for circular-arc graph, where w represets the size of the maximum clique. lso i [31], S maathulla et al. show that ( G) ad 0,1 ( G) 1,1 for circulararc graphs. I [4], Calamoeri et al. have bee studied a lot of result about L( h, ) -labelig of co-comparability graphs, iterval graphs ad circular-arc graphs. They have show that, ( G h ) max ( h, ) for co- p

3 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 871 comparability graphs,, ( G h ) max ( h, ) for iterval graphs. lso they have proved h, ( G) max( h, ) hw for circular-arc graphs. Motivated from this paper we have studied L (3,,1) - ad L(4, 3,,1) -labelig of circular-arc graphs. I this paper, for circular-arc graphs G, it is show that the upper bouds of L(3,,1) -ad L(4, 3,,1) - labelig are 9 6 ad 16 1 respectively, where represets the maximum degree of the vertices. lso two algorithms are desiged to label a circular-arc graph by maitaiig L(3,,1) -ad L(4, 3,,1) - labelig coditios. The time complexities of both the algorithms are O( ), where represet the umber of vertices of G. The remaiig part of the paper is orgaized as follows. Some otatios ad defiitios are preseted i Sectio. I Sectio 3, some lemmas related to our wor ad a algorithm to L(3,,1) -label a circulararc graph are preseted. Sectio 4 is devoted to L(4, 3,,1) -labelig problem of circular-arc graph. I Sectio 5, a coclusio is made.. PRELIMINRIES ND NOTTIONS The graphs used i this wor are simple, fiite without self loop or multiple edges. graph G ( V, E) is called a itersectio graph for a fiite family F of a o-empty set if there is a oe-to-oe correspodece betwee F ad V such that two sets i F have o-empty itersectio if ad oly if there correspodig vertices i V are adacet to each other. We call F a itersectio model of G. For a itersectio model F, we use G( F ) to deote the itersectio graph for F. Depedig o the ature of the set F oe gets differet itersectio graphs. For a survey o itersectio graph see [4]. The class of circular-arc graph is a very importat subclass of itersectio graph. graph is a circulararc graph if there exists a family of arcs aroud a circle ad a oe-to-oe correspodece betwee vertices of G ad arcs, such that two distict vertices are adacet i G if ad oly if there correspodig arcs itersect i. Such a family of arcs is called a arc represetatio for G. lso, it is observed that a arc of ad a vertex v of V are oe ad same thig. circular-arc graph ad its correspodig circular-arc represetatio are show i Figure 1. Figure 1: circular-arc graph ad its correspodig circular-arc represetatio

4 87 S maathulla ad Madhumagal Pal graph G is a proper circular-arc ( PC ) graph if there exists a arc represetatio for G such that o arc is properly icluded i aother. The circular-arc graphs used i this wor may or may ot be proper. It is assumed that all the arcs must cover the circle, otherwise the circular-arc graph is othig but a iterval graph. The degree of the vertex v correspodig to the arc is deoted by d(v ) ad is defied by the maximum umber of arcs which are adacet to. The maximum degree or the degree of a circular-arc graph G, deoted by ( G) or by, is the maximum degree of all vertices correspodig to the arcs of G. Let 1 3 {,,,, } be a set of arcs aroud a circle. While goig i a clocwise directio, the poit at which we first ecouter a arc is called the startig poit of the arc. Similarly, the poit at which we leave a arc is called the fiishig poit of the arc. set C V is called a clique if every pair of vertices of C has a edge. The umber of vertices of the clique represets its size. clique is called maximal if there is o clique of G which properly cotais C as a subset. clique with r vertices is called r-clique. clique is called maximum if there is o clique of G of larger cardiality. The size of the maximum clique is deoted by w(g) or by w. Notatios: Let G be a circular-arc graph with arc set, we defie the followig obects: (i) L( ) : the set of labels which are used before labelig the arc,. (ii) L ( ) : the set of labels which are used to label the vertices at distace i ( i 1,,3, 4 ) from the arc i, before labelig the arc,. (iii) Livl ( ) : the set of all valid labels to label the arc satisfyig the coditio of distace oe, oe ad two, oe, two ad three of L(3,,1) -labelig from the arc, before labelig ( i 1,,3 ) respectively., for (iv) L ivl ( ) : the set of all valid labels to label the arc satisfyig the coditio of distace oe, oe ad two, oe, two ad three, oe, two, three ad four of L(4, 3,,1) -labelig from the arc, before labelig, for ( i 1,,3, 4 ) respectively. (v) f : the label of the arc,. (vi)l: the label set, i.e. the set of labels used to label the circular-arc graph G completely. Defiitio 1. For a circular-arc graph G, for each arc (a) all arcs of (b) o two arcs of (c) each S are adacet to, S is maximal. S are adacet, ad 3. L(3,,1)-LBELING OF CIRCULR-RC GRPHS the set S is defied as I this sectio, we preset some lemmas related to the proposed algorithm. lso, a algorithm is desiged to solve L(3,,1) -labelig problem o circular-arc graph. The time complexity of the algorithm is also calculated.

