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Tohoku University.

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Presentation on theme: "Tohoku University."— Presentation transcript:

1 Tohoku University

2 Tohoku University Sendai 360km Tokyo

3 Inner Rectangular Drawings of Plane Graphs ーApplication of Graph Drawing to VLSI Layout-
Takao Nishizeki Tohoku University

4 Inner Rectangular Drawing
an inner rectangular drawing of G a plane graph G 1:each vertex is drawn as a point 1:each vertex is drawn as a point 2:each edge is drawn as a horizontal or vertical line segment 3:all inner faces are drawn as rectangles

5 Inner Rectangular Drawing
an inner rectangular drawing of G a plane graph G 1:each vertex is drawn as a point 2:each edge is drawn as a horizontal or vertical line segment 2:each edge is drawn as a horizontal or vertical line segment 3:all inner faces are drawn as rectangles

6 Inner Rectangular Drawing
rectilinear polygon Fo F an inner rectangular drawing of G a plane graph G 1:each vertex is drawn as a point 2:each edge is drawn as a horizontal or vertical line segment 3:all inner faces are drawn as rectangles 3:all inner faces are drawn as rectangles

7 Application VLSI floor planning 1 3 4 6 9 5 2 7 8
The outer boundary of a VLSI chip is often an axis-parallel polygon Inner rectangular drawing 1 2 6 3 4 5 7 8 9 Ga G Vertex: module edge : adjacency among modules

8 Application VLSI floor planning The outer boundary of a VLSI chip
is often an axis-parallel polygon 1 2 6 3 4 5 7 8 9 1 2 1 1 3 4 5 6 7 8 9 Ga G Vertex: module edge : adjacency among modules Inner rectangular drawing

9 Application VLSI floor planning The outer boundary of a VLSI chip
is often an axis-parallel polygon 2 2 2 1 2 1 6 3 4 5 7 8 9 3 4 5 6 7 8 9 Ga G Vertex: module edge : adjacency among modules Inner rectangular drawing

10 Application VLSI floor planning The outer boundary of a VLSI chip
is often an axis-parallel polygon 2 1 2 1 2 1 2 1 3 4 5 6 3 4 5 7 8 9 6 7 8 9 Ga G Vertex: module edge : adjacency among modules Inner rectangular drawing

11 Application VLSI floor planning 1 3 4 6 9 5 2 7 8
The outer boundary of a VLSI chip is often an axis-parallel polygon Inner rectangular drawing dual-like graph 1 2 3 4 5 6 7 8 9 2 1 3 4 5 6 7 8 9 Ga G=Ga* G Vertex: module edge : adjacency among modules

12 Application VLSI floor planning The outer boundary of a VLSI chip
is often an axis-parallel polygon dual-like graph 1 3 4 6 9 5 2 7 8 Inner rectangular drawing G 1 2 3 4 5 6 7 8 9 2 1 3 4 5 6 7 8 9 Ga G=Ga* Vertex: module edge : adjacency among modules

13 Application VLSI floor planning The outer boundary of a VLSI chip
is often an axis-parallel polygon dual-like graph 1 3 4 6 9 5 2 7 8 Inner rectangular drawing G 2 1 3 4 5 6 7 8 9 Ga Vertex: module edge : adjacency among modules

14 a rectangular drawing of G
Known Result a necessary and sufficient condition for the existence of a rectangular drawing of G with [T84,RNN98] and a linear algorithm for [RNN98,BS88,KH97] a rectangular drawing of G a plane graph G

15 an inner rectangular drawing of G
Open Problem a necessary and sufficient condition for the existence of an inner rectangular drawing of G (with )? efficient algorithm to find an inner rectangular drawing of G (with )? an inner rectangular drawing of G ? a plane graph G

16 Our Results 1: a necessary and sufficient condition for the existence of an inner rectangular drawing of G.

17 a “sketch” of the outer face
Our Results 2: O(n1.5/log n) algorithm to find an inner rectangular drawing of G if a “sketch” of the outer face is given. a plane graph G a “sketch” of the outer face

18 an inner rectangular drawing of G
Our Results 2: O(n1.5/log n) algorithm to find an inner rectangular drawing of G if a “sketch” of the outer face is given. an inner rectangular drawing of G a plane graph G

19 an inner rectangular drawing of G
Our Results 3: a polynomial time algorithm to find an inner rectangular drawing of G in a general case, where a sketch is not always given. an inner rectangular drawing of G a plane graph G

20 1:A necessary and sufficient condition for the existence of an inner rectangular drawing of G.
2: O(n1.5 /log n) time algorithm to find an inner rectangular drawing of G if a sketch of the outer face is given. 3: a polynomial time algorithm to find an inner rectangular drawing of G in a general case, where a sketch is not always given.

