0 7 8 9 6 1 7 8 5 0 2 7 4 6 1 3 0 6 5 4 7 1 0 6 8 7 2 1 9 8 7 3 1 9 8 7 3 2 9 8 5 4 3 9 2 3 4 5 4 5 6 0 6 0 1 2 9 5 0 2 4 7 8 6 1 3 1 3 5 2 9 2 4 3 8 9 3 4 7 8 9 5 8 6 8 7 1 2 3 7 0 9 8 2 3 4 6 1 6 9 3 4 5 0 7 1 0 4 5 6 1 9 3 2 5 6 0 2 8 5 4 6 0 1 7 6 0 1 2 0 1 7 8 9 2 3 8 9 7 4 5 9 7 8 4 6 5 0 6 1 0 2 1 3 2 4 3 5 8 9 7 8 9 7 COM BIN ATOR I A L MAT HEM A TIC S YEA R February 1969 - June 1970 A GEOMETRIC CHARACTERIZATION OF THE LINE GRAPH OF A SYMMETRIC BALANCED INCOMPLETE BLOCK DESIGN by Martin Aigner Thomas A. Dowling Department of Statistics University of North Carolina Chapel Hill, N.C. Institute of Statistics Mimeo Series No. 600.5 FEBRUARY 1969 Research partially sponsored by the United States Air Force under AFOSR Grant No. 68-1406, under AFOSR Grant No. 68-1415; and the National Science Foundation, under Grant No. 68-8624. A GEOMETRIC CHARACTERIZATION OF THE LINE GRAPH OF A SYMMETRIC BALANCED INCOMPLETE BLOCK DESIGN by Martin Aigner Thomas' A. Dowling I. INTRODUCTION A symmetric balanced incomplete block (SBIB) design is an arrangement of v objects, called treatments, into called blocks, such that each block consists of treatments, each treatment appears in distinct treatments appears in A < k and = k(k-l). A(v-1) the bipartite graph and blocks of H(rr) n(v,k,A) k A blocks. The graph of k v distinct blocks, and each pair of The parameters satisfy n(v,k,A) whose vertices are the 2v is defined as treatments rr, with two vertices adjacent if and only if one is a block and the other is a treatment contained in the block. line graph G(rr) of rr(v,k,A) is the line graph of the graph whose vertices are the edges of in G(rr) sets, The H(rr), i.e. Herr), with two vertices adjacent if and only if the corresponding edges in H(rr) have a common end point. Let {G(rr)} denote the set consisting of the line graphs of all SBIB designs with parameters and Ray-Chaudhuri [4] proved that a regular connected graph on vk V,k,A. G E In a recent paper, Hoffman {G(rr)} if and only if G is vertices and the distinct Research partially sponsored by the United States Air Force under AFOSR Grant No. 68-1406, under AFOSR Grant No. 68-1415; and the National Science Foundation, under Grant No. 68-8624. 2 eigenvalues of the adjacency matrix of unless G are -2, 2k-2, and k_2±(k_A)1/2, 3, A = 2, when the sufficiency of these conditions v = 4, k fails due to the existence of a single exceptional graph. a characterization of We give here in terms of several geometric {G{1T)} properties of these graphs, with again one exceptional case (Figure 1) = for v 7, k = 2. 4, A This result generalizes an earlier charac- terization [2] of the line graph of a finite projective plane but the conditions for those of [2]. A=l given here are slighly different from (See Remarks below.) II. By a graph DEFINITIONS G we mean a finite undirected graph without loops or multiple edges. V(G), E(G) denote, respectively, the vertex set and the edge set of G. vertices and define further x and y {y D. (x) 1. The degree of x x E We denote by E V(G) = IDl(X) n Dl(y) I ~(x,y) and y. E E(G). is written deg x are adjacent. the distance between i 0,1, ... (= IDl(x) I ). is the number of vertices adjacent to both G is regular if A clique d(x,y) V(G) : d(x,y) = i}, and edge-regular if, further, (x,y) (A=l) , ~ x ~(x,y) is constant for all x E V(G), is constant for all K is a set of vertices, any two of which 3 III. THEOREM. Let V,k,A be positive integers with = k(k-J.) , and let A(v-l) THE THEOREM {G(n)} A < k and be the set of line graphs of all symmetric balanced incomplete block designs with parameters If G {G(rr) } , then to V,k,A. G is connected and has the following properties: IV(G)I (P2) deg x (P3) L::.