5 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 873 Lemma 1. For a circular-arc graph G, Li (, i,3, 4 for ay arc. Proof. Case 1: Let i ad let G be a circular-arc graph ad be ay arc of G. lso let L ( m. This implies that m distict labels are used to label the arcs which are at distace two from the arc before labelig the arc. Sice is the degree of the graph G so, is adacet to at most arcs of G. Sice G is a circular-arc graph, amog the arcs those are adacet to, there must exists at most two arcs (the arcs of maximum legth) i opposite directio of the arc, each of which are adacet to at most 1 arcs (except ) of G, obviously these arcs are at distace two from. I figure, all the two distaces vertices of are adacet to either or 1. Except, 1 is adacet at most 1 arcs. Similarly, except most 1 arcs. Hece, m ( 1), i.e. L (.,, is adacet at Case : Let i = 3 ad let G be a circular-arc graph ad be ay arc of G ad let L3 ( r. This implies that r distict labels are used to label the arcs which are at distace three from the arc, before labelig the arc. Sice is the degree of the graph G so, is adacet to at most arcs of G. Sice G is a circular-arc graph, amog the arcs those are adacet to, there must exists at most two arcs (the arcs of maximum legth) i opposite directio of the arc, each of which are adacet to at most arcs of G. I figure, 1 ad are those arcs each of which are adacet to at most arcs of G. mog the arcs those are adacet to ad of distace two from 1, there exists at most oe arc (the arcs of maximum legth) which is adacet to at most 1 arcs (expect p ) obviously these arcs are at distace three from 1. gai amog the arcs which are adacet to ad of distace two from, there exists at most oe arc (the arcs of maximum legth) which is adacet to at most 1 arcs (except p ), obviously these arcs are at distace three from. I figure, all the three distaces arcs are adacet to either p or 1 p. Except, 1 p is 1 adacet at most 1 arcs. Similarly, except, p is adacet at most 1 arcs. Hece, r ( 1), i.e. L3 (. Figure : circular-arc graph

