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1.5-side Boundary Labeling. Type- opo leaders. Type- po leders. Type- s leaders. (Bekos & Symvonis, GD 2005). Boundary labeling (Bekos et al., GD 2004). label. site. leader. 1-side, 2-side, 4-side. Min (total leader length or total bend number) s.t. #(leader crossing) = 0.

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1.5-side Boundary Labeling

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1 5 side boundary labeling

1.5-side Boundary Labeling


Boundary labeling bekos et al gd 2004

  • Type-opo leaders

  • Type-po leders

  • Type-s leaders

(Bekos & Symvonis, GD 2005)

Boundary labeling (Bekos et al., GD 2004)

label

site

leader

1-side, 2-side, 4-side

Min (total leader length or

total bend number)

s.t. #(leader crossing) = 0


Variants

Polygons labeling (Bekos et. al, APVIS 2006)

Multi-stack boundary labeling (Bekos et. al, FSTTCS 2006)

Variants


1 5 side boundary labeling1

#1

#6

#2

indirect

leader

#1

#3

#2

#3

#4

#4

#5

#5

#6

1.5-side Boundary Labeling

  • type-opo: direct leader vs. indirect leader

  • Annotation system for wordprocessing S/W

direct leader


Problem setting

Problem Setting

  • (labelSize, labelPort, Objective)

#1

#3

#2

#1

#2

Uniform label

Nonuniform label


Problem setting1

#1

#2

Problem Setting

  • (labelSize, labelPort, Objective)

#1

#1



#2



#2

Fixed-ratio port (FR)

Fixed-position port (FP)

Sliding port


Problem setting2

Problem Setting

  • (labelSize, labelPort, Objective)

#1

#4

#1

#2

#3

#1

#2

#2

#3

#3

#3

#4

#1

#2

#(bends) = 6

#(bends) = 2

longer length

shorter length

Min (total leader length)

(TLLM for short)

Min (total bend num)

(TBM for short)


Assumptions

#1

#6

#2

#3

#4

#5

pj+1

map

label

map

label

pj

j

pj+1

Aleft

j

pj

Aright

Aright

pi

Aleft

i

pi–1

pi

i

pi–1

Assumptions

  • All the parameters are integers

  • No two sites with the same x- or y- coordinate

  • Map height = label height sum

  • Legal leader


Our contributions

Our Contributions

Solved all the problems of all the combinations of (LabelSize, LabelPort, Objective).

* Pseudo-polynomial time algorithms and fixed-parameter algorithms are designed for those intractable problems.


1 5 side boundary labeling

B

B

p

p

B

lh

B

lv

  • Lemma 1. All direct leaders are optimal for the above concerned case.

leader l

|U|

U

U

B

B

p

p


1 5 side boundary labeling

S(a, b, c) =

// the solution of the problem with pa, pa+1, …, pb connected to label positions c to c+(b-a)+1

// all direct leaders

// downward indirect leader

// upward indirect leader

  • Theorem 2. The above case can be solved by dynamic programming in O(n5) time.

map

label

(c+b-a)-th

pb

# = (b+a)+1

pa

c-th


1 5 side boundary labeling

S(a, b, c) =

// the solution of the problem with pa, pa+1, …, pb connected to label positions c to c+(b-a)+1

// all direct leaders

// downward indirect leader

// upward indirect leader

  • Theorem 2. The above case can be solved by dynamic programming in O(n5) time.

map

label

(c+b-a)-th

pb

pa

c-th


1 5 side boundary labeling

map

label

(c+b-a)-th

S(a+i+1, b, c+i+1)

pb

(c+i+1)-th

pa+i+1

(c+i)-th

pa+i

pa+i-1

S(a+j, a+i-1, c+j+1)

(c+j+1)-th

pa+j

(c+j)-th

pa+j-1

(c+j-1)-th

S(a, a+j-1, c)

pa

c-th

S(a, b, c) =

// the solution of the problem with pa, pa+1, …, pb connected to label positions c to c+(b-a)+1

// all direct leaders

// downward indirect leader

// upward indirect leader

  • Theorem 2. The above case can be solved by dynamic programming in O(n5) time.


