Box-Rectangular Drawing

1. Box-Rectangular Drawing Rida Sadek

2. Overview • Some definitions and observations • Problem: Finding a Box-Rectangular drawing of a graph G • Specific Case: • given 4 vertices corresponding to corner boxes • linear time algorithm • General case: • no given vertices • linear time algorithm Note: For simplicity we are going to assume that G has 3 or more vertices and is 2-connected.

3. Who established these results? • Rahman et al. established: • a necessary and sufficient condition for the existence of a Box-Rectangular drawing of a plane graph, and • a linear time algorithm to find it if it exists.

4. Definitions and observations: • Multigraph is a graph which may have edges sharing both ends. • In a BRD, a vertex may be drawn as a degenerate box. • A degenerate box is called a point and a non-degenerate box is called a real box. • C0(G) is the outer Rectangle which has exactly 4 corners. • A box D is called a corner box if it contains at least one corner. It can be degenerate.

5. Facts • Any box-rectangular drawing has either two, three or 4 corner boxes.

6. Facts (2) • Any corner box contains either one or two corners. • In a box-rectangular drawing D of G, any vertex of degree two or three satisfies one of the following: • Vertex v is drawn as a point containing no corner; • v is drawn as a corner box containing exactly one corner; or • v is drawn as a corner real box containing exactly two corners.

7. Facts (3) • In any box-rectangular drawing D of G, every vertex of degree five or more is drawn as a real box.

8. Lemma • If G has a box-rectangular drawing, then G has a box rectangular drawing in which every vertex of degree four or more is drawn as a real box.

9. Proof of lemma • Assume G has a box rectangular drawing, then by the last fact we know that every vertex of degree five or more in G must be drawn as a real box. • If a vertex v, where d(v) = 4, is drawn as a point in D then modify the drawing D such that v is drawn as follows:

10. Illustration of proof:

11. Finding BRD if it exists: 4 designated vertices • Two major operations: • Removal of a vertex of degree two: • Let v be a vertex where d(v)=2, • we replace the 2 edges u1v and u2v incident to v with a single edge u1u2, • and we delete v. • Replacement of a vertex by a cycle:

12. BRD with designated vertices (contd.) • Idea: Construct G’’ from G using G’ as intermediate. Then make a BRD for G’’. • Construct G’ by removing all non-designated vertices of degree two one by one from G • All vertices of degree 2 in G’ WILL be designated vertices • They must be drawn as corner points in any BRD of G’. • Every designated vertex of degree 3 in G’ must be drawn as a real box since it is a corner. • Every non designated vertex of degree 3 in G’ must be drawn as a point. Result: Using the last Lemma (proved earlier), this implies that: If G’ has a BRD then G’ has a BRD D’ where all designated vertices of degree 3 and all vertices of degree 4 or more in G’ are drawn as real boxes.

13. BRD with designated vertices (contd.) • Construct G’’ from G’: • Replace by a cycle each of the designated vertices of degree 3 and the vertices of degree four or more. • The replaced cycle corresponding to a designated vertex x of degree 3 or more contains exactly one outer edge ex where x is one of the designated vertices. • Put a dummy vertex x’ of degree 2 on ex. Result: G’’ is a simple graph and has exactly four outer vertices a’, b’, c and d’ of degree 2 and all other vertices are of degree 3.

14. BRD with designated vertices (contd.)

15. BRD with designated vertices (contd.) • Theorem: Let G be a plane graph with four designated outer vertices a,b,c and d, and let G’’ be the graph transformed from G as mentioned before. Then, G has a BRD with corner boxes a,b,c, and d IF and only IF G’’ has BRD with designated corners a’,b’,c’, and d’. • Proof: • The necessity:(trivial) From the way G’’ is constructed, if G has a BRD then G’’ will have one. • The Sufficiency: Assume that G’’ has a BRD. • In D’’, each replaced cycle is drawn as a rectangle since it is a face in D’’. PLUS, the 4 outer vertices a’,b’,c’ and d’ of degree 2 in G’’ are drawn as the corners of the rectangle corresponding to C0(G’’)  D’’ immediately gives D’ having the 4 vertices a,b,c and d as corner boxes. • Insert the non-designated vertices of degree 2 on horizontal or vertical line segments in D’, then we obtain the BRD of G having a,b,c, and d as corner boxes.

