Minimizing Intra-Edge Crossings in Wiring Diagrams and Public Transportation Maps

Transcription

1 Minimizing Intra-Edge Crossings in Wiring Diagrams and Pblic Transportation Maps Marc Benkert 1, Martin Nöllenbrg 1, Takeaki Uno 2, and Alexander Wolff 1 1 Department of Compter Science, Karlsrhe Uniersity, Germany. WWW: 2 National Institte of Informatics, Tokyo, Japan. no@nii.jp Abstract. In this paper we consider a new problem that occrs when drawing wiring diagrams or pblic transportation networks. Gien an embedded graph G = (V, E) (e.g., the streets sered by a bs network) and a set L of paths in G (e.g., the bs lines), we want to draw the paths along the edges of G sch that they cross each other as few times as possible. For esthetic reasons we insist that the relatie order of the paths that traerse a node does not change within the area occpied by that node. Or main contribtion is an algorithm that minimizes the nmber of crossings on a single edge, } E if we are gien the order of the incoming and otgoing paths. The difficlty is deciding the order of the paths that terminate in or with respect to the fixed order of the paths that do not end there. Or algorithm ses dynamic programming and takes O(n 2 ) time, where n is the nmber of terminating paths. 1 Introdction In wiring diagrams or pblic transportation networks many paths mst be drawn on the same nderlying graph, see Figres 1 and 2. In order to make the reslting layot as nderstandable as possible it is desirable to (a) aoid crossings whereer possible and (b) insist that the relatie order of the lines that traerse a node does not change in that node. For example, note that the sbway line 5, which passes nder the main station of Cologne (Köln Hbf), crosses lines soth of the station in the clipping of the pblic transport map of Cologne in Figre 2. The crossing is not hidden nder the rectangle that represents the station. This makes it easier to follow the sbway lines isally. We model the problem as follows. We assme that we are gien an ndirected connected graph G = (V, E) together with an embedding in the plane. The graph represents the nderlying strctre of the wiring or the road/tracks in the case of a transportation network. We are also gien a set L of lines in G. They represent the cables in a wiring diagram or the lines in a transportation network. A line l L is an edge seqence e 1 = 0, 1 }, e 2 = 1, 2 },..., e k = k 1, k } E Spported by grant WO 758/4-2 of the German Research Fondation (DFG).

2 2 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff Fig. 1: Clipping of a wiring diagram. Fig. 2: Clipping of the pblic transport network of Cologne. that forms a simple path in G. The stations 0 and k are the terminal stations for line l while 1,..., k 1 are intermediate stations. Or aim is to draw all lines in L sch that the nmber of crossings among pairs of lines in L is minimized. To get a flaor of the problem obsere that the strctre of G enforces certain crossings, see Figre 3a: lines l 1 and l 2 se exactly the path, w 1, w 2, together. The graph strctre (indicated by the first and last line segments of each line) enforces that l 1 enters station aboe l 2 while it leaes below l 2, ths l 1 and l 2 hae to cross somewhere between and. Howeer, fixing the location of the crossing of l 1 and l 2 determines crossings with other lines that hae a terminal stop in w 1 or w 2. If there is a line l that enters between l 1 and l 2 and terminates at w 2 (see Figre 3b), the crossing between l 1 and l 2 shold be placed between w 2 and. Now sppose there is another line l that enters between l 2 and l 1 from the right and terminates at w 2, see Figre 3c. Then at least one of the lines l and l intersects one of the lines l 1 or l 2, no matter where the crossing between l 1 and l 2 is placed. l 1 l 1 l 1 w 1 w 2 w 1 w 2 w 1 w 2 l l l l2 l 2 l 2 (a) (b) (c) Fig. 3: Different placements of the necessary intersection between lines l 1 and l 2 on the path, w 1, w 2,. In (c) at least one of the lines l and l has to intersect one of the lines l 1 or l 2.