1 5 side boundary labeling

map

label

map

label

(c+b-a)-th

(c+b-a)-th

pb

S(a+j+1, b, c+j+1)

S(a+i+1, b, c+i+1)

pb

(c+i+1)-th

(c+j+1)-th

pa+j+1

pa+i+1

(c+i)-th

(c+j)-th

pa+i

pa+j

pa+i-1

S(a+j, a+i-1, c+j+1)

(c+j-1)-th

S(a+i+1, a+j, c+i)

pa+i+1

(c+j+1)-th

pa+j

pa+i

(c+i)-th

(c+j)-th

pa+j-1

pa+i-1

(c+j-1)-th

(c+i-1)-th

S(a, a+i-1, c)

pa

S(a, a+j-1, c)

pa

c-th

c-th

S(a, b, c) =

// the solution of the problem with pa, pa+1, …, pb connected to label positions c to c+(b-a)+1

// all direct leaders

// downward indirect leader

// upward indirect leader

  • Theorem 2. The above case can be solved by dynamic programming in O(n5) time.


Total discrepancy problem is np complete

J3

J1

J0

J2

J4

0

M

Total Discrepancy Problem is NP-complete

J0

J1

J2

J3

J4

  •  job Ji {J0, J1, …, J2n}

    • Execution time length li , where I0 < I1 < … < l2n

    • Preferred midtimeM = (l0 + l1 + … + l2n) /2

  • For a planned schedule

    • Actual midtime of Ji = mi()

    • Min ( |m0() – M| + |m1() – M| + … + |m2n() – M| + |m2n+1() – M’|)

  • Properties for the optimal scheduleopt

    • No gaps between two jobs

    • m0(opt) = M

    • | {Ji : mi < M } | = | {Ji : mi > M } |

    • opt= An, An-1, …, A1, J0, B1, B2, …, Bn where {Ai, Bi} = {J2i-1, J2i}


1 5 side boundary labeling

  • Theorem 3. Total Discrepancy Problem L(nonuniform, FR/FP/sliding, TLLM).

J3

J1

J0

J2

J4

0

M


1 5 side boundary labeling

  • Theorem 4. Subset Sum ProblemL(nonuniform, FR/FP/sliding, TBM).

Subset Sum Problem

Input:A = {a1, …, an} and a num B = (a1 + … + an)/2

Question: find a subset A’A

such that

sum(elements in A’) = B

pn+2

< hmin

< hmin

pn+1

h/2

hmin


1 5 side boundary labeling

* Pseudo-polynomial time algorithms and fixed-parameter algorithms are designed for those intractable problems.


1 5 side boundary labeling

  • Theorem 5 (pseudo-polynomial algorithm).The above two cases can be solved in O(n4h) time, where h is the map height.

S(a, b,t ) =

// the solution of the problem with pa, pa+1, …, pb connected to y-coordinate t

S(a, b, c) =

// the solution of the problem with pa, pa+1, …, pb connected to label positions c to c+(b-a)+1

// all direct leaders

(uniform label case)

// downward indirect leader

// upward indirect leader


1 5 side boundary labeling

type-1, …, type-(k-1)

(i –1) labels using

type-1, type-2, …, type-k

type-k

  • Theorem 6 (fixed-parameter algorithm).The above two cases using k different label heights can be solved in O(nk+4) time.

  • Theorem 5. The above two cases can be solved in O(n4h) time.

  • Lemma 2. num( positions of each label using k different label heights ) = O(nk).

    pf.

    • Induction on k

    • Assume num(…(k-1) …) = O(nk-1)

    • Consider each label, which is the i-th label from the bottom

h = nk

O(nk-1) positions

at most O(n)


Conclusion

Conclusion

Solved all the problems of all the combinations of (LabelSize, LabelPort, Objective).

* Pseudo-polynomial algorithms and fixed-parameter algorithms are designed for those intractable problems.


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