Mximum size of minimum wthing system n the grphs hieving the oun Tille mximum un système e ontrôle minimum et les grphes tteignnt l orne Dvi Auger Irène Chron Olivier Hury Antoine Lostein 00D0 Mrs 00 Déprtement Informtique et Réseux Groupe MIC : Mthémtiques e l Informtion, es Communitions et u Clul
Tille mximum un système e ontrôle minimum et les grphes tteignnt l orne Mximum Size of Minimum Wthing System n the Grphs Ahieving the Boun Dvi Auger, Irène Chron, Olivier Hury, Antoine Lostein {vi.uger, irene.hron, olivier.hury, ntoine.lostein}@teleom-pristeh.fr Résumé Soit G = (V (G), E(G)) un grphe non orienté. Un ontrôleur w e G est une pire w = (l(w), A(w)), où l(w) pprtient à V (G) et A(w) est un ensemle e sommets e G à istne 0 ou e l(w). Si un sommet v pprtient à A(w), on it que v est ouvert pr w. Deux sommets v et v e G sont its séprés pr un ensemle e ontrôleurs si l liste es ontrôleurs ouvrnt v est ifférente e elle e v. On it qu un ensemle W e ontrôleurs est un système e ontrôle pour G si tout sommet v est ouvert pr u moins un ontrôleur e W, et si eux sommets quelonques v, v sont séprés pr W. On énote le nomre minimum e ontrôleurs néessires pour ontrôler G pr w(g). Nous onnons une orne supérieure sur w(g) vlle pour tout grphe onnexe orre n et nous rtérisons les rres qui tteignent ette orne, vnt étuier l rtéristion, plus ompliquée, es grphes onnexes qui l tteignent. Mots-lés : Théorie es grphes, Systèmes e ontrôle, Coes ientifints Institut Téléom - Téléom PrisTeh & Centre Ntionl e l Reherhe Sientifique - LTCI UMR 5, 6, rue Brrult, 756 Pris Ceex - Frne Centre Ntionl e l Reherhe Sientifique - LTCI UMR 5 & Institut Téléom - Téléom PrisTeh, 6, rue Brrult, 756 Pris Ceex - Frne
Astrt Let G = (V (G), E(G)) e n unirete grph. A wther w of G is ouple w = (l(w), A(w)), where l(w) elongs to V (G) n A(w) is set of verties of G t istne 0 or from l(w). If vertex v elongs to A(w), we sy tht v is overe y w. Two verties v n v in G re si to e seprte y set of wthers if the list of the wthers overing v is ifferent from tht of v. We sy tht set W of wthers is wthing system for G if every vertex v is overe y t lest one w W, n ny two verties v, v re seprte y W. The minimum numer of wthers neessry to wth G is enote y w(g). We give n upper oun on w(g) for onnete grphs of orer n n hrterize the trees hieving this oun, efore stuying the more omplite hrteriztion of the onnete grphs hieving this oun. Key-wors: Grph Theory, Wthing Systems, Ientifying Coes Introution Let G = (V (G), E(G)) e n unirete onnete grph (the se of n unonnete grph n lso e trete, y onsiering seprtely its onnete omponents). A wther w of G is ouple w = (l(w), A(w)), where l(w) elongs to V (G) n A(w) is set of verties of G t istne 0 or from l(w); in other wors, A(w) is suset of B(l(w)), the ll of rius entre t l(w). We will sy tht w is lote t l(w) n tht A(w) is its wthing re or wthing zone. If vertex v elongs to A(w), we sy tht v is overe y w. Two verties v n v in G re si to e seprte y set of wthers if the list of the wthers overing v is ifferent from tht of v. We sy tht G is wthe y set W of wthers, or tht W is wthing system for G if: for every v in V (G), there exists w W suh tht v is overe y w; if v n v re two verties of G, v n v re seprte y W. Note tht severl wthers n e lote t sme vertex, n wther oes not neessrily over the vertex where it is lote. The minimum numer of wthers neessry to wth grph G is enote y w(g). We will often represent wthers simply y integers, s for the grph G 0 represente in Figure : the lotion of wther is written insie retngle; for eh vertex v of the grph, the list of wthers overing v is written in itlis, so tht the wthing re of eh wther n e retrieve. In the exmple of Figure, the wther is lote t n overs the verties, n, the wther is lso lote t n overs the verties, n e, the wther is lote t f n overs the verties, e, f n h,
,,, e f, g h Figure : grph G 0 wthe y the wthers,, n n the wther is lote t e n overs the verties f n g. The grph G 0 is wthe y these four wthers n it is esy to verify tht w(g 0 ) =. Let G e grph of orer n. If we hve set W of k wthers, the numer of istint non empty susets of {l(w) : w W } is equl to k. Therefore, it is neessry to hve k n, n so we hve the inequlity: w(g) log (n + ). () Oviously, wthing systems generlize ientifying oes (see the seminl pper [6], n [7] for lrge iliogrphy): inee, ientifying oes re suh tht for ny w = (l(w), A(w)) W, we hve A(w) = B(l(w)), whih mens tht, in this se, wther, or oewor, neessrily overs itself n ll its neighours. See lso [5], [8] for similr ies. Wthing systems were introue in [], where motivtions re expose t lrge, si properties re given, omplexity result is estlishe, n the se of the pths is stuie in etil, with omprison to ientifying oes; see lso []. In Setion, we give n upper oun on w(g) when G is onnete grph with n verties. In Setion, we hrterize the trees of orer n whih hieve this oun: Theorems 7, n re for the ses n = k, n = k + n n = k +, respetively. This helps to stuy, in Setion, the hrteriztion of mximl grphs rehing the oun, tht is, grphs to whih no ege n e e without eresing the minimum numer of neessry wthers: Theorems 5 n 6 give the nswer for n = k n n = k + respetively, n Conjeture 7 is stte for the se n = k +. This in turn gives results for ll the onnete grphs hieving the oun.