(x,y) (P4) ll(x,y) (PS) ID (x) n Dl(y) I 3 (P6) d(x,y) Conversely, if then = vk, (Pl) G E 2k - 2 ~ <; x for all E V(G), k - 2 if (x,y) E E(G) , 2 if (x,y) ¢ E(G), 3 k - A if for all x,y E d(x,y) or else v = = 7, k 1, V(G). G is a connected graph satisfying {G(n)} 2, ll(x,y) 4, A = 2, and (Pl) - (P6), G is the graph shown in Figure 1. Remarks. Properties (Pl) - (PS) are the counterparts of the characterizing properties of the line graph of a finite projective plane (A = 1) given in [2], except that the analogue of in that paper is bound for (P6) here. L::.(x,y) L::.(x,y) ~ 1 if required when That neither (PS) (x,y) ~ E(G). A nor > 1 (P4) The increased upper necessitates the addition of (P6) is redundant in all cases is demonstrated by the example given in [2] and the graph of Figure 2. On comparing our characterization with that of Hoffman and Ray-Chaudhuri [4], we meet with the intriguing problem of the 4 relationship of the eigenvalues of the graph to its geometric properties. Apart from the fact that in the case of regular graphs the dominant eigenvalue is the degree of regularity, very little is known. An important tool is the polynomial of a graph [3], which for regular graphs gives an upper bound to the diameter. the eigenvalues of G(TI) other than the degree are For example, -2, k_2±(k_A)1/2, and the fact that there are just three immediately yields property (P6) of the theorem. the fact that (P4) -2 Using certain "impossible subgraphs" of [4], is the minimal eigenvalue is easily shown to imply except for one possible configuration. directly prove edge-regularity (P3) for of the relationship between eigenvalues (P5) Hoffman and Ray-Chaudhuri k>4. The exact nature k_2±(k_A)1/2 and condition is unknown. IV. The necessity of PROOF OF THE THEOREM (PI) - (P6) when verified and the proof is omitted here. G is a connected graph and satisfies By K in (P3) G. IKI= sense. it is evident that G E {G(A)} is easily We therefore assume that (PI) - (P6). IKj; k for any clique Since we shall only consider cliques K in G such that k, let us agree to use the term "clique" in this restricted We then have LEMMA 1. Each vertex of and each edge of G is contained in exactly two cliques, G is contained in exactly one clique. 5 Proof. If k 4, > the lemma follows from (P2) - theorem of Bose and Laskar [1] on edge-regular graphs. k = 2, 3, 4 that (P4) by a The cases must be examined separately, and it is easily verified (P2) - (P4) imply the lemma except for k =4 , when these three conditions alone do not rule out the possibility that a vertex x ~ V(G) exists such that the subgraph generated by 6-cycle. If this is the case, connectivity implies that the subgraph generated by Dl(y) Dl(x) is a y ~ V(G). is a 6-cycle for all We can then show with little difficulty that no such graph satisfies all the conditions satisfies (PI) - (PI) - (P6) if (P6). (We do encounter one graph that A = 4 ( = k). This graph is the exceptional case in Shrikhande's characterization [5] of the association scheme (square lattice graph). arguments include this additional exception.) A A=k = v k. < provided we As the proof involves a case by case investigation and bears little connection to the main argument, it has been omitted. COROLLARY 1.1. Let C(G) The number of cliques in denote the set of cliques in set of unordered pairs K n L + ~. IK n LI x = (K,L) . 1. (K,L) Then by Letuma. 