6 874 S maathulla ad Madhumagal Pal The proof of the other case is similar. Lemma. For a circular-arc graph G, Li ( ) L( ), for ay arc ofg ad i 1,,3, 4. Proof. y label used to label a circular-arc graph G belog to L( ). So ay label l Li ( ) implies l L( ), for i 1,,3, 4. Hece Li ( ) L( ), for ay arc ofg ad i 1,,3, 4. Lemma 3. L ( ) is the o empty largest set satisfyig the coditio of distace 1,,,..., for 1,, 3, vl of L(3,,1) -labelig, where l p for all l Lvl ( ) ad p max{ L( )} 3, for ay ad 1,,3. Proof. Sice Li ( ) L( ) for i 1,,3 (by Lemma ) ad p max{ L( )} 3, so p l i 3 for ay l L ( ), i 1,,3. Therefore, p L ( ) for 1,,3. Hece L ( ) is o empty set for 1,, 3. i i vl gai, let B be ay set of labels satisfyig the coditio of distace 1,,..., for 1,,3, of L(3,,1) - labelig, where l p for all l B. lso, let b B. The b l 4 i for ay li Li ( ) ad for i, 1,, where i 0. Thus, b Lvl ( ), for 1,,3. So b B implies b Lvl ( ) 1,,3. Therefore, B L ( ), for 1,,3. Sice B is arbitrary, so L ( ) is the largest set of vl labels satisfyig the coditio of distace 1,,..., for 1,,3, of L(3,,1) -labelig, where l p for all l L ( ) for 1,,3. vl Now, we discuss about the bouds of 3,,1 ( G) for a circular-arc graphs. Theorem 1. For ay circular-arc graph G, ( G) 1 where max S 3,,1, 1,,3,,. Proof. Let G be a circular-arc graph ad { 1,, 3,, }. Let such that S max S, the clearly { } S forms a subgraph of G. Thus whe we label this subgraph by L(3,,1) -labelig, the ay member of S ad taes labels so that each differs the other by at least 3 ad all other members get labels so that each label differs from the other by at least. Thus exactly, 1 labels (amely 0, 3,5, 7,, 1) are eeded to label the subgraph S { }. Hece, ( G) 1. 3,,1 Theorem. For ay circular-arc graph G, ( G) 9 6, where 3,,1 is the degree of the graph G. Proof. Let the total umber of arcs of the circular-arc graph G be ad the set of arcs 1 3 {,,,, }. Let L( ) {0,1,,,9 6}, where. The L( 9 5. Now ( G) 9 6 3,,1, if we ca prove that the label i the set L( ) is sufficiet to label all the arcs of G. Suppose, we are goig to label the arc vl i vl, for by L(3,,1) -labelig. We ow that L1 (. So i the extreme ufavorable cases at least ( 9 5) labels of the set L( ) are available satisfyig the coditio of distace oe of L(3,,1) -labelig. lso, sice L (, (by Lemma 1). So i the extreme ufavorable cases at least ( 6 5) ( ) 1 labels of the set L( ) are available satisfyig the

7 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 875 coditio of distace oe ad two of L(3,,1) -labelig. gai, sice L3 (, (by Lemma 1), so i the most ufavorable cases at least oe (viz: ( 1) ( ) 1) label of the set L( ) is available satisfyig L(3,,1) -labelig coditio. Sice is arbitrary, so we ca label ay arc of the circular-arc graph satisfyig L(3,,1) -labelig coditio by usig the labels from the set L( ). If we tae L( ) so that L( ) {0,1,,,9 6} ad we are goig to label the arc by L(3,,1) - labelig, the by similar argumets, it follows that the set L( ) may or may ot cotai a label satisfyig L(3,,1) -labelig coditio. Hece, ( G) ,, lgorithm for L(3,,1)-labelig I this subsectio, a algorithm to L(3,,1) -label a circular-arc graph is desiged. The mai idea of the algorithm is discuss below: First we fid out the set of labels which satisfies the coditio of distace oe of L(3,,1) -labelig. mog these labels we fid the set of labels which also satisfies the coditio of distace two of L(3,,1) - labelig. mog these labels we fid the set of labels which satisfies the coditio of distace three of L(3,,1) -labelig. The we tae the least elemet of that set, which obviously satisfies L(3,,1) -labelig coditio. lgorithm L31 Iput: set of ordered arcs of a circular-arc graph. //assume that the arcs are ordered with respect to clocwise directio i.e. { 1,, 3,, } // Output: f, the L(3,,1) -label of, 1,,3,,. Iitializatio: f1 0 ; L( ) {0} ; for each to 1 compute L ( ) 1, L ( ) ad L ( ) 3 for i 0 to r, //where r max{ L( )} 3 // for 1 to L1 ( if i l 3, the L 1 vl ( ) { i} //where l ( ) L1 // for 1 to for m 1 to Lvl ( for 1 to L 1( if l p 3, the L ( ) { } 1 l m vl m

