Applying the weighted barycentre method to interactive graph visualization

Transcription

1 Applying the weighted barycentre method to interactive graph visualization Peter Eades University of Sydney Thanks for some software: Hooman Reisi Dekhordi Patrick Eades

2 Graphs and Graph Drawings

3 What is a graph?

4 A graph consists of Nodes, sometimes called vertices, and Binary relationships, called edges between the nodes. Example: a Linked-In style social network Nodes: Alice, Andrea, Annie, Amelia, Bob, Brian, Bernard, Boyle Edges: Bob is connected to Alice Bob is connected to Andrea Bob is connected to Amelia Brian is connected to Alice Brian is connected to Andrea Brian is connected to Amelia Boyle is connected to Alice Boyle is connected to Andrea Boyle is connected to Annie Bernard is connected to Alice Bernard is connected to Andrea Bernard is connected to Annie

5 Notation: The degree deg (u) of a node u is the number of edges attached to u. Example: deg Brian = 3; deg (Alice) = 4. Nodes: Alice, Andrea, Annie, Amelia, Bob, Brian, Bernard, Boyle Edges Bob is connected to Alice Bob is connected to Andrea Bob is connected to Amelia Brian is connected to Alice Brian is connected to Andrea Brian is connected to Amelia Boyle is connected to Alice Boyle is connected to Andrea Boyle is connected to Annie Bernard is connected to Alice Bernard is connected to Andrea Bernard is connected to Annie

6 Notation: The neighbour set N (u) of a node u is the set of nodes that are linked to u by an edge. Example: N Brian = {Alice, Andrea, Amelia}; N Alice = {Bob, Brian, Boyle, Bernard}. Nodes: Alice, Andrea, Annie, Amelia, Bob, Brian, Bernard, Boyle Edges Bob is connected to Alice Bob is connected to Andrea Bob is connected to Amelia Brian is connected to Alice Brian is connected to Andrea Brian is connected to Amelia Boyle is connected to Alice Boyle is connected to Andrea Boyle is connected to Annie Bernard is connected to Alice Bernard is connected to Andrea Bernard is connected to Annie

7 Note: Edges can be weighted. Example: an edge-weighted Linked-In style social network Nodes: Alice, Andrea, Annie, Amelia, Bob, Brian, Bernard, Boyle Edges w(bob Alice) = 1 w(bob -- Andrea) = 5 w(bob -- Amelia) = 1 w(brian -- Alice) = 2 w(brian -- Andrea) = 100 w(brian -- Amelia) = 3 w(boyle -- Alice) = 1 w(boyle -- Andrea) = 1 w(boyle -- Annie) = 5 w(bernard -- Alice) = 1 w(bernard -- Andrea) = 9 w(bernard -- Annie) = 1

8 What is a graph drawing?

9 Graph Drawing Graph A graph consists of Nodes, and Edges A graph drawing is a picture of a graph. That is, a graph drawing is a mapping that assigns a location for each node, and a curve to each edge.

10 A drawing of the social network A social network Nodes: Bob, Brian, Bernard, Boyle, Alice, Andrea, Annie, Amelia Edges Bob is connected to Alice Boyle is connected to Alice Bob is connected to Andrea Boyle is connected to Andrea Bob is connected to Amelia Boyle is connected to Annie Brian is connected to Alice Bernard is connected to Alice Brian is connected to Andrea Bernard is connected to Andrea Brian is connected to Amelia Bernard is connected to Annie Alice Brian Amelia Bob Bernard Annie Boyle Andrea

11 Nodes 0, 1, 2, 3, 4, 5, 6, 7 Edges A drawing of the graph A graph

12 A graph drawing is a straight-line drawing if every edge is a straight line segment. A straight-line drawing Alice Brian Amelia Bob Bernard Annie Boyle Andrea NOT a straight-line drawing

13 What is connectivity?

14 Connectivity notions are fundamental in any study of graphs or networks A graph is connected if for every pair u, v of vertices, there is a path between u and v. A graph is k-connected if there is no set of (k-1) vertices whose deletion disconnects the graph. k = 1: 1-connected connected k = 2: 2-connected biconnected k = 3: 3-connected triconnected

15 This graph is connected Connected components This graph is not connected

16 Connectivity notions are fundamental in any study of networks A graph is connected if for every pair u, v of vertices, there is a path between u and v. A graph is k-connected if there is no set of (k-1) vertices whose deletion disconnects the graph. k = 1: 1-connected connected k = 2: 2-connected biconnected k = 3: 3-connected triconnected