1, 2 To avoid some special in the proof of the theorem, we have assumed However, the theorem remains valid when L G is G, and 2v. I(G) the of distinct cliques such that (K,L) E I(G) We denote the single vertex i f and only if x E K n L by 6 C = (K ' K , " ' , Km) O l A clique chain (Ki,K + ) i l cliques such that of is the number C E I(G) Proof. G. Let Suppose Dl(x ) l contradicting (P3). For x V(G), and X 1,2. o 4 K2 , Then ~(xl,x2) > which implies k-l, we define {y : d(x,y) = 2, 2 j KO = Km is a clique cycle in (subscripts modulo 3). D (j) (x) for If C is a clique cycle. = (K ' K , K , K ) 2 O l O C (Ki,K i +l ), xi D (x ) l 2 oE The length G contains no clique cycle of length three. n X i=O,l, ... m-l. xi = (Ki,K i +l >· m of vertices and no other clique appears twice, LEMMA 2. for is a sequence of distinct (P4), the sets Then by ~(x,y) = j} D (l)(x), D (2)(x) 2 2 Then by properly labeling the four cliques K. ,L. , ~ J i,j partition = 0,1, E I(G). we have (1) Y E DO(x) iff K O LO' K = L · 1 1 (ii) y E: Dl(x) iff K O LO' K 1 (iii) y E: (iv) Y E D (2)(x) 2 D (1) (x) 2 Proof. E Ll • iff (KO,L O) E I (G) , (K ,L ) 1 1 iff (KO,L O) E I(G), (K ,L ) 4 I(G) . 1 l Follows immediately from Lemmas 1 and 2. LEMMA 4. (KO,L ) O =1= I(G), If x = (K ,K ), O 1 (K ,L ) 1 l 4 y = (L ,L ), where O 1 I(G), then y E D (l)(x) 2 and there are 7 exactly A cliques, including which intersect L l and KO,K . l one of Proof. there are in L . l If z k-Z vertices Hence the = <Ll,K) Proof. cycle in define y E D (l)(x) 2 Clearly LEMMA 5. and La, r~maining Lemma 3 - (iv). G By A vertices of L l are in (K,K ) 1. = (KO' Kl , K2 , K3 , K4 , KO) is a clique Let xi = (Ki,K + > (subscripts modulo 5) and i l Then K .• 1. xi 5, DZ E (1) we infer that all D (1) (x ) 2 o , a i so by the (x i + 2 ), Since K.1.- 1 (i+l)- (i-I) = a ,say, =Z A = Za. and = ( K,L ) ,where K intersects neither If Y E K nor O K , and L intersects exactly one of KO,K l · D (l)(x ) I A = 2a. Since Zk cliques intersect Z O IK n Then K or O K (including 1 y K ,K ) O 1 and G contains < To show (1) intersecting xl E D Z meeting (1) or C preceeding lemma, I (x). contains no clique cycle of length five. K + , including i 1 is prime to D 2 I(G) E O as the number of cliques intersecting both a. (P5) D (x) n Dl(y) , and these must be 3 (1) is one of these, then either Suppose G. in by is impossible, let KO' but not K ' Z La Then if yO such cliques meeting K or O K and Z La. cliques in all, Zv. be one of the (yO)' and so by Lemma 4, there are K which meet Z Zv (1) k-a = (KO,L O)' Za cliques we find cliques But there are exactly K ' so also there are O a a 1 cliques =a 8 meeting a K 2 and L . O L cliques meeting k-a exactly O and K . l (1) DZ vertices of k-a any of the Then by the same argument, there must be cliques L (x ). O contains O We can repeat the argument for meeting 1 L It follows that K , obtaining K , but not 1 4 finally ID Z(l)(xo)! ~ which is easily seen to contradict COROLLARY 5.1. and L If K ,K O 1 (1). are two disjoint cliques in is a clique intersecting both number of cliques intersecting both A or L • is either Proof. k ~ X o E K O K O and and G K • then the 1 K • including 1 k • xi = {Ki,L). cliques meeting such that are Let Z(k-a)2. K O and K i (1) DZ (Y1). O If there are fewer than ' then there exists Now if K cliques meeting i=O,l. I K O E K 1 \ Y2 = \Kl,L l/ , there K ,L . l 1 which meet one of if such a clique were to meet Yl and L l , G But would contain a clique cycle of length five. H , the clique graph of Consider now a graph by V(H) = C(G) , E(H) = I(G) is a one-to-one function of adjacent edges of H if and only if G onto V(G) such that into adjacent vertices of maps adjacent vertices of follows that ~ : (K,L) ~ <K,L) The mapping E(H) G G of H. Thus G H. € ~ ~-l and into adjacent edges of is the line graph H E {H(rr)}. G, defined It {G(rr)} We shall restrict our attention maps 9 henceforth to R. We first summarize some properties of R is connected and has the following properties: LEMMA 6. 