3 Minimizing Intra-Edge Crossings 3 We can abstract from geometry as follows. Gien an edge, } in G and the orders of the lines in and that traerse and, respectiely, we ask for the order of all lines that enter, } in and for the order in which all lines leae, } in. Then the nmber of crossings on, } is the nmber of transpositions needed to conert one order into the other. The difficlty is deciding the order of the lines that terminate in or with respect to the fixed order of the lines that traerse both and. This is the one-edge layot problem that we stdy in Section 2. In contrast to the well-known NP-hard problem of minimizing crossings in a two-layer bipartite graph [3] the one-edge layot problem is polynomially solable. The main reason is that there is an optimal layot of the lines that pass throgh edge, } sch that no two lines that terminate in intersect and no two lines that terminate in intersect. This obseration allows s to split the problem and to apply dynamic programming. It is then rather easy to come p with an O(n 5 )-time soltion and with some effort we cold redce the rnning time to O(n 2 ), where n is the nmber of lines that do not terminate in or. A soltion of the general line layot problem, i.e., a simltaneos crossingminimal soltion for all edges of a graph, wold be of interest as a second (and mostly independent) step for drawing bs or metro maps, a topic that has receied some attention lately, see the work of Nöllenbrg and Wolff [4] and the references therein. Howeer, in that direction of research the focs has so far been exclsiely on drawing the nderlying graph nicely, and not on how to embed the bs or metro lines along the network. We gie some hints in Section 3 why already the two-edge layot problem seems to be sbstantially harder than the one-edge layot. A agely similar problem has been considered by Cortese et al. [1]. Gien the drawing of a planar graph G they widen the edges and ertices of the drawing and ask if a gien combinatorial cycle in G has a plane embedding in the widened drawing. 2 A Dynamic Program for One-Edge Layot In this section we consider the following special case of the problem. Problem 1. One-edge layot We are gien a graph G = (V, E) and an edge e =, } E. Let L e be the set of lines that se e. We split L e into three sbgrops: L is the set of lines that pass throgh and, i.e., neither or is a terminal station. L is the set of lines that pass throgh and for which is a terminal station and L is the set of lines that pass throgh and for which is a terminal station. We assme that there are no lines that exclsiely se the edge, } as they cold be placed top- or bottommost withot casing any intersections. Frthermore we assme that the lines for which is an intermediate station, i.e., L L, enter in a predefined order S. Analogosly, we assme that the lines for which is an intermediate station, i.e., L L, enter in a predefined order S. The task

4 4 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff is to find a layot of the lines in L e sch that the nmber of pairs of intersecting lines is minimized. Note that the nmber of crossings is determined by inserting the lines in L into the order S and by inserting the lines in L into S. The task is to find insertion orders that minimize the nmber of crossings. Obsere that the orders S and S themseles already determine the nmber of crossings between pairs of lines in L and that the insertion orders of L in S and of L in S do not change this nmber. Ths, we will not take l Fig. 4: The lines in L are drawn solid. In an optimal soltion l L and l L intersect. crossings between lines in L into accont anymore. On the other hand, fixing an insertion order affects the nmber of crossings between lines in L L and L and the nmber of crossings between lines in L and L in a non-triial way. Figre 4 shows that a line l L can indeed cross a line l L in the niqe optimal soltion. Throghot the paper lines in L are drawn solid, lines in L dotted and lines in L dashed. We will now show that no two lines in L (and analogosly in L ) cross in an optimal soltion. This nice property is the key for soling the one-edge layot in polynomial time. Lemma 1. In any optimal soltion for the one-edge layot problem no pair of lines in L and no pair of lines in L intersects. Proof. Assme to the contrary that there is an optimal soltion σ with a pair of lines in L that intersects. Among all sch pairs in σ let l, l } be the one whose intersection point p is rightmost. W.l.o.g. l is aboe l in. Let l p and l p be the parts of l and l to the right of p, see Figre 5a. Since σ is crossing minimal, the corses of l p and l p intersect the minimm nmber of lines in L L in order to get from p to. In particlar, the nmber of crossings between l p and lines of L L and between l p and lines of L L mst be the same otherwise we cold place l p parallel to l p (or ice ersa) which wold redce the nmber of crossings. Howeer, since the nmber of crossings to the right is the same we can easily get rid of the crossing between l and l by replacing l p by a copy of l p infinitesimally close aboe l p, see Figre 5b. The proof for L is analogos. From now on we assme that no two lines of L are consectie in S and analogosly no two lines of L are consectie in S. The reason for this assmption is that a set of consectie lines can simply be drawn parallelly in an optimal layot. Ths a single line sffices to determine the optimal corse for the whole bndle. Technically, we can deal with this case by merging a bndle of k consectie lines of L or L to one line and assigning a weight of k to it. The dynamic program will then rn in a weighted fashion that conts k k crossings for a crossing of two lines with weights k and k. For simplification we only explain the nweighted ersion of the problem in detail. Let n, n and n be the nmber of lines in L, L and L, respectiely. Note that by the aboe assmption n, n n + 1 holds. l

