Weak Unit Disk Contact Representations for Graphs without Embedding - PowerPoint PPT Presentation

Weak Unit Disk Contact Representations for Graphs without Embedding Jonas Cleve, Freie Universität Berlin EuroCG 2020 – March 16–18 – Würzburg, Germany

Unit Disk Contact Representations Unit disk contact representation (UDCR) of G = ( V , E ): graph G 1 2 3 4 5 6 7 8 1

Unit Disk Contact Representations 8 8 7 6 5 4 3 2 1 7 Unit disk contact representation (UDCR) of G = ( V , E ): 6 5 4 3 2 1 • one unit-disk D ( v ) per node v s. t. graph G 1

Unit Disk Contact Representations 7 8 7 6 5 4 3 2 1 8 6 Unit disk contact representation (UDCR) of G = ( V , E ): 5 4 3 2 1 UDCR of G • one unit-disk D ( v ) per node v s. t. graph G 1 • { u , v } ∈ E ⇐ ⇒ D ( u ), D ( v ) touch .

Unit Disk Contact Representations 2 8 7 6 5 4 3 2 1 8 7 6 5 4 3 1 Unit disk contact representation (UDCR) of G = ( V , E ): 8 7 6 5 4 3 2 1 No embedding: neighbor order can be chosen arbitrarily UDCRs of G • one unit-disk D ( v ) per node v s. t. graph G 1 • { u , v } ∈ E ⇐ ⇒ D ( u ), D ( v ) touch .

Unit Disk Contact Representations 7 8 1 2 3 4 5 6 8 6 1 2 3 4 7 8 5 6 7 5 Unit disk contact representation (UDCR) of G = ( V , E ): 4 graph G • one unit-disk D ( v ) per node v s. t. UDCRs of G No embedding: neighbor order can be chosen arbitrarily 1 2 3 4 5 6 7 8 1 2 3 1 • { u , v } ∈ E ⇐ ⇒ D ( u ), D ( v ) touch .

Unit Disk Contact Representations 2 3 4 5 6 7 8 1 3 graph G 4 7 8 5 6 Weak unit disk contact representation (WUDCR) of G = ( V , E ): WUDCRs of G 2 1 8 7 • one unit-disk D ( v ) per node v s. t. No embedding: neighbor order can be chosen arbitrarily 1 2 3 4 5 6 7 8 1 2 3 4 5 6 1 • { u , v } ∈ E = ⇒ D ( u ), D ( v ) touch .

Results Existing Work: NP-hardness and construction algorithms. star caterpillar tree Weak UDCR Weak Emb. UDCR Chiu, Cleve, Nöllenburg; 2019 Cleve; 2020 Cleve; 2020 2

Results Existing Work: NP-hardness and construction algorithms. star caterpillar tree Weak UDCR Weak Emb. UDCR Chiu, Cleve, Nöllenburg; 2019 Cleve; 2020 Cleve; 2020 2

Recognizing Caterpillars in Linear Time • A caterpillar is a tree • backbone path plus leaves Reminder: 3

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G 3

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally Place leaf disks at leftmost position 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally Place leaf disks at leftmost position 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally Place leaf disks at leftmost position 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally Place leaf disks at leftmost position 3 Find longest path v 0 , v 1 , . . . , v k , v k +1

Recognizing Caterpillars in Linear Time Algorithm: Caterpillar G backbone Place backbone disks horizontally Place leaf disks at leftmost position 3 Find longest path v 0 , v 1 , . . . , v k , v k +1 �

NP-hardness of Recognizing Trees 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard.

NP-hardness of Recognizing Trees Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine .

NP-hardness of Recognizing Trees Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine .

NP-hardness of Recognizing Trees Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine .

NP-hardness of Recognizing Trees Proof: Each clause: {0, 0, 1} or {0, 1, 1} 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine .

NP-hardness of Recognizing Trees Proof: Each clause: {0, 0, 1} or {0, 1, 1} 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees Proof: Logic engine Each clause: {0, 0, 1} or {0, 1, 1} 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 2 • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine c 3 c 2 c 1 c 1 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 1 • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine c 3 c 2 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 1 • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine c 3 c 2 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 2 x 2 x 3 x 4 • no literal: both fmags • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine c 3 c 1 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 2 x 2 x 3 x 4 • no literal: both fmags • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine c 3 c 1 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees = 1 x 2 x 3 x 4 • no literal: both fmags • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine = 1 = 1 = 1 c 3 c 3 Proof: x 1 x 2 x 3 x 4 4 c 2 c 1 c 1 c 2 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 3 • no literal: both fmags • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine = 1 = 1 = 1 = 1 c 2 c 1 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

NP-hardness of Recognizing Trees c 3 • no literal: both fmags • negative literal: fmag on top • positive literal: fmag on bottom • variable poles & clause levels Each clause: {0, 0, 1} or {0, 1, 1} Logic engine = 1 = 1 = 1 = 0 c 2 c 1 c 1 c 2 c 3 x 4 x 3 x 2 x 1 Proof: 4 Theorem: Given a tree T , answering “ ∃ weak UDCR for T ? ” is NP-hard. Not-All-Equal-3SAT ≤ P Tree-WUDCR, via logic engine . Example: c 1 = ( x 1 , x 2 , x 3 ) c 2 = ( x 1 , x 2 , x 4 ) c 3 = ( x 1 , x 3 , x 4 )

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