17 2-connected biconnected A cutvertex is a vertex whose removal would disconnect the graph. A graph without cutvertices is biconnected. This graph is biconnected This graph is not biconnected

18 3-connected triconnected A separation pair is a pair of vertices whose removal would disconnect the graph. A graph without separation pairs is triconnected. This graph is triconnected This graph is not triconnected

19 What is a planar graph?

20 A graph is planar if it can be drawn without edge crossings.

21 A graph is planar if it can be drawn without edge crossings. Nodes: 0,1,2,3,4,5,6,7,8,9 Edges

22 Non-planar Nodes: 0,1,2,3,4,5 Edges A graph is non-planar if every drawing has at least one edge crossing.

23 What is a topological embedding?

24 A planar drawing divides the plane into faces. F 0 F 4 F 1 F 3 F 2 F 0 shares a boundary with F 1 F 0 shares a boundary with F 2 F 0 shares a boundary with F 3 F 0 shares a boundary with F 4 F 1 shares a boundary with F 2 F 1 shares a boundary with F 4 F 2 shares a boundary with F 1 F 2 shares a boundary with F 3 The boundary-sharing relationships of the faces defines a topological embedding of the graph drawing F 2 shares a boundary with F 4 F 3 shares a boundary with F 4

25 What is a good graph drawing?

26 Graph Drawing Graph A - B, C, D B - A, C, D C - A, B, D, E D - A, B, C, E E - C, D The input is a graph with no geometry C The output should be a good graph drawing: A B E easy to understand, easy to remember, D beautiful.

27 ~1979 Intuition (Sugiyama et al. 1979; Batini et al 1982; etc ): Planar straight-line drawings make good pictures Bad drawing Good drawing Alice Dunstan Joe Paul Albert Jocelyn Murphy Bob Albert David Paul John David Murphy Jocelyn Joe Dunstan Alice John Bob

28 Purchase et al.,1997: Significant correlation between edge crossings and human understanding More edge crossings means more human errors in understanding

29 Purchase et al., 1997: Significant correlation between straightness of edges and human understanding More bends mean more human errors in understanding

30 What makes a good drawing of a graph? lack of edge crossings straightness of edges (plus some other things) Planar straight-line drawings of graphs are GOOD! (This talk is about planar straight-line drawings of graphs!)

31 Tutte s Barycentre Algorithm

32 W. T. Tutte Code breaker at Bletchley park Pioneer of graph theory William T. Tutte. How to draw a graph. Proc. London Math. Society, 13(52): , 1963.

33 Tutte s barycentre algorithm is the original graph drawing algorithm

34 Run VinciTest1

35 Tutte s barycentre algorithm Input: A planar graph G = (V, E) Output: A straight-line drawing p of G Step 1. Choose a face f of G. Step 2. Draw f as a convex polygon. Step 3. For all u not in f, choose p(u) by p u = 1 p(v) deg (u) v N(u) where the sum is over all neighbors v of u. Vertex u is placed at the barycenter of its neighbors

36 Example output

37 Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3. For all u not on f, choose p(u) by p u = 1 p(v) deg (u) v N(u) where the sum is over all neighbors v of u x u = 1 deg (u) and y u = 1 deg (u) v N(u) v N(u) x(v) y(v) 2 V f = O( V ) linear equations barycentre equations

38 Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3. For all u not on f, choose p u = x u, y u by x u = 1 deg (u) v N(u) x(v) and y u = 1 8 y(v) deg (u) v N(u) where the sum is over all neighbors v of u The critical part is step 3.

39 Example Step 1. f = {4, 5, 6, 7, 8} Step 2. For all i = 4, 5, 6, 7, 8, choose the location (x i, y i ) of i so that they form a convex polygon. Step 3. Find x 1, x 2, x 3 such that: x 1 = 1 4 (x 2 + x 3 + x 4 + x 8 ) x 2 = 1 4 (x 1 + x 3 + x 5 + x 6 ) x 3 = 1 3 (x 1 + x 2 + x 7 ) 8 7 and find y 1, y 2, y 3 such that: y 1 = 1 4 (y 2 + y 3 + y 4 + y 8 ) y 2 = 1 4 (y 1 + y 3 + y 5 + y 6 ) y 3 = 1 3 (y 1 + y 2 + y 7 ) where x i and y i are the values chosen at step 2.

