Rectangular Drawing (continue)

1. Rectangular Drawing (continue) Harald Scheper

2. Overview • algorithm (directions) • algorithm in linear time • outline of algorithm (placement) • Rect. Drawings without Designated Corners

3. Algorithm • To find directions of edges in rectangular drawing of G (vertical, horizontal) • Later decide the integer coordinates of vertices

4. Algorithm (outline) • Assume plane graph G has no bad cycle. • Each C0(G) component treated independently. • If there exists a boundary NS-, SN-, WE-, or EW-path then choose it as a partition path • Otherwise find partition path PC and PCC from westmost NS-path, and recurse over subgraphs

5. Algorithm Main • Algorithm Rectangular-Draw(G) • Draw the outer cycle C0(G) as a rectangle by 2 horizontal line segments PN, PS and 2 vertical line segments PE and PW • Find all C0(G) components J1,..,JP • For each component Ji Gi = C0(G) Ji Draw(Gi,Ji)

6. Draw(G,J) if part • If G has boundary NS-, SN-, WE-, or EW-path P • assume without loss of generality that P is a boundary NS-path • Draw all edges of P on vertical line (directions are vertical) • If |E(P)| ≥ 2 then • F1,..,Fq are C0 components of GP • For each Fi, i ≤ 1 ≤ q do • Draw (C0(GP ) Fi, Fi)

7. Draw(Gi,Ji) else part • No boundary NS-,SN-,EW-, or WE path • Find westmost NS-path P • Find partition-pair Pc and Pcc from P • If Pc = Pcc then • Draw all edges of Pc on a vertical line segment • Recursive • Draw(C0(Gi) Fi,Fi)

8. Draw(Gi,Ji) else, else part • If Pc≠ Pcc then • Draw all edges of Pc andPcc on alternating sequences of horizontal, vertical line segments • G1 is graph obtained form GW by contracting all edges of PCC that are on horizontal sides of rectangular embeddings of C1,C2,..Ck G2 = of Pc G3 = G(C1), G4 = G(C2)…G(Ck) • Recursive: Draw(C0(Gj) Fj,Fj), 1 ≤ j ≤ q for each graph Gi, 1 ≤ i ≤ k+2 Pcc

9. Linear time • Find all C0 components • For each C0 we find boundary NS-,SN-,EW- and WE-paths if exist, by traversing all boundary faces of G by cc depth-first search. • Each boundary path gets a label, NN, NE,…, WW (start/end points). So NS-, SN-, EW-, WE-paths are found in constant time

10. Linear time • no boundary NS-, SN-, EW-, WE paths • find partition-pair Pc and Pcc • give labels to the newly created paths by traversing them (once) • problem if a subpath of the westmost NS-path is chosen as westmost NS-path P’ in a later recursive stage, which is not on Pc or Pcc.

11. Linear time • then again have to traverse the facial cycles attached to P’, so time complexity not linear • so store: • list edges ei elemof E(P) contained in boundary NN- and EN-paths • list edges ei elemof E(P) contained in boundary SS- and SE-paths • to find Pst and Pen in later recursive stages

12. Linear time • store array of length n = whether the vertex is a head or a tail vertex of a clock wise critical cycle C attached to P and whether ncc(C) = 1 or ncc(C) >1 • indicate existence of critical cycles attached to P’ (not to find critical cycles again) • every face become boundary face, and not again. so traversed constant time. => linear time.

