Flow and Upward Planarity

1. Flow and Upward Planarity Graph Drawing, chapter 6 by: Ami Hauptman Masha Igra

2. Introduction • Upward planar: inclusion in a Planar st-Graph • Angles in Upward Drawing • Upward Planarity Testing of Embedded Digraphs • Optimal Upward Planarity Testing of Single-Source Embedded Digraphs

3. Introduction • A digraph is upwardplanar if it admits a planar and upward drawing. • upward – edges are monotonically increasing in the vertical direction • planar – no two edges intersect

4. Introduction • An embedded graphdescribed by circular order of the neighbors of each vertex.

5. Inclusion in a Planar st-Graph • A planar st-graph is an st-graph that is planar and embedded with vertices s and t on the boundary of the external face. • St-graph is an acyclic graph with a single source s and single sink t t S

6. Theorem 6.1 Let G be a digraph. The following statements are equivalent: • G is upward planar • G admits an upward planar straight-line drawing • G is the spanning subgraph of a planar st-graph

7. 1(upward planar)3(spanning subgraph of a planar st-graph) • s,t respc. lowest and highest y coord. • At sink v ≠ t draw new edge upward • Add new edges (v, dest(e)) • Able to cancel all sinks, except t • Special cases • Similar process cancels all sources, except s • Statement shown.

8. 3 (G is spanning subgraph of a planar st-graph G’)2 (G admits an upward planar straight-line drawing) 3 steps: • Add edges to G′. Resulting digraph G′′ is planar st-graph with all faces consisting of three edges • Construct an upward planar straight-line drawing of G′′ • Remove edges that don’t belong to G from drawing of G′′

9. Lemma 6.1: P = or , P =, k >= 4. Edge or can be added, such a resulting digraph is a planar st-graph Lemma 6.2: and has at least 3 vertices. Edge can be added, such a resulting digraph is a planar st-graph Step 1: faces triangulation dest(f) f : a face of planar st-graph G’ (Lemma 4.1): f origin(f)

10. t w v G2 u G1 s Step 2: Upward planar straight-line drawing Lemma 6.4:Let G be a planar st-graph with all faces consisting of three edges. Given any upward planar straight-line drawing Δ for the external face of G, there exists an upward planar straight-line drawing of G with the external face drawn as Δ. Proof:by induction on the number n of vertices of G. n = 3, holds for graphs with fewer than n vertices: v – not on external face, χ – undirected cycle of the neighbors of v Case 1: χ – has a chord (u, w) λ– undirected cycle (u, v, w) G1 and G2 – planar st-graphs By induction: Γ1 of G1, with external face Δ. Γ2 of G2, with external face λ

11. u Step 2: cont. • Case 1: χ – has no chord • u – a predecessor of v, such that there is no directed path from u to any other predecessor of v (highest topological numbering). • *Contract edge (u,v) into vertex u: • * The resulting G’ is a planar st-graph with n-1 vertices. • * By induction: Γ’ of G’ with external face Δ. • * Reinsert v: - every vertex properly visible from v, • - below its successors and above its predecessors v u

12. 2 (G admitsupward planar straight-line drawing) 3 (G isspanning subgraph of a planar st-graph) 3 (G isspanning subgraph of a planar st-graph)1 (G is upward planar):- Given a planar st-graph G’ including G- Construct a planar polyline drawing of G’ (Ch. 4) - Remove edges that do not belongs to G

13. Planar n-vertex st-graph testing – O(n):- G has a single source s and a single sink t – O(n) - G is planar – O(n) (ch. 3)- G is acyclic – O(n) (DFS) Upward planarity testing algorithm – exponential-timeAdding all the possible subsets of edges and testing whether each of the resulting digraphs is a planar st-graph

14. 6.2 – Angles in Upward Drawings • Target: polynomial time UP testing for embedded digraphs • Drawing = points to Vs; curves to edges • Embedding = eqv. class of Drawings (same clockwise edges) • Definitions • Bimodal vertex: cyclic edge sequence can be partitioned to incoming, outgoing (possibly empty) • Bimodal graph: all edges bimodal • Upward planarity  Bimodality (Lemma 6.5) • Planar st-graphs are UP  they are also bimodal • NOT true in other direction: e.g. simple 3-cycle

