Two-Sided Boundary Labeling with Adjacent Sides Philipp Kindermann - - PowerPoint PPT Presentation

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Two-Sided Boundary Labeling with Adjacent Sides Philipp Kindermann - - PowerPoint PPT Presentation

Aug 17, 2023 440 likes •1.34k views

Two-Sided Boundary Labeling with Adjacent Sides Philipp Kindermann Lehrstuhl f ur Informatik I Universit at W urzburg Joint work with Benjamin Niedermann, Ignaz Rutter, Marcus Schaefer, Andr e Schulz & Alexander Wolff Problem

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SLIDE 1

Two-Sided Boundary Labeling with Adjacent Sides

Philipp Kindermann Lehrstuhl f¨ ur Informatik I Universit¨ at W¨ urzburg

Joint work with Benjamin Niedermann, Ignaz Rutter, Marcus Schaefer, Andr´ e Schulz & Alexander Wolff

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SLIDE 2

Problem Statement

enclosing rectangle

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SLIDE 3

Problem Statement

sites enclosing rectangle

  • n sites in general position
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SLIDE 4

Problem Statement

labels enclosing rectangle sites

  • n sites in general position
  • labels on the boundary
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SLIDE 5

Problem Statement

enclosing rectangle sites labels

  • n sites in general position
  • labels on the boundary

ports

  • ports are fixed or sliding
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SLIDE 6

Problem Statement

enclosing rectangle sites labels

  • n sites in general position
  • labels on the boundary
  • ports are fixed or sliding

ports

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SLIDE 7

Problem Statement

leaders ports enclosing rectangle sites labels

  • n sites in general position
  • labels on the boundary
  • ports are fixed or sliding

Find a site-label matching with crossing-free leaders.

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SLIDE 8

Why Boundary Labeling?

c Google Maps

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SLIDE 9

Why Boundary Labeling?

Hotel Ibis Kulturspeicher Main Station Bus Terminal Main Station West Hotel Poppular Barbarossaplatz Juliuspromenade Escalera Ulmer Hof Hotel Urlaub

c Google Maps

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SLIDE 10

Why Boundary Labeling?

Hotel Ibis Kulturspeicher Main Station Bus Terminal Main Station West Hotel Poppular Barbarossaplatz Juliuspromenade Escalera Ulmer Hof Hotel Urlaub

c Google Maps

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SLIDE 11

Why Boundary Labeling?

c DW-TV

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SLIDE 12

Why Boundary Labeling?

c DW-TV Henry Vandyke Carter, via Wikimedia Commons

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SLIDE 13

Why Boundary Labeling?

c DW-TV Henry Vandyke Carter, via Wikimedia Commons c friv5games.me

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SLIDE 14

Previous Work

1 2 (opp.) 4 2 (adj.) style sides

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SLIDE 15

Previous Work

1 2 (opp.) 4 s O(n2+ǫ)[1] 2 (adj.) style sides

[1] Bekos et al. CGTA’07

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SLIDE 16

Previous Work

1 2 (opp.) 4 s

  • po

O(n2+ǫ)[1] O(n2 log3 n)[1] O(n log n)[1] 2 (adj.) O(n2)[1] style sides

[1] Bekos et al. CGTA’07

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SLIDE 17

Previous Work

1 2 (opp.) 4 s

  • po

do/pd O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] 2 (adj.) O(n2)[1] O(n3)[3] style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

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SLIDE 18

Previous Work

1 2 (opp.) 4 s

  • po

do/pd po O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] O(n log n)[2] O(n2)[1] 2 (adj.) O(n2)[1] O(n3)[3] style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

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SLIDE 19

Previous Work

1 2 (opp.) 4 s

  • po

do/pd po O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] O(n log n)[2] O(n2)[1]

?

2 (adj.) O(n2)[1] O(n3)[3]

?

style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

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SLIDE 20

Previous Work

1 2 (opp.) 4 s

  • po

do/pd po O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] O(n log n)[2] O(n2)[1]

?

2 (adj.) O(n2)[1] O(n3)[3]

?

style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

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SLIDE 21

Previous Work

1 2 (opp.) 4 s

  • po

do/pd po O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] O(n log n)[2] O(n2)[1]

?

2 (adj.) O(n2)[1] O(n3)[3]

?

style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

2-Approximation on number of bends[4]

[4] Speckmann & Verbeek LATIN’10

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SLIDE 22

Previous Work

1 2 (opp.) 4 s

  • po

do/pd po O(n2+ǫ)[1] O(n2 log3 n)[1] O(n2)[2] O(n log n)[1] O(n log n)[2] O(n2)[1]

?

