Computing Storyline Visualizations with Few Block Crossings Thomas - - PowerPoint PPT Presentation

Computing Storyline Visualizations with Few Block Crossings

Thomas C. van Dijk Fabian Lipp Peter Markfelder Alexander Wolff

Storyline Visualizations

[https://xkcd.com/657/]

Storyline Visualizations

[https://xkcd.com/657/]

Design considerations for storyline drawings:

• Line wiggles
• White space gaps
• Line crossings

[Tanahashi & Ma, TVCG12]

Storyline Visualizations

[https://xkcd.com/657/]

Design considerations for storyline drawings:

• Line wiggles
• White space gaps
• Line crossings

[Tanahashi & Ma, TVCG12]

Related Work: Simple Crossings

Aim: Drawing with a minimum number of crossings

Related Work: Simple Crossings

[Kostitsyna et al., GD’15]

• FPT for #characters
• upper and lower bounds for some cases with pairwise

meetings

• NP-hardness

Aim: Drawing with a minimum number of crossings

Related Work: Simple Crossings

[Kostitsyna et al., GD’15]

• FPT for #characters
• upper and lower bounds for some cases with pairwise

meetings

• NP-hardness

Aim: Drawing with a minimum number of crossings [Gronemann et al., GD’16]

• ILP solving the problem

Related Work: Block Crossings

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Alam, Fink, Pupyrev, GD 2016]

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Alam, Fink, Pupyrev, GD 2016] Our Aim: Drawing with a minimum number of block crossings

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Van Dijk et al., GD 2016]

• Storyline Block Crossing Minimization

– NP-hardness – Approximation algorithm – FPT [Alam, Fink, Pupyrev, GD 2016] Our Aim: Drawing with a minimum number of block crossings

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Van Dijk et al., GD 2016]

• Storyline Block Crossing Minimization

– NP-hardness – Approximation algorithm – FPT [Alam, Fink, Pupyrev, GD 2016] Now: Solve SBCM via a SAT formulation Our Aim: Drawing with a minimum number of block crossings

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Van Dijk et al., GD 2016]

• Storyline Block Crossing Minimization

– NP-hardness – Approximation algorithm – FPT [Alam, Fink, Pupyrev, GD 2016] Now: Solve SBCM via a SAT formulation Our Aim: Drawing with a minimum number of block crossings

Related Work: Block Crossings

• Block crossings for metro lines

[Fink, Pupyrev, Wolff; 2015]

• Bundled Crossings

[Fink et al., LATIN 2016] [Van Dijk et al., GD 2016]

• Storyline Block Crossing Minimization

– NP-hardness – Approximation algorithm – FPT [Alam, Fink, Pupyrev, GD 2016] Now: Solve SBCM via a SAT formulation Our Aim: Drawing with a minimum number of block crossings

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7 begin of meeting end of meeting

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7 birth of character

Events

begin of meeting end of meeting death of character

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

No births and deaths in this talk – but in the paper.

birth of character

Events

begin of meeting end of meeting death of character

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

π

Permutation π supports meetings {2, 3, 4} and {5, 6} Sets {2, 3, 4} and {5, 6} each form a contiguous group in π

=

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

π

Permutation π supports meetings {2, 3, 4} and {5, 6} Sets {2, 3, 4} and {5, 6} each form a contiguous group in π

=

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

π

Permutation π supports meetings {2, 3, 4} and {5, 6}

Find a shortest sequence of permutations such that . . . . . . meetings are supported in the right order . . . adjacent permutations differ by at most one block crossing

Sets {2, 3, 4} and {5, 6} each form a contiguous group in π

=

Problem Definition

Meeting: set of characters, start and end time

1 2 3 4 5 6 7

π

Permutation π supports meetings {2, 3, 4} and {5, 6}

Find a shortest sequence of permutations such that . . . . . . meetings are supported in the right order . . . adjacent permutations differ by at most one block crossing

Sets {2, 3, 4} and {5, 6} each form a contiguous group in π

=

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints)

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints) Each event corresponds to a layer

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints) Each event corresponds to a layer Trees describe neighborship between characters

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints) Each event corresponds to a layer Trees describe neighborship between characters

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints) Each event corresponds to a layer Trees describe neighborship between characters Find a permutation for each layer

Approach of Gronemann et al.

Reduce to MLCM-TC (Multi-Layer Crossing Minimization Problem with Tree Constraints) Each event corresponds to a layer Trees describe neighborship between characters Use ILP to solve MLCM-TC Find a permutation for each layer

