UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIG.N CENTRAL CIRCULATION BOOKSTACKS The person charging this material is re- sponsible for its renewal or its return to the library from which it was borrowed on or before the Latest Date stamped below. You may be charged a minimum fee of $75.00 for each lost book. Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. TO RENEW CALL TELEPHONE CENTER, 333-8400 UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN JHI ! v When renewing by phone, write new due date below previous due date. L162 Digitized by the Internet Archive in 2013 http://archive.org/details/deficientgeneral947perl UIUCDCS-R-78-947 DEFICIENT GENERALIZED FIBONACCI MAXIMUM PATH GRAPHS BY Y. Perl and S. Zaks UILU-ENG 78 1739 November 1978 "j DEPARTMENT OF COMPUTER SCIENCE UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN URBANA, ILLINOIS DEFICIENT GENERALIZED FIBONACCI MAXIMUM PATH GRAPHS Y. Perl + and S. Zaks ++ October 1978 f On leave from the Department of Mathematics and Computer Science, Bar- II an University, Ramat-Gan, Israel. tt Department of Computer Science, University of Illinois at Urbana- Champaign, Urbana, Illinois 61801. The work of this author was supported in part by the National Science Foundation under grant NSF MCS 77-22830. TABLE OF CONTENTS I. Introduction 1 II. Presentation of the problem 2 III. Description of the solution 5 IV. Local properties 8 V. Phase 1 10 VI. Phase 2 15 VII. Phase 3 23 Appendices 26 References 30 ABSTRACT The structure of an acyclic directed graph with n vertices and m edges, maximizing the number of distinct paths between two given vertices, is studied. New techniques for solving this problem are developed, thus enabling us to extend previous results. I. INTRODUCTION 1 .1 In this work we continue to investigate the structure of an acyclic directed graph with n vertices and m edges, maximizing the number of distinct paths between two given vertices. In previous work it was shown that there exists such a graph containing a Hamiltonian path, thus uniquely ordering the vertices. The length of an edge (i,j) is defined now as k=j-i and it belongs to level k. It was further shown that such a graph contains k-1 full levels and some edges of level k - a deficient k-generalized Fibonacci graph. 1 .2 We investigate the distribution of the edges in level k. In particular, we find that even for k=3 the structure is quite complicated. The behaviour of the solution is different in three ranges of the number of the edges (called "phases"). We show some forbidden configurations for the solution. These observations, combined with some local properties, lead to the solution for the different phases. The proof techniques which were developed can help to investigate the more complicated distribution of the edges in levels k, k>3. 1 .3 Section 2 presents the problem, and the solution is summarized in section 3. The analysis of the solution relies on certain local properties studied in section 4. The solution is partitioned into three phases, and these phases 1, 2 and 3 are the subject of sections 5, 6 and 7, respectively. General description of the proof techniques appear in the beginning of section 5. II. PRESENTATION OF THE PROBLEM 2.1 Let G be an acyclic directed graph without multiple edges or isolated vertices, containing two distinguished vertices s and t. Define N(G) to be the number of distinct paths from s to t. We consider the following problem: given integers n and m, find a graph G with n vertices and m edges maximizing N(G). We call such a graph G a maximum path graph and define N =N(G). Clearly N is defined in the domain n-1 < m < ( ) . n,m — — V l I 2.2 In Perl [1] several cases of maximum path graphs are studied, given only the number of edges m. It is shown that for acyclic directed graphs without multiple edges the (almost) Fibonacci graphs are maximum path graphs for (even) odd number of edges. Examples are shown in figure 1. It is clear that for a (almost) Fibonacci grapK G of n vertices