2 (adj.) O(n2)[1] O(n3)[3]

?

style sides

[1] Bekos et al. CGTA’07 [2] Benkert et al. JGAA’09 [3] Bekos et al. Algorithmica’10

2-Approximation on number of bends[4]

[4] Speckmann & Verbeek LATIN’10

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SLIDE 23

Structure of Planar Solutions

R

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SLIDE 24

Structure of Planar Solutions

xy-separating curve C R C

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SLIDE 25

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone

R C

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SLIDE 26

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear

R C

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SLIDE 27

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear
  • connects the top-right to the

bottom-left corner of R R C

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SLIDE 28

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear
  • connects the top-right to the

bottom-left corner of R R C xy-separated solution

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SLIDE 29

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear
  • connects the top-right to the

bottom-left corner of R R C xy-separated solution

  • top sites and leaders lie above C
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SLIDE 30

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear
  • connects the top-right to the

bottom-left corner of R R C xy-separated solution

  • top sites and leaders lie above C
  • right sites and leaders lie below C
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SLIDE 31

Structure of Planar Solutions

xy-separating curve C

  • xy-monotone
  • rectilinear
  • connects the top-right to the

bottom-left corner of R R C xy-separated solution

  • does not contain any of the following patterns

C C C C

  • top sites and leaders lie above C
  • right sites and leaders lie below C
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SLIDE 32

Structure of Planar Solutions

xy-separated solution

  • does not contain any of the following patterns

C C C C

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SLIDE 33

Structure of Planar Solutions

xy-separated solution

  • does not contain any of the following patterns

Planar solution ⇒ xy-separated planar solution C C C C

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SLIDE 34

Structure of Planar Solutions

xy-separated solution

  • does not contain any of the following patterns

Planar solution ⇒ xy-separated planar solution C C C C C C C C

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SLIDE 35

Structure of Planar Solutions

xy-separated solution

  • does not contain any of the following patterns

Planar solution ⇒ xy-separated planar solution C C C C Eliminate local crossings C C C C

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SLIDE 36

The Strip Condition

R C Given: xy-monotone curve C

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SLIDE 37

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk h1

  • h0, . . . , hk: horizontal segments of C extended to the left
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SLIDE 38

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk

  • S1, . . . , Sk:

S1 Si Sk h1

  • h0, . . . , hk: horizontal segments of C extended to the left

strips of R partitioned by h0, . . . , hk

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SLIDE 39

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk

  • S1, . . . , Sk:

S1 Si Sk h1

  • h0, . . . , hk: horizontal segments of C extended to the left
  • pi:

any point on hi \ C

pi

strips of R partitioned by h0, . . . , hk

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SLIDE 40

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk

  • S1, . . . , Sk:

S1 Si Sk h1

  • h0, . . . , hk: horizontal segments of C extended to the left
  • pi:

any point on hi \ C

  • Rpi:

polygon spanned by pi, hi and C

Rpi pi

strips of R partitioned by h0, . . . , hk

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SLIDE 41

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk

  • S1, . . . , Sk:

S1 Si Sk h1

  • h0, . . . , hk: horizontal segments of C extended to the left
  • pi:

any point on hi \ C

  • Rpi:

polygon spanned by pi, hi and C

Rpi pi

strips of R partitioned by h0, . . . , hk

  • Rpi is valid: number of sites ≥ number of ports
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SLIDE 42

The Strip Condition

R C Given: xy-monotone curve C

h0 hi hk

  • S1, . . . , Sk:

S1 Si Sk h1

  • h0, . . . , hk: horizontal segments of C extended to the left
  • pi:

any point on hi \ C

  • Rpi:

polygon spanned by pi, hi and C

Rpi pi

strips of R partitioned by h0, . . . , hk

  • Rpi is valid: number of sites ≥ number of ports
  • Condition. strip condition of Si is satisfied

⇔ ∃pi ∈ hi \ C : Rpi is valid.

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SLIDE 43

The Strip Condition - An Example

C

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SLIDE 44

The Strip Condition - An Example

C

h1

S1

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SLIDE 45

The Strip Condition - An Example

C

h1

S1

p1

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SLIDE 46

The Strip Condition - An Example

C

h1

S1

p1

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SLIDE 47

The Strip Condition - An Example

C

h1

S1

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SLIDE 48

The Strip Condition - An Example

C

h1

S1

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SLIDE 49

The Strip Condition - An Example

C

h1

S1

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SLIDE 50

The Strip Condition - An Example

C

h1

S1

p1

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SLIDE 51

The Strip Condition - An Example

C

h1

S1

p1

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SLIDE 52

The Strip Condition - An Example

C

h1

S1

p1 The strip conditions are sufficient and necessary

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SLIDE 53

The Algorithm

  • Find an xy-separating curve C

that satisfies the strip conditions.

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SLIDE 54

The Algorithm

  • Find an xy-separating curve C

that satisfies the strip conditions.

  • Consider the dual of the grid

induced by sites and ports

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SLIDE 55

The Algorithm

  • Find an xy-separating curve C

that satisfies the strip conditions.