8 876 S maathulla ad Madhumagal Pal //where l L ( ) ad p L ( ) 1 // f mi 3vl { L ( )}; m vl L( 1) L( ) { f } ; for i 0 to s, where s max{ L( )} 3 for 1 to L1 ( if i l 3, the L 1 vl ( ) { i} //where l ( ) L1 // for = 1 to for m 1 to L ( vl for q 1 to L 1( if l p 3, the L ( ) { } 1 l f mi 3vl m q vl m //wherel L ( ) 1 ad p L ( ) 1 // { L ( )}; L L( ) { f } ; ed L31. m vl q Theorem 3. The lgorithm L31 correctly labels the vertices of a circular-arc graph usig L31 L(3,,1)- labelig coditio. Proof. Let { 1,, 3,, }, also let f1 0, L( ) {0}. If the graph has oly oe vertex the L( ) is sufficiet to label the whole graph ad obviously, ( G) 0. 3,,1 If the graph has more tha oe vertex the the set L( ) is isufficiet to label the whole graph G, because i this case more tha oe label is required adl( ) cotais oly oe label. Suppose, we are goig to label the arc. L ( ) is the o empty largest set satisfyig the coditio of distace vl 1,,..., for 1,,3, of L(3,,1) -labelig, where l p for all l Lvl ( ) ad p max{ L( )} 3, for ay ad 1,,3 (by Lemma 3). lso o labell L3 ( ) adl p satisfyig the coditio of vl

9 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 877 L(3,,1) -labelig of graph. So the labels o the set are the oly valid labels for, which is less tha or equal to p ad satisfyig L(3,,1) -labelig coditio. So Our aim is to label the arc by usig as few labels as possible, satisfyig L(3,,1) -labelig coditio. f q, where q mi{ L3vl ( )}. Now q is the least label for, because o label less tha q satisfies L(3,,1) -labelig coditio. Sice is arbitrary so this algorithm spet miimum umber of labels to label ay arc of a circular-arc graph satisfyig L(3,,1) -labelig coditio ad 3,,1 ( G) max{ L( ) { f }}. Theorem 4. circular-arc graph ca be L(3,,1) -labeled usig O( ) time, where, ad represet umber of vertices ad the degree of the graph G. Proof. Let L be the label set ad L be its cardiality. ccordig to the algorithm L31, L ( L for i=1,,3, for ay, ad also 9 3 r, where r max{ L( )} 3. So we ca compute L 1 vl ( ) usig at most L (9 3) time, i.e. usig at most O( L ) time. lso, L ( 9 6 for 1,, so for each 1,, L ( ) 1 vl ca be computed usig at most L (9 6) time, i.e. usig at most O( L ) time. This process is repeated for 1 times. So the total time complexity for the algorithm L31 is O(( 1) L ) O( L ). Sice, L 9 5, therefore the ruig time for the algorithm L31 is O( ). Illustratio of the algorithm L31 Let us cosider a circular-arc graph of Fig. 3 to illustrate the algorithm L31. vl i Figure 3: Illustratio of lgorithm L31