The mximum minimum numer of wthers The following three esy lemmt will prove effiient. We rell tht H = (V (H), E(H)) is prtil grph of G = (V (G), E(G)) if V (H) = V (G) n E(H) E(G). Lemm Let G e grph n H e prtil grph of G. Then w(h) w(g). Proof. If H is wthe y set W of wthers, the sme set W wthes G, sine two jent verties in H re lso jent in G. Note tht this monotony property oes not hol in generl for ientifying oes. Lemm Let T e tree, x e lef of T, n y e the neighour of x. () There exists minimum wthing system for T with one wther lote t y. () If y hs egree, there exists minimum wthing system for T with one wther lote t z, the seon neighour of y. Proof. () A wthing system must over x, so there is wther w lote t x or y, with x A(w ). If w = (x, A(w )), then we n reple it y w = (y, A(w )), sine B (y) B (x). () If y / A(w ), then one other wther must over y, n if y A(w ), then one must seprte x n y, sine x A(w ). In oth ses, the tsk n e one y wther lote t z. Lemm Let T e tree of orer n let v e vertex of T ; there exists set W of two wthers suh tht the verties in V (T ) \ {v} re overe n pirwise seprte y W in this se, we shll sy, with slight use of nottion, tht V (T ) \ {v} is wthe y W ; the vertex v is overe y t lest one wther. Proof. On Figure, we give ll possiilities: the two trees of orer, n for eh of them, the two lotions for v (v is lef, or v is not lef). We re now rey to give n upper oun for w(g) with respet to n, the orer of G. Note in ontrst tht the upper oun for ientifying oes, when they exist, is n, see [], [], n is rehe, mong other grphs, y the str. Theorem Let G e onnete grph of orer n. If n =, w(g) =.
v or v, or v v, Figure : trees of orer,,,,,,, Figure : the se n=5 in Theorem If n = or n =, w(g) =. If n = or n = 5, w(g) =. If n / {,, }, w(g) n. The proof n e foun in [], [], ut we give it here, euse the results of the four ses into whih it is ivie will e frequently use in the sequel. Proof. For n =, n =, or n =, the result is iret. For n =, it is neessry to hve t lest log (5) = wthers n it is esy to verify tht this is suffiient. For n = 5, ll possiilities re given y Figure n we n see tht we lwys hve w(g) =. We proee y inution on n. We ssume tht n 6 n tht the theorem is true for ny onnete grph of orer less thn n. Let G e onnete grph of orer n. Let T e spnning tree of G; we will prove tht w(t ) n n then the theorem will result from Lemm. We enote y D the imeter of T n we onsier pth v 0, v, v,..., v D, v D of T, with length D. We istinguish etween four ses, oring to some onitions on the egrees of v D n v D. First se: the egree of v D is equl to The vertex v D is jent to vertex x other thn v D n v D ; euse D is the imeter, lerly x n v D re leves of T (see Figure ). We onsier the tree otine y removing x, v D n v D from T ; this new tree T hs orer n. If n 8 or if n = 6, we onsier minimum set W of wthers wthing T ; if n = 7, then T is of orer, n, using Lemm, we hoose 5
x v 0 v v D v D,,... v D Figure : first se of Theorem : the egree of v D is equl to v 0 v v v D D v D,..., Figure 5: seon se of Theorem : the egrees of v D n v D re equl to v D set W of two wthers to wth V (T ) \ {v D } n over the vertex v D. Then for T, in oth ses, we to W two wthers w = (v D, {v D, v D, v D }) n w = (v D, {v D, x}). On Figure, we renme n these wthers. Then T is wthe y W {w, w }. So, w(t ) W + w(t ) +. Now we use the inution hypothesis: if n 8 or n = 6, then w(t ) (n ) + = n ; n if n = 7, then w(t ) + = < 7. Seon se: the egrees of v D n v D re equl to The neighours of v D re v D n v D, the neighours of v D re v D n v D (see Figure 5). We onsier the tree otine y removing v D, v D n v D from T ; this new tree T hs orer n. If n 8 or if n = 6, we onsier minimum set W of wthers wthing T ; if n = 7, T is of orer ; gin using Lemm, we hoose set W of two wthers to wth V (T ) \ {v D } n over the vertex v D. As in the first se, we to W two wthers: w = (v D, {v D, v D, v D }) n w = (v D, {v D, v D }), n T is wthe. So, w(t ) W + w(t )+. The en of this se is the sme s in the first se. Thir se: the egree of v D is t lest The vertex v D is jent to t lest two verties other thn v D n v D : let x n y e two neighours of v D istint from v D n v D ; these two verties re leves of T (see Figure 6). We onsier the tree T otine y removing x n y from T. By Lemm, there exists minimum set W of 6
x, v 0 v v v D v D D y Figure 6: thir se of Theorem : the egree of v D is t lest wthers wthing T with wther w lote t v D. For T, we tke the set W n we the wther w = (v D, {x, y}); we lso the vertex x to the wthing re of w. The tree T eing wthe y W, the tree T is wthe y W {w }. So, w(t ) w(t ) +. If n 7, the orer of T is t lest 5 n, using the inution hypothesis, w(t ) (n ) + < n. If n = 6, then n = n w(t ) + = = 6. Fourth se: the egree of v D is equl to n the egree of v D is t lest The neighours of v D re v D n v D. The vertex v D is jent to v D n v D ut lso to t lest one other vertex x (see Figure 7); if the egree of x is t lest, using the ft tht the imeter of T is equl to D, we n use the first or thir se to onlue, with x plying the prt of v D. So, we ssume tht the egree of x is or ; if its egree is, it hs neighour y other thn v D. We onsier the tree T of orer n otine y removing v D n v D from T. By Lemm, there exists minimum set W of wthers wthing T with wther w lote t v D. To wth T, we tke the set W n the wther w = (v D, {v D, v D }); we lso the vertex v D to the wthing re of w. Then T is wthe y W {w }. The en of this se is extly the sme s in the previous se. Remrk 5 In the proof of Theorem, we hve onstrute, oring to the ses, tree T with orer n suh tht w(t ) w(t )+, or tree T with orer n suh tht w(t ) w(t ) +. These two onstrutions, from T to T, will e use severl times in the sequel, e.g., in the proof of Theorem 7. 7
x y, v 0 v v D v D v D Figure 7: fourth se of Theorem : the egree of v D is equl to n the egree of v D is t lest Trees T of orer n hieving w(t ) = n In this setion, we hrterize the trees T with n verties n w(t ) = n. Our stuy is ivie into three ses, n = k, n = k + n n = k +. We first efine some prtiulr trees, of orer to 5, tht we nme ggets. For eh gget, we hoose one or two prtiulr vertex(ies) nme ining vertex(ies), through whih the ifferent ggets will e exlusively onnete etween themselves; vertex whih is not ining vertex is si to e orinry. In the sequel, we will sometimes enote gget of orer i y gi, i 5, n use the revition. v. for ining vertex. The ggets re epite in Figure 8; we represent the ining verties with squres n orinry verties with irles. We will use the following esy lemm, whose proof we omit. Lemm 6 Let T e tree of orer, n v n v e two istint verties in T. It is possile to wth T with one wther lote t v n one wther lote t v. As onsequene, if T is tree of orer n x is lef of T, there exists set W of two wthers suh tht T \ {x} is wthe y W n x is overe y W. The following theorem hrterizes the trees T with orer n = k n w(t ) = k. Theorem 7 Let T e tree of orer n = k for k. We hve: w(t ) = k T n e otine y hoosing k ggets of orer n joining these ggets y their ining verties to otin tree. The tree T 5 in Figure 9 is n exmple of tree rehing this mximum. Proof. Assume tht tree T of orer n = k is otine y hoosing k ggets of orer n joining these ggets y their. v. s to form tree. It is ler tht, to wth T, it is neessry to lote two wthers on eh gget. So T rehes the oun k. 8