21 an inner rectangular drawing of G
Definition of Labeling 3 2 1 an inner rectangular drawing of G a plane graph G Consider a labeling which assigns label 1,2 or 3 to every angle of G

22 an inner rectangular drawing of G
Definition of Labeling     3 3 2 1 2 1 3 an inner rectangular drawing of G a plane graph G Consider a labeling which assigns label 1,2 or 3 to every angle of G

23 Regular labeling A regular labeling satisfies the following three conditions (a)-(c) (a) the labels of all the angles of each vertex v total to 4; 2 3 1 3 2 1 1 1 1 1 1 2 1 2 1 1 1 3 2 1 1 2 1 2 1 2 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 3 3 2 1 1 1 1 1 3 1 3 2 a plane graph G an inner rectangular drawing of G

24 an inner rectangular drawing of G
Regular labeling (b) the labels of any inner angles is 1 or 2, and any inner face has exactly four angles of label 1; 3 2 1 1 a plane graph G an inner rectangular drawing of G

25 an inner rectangular drawing of G
Regular labeling (c) ncv - ncc = 4. ncv: the number of outer angles having label 3 ncc : the number of outer angles having label 1 rectilinear polygon 3 2 1 convex corners ncv=6 a plane graph G an inner rectangular drawing of G

26 an inner rectangular drawing of G
Regular labeling (c) ncv - ncc = 4. ncv: the number of outer angles having label 3 ncc : the number of outer angles having label 1 3 2 1 convex corners ncv=6 ncc=2 concave corners a plane graph G an inner rectangular drawing of G

27 Regular labeling (c) ncv - ncc = 4.
ncv: the number of outer angles having label 3 ncc : the number of outer angles having label 1 ncv = 4 ncc = 0 ncv = 5 ncc = 1 ncv = 6 ncc = 2

28 an inner rectangular drawing of G
A necessary and sufficient condition for the existence of an inner rectangular drawing of G A plane graph G has an inner rectangular drawing a plane graph G G has a regular labeling 3 2 1 an inner rectangular drawing of G a plane graph G

29 an inner rectangular drawing of G
A necessary and sufficient condition for the existence of an inner rectangular drawing of G an inner rectangular drawing of G A plane graph G has an inner rectangular drawing G has a regular labeling 3 2 1 a plane graph G

30 an inner rectangular drawing of G
A necessary and sufficient condition for the existence of an inner rectangular drawing of G A plane graph G has an inner rectangular drawing G has a regular labeling 3 2 1 an inner rectangular drawing of G a plane graph G

31 1:A necessary and sufficient condition for the existence of an inner rectangular drawing of G.
2: O(n1.5 /log n) time algorithm to find an inner rectangular drawing of G if a sketch of the outer face is given. 3: a polynomial time algorithm to find an inner rectangular drawing of G in a general case, where a sketch is not always given.

32 1:A necessary and sufficient condition for the existence of an inner rectangular drawing of G.
2: O(n1.5 /log n) time algorithm to find an inner rectangular drawing of G if a sketch of the outer face is given. 3: a polynomial time algorithm to find an inner rectangular drawing of G in a general case, where a sketch is not always given.

33 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 2 1 a sketch of the outer face of G 2 1 3 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3

34 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 3 3 2 1 a sketch of the outer face of G 2 1 3 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3

35 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 2 1 1 a sketch of the outer face of G 2 1 3 1 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3

36 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 2 1 a sketch of the outer face of G 2 1 3 2 2 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3

37 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 a sketch of the outer face of G 2 1 3 1 2 3 2 2 2 3 3 2 2 2 3 2 3 1 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3 Find an inner rectangular drawing with a prescribed sketch of the outer face

38 a sketch of the outer face of G
Inner rectangular drawing with sketched outer face 3 a sketch of the outer face of G 2 1 3 2 1 3 1 2 3 2 2 2 3 3 2 2 2 3 2 3 1 a plane graph G Suppose that a sketch of the outer face of G is prescribed, that is, all the outer angles of G are labeled with 1, 2 or 3 Find an inner rectangular drawing with a prescribed sketch of the outer face