40 5 Re-write the equations for step 3: and 4x 1 x 2 x 3 = x 4 + x 8 x 1 + 4x 2 x 3 = x 5 + x 6 x 1 x 2 + 3x 3 = x 5 4y 1 y 2 y 3 = y 4 + y 8 y 1 + 4y 2 y 3 = y 5 + y 6 y 1 y 2 + 3y 3 = y That is: 4x 1 x 2 x 3 = c 1 x 1 + 4x 2 x 3 = c 2 x 1 x 2 + 3x 3 = c 3 and 4y 1 y 2 y 3 = d 1 y 1 + 4y 2 y 3 = d 2 y 1 y 2 + 3y 3 = d 3 where c 1, c 2, c 3, d 1, d 2, d 3 are constants

41 Re-writing the equations using a matrix: x 1 y 1 We want vectors x = x 2 and y = y 2 x 3 y 3 such that Mx = c and My = d where 8 7 M = The essence of Tutte s barycentre algorithm is inverting a matrix

42 Tutte s barycentre algorithm Input: A triconnected planar graph G = (V, E) Output: A planar straight-line drawing p of G Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3. (a) Let M be the square symmetric matrix indexed by the vertices of G f defined by M uv = deg (u) if u = v 1 if u, v E 0 otherwise The matrix of the barycentre equations is a Laplacian submatrix. (b) Let c and d be vectors indexed by the vertices of G f defined by c u = w f x w and d u = w f y w (c) Let x and y be vectors indexed by the vertices of G f defined by x = M 1 c and y = M 1 d (d) Let p(u) = (x u, y u ) for all u not on f.

43 The subset A of V is called the set of fixed nodes. A more general barycentre algorithm Step 1. Choose a subset A of V Step 2. Choose a location p(a) = (x(a), y(a)) for each vertex a A Step 3. Solve the barycentre equations to give a position p u = x u, y u for each vertex u V A

44 A few lemmas and theorems about the barycentre equations

45 Definition: If A V then the A-contraction G A of G is the graph obtained from G by identifying all vertices in A (ie, contracting A to a single vertex) G 1 G A Determinant Lemma: If G A is connected then the matrix M of the barycentre equations has a positive determinant. Proof: Follows from Kirchhoff's spanning tree theorem. (actually, even more is true: The barycentre matrix is positive definite)

46 A corollary: Uniqueness Lemma: If G A is connected then the barycentre equations have a unique solution.

47 Tutte s amazing theorem Theorem (Tutte 1960/63) If the input graph is planar and triconnected, then the drawing output by the barycentre algorithm is planar, and every face is convex.

48 Implementation The main part of the barycentre algorithm inverting the nxn barycentre matrix M. The usual Gaussian elimination: O(n 3 ) Williams algorithm: O(n ) If the graph is planar, then the matrix M is sparse, then we can use Lipton-Tarjan nested dissection: O(n 1. 5 ) In practice we can use numerical methods (such as Gauss- Seidel / Jacobi iterations); these converge rapidly since M is diagonally dominant.

49 Example (planar graph, 100 vertices)

50 Example (planar graph, 100 vertices)

51 Example (planar graph, 200 vertices)

52 Example (planar graph, 1000 vertices)

53 The Weighted Barycentre Algorithm

54 Weighted barycentre algorithm Input: A positive-edge-weighted triconnected planar graph G = (V, E, w) Output: A straight-line drawing p of G Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3. For all u not on f, choose p(u) by 1 p u = w wdeg (u) uv p(v) v N(u) where wdeg u = w uv v N(u) Vertex u is placed at the weighted barycenter of its neighbors

55 Weighted barycentre algorithm Input: A weighted triconnected planar graph G = (V, E, w) Output: A planar straight-line drawing p of G Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3. (a) Let M be the square symmetric matrix indexed by the vertices of G f defined by M uv = v N(u) v N(u) w uv if u = v w uv p(v) if u, v E 0 otherwise The matrix is diagonally dominant and has positive entries. (c) Let x and y be vectors indexed by the vertices of G f defined by x = M 1 c and y = M 1 d (d) Let p(u) = (x u, y u ) for all u not on f.

56 A few lemmas and theorems about the weighted barycentre equations

57 Determinant Lemma: If G A is connected then the matrix M of the barycentre equations has a positive determinant. Uniqueness Lemma: If G A is connected then the barycentre equations have a unique solution.

58 Corollary to Tutte s amazing theorem (Floater 1997, maybe earlier) If the input graph is planar and triconnected, then the drawing output by the weighted barycentre algorithm is planar, and every face is convex.