one gget of orer type type one gget of orer two ggets of orer type type three ggets of orer type type type type type type e five ggets of orer 5 Figure 8: ll the ggets Figure 9: the tree T 5 9
,,,,,,,,, Figure 0: the trees of orer 6 for the proof of Theorem 7 Figure : two representtions for g of type or We will prove the onverse y inution on k. For k =, it is immeite. We lso exmine the se k =, tht is to sy n = 6. We rw on Figure 0 the six ifferent trees T on six verties; when tree is not of type esrie in the right prt of the equivlene, we expliitly give the wthers showing tht w(t ) = n, in the other ses, we simply inite the. v. s of the two ggets involve. We will sometimes represent g of type or with tringle, s on Figure : she ege mens tht the ege my exist or not, with lwys extly two eges in eh g. A wther inite insie the tringle mens tht this wther is lote t one of the three verties of the tringle, t n pproprite vertex oring to the se. We ssume now tht k n tht the theorem is true for k < k. Let T e tree of orer n = k with w(t ) = k. We onsier gin the proof of Theorem using pth v 0, v, v,..., v D, v D of length D, where D is the imeter of T. Here, the thir n fourth ses re impossile, euse they imply tht w(t ) < n = k, unless n = 6, whih hs just een elt with. In the first se of Theorem, we renme y,, n respetively, the verties v D, v D, x n v D ; in the seon se, we renme y,, n respetively, the verties v D, v D, v D n v D ; in oth ses, we remove the verties, n from T 0
g,, 5 5, 5 Figure : k wthers re suffiient in T (en of proof of Theorem 7) n otin tree T of orer (k ); y Remrk 5, it ppers tht T nees t lest w(t ) = k wthers n so w(t ) = (k ) n we n pply the inution hypothesis to T : the vertex elongs to g, g. Assume tht is not the ining vertex of g. The. v. of g is jent to the. v. of nother g in T (f. Figure ). By Lemm 6, we n lote wthers w n w t n, so tht is overe y w n is overe y w ; it is then possile to wth T with only one wther lote on the gget g, s we n see on Figure, y hoosing the pproprite vertex of g t whih we lote the wther enote y. This les to ontrition on w(t ), n shows tht is the. v. of g, in whih se the result is immeitely otine, sine {,, } n e seen s g, with its. v. in, onnete to. The following lemmt n efinition will e use repetely in the sequel. Lemm 8 Let T e tree of orer 5 n v e vertex of T. It is possile to wth T with three wthers, one of the three wthers eing lote t v. As onsequene, if T is tree of orer 6 n x is lef of T, there exists set W of three wthers suh tht T \ {x} is wthe y W n x is overe y W. Proof. The result for T is strightforwr, y exmining ll the ifferent possiilities, s we n see on Figure ; the onsequene on T is immeite. Lemm 9 Consier g5 with ining vertex n orinry verties v, x, y n z; there exists set W of two wthers suh tht {x, y, z} is wthe y W ; the vertex v is overe y W. Proof. If the g5 is of type,,, or, then the four verties v, x, y, z form tree, n y Lemm, we re one. If the g5 is of type e, then it is lso possile, with two wthers lote t, the entre of the str, to wth {x, y, z} n over v.
or or or,,,,,, Figure : illustrtion for Lemm 8 5,5, 5,,,, v, v,, v 5, () () Figure : illustrtion for Lemm () Definition 0 Let H = (V (H), E(H)) e onnete grph n v e vertex in V (H); let H e the grph otine y removing the vertex v from H (H is onnete or not). We sy tht v is free of hrge, or free, in H if there exists minimum wthing system for the grph H whih is lso wthing system for H. Lemm Let p e n integer verifying p. Let F e forest otine y hoosing p ggets of orer or 5 n possily, if esire, y ing eges etween the ining verties of the p ggets. Let v e new vertex, whih is jent to t lest one ining vertex n nnot e jent to orinry verties; we ssume tht the grph otine y ing v to F is tree, T. Then, the vertex v is free in T. Proof. If v is jent to only one. v., let e this vertex; sine T is onnete n p, the vertex is jent to nother. v.,. If v is jent to t lest two. v. s mong the p ggets, let n e two suh verties. Figure illustrtes the lemm in etil in three ses: () v is linke to the. v. of g5 n is linke to the. v. of g; () v is linke to the. v. of g n is linke to the. v. of g; () v is linke to the. v. s of g5 n of g. The other ses, using repetely Lemmt 6 n 8, n e trete extly in the sme wy. We re now rey to hrterize the trees T with orer n = k + n w(t ) = k +.
one g5 n four g s one g n five g s Figure 5: the trees T 7 n T 7 Theorem Let T e tree of orer n = k + for k. We hve: w(t ) = k + T n e otine y hoosing one gget of orer n k ggets of orer, or one gget of orer 5 n k ggets of orer, n joining these ggets y their ining verties to otin tree. The trees T 7 n T 7 of Figure 5 re exmples of trees whih hieve this mximum. Proof. Assume tht tree T of orer n = k + is otine y hoosing one g n k g s, or one g5 n k g s, n finlly joining these ggets y their ining verties, in orer to otin tree. It is neessry to lote one wther on g, two wthers on g n, euse g5 hs four orinry verties, three wthers on g5. So T hieves the oun k + : if there is g, we nee one wther for the g n k wthers for the k g s, if there is g5, three wthers for the g5 n k wthers for the k g s. We will prove the onverse y inution on k. For k =, n = 5 n the result is ler, see Figure : T is g5 (n in two out of three ses, it n lso e seen s the onnexion of g n g). We ssume now tht k n tht the theorem is true for k < k. Let T e tree of orer n = k + with w(t ) = k +. We onsier gin the proof of Theorem, using pth v 0, v, v,..., v D, v D of length D, where D is the imeter of T. Prt (): we ssume tht we re in the first or seon se in the proof of Theorem In the first se, we renme y,, n respetively, the verties v D, v D, x n v D ; in the seon se, we renme y,, n respetively, the verties v D, v D, v D n v D ; we remove the verties, n from T n otin tree T of orer (k ) + ; y Remrk 5, it ppers tht T nees t lest w(t ) = k wthers n so w(t ) = (k )+ n we n pply the inution hypothesis to T : T is of one of the two types esrie in the right prt of the equivlene, n the vertex elongs to gget g, whose. v. we enote y. Assume first tht. (i) If g is of orer, the sutree inue y the verties of g n the verties, n yiels g5 of type or, n the result is prove for T.
