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Rectangular Drawing (continue)

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Rectangular Drawing (continue)

Harald Scheper

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

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

- 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

- 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)

- 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)| â‰¥ 2 then
- F1,..,Fq are C0 components of GP
- For each Fi, i â‰¤ 1 â‰¤ q do
- Draw (C0(GP ) Fi, Fi)

- 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

- Draw all edges of Pc
- Recursive
- Draw(C0(Gi) Fi,Fi)

- If Pcâ‰ 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)â€¦G(Ck)

- Recursive: Draw(C0(Gj) Fj,Fj),
1 â‰¤ j â‰¤ q

for each graph Gi, 1 â‰¤ i â‰¤ k+2

- Draw all edges of Pc andPcc on

Pcc

- 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,â€¦, WW (start/end points). So NS-, SN-, EW-, WE-paths are found in constant 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 Pâ€™ in a later recursive stage, which is not on Pc or Pcc.

- then again have to
traverse the facial cycles

attached to Pâ€™, 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

- 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 Pâ€™ (not to find critical cycles again)
- every face become boundary face, and not again. so traversed constant time. => linear time.

- 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

- Graph Ty spanning tree obtained from G

delete

not deleted

16 maximal vertical lseg

15 maximal horizontal lseg

- 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(Lâ€™)+1)
- v has no neighbor u below v (temp(v) = 0)

- y(L) = max{temp(v)}

v

- All maximal horizontal line segments with depth-first search = linear time
- Upperbounds on area of grid, W+H â‰¤ n/2 and W Â· H â‰¤ 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 (PNâ€¦) so n-4 = 2(l-4) => l-2 = n/2 because compact: H â‰¤ lh -1, W â‰¤ lv -1
= W+H â‰¤ lh + lv â€“ 2 = l â€“ 2 = n/2 =>

W Â· H â‰¤ n2/4

- until now considered rect. plane graph G with Î” â‰¤ 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

- independent: no common vertex S1 = {C1,C2}, S2 = {C2,C3,C4}

- Theorem 6.4.1:
G is 2 connected graph, Î” â‰¤ 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 â‰¤ 4

- 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

- 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+c3â‰¤ 4

- assume G has rectangular drawing D with 4 corners. by fact 6.3.1:

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

- 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

- show if the 3 conditions hold, then can choose 4 outer vertices of degree 2 as corners a,b,c,d, such that (theorem):

- outline = (complete in article)
- let J1,..,Jp p â‰¥ 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.

- example

- until now considered rectangular drawings of plane graphs (with fixed embedding), now of planar graphs with Î”â‰¤ 3 (without fixed embedding)
- say planar graph G has rect. drawing if at least one of the plane embeddings has rect. drawing

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

- 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

- Questions?