Describing Permutations

Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

x1

23 = 1

x1

13 = 1

3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

x1

23 = 1

x1

13 = 1

x1

21 = 0

x1

32 = 0

x1

31 = 0

3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

x1

23 = 1

x1

13 = 1

x1

21 = 0

x1

32 = 0

x1

31 = 0

3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

x1

23 = 1

x1

13 = 1

x1

21 = 0

x1

32 = 0

x1

31 = 0

x2

12 = 1

x2

23 = 0

x2

13 = 0

x2

21 = 0

x2

32 = 1

x2

31 = 1

3 1 2 Variables describing the order in each permutation: xr

ij

characters layer

Describing Permutations

1 2 3 π1 π2 x1

12 = 1

x1

23 = 1

x1

13 = 1

x1

21 = 0

x1

32 = 0

x1

31 = 0

x2

12 = 1

x2

23 = 0

x2

13 = 0

x2

21 = 0

x2

32 = 1

x2

31 = 1

3 1 2 Antisymmetry: xr

ij ⇔ ¬xr ji

Transitivity: xr

ij ∨ xr jk ∨ xr ki

¬xr

ij ∨ ¬xr jk ∨ ¬xr ki

Variables describing the order in each permutation: xr

ij

characters layer

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

)

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

Our approach for block crossings: Allow at most one block crossing between two permutations i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

)

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

Our approach for block crossings: Allow at most one block crossing between two permutations Add more permutations i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

)

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

Our approach for block crossings: Allow at most one block crossing between two permutations Add more permutations i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

)

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

Our approach for block crossings: Allow at most one block crossing between two permutations Add more permutations i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

)

Counting Block Crossings

Easy to count pairwise crossings using xr

ij:

Our approach for block crossings: Allow at most one block crossing between two permutations Add more permutations i and j cross after layer r: χr

ij ⇔ (xr ij = xr+1 ij

) Prescribe maximum number of permutations

Describing Block Crossings

F G H F Blocks G and H cross

Describing Block Crossings

F G H F Blocks G and H cross Constraints: Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are contiguous blocks xr

ij ∧ xr jk ∧ g r i ∧ g r k ⇒ g r j

xr

ij ∧ xr jk ∧ hr i ∧ hr k ⇒ hr j

G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Meeting Groups

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Minimize the Number of Block Crossings

Choose a number of permutations λ and construct the clauses.

Minimize the Number of Block Crossings

Choose a number of permutations λ and construct the clauses. There is a satisfying assignment ⇔ ∃ solution for the SBCM instance using λ permutations.

Minimize the Number of Block Crossings

Choose a number of permutations λ and construct the clauses. O(λ(κ2 + µ)) variables, O(λµ(λ + κ3)) clauses λ: number of permutations κ: number of characters µ: number of meeting groups There is a satisfying assignment ⇔ ∃ solution for the SBCM instance using λ permutations.

Minimize the Number of Block Crossings

Choose a number of permutations λ and construct the clauses. Finding the optimum: Repeatedly run the SAT solver with different values for λ (exponential search) O(λ(κ2 + µ)) variables, O(λµ(λ + κ3)) clauses λ: number of permutations κ: number of characters µ: number of meeting groups There is a satisfying assignment ⇔ ∃ solution for the SBCM instance using λ permutations.

Experiments

FPT Breadth-first search a smarter state space; runtime: O(k! · k3 · n)

• FPT are implemented in C++
• Concurrent meetings not implemented for FPT
• SAT clauses generated by Python and solved using MiniSat

Experiments

FPT Breadth-first search a smarter state space; runtime: O(k! · k3 · n)

• FPT are implemented in C++
• Concurrent meetings not implemented for FPT
• SAT clauses generated by Python and solved using MiniSat
• Real-World instances (movies used by Gronemann et al.):

The Matrix, Inception, Star Wars

• Random instances
• Random instances having a solution with few block

crossings Test Data:

SAT: Runtime vs Number of Permutations

20 40 1 2

OPT

↓ The Matrix Permutations Time [s]

SAT: Runtime vs Number of Permutations

20 40 1 2

OPT

↓ The Matrix Permutations Time [s] 20 40 1 2

OPT

↓ Star Wars Permutations

SAT: Runtime vs Number of Permutations

20 40 1 2

OPT

↓ The Matrix Permutations Time [s] 20 40 1 2

OPT

↓ Star Wars Permutations 20 40 1 2

OPT

↓ Inception Permutations

Uniform Random Instances: FPT

200 400 600 800 1,000 10 20 κ = 5 κ = 6 κ = 7 Number of meetings Time on 100 instances [s]

Uniform Random Instances: FPT

200 400 600 800 1,000 10 20 κ = 5 κ = 6 κ = 7 Number of meetings Time on 100 instances [s]

Uniform Random Instances: SAT

20 40 20 40 60 80 100 κ = 9 Number of meetings Time on 1 instance [s]

Small-OPT Random Instances: SAT

50 100 150 200 0.5 1 1.5 2 κ = 10, β ≤ 10 Number of meetings Time on 1 instance [s]

Results

Our approach Gronemann et al. Instance cr bcOPT Time [s] crOPT bc Time [s] Star Wars 54 10 3.77 39 18 0.99 The Matrix 21 4 2.86 12 8 0.77 Inception 51 12 1.54 35 20 2.02

Movie Instances:

Results

Our approach Gronemann et al. Instance cr bcOPT Time [s] crOPT bc Time [s] Star Wars 54 10 3.77 39 18 0.99 The Matrix 21 4 2.86 12 8 0.77 Inception 51 12 1.54 35 20 2.02

Movie Instances:

Results

Our approach Gronemann et al. Instance cr bcOPT Time [s] crOPT bc Time [s] Star Wars 54 10 3.77 39 18 0.99 The Matrix 21 4 2.86 12 8 0.77 Inception 51 12 1.54 35 20 2.02

Movie Instances:

Results

Our approach Gronemann et al. Instance cr bcOPT Time [s] crOPT bc Time [s] Star Wars 54 10 3.77 39 18 0.99 The Matrix 21 4 2.86 12 8 0.77 Inception 51 12 1.54 35 20 2.02

Movie Instances:

Example: The Matrix

Gronemann et al. 12 crossings / 8 block crossings Our approach 21 crossings / 4 block crossings

Example: The Matrix

Gronemann et al. 12 crossings / 8 block crossings Our approach 21 crossings / 4 block crossings

Example: The Matrix

Gronemann et al. 12 crossings / 8 block crossings Our approach 21 crossings / 4 block crossings

Example: The Matrix

Gronemann et al. 12 crossings / 8 block crossings Our approach 21 crossings / 4 block crossings

Conclusion

• Our SAT approach is usable for real-world instances.
• Use SAT instead of ILP – turned out to be much faster!
• Source code is available online.

Conclusion

• Try other (parallel) SAT solvers.
• Find more efficient way to model lifespans.
• Consider additional quality criteria of the drawing,

e.g., minimize wiggles. [Fr¨

• schl & N¨
• llenburg, GD17]
• Perform a user study on the effect of block crossings,

especially for storyline visualizations.

• Our SAT approach is usable for real-world instances.
• Use SAT instead of ILP – turned out to be much faster!
• Source code is available online.

Future work

Appendix

Describing Block Crossings

F G H F Blocks G and H cross

Describing Block Crossings

F G H F Blocks G and H cross Constraints: Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are contiguous blocks xr

ij ∧ xr jk ∧ g r i ∧ g r k ⇒ g r j

xr

ij ∧ xr jk ∧ hr i ∧ hr k ⇒ hr j

G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are contiguous blocks xr

ij ∧ xr jk ∧ g r i ∧ g r k ⇒ g r j

xr

ij ∧ xr jk ∧ hr i ∧ hr k ⇒ hr j

G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are contiguous blocks xr

ij ∧ xr jk ∧ g r i ∧ g r k ⇒ g r j

xr

ij ∧ xr jk ∧ hr i ∧ hr k ⇒ hr j

G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Describing Block Crossings

F G H F Blocks G and H cross Constraints: G and H are contiguous blocks xr

ij ∧ xr jk ∧ g r i ∧ g r k ⇒ g r j

xr

ij ∧ xr jk ∧ hr i ∧ hr k ⇒ hr j

G and H are adjacent xr

ij ∧ xr jk ∧ g r i ∧ hr k ⇒ ¬f r j

G is above H g r

i ∧ hr j ⇒ xr ij

Exactly characters of G and H cross each other χr

ij ⇔ g r i ∧ hr j

Meeting Groups

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4 π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4

q1

1 = 1

qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4

q1

1 = 1

q2

1 = 1

qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4

q1

1 = 1

q2

1 = 1

q3

1 = 1

qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Meeting Groups

{{o, g}} {{p, r, b}, {o, g}} {{p, r, b}} {{o, b}} {{p, g, o}, {b, r}}

π1 π2 π4

q1

1 = 1

q2

1 = 1

q3

1 = 1

q4

4 = 1

q5

4 = 1

qr

ℓ = 1 ⇔

meeting group ℓ is assigned to permutation πr π3

Mapping the Meeting Groups

• Map the meeting groups in the right order:

qr

ℓ ⇒ r u=1 qu ℓ−1

• Map every meeting group exactly once:

λ

r=1 qr ℓ and ¬(qr ℓ ∧ qp ℓ )

• Force meeting characters to be next to each other:

If i and k are part of the same meeting in meeting group ℓ and j is not: qr

ℓ ⇒ (xr ij = xr kj)

Mapping the Meeting Groups

• Map the meeting groups in the right order:

qr

ℓ ⇒ r u=1 qu ℓ−1

• Map every meeting group exactly once:

λ

r=1 qr ℓ and ¬(qr ℓ ∧ qp ℓ )

• Force meeting characters to be next to each other:

If i and k are part of the same meeting in meeting group ℓ and j is not: qr

ℓ ⇒ (xr ij = xr kj)

Mapping the Meeting Groups

• Map the meeting groups in the right order:

qr

ℓ ⇒ r u=1 qu ℓ−1

• Map every meeting group exactly once:

λ

r=1 qr ℓ and ¬(qr ℓ ∧ qp ℓ )

• Force meeting characters to be next to each other:

If i and k are part of the same meeting in meeting group ℓ and j is not: qr

ℓ ⇒ (xr ij = xr kj)

Mapping the Meeting Groups

• Map the meeting groups in the right order:

qr

ℓ ⇒ r u=1 qu ℓ−1

• Map every meeting group exactly once:

λ

r=1 qr ℓ and ¬(qr ℓ ∧ qp ℓ )

• Force meeting characters to be next to each other:

If i and k are part of the same meeting in meeting group ℓ and j is not: qr

ℓ ⇒ (xr ij = xr kj)