2v, \V(R)I (Ql) deg K (Q2) K V(R) , = 2, (Q4) R contains no cycles of length three or five. A or (Ql) k is Corollary 5.1, and LEMMA 7. d(K,L) ~(K,L) < If VI = {K} (Q4) k (Q3) = 2. Then VI If two vertices of VI edges of V 2 R by V = VCR) - VI 2 vertex sets (Ql) R . VCR) K,L we have d(K,L) Let K VCR) E E such that =A for all and define (Q2) - (Q4) imply = 1 + k(k-l)/A are adjacent, then are all distinct. and = v. R contains as-cycle. vk edges incident Since these are all the (Q2) , the complementary set is also independent. Thus V is bipartite with V 'V 2 . l Finally let VI follows from Lemmas 2 and 5, since is an independent set, and so the with vertices of follows from Lemma 1, for all Ivll Thus (Q2) {R(TI)}. d(K,L) D (K). 2 u E In view of such that d(K,L) G corresponds to a cycle in = 2, then R Proof. if is Corollary 1.1, a clique cycle in in E ~(K,L) (Q3) K,L for all k (Q3) Proof. R in such that n be the number of unordered pairs ~(KO,Kl) = A. Then KO,K l nA = vk(k-l)/2, since 10 each of the k(k-1)/2 v vertices in such pairs V 2 K ,K . O 1 for all KO,K 1 € Vl,KO + K1 . is in D (K ) n D (K ) 1 O 1 1 for exactly i1(K ,K ) O 1 n = v(v-1)/2, Le. Thus If we identify the vertices of with treatments and those of V 2 >.. V 1 with blocks, and define the treatment to be contained in a block if and only if the corresponding vertices are adjacent in H , then it is clear that graph of an SBIB qesign rr(v,k,>") ,i.e. H E H is the bipartite {H(n)}. 1) We consider now the case where there exists two vertices, K,L in H such that vertices K ,K O l d(K,L) to be equiva1en~ L1(K ,K ) = k. O 1 d(KO,K ) = 2, 1 if and only if = 2, By class containing K € V(H) and write (Q2) D (K ) = Dl(K ). 1 O l = k. i1(K,L) Let Let us define two K :: K , if KO=K l O 1 we then have K :: K O 1 K denote the equivalence (:: is readily seen to be, in fact, an equivalence relation.) LEMMA 8. same number to ~ t Proof. first that Any two equivalence classes in Let 2 contain the of vertices. K ,K O 1 (KO,K ) l V(H) E € V(H) , E(H). t. 1 = IK.1 I , i =0 , 1. Suppose If we can show that in this case = t 1 = t , then the lemma will follow by the connectedness of H. Consider then the number (KO,L ) O that E E(H), (K ,L ) 1 1 n E E(H), L O L can be chosen in O ~ (L ,L ) O 1 of edges ~ E E(H) + KO' K , 1 L 1 k-1 ways, while K - {K } ;:',:L can be chosen. in",·",>..-l O O O Hence if L l 1) It is interesting to note that so far we have not made use of (P6). such wa~s2. or 11 n = (to-l)(k-l) + (k-tO)(A-l). choose k-2 L , we obtain l 0 , this implies > t ~ 2 (Note Suppose L l n to (tl-l)(k-l) + (k-t ) (A-l). l = (K,L) E (K,L ) l E(H). +L , and if t + t V(H) K l (Kl,L ) l K u L vertices. E E K E(H). is the The equivalence induces a homomorphism H+ H , H is the graph defined by where v VCR) {K : K E(H) {(K, i) V(H)}, E (K, L) E E(H)}. H is connected and has the following properties, = v/t, k = kit, (Ql) l"v(R) I (Q2) deg K A = A/t: 2v, = k for all A if K V(H), = 2, t,(K,L) (Q4) H contains no cycles of length three or five. (Ql) and Lemma 6 , since now = L. (Q2) d(K,L) E (Q3) Proof. K K H on the vertices of defined on LEMMA 9. of Since ' and the proof is complete. and therefore complete bipartite graph on where l Then E(H) E Hence the sub graph of relation t L and O by hypothesis.) L ,then E = Similarly if we first fix are obvious. ~(K,L) (Q3) follows from = k would imply K = L We finally observe that if length three or five, then so would (Q3) and hence H were to contain a cycle of H , contradicting (Q4) of Lemma 6. lZ v, k, A reads The equation relating A(vt-l) (Note that v need not be an integer.) Consider now a fixed edge A a B l A Z (Z) = k(kt-l). (Ka,La) {K } Da(K ), a a D (K ) Da(La), l a