1 Orthogonal Drawing (continued)  Sections 8.3 – 8.5 from the book  Bert Spaan

2 Orthogonal Drawing (continued) 1.Linear Algorithm  Without Bad Cycles  With Bad Cycles 2.Orthogonal Grid Drawing 3.Orthogonal Drawings without Bends

3 Linear Algorithm for Bend-Optimal Drawing  Linear time algorithm for graphs with the following properties: 3-connected Δ ≤ 3 Four or more outer edges

4 k-Connectivity  A graph G is connected if for any distinct vertices u and v there is a path between u and v in G.  A graph is k-connected if the minimum number of vertices whose removal results in a disconnected graph is k.

5 Examples  1-connected:  2-connected:  3-connected:

6 Bad Cycles  A graph has no bad cycles and thus a rectangular drawing if and only if Any 2-legged cycle contains two or more corners Any 3-legged cycle contains one or more corners  A bad cycle is a 2-legged cycle which contains at most one corner, or a 3- legged cycle which contains no corners.

7 Maximal Bad Cycles  A bad cycle C in G is defined to be maximal if C is not contained in any other bad cycle in G

8 Linear Algorithm Algorithm Orthogonal-Draw(G) begin  Add four dummy vertices of degree two on four distinct outer edges as the corners;  Let G’ be the resulting graph;  Let C 1, C 2, …, C l, be the maximal bad cycles in G’;  for each i, 1 ≤ i ≤ l, construct a genealogical tree T C i and determine green paths and red paths for every cycle in T C i ;  for each i, 1 ≤ i ≤ l, find an orthogonal drawing D(G(C i )) of G(C i ) feasible for an arbitrary green path of C i by Feasible-Draw;  Let G’’ be a plane graph derived from G’ by contracting each for each G(C i ), 1 ≤ i ≤ l, to a single vertex v i ;  Find a rectangular drawing D(G’’) of G’’ by Rectangular-Draw;  Patch the drawings D(G(C 1, D(G(C 2 )), …, D(G(C l )) into D(G’’) to get an orthogonal drawing D(G’) of G’.  Obtain an orthogonal drawing D(G) of G by replacing the four dummy vertices a, b, c and d in D(G’) with bends end.

9 Linear Algorithm Without Bad Cycles Algorithm Orthogonal-Draw(G) begin  Add four dummy vertices of degree two on four distinct outer edges as the corners;  Let G’ be the resulting graph;  Obtain an orthogonal drawing D(G) of G by replacing the four dummy vertices a, b, c and d in D(G’) with bends end.

10 Example  Start:  Step 2:  Step 1:  Step 3:

11 Linear Algorithm With Bad Cycles, same Algorithm  Start:  Step 2:  Step 1:  Step 3: Doesn’t lead to orthogonal drawing due to bad cycle

12 Bad Cycles in G’  G is 3-connected and thus has no 2- legged cycles  G’ has four 2-legged cycles, but none of those are bad cycles  Every bad cycle in G’ is a 3-legged cycle containing no corner

13 Genealogical Tree  Tree of cycles in G, T C  In algorithm: tree of 3-legged bad cycles  T C can be found in linear time

14 Example  3-legged cycles in C and genealogical tree T C  C 2, C 3, C 4 and C 5 are the maximal bad cycles in C  C 7 is the maximal bad cycle in C 4

15 Assignment & Labeling  For each 3-legged cycle C in graph G we define the following terms in a recursive manner, based on a genealogical tree. C is divided into three contour paths P 1, P 2, and P 3 by the three leg- vertices x, y and z Each contour path is classified as either a green path or a red path C is assigned an integer bc(C), the bend-count of C

16 Three cases 1.C has no child-cycle, that is, l = 0 2.C has child-cycles C 1, C 2, …, C l, l ≥ 1, but none of them has a green path on C 3.Otherwise

17 Case 1 C has no child-cycles:  T C consists of a single vertex  Classify all the three contour paths of C as green paths  Set bc(C) = 1

18 Case 2 C has one or more child-cycles, none of them with a green path on C:  Classify all the three contour paths of C as green paths  Set bc(C) = 1 + the sum of the bend-counts of all child- cycles