N(G)=F (N(G)= 2F „), where [f] is the Fibonacci sequence defined by F l ■ F 2 * ! • F i + 1 " F i + F i-1 for i > 1 Hence N , - = ¥ n and N 0n . = 2V n . The Fibonacci graph for n=7 The almost Fibonacci graph for n=7 2.3 In Gol umbic and Perl [2] maximum path acyclic directed graphs are studied, given the number of vertices n and the number of edges m. It is shown that there exists such a maximum path graph containing a Hamiltonian path, thus uniquely ordering the vertices {1,2,.. .n} . Hence, all the edges are of the form (i,j), i k . I j=i-k A graph of the second kind is called deficient k-generalized Fibonacci graph Examples are shown in figure 2. (1) A 3-generalized Fibonacci graph (2) A deficient 3-generalized Fibonacci graph Finiiro 9 2.4 In this paper we investigate the distribution of the edges of level k in a deficient k-generalized Fibonnaci graph. We refer to such a graph as the solution. The solution for k=2 is given in [2] . We find that even for k=3 the structure is quite complicated, and this work is dealing only with this case. The solution for k>3 seems to be more complicated, but of the same spirit as the one for k=3. Thus the understanding and the proof tech- niques which were developed here can help to investigate the solution for k>3, For convenience we will refer and draw only edges of level 3, and also describe level 3 by a 0,1-sequence c i c 2"* c n-3 ' wnere c i = ^ if (1 »1 + 3) is an edge and otherwise. For example, the sequence 100101 describes the graph of figure 2(2). A compact representation of this binary sequence is used whenever possible. For example, (100) 3 01 (Oil ) 2 means 10010010001011011, [x] means that x is optional; for example, 10[0](100) represents both 100(110) k and 10(110) k . III. DESCRIPTION OF THE SOLUTION 3.1 Let us note first that the solution is not unique, and thus we only claim that the graph described is a solution . The behaviour of the solution is different for three ranges, where the number of edges in level 3 is about (1) up to one third, (2) between one third and two thirds, and (3) more than two thirds of the n-3 possible edges in this level. These ranges are called phase 1 , phase 2 and phase 3 , respectively. 3.2 First we mention local properties of the solution. Let I denote the number of edges in level 3, < I < n-3. If I>1 then the solution contains both end-edges (1,4) and (n-3,n). Furthermore, if Il. If a graph does not contain the edge (i,j) then (i,j) is a non-edge . If the graph contains neither of (i,i+3), (i+l,i+4) and (i+2,i+5) then (i+2,i+3) is a gap . 3.3 We describe the three phases of the solution. Phase 1 : I < I ^-1 . ( ttj denotes the largest integer not larger than t.) The graph is composed of two runs of consecutive disjoint edges, one starting and the other ending with end-edges. The number of edges in each run is arbitrary. In case I=(n-l)/3, where n= l(mod 3), the two runs merge into one. Examples: (100)V(001 ) 2 , (100) 4 1. Phase 2: n+2| . | 2n-6 |. n case n ^ l(mod 3) the solution is composed of two runs of con- secutive disjoint edges, one starting and the other ending with both end- edges. The number of the edges in the two runs is arbitrary, except that the two near-end edges are not contained in a run. Examples: 100(110) 4 01 (which is composed of (100) 5 2 and 2 (001) 5 ), (100) 2 10(110) 2 1. In case n = l(mod 3) the solution is composed of two runs of con- secutive disjoint edges, one starting and ending with both end-edaes , and the other run starting at any vertex but not containing a near-end edae. Examples: (100) 2 (1 10) 2 1 , 10010(1 10) 3 01 . There are two border cases discussed later. Phase 3 : I > 2 I jl - 1 . The structure of the graph is quite complicated in terms of runs of consecutive disjoint edges. On the other hand, the solution contains all n-3 edges except two runs of consecutive disjoint non-edges , one starting at one near-end edge and the other ending with the other near-end edge, (compare with phase 1). We conjecture that the number of non-edges in these runs differs by at most one. Examples: 1 (01 1 ) 2 1 2 (011 ) 2 01 , 1 (011) 3 1 (Oil ) 2 01 . 