The Random Edge Rule on Three-Dimensional Linear Programs 1 (extended abstract)

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1 arxi:math/ [math.co] 9 Jan 2003 The Random Edge Rule on Three-Dimensional Linear Programs 1 (extended abstract) Volker Kaibel 2 Rafael Mechtel 2 Micha Sharir 3 Günter M. Ziegler 2 October 10, 2017 Abstract The worst-case expected length of the path taken by the simplex algorithm with the Random Edge piot rule on a 3-dimensional linear program with n constraints is shown to be bounded by for large enough n. 1 Introduction n f(n) n The Random Edge piot rule is undoubtedly the most natural, and simplest (randomized) piot rule for the simplex algorithm: At each iteration, proceed from the current ertex of the polyhedron P of feasible solutions to an improing neighbor, chosen uniformly at random in the one-dimensional skeleton (i.e., the graph) of P. 1 Work on this paper by Micha Sharir was supported by NSF Grants CCR and CCR , by a grant from the U.S.-Israeli Binational Science Foundation, by a grant from the Israel Science Fund (for a Center of Excellence in Geometric Computing), and by the Hermann Minkowski MINERVA Center for Geometry at Tel Ai Uniersity. Volker Kaibel and Günter M. Ziegler were supported by the DFG-Forschergruppe Algorithmen, Struktur, Zufall (FOR 413/1-1, Zi 475/3-1). Günter M. Ziegler was supported by a DFG Leibniz grant. Part of the work was done during the workshop Towards the Peak at La Claustra, Switzerland, August MA6 2,TUBerlin, 10623Berlin, Germany, {kaibel,mechtel,ziegler}@math.tu-berlin.de 3 SchoolofComputerScience,TelAiUniersity,Tel-Ai69978,IsraelandCourantInstituteof Mathematical Sciences, New York Uniersity, New York, NY 10012, USA, michas@post.tau.ac.il 1

2 Despite its simplicity, this algorithm until now has resisted almost all attempts to analyze its worst-case behaior, with a few exceptions for special cases, among them the linear assignment problem(toey [6]), certain linear programs on cubes, including the Klee-Minty cubes (Kelly [4], Gärtner, Henk & Ziegler [2]), and d-dimensional linear programs (i.e., dim(p) = d) with at most d+2 constraints (Gärtner et al. [3]). All known results leae open the possibility that the Random Edge piot rule yields a strongly polynomial time algorithm it might be een quadratic. In particular, it is not fooled by the deformed products (defined by Amenta and Ziegler [1]), which yield the well-known exponential examples for all the classical deterministic piot rules. Here, we only treat the case of 3-dimensional linear programs, which, of course, is soled in linear time by eery (een deterministic) finite ariant of the simplex algorithm. Neertheless, due to the remarks made aboe, it seems interesting to analyze the Random Edge piot rule for this case and here too, it seems that accurate analysis of Random Edge is quite hard. With the usual reductions (see, e.g., Ziegler [7, Lect. 3]), we may assume that our linear program is min{x 3 : x P}, where P is a 3-dimensional simple polytope (its graph is 3-regular) with exactly n facets, and hence 2n 4 ertices and 3n 6 edges, and no two ertices hae the same objectie function alue x 3. Thus we hae an ordering of the ertices 2n 5, 2n 6,..., 1, 0 by decreasing objectie function (i.e., by height). Here 0 = min is the unique minimal (lowest) ertex of the linear program, while 2n 5 = max is the unique maximal (highest) ertex. The expected length of the path (i.e., the number of piot steps) taken by the simplex algorithm on the linear program, starting at ertex of P and using the Random Edge rule, is then gien by E( 0 ) = 0 and E() = 1+ 1 d E(w j ) ( 0 ), d j=1 where w 1,...,w d are the lower neighbors of. It is easy to see that (in addition to the unique maximal ertex and the unique minimal ertex) there are n 3 ertices with d = 1 (1-ertices) and n 3 ertices with d = 2 (2-ertices); this is the 3-dimensional case of the Dehn-Sommerille equations [7, Thm. 8.21]. Define f(n) to be the maximum expected number of piot steps taken by the Random Edge algorithm on any 3-dimensional linear program with n constraints. While it is quite straightforward to construct a sequence of examples with E() 4 3 n const, our results in Section 2 (Theorem 2.2) show f(n) n for infinitely many n in arithmetic progression ( 1721 = ). In Section 3 we proe f(n) n