16. Time Complexity • Constructing G’’ from G takes linear time. • Removing vertices of degree 2 takes O(n). • Each replacement takes constant time, and • there is at most a linear number of vertices to be replaced. • Replacing the vertices, that need to be replaced, by a cycle takes O(n) • The replacement of a vertex takes O(d) where d is the degree of the vertex. • But the sum of the degrees of the vertices is twice the number of edges since each edge contribute one degree to each vertex at its ends  O(n) overall.

17. Grid size • The half perimeter of the BRD D of G is bounded by m+2. • Let n2 be the non-designated vertices of degree 2 in G. • Let n’ = |V(G’)| and m’ = |E(G’)|  m’ = m – n2 (because we remove 2 edges and add one for each of the n2 vertices). • We replace some vertices in G’ by cycles and add at most 4 dummy vertices to construct G’’  G’’ has at most 2m’ + 4 vertices.

18. Grid Size (contd) • By theorem 6.3.8 which states that if all vertices of a plane graph G have degree 3 except the 4 corners, then the sizes of any compact rectangular grid drawing D of G satisfy W + H <= n/2. • In our case, the half perimeter (W+H) of the produced rectangular drawing D’’ is bounded then by: (2m’ + 4)/2 = m’ + 2 • The insertion of each of the n2 vertices of degree 2 back to get D increases the half perimeter at most by one  half perimeter of D is bounded by: m’ + 2 + n2 = m + 2

19. Finding BRD if it exists: Problem Recall • Case 1: (solved) a set of outer vertices of a plane graph G are designated as corner boxes • Case 2: the general case (to be solved) no designated vertices in advance • Problem definition: • See whether G has some set of outer vertices such that there is a BRD of G having them as the corner boxes. • How to find if they exist.

20. BRD without designated corners (contd) • First, we will derive a necessary and sufficient condition for a plane graph G with maximum degree Δ ≤ 3 to have a BRD and give a linear time algorithm to find it. • Second, we reduce BRD problem of a plane graph with Δ ≥ 4 to that of a new plane graph J with Δ ≤ 3.

21. BRD without designated corners (contd) • MAIN RESULT: A plane graph G with Δ ≤ 3 has a BRD (for some set of outer vertices designated as corner boxes) if and only if G satisfies the following 2 conditions: • Every 2-legged or 3-legged cycle in G contains an outer edge; and • 2c2 + c3≤ 4 for any independent set of cycles in G, where c2 and c3 are the numbers of 2-legged and 3-legged cycles respectively.

22. Fact • In a box-rectangular drawing D of G, • any 2-legged cycle contains at least two corners of the outer rectangle, • any 3-legged cycle of G contains at least one corner, and • any cycle with four or more legs may contain no corner.

23. Necessity of the main result theorem • First condition: Assume that G has a BRD. • By the previous fact, • any 2-legged or 3-legged cycle contains a corner of the outer rectangle  contains an outer edge. • Second condition: Let S be any independent set of cycles in G. • By the previous fact, • each of the c2 2-legged cycles in S contains at least 2 corners, and • each of the c3 3-legged cycles in S contains at least one corner. • Since the cycles in S are independent (vertex-disjoint with each other) THIS MEANS there are at least 2c2 + c3 corners of the outer rectangle • The outer rectangle has EXACTLY four corners  2c2 + c3≤ 4

24. Sufficiency of the main result theorem • The proof for the sufficiency is a constructive proof • The proof leads to a linear time algorithm to find a BRD if it exists.

25. Important Definition • Cubic Graphs: They are connected graphs having the property that each node has degree EXACTLY 3.

26. Sufficiency of the main result theorem (contd) • Lemma (will not proof it here): • Let G be a plane graph with Δ ≤ 3. • Assume G satisfies the condition of the MAIN RESULT, and that G has at most 3 outer vertices of degree 3. • THEN G has a box-rectangular drawing.