5 Minimizing Intra-Edge Crossings 5 l l p (a) l p l p l l p (b) l p Fig. 5: Two lines of L do not intersect in an optimal soltion. Recall that by assmption the lines in L L enter in a predefined order and the lines in L L leae in a predefined order. Let S = (s 1 < < s n+n ) be the bottom-p order of lines in L L in and S = (s 1 < < s n+n ) be the bottom-p order of lines in L L in. A line l in L can terminate below s 1, between two neighboring lines s i, s i+1, or aboe s n+n. We denote the position of l by the index of the lower line and by 0 if it is below s 1. Let S L = (s π(1) <... < s π(n ) ) denote the order on L indced by S and let S L = (s µ(1) <... < s µ(n ) ) denote the order on L indced by S. Here, π and µ are injectie fnctions that filter the lines L ot of all ordered lines L L in S and the lines L ot of all ordered lines L L in S, see Figre 6. Preprocessing. The orders S and S already determine the nmber of necessary crossings between pairs of lines in L. Let cr denote this nmber. Since cr is fixed there is no need to consider the corresponding crossings in the minimization. We will now fix the corse of a line in L. This line, say l = s π(j), has index π(j) in S, and we fix the corse of l by choosing its terminal position i in the order S. We denote the nmber of crossings between l and all lines in L by cr (i, j). This nmber is determined as follows. The line l crosses a line l L with left index i and right index j if and only if either it holds that i i and j > π(j) or it holds that i > i and j < π(j). The table cr for all lines in L has (n + n + 1) n = O(n 2 ) entries. For fixed i we can compte the row cr (i, ) as follows. We start with j = 1 and compte the nmber of lines in L that intersect line l with indices i and π(j). Then, we increment j and obtain cr (i, j +1) by cr (i, j) mins the nmber of lines in L that are no longer intersected pls the lines that are newly intersected. As any of the n lines in L receies this stats no longer or newly at most once and this stats can easily be checked by looking at S, this takes O(n) time per row. Ths, in total we can compte the matrix cr in O(n 2 ) time. We define cr (i, j) analogosly to be the nmber of crossings of the lines in L with a line in L that has index µ(i) in S and position j in S. Compting cr is analogos to cr. Dynamic Program. Assme that we fix the destination of s π(j) to some i 0,...,n + n }. Then we define F(i, j) as the minimm nmber of crossings of

6 6 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff 6 5 µ(2) = 4 3 µ(1) = 2 1 s i i s π(j) Fig. 6: The order S and the indced sborder S L = (s 2 < s 4). Fig. 7: Configration corresponding to F(i, j): line s π(j) terminates at position i in. (a) the lines in s 1,...,s i } L with the lines in L L and (b) the lines in s 1,...,s π(j) } L with the lines in L L. This sitation is depicted in Figre 7, where only the crossings indicated by gray disks are conted in F(i, j). Then the ales F(i, j) define an (n + n + 1) n -matrix F. Once the last colmn F(, n ) of the matrix F is compted, i.e., all lines in L are placed, we can determine the optimal soltion for L e as F := minf(i, n ) + C(i, n + n, n + 1) i = 0,...,n}, where C(i, n + n, n + 1) is the remaining nmber of crossings of lines in L s i+1,..., s n+n } with L L, which are not yet conted in F(i, n ). Before trning to the recrsie comptation of F(i, j) we introdce another notation. Let s assme that s π(j 1) terminates at position k and s π(j) terminates at position i, where 0 k i n + n and j 1,..., n }. Then let C(k, i, j) denote the minimm nmber of crossings that the lines L k,i := s k+1,...,s i } L case with L L. In other words C(k, i, j) conts the minimal nmber of crossings of all lines of L in the interal defined by the endpoints of the two lines s π(j 1) and s π(j). This sitation is illstrated in Figre 8, where those crossings that are marked with gray disks are conted in the term C(k, i, j). The following theorem gies the recrsion for F and shows its correctness. Theorem 1. The ales F(i, j), i = 0,...,n + n, j = 1,...,n, can be compted recrsiely by min k i F(k, j 1) + C(k, i, j) + cr (i, j)} if i 1, j 2 F(i, j) = j l=1 cr (0, l) if i = 0, j 1 (1) C(0, i, 1) + cr (i, 1) if i 1, j = 1. Proof. The base cases of Eqation (1) consist of two parts. In the first row, an entry F(0, j) means that all lines s π(1),..., s π(j) terminate at position 0 in and hence the reqired nmber of crossings is jst the nmber of crossings of these lines with L, which eqals the sm gien in Eqation (1). In the first colmn,

7 Minimizing Intra-Edge Crossings 7 an entry F(i, 1) reflects the sitation that line s π(1) terminates at position i. The reqired nmber of crossings in this case is simply cr (i, 1), the nmber of crossings of s π(1) with L, pls C(0, i, 1), the nmber of crossings of L 0,i with L L. The general case of Eqation (1) means that the ale F(i, j) can be composed of the optimal placement F(k, j 1) of the lines below and inclding s π(j 1) (which itself terminates at some position k below i), the nmber C(k, i, j) of crossings of lines in L in the interal between k and i, and the nmber of crossings cr (i, j) of s π(j) at position i. De to Lemma 1 we know that s π(j) cannot terminate below s π(j 1) in an optimal soltion. Hence, for s π(j) terminating at position i, we know that s π(j 1) terminates at some position k i. For each k we know by the indction hypothesis that F(k, j 1) is the correct minimm nmber of crossings as defined aboe. In order to extend the configration corresponding to F(k, j 1) with the next line s π(j) in L we need to add two terms: (a) the nmber of crossings of L k,i with L L, which is exactly C(k, i, j), and (b) the nmber cr (i, j) of crossings that the line s π(j) (terminating at position i) has with L. Note that potential crossings of s π(j) with lines in L k,i are already considered in the term C(k, i, j). Figre 8 illstrates this recrsion: s π(j) is placed at position i in the order S, and s π(j 1) terminates at position k. The crossings of the configration corresponding to F(i, j) that are not conted in F(k, j 1), are the C(k, i, j) crossings of the marked, dotted lines of L (indicated by gray disks) and the cr (i, j) encircled ones of s π(j) with L. Finally, we hae to show that taking the minimm ale of the sm in Eqation (1) for all possible terminal positions k of line s π(j 1) yields an optimal soltion for F(i, j). Assme to the contrary that there is a better soltion F (i, j). This soltion indces a soltion F (k, j 1), where k is the position of s π(j 1) in S. Lemma 1 restricts k i and hence s π(j 1) rns completely below s π(j). Therefore we hae F (k, j 1) F (i, j) C(k, i, j) cr (i, j) < F(i, j) C(k, i, j) cr (i, j) F(k, j 1). This contradicts the minimality of F(k, j 1). If we store in each cell F(i, j) a pointer to the corresponding predecessor cell F(k, j 1) that minimizes Eqation (1) we can reconstrct the optimal edge layot: starting at the cell F(i, n ) that minimizes F, we can reconstrct the genesis of the optimal soltion sing backtracking. Obiosly, sing the combinatorial soltion to place all endpoints of L e in the correct order and then connecting them with straight-line segments reslts in a layot that has exactly F crossings in addition to cr, the inariable nmber of crossings of L. Now, we can gie a first, naie approach: As mentioned earlier the tables cr and cr can be compted in O(n 2 ) time. For the comptation of one cell entry C(k, i, j) we only hae to look at the at most n lines L k,i and their possible n+1 terminal positions in. Once we hae fixed a terminal position of a line l L k,i, we hae to compte the nmber of crossings that l has with L L. For the crossings with L we simply hae to look at the corresponding ale of cr. For