3 (Theorem 3.2). Both results taken together, this yields that n f(n) n holds for all large enough n. In particular, asymptotically f(n) lies between ( 4+ε) n 3 and ( 3 ε) n for some ε > 0. Determining the right coefficient seems, howeer, 2 to be ery hard. 2 Lower Bounds The expected number of piot steps required by the simplex algorithm using Random Edge only depends on the graph of the polytope, directed ia the objectie function. Therefore, we will describe our examples yielding lower bounds on f(n) by the corresponding directed graphs. The following result proides a nice certificate for a directed graph to come from a 3-dimensional linear program. Theorem 2.1 (Mihalisin and Klee [5]). A directed graph D (without loops and parallel arcs) is induced by a 3-dimensional linear program if and only if it is planar and 3-connected (as an undirected graph), it is acyclic with a unique source and a unique sink, it has a unique local sink in eery face cycle (these are the non-separating induced cycles), and it admits three directed paths from its source to its sink that hae disjoint sets of interior nodes. 2.1 Duals of Cyclic Polytopes Example 1. Our first sequence of examples are wedges, i.e., they are combinatorially equialent to duals of cyclic polytopes. Figure 1 depicts the orientations of the edges. Here, as well as in the sequel, our conention is that the ordering of the ertices from left to right in the figure defines the (decreasing) ordering of the ertices according to the objectie function. It is easy to see that the conditions of the Mihalisin Klee theorem are satisfied. 2n 5 2n 6 2n 8 2j+1 2j 2 2n j 2j = min Figure 1: The example on the dual cyclic polytope. 3

4 For the expected number of piot steps E( i ), we then hae the starting alues E( 0 ) = 0 and E( 1 ) = 1, and the recurrences E( 2j ) = E( 2j 1 )+1, E( 2j+1 ) = 1 2( E(2j )+E( 2j 2 ) ) +1 for 0 < j n 4. Thus, using induction, we obtain E( 2j )+2E( 2j+1 ) = 4j +2 for 0 j n 4. In particular, for j = n 4 this yields { } max E( 2n 8 ),E( 2n 7 ) 2.2 Improed Lower Bounds 4n The next examples are based on the construction of a backbone polytope : This will be a simple 3-polytope P k with k +2 facets and 2k ertices, of which k ertices k 1,..., 0 form a decreasing chain, such that 0 is the minimal ertex, and i 1 is the only lower neighbor of i, for i > 0. Constructing the backbone. We start with the simplex, obtained for k = 2 with ertices w 0,w 1, 1, 0, as shown on the left. We then inductiely cut off the ertex k 1 by a plane, replacing it by a small triangle, as shown on the right w 0 w = min w 0 w 1 w 2 w k 1 k 1 k = min Starting with the simplex for k = 2 Example 2. Our second sequence of examples is obtained from the backbone polytopes P k by performing three specific ertex cuts at each ertex i, for i = 0,...,k 1. Before cutting (in P k ), each ertex i (i > 0) has indegree 2, while 0 has indegree 3. The two ertex cuts are supposed to create at each ertex i the following configuration (where again all edges are directed to the right ): i,4 i,3 i,2 i,6 i,5 i,1 i,0 This creates a simple polytope P k with n = k +2+3k = 4k +2 facets. 4