10 878 S maathulla ad Madhumagal Pal For this graph, { 1,, 3,, 10 } ad 4. f the label of the arc f1 0, L( ) {0}. Iteratio 1: For., for 1,,3,,10. L1 ( ) {0}, L ( ), L3 ( ). L1 vl ( ) {3}, vl L ( ) {3}, L3vl ( ) {3}. Therefore, f mi{ L3vl ( )} 3 ad L( 3 ) L( ) { f} {0} {3} {0,3}. Iteratio : For 3. L1 ( 3 ) {0}, L ( 3 ) {3}, L3 ( 3 ). L1 vl ( 3 ) {3, 4,5,6}, vl 3 So 3 3 vl 3 L ( ) {5,6}, L3vl ( 3 ) {5,6}. f mi{ L ( )} 5 ad L( 4 ) L( 3 ) { f3} {0,3,} {5} {0,3,5}. Iteratio 3: For 4. L1 ( 4 ) {0,5}, L ( 4 ) {3}, L3 ( 4 ). L1 vl ( 4 ) {8}, vl 4 L ( ) {8}, L3vl ( 4 ) {8}. Therefore, f4 mi{ L3vl ( 4 )} 8 ad L( 5 ) L( 4 ) { f4} {0,3,5} {8} {0,3,5,8}. Iteratio 4: For 5. L1 ( 5 ) {5,8}, L ( 5 ) {0}, L3 ( 5 ) {3}. L1 vl ( 5 ) {0,1,,11}, vl 5 L ( ) {,11}, L3vl ( 5 ) {,11}. Therefore, f5 mi{ L3 vl ( 5 )} ad L( 6 ) L( 5 ) { f5} {0,3,5,8} {} {0,,3,5,8}. I this way f6 11, f7 14 f8 6, f9 9, ad fially, f10 1. The vertices ad the label of the correspodig vertices its are give below: Vertices L (3,, 1)-labels L(4,3,,1)-LBELING OF CIRCULR-RC GRPH By extedig the idea of L(3,,1) -labelig, we desig a algorithm for L(4, 3,,1) -labelig of circular-arc graph. I this sectio, we preset some lemmas related to our wor, bouds of L(4,3,,1) -labelig, the algorithm L431 ad time complexity of the proposed algorithm L431

11 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 879 Lemma 4. L ( ) is the o empty largest set satisfyig the coditio of distace 1,,..., vl for 1,,3,4 of L(4,3,,1) -labelig, where l p for all l L vl ( ) ad p max{ L( )} 4, for ay ad 1,,3,4. Proof. Sice Li ( ) L( ) for i 1,,3, 4 (by Lemma 3) ad p max{ L( )} 4, so p l i 4 for ay l L ( ), i 1,,3, 4. Therefore, p L ( ) for 1,,3,4. Hece, L ( ) is o empty set for i i 1,,3,4. vl gai, let B be ay set of labels satisfyig the coditio of distace 1,,..., for 1,,3,4 of L(4,3,,1) - labelig, where l p for all l B. lso, let b B. The b l 5 i for ay li Li ( ) ad for i 3,, 1,, where i 0. Thus, b L vl ( ), for 1,,3,4. So b B implies b L vl ( ), for 1,,3,4. Therefore, B L ( ), for 1,,3,4. Sice, B is arbitrary, so L ( ) is the largest set of vl labels satisfyig the coditio of distace 1,,..., for 1,,3, 4 of L(4,3,,1) -labelig, where l p for all l L ( ), for 1,,3, 4. vl Hece the lemma. Now, we discuss about the upper boud of ( G) 4,3,,1 of a circular-arc graphs. Theorem 5. For ay circular-arc graph G, ( G) 3 1 where max S 4,3,,1, 1,,3,,. Proof. Let G be a circular-arc graph ad { 1,, 3,, }. Let such that S max S, the clearly { } S forms a subgraph of G. Thus, whe we label this subgraph by L(4, 3,,1) -labelig, the ay oe member of S ad taes labels so that each differs the other by at least 4 ad all other members get labels so that each label differs from the other by at least 3. Thus exactly, 3 1 labels (amely 0, 4,7,9,,3 1) are eeded to label the subgraph S { }. Hece, ( G) ,,1 Theorem 6. For ay circular-arc graph G, 4,3,,1 ( G ) 16 1, where is the degree of the graph G. Proof. Let G be a circular-arc graph havig arcs ad the set of arcs be { 1,, 3,, }. lso, let L( ) {0,1,,,16 1}, where Now 4,3,,1 i vl vl. The L( ( G ) 16 1, if we ca prove that the label i the set L( ) is sufficiet to label all the arcs of G. Suppose, we are goig to label the arc by L(4, 3,,1) -labelig. We ow that L1 (. So i the extreme ufavorable cases at least (16 11) labels of the set L( ) are available satisfyig the coditio of distace oe of L(4, 3,,1) -labelig. lso, sice L (, (by Lemma 1). So i the worst case at least (1 11) 3( ) 6 5 labels of the set L( ) are available satisfyig

