1 / 22

Proving Lower Bounds on Graph Drawing Problems

Proving Lower Bounds on Graph Drawing Problems. Rajat Anantharam Department of Gaming and Media Technology Utrecht University. Objective. Technique for proving exponential area lower bounds The Logic Engine Description Illustration Simulation Extension. Tune ups. Planar acyclic graphs

wayde
Download Presentation

Proving Lower Bounds on Graph Drawing Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Proving Lower Bounds on Graph Drawing Problems Rajat Anantharam Department of Gaming and Media Technology Utrecht University

  2. Objective Technique for proving exponential area lower bounds The Logic Engine Description Illustration Simulation Extension

  3. Tune ups Planar acyclic graphs Exponential variables Planar straight line Upward drawing Upper and Lower bounds

  4. Digraphs G1 and Gn (a) (b)

  5. Theorem 1 Given any resolution rule, a planar straight line upward drawing of digraph Gn (with 2n + 2 vertices) has area p(2n)

  6. Approach to Proving With An as the minimum area of a planar straight line upward drawing Gn We use induction to prove that An >= 4. An -2 Since A1 >= c for some constant c depending on the resolution rule, this implies the claimed result.

  7. Variable Instantiation Gn straight line drawing of Gn with min area Gn-1 straight line drawing of Gn-1 by removing vertices and edges sn and tn Gn-2 for Gn-2 and so on ..

  8. Variable Instantiation – Part 2 s  Horizontal line through vertex sn-2 t  Horizontal line through vertices tn-2 r1  Line extending edge (tn-3, tn-2) q1  Angle formed by edge (tn-3, tn-2) and the x-axis q2  Angle fomed by edge (sn-2, sn-3) and the x-axis l1  Line paralle to r1 through vertex sn-2 l2  Line extending edge (sn-2, sn-3)

  9. Observations Leading to Inference Area(P) >= 2 * Area(Gn-2) >= 2 * An-2 An = Area(Gn) >= Area(P) + Area(D1) + Area(D2) An = 2 * Area(P) >= 4 * Area(Gn-2) = 4 * An-2

  10. Logic Engine

  11. Links in a Chain Each armature is connected to the shaft by a chain When the engine lies flat then the chain can be in two possible positions – viz.aj and aj* If xj = 1, aj = xj If xj = 0, aj* = xj For clauses c1, c2, ... cm the chain links are numbered in outward order as 1,2, ... m

  12. Flags and Flag Collision The Flag Two flags attached along the same row across two armatures collide with each other if they point towards each other Flag connected to the outermost armature An collides with the frame if it points towards it If literal xj appears in the clause ci then link i of aj is unflagged If the literal xj * appears in the clause ci then the link i of aj * is unflagged

  13. Theorem An instance of NAE3SAT is a “yes” instance if and only if the corresponding logic engine has a flat collission free configuration.

  14. And .. The proof Assume “yes” instance of NAE3SAT Rotate armatures such that if truth assignment t(xj) = 1, aj is on top and if t(xj) = 0, aj is in bottom Since clause ci contains atleast one literal y with t(y) = 1 and atleast one literal z with t(z) = 0 there’s atleast one unflagged link in each row. So we allign the chain such that the flaggs from the remaining link point towards the unflagged link leading to “no collission” A similar analogy can be drawn from a flat collission free configuration thereby implying the validity of the theorem

  15. Logic engine and a graph Drawing Problem To prove that the following problem is NP-Hard Theorem : “Unit length planar straight line drawing of a graph is NP-Hard” The question: - Is there a straight line planar drawing of G such that every edge is of length one ?

  16. Prelude How to construct the unversal part of a logic graph? A graph G is uniquely drawable if all the unit length planar drawings f G can be obtained through translation, rotation, scaling, mirroring

  17. Construction of the Logic Engine Each unit length of the constructed Logic Engine corresponds to the link graph Armature and Outer frame is necessarily unique The max Euclidean distance b/w the extremal endpoints to the shaft gives the number of edges in the shaft making its construction unique as well

  18. Hence the Proof The logic graph corresponding to an instance of NAE3SAT with n variables and m clauses has O((m+n)2) vertices and edges and the time taken to construct the logical graph is linear in the size of the graph

  19. Extending Usage of Logic Graphs Is there a grid drawing of a tree T, such that each edge has length one ? Is there a grid drawing of tree T of area at most K ? – where K is an integer. Is there a drawing of tree T such that T is a minimum spanning tree of the vertex locations? Is there a drawing of graph G such that G is the mutual nearest neighbour graph of the vertex location?

  20. Thank You

More Related