5 Our starting ertex for Random Edge on this example will be k 1,6 ; the expected number of steps taken by Random Edge is the sum of the probabilities p e that the edge e is traersed. We think of these probabilities as a flow from k 1,6 to 0,0. Our figure indicates the flow alues on the edges, for a flow of total alue 8; equialently, these are the transersal probabilities in units of i,4 i,2 i, i,6 i,5 i,1 i,0 We get the same alues for each of the triple-ertex-cut-off configurations, except for the last one, which has no edge leaing the global sink 0,0. Thus the expected number of Random Edge steps, starting from k 1,6, is E( k 1,6 ) = k = n 32 16, with k = 1 43 (n 2). Asymptotically, this yields a better lower bound, due to > The graph of P k looks like this: 1,3 0, w 0 w 1 w 3 k 1,6 k 1,0 1,6 1,0 0,6 0,0 k 1,3 = min Example 3. The last examples produced k = 1 (n 2) ertices which are not 4 used. We will further improe the lower bound by using more facets in each local configuration and thus reducing the number of unused ertices (though our final example will still hae linearly many unused ertices). We use the same backbone polytope as before, but we replace each ertex i, for i = 0,...,k 1, with the following graph. (We do not gie an explicit polytopal construction for this example, but it can be constructed by cutting off ertices and edges of the backbone polytope. Alternatiely, such a construction is proided by the Mihalisin-Klee Theorem.) To analyze the random path length through this graph we send from i, units of flow through the network. 5

6 90 i, i, The total flow through all edges is For each configuration 9 facets are required. Together with the facets from the backbone construction, this yields n = 10k + 2. Hence we get E( k 1,18 ) = k = n , where = = > In contrast to the preceeding examples, this example does not contain a directed Hamiltonian path. Summarizing, we hae proed the following bound. Theorem 2.2. For n = 10k +2 12, holds. f(n) n Starting from the source. By splitting the maximal ertex, one can also construct examples where the expected number of steps starting at the maximal ertex is at least ( 1721 ε)n. This obseration is due to Günter Rote Upper Bounds Consider any linear program on a simple 3-polytope with the notations as described in the introduction. For a ertex, let N 1 () (resp., N 2 ()) denote the number of 1-ertices (resp., 2-ertices) that are not higher than (including itself). Put N() = N 1 ()+N 2 (). This is the number of ertices lower than. The core of our upper bound on f(n) is the following result. Theorem 3.1. For each ertex, other than the maximal ertex 2n 5, we hae Theorem 3.1 implies E() N 1() N(). E() 130(n 3) = n

7 for all. Here, the 15 comes from the fact that Theorem 3.1 is proed only for 29 2n 5 ; therefore, we bound E( 2n 5 ) by i=1 E(w i), where w 1,w 2,w 3 are the neighbors of 2n 5, and we exploit N(w 1 )+N(w 2 )+N(w 3 ) 6n 21. Theorem 3.2. For eery 3-dimensional linear program with n constraints, the expected number of piot steps taken by Random Edge is not more than n In the remaining part of this section, we briefly sketch the proof of Theorem 3.1. It proceeds by deriing the generic inequality E() αn 1 ()+βn(), (1) where, for most of the proof, α and β are treated as indeterminates. Each step of the proof yields a linear inequality on α and β that needs to be satisfied in order to imply (1). The proof is then completed once it is shown that (α,β) = ( 46, 42) satisfies all inequalities; in fact, it is optimal with respect to the objectie function α + 2β (see below for more details). Inequality (1) is proed by induction on N(). The base case N() = 0 is obious, since is the optimum in this case, and E() = 0. Suppose the theorem holds for all ertices lower than some ertex. We express E() in terms of the expected costs E(w) of certain ertices w that are reachable from ia a few downward edges. The general form of such a recursie expression will be k E() = c+ λ i E(w i ), i=1 where λ i > 0 for each i = 1,...,k, and i λ i = 1. Since we assume by induction that E(w i ) αn 1 (w i ) + βn(w i ), for each i, it suffices to show that k k αλ i (N 1 () N 1 (w i ))+ βλ i (N() N(w i )) c. Write i=1 i=1 1 (w i ) = N 1 () N 1 (w i ), (w i ) = N() N(w i ), for i = 1,...,k. (Of course, these terms are defined with respect to the currently considered ertex.) Note that (w i ) is the distance between and w i, that is, one plus the number of ertices between and w i. We thus need to show that k αλ i 1 (w i )+ i=1 k βλ i (w i ) c. (2) i=1 7