12 880 S maathulla ad Madhumagal Pal the coditio of distace oe ad two of L(4,3,,1) -labelig. gai sice L3 (, (by Lemma 1), so i the most ufavorable cases at least ( 6 5) ( ) 1 labels of the set L( ) are available satisfyig the coditio of distace oe, two ad three of L(4, 3,,1) -labelig. Fially, sice L (, (by Lemma 1), so i the most ufavorable cases at least oe (viz: ( 1) ( ) 1) 4 label of the set L( ) is available satisfyig L(4, 3,,1) -labelig coditio. Sice is arbitrary, so we ca label ay arc of the circular-arc graph satisfyig L(4, 3,,1) -labelig coditio by usig the labels of the set L( ). If we tae L( ) so that L( ) {0,1,,,16 1} ad we are goig to label the arc by L(4, 3,,1) - labelig, the by similar argumets, it follows that the set L( ) may or may ot cotai a label satisfyig L(4.3,,1) -labelig coditio. Hece, 4,3,,1 ( G ) lgorithm for L(4, 3,, 1)-labelig I this subsectio we preset a algorithm to solve L(4, 3,,1) -labelig of circular-arc graph. lgorithm L431 Iput: set of ordered arcs of a circular-arc graph. //assume that the arcs are ordered with respect to clocwise directio amely 1,, 3,, where 1 3 {,,,, } // Output: f, the L(4,3,,1) -label of, 1,,3,,. Iitializatio: f1 0 ; L( ) {0} ; for each to 1 compute L ( ) for p 1,,3, 4 for i 0 to r, where r max{ L( )} 4 for 1 to L1 ( for 1 to 3 p if i l 4, the L 1 vl ( ) { i} //where l ( ) L1 // for m 1 to L vl ( for 1 to L 1( if l p 4, the L ( ) { } 1 l m vl m //where l L ( ) ad p L ( ) 1 // m vl

13 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 881 f mi L4vl { ( )} ; L( 1) L( ) { f } ; for i 0 to s, where s max{ L( )} 4 for 1 to L1 ( if i l 4, the L 1 vl ( ) { i} //where l ( ) L1 // for 1 to 3 for m 1 to L ( vl for q 1 to L 1( f mi L4vl if l p 4, the L ( ) { } 1 l m q vl m //where l L ( ) ad p L ( ) 1 // { ( )} ; L L( ) { f } ; ed L431. m vl Theorem 7. The lgorithm L431 correctly labels the vertices of a circular-arc graph usig L(4,3,,1)- labelig coditio. Proof. Let { 1,, 3,, }, also let f1 0, L( ) {0}. If the graph has oly oe vertex the L( ) is sufficiet to label the whole graph ad obviously, ( G) 0. 4,3,,1 If the graph has more tha oe vertex the the set L( ) is isufficiet to label the whole graph G. Suppose, we are goig to label the arc. L ( ) is the o empty largest set satisfyig the coditio vl of distace1,,..., for 1,,3,4 of L(4,3,,1) -labelig, where l p for all l L ( ) ad p max{ L( )} 4, for ay ad 1,,3,4 (by Lemma 4). So, the labels i the set L 4 vl ( ) are the oly valid labels for, which is less tha or equal to p ad satisfyig L(4,3,,1) -labelig coditio. Our aim is to label the arc 4vl by usig least possible label by L(4, 3,,1) -labelig. So f vl q, where q mi{ L ( )}. Now q is the least label for, because o label less tha q satisfies L(4, 3,, 1)-labelig coditio. Sice is arbitrary, so this algorithm spet miimum umber of labels to label ay arc of a circular-arc graph by L(4, 3,,1) -labelig ad ( ) { ( ) { }} 4,3,,1 G max L f.

