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Chapter 4: Straight Line Drawing

Chapter 4: Straight Line Drawing. Ronald Kieft. Contents. Introduction Algorithm 1: Shift Method Algorithm 2: Realizer Method Other parts of chapter 4 Questions?. Introduction. Straight Line Drawing deals with plane graphs

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Chapter 4: Straight Line Drawing

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  1. Chapter 4: Straight Line Drawing Ronald Kieft

  2. Contents • Introduction • Algorithm 1: Shift Method • Algorithm 2: Realizer Method • Other parts of chapter 4 • Questions?

  3. Introduction • Straight Line Drawing deals with plane graphs • Plane graph: can be drawn without edge crossings, given a fixed embedding • However, the edges can be curved • It was proved that every plane graph G has a straight line drawing • Which can be found in polynomial time • But the area of the drawing was not bounded • Impractical

  4. Introduction cont’d • In 1990 two methods were found which did have a bound on the area • Fraysseix: area is a grid of (2n - 4) x (n – 2) • Scnhyder: area is a grid of (n – 2) x (n – 2) • Both method have O(n) implementations • Question: what is the minimum size of a grid required for a Straight Line Drawing? • Probably ceil(2n / 3) x ceil(2n / 3) • Has not yet been proven

  5. Algorithm 1: Shift Method • The method by Fraysseix is called the “Shift Method” • Finds a Straight Line Drawing for every plane graph G with n ≥ 3 • Area is bouned by a grid of size (2n – 4) x (n – 2) • Shift Method can be divided into two steps: • Step 1: Construct a Canonical Ordering • Step 2: Insert vertices one by one, according to the Canonical Ordering (and make sure the graph stays a Straight Line Drawing)

  6. Algorithm 1: Shift Method • Step 1: Construct a Canonical Ordering • What is a Canonical Ordering? • Cycle: path through G which starts and ends at the same vertex • Chord: an edge between two non-consecutive vertices on a Cycle • Outer Cycle; Co(G): the boundary of the outer face • π(v1, v2,…, vn): ordering of all vertices of G • Gk: a subgraph of G formed by the first k vertices of ordering π

  7. Algorithm 1: Shift Method • Examples of all definitions:

  8. Algorithm 1: Shift Method • The ordering π is a Canonical Ordering if for every k, 3 ≤ k ≤ n: • Gk is 2-connected and internally triangulated • (v1, v2) is an outer edge of Gk • If k + 1 ≤ n, then vertex vk+1 is located in the outer face of Gk and all neighbors of vk+1 in Gk appear on Co(G)

  9. Algorithm 1: Shift Method • Example Canonical Ordering:

  10. Algorithm 1: Shift Method • Proof: every triangulated plane graph G has a Canonical Ordering • Proof with induction on the number of vertices • Two base cases: • n = 3 and n = n • Induction step: • Start with n =n • Then there exists a vertex w (w ≠ v1, v2) on the cycle Co(Gk) and w is not end of a chord on Co(Gk) • Then Gk-1 is Gk without vertex w

  11. Algorithm 1: Shift Method • Conclusion: it is enough to show that there always is such a vertex w • Take Co(Gk) = w1, w2,…, wt (with w1 = v1 and wt = v2) • Then two cases: • Co(Gk) has no chords  any vertex of w2, w3,…, wt is such a vertex w • Co(Gk) has one or more chords: • Then Co(Gk) has a minimal chord (wp, wq) such that the vertices wp+1, wp+2,…, wq-1 are not end of a chord • Then any of the vertices wp+1, wp+2,…, wq-1 are such a vertex w

  12. Algorithm 1: Shift Method • Illustration of the proof:

  13. Algorithm 1: Shift Method • Step 2: Construct Straight Line Drawing using the Canonical Ordering • L(vi): set of vertices which need to be moved when vertex vi changes position • μ(P1, P2): new gridpoint defined by a line +1 through P1 and a line -1 through P2 • Manhattan distance between P1 and P2 should be even • Example:

  14. Algorithm 1: Shift Method • G3 is drawn in a standard way: • Furthermore assume the following holds for Gk-1 with k-1 ≥ 3: • P(v1) = (0, 0) and P(v2) = (2k – 6, 0) • The x-coordinates are increasing for the vertices on Co(Gk) • Each edge (wi, wi+1) on Co(Gk) is drawn by a straight line with slope +1 or -1