8 This requires a quite extensie case analysis, of which we present only the beginning in this extended abstract in order to indicate the kinds of arguments used. The complete case analysis is gien in the appendix. Case 1: is a 1-ertex. Let w 1 denote the target of the unique downward edge emanating from as in the following figure, where (here and in all subsequent figures) each edge is labelled by the probability of reaching it from. 1 w 1 In this case, E() = 1+E(w 1 ) holds. In the setup presented aboe, we hae c = 1, 1 (w 1 ) 1, and (w 1 ) 1, thus (2) is implied by α+β 1. (3) Case 2: is a 2-ertex. Let w 1 andw 2 denote thetargets ofthe two downward edges emanating from, where w 2 is lower than w 1. w 1 w 2 We hae E() = E(w 1)+ 1 2 E(w 2), hence we need to require that Note that (w 1 ) 1. α 2 1(w 1 )+ α 2 1(w 2 )+ β 2 (w 1)+ β 2 (w 2) 1. Case 2.a: (w 2 ) 4. Ignoring the effect of the 1 (w j ) s, it suffices to require that which will follow if β 2 (w 1)+ β 2 (w 2) 1, β 2 5. (4) 8

9 Case 2.b.i: (w 2 ) = 3 and one of the two ertices aboe w 2 and below is a 1- ertex. In this case 1 (w 2 ) 1 and (w 1 )+ (w 2 ) 4, so (2) is implied by 1 α+2β 1. (5) 2 We skip the remaining cases 2.b.ii and 2.c ( (w 2 ) = 2) in this extended abstract; the latter one splits into quite a large number of subcases, which become slightly more inoled. One ends up with roughly 24 linear inequalities in addition to (3), (4), (5). Assuming we hae α and β that satisfy all these inequalities, each of the induction steps is justified, and the inequality (1) follows. Since we always hae N 1 (),N 2 () n 3, we obtain E() αn 1 ()+β(n 1 ()+N 2 ()) = (α+β)n 1 ()+βn 2 () (α+2β)(n 3). Hence we choose (α,β) to minimize α + 2β, subject to all the deried inequalities. This is indeed the choice appearing in the statement of Theorem Discussion The improed lower bounds of Section 2.2 arose from complete enumerations for small n. In particular, the lower bounds proided by examples 2 and 3 are tight for n = 10,12, respectiely. Thus, we hae f(10) = = 9.75 and f(12) = Weareconinced thattheboundintheorem3.2isnottight. Infact, precisely two of the inequalities of the proof of Theorem 3.1 are tight for (α,β) = ( 46, 42 ). The two corresponding subcases thus constitute the bottleneck for the current upper bound. In order to improe the bound, one should expand these two subcases further, aiming at replacing those two inequalities by weaker ones (at the cost of a longer proof). As a matter of fact, in an earlier (unpublished) ersion of this manuscript we had obtained an upper bound of 1.5 n, using a somewhat more compact enumeration scheme. The current scheme is a refinement, based on further expansion of the preceding one. At this point, we hae no real sense of what the exact bound should be. The refinement of the approach in this section, as just outlined, is not likely to yield substantial improements in the upper bound, so a radically different approach is probably called for. Such an improement might be based on the obseration that certain local structures inole 1-ertices with one of its upward neighbors lying aboe. In fact, if the portion below contains k such ertices, there must exist at least k ertices higher than, so the upper bound αn 1 ()+βn() is much smaller than (α+2β)(n 3). As a matter of fact, the lower bounds deried in Section 2 do take this constraint into consideration. Another obseration is that the proof of Theorem 3.1 (as detailed in the appendix) uses (twice) the 3-connectiity of the edge graph of P, but it does not use its planarity at all, although it does occassionally run into nonplanar configurations. It 9

