Short Plane Supports for Spatial Hypergraphs Thom Castermans
- Slides: 42
Short Plane Supports for Spatial Hypergraphs Thom Castermans Mereke van Garderen Wouter Meulemans Martin Nöllenburg Xiaoru Yuan
A hypergraph H = (V, S) where S contains hyperedges which are subsets of V
Drawing a hypergraph Hyperedge as polygon containing its members (Gestalt theory)
Drawing a hypergraph Hyperedge as lines connecting its members (Tufte’s minimal ink)
Support graphs A support graph G = (V, E) of a hypergraph H = (V, S) G uses the same vertices as H Each hyperedge in S induces a connected subgraph in G
Criteria
Criteria Short total edge length
Criteria Short total edge length Planar
Criteria Short total edge length Planar Kelp-style rendering [Dinkla et al 2012, Meulemans et al 2013]
Criteria Short total edge length Planar
Criteria Short total edge length Planar Possibly a tree
Known results on plane supports Support |S| Tree 2 3+ Graph 2 3+ Existence Length Minimization
Known results on plane supports Support |S| Existence Tree 2 P [Bereg et al 2014] Length Minimization 3+ Graph 2 P [Bereg et al 2014] 3+ NP-hard [Buchin et al 2011]
Related work Nonplanar Existence of support trees in P Length Min for 2 hyperedges in P Length Min NP-hard for 3 hyperedges Combinatorial Existence is NP-hard for many hyperedges Variants Hamiltonian induced subgraphs Steiner setting of disjoint hyperedges Stricter planarity than supports [Klemz et al 2014] [Hurtado et al 2018] [Akitaya et al, 2016] [Buchin et al 2011] [Brandes et al 2010] [Bereg et al 2015] [Van Goethem et al 2018]
Results We study plane support trees (and graphs) Existence A simple sufficient condition Length minimization Why simple ideas do not lead to an approximation algorithm Computational complexity Integer linear program Two heuristic approaches Experiments
Results Support |S| Existence Length Minimization Tree 2 P [Bereg et al 2014] NP-hard yes NP-hard 3+ Sufficient condition NP-hard 2 P [Bereg et al 2014] NP-hard yes NP-hard [Buchin et al 2011] Graph 3+
A sufficient condition
A sufficient condition
Improving the length
Can we use this?
Can we use this?
Can we use this?
Can we use this?
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. a or b or d b or c or d a b c not a or not c or not d d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. a or b or d b or c or d b c not a or not c or not d d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. a or b or d b or c or d b c not a or not c or not d d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. a or b or d b or c or d b c not a or not c or not d d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. a or b or d b or c or d not a or not c or not d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT. b or c or d not a or not c or not d
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT.
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT.
The computational problem Given H = (V, S) and L > 0, is there a plane support tree with total length at most L? This is NP-hard, via a reduction from planar monotone 3 -SAT.
Heuristics • • MST Iteration Local. Search (hill climbing)
Heuristics • • MST Iteration Local. Search (hill climbing)
Heuristics • • MST Iteration Local. Search (hill climbing)
Generating random hypergrahs Number of vertices: Number of hyperedges: Degree distribution: n k d EVEN HIGH LOW MID
Experiments: heuristics comparisons n = 20, 40, 60, 80, 100 k = 2, 3, 4, 5, 6, 7 d = EVEN, LOW, MID, HIGH 1000 random hypergraphs MST Approximation ; MST Iteration ; Local. Search U / T / PT 1. 2. MST Iteration better than MST Approximation for higher k. MID and EVEN benefit more from iteration than LOW and HIGH. 3. 4. Local. Search is on average 12% shorter than MST Iteration. Requiring planarity affects LOW and MID. 5. 6. MST Iteration is on average 95. 11% faster than Local. Search U. Requiring planarity makes Local. Search on average 272. 64% slower, 354. 06% for n = 100.
Experiments: optimality comparisons n = 10, 15, 20 k = 2, 3 d = LOW, MID 1000 random hypergraphs Local. Search U / T / PT ; OPT U / T / PT 1. 2. Local. Search T is always optimal. Local. Search is close to optimal: ratio less than 1. 61 in all cases, ratio less than 1. 2 in 99% of the cases.
Theoretical results We study plane support trees (and graphs). Existence A vertex that is in all hyperedges ensures a plane support tree. Length minimization EMST on common vertices is not always part of an o(n)-approximation. Deciding whether a plane support exists is NP-hard. n Two hyperedges, one containing the other n Integer linear program
Experimental results Heuristics: MST Iteration and Local. Search • MST Iteration is fast • Local. Search is close to optimal • Having a nonplanar tree is always optimal
Future work Can we efficiently decide whether a plane support tree exists? (We know this only for k = 2. ) How many iterations are needed for MST Iteration with k > 2 before the solution stabilizes? What is the effect of initializing Local. Search with MST Iteration? Do other search techniques (e. g. simulated annealing) work better? Explore real-world data.
Short Plane Supports for Spatial Hypergraphs Thom Castermans Mereke van Garderen Wouter Meulemans Martin Nöllenburg Xiaoru Yuan
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