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## PowerPoint Slideshow about ' 1.5-side Boundary Labeling' - violet

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- 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

Polygons labeling (Bekos et. al, APVIS 2006)

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

Variants#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

#2

Problem Setting- (labelSize, labelPort, Objective)

#1

#1

#2

#2

Fixed-ratio port (FR)

Fixed-position port (FP)

Sliding port

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)

#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

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.

// 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

// 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

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.

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.

J1

J0

J2

J4

0

M

Total Discrepancy Problem is NP-completeJ0

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 scheduleopt
- 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}

Theorem 4. Subset Sum ProblemL(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

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

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

(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

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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