15. Angles of a Graph • Formed in embedding between 2 consecutive edges of same vertex • straight-line drawing • Single edge  angle = 2*PI

16. Large and Small Angles • Only between pairs of incoming (or outgoing) edges • Intuition: Small < PI; Large > PI • Used in UP straight-line drawing • If p = vertex OR face • L(p) = number of Large angles of p • S(p) = number of Small angles of p

17. Example-Small and Large Angles • Here edges are undirected; • “F” means a FLAT angle – between incoming AND outgoing edges (not needed)

18. Angle Consistency Properties • The following holds for any UP straight-line drawing (for all fs, vs): • Follows from elementary geometry • Intuition: external face has more large angles; Start from triangle (S=2,L=0) and add points

19. Angles Consistency - Example • Only Large anglescolored • Difference between Small & Large is always 2 • External (h): more Large angles (due to Ss, Ts)

20. Incoming OR Outgoing angles • A(f)= number of incoming (OR outgoing) angles of face f • Small AND Large • For a given face: |Incoming| = |Outgoing| • A is determined by Embedding; 2A(f) = L(f)+S(f)

21. Lemma 6.6 Revisited • Lemma 6.6 + simple algebra using 2A(f) = L(f)+S(f) WE GET: • Lemma 6.7 • As before, holdsfor any UP straight line drawing of G

22. Assignments • Vertices to faces -  : v  f • (v): map each source\sink to one incident face • At least 2 faces possible • Only non-internal vertices (only sources\sinks) •  (f) = group of vertices assigned to f -1

23. Assignments - Example A(f)-1 next to each f; A(f)+1 next to h • Squares = faces • A(f) -1 shown next to internal faces • A(f)+1 shown next to external faces

24. Consistent Assignments •  is consistent if there exists a face h s.t. • h is the external face • Relative to Lemma 6.7: • 6.7: |Large angels| on the left (here: |vs assigned|) • 6.7: Also mentions angles of vertices • Assigning v to f v will be Large in f • Hence this is (essentially) Lemma 6.7

25. Consistency and Upward Planarity Lemma 6.8: • An embedded bimodal digraph is upward planar only if it admits a consistent assignment of sources and sinks to faces • A(f)-1 to all internal edges; A(f)+1 to external edge • This is merely a reformulation of previous lemmas • Conditions in Lemmas 6.5-6.8 are also sufficient • GIVEN embedded, bimodal digraph G with consistent  THEN G spans an UP st-graph (and hence G is UP) • Proved by correctness of Algorithm 6.1 below

26. Algorithm 6.1 • Mainly: apply 6.2 to each face; (except for h:) connect to s\t

27. Switches • Source-switch (or sink-switch) of a face f= source (sink) of f • Switch of f = source OR sink of f • A source (sink) in G • is a source(sink)-switch in all incident faces • Source in G =\= source in f • An internal vertex in G • Is a source or sink in all incident fs but two • E.g. central v is a switch in all fs but 2 lower ones

28. Sequences σ of Switches • Using , assign to every source\sink: • SL, tL = “Large” source\sink of f • Ss, ts = “Small” source\sink of f • σf = circular symbol sequence, obtained by • traversing f clockwise • Gathering all SL, tL, Ss, ts symbols on the path • E.g. : σf = ts, SL, ts, SL, ts, Ss, ts… • S\L not determined by drawing

29. Algorithm 6.2 – Preliminaries • Canonicalsub-sequences: “LSS” sequences of σf • We want to cancel sinks\sources (canceling Large angels) • by adding edges in LSS subsequences • Two cases are possible: • SL, ts, Ss  add edge from Ss to SL • tL, Ss, ts  add edge from tL to ts • (cannot have 2 sinks\sources in a row) • New edge splits f (and hence splitsσf) • Ss  SL cancels SL; tLts cancels tL

30. Algorithm 6.2 – Saturate Face

31. Example 1 for Algorithm 6.2 • f splits to f’ (next to split) and f’’ (containing LSS) • σf splits to σf’ and σf’’ • Continue recursively with f’, σf’

32. Splitting – Additional Examples

33. Correctness Proof of Algorithm 6.1 • Prove by showing: • Edge insertions preserve planarity, acyclicity and bimodality, as well as consistency of ` • After all insertions, G’ has only one source and one sink on same f (whichis h)

34. Planarity, Acyclicity, Bimodality • Planarity – each edge is inserted inside a face, and hence no crossings • Acyclicity – assume a cycle exists; prove by contradiction to  properties[omitted] • Bimodality • Assume edge (z,x) is inserted between two source-switches (for sinks – analogous) • For x – (z,x) is the only incoming edge bimodal • For z – outgoing edge (z,x) inserted between 2 outgoing bimodal

35. Assignment Consistency Invariant • After inserting edge (z, x)--  `: • All sources\sinks which were not canceled are assigned by ’ to f’ • For all other sources\sinks = ` • Hence ` is consistent • A  A-1 so we need one less angle assigned

36. Single Source and Sink Internal Faces • All are now with A(f)=1 • (no vertex assigned) • All fs created contain exactly onesource and one sink, both labeled S • Otherwise, there are still more canonical subsequences • A(f)+1 S symbols and A(f)-1 L  “LSS” still exists and we can split

37. Single Source and Sink [2] The External Face • A(h)+1 assigned sources\sinks • Stop when we have • k≥0 S symbols ; k+2 L symbols; no two S symbols in a row • Final sequence has following structure: • σh = L1, σ1, L2, σ2; σ1=S,L,S,...,L,S; σ2=S,L,S,...,L,S • L symbols in one of σ1\2 referes to sources, the other to sinks

38. Final Step Either connect L1 to highest and L2 to lowest OR add s,t

39. Section Conclusion • Time complexity of Saturate Face is linear in f’s vertices • Simple manipulations of σs  Algorithm 6.2 takes O(n) time • Proof: • “only if” : from Lemmas 6.5-6.8 • “if” and O(n) : Lemma 6.9 and Theorem 6.1

40. 6.3 – Upward Planarity Testing How to construct consistent assignments? An algorithm for testing UP of an embedded digraph Need to test if a consistent assignment of sources\sinks to faces exists given h MAINIDEA: Construct a biparitite flow network Bh to represent assignments

41. Bipartite Flow Network Bh • Nodes of Bh are sources, sinks and faces of G • Sources of Bh = Sources\sinks of G • Each supplies 1 unit of flow • Sinks of Bh = Faces of G • Demand = A(f)-1 for internal faces; A(f)+1 for h • Arcsof Bh • (v,f) exists if v is a source\sink of G on face f

42. Flow in Bh • An assignment of values {0,1} to arcs of Bh • For each source v of Bh : • Sum of values assigned to outgoing arcs of v supply of v • For each sink f of Bh : • Sum of values assigned to incoming arcs of f demand of f • The Value of the flow • = sum of values assigned to all arcs in Bh • Multi source\sink network:

43. Example a.

44. Network Properties • G with n vertices  O(n) vertices in Bh • G admits a consistent assignment  Bh admits a flow of value r with correct demands • r is number of sources & sinks in G • Demands: A(f)+1 for External face; A(f)-1 for internal

45. Complexity of Basic UP Testing 2 3 • Choose h • O(n) possible choices • Construct Bh • O(n) time [O(n) edges] • Test existence of flow |r| for Bh • O(rn) time using r flow augmentations • If exists return #t ELSE choose next h • Overall  O(r*n ) = O(n )

46. Improvement - O(n ) time 2 • Algorithm 6.3 – Embedded-UP-Test • Construct a network B • Same as Bh but demands are [A(f) – 1] for ALL. • Independent of h O(n) • Test whether B admits a flow of |r-2|. O(n-2) • Tryto increase demand by two for each face • test if flow of |r-2| can be augmented by two units • Total: (r-1+r)*n = O(nr) • Return the set of all faces of G for which test was successful

47. Optimal Upward Planarity Testing of Single-Source Embedded Digraphs • Face-sink graph F of G: • Vertices of F: faces and vertices of G that are sink-switches • Edges (f,v): v is a sink-switch of face f • F is a forest

48. Thereom 6.6 • Let G be an embedded single-source digraph, and h a face of G. Digraph G is upward planar, subject to h being the external face, if and only if all the following conditions are satisfied: • The source of G is on the boundary of face h. • The face-sink graph F of G is a forest. • One tree T of F has no internal vertices of G, while the remaining trees have exactly one internal vertex. • Face h is a vertex of tree T.

49. Algorithm 6.4 Embedded-Single-Source-Upward-Planar-Test O(n) time

50. Thankyou!