8 8 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff s i C(k, i,j) s k i k cr (i,j) s π(j) s π(j 1) Fig. 8: The recrsion for F(i, j): lines s π(j 1) and s π(j) terminate at pos. k and i, resp. C(k, i, j) = min. # crossings of the marked dotted lines L k,i with L L = here 4 cr (i, j) = nmber of crossings of s π(j) with L = here 2 the crossings with L it is sfficient to look at the index of the terminal position becase we know that position k is the terminal position of s π(j 1) and position i is the terminal position of s π(j). Ths, compting one of the O(n3 ) cells of the table C reqires O(n 2 ) time, so in total we need O(n 5 ) time for filling C. This dominates the comptation of the table F. In the remainder of this section we show how to speed p the comptation of F and C. Improing the Rnning Time. Let s for the moment assme that the ales C(k, i, j) and cr (i, j) are aailable in constant time. Then the comptation of the (n + n + 1) n matrix F still needs O(n 3 ) time becase the minimm in Eqation (1) is oer a set of O(n) elements. The following series of lemmas shows how we cold bring the rnning time down to O(n 2 ). First we show a relation for the entries of the matrix C. Lemma 2. C(k, i, j) is additie in the sense that C(k, i, j) = C(k, l, j)+c(l, i, j) for k l i. Proof. Since C(k, i, j) denotes the nmber of crossings of the lines in L s k+1,..., s i } and no two of these lines intersect each other (recall Lemma 1) we can split the layot corresponding to C(k, i, j) at any position l, k l i and get two (possibly non-optimal) configrations for the indced sbproblems. This implies C(k, i, j) C(k, l, j) + C(l, i, j). Conersely, we can get a configration for C(k, i, j) by ptting together the optimal soltions of the sbproblems. W.l.o.g. this introdces no additional crossings (they cold be remoed as in the proof of Lemma 1). Hence we hae C(k, i, j) C(k, l, j) + C(l, i, j). Now we show that we do not need to compte all entries of C. Lemma 3. Gien the matrix C, the matrix F can be compted in O(n 2 ) time. Proof. Haing compted entry F(i 1, j) we can compte F(i, j) in constant time as follows:

9 Minimizing Intra-Edge Crossings 9 F(i 1, j) + C(i 1, i, j) cr (i 1, j) + cr F(i, j) = min (i, j), F(i, j 1) + cr (i, j). (2) The correctness follows from Eqation (1), Lemma 2, and the fact that C(i, i, j) anishes: min F(i, j) (1) = min F(k, j 1) + C(k, i, j) + cr (i, j)}, k<i F(i, j 1) + C(i, i, j) + cr (i, j) L. 2 = min (1) = min min F(k, j 1) + C(k, i 1, j) + C(i 1, i, j) + cr (i, j)}, k i 1 F(i, j 1) + cr (i, j) F(i 1, j) cr (i 1, j) + C(i 1, i, j) + cr (i, j), F(i, j 1) + cr (i, j) In the first colmn, we can reformlate the recrsion for i 1 as follows: F(i, 1) (1) = C(0, i, 1) + cr (i, 1) Lemma 2 = C(0, i 1, 1) + C(i 1, i, 1) + cr (i, 1) (1) = F(i 1, 1) cr (i 1, 1) + C(i 1, i, 1) + cr (i, 1) Hence the whole matrix F can be compted in O([n+n ] n ) = O(n 2 ) time. Obsere that de to the reformlation in Lemma 3 we only need the ales C(i 1, i, j) explicitly in order to compte F. Now we will show that we can compte these releant ales in O(n 2 ) time. For simplification we introdce the following notation: C (i, j) := C(i 1, i, j). Lemma 4. The ales C (i, j) (i = 1,...,n + n, j = 1,...,n ) can be compted in O(n 2 ) time. Proof. We compte C (i, j) row-wise, i.e., we fix i and increase j. Recall that C(k, i, j) was defined as the minimal nmber of crossings of the lines in L k,i = s k+1,..., s i } L with the lines in L L nder the condition that s π(j 1) ends at position k and s π(j) ends at position i. For C (i, j) = C(i 1, i, j) this means we hae to consider crossings of the set L i 1,i = s i } L. Hence, we distingish two cases: either s i is a line in L and then L i 1,i is empty or s i L and we hae to place the line s i optimally. Clearly, in the first case we hae C (i, j) = 0 for all j as there is no line to place in L i 1,i. Now we consider the case that s i L. For each j we split the set of candidate terminal positions for s i into the interals [0, π(j 1)), [π(j 1), π(j)), and [π(j), n + n ]. Let LM(i, j), MM(i, j), and UM(i, j) denote the minimm nmber of crossings of s i with L L for terminal positions in [0, π(j 1)), [π(j 1), π(j)), and [π(j), n + n ], respectiely. Now we hae C (i, j) = minlm(i, j), MM(i, j), UM(i, j)} as the minimm of the three distinct cases. The sitation is illstrated in Figre 9.