D (K ) - D (La)' Z a l E(H) E and define Al {La} = Da(La), D (La) - D (Ka), a l B Z D (La) - Dl(Ka )· Z B a Then Da(Ka) = Aa , D (K ) = B U l a a D (K ) = A U Z a 1 It follows from (Q4) Da(La) = Ba, Dl (La) = Aa U AI' DZ(La) = B1 U BZ· B , l A . Z that the six sets are pairwise disjoint, and thus the two sets disjoint for each vertices of Using (QZ) i=a,l,Z. Di(Ka ) and or of (Q3) Also by (Q4) A.,B. 1 1 (i=a,l,Z) Di(K ), a are we see that no two can be adjacent for Di(La ) Di(L ) a i=a,l,Z. we can then easily determine the number of vertices in each set, obtaining Let using IAal IBal 1, IBII JAIl k"-l, IAZI IBZI (k-l) (k-A) / A. C denote the set of vertices not in any of these sets. (Ql), (Z), and (3) (3) Then we have (4) 13 Thus t 2(k-~), divides and so 2 LEMMA 10. Proof. K '0 Let C. x Clearly we have C and E Then by = K,L, say K E: C . If Proof. Let Let < 3 (K,L) , in G. n and let If D (K) l . , we have K be a fixed vertex of ID (K) n BZI l H , is even simpler. ~ A u B Z Z s for all = ID I (K) n DZ(L) I DZ(K) n D (1) l n = rA = sA , =s . b KO,L O = 3 , then (Q3) = IDl (K) nAZi E(H) . a contra- of edges joining a for all 3 E meeting one of l r = IDZ(K) n D (1) I l Using r = = Z in Lemma 10 that d(K,1) Consider the number x K C ~ D (K ) u D (1 ) 3 O 3 0 If L (K,L) then d(xO'x) o = K or f Z , the result follows from LEMMA 11. i.e. (P6) , d(LO,K) K d(xO'x) Hence 3 defined by K and O which implies either It G = E(H) , E (5) C are adjacent. 3 , there exists a clique and one of diction. z(iZ-i) . d(K ,K) O (K,L) be the vertex of d(xO'x) < t No two vertices of K,L If f C and define and 14 a + b Then = 0. k l IDZ(K) n Bli bl IDZ(K) nAzi a If we define = a , since clearly then it follows from Lemma 11 that a DI(K) n DZ(K O) DZ(K) n DI(KO) b = b Similarly b ~ I a ,;, I D (K) n A I Z . 1 Since Hence implies . C = D/KO) n D3 (LO). Again, and hence there exists =A with be in DZ(K) n BI < a = I DZ(K) n Bli = IB11 = k-l , we have > I a = Similarly and l Dl(K) n A ' and therefore Z Each of these vertices must > = A • Similarly b a ~ ~ and we have < a,b where a + b k-A (6) k We wish to establish an upper bound on the number of vertices Dl(K) L = (DICK) n AZ) E U DZ(K) (DI(K) n B ) , Z DI(K) n A Z some vertex of which are not in such an C L or to some vertex of Since is adjacent to DI(K) n B . Z Assume without loss of generality the former and call the vertex K · Z It then follows that for some vertex Kl E L Dl(L ) . O tains a cycle of length five. set of a vertices. IDZ(K ) n D (LO) I Z I Thus at most a < E DZ(L ) , O If Hence K l E By Lemma of the k L E Dl(K ) l DZ(K) n AI' then K E A U (Al-DZ(K» I O Thus we have, since a and hence 11, K I E D (K ) Z Z H , a , ID (K ) n DZ(L ) I I Z O vertices adjacent to KZ E con- D (K) I < = a n A Z 15 are in DZ(K) n DZ(L ) O and since any vertex of DlCK) 0 D (L ) 2 we have at most DI(K ) Z D (L 1 D (K) n D (L ) 2 2 ) A K € 2 I [2 lei > € 2 <'K) 2 which are also in 2 (K) ~ E D vertices E D Dl(K) n B ) , A similar argument shows that 2 b2 / for some La Dl(K) n A " and thus together with D (K) n B which are also This now 2 1 gives us < (3) 1 + I -1 [k(k-1) - (a 2 + b 2 )] " (7) (4) z(k-I) (1-l/t) > The inequality remains valid on replacing 2 (k-:) (by (Q4), of them I cl Then by adjacent to one vertex O a2 / vertices DI(K) n A ' Z vertices in cannot be adjacent to any vertex of for some 2 € a O there are at least in La AZ is adjacent to exactly n La ~ DZ(K) Since there are of by the upper bound (5) , and we then have k2 Substituting t k-a for b 3k + 3). + 1 " and simplifying, a 2 - ka + 1/2[3(k-I) --1] > o. ( 8) 16 Assuming with no loss of generality that verified that the only values of (6) and (8) k ~ 2, (ii) k = from (3) sets A. ,B. 1 4, 2, k and 1 a,b,k,A satisfying both are (i) If < = b , it is easily a (i A = A = 1, a = 1, b A = = 1, 1, a a = b b = = k-l, 2. 