g one g n one g5 (of type or ) g two g s n one g Figure 6: ses for n = 8 in prt () of Theorem, when g is of orer, 5, 6 5 γ 5 6 5 5,, 5, 6,, 5 Figure 7: ses for n in prt () of Theorem, when g is of orer (ii) Assume next tht g is of orer. If T is of orer 8, the two possiilities re given y Figure 6. If T is of orer t lest, there exists in T gget g onnete to, n we enote y the. v. of g. If g is g, whih is itself onnete to nother g with ining vertex γ, then the left prt of Figure 7 shows how to use only one wther for g, whih les to ontrition on w(t ); the sme is true if g is g, see the right prt of Figure 7, whih tully is the sme s Figure ; the se when g is g5 works similrly, using Lemm 8. So the only se left is when g is g onnete only to g, n Figure 8 shows how to solve it. (iii) Finlly, ssume tht g is of orer 5. If T is of orer 8, the reer will onvine himself tht loting t ll the ifferent verties, exept t the. v., of ll the ifferent types for g5 les to the six ptterns given y Figure 9. If T is of orer, the. v. is jent to the. v. of g. When one exmines the ifferent possiilities, it ppers tht if T rehes the oun, it is of the wishe shpe: this is shown y Figure 0. To lose the se when g is of orer 5, we stuy the se when T is of orer t lest ; then the tree T otine from T y removing the four verties of g other thn hs orer t lest 7 n we n pply Lemm to it, whih shows tht the vertex is free in T. Using Lemm 9, we n use two wthers on g to wth V (g) \ {, } n over the vertex. With one
(),,, 5 one wther n e sve 5 () () one g5 n g s one g n g s Figure 8: more ses for n in prt () of Theorem, when g is of orer one g n one g5 one g n one g5 two g s n one g two g s n one g one g n one g5 one g n one g5 Figure 9: ses for n = 8 in prt () of Theorem, when g is of orer 5 5
, 5,, 5, 6 6 5 5,, oes not reh the oun oes not reh the oun, 5, 5, 6 6 5,,, 5 oes not reh the oun oes not reh the oun two g s n one g5 two g s n one g5 Figure 0: ses for n = in prt () of Theorem, when g is of orer 5 6
v v D v v D vd D D v D v D g g γ x Figure : illustrtion for prt () of Theorem wther t overing, we n seprte from ll the other verties: so, we n o with only two wthers on g, n T oes not hieve the oun. This shows tht if, then either the tree oes not hieve the oun, or it is of the esire form. On the other hn, if =, then the result is immeitely otine. This ens prt (). Prt (): we ssume tht we re in the thir or fourth se in the proof of Theorem If we re in the thir se, we remove the verties x n y n if we re in the fourth se, we remove the verties v D n v D ; we otin tree T of orer k. By Remrk 5, we hve w(t ) = k n Theorem 7 my e use: T n e otine s olletion of g s linke y some eges etween their ining verties. So, the vertex v 0 is lef of g, g; now we reverse the longest pth v 0, v,..., v D in T. If g is of type (see the left prt of Figure ), then v D is linke to only one. v., v D, n hs egree, euse D is the imeter of the tree, n we re rought k to the first se. An if g is of type, then v D hs egree, n either v D hs egree n we re in the seon se, or v D hs egree t lest n we re in the fourth se, with t lest one. v. x linke to v D n x of egree t lest (see the right prt of Figure ); however, x nnot e linke to nother.v. γ, sine this woul inrese the imeter of the tree, n for the sme reson the g of x is of type, so tht neessrily x hs egree. With x plying the prt of v D, we re gin in the first se. In ll ses, we n re-use the result otine in prt (). The lst se, n = k + n w(t ) = k, offers the gretest numer of possiilities for the ggets. Theorem Let T e tree of orer n = k + for k. We hve: w(t ) = k T n e otine y hoosing (i) two ggets of orer n k ggets of orer, (ii) or one gget of orer, one gget of orer 5 n k ggets of orer, 7
two g s, one g n one g5 three g s n one g Figure : the trees T, T n T (iii) or two ggets of orer 5 n k ggets of orer, (iv) or one gget of orer n k ggets of orer, (v) or one gget of orer n k ggets of orer, n joining these ggets y their ining verties to otin tree. It my e tht one of the two ining verties of g is not linke to ny (ining) vertex. The trees T, T n T of Figure re exmples of trees hieving the oun k for n = k + (with k = ). Proof. Assume tht tree T of orer n = k + is otine s speifie in the right prt of the ove equivlene. It is neessry to lote one wther on g, two wthers on g n two on g (euse g hs two orinry verties), n three wthers on g5. So T rehes the oun k: if we re in (i), ( ) + ((k ) ) = k; in (ii), ( ) + ( ) + ((k ) ) = k; in (iii), ( ) + ((k ) ) = k; in (iv), ( 0) + (k ) = k; in (v), ( ) + ((k ) ) = k. We will prove the onverse y inution on k. For n = 7, the ifferent possiilities re exmine on Figure. Now, we ssume tht n 0. We use the sme sheme of proof s for Theorem : we ssume tht k n tht the theorem is true for k < k, we let T e tree of orer n = k + with w(t ) = k, n we onsier the proof of Theorem, using pth v 0, v, v,..., v D, v D of length D, where D is the imeter of T. Prt (): we ssume tht we re in the first or seon se in the proof of Theorem If in T there is g, n if only one of its ining verties is onnete to nother gget, this g n e viewe s two g s or s one g n one g. So we n ssume tht, if there is g, i.e., if we re in se (v) of the theorem, then eh of the two. v. s of the g is onnete to t lest one g. In the first se, we renme y,, n respetively, the verties v D, v D, x n v D ; in the seon se, we renme y,, n 8