3.4 Border cases : There are two border cases. One occurs between phases 1 and 2, and is of the form ( 100) J 101 (001 ) k , j,k>0, in the case when n=0(mod 3) and I=n/3; the second one is between phases 2 and 3, and is of the form l(011) J 010(110) k l, j,k>0 (and we conjecture j-ki , we have then N(G 2 ) > N^). Furthermore, if c-=0 (i.e. (i,i+3) is a non-edge) and P 2 (i+2) > P 1 (1+2) , P 2 (i+3) >_ P 1 (1+3) then N(G 2 ) > N(G-|). In case c- + ,=0 this is trivial. Otherwise P 2 (i+3) = P 2 (i+2) + P 2 (i+1) since c.=0. Therefore P 2 (i + 4) = P 2 (i+3) + P 2 (i+2) + P 2 (i+1) = 2P 2 (i+3) > 2P 1 (i+3) = P 1 (i+3) + P 1 (i+2) + P 1 (i+l) = P^i+4) . Hence N(G«) >.N(G,) by the first observation. 4.2 Theorem Every solution contains both end-edges (if 1=1 then only one end-edge), Proof : We prove that the solution contains the edge (1,4). The proof for (n-3,n) follows by symmetry. Suppose that a solution G, starts with 1 , i>l. Let G 2 be the graph obtained from G, by replacing this configuration by lO 1 "" 1 . P,(i+3) is equal to the number of paths in the opposite direction from i+3 to 1 in G,. Thus P 2 (i+3) = P,(i+3). It is clear that P 2 (i+2)> P 1 (i+2) and P 2 (i+1) > P,(i+1). Thus N(G 2 ^ > N(G-,), contradicting the optimal ity of G, . Hence G, must contain the edge (1,4). □ Corollary: N n>2n _ 2 = F n + F^ . N^., ■ F p ♦ 2F n _ 3 + F p . 6 . 4.3 Theorem There exists a solution not containing the near-end edges (unless I>n-4) Remark : A stronger version of the theorem is false; for example, 11001 is a solution containing a near-end edge. Proof : We prove it for the edge (2,5); the proof for (n-4,n-l) follows by symmetry. Suppose that a solution G, contains the edge (2,5). Then G ] beoins with 1 1_1 , i>2. Let G 2 be the graph we get by replacinq this configuration by 101 1 " 2 . We compute A(j) = P 2 (j)-P-|(j) for j=i+2,i+3 : A(1+2) = P 2 (i+2) - P^i+2) = (F^ +2 - F^_ 2 ) - F^ +2 = -F^ 2 < , A(i+3) = P 2 (i+3) - P^i+3) = (F* +3 - F^ ) - ( Fi 3 +2 + F^ ) = F l 2 + F i 3 -3 > °« se If c + ,=l then P 1 (i+4) = P 2 (i+4) by symmetry, otherwi A(i+4) = A(i+3) + A(i+2) = F?_ 3 . Thus a( i+4) ^0 in both cases. A(i+5) > A(i+4) + A(i+3) + A ( i +2 ) since in case c +2 = A ( i +2 ) < is omitted. Thus a(i'+5) > F i _ 3 > , and N(G 2 ) > N(G-j) , and G 2 is also a solution. □ The next lemma presents inequalities used later. 4.4 Lemma 1. |p(i) < P(i+D < 2P(i) for 1 > 2 . 2. |f(i) < P(i+2) < l^P(i) for i > 3 . 3. 4P(i) < P(i+3) < ^P(i) f or i > 3 . 4. §P(i+2) < f|p(i+2) < P(i+3) < 2P(i+2) if c^l , for 1 > 3 5. P(i+4) = 2(P(i+3) iff c.=0 and c i+ i=l , for i > 1 . Prnnf • coo AnnonrH y A 10 V. PHASE 1 5. 1 We first describe Droof techniques, which are used for all three phases, and then demonstrate them in this section for phase 1, which is simpler than both phases 2 and 3. An essential tool in our proofs is to express the number of paths to any vertex j, in a special configuration of a graph, as a function of P(a) and P(b), where a and b are two vertices at the beginning of this configuration. Such special structures are (100) 1 , (HO) 1 , i>0, and 1 , i>3, for phases 1, 2 and 3, respectively. This enables us to measure the effect of some local transformations on N(G). We use this technique to prove that some configu- rations are forbidden for the different phases. By eliminating these forbidden configurations we establish some order in the structure. of the solution. Thereafter, more local transformations are used to determine the exact structure of the solution, many technically involved calculations were required in the proofs, and only few of them are given in the text; more proof details are found in the Appendices. 