14 88 S maathulla ad Madhumagal Pal Hece the theorem. Theorem 8. circular-arc graph ca be L(4, 3,,1) -labeled usig O( ) time, where ad represet umber of vertices ad the degree of the graph G respectively. Proof. Let L be the label set ad L be the cardiality of L. ccordig to the algorithm L431, L ( L for i = 1,, 3, 4 ad for ay, ad also 16 8 r, where r max{ L( )} 4. So L 1 vl ( ) is computed usig at most L (16 8) time, i.e. usig at most O( L ) time. lso L ( for 1,,3, so L ( ) 1 ca be computed usig at most L (16 11) time, i.e. usig at most O( L ) time for each vl 1,,3. This process is repeated for 1 times. So, the total time complexity for the algorithm L431 is O(( 1) L ) O( L ). Sice, L 16 11, therefore the ruig time for the algorithm L431 is O( ). Illustratio of the algorithm L431 To illustrate the algorithm we cosider a circular-arc graph of Fig. 4. vl i For this graph, V { v1, v, v3,, v10 } ad 4. Figure 4: Illustratio of lgorithm L431 f The label of the vertex v, for 1,,3,,10. f1 0, L( v ) {0}. Iteratio 1: For. L1 ( v) {0}, L ( v), L3 ( v), L4 ( v). L 1vl ( v) {4}, L vl ( v ) {4}, L 3vl ( v) {4}, L 4vl ( v ) {4}. Therefore, f mi{ L 4vl ( v )} 4 ad L( v3) L( v) { f} {0} {4} {0, 4}.

15 L(3,, 1)-ad L(4, 3,, 1)- Labelig Problems o Circular-RC Graphs 883 Iteratio : For 3. L1 ( v3) {0}, L ( v3) {4}, L3 ( v3), L4 ( v3) L 1vl ( v3) {4,5,6,7,8}, L vl ( v3 ) {7,8}, L 3vl ( v3) {7,8}, L 4vl ( v3) {7,8}. Therefore, f3 mi{ L 4vl ( v3)} 7 ad L( v4 ) L0 ( v3) { f3} {0, 4} {7} {0,4,7}. Iteratio 3: For 4. L1 ( v4) {0,7}, L ( v4) {4}, L3 ( v4 ), L4 ( v4) L 1vl ( v4) {11}, L vl ( v4) {11}, L 3vl ( v4) {11}, L 4vl ( v4) {11}. Therefore, f4 mi{ L 4vl ( v4 )} 11 ad L( v5 ) L( v4 ) { f4} {0, 4,7} {11} {0,4,7,11}. Iteratio 4: For 5. L1 ( v5) {7,11}, L ( v5 ) {0}, L3 ( v5 ) {4}, L4 ( v4). L 1vl ( v5) {0,1,,3,15}, L vl ( v5 ) {3,15}, L 3vl ( v5 ) {15}, L 4vl ( v5) {15}. Therefore, f5 mi{ L 4vl ( v5 )} 15 ad L( v6 ) L( v5) { f5} {0, 4,7,11} {15} {0,4,7,11,15}. I this way f6 19, f7 f8 9, f9, ad fially, f The vertices ad the label of the correspodig vertices are show below: Vertices v 1 v v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 L (4, 3,, 1)-labels CONCLUSION I this paper, we determie the upper bouds for 3,,1 ad 4,3,,1 for a circular-arc graph G, ad have show that ( G) 9 6 ad 3,,1 4,3,,1 ( G ) These are the first bouds for the problems o circular-arc graphs. lso, two algorithms are desiged to L(3,,1) -label ad L(4, 3,,1) -label for circular- arc graphs. The ruig time for both the algorithm is O( ). Sice the upper bouds are ot tight, so there is a chace for ew upper bouds for the problems. lso the time complexities of the proposed algorithms may be reduced. REFERENCES [1] S. tta, P. R. S. Mahapatra, L(4, 3,, 1)-labelig for Simple Graphs, Iformatatio Systems Desig ad Itelliget pplicatios, dvaces i Itelliget Systems ad Computig, 015. [].. Bertossi ad C. M. Piotti, pproximate L( 1,,..., t )-colorig of trees ad iterval graphs, Networs, 49(3), 04-16, 007. [3] T. Calamoeri, Emauele G, Fusco, Richard B. Ta, Paola Vocca, L(h, 1, 1)-labelig of outerplaar graphs, Math Meth Operatio Research, 69, , 009.

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