10 is conceiable that further refinement stages might reach nonplanar configurations, whose exclusion would allow us to further improe the bound. What if we also drop the 3-connectiity assumption? Then we need to consider additional cases, which cause our upper bound to increase. The best upper bound we hae at the moment for this relaxed situation is 13n/8 = n, but we are coninced that it too can be further improed. Acknowledgements. We are grateful to Emo Welzl and Günter Rote for inspiring discussions and helpful comments. References [1] N. Amenta and G. M. Ziegler, Deformed products and maximal shadows, in Adances in Discrete and Computational Geometry, B. Chazelle, J. E. Goodman, and R. Pollack, eds., ol. 223 of Contemporary Mathematics, Amer. Math. Soc., Proidence RI, 1998, pp [2] B. Gärtner, M. Henk, and G. M. Ziegler, Randomized simplex algorithms on Klee-Minty cubes, Combinatorica, 18 (1998), pp [3] B. Gärtner, J. Solymosi, F. Tschirschnitz, P. Valtr, and E. Welzl, One line and n points, STOC, (2001), pp [4] D. G. Kelly, Some results on random linear programs., Methods Oper. Res., 40 (1981), pp [5] J. Mihalisin and V. L. Klee, Conex and linear orientations of polytopal graphs, Discrete Comput. Geometry, 24 (2000), pp [6] C. A. Toey, Low order polynomial bounds on the expected performance of local improement algorithms, Math. Programming, 35 (1986), pp [7] G. M. Ziegler, Lectures on Polytopes, ol. 152 of Graduate Texts in Mathematics, Springer-Verlag, New York, Reised edition, 1998; Updates, corrections, and more at 10

11 Appendix This appendix contains the complete case analysis of the proof of Theorem 3.1 in the extended abstract (including the cases presented there). Case 1: is a 1-ertex. Let w 1 denote the target of the unique downward edge emanating from as in the following figure, where each edge is labelled by the probability of reaching it from. 1 w 1 In this case, E() = 1+E(w 1 ) holds. In the setup presented aboe, we hae c = 1, 1 (w 1 ) 1, and (w 1 ) 1, thus (2) is implied by α+β 1. (3) Case 2: is a 2-ertex. Let w 1 andw 2 denote thetargets ofthe two downward edges emanating from, where w 2 is lower than w 1. w 1 w 2 We hae E() = E(w 1)+ 1 2 E(w 2), hence we need to require that Note that (w 1 ) 1. α 2 1(w 1 )+ α 2 1(w 2 )+ β 2 (w 1)+ β 2 (w 2) 1. Case 2.a: (w 2 ) 4. Ignoring the effect of the 1 (w j ) s, it suffices to require that which will follow if β 2 (w 1)+ β 2 (w 2) 1, β 2 5. (4) 11