  • Consider the dual of the grid

induced by sites and ports

  • Define grid points G(s, t)

and top-right corner r

G(s, t) s t G(0, 0) r s−1 r

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SLIDE 56

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] .

r s t s−1

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SLIDE 57

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] . T[(s, t), u] = true ⇔ ∃xy-monotone chain C:

r s t s−1

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SLIDE 58

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] . T[(s, t), u] = true ⇔ ∃xy-monotone chain C:

  • C starts at r and ends at G(s, t)

r C s t s−1

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SLIDE 59

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] . T[(s, t), u] = true ⇔ ∃xy-monotone chain C:

  • C starts at r and ends at G(s, t)

r C s t s−1

  • u sites above C
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SLIDE 60

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] . T[(s, t), u] = true ⇔ ∃xy-monotone chain C:

  • C starts at r and ends at G(s, t)

r C s t s−1

  • u sites above C
  • strip conditions are satisfied
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SLIDE 61

The Algorithm

G(0, 0) G(s, t)

Dynamic Program: Compute table T[(s, t), u] . T[(s, t), u] = true ⇔ ∃xy-monotone chain C:

  • C starts at r and ends at G(s, t)

r C s t s−1

  • u sites above C
  • strip conditions are satisfied

Instance solvable ⇔ T[(0, 0), u] = true for some u.

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SLIDE 62

One Step of the Algorithm

G(0, 0) G(s+1, t) r

Assume T[(s + 1, t), u] = true.

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SLIDE 63

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s.

G(s, t)

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SLIDE 64

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event

G(s, t) p

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SLIDE 65

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t)

G(s, t) p

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SLIDE 66

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true.

G(s, t) p

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SLIDE 67

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true. y(p) < Gy(t)

G(s, t) p

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SLIDE 68

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true. y(p) < Gy(t) ⇒ T[(s, t), u] = true.

G(s, t) p

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SLIDE 69

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true. y(p) < Gy(t) ⇒ T[(s, t), u] = true. Case 2: port event

G(s, t) ℓ

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SLIDE 70

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true. y(p) < Gy(t) ⇒ T[(s, t), u] = true. Case 2: port event

G(s, t) ℓ

Check strip condition

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SLIDE 71

One Step of the Algorithm

G(0, 0) r

Assume T[(s + 1, t), u] = true. Go from s + 1 to s. Case 1: site event y(p) > Gy(t) ⇒ T[(s, t), u + 1] = true. y(p) < Gy(t) ⇒ T[(s, t), u] = true. Case 2: port event

G(s, t) ℓ

Check strip condition satisfied ⇒ T[(s, t), u] = true

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SLIDE 72

Running Time

  • Three table entries ⇒ O(n3) steps
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SLIDE 73

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time
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SLIDE 74

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

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SLIDE 75

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

  • Entries of T depend only on previous row and column
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SLIDE 76

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

  • Entries of T depend only on previous row and column
  • Use algorithm of [Hirschberg, 1975] to backtrack a solution
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SLIDE 77

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

  • Entries of T depend only on previous row and column

→ O(n3) time, O(n2) space

  • Use algorithm of [Hirschberg, 1975] to backtrack a solution
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SLIDE 78

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

  • Entries of T depend only on previous row and column

→ O(n3) time, O(n2) space

  • Use algorithm of [Hirschberg, 1975] to backtrack a solution
  • Valid values of u form an interval

⇒ We only need to save the boundaries of u

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SLIDE 79

Running Time

  • Three table entries ⇒ O(n3) steps
  • O(n2) preprocessing ⇒ check strip condition in O(1) time

→ O(n3) time, O(n3) space

  • Entries of T depend only on previous row and column

→ O(n3) time, O(n2) space

  • Use algorithm of [Hirschberg, 1975] to backtrack a solution
  • Valid values of u form an interval

⇒ We only need to save the boundaries of u → O(n2) time, O(n) space

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SLIDE 80

Extensions

  • Sliding Ports:

O(n2) time, O(n) space

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SLIDE 81

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Sliding Ports:

O(n2) time, O(n) space

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SLIDE 82

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Sliding Ports:

O(n2) time, O(n) space

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SLIDE 83

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space

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SLIDE 84

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space

Guess ”extremal“ leader for all sides

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SLIDE 85

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space

Guess ”extremal“ leader for all sides Partition rectangle into four areas

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SLIDE 86

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space

Guess ”extremal“ leader for all sides Partition rectangle into four areas Solve each area independantly

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SLIDE 87

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space O(n10) time, O(n) space

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SLIDE 88

Extensions

  • Maximize number of labeled sites:

O(n3 log n) time, O(n) space

  • Leader-length minimization:

O(n8 log n) time, O(n6) space

  • Three-Sided Boundary Labeling:

O(n4) time, O(n) space

  • Four-Sided Boundary Labeling:
  • Sliding Ports:

O(n2) time, O(n) space O(n10) time, O(n) space