27. Sufficiency of the main result theorem (contd) • By the previous Lemma, we may assume that G has 4 or more outer vertices of degree 3  we can choose 4 distinct outer vertices of degree 3 as corner boxes. • No vertex v of degree 2 is chosen as corner box,  the edges incident to v are drawn on a common straight line segment in a BRD of G. • So, let G’ be the cubic graph obtained from G by removing all vertices of degree 2 one by one.  we can construct D from D’.

28. Sufficiency of the main result theorem (contd) • Lemma: • Let G be a plane cubic graph. • Assume that G satisfies: • Condition (i) from the main result (every 2-legged and 3 legged cycles contain an outer edge); and • G has four or more outer vertices; and • there is EXACTLY one C0(G)-component (A C0(G)-component is a subgraph J of G that consists of a single inner edge joining 2 outer vertices). • Then • G has a 3-legged cycle, and • If G has 2 or more independent 3-legged cycles, then the set of all minimal 3-legged cycles in G is independent (A k-legged cycle is minimal if G(C) does not contain any other k-legged cycle of G).

29. Sufficiency of the main result theorem (contd) • Proof ofpart (a) • Let w be any outer vertex, and • Let e be the inner edge incident to w. • Let x be the other end of e  then x is an inner vertex sinceG has ONE C0(G)-component and 4 or more outer vertices. • e belongs to exactly 2 faces say F1 and F2. • Since G has ONE C0(G)-component and since G is a cubic graph then: The contour of F1 contains 2 outer vertices: w and another say y. The same goes for F2. It has w and say z. • y ≠ z since d(y) = d(z) = 3. •  G has a 3-legged cycle C with leg-vertices x, y and z. Note: the legs of this 3-legged cycle meet at common vertex w.

30. Sufficiency of the main result theorem (contd) • Proof sketch of part (b) • Assume for contradiction that G has 2 or more 3-legged cycles but 2 minimal 3-legged cycles C and C’ are NOT independent  G(C) and G(C’) share a common vertex (let e1..3 and e’1..3 be the legs of C and C’ respectively). • Claim that G(C) and G(C’) do not share a common face. • Suppose for contradiction that G(C) and G(C’) do share a common face. Get the conclusion that G does not have 2 or more independent 3-legged cycles  contradiction • And then use this with the fact that G is cubic and observe that G(C) and G(C’) share a common edge on C and C’. to contradict with the first assumption by arriving to the conclusion that G has exactly 3 outer vertices contradiction.

31. Sufficiency of the main result theorem (contd) • NOW, we can prove the following Lemma: • Let G be a planar cubic graph. • Assume that G satisfies the 2 conditions of the MAIN RESULT and that G has four or more outer vertices. • Then G has a box-rectangular drawing.

32. Sufficiency of the main result theorem (contd) • First lets observe that: If G has a 2-legged cycle C, then G has a pair of independent 2-legged cycles. • By condition (i): C contains an outer edge  the 2 legs of C are outer edges, say (v, v’) and (w, w’). • Let v and w be the leg vertices of C. • It is trivial to say that v’ ≠ w’ •  G has a 2-legged cycle C’ which has v’ and w’ as leg vertices AND C and C’ are independent.

33. Sufficiency of the main result theorem (contd) • SO, we will consider 2 cases: • Case 1: G has no 2-legged cycle. • Case 2: G has a pair of independent 2-legged cycles.

34. Case 1: G has no 2-legged cycles. • This case means that G has ONE C0(G)-component, otherwise G would have a 2-legged cycle. •  By the previous Lemma, G has a 3-legged cycle.

35. Case 1: G has no 2-legged cycles (contd) • Choose the 4 corner boxes as follows: • 2 cases: • G has no pair of independent 3-legged cycles • G has a pair of independent 3-legged cycles

36. Case 1: G has no 2-legged cycles: sub-case (a) • Choose 4 outer vertices arbitrarily and regard them as the 4 designated vertices for a BRD of G. • Claim: Every 3-legged cycle C in G has at least one designated vertex. • Since C has an outer edge by condition (i) of the main result  EXACTLY 2 of the 3 legs of C are outer edges. Say x and y are the leg vertices of these 2 legs.