39 inner rectangular drawing G has a regular labeling
A plane graph G has an inner rectangular drawing G has a regular labeling G has a regular labeling Gd has a perfect matching 1 3 2 a decision graph Gd of G a plane graph G

40 Construct a decision graph Gd
Labels of some of the inner angles of G can be immediately determined 3 2 1 2 3 1 a plane graph G

41 Construct a decision graph Gd
degree 2 3 2 1 2 3 1 2 2 a plane graph G

42 Construct a decision graph Gd
degree 2 3 2 1 3 2 3 1 3 2 2 2 2 2 3 1 3 3 2 a plane graph G

43 Construct a decision graph Gd
degree 4 3 2 1 2 3 1 2 2 1 1 a plane graph G

44 Construct a decision graph Gd
degree 4 3 2 1 2 3 1 2 2 1 1 a plane graph G

45 Construct a decision graph Gd
degree 4 3 2 1 2 3 1 1 1 2 2 1 1 a plane graph G

46 Construct a decision graph Gd
degree 4 3 2 1 2 3 1 1 1 2 2 1 1 a plane graph G

47 Construct a decision graph Gd
outer vertex of degree 2 3 2 1 2 3 2 1 2 1 1 2 2 1 1 1 a plane graph G

48 Construct a decision graph Gd
outer vertex of degree 2 3 2 1 2 3 1 2 1 2 1 1 1 2 2 1 1 a plane graph G

49 Construct a decision graph Gd
outer vertex of degree 2 3 2 1 2 3 1 1 2 1 2 1 1 1 1 2 2 1 1 1 1 1 a plane graph G

50 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 1 2 2 1 1 a plane graph G

51 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 1 1 2 2 1 1 a plane graph G

52 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 a plane graph G

53 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 a plane graph G

54 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 a plane graph G

55 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 2 a plane graph G

56 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 1 1 2 2 a plane graph G

57 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 2 1 a plane graph G

58 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 2 2 1 1 a plane graph G

59 Construct a decision graph Gd
outer vertex of degree 3 3 2 1 2 3 1 1 2 1 2 x x x = 1 or 2 a plane graph G

60 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 x x x = 1 or 2 a plane graph G

61 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 2 1 1 x x x = 1 or 2 a plane graph G

62 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 2 2 1 1 1 1 x x x = 1 or 2 a plane graph G

63 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 1 1 2 x x x = 1 or 2 a plane graph G

64 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 2 1 1 1 1 1 1 2 1 1 1 2 x x 1 1 x = 1 or 2 a plane graph G

65 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 1 2 1 x x x = 1 or 2 a plane graph G

66 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 1 1 1 2 1 1 1 1 1 1 1 2 1 1 2 x x 1 1 x = 1 or 2 a plane graph G

67 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 x x x x x = 1 or 2 a plane graph G

68 Construct a decision graph Gd
degree 3 3 2 1 2 3 1 1 2 1 2 x x x x x x x x x = 1 or 2 a plane graph G

69 Construct a decision graph Gd
3 2 1 a decision graph Gd of G 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a plane graph G

70 Construct a decision graph Gd
3 2 1 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

71 Construct a decision graph Gd
3 2 1 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

72 Construct a decision graph Gd
2 of x’s must be 1’s. 3 2 1 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

73 Construct a decision graph Gd
3 2 1 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

74 Construct a decision graph Gd
3 2 1 1 2 1 1 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

75 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

76 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

77 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x =2 x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

78 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 1= x =1 x 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

79 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 2 = x 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

80 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1= x 1 = 1 x 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

81 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 2 x = 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

82 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

83 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

84 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

85 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

86 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 2 of x’s are 1’s. 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 x =1 2 x 1= x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

87 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching 3 2 1 1 2 1 1 =1 2 = 1 =2 1= x 1 2 x 1 1 1 x 1 x 1 1 x 1 x 1 1 1 1 a decision graph Gd of G a plane graph G

88 A necessary and sufficient condition for the existence of
a regular labeling G has a regular labeling Gd has a perfect matching a plane graph G x 1 3 2 =1 2 = 1 =2 1= a decision graph Gd of G