12 Case 2.b.i: (w 2 ) = 3 and one of the two ertices aboe w 2 and below is a 1- ertex. In this case 1 (w 2 ) 1 and (w 1 )+ (w 2 ) 4, so (2) is implied by 1 α+2β 1. (5) 2 Case 2.b.ii: (w 2 ) = 3 and the two ertices between and w 2 are 2-ertices. Denote the second intermediate ertex as. We may assume that is reachable from, otherwise we can ignore it and reduce the situation to Case 2.c treated below (be choosing another ordering of the ertices producing the same oriented graph). Three subcases can arise. First, assume that none of the three edges that emanate from w 1 and further down reaches w 2. Denote by x,y the two downward neighbors of and by z the downward neighbor of w 1 other than. The ertices x,y,z need not be distinct but none of them coinicdes with w 2. w 1 1/8 w 2 1/8 x y c = 7/4 z We hae here c = 7/4. To make the analysis simpler to follow isually, we present it in a table. Each row denotes oneofthetarget ertices w 2,x,y,z, multiplied by theprobability ofreaching it from. The left (resp., right) column denotes a lower bound on the corresponding quantities 1 ( ) (resp., ( )). To obtain an inequality that implies (2), one has to multiply each entry in the left (resp., right) column by the row probability times α (resp., times β), and require that the sum of all these terms be c. w /8x 0 4 1/8y 0 5 z 0 4 Note the following: (a) We do not assume that the rows represent distinct ertices (in fact, x = z is implicit in the table); this does not cause any problem in applying the rule for deriing an inequality from the table. (b) We hae to squeeze the ertices so as to make the resulting inequality as sharp (and difficult to satisfy) as possible; thus we made one of x,y the farthest ertex, because making z the farthest ertex would hae made the inequality easier to satisfy. 12

13 or We thus obtain ( ) β 7 4 4, β (6) Next, assume that w 2 is connected to. In this case w 2 is a 1-ertex, and we extend the configuration to include its unique downward neighbor w 3. w 1 w 2 w 1/8 5/8 3 c = 19/8 1/8 x y Let x denote the other downward neighbor of and let y denote the other downward neighbor of w 1. In the following table, the worst case is to make w 3 and y coincide, and make x the farthest ertex. We then obtain or α+ 5/8w /8x 1 5 y 1 4 ( ) β , α β (7) Finally, assume that w 2 is connected to w 1. Here too w 2 is a 1-ertex, and we extend the configuration to include its unique downward neighbor w 3. c = 5/2 w 1 w 2 1/8 3/4 w 3 Denoting by x,y the two downward neighbors of, our table and resulting inequality become 3/4w α+ 33 1/8x β 5 2, (8) 1/8y /8 x y

14 which, by the way, is stronger than (7). Case 2.c: (w 2 ) = 2. Hence, the only remaining case is that w 1 and w 2 are the two ertices immediately following. Case 2.c.i: w 1 is a 1-ertex (whose other upward neighbor lies aboe ). Its unique downward edge ends at some ertex which is either w 2 or lies below w 2. Assume first that this ertex coincides with w 2, which makes w 2 a 1-ertex, whose unique downward neighbor is denoted as. The local structure, table, and inequality are w 1 w 2 1 c = 5/ α+3β 5 2. (9) Suppose next that the downward neighbor w 3 of w 1 lies below w 2. We get w 1 w 2 w3 c = 3/2 w w α+ 5 2 β 3 2. (10) Case 2.c.ii: w 1 is a 2-ertex, both of whose downward neighbors lie strictly below w 2. Denote these neighbors as w 3,w 4, with w 3 lying aboe w 4. w 1 w 2 w3 w 4 c = 3/2 We may assume (w 3 ) = 3 (i.e., there is no ertex between w 2 and w 3 ), since (w 3 ) 4 requires β 6 as the sharpest inequalitity, which is already implied 13 by (6). Case 2.c.ii.1: w 2 is a 1-ertex. Then the table and inequality become w w w α β 3 2. (11) 14