37. Case 1: G has no 2-legged cycles: sub-case (a)(cntd.) • Let P be the path from x to y on C0(G). Then P has exactly one intermediate vertex z: otherwise either G would have more than one C0(G)-component or a pair of independent 3-legged cycles  contradiction. (Remember the NOTE proof (a) of Lemma on slide 28 where we said that these 3 legs meet at one point w and this point is exactly the vertex z we are mentioning). •  We can know that all 3 legs are incident to z •  all outer vertices except z lie on C. •  C contains at least one of the four designated vertices (whether z is one of them or not) •  Claim is proven.

38. Case 1: G has no 2-legged cycles: sub-case (b) • G has a pair of independent 3-legged cycles. • Let M be the set of all minimal 3-legged cycles in G. • By the Previous Lemma (slide 28), M is independent. • Let k = |M|, then k ≤ 4 by condition (ii) of main result (2c2+c3≤4) that this theorem satisfies. • For each 3-legged cycle Cj; where 1 ≤ j ≤ 4; in M, we arbitrarily choose an outer vertex on Cj. If k<4 we arbitrarily chose 4-k outer vertices which are not chosen so far. •  We have chosen exactly four outer vertices, and we take these vertices to be the 4 designated vertices of a BRD of G.

39. Illustration • Vertices a, b, c and d are chosen as the designated vertices. • a, b and d are chosen on 3 independent 3-legged cycles AND vertex c is chosen arbitrarily on C0(G).

40. Illustration (cntd.) • Claim: Every 3-legged cycle C of G has at least one designated vertex: • C has a designated vertex if C is minimal (since C belongs to M and this is how we chose the designated vertices). • If C is not minimal then G(C) contains a minimal 3-legged cycle Cj that belongs to M and that Cj has a designated outer vertex. • SO WE HAVE CHOSEN 4 DESIGNATED OUTER VERTICES

41. Construction • Replace each designated vertex by a cycle and add a dummy vertex of degree 2 on the edge of the cycle that belongs to the OUTER cycle. • The new graph is G’ where all vertices have degree 3 except the 4 dummy outer vertices. • From chapter 6 there is a theorem that says: • Assume that G is a 2-connected plane graph with Δ≤ 3, and 4 outer vertices of degree 2 are designated as the corners, then G has a rectangular drawing if and only if • Any 2-legged cycle contains 2 or more corners, and • Any 3-legged cycle contains one or more corners

42. Construction (contd) • G’ satisfies the conditions of that theorem and we can draw a rectangular drawing D’ as shown in the figure. • Replace the four faces in D’ corresponding to the replaced cycles BY BOXES  the resulting graph D is a BRD of G.

43. Case 2: G has a pair of independent 2-legged cycles • Will not be discussed because of time constraints.

44. Sufficiency of the main result theorem (contd) • Using these results one can find a box rectangular drawing of G if G satisfies the conditions of the MAIN RESULT (stated at first)  proved constructively the sufficiency of the MAIN RESULT • FINAL RESULT that is a direct result from the discussed proofs: • A plane graph G with Δ≤ 3 has a BRD iff G satisfies the 4 following conditions: • Every 2-legged or 3-legged cycles has an outer edge • At most 2-legged cycles of G are independent of each other • At most four 3-legged cycles are independent of each other • If G has a pair of independent 2-legged cycles C1 and C2, then {C1, C2, C3} is not independent for any 3-legged cycle C3 in G, and neither G(C1) nor G(C2) has more than 2 independent 3-legged cycles of G. (this is actually proven in the proof that we didn’t discuss where there exists a pair of independent 2-legged cycles).

45. Result of time complexity • Result: • To see if G has a BRD, this takes O(m) where m is the number of edges in G. • To find D if it exists, this takes O(m) where m is the number of edges in G. NOTE: The Proof will not be discussed but it is very clear in the book.

46. Reducing the problem of Δ≥4 to the one of Δ≤3 (Sketch) • Let G be a plane graph with Δ≥4 • Construct J by replacing each vertex of degree 4 or more by a cycle. • A replaced cycle corresponds to a real box in the BRD of G. • Clearly Δ(J) ≤ 3.

47. Reducing the problem of Δ≥4 to the one of Δ≤3 (Sketch).

48. Questions?