() () () two g s n one g () two g s n one g one g n two g s (e) one g n one g5 one g n two g s,,,,, three wthers lote t the entre re suffiient for the str Figure : ll the possiilities for n = 7 in Theorem respetively, the verties v D, v D, v D n v D. In oth ses, we remove the verties, n from T n otin tree T with orer (k ) + ; y Remrk 5, it ppers tht T nees t lest w(t ) = k wthers n so w(t ) = (k ): we n pply the inution hypothesis to T, whih is of one of the five types esrie in the right prt of the equivlene. The vertex elongs to gget g; s efore, if is ining vertex, we re one, so we ssume from now on tht is orinry, so tht g is of orer or more, n we hve four ses, oring to the orer of g. () If g is of orer, the sutree inue y the verties of g n the verties, n yiels g5 n the result is prove: inee, if T hs two g s n k g s (se (i)), or one g, one g5 n k g s (se (ii)), then T n e otine with one g, one g5 n k g s (se (ii)), or two g5 s n k g s (se (iii)), respetively. () Assume next tht g is of orer, with ining vertex. If n = 0, we onsier for T the ses in Figure where there is t lest one g, tht is, the ses () (). The ses () n (), where there is g, will e stuie elow, for generl vlues of n. If, in the se (), g is of type, then, see Figure (), five wthers re suffiient to wth T, wheres if g is of type, then, oring to the lotion of in g, T onsists of two g5 s, or of one g, one g n one g5, see Figure () n (). Similrly, if in the se () of Figure, g is of type, then five wthers re suffiient, wheres if g is of type, then T onsists of one g, one g n one g5, or of two g s n two g s f. Figure 0() elow. We onsier now the se when there is g, with vertex δ, in T, with n 0: we re in se (iv) of Theorem n ll the other ggets in T re g s. If δ is linke neither to nor to ny neighour of, we re in the sitution epite y Figure 5() n k wthers re suffiient for T : sine γ or δ is linke to nother g, these two verties n e seprte y nother wther. So we n ssume from now on tht δ is linke either to 9
, 5 (),,, T ontins two g s n one g of type, five wthers re suffiient in T () T ontins two g s n one g of type two g5 s () T ontins two g s n one g of type one g, one g, one g5 Figure : ses for n = 0 in prt () of Theorem, when g is g or to one of its neighours. First, we ssume tht is linke to t lest one g, f. the left prt of the tree in Figure 5(). Agin, we n sve one wther, so tht T oes not hieve the oun k, unless we re in one of the following three ses: (i) δ is linke only to, see Figure 5(). Then either δ is not overe y ny wther, or it is overe y the wther lote t, in whih se no wther seprtes e n δ. This gives the three possiilities etile in Figure 6. (ii) δ is linke to n to extly one other g, whih is not linke to ny other g, n the wther nnot e lote t, whih mens tht g is of type, see Figure 5(), where γ n δ re not seprte. Then there re two possiilities, given in Figure 7. (iii) δ is linke to neighour of, n neither δ nor is linke to other. v. s, n g is of type, see Figure 5(), where n f re not seprte. Figure 8 shows then tht we still re in the onitions of Theorem. Now we n ssume tht is not linke to ny g, whih mens tht it is linke to δ, whih in turn is linke to t lest one g; then g ppers, ontining g n δ, n with ining verties n δ. So the se when there is g in T is lose, lso ompleting the se n = 0. From now on, we ssume tht n n tht there is no g in T. We n remrk the following: if in T we hve g, n if one of its ining verties is onnete to g, it is lwys possile to lote w(t ) wthers on T with one wther t the seon ining vertex v of the g, see Figure 9. Sine we hve require tht eh of the two. v. s of the g 0
() 5,, sving one wther δ g 6 6 γ 7 6 7 6, 7 () (),, 5 e 5 7 6, δ, 6 g δ g 7 γ filing to sve one wther filing to sve one wther 6 6, 7 f 5,, 5 5 (), 5 δ g, filing to sve one wther Figure 5: prt () of Theorem : g is g n there is g one g n g s one g n g s two g s n g s Figure 6: ses with g in prt () of Theorem, when g is g
one g, one g5 n g s two g5 s n g s Figure 7: more ses with g in prt () of Theorem, when g is g two g5 s n g s one g, one g5 n g s Figure 8: lst ses with g in prt () of Theorem, when g is g
type, v,, v, type type v,, v,, type Figure 9: hoie of wther t ining vertex of g f γ () one g, one g5 n g s OR two g s n g s () two g5 s n g s OR one g, one g5 n g s Figure 0: ses with two g s in prt () of Theorem, when g is g e onnete to t lest one. v. of g, this mens tht we n lote one wther t one. v. of the g in orer to possily over. This remrk or Lemm 8 or 6 llows us to sve one wther on g when there is one g or (t lest) one g5 in T, s we i in the left prt of Figure 5(), with overing. Therefore, we hve only one se left when g is g: when T ontins t lest one gget of orer. If there is extly one g (se (ii) of Theorem ), the sitution is very lose to tht of Theorem (see Figure 8 n the left prt of Figure 7), the ifferene eing the existene of g5. So we ssume tht T ontins two g s, n g s (se (i) of Theorem ). In generl, one n still sve one wther on g; the two ritil situtions re given y Figure 0, in whih single wther lote on g nnot simultneously over, f n (n γ when γ is onnete to ) f. Figure. () Assume now tht g is of orer 5, with ining vertex ( ). The other ggets in T re either one g n k g s, or one g5 n k g s. We illustrte the ses ourring when in T, is linke to the g n only to this gget: Figure is for n = 0 n uses the only
,, 5,, Figure : se n = 0 in prt () of Theorem, when g is g5: five wthers re suffiient representtion with one g5 for tree of orer 7, f. Figure (e); Figure is for n n is otine y loting t ll the ifferent verties, exept t the. v., in ll the ifferent types for g5, f. Figure 0. The other ses, very similr to Figure or to previous stuies involving g5 s, often use Lemm 9 n re left to the reer. () The finl se of this prt () is when g is g, with its two ining verties n x (, x ) onnete to other ggets, n in T ll ggets exept g re g s (se (v) of Theorem ). Denote y T the onnete omponent ontining x in the forest otine from T y removing the eges of g, see Figure. Assume tht T is of orer t lest 7; then y Lemm, x is free in T. In minimum wthing system for T, we n ssume tht there is one wther lote t whih overs, n one wther lote t whih overs. Then (see Figure ), either only one more wther, enote y, is neessry insie g to wth T, n T oes not reh the oun k, or T onsists of one g, one g5 n g s, or of two g5 s n g s. The sme rgument with shows tht we n ssume tht eh. v. of g is linke to extly one g; then n = n Figure 5 gives ll the possile ses. Prt (): we ssume tht we re in the thir or fourth se in the proof of Theorem If we re in the thir se, we renme y, n respetively, the verties x, y n v D, n if we re in the fourth se, we renme y, n respetively, the verties v D, v D n v D ; in oth ses, we remove n, otining tree T of orer (k ) +. By Remrk 5, w(t ) = (k ) + n Theorem my e use: T n e otine y hoosing one g or one g5 n olletion of g s linke y their ining verties. Note tht the vertex hs egree t lest in T ; it elongs to gget g with. v.. We first ssume tht. Beuse of the egree of, the gget g nnot e of orer, n if it is of orer, with vertex set {,, }, then its ege set is {{, }, {, }}, n {,,,, } is g5 of type or : T is of the esire form. We re left with the se when g is g5, in whih hs