15 Case 2.c.ii.2: w 2 is a 2-ertex but w 3 is a 1-ertex. Then w 3 (which satisfies (w 3 ) = 3) is connected either to w 2 or to a ertex aboe. In the former case, let x denote the other downward neighbor of w 2, and let y denote the unique downward neighbor of w 3. The local structure looks like this (with x,y,w 4 not necessarily distinct): x y w 1 w2 w 3 w4 c = 5/2 The table depends on whether x precedes or succeeds w 3. In the former case the (worst) table and inequality are x 0 3 y 1 5 w α+ 9 2 β 5 2. (12) In the latter case the (worst) table and inequality are x 1 4 y 1 4 w α β 5 2. (13) The next case is where the other upward neighbor of w 3 lies aboe. Let x,y denote the two downward neighbors of w 2, and let z denote the unique downward neighbor of w 3. The local structure is: x y w 1 z w 2 w 3 w4 c = 9/4 The table depends on how many of x,y precede w 3. If both precede w 3, the table and inequality become x 0 3 y 0 4 z 1 6 w α β 9 4. (14) 15

16 If only one of x,y precedes w 3, say x, the table and inequality become which is weaker than (14). x 0 3 y 1 5 z 1 5 w α β 9 4, (15) Finally, if none of x,y precedes w 3, the table and inequality become x 1 4 y 1 4 z 1 5 w α+ 9 2 β 9 4. (16) Case 2.c.ii.3: Both w 2 and w 3 are 2-ertices. We hae to consider the following type of configuration (where x,y,z,t,w 4 need not all be distinct, but x y and z t, and we may assume x t, y z; also, because (w 3 ) = 3, both x and y are lower than w 3 ): c = 9/4 1/8 w 1 1/8 w 2 w w 4 3 Intuitiely, a worst table is obtained by squeezing x,y,z,t, and w 4 as much to the left as possible, placing two of them at distance 4 from, two at distance 5, and one at distance 6. Howeer, squeezing them this way will make some pairs of them coincide and form 1-ertices, which will affect the resulting tables and inequalities. Suppose first that among the three heaier targets x,y,w 4, at most one lies at distance 4 from. The worst table and the associated inequality are (recall that x y): x 0 4 y 0 5 1/8z 0 4 1/8t 0 6 w x y z t 19 4 β 9 4. (17) Suppose then that among {w 4,x,y}, two are at distance 4 from, say w 4 and y. Then w 4 = y is a 1-ertex, and we denote by w its unique downward neighbor. The local structure is: 16

17 x z 1/8 t w 1 1/8 w 2 w3 w4 c = 1 w Two equally worst tables, and the resulting common inequality are x 1 5 1/8z 1 6 1/8t 1 7 w 1 5 x 1 6 1/8z 1 6 1/8t 1 5 w 1 5 α β (18) Case 2.c.iii: w 1 is a 2-ertex that reaches w 2. Then w 2 is a 1-ertex, and we denote by x its unique downward neighbor. 3/4 x w 1 w 2 = w 3 w 4 c = 9/4 A crucial obseration is that x cannot be equal to w 4. Indeed, if they were equal, then w 4 would be a 1-ertex. w 1 w 2 w 4 In this case, cutting the edge graph G of P at the downward edge emanating from x and at the edge entering would hae disconnected G, contradicting the fact that G is 3-connected. We first dispose of the case where x lies lower than w 4. The table and inequality are 3/4x 1 4 α β 9 4. (19) w In what follows we thus assume that x lies aboe w 4. 17

18 Case 2.c.iii.1: x is a 1-ertex that precedes w 4. Suppose first that w 4 is the unique downward neighbor of x. Then w 4 is a 1-ertex, and we denote its unique downward neighbor by z. The local structure, table and inequality are: w 1 w 2 x w 4 3/4 3/4 1 z c = 4 z 3 5 3α+5β 4. (20) Suppose next that the unique downward neighbor y of x is not w 4. The local structure, table and inequality look like this (y is drawn aboe w 4 because this yields a sharper inequality): w 1 w 2 x y w 4 3/4 3/4 c = 3 3/4y 2 4 w α+ 17 β 3. (21) 4 Case 2.c.iii.2: x is a 2-ertex that precedes w 4. This subcase splits into seeral subcases, where we assume, respectiely, that (w 4 ) 6, (w 4 ) = 4, and (w 4 ) = 5. Case 2.c.iii.2(a). Suppose first that (w 4 ) 6. The configuration looks like this: w 1 w 2 x w4 3/4 c = 9/4 The table and inequality are 3/4x 1 3 w α β 9 4. (22) Case 2.c.iii.2(b). Suppose next that (w 4 ) = 4, and that one of the downward neighbors of x is w 4. Let z denote the other downward neighbor. w 4 is a 1-ertex, and we denote by w its unique downward neighbor. 18