1 ,then t = 2 by (5), and (4) , we have exactly one vertex in each of the = 0,1,2) and C. In this case H is a cycle of length seven, v = 7, k = 4, A = 2, Figure 1 The lines in the figure represent cliques, two lbel~. vertices being adjacent in Figure 1. perties and G is the graph 01 G if and only if they are collinear. A graph (PI) - (P6) G 4 {G(n)} with v satisfying the pro- = 7, k = 4, A = 2 . 17 -> Consider next the case is the vertex of A Z b = k-1 or k-1 of Co Co = Dl (12 ) a = k-l, b K for some K Z C ,C O 1 (9) b. 1. a. 1 C. = = k-1 A Z and For = k is adjacent to Z Since C is contained in at least one E B Z < Ici . By K c such that either (4) ~ C (KZ) Co D l and (5) or we have (9) Zk-3 (Q3) A = 1. ,since CO,C l ~ C K k-1 ~ 3 ICo n Cli with , while in the latter C Hence it follows o~ C . = 1 C ' , and let Then since a Ici = Zk-3 . (Q3) , that the number of edges = i !Dl(K ) nAZi i be the number of a or else In the former is equal to the number of edges joining E C ,set i ta i = \ Lb.1. , B Z and K.1. 's flor whi{!h we obtain ;:(k-l) + (I C 1-;:) + (I,c 1-;:) (k-l) a and since K If ID (K ) n BZI 2 i 1, b since k-l for every choice of We can easily verify, using and there are exactly that either there is only one such set ICI joining K Then if k-l are two such sets, they must have a non-empty there are two sets case, E = 1 C < 6(K,L) ~ Z ,contradicting from C A , L Z 2 E k-l intersection. E vertices in ,~. so that if K b by our earlier argument. DZ(L ) O it follows that each vertex set = 1, adjacent to exactly one vertex of = 1, 3, A D (K ) n C , including l 2 vertices in a - = k , Ici = Za. (10) 18 Ici = 2k-3 This rules out the possibility = ICI k-l , it follows from K ,K c C 2 I LI E B 2 with, say a = l (10) 1 that we must have two vertices and b l. 2 1 ,12 1 Let Kl ,K2 be the vertices adjacent to by our earlier argument As to the case L A 2 E 2 and ' respectively, then are both adjacent to all of C, ~(Kl,K2) ~ 2 , a contradiction. which implies k - 4, A The remaining case = 2, a =b = 2 can be shown to be impossible by a rather involved argument which demonstrates that H must cpntain a cycle of length five. We have omitted the proof here. Concluding Remarks. (PI) - (P6) The question whether the conditions are redundant is difficult to answer in general. The example given in that (PS) [2] with 7, k = 3, A = 6, A = 2 Figure 2 H shows with v demonstrates that the same holds true for t = The two vertices of in 1 below with For ease of exposition we have drawn the graph Figure 2 = cannot be dropped without admitting additional exceptions, while the graph in k = v H in 3, A 1 corresponding to the edges e 2 , and thus G v = 8, k are readily seen to be at distance = 4 = 16, (P6). and f from each other. 19 e Figure 2. G ~ {G(n)} (PI) - (PS) A graph H whose corresponding graph and satisfies the properties with v = 16, k = 6, A =2 . 20 REFERENCES [1] Bose, R. C. and Laskar, R.: Tetrahedral Graphs, 3 [2] A Characterization of J. Combinatorial Theory (1967), 366-385 Dowling, T. A. and Laskar, R.: A Geometric Characteri- zation of the Line Graph of a Projective Plane, J. Combinatorial Theory [3) Hoffman, A. J.: 3, (1967), 402-410 On the Polynomial of a Graph, Amer. Math. Monthly 1Q (1963), 30-36 (4] Ho ffman, A. J. and Ray-Chaudhuri, D. K.: On the Line Graph of a Symmetric Balanced Incomplete Block Des~gn, [5) Trans. Amer. Math. Soc. Shrikhande, S. S.: Scheme, 11 (1965), 238-252. The Uniqueness of the Ann. Math. Stat. L Association 2 30 (1959), 39-47.

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