89 an inner rectangular drawing of G
A necessary and sufficient condition for the existence of a regular labeling A necessary and sufficient condition for the existence of an inner rectangular drawing G has an inner rectangular drawing with sketched outer face G has a regular labeling Gd has a perfect matching a plane graph G x 1 3 2 a plane graph G an inner rectangular drawing of G =1 2 = 1 =2 1= a decision graph Gd of G

90 a perfect matching of Gd
Running time 1 a regular labeling of G 3 2 nd = O(n) md = O(n) A perfect matching of Gd can be found in time O( ) [HK73,MV80] or in time O( ) [FM91,Hoc04,HC04] a perfect matching of Gd A perfect matching of Gd can be found in time O(n1.5/log n)

91 Running time nd = O(n) md = O(n) 3
1 a regular labeling of G 3 2 nd = O(n) md = O(n) A perfect matching of Gd can be found in time O( ) [HK73,MV80] or in time O( ) [FM91,Hoc04,HC04]

92 an inner rectangular drawing of G
Running time nd = O(n) md = O(n) an inner rectangular drawing of G A perfect matching of Gd can be found in time O( ) [HK73,MV80] or in time O( ) [FM91,Hoc04,HC04] An inner rectangular drawing of G can be found in time O(n1.5/log n)

93 1:A necessary and sufficient condition for the existence of an inner rectangular drawing of G.
2: O(n1.5 /log n) time algorithm to find an inner rectangular drawing of G if a sketch of the outer face is given. 3: a polynomial time algorithm to find an inner rectangular drawing of G in a general case, where a sketch is not always given.

94 ncv = 6, ncc = 2 Z-type U-type T-type stair-type
Case 1: the numbers of convex and concave outer vertices are given. Z-type U-type T-type stair-type ncv = 6, ncc = 2

95 Case 1: the numbers of convex and concave outer vertices are given.
a decision graph Gd of G ncv ncc - no4 ncv = 6, ncc = 2 a plane graph G

96 an inner rectangular drawing of G
Case 1: the numbers of convex and concave outer vertices are given. ncv ncc - no4 ncv = 6, ncc = 2 an inner rectangular drawing of G 2 1 3 a plane graph G a decision graph Gd of G

97 Case 1: the numbers of convex and concave outer vertices are given.
Running time nd = O(n) md = O(N), N=n + ncvno (no: the number of outer vertices). An inner rectangular drawing of G can be found in time O( ).

98 Case 2: neither the outer sketch nor the numbers of corners are given.
ncv , ncc are arbitrary

99 Case 2: neither the outer sketch nor the numbers of corners are given.
a decision graph Gd of G no2 - no4 -4 a plane graph G

100 an inner rectangular drawing of G
Case 2: neither the outer sketch nor the numbers of corners are given. no2 - no4 -4 an inner rectangular drawing of G 1 2 3 a plane graph G a decision graph Gd of G

101 Case 2: neither the outer sketch nor the numbers of corners are given.
Running time nd = O(n) md = O(N’), N’ =n + (no2 – no4 – 4)no (no: the number of outer vertices, no2 and no4 : the numbers of outer vertices of degrees 2 and 4 ) An inner rectangular drawing of G can be found in time O( ).

102 Conclusion G has an inner rectangular drawing
Gd has a perfect matching (1) (2) An inner rectangular drawing can be found in time O(n1.5 / log n) if the outer face is sketched. O( ) if (ncv, ncc) is prescribed. N=n + ncvno no: the number of outer vertices O( ) for a general case. N’ =n + (no2 – no4 –4)no no2 and no4 : the numbers of outer vertices of degrees 2 and 4 (3) Linear algorithm ?

103

104 Network Flow 1 2 4 Network N G 2 3 1 2 1 x 3 1 x 3 2 x 1 1 2 x 3 1 1 2

105 Network Flow 1 1 2 2 4 2 2 2 1 1 Network N G 2 3 1 2 1 x 1 1 3 1 x 3 2
1 2 1 x 1 1 1 3 1 x 3 2 2 x 1 2 1 1 1 1 2 4 2 1 1 1 x 3 1 2 1 2 2 2 1 1 1 2 2 1 2 1 2 1 1 3 2 Network N G

106 Network Flow 1 1 2 2 4 2 2 2 1 1 Network N G 2 3 1 2 1 2 1 1 1 3 1 3 1
1 2 1 2 1 1 1 1 3 1 3 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 4 2 1 1 2 3 1 1 1 1 2 1 2 2 2 1 1 2 1 1 1 2 2 1 2 1 2 1 1 3 2 Network N G

