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 Exponential variables Planar straight line Upward drawing Upper and Lower bounds

(a) (b) Digraphs G 1 and G n

Theorem 1 Given any resolution rule, a planar straight line upward drawing of digraph G n (with 2n + 2 vertices) has area  2 n )

Approach to Proving With A n as the minimum area of a planar straight line upward drawing G n We use induction to prove that A n >= 4. A n -2 Since A 1 >= c for some constant c depending on the resolution rule, this implies the claimed result.

Variable Instantiation  n  straight line drawing of G n with min area  n-1  straight line drawing of G n-1 by removing vertices and edges s n and t n  n-2 for G n-2 and so on..

Variable Instantiation – Part 2    Horizontal line through vertex s n-2    Horizontal line through vertices t n-2    Line extending edge (t n-3, t n-2 )    Angle formed by edge (t n-3, t n-2 ) and the x-axis    Angle fomed by edge (s n-2, s n-3 ) and the x-axis    Line paralle to  through vertex s n-2    Line extending edge (s n-2, s n-3 )

Observations Leading to Inference Area(P) >= 2 * Area(  n-2 ) >= 2 * A n-2 A n = Area(  n ) >= Area(P) + Area(    Area    A n = 2 * Area(P) >= 4 * Area(  n-2 ) = 4 * A n-2

Logic Engine

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.a j and a j * If x j = 1, a j = x j If x j = 0, a j * = x j For clauses c 1, c 2,... c m the chain links are numbered in outward order as 1,2,... m

Flags and Flag Collision 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 x j appears in the clause c i then link i of a j is unflagged If the literal x j * appears in the clause c i then the link i of a j * is unflagged The Flag

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

And.. The proof Assume “yes” instance of NAE3SAT Rotate armatures such that if truth assignment t(x j ) = 1, a j is on top and if t(x j ) = 0, a j 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

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 ?

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

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

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

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?

Thank You