5.2 Lemma The number of paths P(b+j) to the j-th vertex from b, j>l, in a configuration OIOO(IOO) 1 , i^O, can be expressed as a function of P(b) and abc P(c) (and as a function of P(a) and P(b) if the configuration is preceded by a 0), where a,b and c are three consecutive vertices at the beginning of this configuration, as follows: 3-5 1 " 1 - z j = 31+1 , i > 1 , P(b+j) = < ■ i 5 ■ i j = 3i+2 , i _>_ , 2-5'- z j = 3i+3 , i > , where z = P(b) + P(c) (and, if precedes the configuration, z is also P(a) +2P(b) ). Proof: By a straightforward induction on j, using the corresponding formulae 11 P(b+3i+l) = P(b+3i) + P(b+3i-l), P(b+3i+2) = P(b+3i+l) + P(b+3i) , and P(b+3i+3) = P(b+3i+2) + P(b+3i+l) + P(b+3i) . D We turn now to the structure of phase 1. 5.3 Theorem In phase 1 of the solution the edges are disjoint (i.e. no two edges intersect). Proof : We show that if the solution G contains non-dis joint edges, namely G contains either a 11 or a 101 configuration, then it is possible to transform G into a graph G, such that N(G,) > N(G) , contradicting the optimality of G. Since i phase 1 I ,<|n/3J > the existance of either 11 or 101 implies the existance of a gap, which corresponds to a 000 configuration, in G. Without loss of generality (wlog) we consider a first occurrence of either 11 or 101, to the left of a 000 configuration. Case 1 : 101 . k-1 Wlog we assume the configuration 101(001) 000 in G. The graph G-, k+1 is obtained from G by replacing this configuration by (100) G : -10 10 (100) k_1 u v w a b i i+1 i+5 G 1 : -10 10 (100) k_1 u v w a b i+1 i+5 We denote y = P(a) + P(b) and z = P(a) + 2P(b) (note that P^a) = P(a) and P.|(b) = P(b) ). P, P, and a = P 1 - P for the vertices i - i+5 are summarizes in table 1. The calculations are based on lemmata 4.4 and 5.2, and some of them are given here; the rest, being simpler, are left to the reader. A(i+3) = 5-5 k - z - 8-5 k - y = 5 k C-3P(a) +2P(b)] > , by lemma 4.4(4). If c i+1 = 1 then lemma 4.4(5) implies that a(i+4) = 2 a(i'+3) > 0. 12 P p l A = P, - P i 2-5 k - y 5 k - z - 5 k P(a) < i+1 3-5 k - y 2-5 k - z 5 k ( P(b) - P(a) ) > i+2 5-5 k - y 3-5 k - z 5 k ( P(b) - 2P(a)) < i+3 8-5 k - y 5-5 k - z 5 k (2P(b) - 3P(a)) > c 1+l -l , i+4 16-5 k - y 10-5 k - z 5 k (4P(b) - 6P(a)) > c i+1 =0 , i+4 13-5 k - y 8-5 k - z 5 k (3P(b) - 5P(a)) > c i+2 =c i+1 =l , i+5 29-5 k - y 18-5 k - z 5 k (llP(a)+ 7P(b)) > C i+2 =1 ' c i+l =0 ' i+5 26-5 k - y 16-5 k - z 5 k ( 6P(a)-10P(b)) > Table 1 If c i+1 = then a ( i +4 ) = 8-5 k - z - 13-5 k . y = 5 k [-5P(a)+3P(b)]> by lemma 4.4(4). If c. + « = then the previous inequlities imply N(G,) > N(G) . If c.+ == c. xl = 1 then a ( i +5 ) = 18-5 k - z - 29-5 k - y = 5 k [-HP(a)+7P(b)] > 5 k [6P(b)-10P(a)] > 0, by lemma 4.4(4). If c i+2 = 1 and c i + 1 =0 then by lemma 4.4(5) a(i'+5) = 2a(i'+4) > , and again N(G ] ) > N(G) . 13 Case 2 : 11 . Wlog we assume the configuration 11(001) 000 in G. Replacing this k+1 configuration by (100) 10 can be shown to increase N(6), a contradiction. □ From now on we omit long technical proofs in case when the proof technique has already been demonstrated, as we did in case 2 above. 5.4 Theorem In phase 1 the solution contains exactly two runs of consecutive disjoint edges; one of j edges, j >1 , starting at one end-edge, and the other of k edges, k>l, ending at the other end-edge, where j + k = I. i.e. the solution is of the form (100) J n ~ 3 ~ 3I (001 ) k . Proof : By theorem 5.3 the edges in the solution are disjoint, and by theorem 4.2 both end-edges belong to the solution. We could obtain the solution by applying local transformations and measuring their effect, using previous techniques (similar to the proof technique of theorem 5.3). Instead, we present a different proof, based on a global approach. We could not find such proofs for the other phases, which have a more complicated structure. We first calculate N(6) for the solution G (100) J n " 3 " 31 (001 ) k , writing G as 1 (100)J -2 10 o n-3I+1 10 1 (001) k ~ 2 . u v a b c d By lemma 5.2 P(a) = 5 J_1 . z and P(b) = 2»5 J ~ . z , where z = P(u) + P(v) = 1+1 =2. Note that the coefficients of 5 J ' _1 - z , 1 and 2 , are the Fibonacci numbers F ? and F„ , respectively. Hence P(c) = 2.5J- 1 - F n . 3I+2 and P(d) = 2.5^- F n . 