10 10 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff [ π(j),n + n ] UM(i,j) s i+1 L s s i L i+1 L s π(j) L [ ) π(j 1),π(j) MM(i,j) s π(j 1) L [ 0,π(j 1) ) LM(i,j) Fig. 9: Splitting the candidate terminal positions for s i π(j 1) and π(j). into three interals w.r.t. Next, we hae to show how to compte MM, LM, and UM. First, we consider MM. Recall that cr (i, j) was defined as the nmber of crossings of the lines in L with the line s µ(i) that terminates at position j in. It follows that MM(i, j) = mincr (µ 1 (i), k) π(j 1) k < π(j)}, (3) where π(0) is defined as 0. Becase of Lemma 1 there are no lines of L intersecting the tnnel between s π(j 1) and s π(j). Hence, if the line s i terminates at position k [π(j 1), π(j)) it does not cross any line of L and Eqation (3) is correct. We can calclate MM(i, j) by a straight-forward minimm comptation throgh j = 1,...,n which takes O(n + n ) = O(n) time for each ale of i. Secondly, we consider LM. Initially, in the case that j = 1 there is no line s π(j 1) and the corresponding interal is empty. Hence we set LM(i, 1) =. Then we recrsiely compte LM(i, j + 1) = minlm(i, j) + 1, MM(i, j) + 1}. (4) Obsere that for LM(i, j + 1) we merge the preios interals corresponding to LM(i, j) and MM(i, j). Moreoer the line s π(j), which preiosly ended at position i, now terminates at position i 1. Hence, in order to reach its terminal position in the interal [0, π(j)), the line s i has to cross s π(j) in addition to the crossings conted before by MM(i, j) and LM(i, j). This explains the recrsion in Eqation (4). The comptation again reqires O(n) time for each ale of i. Finally, we initialize UM(i, n ) = 1 + mincr (µ 1 (i), k) π(n ) k n + n } as for j = n the line s i crosses the line s π(n bt no other line of L ). In decreasing order we compte UM(i, j 1) = minum(i, j) + 1, MM(i, j) + 1} (5) analogosly to LM, which again reqires O(n) time. As the whole procedre needs linear time for each i = 1,...,n + n the total rnning time is O(n 2 ). Ptting the intermediate reslts in Lemmas 3 and 4 together, we conclde:

11 Minimizing Intra-Edge Crossings 11 Theorem 2. The one-edge layot problem can be soled in O(n 2 ) time. So far, the algorithm reqires O(n 2 ) space to store the tables F, C, cr, and cr. If we are only interested in the minimm nmber of crossings this can easily be redced to O(n) space as all tables can be compted row-wise: in F we need only two consectie rows at a time and we can discard preios rows; in the other tables the rows are independent and can be compted on demand. This does not affect the time complexity. Howeer, to restore the optimal placement we need the pointers in F to do backtracking and hence we cannot easily discard rows of the matrix. Bt we can still redce the reqired space to O(n) with a method similar to a diide-and-conqer ersion of the Needleman-Wnsch algorithm for biological seqence alignment [2] adding a factor of 2 to the time complexity. Basically, the idea is to keep only the pointers in one colmn of the matrix to reconstrct the optimal position of the corresponding line. This line cts the problem into two smaller sbproblems which are soled recrsiely. 3 Generalization to a Path Srely it is desirable to draw lines on a more general fraction of the graph than only on a single edge. Howeer, the problem seems to become significantly harder een for two edges. Let s first define the problem on a path. Problem 2. Path layot Gien a graph G = (V, E) and a simple path P =, w 1,...,w m, in G. Let L P be the set of lines that se at least one edge in P. We split L P into three sbgrops: L P is the set of lines that hae no terminal station in, w 1,..., w m, }, L P is the set of lines for which is an intermediate station and that hae a terminal station in w 1,..., w m, }, and L P is the set of lines for which is an intermediate station and that hae a terminal station in, w 1,...,w m }. We assme that there are no lines haing both terminal stations in, w 1,...,} as these cold be placed top- or bottommost casing the minimm nmber of crossings with lines in L P. We also assme that any two lines l 1, l 2 that se exactly the sbpath x,..., y P together and for which neither x nor y is a terminal station enter P in x in a predefined order and leae P in y in a predefined order. The task is to find a layot of the lines in L P sch that the nmber of pairs of intersecting lines is minimized. We tried to apply the same dynamic-programming approach as for the oneedge case. Howeer, the dilemma is that the generalized ersion of Lemma 1 does not hold, namely that no two lines in L P intersect. Ths, the problem instance cannot be separated into two independent sbproblems, which seems to forbid dynamic programming. In Figre 10a we gie an instance where two lines of L P cross in the optimal soltion. Here, the lines in LP are drawn solid. Recall that we do not hae to take the intersections of lines in L P into accont as these are again gien by the orders in S and S. The two lines l and l L P are only needed to force the bndle crossings between the bndles L L P and

12 12 M. Benkert, M. Nöllenbrg, T. Uno, and A. Wolff w 1 w 1 L L L l,l (a) Optimal soltion (7 crossings). l l L l,l L L (b) Maniplating l (8 crossings). l l Fig. 10: The two lines l,l L P cross in the optimal soltion. L L P to be on the edge w 1, }. In the optimal soltion the lines l, l LP intersect casing a total nmber of 7 crossings between lines in L P with lines in L P LP. We hae to arge that any soltion in which l and l do not cross prodces more than 7 crossings. We look at the optimal soltion and arge that getting rid of the crossing between l and l by maniplating the corse of either l or l prodces at least 8 crossings. First, we consider maniplating the corse of l, see Figre 10b. Howeer then, as l has to be aboe l, it has to cross the 4 lines in the bndle L reslting in a total nmber of 8 crossings. Similarly, maniplating the corse of l wold also reslt in at least 8 crossings. Conclding Remarks Clearly or work is only a first step in exploring the layot of lines in graphs. What is the complexity of the problem if two edges of the nderlying graph are considered, what abot longer paths, trees and finally, general plane graphs? A ariant of the problem where lines mst terminate bottom- or topmost in their terminal stations is also interesting. This reqirement preents gaps in the corse of contining lines. References 1. P. F. Cortese, G. D. Battista, M. Patrignani, and M. Pizzonia. On embedding a cycle in a plane graph. In P. Healy and N. S. Nikolo, editors, Proc. 13th Int. Symp. Graph Drawing (GD 05), olme 3843 of LNCS, pages Springer-Verlag, R. Drbin, S. Eddy, A. Krogh, and G. Mitchison. Biological Seqence Analysis. Cambridge Uniersity Press, M. R. Garey and D. S. Johnson. Crossing nmber is NP-complete. SIAM J. Alg. Disc. Meth., 4: , M. Nöllenbrg and A. Wolff. A mixed-integer program for drawing high-qality metro maps. In P. Healy and N. S. Nikolo, editors, Proc. 13th Int. Symp. Graph Drawing (GD 05), olme 3843 of LNCS, pages Springer-Verlag, 2006.