  15. Algorithm 1: Shift Method • Then vertex vk can be inserted into Gk-1 as follows: • Let wp, wp+1,…, wq be the neighbors of vk on Co(Gk) • Vertex vk covers vertices wp+1, wp+2,…, wq-1 • If vertex vk is inserted at μ(wp, wq), then the edge (vk, wp) might overlap with edge (wp, wp+1) • Therefore all vertices w1, w2,…, wp are shifted to the left by one • Same thing holds for the edge (vk, wq) and therefore all vertices wq, wq+1,…, wt are shifted to the right by one • After both shifts a correction is made to ensure that v1 stays located at (0, 0) • Because of the shifts, vertex vk can see all vertices wp, wp+1,…, wq • Therefore all egdes starting at vk can be drawn as straight lines, without edges crossings

  16. Algorithm 1: Shift Method

  17. Algorithm 1: Shift Method • When all vertices are inserted, it holds that: • P(v1) = (0, 0) • P(v2) = (2n – 4, 0) • P(vn) = (n – 2, n – 2) • Thus the final Straight Line Drawing has a grid of size (2n – 4) x (n – 2) • The algorithm can be implemented in O(n) time

  18. Algorithm 1: Shift Method • Example Straight Line Drawing:

  19. Algorithm 2: Realizer Method • The method by Schnyder is called the “Realizer Method” • Finds a Straight Line Drawing for every plane graph G • Area is bounded by (n – 2) x (n – 2) • The Realizer Method can be divided into four steps: • Step 1: Proof a Barycentric Representation gives a Straight Line Drawing • Step 2: Make a Schnyder Labeling of G • Step 3: Make a Realizer for G • Step 4: Proof that a Realizer can be seen as a WeakBarycentric Representation

  20. Algorithm 2: Realizer Method • Step 1: Barycentric Representation gives a Straight Line Drawing • What is a Barycentric Representation? • An injective function which assigns three values v1, v2 and v3 to each vertex v, such that: • v1 + v2 + v3 = 1 • For each edge (x, y) and each vertex z ≠ x, y there is some index k (with k = 1, 2 or 3) such that vk from x and y < vk from z

  21. Algorithm 2: Realizer Method • Proof: Barycentric Representation gives a Straight Line Drawing on the plane spanned by (1,0,0), (0,1,0) and (0,0,1) • Follows from the definition that each vertex is mapped onto this plane • What is left: prove that edges do not intersect

  22. Algorithm 2: Realizer Method • Proof: Two edges (x, y) and (u, v) do not intersect: • From the definition of the barycentric representation: • ui, vi < xi and uj, vj < yj • xk, yk < uk and xl, yl < vl • Solving these equations gives: • i = j or k = l • Assume i = j = 1 (since i, j, k, l = 1, 2, 3) • Then u1, v1 < x1 and u1, v1 < y1 which says that the x-coordinate of both vertex u and v < the x-coordinate of both vertex x and y • This implies that the edges (x, y) and (u, v) are separated by a line through (0,1,0)(0,0,1) • This implies that the edges (x, y) and (u, v) do not intersect

  23. Algorithm 2: Realizer Method • Thus Barycentric Representation gives a Straight Line Drawing • Grid size is (2n – 5) x (2n – 5) • Using a Weak Barycentric Representation, the grid size can shrink to (n – 2) x (n - 2) • A Weak Barycentric Representation is again an injective function assigning three values, such that: • v1 + v2 + v3 = 1 • For each edge (x, y) and each vertex z ≠ x, y there is some index k (with k = 1, 2 or 3) such that (vk, vk+1) from x and y <lex (vk, vk+1) from z • Where (a, b) <lex (c, d) is true if a < c or when a = c and b < d • Proof: Weak Barycentric Representation gives a Straight Line Drawing • Same idea as proof for normal Barycentric Representation

  24. Algorithm 2: Realizer Method • Step 2: Make a Schnyder Labeling • What is a Schnyder Labeling? • An assignment of a label 1, 2 or 3 to all angles of inner facial triangles, such that: • Each inner facial triangle has an angle labeled 1, another one labeled 2 and another one labeled 3, in a counter-clockwise order • The labels of the angles at each inner vertex of G form, in counter-clockwise order, a nonempty interval of 1’s, a nonempty interval of 2’s and a nonempty interval of 3’s

  25. Algorithm 2: Realizer Method • Proof: every triangulated plane graph G has a Schnyder Labeling • N(x): the set of neighbors of x in G • G/(x,y): the graph obtained by contracting edge (x,y) from G (that is, remove y from G and connect al neighbors from y to x) • Contractible(x, y): whether or not edge (x, y) is contractible. An edge (x, y) is only contractible if N(x) and N(y) share two vertices • Separating Triangle: a triangle in G which has both interior and exterior vertices • An edge (x, y) is contractible if there is no separating triangle containing (x, y), otherwise the resulting graph will no longer be triangulated