3I+3 . Then, by lemma 5.2, we get . P(n) = 2.5 k_1 .(P(c) + P(d)) = 2.5 k - 1 .2-5 j - 1 .(F n _ 3I+2 + F n _ 3I+3 ) = 4- 5 j+k - 2 . F n-3T+4 » 14 hence N(G) = P(n) = 4-5 1 " 2 - F , T± , n-3I+4. 1 1-1 I Suppose G, contains I arbitrary disjoint edges: 10 1 - - - 10 10 , I ' where z t. = n-3-I. Write G, as i=0 ] ' t/-> L-i~0 Lp-J L t -I tj"0 U 1 ' 10000^ 1 ••• 1 1 " 1 1 ' a b c d e f P(a) = F ~ , and by lemma 4.4(5) P(b) = 2P(a) = 2F. ., . The coefficients x J r J of F. +3 in P(a) and P(b), namely 1 and 2, are the Fibonacci numbers F« and F_ , respectively. Hence P(c) = F + -« F. +3 , and by lemma 4.4(5) we get P(d) = 2F. +3 « F. +3 . It can be proved by induction on I that p( 9 ) = p • F • F • • • F ne; r t Q +3 r ^+3 r t 2 +3 r t I _- | +3 ' P(f) = 2P(e) , and N(G) = P(n) = F tQ+3 - F ti+3 . F^^ .-• F^^- F^ . It remains to prove that 1-2 F • F ••• F • F < 4 • l a 3i = a 3i-l + a 3i-2 + a 3i-3 = 2a 3i-l ' a 3i + l = a 3i + a 3i-l +a 3i-2 ' a 3i+2 a 3i+l + a 3i * Proof : by induction. □ Next we will use the fact that in phase 2 pyM< I < pxj to show forbidden configurations for this phase. 6.2 Theorem The solution in phase 2 contains no configuration 111 . Proof : We assume that a solution G contains a configuration 111, and show that it is possible to transform G into G, such that N(G,)>N(G), contradicting the optimal ity of G. Wlog we consider the rightmost occurrence (in G) of either lll(011) k 00 or in(on) k oio , k>0. Case 1 : lll(011) k 00 . k+1 construct G-, by replacing this configuration by 1(011) 0. Applying lemma 6.1 we calculate the number of paths to two vertices i and i+1 in both G and G-, . 16 G : 1 1 1 1 1 w a b c cl e 1 1 1 1 a b c (011) k_1 i i+1 i+2 1 (Oil) k-1 • • i i+1 i+2 i+1 a 3k+1 ((P(d)+P(e))-a 3k _ 2 P(d) a 3k+2 ((P(d)+P(e))_a 3k-l P(d) a 3k+3 ((p(b)+P(c))-a 3k P(b) a 3k+4 ((P(b)+P(c))-a 3k+1 P(b) A(i) = a 3k _ 1 (P(c)-P(b)-6P(a))-a 3k _ 2 P(a) < A(i+1) = a 3k+1 (P(c)-2P(a))+a 3k _ 1 (P(c)+P(b)-3P(a)) > Table 2 The details for A(i) and A(i+1) can be found in Appendix C, A ( i +2 ) = A(i + 1) + A(i) since c. -,=0, hence A(i+2) = a 3k _ 1 (P(c)-P(b)-6P(a)+P(c)-P(b)-3P(a)) + a 3k+1 (P(c)-2P(a)) - a 3k _ 2 P(a) - P(c)(a 3k+1 + 2a 3k-1 ) - P(a)( 2a 3k+] + Qa^ + a 3k _ 2 ) = P(c)(a 3k + 3a 3k _ 1 + a 3k _ 2 ) - P(a)(2a 3k + lla 3k _ 1 + 3a 3k _ 2 ) =P(c)(5a 3k _ 1 +a 3k _ 2 ) - P(a)(15a 3k _ 1 + 3a 3k _ 2 ) = ( P(c) - 3P(a) )( Sa^^ + a 3k _ 2 ) > , since P(c) > 3P(a). Then N(6 ] )> N(G) , since A(i+1)>0, A(i+2)>0, and 0^=0. 17 Case 2 : lll(0n) k 010 . k+1 Replacing this configuration by 11(011) can be shown to increase N(G), a contradiction. D 6.3 Theorem The solution in phase 2 contains no configuration 000 . Proof : The proof is similar to that of theorem 6.2 . The computations are even simpler, because the a.'s in the formulae are replaced by coefficients of the form c-5 1 (for a constant c) from phase 1, since configurations (100) replace the configurations (HO) 1 in the analysis. □ 6.4 Theorem The solution in phase 2 contains no configuration 01010 ; this configuration occurs only in the border case between phases 1 and 2, in the case n e 0(mod 3) and I=n/3, where the solution is of the form (100) J 101 (001 ) k , j,k > 0. Proof : Assume that the solution G contains a configuration 01010. Construct a graph G, by replacing this configuration with 10010. We compare P and P-, for several vertices in both G and G, . G : 6 1 : 10 10- a-3 a-2 a-1 a b c i i+1 10 10 a-3 a-2 a-1 a i i+1 1 i-2 i-1 i P(b)+P(c) 2(P(b)+P(c)) 3(P(b)+P(c)) P(a)+P(b)+P(c) P(a)+P(b)+2P(c) 2( D (a)+P(b))+3P(c) P(a) > P(a)-P(b) < 2P(a)-P(b) > Table 3 18 By lemma 4.4(5) A(i+l )=2a(i ) , since c. 3 =0 and c. « s l in both graphs. This, together with c. ^0, implies N(G-, ) > M(G). If a(i' )=2P(a)-P(b) > then N(G-| ) > N(G) S contradiction the optimality of G. If A(i)=0 and N(G,)=N(G) then P(b)=2P(a). This, by lemma 4.4(5), implies c ,=0 and c ~=1. Consider a~ j a- c. the two possibilities for c , . a - 1 c a _i = 1 • In this case G-. contains a 111 starting at a-2, and thus (by