59 The Energy View

60 Tutte s barycentre algorithm is force-directed

61 Tutte s barycenter algorithm: The energy view 1. Choose a set A of vertices. 2. Choose a location p(a) for each a A 3. Place all the other vertices to minimize energy What is the energy of a drawing p? 8 7 The Euclidean distance between u and v in the drawing p is: d u, v = x u x v +(y u y v ) 2 The energy in the edge (u, v) is d u, v 2 = (x u x v ) 2 +(y u y v ) 2 The energy in the drawing p is the sum of the energy in its edges, ie, energy p = d(u, v) 2 (u,v) E = (x u x v ) 2 +(y u y v ) 2 (u,v) E

62 Tutte s barycenter algorithm: The energy view 5 We represent each vertex by a steel ring, and represent each edge by a spring of natural length zero connecting the rings at its endpoints. 1. Choose a set A of vertices For each a A, nail the ring representing a to the floor at some position The vertices in V A will move around a bit; when the movement stops, take a photo of the layout; this is the drawing. Note: for Tutte s barycentre algorihm: The springs have zero natural length Similar to elastic bands

63 How to minimize energy: Choose a location p u = x u, y u for each u V A to minimize energy p = d(u, v) 2 (u,v) E = (x u x v ) 2 +(y u y v ) 2 (u,v) E Note that the minimum is unique, and occurs when (energy p ) = 0 and x u for each u V A. (energy p ) y u = 0

64 For x u, the minimum occurs when: (energy p ) x u = 0 x u (x u x v ) 2 +(y u y v ) 2 (u,v) E = 0 v V s.t.(u,v) E 2(x u x v ) = 0 Barycentre equations x u = 1 deg (u) v V s.t.(u,v) E x v

65 Tutte s barycenter algorithm: The energy view 4 5 Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3: Place all the other vertices to minimize energy The weighted barycentre algorithm is a force directed algorithm Edges are Hooke s law springs with zero natural length. The weights define strengths for the springs. The fixed set are nodes nailed down (i.e., they don t move).

66 Tutte s barycenter algorithm: The energy view 4 5 Step 1. Choose a face f of G Step 2. Draw f as a convex polygon Step 3: Place all the other vertices to minimize energy The biggest differences between the weighted barycentre algorithm and other force-directed methods are: For the barycentre algorithm, the springs have zero natural length. For the barycentre algorithm, the equations are easy to solve.

67 Animation

68 The animation problem: Given two drawings of a graph G, find a nice animation that takes one to the other. animate Problem: how to compute the frames in between Key frame 0 Key frame 1

69 Run VinciTest2

70 The animation problem: Given two planar drawings p 0 and p 1 of a graph G: Find a nice animation that takes p 0 to p 1. Here we represent the drawing p 0 as a 2Xn vector with a row x 0 (u), y 0 (u) for each vertex u; similarly for p 1. In between frames p t for 0 < t < 1 p 0 p 1 The animation consists of a drawing p t for each t with 0 t 1. We want two properties of the animation: 1. Planarity: The drawing p t is planar for 0 t 1, 2. Smoothness: The animation is smooth, that is, the animation function a: t p t is continuous and differentiable.

71 The animation problem: Given two planar drawings p 0 and p 1 of a graph G: Find an animation a: t p t with two properties: Planarity Smoothness Solution (Floater and Gotsman, 1999, maybe others): If p 0 and p 1 are weighted barycentre drawings, then the animation problem is easy: just interpolate the weights.

72 Say drawings p 0 and p 1 are barycentre drawings with barycentre matrices M 0 and M 1, that is, M 0 p 0 = c and M 1 p 1 = c for a constant 2Xn vector c. Let M t = 1 t M 0 + tm 1. Theorem: Suppose that p t is a solution to the barycentre equation with matrix M t, that is, M t p t = c. Then the animation function a: t p t is planar and smooth. Proof It is planar by Tutte s amazing Theorem. It is smooth by the Determinant Lemma.

73 The Bi-Weighted Barycentre Algorithm

74 Run VinciTest3

75 The bi-weighted barycentre algorithm with weak parameter δ Suppose that G = (V, E) is a triconnected planar graph and f is a face of G. Suppose that G = (V, E ) is a subgraph of G. The f-contractions G f and G f are connected. δ > 0 is small. We define weights w uv = 1 if u, v E δ otherwise. Edges in E are strong springs Edges in E E are weak springs. The strong graph G = (V, E ) is the subgraph with all edge weights 1, plus the vertices and edges of f. The weak graph G = (V, E ) is the subgraph with all edge weights δ, plus the vertices and edges of f.