107 An inner rectangular drawing of G
Network Flow 2 3 An inner rectangular drawing of G 1 2 1 2 1 3 1 1 3 2 1 2 1 1 2 1 2 1 3 1 1 1 1 2 2 2 1 1 2 1 2 2 2 1 2 1 1 3 2 G

108

109 Case 1: the numbers of convex and concave outer vertices are given.
ncv = 6, ncc = 2 Case 2: general case

110 Inner rectangular drawing with prescribed numbers ncv and ncc
ncv = 6, ncc = 2 a plane graph G

111 a decision graph Gd* of G
Inner rectangular drawing with prescribed numbers ncv and ncc ncv ncc - no4 ncv = 6, ncc = 2 y x 1 x 2 a plane graph G a decision graph Gd* of G

112 a decision graph Gd* of G
Inner rectangular drawing with prescribed numbers ncv and ncc ncv ncc - no4 ncv = 6, ncc = 2 y y y x x 1 x 1 1 1 x x x x x x 2 2 y x x x x x x x x x x 1 1 x x x x 1 y 1 x x x x x x y x y x a plane graph G a decision graph Gd* of G

113 a decision graph Gd* of G
Inner rectangular drawing with prescribed numbers ncv and ncc ncv ncc - no4 ncv = 6, ncc = 2 2 1 3 y y y x x 1 x 1 1 1 x x x x x x 2 2 y x x x x x x x x x x 1 1 x x x x 1 y 1 x x x x x x y x y x a plane graph G a decision graph Gd* of G

114 an inner rectangular drawing of G
Inner rectangular drawing with prescribed numbers ncv and ncc ncv = 6, ncc = 2 y 3 y 3 an inner rectangular drawing of G 1 y 3 x x 1 x 1 1 1 2 x x x x x x 2 2 y 3 x x x x x 2 x x x x x 1 1 1 x x x x 1 1 3 y x x 2 x x x x y x y 3 x a plane graph G

115 Running time Gd* has an O(n) number of vertices and O(N) (N=n + ncvno no: the number of outer vertices) number of edges. An inner rectangular drawing D of G can be found in time O( ).

116 Case 1: the numbers of convex and concave outer vertices are given.
ncv = 6, ncc = 2 Case 2: in general case

117 Inner rectangular drawing
no2 - no4 -4 y y y x x 1 x 1 1 1 x x x x x x 2 2 y x x x x x x x x x x 1 1 x x x x 1 1 y x x x x x x y x y x a plane graph G a decision graph Gd of G

118 Inner rectangular drawing
no2 - no4 -4 y 3 2 1 y y x x 1 x 1 1 1 x x x x x x 2 y 2 x x x x x x x x x x 1 1 x x x x 1 1 y x x x x x x y x y x a plane graph G a decision graph Gd of G

119 Inner rectangular drawing
an inner rectangular drawing of G a plane graph G x y 1 2 3

120 Running time O( ). Gd has an O(n) number of vertices and O(N’)
(N’=n + (no2 - no4 – 4)no no: the number of outer vertices no2 and no4 : the numbers of outer vertices of degrees 2 and 4 ) number of edges. An inner rectangular drawing D of G can be found in time O( ).

121

122 Related result 2 3 1 a plane graph G If a sketch of several faces of G including the outer face is prescribed, then one can examine whether G has a drawing such that each of the other face is a rectangle.

123 Related result a plane graph G
x 1 y ncv = 4, ncc = 0 ncv = 5, ncc = 1 a plane graph G If faces F0,F1, … Fi of G are vertex-disjoint and the numbers of convex and concave vertices are prescribed, then one can examine whether G has a drawing such that each of F0,F1, … Fi has prescribed numbers of convex and concave vertices and each of the other faces is a rectangle.

124 (a) the labels of any vertex in G total to 4;
Regular labeling We call f a regular labeling of G if f satisfies the following three conditions (a)-(c) (a) the labels of any vertex in G total to 4; (b) the labels of any inner angles is 1 or 2, and any inner face has exactly four angles of label 1; (c) ncv - ncc = 4. ncv: the number of outer angles having label 3 ncc : the number of outer angles having label 1 3 2 1 2 1 3


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