theorem 6.2) G, , and consequently G, are not solutions, a contradiction. c , = , In this case G and G, contain 01001010 and 01010010 starting a~ I at a-3, respectively. But now G, contains 01010 shifted three positions to the left. Applying the same argument yields the configuration 01001010010 in G-j (corresponds to 01001001010 in G), and 01010010010 in another solution G ? . By repeatedly applyina this argument, and similarly to the right of the original configuration 01010 in G, we can prove the structure (100) J 101(001) k , j,k > 0, of the border case between phases 1 and 2, as the only exception. 6.5 Theorem The solution in phase 2 contains no configuration 10101 ; this configuration occurs only in the border case between phases 2 and 3, in the case n = 2 (mod 3) and I = n I , where the solution is of the form l(0n) j 010(110) k l, j,k > 0. Proof : Assume a solution G contains 10101. If the solution is not of he desired form then it contains either 11010(110) 01, k >. 1 , or C0J1 (011 )^010(110)10, j,k >_ (or symmetric configurations). It can be shown that replacing these configurations by (110) k+1 1001 or [0]100(110) J+k+1 , respectively, increases N(G), a contradiction. O 6.6 Conjecture : The solution for the border case between phases 2 and 3 is of the form 10(110) J 1 (011 ) k 01 , j,k> 0, where |j-k| < 1. (see also conjec- ture 7.4) 19 6.7 Corollary A solution in phase 2 contains the configuration 111 . Proof : Follows from theorems 6.3, 6.4 and 6.5. □ In phase 2 the forbidden configurations are not sufficient for a straightforward proof of the structure of the solution, as was the case in phase 1. A detailed "case analysis", given in the following theorem, is required. 6.8 Theorem The solution in phase 2 contains none of the following configurations (j,k > 0): a. 0100(110) J '[0](100) k 1011 b. 0100(110) J '[0](100) k ll c. 010(110) J '[0](100) k 1011 d. 010(110) j [0](100) k ll . Proof : We prove only (a) (with the optional 0); the rest of the cases are similar. Suppose the solution G contains 0100(110) J 'o(100) k 1011 , j,k > 0. k+2 i Construct G, by replacing this configuration with 0(100) (110) J 11. Applying lemmata 5.2 and 6.1 we calculate F and P, for several vertices in both G and V G : 1 1 1 (110) J ' _1 10 (100) k-1 10 11 a b c d e x y i i+1 i+2 i+3 i+4 G,: 1 (100) k+1 1 1 (110) J ' _1 1 1 • • • a b c w x i i+1 i+2 i+3 i+4 20 w i+1 a 3j+1 (P(d)+P(e))-a 3j _ 2 P(d) i3 j+2 (P(d)+P(e))-a 3j _ 1 P(d) 5 K (P(x)+P(y)) 2-5 K (P(x)+P(y)) 2-5 k+1 (P(a)+P(b)) 3-5 k+1 (P(a)+P(b)) a 3j (P(w)+P(x))-a 3i _ 3 P(w) a 3 j +1 (P(w)+P(x))-a 3j . 2 P(w) A(i) ■ 2-5 K (a 3j _ 1 + a 3j _ 2 )(P(a) + P(b)) > A(i+1) = -5 K (a 3j _-, + a 3j _ 2 )(P(a) + P(b)) < Table 4 A(i) and A(i+1) are discussed in Appendix D. Let A = P(a) + P(b). Then A ( i +2 ) = A(i + 1) + A(i) = 5 -A-(a 3 . 1 + a 3 - 2 ) > , and A(i+3) = 2A(i+2) = 2.5 k .A-(a 3 ._ 1 + a 3 - 2 ) > by lemma 4.4(5). Hence A(i+4) = A(i+3) + A(i+2) + A(i+1) = a(i'+3) > (since A(i+2) = -A(i+1) ). Hence N(G,) > N(G), a contradiction. D 21 6.9 Theorem The solution in phase 2 does not begin with the configuration 1010 and does not end with the configuration 0101 . Proof : Assume that the solution G begins with 1010. Then G begins with either 1010011 or 1010010, since any other configuration , that starts with 1010, must contain a forbidden configuration. If G starts with 1010011 it can be easily shown that replacing this by 1001011 increases M(G), a contradiction. If G starts with 1010010 it can be easily shown that a graph G, , obtained from G by replacing this by 1001010, satisfies N(G 1 )=N(G). But G 1 is not a solution since it contains the forbidden configuration 01010. Hence G cannot be a solution. The proof for the second part follows by symmetry. □ 6.10 Theorem There exists a solution for phase 2 of the form (100) 1 10[0](110) J '[0]l(001) k , i , j , k > 0. (The options are according to n(mod 3) ). Proof : Consider the rightmost occurrence of 11 in a solution for phase 2. This 11 is followed by [0]01(001) k , k >_ 0, since 111,000,01010 are forbidden con- figurations for phase 2, and 11 cannot appear in this configuration (note that this solution is not unique, since we are using here theorem 4.3 saying that there exists a solution not containing near-end edges). The corresponding analysis applies for the configuration to the left of the leftmost occurrence of 11 in the solution. Therefore it remains to show that the configuration between the right- most and the leftmost 11 is 0(110) J " . This configuration almost follows from the forbidden configuration shown in theorems 6.2 - 6.5 and 6.8 . But the solution may still contain a 110011. Consider the leftmost occurrence of 110011, 22 Case 1 : The solution starts with (100) 1 10(110) J 0(110) k . It can be shown that replacing it by (100) (110) J+k increases N(G), a contradiction. Case 2 : The solution starts with (100) 1 (110) J 0(110) k . Consider now the second leftmost 110011. If there is no second 110011 then, by symmetry, the solution is of the form (100) 1 (1 10) J (011 ) k (001 ) t . But it can be shown that replacing it by (100) 1+1 10(110) j+k ~ 1 l(001) t ~ 1 increases N(G), a contradiction. If there exists a second 110011 then the solution starts with (100) 1 (n0) J 0(110) k 0(110) t . It can be shown that replacing it by (100) 1 + 1 10(110) J ' +k+t ~ 1 increases M(G), a contradiction. 23 VII. PHASE 3 7.1 Lemma The number of paths P(a+j) to the j-th vertex from a in a configuration 111, k>0, in a qraph G can be expressed as a function of P(a) and P(b), a b where a and b are the two consecutive vertices at the beginning of the configura- tion, as follows: P(a+j) ■ F^(P(a) + P(b)) - F]_ 3 P(a) , j > 2 , 3 3 where F. denotes the t-th 3-generalized Fibonacci number, and F =0 for t <_ 0. Proof : By induction on j (see Appendix E). 7.2 Theorem In the solution in phase 3 the non-edges are disjoint. Remark : Compare this theorem, and the "case analysis" in its proof, with those of theorem 5.3 in phase 1 (concerning edges rather than non-edges). Proof : Assume that a solution G contains either 00 or 010. Case 1 : 010 . Since we are in phase 3 G must contain a 111. Wlog we assume the k k+? configuration 010(110) 111, k > 0, in G. Replacing this by (011) increases N(G), (this is shown by applying lemma 6.1 ), a contradiction. Case 2 : 00 . As in case 1 we may assume the configuration 00(110) 111, k >, 0, in G. k+1 Replacing this by 0(110) 1 increases N(G), (as before, by applying lemma 6.1 ), a contradiction. n 7.3 Theorem The solution in phase 3 is composed of two runs of consecutive disjoint non-edges, one starting with one near-end edge and the other ending with the other near-end edge, 10(110)V(011) k 01, i,k > 0, j > 3. 24 Proof : In phase 3 we have a stronger version of theorem 4.3, where we showed that the solution contains both near-end edges. By the previous theorem the non-edges are disjoint, hence the solution contains at least two runs of consecutive disjoint non-edges, one starting with one near-end edge and the other endinq with the other near-end edge (these runs may contain only one non-edge). Therefore it remains to show that there are no more runs of consecutive disjoint non-edges. Suppose G contains more such runs; then G contains at least two runs of consecutive edges (i.e. 1 J for some j > 3). Hence G contains a configura- i k t tion 01 J 0(110) 1 , j,t > 3, k > 0. Construct a graph G, by replacing it by 0(110) k+1 l J+t " 2 . We apply lemmata 6.1 and 7.1 to calculate P and P ] for several vertices in G and G, . G : 1 1 1 J ~ 2 110 (110) k_1 1 a b x y G, : 1 10 (110) k l 1 l j " 2 a b x y i-2 i-1 i 1+1 i+2 • • • • • i-2 i-1 i i+1 i+2 i-2 i-1 F j+1 (P(a)+P(b))-F j _ 2 P(a) F j+2 (P(a)+P(b)) -F j _ ] P(a) a 3k+2 (P(x)+P(y))-a 3k _ 1 P(x) a 3k+3 (P(x)+P(y))-a 3k P(x) a 3k+4 (P(x)+P(y))-a 3k+1 P(x) a 3k+3 (P(a)+P(b))-a 3(< P(a) a 3k+4 (P(a)+P(b))-a 3k+1 P(a) F 1 (P l (x)+P l (y))-F J-3 P(x) F j+1 (P 1 (x)+P 1 (y))-F j _ 2 P(x) F j+2 (P 1 (x)+P 1 (y))-F P(x) F j _ 2 (a 3k _ 1 P(b) + (a 3k + 2 +a 3k + l )P(a) ) > ° " F j-3 a 3k + 2 P(a ) * n F j-3 a 3k + 2^ P ( b )- P ( a ^ Table 5 25 The details for a(i") > are found in Appendix F. In a way similar to A(i) it can be shown that A(i+D = Fj_ 2 ( (a 3k+1 +a 3(< _ 1 )(P(a)+P(b)) + a 3k _.,P(b) ) + F j-3 (3a 3k+l " a 3k-2 )(P(b) " P(a)) > ° • and A ( i +2 ) > A(i+1) + A(i) + A(i-l) . Therefore A(i+2) 1 Fj_ 2 ( (a 3k+1 + a 3k _ 1 )(P(a)+P(b)) + a 3k _.,P(b) ) + F j-3^ 3a 3k + l " a 3k _ 2 )(P(b)-P(a)) ) + F j-3 a 3k + 2 (P(b) - P ^ a ^ - F j-3 a 3k + 2 P(a) > F j-3 (a 3k+2 + a 3k+l + a 3k . 