76 The strong graph theorem and The weak graph theorem

77 Strong Graph Theorem Suppose that the G f is connected, and the strong graph is G. Suppose that G f is connected, p G is the bi-weighted barycentre drawing of G restricted to the vertices of G, with weak parameter δ, and p is the barycentre drawing of G. Then for every ε > 0 there is a δ > 0 such that p G p < ε. Informally: If the strong edges are connected to the fixed set f, then the barycentre drawing of the whole graph is arbitrarily close to the barycentre drawing of the strong graph.

78 Strong Graph Theorem Suppose that the G f is connected, and the strong graph is G. Suppose that G f is connected, p G is the bi-weighted barycentre drawing of G restricted to the vertices of G, with weak parameter δ, and p is the barycentre drawing of G. Then for every ε > 0 there is a δ > 0 such that p G p < ε. The drawing of the whole graph is similar to the drawing of the strong graph.

79 Strong Graph Theorem Suppose that the G f is connected, and the strong graph is G. Suppose that G f is connected, p G is the bi-weighted barycentre drawing of G restricted to the vertices of G, with weak parameter δ, and p is the barycentre drawing of G. Then for every ε > 0 there is a δ > 0 such that p G p < ε. Proof. Use two results: 1. The Hoffman principle : if Mx b is small then x is close to a solution of Mx = b 2. The uniqueness lemma.

80 Weak Graph Theorem Suppose that the G f f is connected, and the weak graph is G. Suppose that the G strong f is connected, graph G pis connected G is the bi-weighted but has barycentre no vertex in f. Let drawing p G be the of G barycentre restricted drawing to the vertices of the graph of G, formed with weak from parameter G by δ, and collapsing p is the all barycentre of G to a single drawing node. of G. Then Suppose for every that G ε f > is 0 connected, there is a δ p> G 0 such is the bi-weighted that p G p barycentre < ε. drawing of G restricted to the vertices of G, with weak parameter δ. Then for every ε > 0 there is a δ > 0 such that p G p G < ε. Informally: If the strong graph is connected but not connected to the fixed set f, then the barycentre drawing of the whole graph is arbitrarily close to the barycentre drawing where the strong graph is collapsed to a single node.

81 The drawing of the whole graph is similar to the drawing of the weak graph.

82 Clustered Planar Graphs

83 Clustered graphs

84 Run VinciTest3 (again)

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87 Semantic Zoom

88 Focus + context display: Focus: we want to zoom in to the neighbourhood of u, to see it in detail. Context: we can reduce detail for parts of G that are further from u. This can be achieved via geometric or semantic zooming. Geometric zoom: neighbourhood and distance are defined geometrically Semantic zoom: neighbourhood and distance are defined graph-theoretically Semantic zoom is better!

89 Geometric zoom

90 VinciTest4

91 Weighted barycentre for semantic zoom for focus vertex r: For each vertex u, let d(u) = graph theoretic distance between r and u. For each edge (u, v), w uv = k d(u) + kd v That is, edges become weaker as they get close to the focus node.

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93 Finishing Up

94 The weighted barycentre algorithm can do many things Animation Clustered graph drawing Semantic zoom But there is more

95 there s more. Emphasize a specified spanning tree Emphasize a specified path (run VinciTest10) Adjust layout after adding/deleting and edge/vertex

96 The weighted barycentre algorithm can do many things Animation Clustered graph drawing Semantic zoom Emphasize a specified spanning tree Emphasize a specified path Adjust layout after adding/deleting and edge/vertex But some things are still a problem

97 Consider this graph G k with k + 2 nodes In a barycentre drawing of graph G k, nodes u k 1 and u k are exponentially close together (distance less than 1 2 k) u 1 u 2 u 3 u k 1 u k

98 Run VinciTest7

99 Important note: For practical graph drawing, vertex resolution is very important, because nontrivial labels need to be attached to nodes.

100 u 1 But by weighting the edges, we can get a better drawing: u 2 w u i, u i+1 = 2 k i u 3 u k 1 u k

101 Run VinciTest8, then 9

102 Open problem Given a triconnected planar graph G = (V, E) Find weights on the edges so that the vertex resolution is good. That is: Given a graph G = (V, E), find a diagonally dominant matrix M such that and M uv > 0 if u v and (u, v) E M uv = 0 if u v and (u, v) E min u, v V p u p v is maximal.

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