19 Case 3 Otherwise:  Classify the contour paths in which a child-cycle has a green path as green paths  Classify the other contour paths (if any) as red paths  Set bc(C) = the sum of the bend-counts of all child- cycles

20 Properties  Each cycle has at least one green contour path  Assignment & Labeling can be done in linear time

21 Feasible-Draw  Find an orthogonal drawing G(C) for a 3-legged cycle C  Feasible if the drawing has the following properties: Drawing has bc(C) bends At least one bend appears on the green path The drawing is bound by six open halflines, two at each point x, y and z

22 Case 1 C has no child-cycles:  bc(C) = 1  Insert a dummy vertex t of degree two on an arbitrary edge on the green path in C  Now there are four corners: Dummy vertex t Vertices x, y and z  No bad cycles, algorithm Rectangular-Draw can be used to find a rectangular drawing  After replacing t with a bend, an orthogonal drawing is created

23 Case 2 C has one or more child-cycles, none of them with a green path on C:  Recursion: first find an orthogonal drawing for each child-cycle  Construct a plane graph J from G(C) by contracting each child- cycle to a single vertex  Add a dummy vertex t  Now, with Rectangular-Draw and removal of t, an orthogonal drawing can be found  Replace the contracted vertices with the original child-cycles. This is always possible, there are 12 distinct embeddings, depeding on the three legs x, y and z and the chosen green path

24 Case 3 Otherwise  Construct a plane graph J from G(C) by contracting each child- cycle to a single vertex  Replace one of the contracted cycles with a quadrangle (x 1, t, y 1, z 1 ) where t is a dummy vertext of degree two  Recursion: find an orthogonal drawing for each child-cycle  Now, with Rectangular-Draw and removal of t, an orthogonal drawing can be found  Replace the contracted vertices with the original child-cycles and the quadrangle with its corresponding cycle.

25 Linear Algorithm Algorithm Orthogonal-Draw(G) begin  Add four dummy vertices of degree two on four distinct outer edges as the corners;  Let G’ be the resulting graph;  Let C 1, C 2, …, C l, be the maximal bad cycles in G’;  for each i, 1 ≤ i ≤ l, construct a genealogical tree T C i and determine green paths and red paths for every cycle in T C i ;  for each i, 1 ≤ i ≤ l, find an orthogonal drawing D(G(C i )) of G(C i ) feasible for an arbitrary green path of C i by Feasible-Draw;  Let G’’ be a plane graph derived from G’ by contracting each for each G(C i ), 1 ≤ i ≤ l, to a single vertex v i ;  Find a rectangular drawing D(G’’) of G’’ by Rectangular-Draw;  Patch the drawings D(G(C 1, D(G(C 2 )), …, D(G(C l )) into D(G’’) to get an orthogonal drawing D(G’) of G’.  Obtain an orthogonal drawing D(G) of G by replacing the four dummy vertices a, b, c and d in D(G’) with bends end.

26 Orthogonal Grid Drawing  An orthogonal drawing is called an orthogonal grid drawing if all vertices and bends are located on integer grid points. DgDg D’ g

27 Compact  D g is compact: There is at least one vertical line segment of x-coordinate i for each integer i, 0 ≤ i ≤ W (W is the width of the grid) There is at least one horizontal line segment of y-coordinate j for each integer j, 0 ≤ j ≤ H (H is the height of the grid)

28 Bend-count and grid size relation  Let D g be a compact orthogonal grid drawing of a plane graph G with b bends. Then W + H ≤ b + 2 * |E| – |V| – 2

29 Orthogonal Drawings without Bends  Plane graphs with Δ ≤ 3 may have an orthogonal drawing without bends

30 Conditions  Assume that G is a plane 2-connected graph with Δ ≤ 3 and four or more outer vertices of degree two. G has an orthogonal drawing without bends if and only if Every 2-legged cycle contains at least two vertices of degree two Every 3-legged cycle contains at least one vertex of degree two

31 Example  2-legged cycles:  3-legged cycles: