Flow and Upward Planarity

Graph Drawing, chapter 6 by:

Ami Hauptman Masha Igra

IntroductionUpward planar: inclusion in a Planar st-

GraphAngles in Upward DrawingUpward Planarity Testing of Embedded

DigraphsOptimal Upward Planarity Testing of Single-

Source Embedded Digraphs

IntroductionA digraph is upward planar if it admits a

planar and upward drawing.

• upward – edges are monotonically increasing in the vertical direction• planar – no two edges intersect

IntroductionAn embedded graph described by circular order

of the neighbors of each vertex.

Inclusion in a Planar st-GraphA 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

S

t

Theorem 6.1Let G be a digraph. The following statements are

equivalent:1. G is upward planar2. G admits an upward planar straight-line

drawing3. G is the spanning subgraph of a planar st-graph

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 upwardAdd new edges (v, dest(e))Able to cancel all sinks, except tSpecial casesSimilar process cancels all sources, except sStatement shown.

3 (G is spanning subgraph of a planar st-

graph G’) 2 (G admits an upward planar straight-line

drawing)

3 steps:1) Add edges to G′. Resulting

digraph G′′ is planar st-graph with all faces consisting of three edges

2) Construct an upward planar straight-line drawing of G′′

3) Remove edges that don’t belong to G from drawing of G′′

Step 1: faces triangulation

f : a face of planar st-graph G’ (Lemma 4.1): r

Pf

origin(f)

dest(f)

lP

1v

2v

3v

4v

lP

Lemma 6.1: P = or , P = , k >= 4. Edge or can be added, such a resulting digraph is a planar st-graph

),...,(1 k

vv ),(11 k

vv ),(2 k

vvl

Pr

P

Lemma 6.2: and has at least 3 vertices. Edge can be added, such a resulting digraph is a planar st-graph

),(rl

vvl

Pr

P

lv

rv

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

v

s

t

Case 1: χ – has a chord (u, w)

λ – undirected cycle (u, v, w)

G1 and G2 – planar st-graphsBy induction: Γ1 of G1, with external face Δ. Γ2 of G2, with external face λ

u

w

G1

G2

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 u

2 (G admits upward planar straight-line drawing) 3 (G is spanning subgraph of a planar st-graph)

3 (G is spanning 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

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

Target: polynomial time UP testing for embedded digraphsDrawing = points to Vs; curves to edgesEmbedding = 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 bimodalUpward planarity Bimodality (Lemma 6.5)Planar st-graphs are UP they are also bimodal

NOT true in other direction: e.g. simple 3-cycle

6.2 – Angles in Upward Drawings

Angles of a Graph

Formed in embedding between 2 consecutive edges of same vertex straight-line drawing

Single edge angle = 2*PI

Large and Small Angles

Only between pairs of incoming (or outgoing) edges

Intuition: Small < PI; Large > PIUsed in UP straight-line drawing

If p = vertex OR faceL(p) = number of Large angles of pS(p) = number of Small angles of p

Example-Small and Large Angles

• Here edges are undirected; • “F” means a FLAT angle – between incoming AND outgoing edges (not needed)

Angle Consistency Properties

The following holds for any UP straight-line drawing (for all fs, vs):

Follows from elementary geometryIntuition: external face has more large angles; Start from triangle (S=2,L=0) and add

points

Angles Consistency - Example

Only Large angles coloredDifference between Small & Large is always 2External (h): more Large angles (due to Ss, Ts)

Incoming OR Outgoing anglesA(f)= number of incoming (OR outgoing)

angles of face fSmall AND Large

For a given face: |Incoming| = |Outgoing|A is determined by Embedding; 2A(f) = L(f)+S(f)

Lemma 6.6 RevisitedLemma 6.6 + simple algebra using 2A(f) =

L(f)+S(f)

WE GET:Lemma 6.7

As before, holds for any UP straight line drawing of G

Vertices to faces - : v f (v): map each source\sink to one incident

faceAt least 2 faces possibleOnly non-internal vertices (only sources\sinks)

(f) = group of vertices assigned to f

Assignments

-1

A(f)-1 next to each f; A(f)+1 next to h

Assignments - Example

Squares = facesA(f) -1 shown next to internal faces A(f)+1 shown next to external faces

Consistent Assignments

is consistent if there exists a face h s.t.

h is the external faceRelative to Lemma 6.7:

6.7: |Large angels| on the left (here: |vs assigned|)

6.7: Also mentions angles of verticesAssigning v to f v will be Large in f

Hence this is (essentially) Lemma 6.7

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 facesA(f)-1 to all internal edges; A(f)+1 to external edgeThis is merely a reformulation of previous lemmas

Conditions in Lemmas 6.5-6.8 are also sufficientGIVEN 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

Algorithm 6.1

Mainly: apply 6.2 to each face; (except for h:) connect to s\t

Switches

Source-switch (or sink-switch) of a face f = source (sink) of f

Switch of f = source OR sink of fA source (sink) in G

is a source(sink)-switch in all incident faces

Source in G =\= source in fAn internal vertex in G

Is a source or sink in all incident fs but twoE.g. central v is a switch in all fs but 2

lower ones

Sequences σ of Switches

Using , assign to every source\sink:SL, tL = “Large” source\sink of fSs, ts = “Small” source\sink of f

σf = circular symbol sequence, obtained bytraversing f clockwiseGathering 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

Algorithm 6.2 – Preliminaries

Canonical sub-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 splits σf )Ss SL cancels SL; tL ts cancels tL

Algorithm 6.2 – Saturate Face

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’

Splitting – Additional Examples

Correctness Proof of Algorithm 6.1

Prove by showing:1) Edge insertions preserve planarity,

acyclicity and bimodality, as well as consistency of `

2) After all insertions, G’ has only one source and one sink on same f (which is h)

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]

BimodalityAssume edge (z,x) is inserted between two

source-switches (for sinks – analogous)For x – (z,x) is the only incoming edge

bimodalFor z – outgoing edge (z,x) inserted between 2

outgoing bimodal

Assignment Consistency Invariant

After inserting edge (z, x)-- `:1) All sources\sinks which were not canceled

are assigned by ’ to f’2) For all other sources\sinks = ` Hence ` is consistent

A A-1 so we need one less angle assigned

Single Source and Sink Internal FacesAll are now with A(f)=1

(no vertex assigned)

All fs created contain exactly one source and one sink, both labeled S

Otherwise, there are still more canonical subsequencesA(f)+1 S symbols and A(f)-1 L “LSS” still exists and we can split

Single Source and Sink [2] The External FaceA(h)+1 assigned sources\sinksStop 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,SL symbols in one of σ1\2 referes to sources, the

other to sinks

Final Step

Either connect L1 to highest and L2 to lowest OR add s,t

Section Conclusion

Time complexity of Saturate Face is linear in f’s verticesSimple 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

6.3 – Upward Planarity Testing How to construct consistent assignments?An algorithm for testing UP of an

embedded digraphNeed to test if a consistent assignment of

sources\sinks to faces exists given h

MAIN IDEA:Construct a biparitite flow network Bh

to represent assignments

Bipartite Flow Network Bh

Nodes of Bh are sources, sinks and faces of G

Sources of Bh = Sources\sinks of GEach supplies 1 unit of flow

Sinks of Bh = Faces of GDemand = A(f)-1 for internal faces; A(f)+1 for h

Arcs of Bh(v,f) exists if v is a source\sink of G on face f

Flow in BhAn assignment of values {0,1} to arcs of BhFor 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 fThe Value of the flow

= sum of values assigned to all arcs in Bh

Multi source\sink network:

Example

a.

Network Properties

G with n vertices O(n) vertices in BhG admits a consistent assignment Bh admits

a flow of value r with correct demandsr is number of sources & sinks in GDemands: A(f)+1 for External face; A(f)-1 for

internal

Complexity of Basic UP Testing

1. Choose h O(n) possible choices

2. Construct Bh O(n) time [O(n) edges]

3. Test existence of flow |r| for Bh O(rn) time using r flow augmentations

4. If exists return #t ELSE choose next h Overall O(r*n ) = O(n )2 3

Improvement - O(n ) timeAlgorithm 6.3 – Embedded-UP-TestConstruct 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)Try to 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

2

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

Thereom 6.6Let 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:

1. The source of G is on the boundary of face h.

2. The face-sink graph F of G is a forest.

3. One tree T of F has no internal vertices of G, while the remaining trees have exactly one internal vertex.

4. Face h is a vertex of tree T.

Algorithm 6.4 Embedded-Single-Source-Upward-Planar-Test

O(n) time

Thank you!