13. Rectangular Grid Drawing 1 • Algorithm finds only directions of all edges in G • Now coordinates of vertices in G determined in linear time • Assume for simplicity not-corner vertices have degree 3

14. Rectangular Grid Drawing 2 • Graph Ty spanning tree obtained from G delete not deleted

15. Rectangular Grid Drawing 3 16 maximal vertical lseg 15 maximal horizontal lseg

16. Rectangular Grid Drawing 4 • to each maximal horizontal line segment L, we assign y(L) as y-coordinate of every vertex on L • PS is lowermost, PN is topmost • y(PS)=0 • compute y(L) bottom to top • For each vertex v assign temp(v) as temporary y-coordinate of v • For every vertex v on L two cases: • v has neighbor u below v (temp(v) = y(L’)+1) • v has no neighbor u below v (temp(v) = 0) • y(L) = max{temp(v)} v

17. Rectangular Grid Drawing 5 • All maximal horizontal line segments with depth-first search = linear time • Upperbounds on area of grid, W+H ≤ n/2 and W · H ≤ n2/16 • coordinate of south-west corner = (0,0), northeast = (W,H), at least one hor., vert. line segment • lv = #vertical linesegments, lh = #horizontal linesegments, l = lv + lh • each vertex (- 4) = one of the l-4 max. line seg (PN…) so n-4 = 2(l-4) => l-2 = n/2 because compact: H ≤ lh -1, W ≤ lv -1 = W+H ≤ lh + lv – 2 = l – 2 = n/2 => W · H ≤ n2/4

18. RD no designated corners 1 • until now considered rect. plane graph G with Δ ≤ 3 with 4 outer vertices of degree 2 as corners • now corners not designated • efficiently find whether G has 4 outer vertices (degree 2) such that there is a rect. drawing

19. RD no designated corners 2 • independent: no common vertex S1 = {C1,C2}, S2 = {C2,C3,C4}

20. RD no designated corners 3 • Theorem 6.4.1: G is 2 connected graph, Δ ≤ 3, has min. 4 outer vertices of degree 2, then rect. draw with 4 corners desig. corners if G satisfies: • every 2-legged cycle in G contains at least 2 outer vertices of degree 2 • every 3-legged cycle in G contains at least 1 outer vertex of degree 2 • if an independent set S of cycles in G consists of c2 2 legged cycles and c3 3-legged cycles then 2c2 + c3 ≤ 4

21. RD no designated corners 2 • independent: no common vertex S1 = {C1, C2}, c2 = 2, c3 = 0, 2*2+0 = 4 S2 = {C2, C3, C4}, c2 = 1, c3 = 2, 2*1+2 = 4

22. RD no designated corners 3 • Necessity of Th. 6.4.1 : • assume G has rectangular drawing D with 4 corners. by fact 6.3.1: an indep. set S has c2 2-legged C contains at least 2 corners, c3 3-legged C contains at least 1 corner. All cycles independent, so at least 2c2 + c3 corners in D. Since there are 4 corners in D, 2c2+c3≤ 4 In any rectangular drawing D of G, every 2-legged cycle of G contains 2 or more corners, every 3-legged cycle of G contains 1 or more corner, more than 3-legged has no corner

23. RD no designated corners 4 • prove of 6.4.1: • show if the 3 conditions hold, then can choose 4 outer vertices of degree 2 as corners a,b,c,d, such that (theorem): • any 2-legged cycle contains 2 or more corners • any 3-legged cycle contains 1 or more corners

24. RD no designated corners 5 • outline = (complete in article) • let J1,..,Jp p ≥ 1 be C0(G) components of G in series • C1 and C2 in J1 and Jp • if 4 corners not been chosen, we choose a corner from each of innermost 3-legged cycles.

25. RD no designated corners 6 • example

26. RD of Planar Graphs • until now considered rectangular drawings of plane graphs (with fixed embedding), now of planar graphs with Δ≤ 3 (without fixed embedding) • say planar graph G has rect. drawing if at least one of the plane embeddings has rect. drawing

27. RD of Planar Graphs • b, c, d are embeddings of planar graph b = rd, c and d not (3-legged cycle with no outer vertex of degree 2)

28. RD of Planar Graphs • finding not trivial because exponential number of plane embeddings by algorithm as before. • now linear algorithm to examine if there is a plane embedding with rect. drawing

29. Questions?