Approximation Algorithms for Contact Representations of Rectangles - PowerPoint PPT Presentation

Our Results – Approximation Factors Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt, Wolff – LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Our Results – Approximation Factors Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt, Wolff – LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Our Results – Approximation Factors Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt, Wolff – LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Our Results – Approximation Factors Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt, Wolff – LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Tool #1: Gap

Tool #1: Gap Knapsack items 2 e 3 e 4 e 2 e 1 e 1 e – size s i – value v i

Tool #1: Gap Knapsack items bin 2 e 3 e 4 e 2 e 1 e 1 e – size s i – bin has capacity c – value v i

Tool #1: Gap Knapsack items bin 2 e 3 e 3 e 4 e 2 e 2 e 4 e 1 e 1 e – size s i – bin has capacity c – value v i

Tool #1: Gap Knapsack items bin 2 e 3 e 3 e 4 e 2 e 2 e 4 e 1 e 1 e – size s i – bin has capacity c – value v i – maximize total value packed

Tool #1: Gap Knapsack Generalized Assignment Prob. items bin 2 e 3 e 3 e 4 e 2 e 2 e 4 e 1 e 1 e – size s i – bin has capacity c – value v i – maximize total value packed

Tool #1: Gap Knapsack Generalized Assignment Prob. items items bin bin 1 2 2 e 4 e 3 e 3 e 3 e 3 e 4 e 2 e 4 e 2 e 2 e 2 e 4 e 1 e 4 e 1 e 2 e 4 e – size s i – bin has capacity c j j j – value v i – maximize total value packed j

Tool #1: Gap Knapsack Generalized Assignment Prob. items items bin bin 1 2 2 e 4 e 3 e 3 e 3 e 3 e 4 e 2 e 4 e 2 e 2 e 2 e 4 e 1 e 4 e 1 e 2 e 4 e – size s i – bin has capacity c j j j – value v i – maximize total value packed j Theorem. Gap admits an approximation algorithm with ratio α = e / ( e − 1) ≈ 1.58. [Fleischer et al., MOR’11]

Max-Crown for Stars u 1

Max-Crown for Stars u 1 B 1

Max-Crown for Stars u 4 u 2 u 1 u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 u 2 u 1 u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 u 1 u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) u 1 u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i B 3 B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: B 1 B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. Each contact may be horizontal or vertical. B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. Each contact may be horizontal or vertical. Try all 2 4 possibilities B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. Each contact may be horizontal or vertical. Try all 2 4 possibilities by calling α -approx. for Gap . B 5 B 2

Max-Crown for Stars u 4 Set up Gap : u 2 – eight bins (for the 4 sides and the 4 corners of B 1 ) – corner bins have capacity 1/2 u 1 – the capacity of side bins is their “free” length u 3 u 5 – items 2, . . . , n ; one for each leaf – the value of item i is p ( v 1 v i ), the profit of edge v 1 v i – item i has size 1/2 in corner bins, B 3 w i in top/bottom side bins, h i in left/right side bins B 4 Algorithm: Assume that the 4 corner rectangles have contacts B 1 of length 1 2 in a fixed optimal solution. Each contact may be horizontal or vertical. Try all 2 4 possibilities by calling α -approx. for Gap . B 5 B 2 ⇒ α -approx. algorithm for Max-Crown on stars �

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 � star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 � star α 1 + ε tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 α 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma.

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i ,

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G .

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Analysis.

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Analysis.

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis.

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 ,

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits,

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg.

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i .

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 .

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 .

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥ ALG 1 ≥

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥ ALG 1 ≥ OPT 1 ≥ α 1

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥ ALG 1 ≥ OPT 1 ≥ OPT 1 + OPT 2 ≥ α 1 α 1 + α 2

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥ ALG 1 ≥ OPT 1 ≥ OPT 1 + OPT 2 OPT ≥ . � α 1 α 1 + α 2 α 1 + α 2

Tool #2: The Combination Lemma Let G 1 = ( V , E 1 ), G 2 = ( V , E 2 ), G = ( V , E 1 ∪ E 2 ). Lemma. If Max-Crown admits an α i -approx. on G i , then Max-Crown admits ( α 1 + α 2 )-approx. on G . Algorithm. Proof. Apply α 1 -approx. to G 1 and α 2 -approx. to G 2 . Return result with larger profit for G . Analysis. For G , G 1 , G 2 , – let OPT, OPT 1 , OPT 2 be the optimum profits, – let ALG, ALG 1 , ALG 2 be the profits of the approx. alg. By def., ALG i > OPT i /α i . Clearly, OPT ≤ OPT 1 + OPT 2 . Assume OPT 1 /α 1 ≥ OPT 2 /α 2 . Then ALG ≥ ALG 1 ≥ OPT 1 ≥ OPT 1 + OPT 2 OPT ≥ . � α 1 α 1 + α 2 α 1 + α 2

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs. Proof.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs. Proof. Can cover any tree by 2 star forests (“star arboricity 2” ).

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs. Proof. Can cover any tree by 2 star forests (“star arboricity 2” ).

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs. Proof. Can cover any tree by 2 star forests (“star arboricity 2” ). Now apply the combination lemma.

Star Forests, Trees, (Outer-) Planar Graphs A star forest is the disjoint union of a set of stars. Def. Thm. Max - Crown admits an α -approx. on star forests. Proof. Use the α -approx. alg. for stars. Treat each star indep. Thm. Max - Crown admits – a 2 α -approx. on trees, – a 3 α -approx. on outerplanar graphs, – a 5 α -approx. on planar graphs. Proof. Can cover any tree by 2 star forests (“star arboricity 2” ). Now apply the combination lemma. Outerplanar | planar graphs have star arboricity 3 | 5. [Hakimi et al., DM’96] �

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε � tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε outerplanar 3 α 3 α 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε � � tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε 3 α � � outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Overview Weighted Unweighted new ◦ new ◦ Graph class old ⋆ cycle, path 1 star α 1 + ε � � tree 2 α , NP-hard 2 + ε 2 max-degree ∆ ⌊ ( ∆ + 1) / 2 ⌋ planar max-deg. ∆ 1 + ε 3 α � � outerplanar 3 + ε planar 5 α 5 + ε bipartite 16 α/ 3 ≈ 8.4 APX-hard general rand.: 32 α/ 3 ≈ 16.9 5 + 16 α/ 3 ≈ 13.4 det.: 40 α/ 3 ≈ 21.1 ⋆ ) [Barth, Fabrikant, Kobourov, Lubiw, N¨ ollenburg, Okamoto, Pupyrev, Squarcella, Ueckerdt & Wolff, LATIN’14] ◦ ) [Bekos, van Dijk, Fink, Kindermann, Kobourov, Pupyrev, Spoerhase, Wolff – submitted] α = e / ( e − 1) ≈ 1.58

Tool #1 ++ : PTAS for Gap with O (1) bins Theorem. Gap with O (1) bins admits a PTAS. [Briest, Krysta V¨ ocking: SIAM J. Comput.’11]