  26. Algorithm 2: Realizer Method • Edge contraction examples:

  27. Algorithm 2: Realizer Method • To prove that every graph G has a Schnyder Labeling, first prove that G has a outer vertex a, which has a neighbor x, such that edge (a, x) is contractible • Proof by induction on the number of vertices • Base case: 4 vertices, K4 inner vertex as x • Induction Step: two cases: • G has no separating triangles containing a  edge (a, x) is contractible for any neighbor x of a • G has a separating triangle T containing a  by induction the subgraph inside T has a vertex x such that (a, x) is contractible

  28. Algorithm 2: Realizer Method • Proof: every graph G has a Schnyder Labeling: • This is trivial for n = 3 • For n ≥ 4, then there exists an inner vertex x adjacent to outer vertex a, such that edge (a, x) is contractible (previous proof) • Contracting this edge gives a graph which has one node less and (by induction) has a Schnyder Labeling • Take this Schnyder Labeling and perform a reverse edge contraction (that is, adding a vertex), then the Schnyder Labeling can extended to a Schnyder Labeling of graph G • The order in which the vertices are added is given by the expansion sequence • Canonical Ordering is an expansion sequence

  29. Algorithm 2: Realizer Method • The way in which the Schnyder Labeling is adjusted:

  30. Algorithm 2: Realizer Method • Orientation of an edge: • Each edge has two distinct labels at one end and two identical labels at the other end • Follows from the conditions of the Schnyder Labeling • The side with the two identical labels is called the “label of the edge” • The edge is oriented in the direction of the “label of the edge”

  31. Algorithm 2: Realizer Method • Proof: the angles at one outer vertex have all label 1, all label 2 for another and all label 3 for the last outer vertex • G is triangulated  (n – 3) inner vertices and (3n – 9) inner edges • Each inner vertex has three outgoing edges (condition 2 Schnyder Labeling) • Together: 3(n – 3) outgoing edges for all inner vertices • Thus: no outgoing edges leaves a outer vertex, so all edges incident on a outer vertex are incoming • Thus: the angles at outer vertices all have the same label (definition of the orientation of an edge)

  32. Algorithm 2: Realizer Method • Step 3: Make a Realizer for G • What is a Realizer? • A Realizer of a triangulated graph G is a partition of the inner edges into three sets T1, T2 and T3, such that for each inner vertex v: • v has an outgoing edge in each of T1, T2 and T3 • The counter-clockwise order of the edges incident on v is: • Leaving in T1 • Entering in T3 • Leaving in T2 • Entering in T1 • Leaving in T3 • Entering in T2

  33. Algorithm 2: Realizer Method • Realizer definitions: • Root ri: the outer vertex of Ti • Pi(v): the path in Ti from vertex v to the root ri of Ti • Since Pi(v) and Pj(v) for i ≠ j share only vertex v, the path P1(v), P2(v) and P3(v) divide the graph into three regions R1(v), R2(v) and R3(v) • Ri(v): the closed region opposite to root ri of Ti

  34. Algorithm 2: Realizer Method • Proof: Realizer is a Weak Barycentric Representation • Pi(v): the number of vertices in path Pi(v) • Ri(v): the number of vertices in region Ri(v), which includes both roots and vertices on both paths • ni(v): all vertices in region Ri(v) which are not on the path Pi-1(v) • Then n1(v) + n2(v) + n3(v) = n – 1, since vertex v is in all regions, but also on all path and is therefore never counted

  35. Algorithm 2: Realizer Method • Take function f, which for every vertex v gives: 1 / (n – 1) (n1(v), n2(v), n3(v)) • Satisfies condition 1, since n1(v) + n2(v) + n3(v) = n – 1 • Also satisfies condition 2 • A lot of math and definitions • This part of the proof is omitted • Conclusion: • Weak Barycentric Representation  Straight Line Drawing on a (n – 2) x (n – 2) grid • Schnyder Labeling gives a Realizer • Realizer is a Weak Barycentric Representation • All together: Realizer Method gives a Straight Line Drawing on a (n – 2) x (n – 2) grid

  36. Algorithm 2: Realizer Method • Example Straight Line Drawing:

  37. Other parts of chapter 4 • Special types of graphs can be drawn on smaller grid (than those by the previous two methodes) • 4-connected graph: grid of ((n / 2) – 1) x (n / 2) • Algorithm of Miura • Idea: • Use a 4-Canonical Ordering to split the graph G into two pieces • Draw both pieces inside a triangle • Connect both triangles

  38. Questions?

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