5 5, 6 5, 5 5 5, 6, 5 7 7 6,, 6,, oes not reh the oun oes not reh the oun, 6 5, 5 5 5 6, 7 7,, 6 5, 5,, 6 oes not reh the oun oes not reh the oun one g n g s Figure : ses when g is g5 in prt () of Theorem, for n T" x g Figure : the efinition of T in prt () of Theorem 5
g is of type g is of type x x,,,, g is of type g is of type T,, x x, t lest 7 verties is lso free in T g is of type x only one g one g, one g5 n g s, g is of type g is of type T x x t lest 7 verties,, is lso free in T g is of type x only one g two g5 s n g s g is of type g is of type x, x,,, Figure : prt () of Theorem : T is of orer t lest 7 n x is free in T 6
two g5 s n one g initil g one g, one g5 n two g s, x x,,,,, the tree oes not reh the oun: seven wthers re suffiient Figure 5: the remining ses when g is g in prt () of Theorem egree or more, n the other ggets in T re ll g s; this is epite in Figure 6, where we give the lotions of the wthers showing tht T oes not reh the oun k, or show the. v. s of the ggets involve; note tht if n, then y Lemm, is free in T eprive of the four orinry verties of g. Finlly, if =, then Figures 7 9 give the ifferent ses, oring to the orer of g. This ompletes the proof of Theorem. Grphs G rehing the mximum vlue of w(g) We first give the following efinition. Definition A onnete grph G is si to e mximl if, when we ny ege to G, we otin grph G verifying: w(g ) < w(g). We enote y ω(n) the mximum minimum numer of wthers neee in onnete grph of orer n, i.e., ω(n) = mx{w(g) : G onnete of orer n}. In the previous setion, we hve estlishe tht ω(n) = n for n / {,, }, n we hve hrterize the trees of orer n rehing ω(n). In this setion, we wnt to esrie ll the mximl onnete grphs of orer n whih reh ω(n). Using Lemm, the grphs of orer n whih reh ω(n) re extly the onnete prtil grphs of the mximl onnete grphs of orer n rehing ω(n). We rell tht K p enotes the omplete grph (or lique) of orer p. Agin, we ivie our stuy into three ses, n = k, n = k + n n = k +. 7
type type type n =0: two g5 s n>0: oun not rehe n=0: one g, one g, one g5 type type or type n>0: oun not rehe one g n g s n=0: two g5 s type type type n=0: one g, one g, one g5 n>0: oun not rehe oes not reh the oun type type oes not reh the oun oes not reh the oun Figure 6: illustrtion for prt () of Theorem, when g is g5 n one g n g s in T one g n g s Figure 7: illustrtion for prt () of Theorem, when = n g is g 8
OR one g or one g5, n g s in T one g5 ppers one g ppers Figure 8: illustrtion for prt () of Theorem, when = n g is g one g5 of type or, n g s in T one g n g s one g5 of type n g s in T two g s n g s one g5 of type n g s in T one g5 of type e n g s in T : the tree,, one g n g s,,, T oes not hieve the oun one g5 of type e n g s in T : g ppers Figure 9: illustrtion for prt () of Theorem, when = n g is g5 9
Figure 0: G 5, the mximl grph of orer 5 rehing the oun Theorem 5 Let k e n integer, k, n G e mximl grph of orer k. We hve: w(g) = k G is otine y tking olletion of k K s, hoosing one vertex nme ining vertex in eh K, n onneting these k ining verties y K k. For instne, the grph G 5 of Figure 0 is the unique mximl grph of orer 5 rehing the oun ω(5) = 0. Proof. The implition from the right to the left is iret. So, given mximl grph G of orer k stisfying w(g) = k, we hve to prove tht G is of the form esrie in the theorem. Let T e spnning tree of G. Using Lemm n Theorem, we n see tht w(t ) = k. By Theorem 7, T is olletion of k ggets of orer onnete y their ining verties. We shll show tht in G ny ege whih is not in T is n ege etween two. v. s of T, or is the missing ege of g; to o this, we ssume tht there is in G n ege e whih is not n ege etween two. v. s of T, nor the missing ege of g. In Figure, we onsier the four possiilities: () The ege e links n orinry vertex of g, enote y g, whose. v. is enote y, n the. v. of nother g, n the ege {, } exists; then, whtever the type of g, we n lote wther on g overing, n, n the six verties re overe n seprte y three wthers only. () e links two orinry verties of two g s whih re linke y their. v. s. Agin, the six verties involve n e wthe y three wthers. In pssing, these two ses show how to hnle the se n = 6, so from now on we ssume tht n 9. () e links n orinry vertex of g, whose. v. is, n the. v. of nother g, n {, } oes not exist. Then n re linke to t lest one other g (possily the sme), euse in the spnning tree T, there is onnexion etween ny two. v. s. () This is lso true when e links two orinry verties of two g s whih re not linke y their. v. s. In eh of these two ses, we n see tht we re le to lote only one wther on g, so there is ontrition with the vlue of w(g). 0
n, e on the three verties: n n, e, n,, () () () (),, δ 6 e 5, 5 γ γ = δ ( = 5) is possile e, γ 5 Figure : forien eges etween two g s in the proof of Theorem 5 Furthermore, if we to T the missing ege on eh g n ll the missing eges etween the. v. s of T, the numer of neee wthers remins equl to k: we hve otine the unique mximl grph ontining T. Theorem 6 () Let k e n integer, k, n G e mximl grph of orer k +. We hve: w(g) = k + G is otine y tking olletion of k K s n one K, or k K s n one K 5, hoosing one vertex nme ining vertex in eh of these omplete grphs, n onneting these ining verties y K k+ if we hve tken K, n y K k if we hve tken K 5. () If G is mximl grph of orer 8, then we hve: w(g) = 5 G is the grph given y Figure, or G is otine y following the rules given in Cse (), for k =. () The only mximl grph G of orer 5 with w(g) = is the lique K 5. For instne, the grphs G 7 n G 7 of Figure re the two mximl grphs of orer 7 rehing the oun ω(7) =. Proof. The implitions from the right to the left re iret. So, given mximl grph G of orer k + stisfying w(g) = k +, we hve to prove tht G is of the form(s) esrie in the theorem. By inequlity () from the Introution n Theorem, ll onnete grphs G of orer 5 re suh tht w(g) =, K 5 is the unique mximl grph of orer 5, n Cse () is true. The se n = 8, whih oes not fit the generl frmework either, is rther teious to hek, n is not given here. We ssume from now on tht n. Let T e spnning tree of G. Using Lemm n Theorem, we n see tht w(t ) = k +. From Theorem, T n e otine s one g or one g5 plus olletion of g s,
Figure : the two mximl grphs of orer 7 rehing the oun Figure : mximl grph of orer 8 rehing the oun with the ggets onnete y their ining verties to form tree. If, mong the spnning trees of G, there is one with g5, we hoose this tree; n if, in ll the spnning trees, we nnot voi g, then we hoose tree in whih the. v. of the g hs mximum egree (in the tree). We shll list pirs of verties whih nnot e jent in the mximl grph G: etween g s, etween the g5 n g, n etween the g n g (the most elite se). () Assume first tht there is n ege etween two g s, with t lest one of its ens ifferent from. v. This se hs een trete for Theorem 5, f. Figure. If now, the. v. of g, is not linke to ny. v. other thn, or if is linke to the. v. of g other thn, then we n sve one wther in extly the sme wy s on Figure. If is linke to the. v. γ of the g5, y Lemm 8 we n hve wther lote t γ n overing, thus still sving one wther on g. So we n ssume tht is linke to the. v. γ of the g. In ses () n () of Figure, we n sve one wther on the g with. v., sine n ply symmetril prts. In se (), in ll ses, ut one, we n still sve one wther on g : the ritil se (see Figure ) is when the g hs no onnexion other thn, n moreover the wther, whih is use to over, nnot e lote t, so tht the two verties of the g re not seprte, n we nnot sve one wther; in this se however, sine the. v. s,, γ, δ,... in T re onnete, it is possile to in T the ege e = {, } n elete one ege etween two. v. s, so tht the result is spnning tree of G, in whih γ eomes n orinry vertex in g, n eomes the. v. of the g, now onnete to two g s. This mens tht the spnning tree in the left prt of the figure nnot hve een hosen, sine the. v. of its g, γ, oes not hve
n,, 5 e γ 5, 5 5 δ δ e γ Figure : forien eges etween two g s: ritil se in prt () of Theorem 6 g 5 e g 5, n 5, 6 n 6, 6 e 6, 5 5 γ, 5 5 γ, 5 three wthers in g 5 two wthers in g 5 Figure 5: prt () of Theorem 6: forien eges etween g5 n g mximum egree mong the spnning trees of G. Cse () of Figure n e elt with in the sme wy, with ritil sitution similr to Figure, where we n the ege {, } n elete the ege {, }. () Assume next tht there is one g5, nme g 5, in T, n tht there is in G n ege e etween g 5 n g, g, with t lest one of its ens ifferent from. v. Let n e the. v. s of g 5 n g, respetively. If the ege {, } oes not exist, then n re onnete to t lest one other g (possily the sme), n we n sve one wther, using in prtiulr Lemm 8: see Figure 5, where n e equl to in the left prt. In the right prt, sine is free in the tree T onsisting of the spnning tree T eprive of the four orinry verties of g 5 (even if is linke only to γ, in whih se is overe y the wther 5), we re left with the prolem of tking re of the three verties x, y, z of g 5 other thn n, with only two wthers; this n e one using Lemm 9. So from now on we ssume tht we hve the ege {, } in T. Beuse n, is still free in T, n oviously, if oth n re still onnete to other g s, the rgument ove still works. So we ssume tht only one of n is onnete to (t lest) one (other) g. We first onsier the se when it is. Figure 6 epits the sitution, where n e equl to in the left prt. In this left prt, thnks to Lemm 8, the sitution is the sme s previously (without the ege {, }). An if e links n (see the right prt), then,
g 5 e g 5, n, 6 e, 5 5 γ, 5 5 6 γ Figure 6: prt () of Theorem 6: forien eges etween g5 n g, with {, } in T g 5 e g 5 δ,, 6 e 6 6 Figure 7: prt () of Theorem 6: more forien eges etween g5 n g, with {, } in T still enoting y x, y n z the verties in g 5 other thn n, we n use Lemm 9: two wthers re suffiient to wth {x, y, z} n over, so tht n re now seprte y wther. Thus, one wther n e sve on g 5. We n now ssume tht is linke to no gget other thn g 5. Then the sitution is esrie y Figure 7, with free in T, n or = in the left prt (in the ltter se, lote the wther t ). When is one extremity of e, we use Lemm 9 n sve one wther on g5, so we re left with the se e = {, } with (see the right prt of the figure), whih is solve lso using Lemm 9 n sving one wther on g. () We finlly stuy the se when there is one g, nme g, with. v. n orinry vertex, in the spnning tree T. The sitution is now slightly ifferent from the previous ses, euse we my, without ontrition, hve in G n ege etween, for instne, n vertex of g, sine T my hve een originlly proue from K 5 in G. We onsier in T g nme g, with. v., n investigte whih ege(s) n exist in G etween g n g. First, we ssume tht {, } is not in T. Then in T, is linke to the. v. γ of g, n is linke to the. v. δ of g, possily with γ = δ. Using Lemm 6, we lote wthers t γ n δ, n Figure 8()() shows how to routinely sve one wther on g when in G there is n ege etween or n n orinry vertex of g, even if the wthers n oinie. Assume now tht it is the ege {, } whih is in G. If in T,
,, γ γ, δ,,, () () δ () γ γ Figure 8: prt () of Theorem 6: forien eges etween g n g neither nor γ is onnete to ny g, we re in se () of Figure 8 n we onsier tht there is g5 of type or in T rther thn g. So either or γ is onnete to g, with. v. φ. If γ δ, then φ = δ is possile, or (if φ is not linke to ) φ = ; if γ = δ, then φ = is possile (if φ is not linke to ). All this is epite in Figure 9, where it n esily e seen how to sve one wther on g in ll ses; we give only the full esription of the lst se, (e). So we hve just estlishe tht if is not onnete to the. v. of g, then there exists no ege etween this g n g in G. Wht hppens now if is onnete to the. v. of g, g, tht is to sy if there is the ege {, } in T? If in T, is still linke to the. v. γ of g (with γ ) n is still linke to the. v. δ of g (with γ δ euse there is no yle in T ), we n re-run the rgument use in the sene of {, }: the first two ses of Figure 8 re extly the sme with or without {, }, n in the thir se, we hve the eges {, } n {, } in G, from whih we n still pik spnning tree with g5; n, euse φ n φ δ, Figure 9 reues to its first se (), whih n e trete similrly. Therefore, in T, either is not linke to the. v. of ny g other thn, or is not linke to the. v. of ny g. If in T, is linke to only, then we hve seen tht no ege exists in G etween g n ny g other thn g. But eges n exist etween g n g, n inee, we n ll the missing eges etween these two ggets, plus the missing ege in eh g, plus ll the missing eges etween the. v. s of T, the numer of neee wthers remins equl to k +, n we hve otine the only mximl grph ontining T, whih is of the form esrie in the theorem; see Figure 50. Note tht in some ses, the rgument of the hoie of g5 in T n lso e use, for instne if g is of type n there is the ege {, }. If is linke only to n if is the only. v. whih is linke only to, then ny g other thn g, with. v. γ, n e linke to g uniquely 5
() φ γ δ φ γ δ () γ δ γ φ δ φ () γ φ δ () φ δ γ φ δ γ (e) n, n, δ,, γ φ,, Figure 9: prt () of Theorem 6, with the ege {, } not in T : () γ δ, φ δ, φ ; () γ δ, φ = δ; () γ δ, φ = ; () γ = δ, φ ; (e) γ = δ, φ =. G mximl: Figure 50: prt () of Theorem 6, possile eges etween g n g: tully, eges insie K 5 6
γ γ γ Figure 5: prt () of Theorem 6: is the only ining vertex linke only to () γ (),,,,, Figure 5: prt () of Theorem 6: the ining verties of (t lest) two g s re onnete only to through the ege {, γ}, n so in G, the possile eges etween the orinry vertex of g n g must ffet g only. In the first two ses in Figure 5, g5 shoul hve een tken when hoosing T, or, s in the thir se n s in the previous figure, we n ll the eges etween g n g n otin K 5. So we re left with the se when there re two (or more) g s with. v. s linke only to in T, see Figure 5(). If in G there is the ege {, }, {, } or {, }, then gin spnning tree with g5 oul hve een hosen, n if there is the ege {, } or {, } n neither {, } nor {, }, we n to T ll the eges etween g n g in orer to otin K 5 in mximl grph. So the only possiility not rule out yet is if there re the eges, sy, {, } n {, } (more eges in G n only help). Then Figure 5() shows how to sve (t lest) one wther, y loting two wthers t. Now we re in position to onlue. If in T there re eges etween orinry verties of ifferent ggets or etween the. v. of gget n n orinry vertex of nother gget, then nother spnning tree shoul hve een hosen, ontining g5 inste of g, or ontining g with ining vertex of higher egree, or these eges re prt of K 5, or we n sve wthers. Furthermore, if we to T the missing ege on eh g, the missing eges on the possile g5, n ll the missing eges etween the. v. s in T, the numer of neee wthers remins equl to k+: we hve otine the only mximl grph ontining T. The proof of Theorem 6 is omplete. 7
Figure 5: the three new ggets of orer 7 The proof of the previous theorem, for n = k+, is not very enourging in view of the se n = k +. Inee, lthough we hve some insight into the sitution, we n only onjeture the following result, in whih, to esrie the grphs, we nee three new ggets of orer 7 (whih re not trees), with one or two ining vertex(ies), see Figure 5. Conjeture 7 Let k e n integer, k 6, n G e mximl grph of orer k +. We hve: w(g) = k G is otine y: (i) tking two K s n k K s, (ii) or tking one K, one K 5 n k K s, (iii) or tking two K 5 s n k K s, (iv) or tking one K n k K s, (v) or tking one g7 n k K s, hoosing one vertex nme ining vertex on eh of the omplete omponents K i, exept on K for whih we hoose two ining verties, tking for the g7 one or two ining vertex(ies) oring to its type, n onneting these ining verties to form omplete grph with them. The grphs of Figure 5 re grphs of orer 9 rehing the oun ω(9) = : () with one K, one K 5 n four K s; () with one K n five K s; () with one g7 n four K s; oring to Conjeture 7, they woul e mximl. For n = k+ with k 5, there re mximl grphs neeing k wthers whih re not of the form esrie in the onjeture. We give ertifie exmple for n = 6 in Figure 55. Referenes [] D. Auger, I. Chron, O. Hury, A. Lostein, Systèmes e ontrôle ns les grphes : une extension es oes ientifints, Rpport Interne Téléom PrisTeh (in English), forthoming. 8
() () () Figure 5: three grphs rehing the oun ω(9) = Figure 55: mximl grph rehing the oun ω(6) = 0 9
[] D. Auger, I. Chron, O. Hury, A. Lostein, Wthing systems in grphs: n extension of ientifying oes, forthoming. [] I. Chron, O. Hury, A. Lostein, Extreml rinlities for ientifying n loting-ominting oes in grphs, Rpport interne Téléom Pris-00D006, Pris, Frne, 8 pges, 00. [] S. Grvier, J. Monel, On grphs hving V \ {x} set s n ientifying oe, Disrete Mth. 07 (007),. [5] I. Honkl, A. Lostein, On ientifition in Z using trnsltes of given ptterns, J. Universl Comput. Si. 9(0) (00), 0 9. [6] M. G. Krpovsky, K. Chkrrty, L. B. Levitin, On new lss of oes for ientifying verties in grphs, IEEE Trns. Inform. Theory (998), 599 6. [7] http://www.infres.enst.fr/ lostein/ilocdometid.html [8] P. Rosenhl, On the ientifition of verties using yles, Eletron. J. Comin. 0() (00), R7. 0
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