1 )(P(b)-P(a)) + F j . 3 a 3k+2 (P(b)-2P(a) ) F. 3 a 3|<+2 (2P(b) - 3P(a)) > , by lemma 4.4(1). > therefore N(G-.) > N(G), a contradiction. D 7.4 Conjecture : The numbers of non-edges in the two runs of consecutive disjoint non-edges, in the solution of phase 3, differ by at most one (compare with conjecture 6.6). We couldn't prove this, since multiplications of a.'s are involved in the computation of N(G); however, experimental results seem to support our conjecture. 26 APPENDICES Appendix A Proof of lemma 4.4 : 1. P(i+1) < P(i) + P(i-l) + P(i-2) < 2P(i) . P(i + 1) > P(i-l) + P(1) >.|P(i) by the preceding line. 2. P(i+2) * P(i) + P(i+1) > |p(i) by (1). P(i+2) < P(i+1) + P(i) + P(i-l) < ZP(1) + P(i) + |p(i) = ilf(i) by (1). 3. P(1+3) > P(1+2) + P(1+l) > |p(1) + |p(i) = 4P(i) . P(1+3) < P(i+2) + P(i+1) + P(i) < (11 + 2 + l)P(i) = ^P(i) . 4. P(i+3) < 2P(i+2) by (1). If c.=l then P(i+3) > P(i+2) + P(i+1) + P(i) > |p(i+2) + P(i) by (1) > |p(i+2) + ^P(i+2) = §P(i+2) by (2) . 5. P(i+3) = P(i+2) + P(i+1) since c.=0, hence P(i+4) = P(i+3) + P(i+2) + P(i+1) = 2P(i+3) (since c +1 =l), and in any other case P(i+4) < 2P(i+3). Appendix B End of proof of theorem 5.4 : We prove that F t +3* ^+3 •*•• F t I+3 - 4 * 5 * F n-3I+4 for any n, < I 5 n m — 5 n+m-5 n+m-5 (which is easily proved by induction), to show that 4 ' 5 "' F n'-3I+4' F t I+1 +3 - 4 ' 5 ' F n-3I+1 . Hence F • F • • • F • F < 4.5 • F r t Q +3 t^3 r t T +3 t I+1 +3 - 4 D n-3(I+l)+4 Appendix C a(i') and a(i' + 1) in the proof of theorem 6.2 : P(d) = P(c) + P(b) + P(a) , P(e) = P(d) + P(c) + P(b) = 2P(c) + 2P(b) + P(a) , P(e) + P(d) = 3P(c) + 3P(b) + 2P(a). A(i) = P^i) - P(i) =P(c)(a 3k+ 3-3a3 k+1+ a 3k _ 2 ) + P(b) (a3 k+ 3-a 3k -3a 3k+1+ a 3k _ 2 ) + P(a)(-2a 3k+1+ a 3k _ 2 ) =P(c)(a 3k+2 -2a 3k+1+ a3 k+ a3 k _ 2 ) + P(b)(a 3k+2 -2a 3k+1+ a 3k _ 2 ) + P(a)(-2a 3k -2a 3k _ r a 3k _ 2 ) =P(c)(-a 3k+1+ 2a 3k+ a 3k _ 2 ) + P(b)(-a 3k+1+ a 3k+ a 3k _ 2 ) + P(a)(-6a 3k _ r a 3k _ 2 ) =P ( c ^ a 3k- a 3k-l^ + p ( b )(" a 3k-l) + P ( a ^- 6a 3k-r a 3k-2 } =a 3|< _ 1 ( P(c)-P(b)-6P(a) ) - a 3k _ 2 P(a) =a 3k _ 1 (P(w)-5P(a))-a 3k _ 2 P(a) <(-4a 3k _ r a 3k _ 2 )P(a) < since P(w) < P( a ) . 28 A(i-M) = P(c)(a 3k+4 -3a 3k+2 +a 3k _ 1 ) + W^^"^-^^^..,) + ■ P(a)(-2a 3k+2+ a 3k _ 1 ) = P(c)(a 3k+3 -2a 3k+2+ a 3k+1+ a 3k _ 1 ) + PMUg^ag^+a^) + P(a)(-2a 3k+1 -2a 3k+ a 3k _ 1 ) = P(c)(a 3k+1+ a 3k _- 1 ) + P(b)a 3k . 1 + P(a)(-2a 3k+1 -3a 3k . 1 ) =a 3k+1 (P(c)-2P(a)) + a 3k . 1 (P(c)+P(b)-3P(a)) > , since P(c) = P(b)+P(a)+P(w) > 2P(a)+2P(w) > 3P(a) by lemma 4.4(1). Appendix D A(i) and A(i+1) in the proof of theorem 6.8 : Note that P(d) = 2(P(a)+P(b)), P(e) = 3(P(a)+P(b)) and P(d)+P(e) = 5(P(a)+P(b)), and let A = P(a)+P(b). 4(1)- a 3j 5 K ^A - a 3j _ 3 2-5 K 7, - 5 K ( (a 3j+1 + a 3j+2 )5A - (a 3j ._-| +a 3j _ 2 )2A ) =5 k A(25a 3j -10a 3j _ 3 -5a 3j+r 5a 3j+2+ 2a 3j . 1+ 2a 3j . 2 ) =5 k A(-10a 3j+1+ 20a 3j+ 2a 3j . 1+ 2a 3j . 2 -10a 3j . 3 ) =5 k A(10a 3 .-8a 3 .. ] -8a 3j _ 2 -10a 3j _ 3 ) =5 k A(2a 3j . 1+ 2a 3j . 2 ) > . 4(i+l)=a 3j+1 5 k+2 A - a 3j _ 2 2-5 k+1 A - 2-5 k ( (a 3j+1 +a 3j+2 )5A - (a 3j . 1 +a 3j . 2 )2A ) =5 k A(25a 3j+1 -10a 3j . 2 -10a 3j+1 -10a 3j+2+ 4a 3j . 1+ 4a 31 _ 2 ) =5 k A(-10a 3j+2+ 15a 3j+1+ 4a 3j _ r 6a 3l ._ 2 ) =5 k A(5a 3j+1 -10a 3j+ 4a 3j . r 6a 3j _ 2 ) =5 A(-5a 3j+ 9a 3j . r a 3j _ 2 ) =-5 k A(a 3J _ 1 +a 3J _ 2 ) < (since a 3j = Za^.,). Appendix E Proof of lemma 7.1 : P(a+2) = P(a) + P(b) and P(a+3) = 2(P(a) + P(b)). Assume the formula holds for j