19 3/8 w 1 w 2 x 5/8 3/4 3/8 w 4 z w c = 29/8 The 3-connectiity of the edge graph of P implies, as aboe, that w z. Since we assume that (w 4 ) = 4, z also lies below w 4, and the table and inequality are 5/8w 2 5 3/8z 2 6 2α β (23) Suppose next that (w 4 ) = 4 and w 4 is not a downward neighbor of x. Denote those two neighbors as w and z, both of which lie lower than w 4, by assumption, and are clearly distinct. The configuration, table and inequality look like this: 3/4 x 3/8 3/8 w z c = 3 w 1 w 2 w 4 α 1 β w /8w 1 5 3/8z 1 6 α+ 41 β 3. (24) 8 Case 2.c.iii.2(c). It remains to consider the case (w 4 ) = 5. Let z denote the unique ertex lying between x and w 4. We may assume that z is connected to x, for otherwise z is not reachable from, and we might as well reduce this case to the case (w 4 ) = 4 just treated. Consider first the subcase where the other downward neighbor of x is w 4 itself. Then w 4 is a 1-ertex, and we denote by w its unique downward neighbor. This subcase splits further into two subcases: First, assume that z is a 1-ertex, and let y denote its unique downward neighbor. Clearly, y must lie below w 4 (it may coincide with or precede w). The configuration looks like this: 3/8 3/8 w 1 w 2 x 5/8 3/4 3/8 z w4 y w c = 4 The table and inequality are 3/8y 3 6 5/8w 3 6 3α+6β 4. (25) 19

20 In the other subcase, z is a 2-ertex; we denote its two downward neighbors as y and t. The ertices w,y,t all lie below w 4 and may appear there in any order. The configuration looks like this: w 1 w 2 3/16 3/8 3/16 x 5/8 3/4 3/8 z w4 t y w c = 4 The table and inequality are 3/16y 2 6 3/16t 2 7 5/8w 2 6 2α+ 99 β 4. (26) 16 Consider next the subcase where w 4 is not a downward neighbor of x. Denote the other downward neighbor of x as y, which lies strictly below w 4. This subcase splits into three subcases. First, assume that z is a 1-ertex, and denote its unique downward neighbor as w. The configuration looks like this: 3/8 y 3/8 w w 1 w 2 x 3/4 3/8 z w4 c = 27/8 The table and inequality are w /8y 2 6 3/8w 2 5 2α β (27) Second, assume that z is a 2-ertex, so that none of its two downward neighbors is w 4. Denote these neighbors as w and t. All three ertices y,t,w lie strictly below w 4. The configuration looks like this: 3/8 3/16 3/16 w 1 w 2 x 3/4 3/8 z w4 y t w c = 27/8 20

21 The table and inequality are w /8y 1 6 3/16w 1 6 3/16t 1 7 α β (28) Finally, assume that z is a 2-ertex, so that one of its two downward neighbors is w 4. Denote the other neighbor as w. In this case w 4 is a 1-ertex, and we denote its unique downward neighbor as t. All three ertices y,t,w lie strictly below w 4. The configuration looks like this: 3/8 3/16 w 1 w 2 x 3/16 7/16 3/4 3/8 z w4 y w t c = 61/16 The table and inequality are 3/8y 2 6 3/16w 2 7 7/16t 2 6 2α β 61 16, (29) which, by the way, is weaker than (26). 21