Octagonal Drawing

Johan van Rooij

Overview

What is an octagonal drawing Good slicing graphs Octagonal drawing algorithm for

good slicing graphs Correctness Time Complexity

Definition

Octagonal drawing: Orthogonal drawing Outer cycle is a rectangle Each inner face has at most eight

corners The area of each inner face is

equal to a prescribed number

Octagonal Drawing Example

Applications

VLSI floorplanning Modules with fixed space requirements Adjacency requirements for

communication Often not possible with rectangular

drawings

Often “good slicing floorplans”

Slicing Graph

For a 2-3 plane graph, with designated corners and N,E,S,W boundary we have:Slicing Path:Path from N to S or E to W boundary that does not pass through other outer vertices or edgesSlicing Graph:One inner face or has a slicing path which divides the graph in two slicing graphs

Slicing Graph Example

Slicing Tree

We can make a tree using the recursive definition of a slicing graph Leaf: face of the graph Internal node: slicing path

V-Node: Vertical (N-S) slicing path H-Node: Horizontal (W-E) slicing path

Slicing Tree Example

Good Slicing Graph

Good Slicing Tree:Every H-Node has at least one child which is a leaf.Good Slicing Graph:Graph corresponding to a good slicing treeTherefore: north or south subgraph of WE-slicing path never sliced

Octagonal Drawing Algorithm

Given a good slicing graph with a given good slicing tree we will create an octagonal drawing for the graph.

Preprocessing step:Add to each node of the slicing tree the total required area of the faces corresponding to its descendants.

Algorithm Outline

Choose initial height and width and draw the outer cycle and arbitrarily fix all vertices on east boundary.

Traverse tree: root, right, left subtree Draw slicing paths corresponding to

the nodes of the tree from east to west.

For Leaf: do nothing Always: east boundary vertices fixed

Face Shapes

Feasible Face

1. Area equals prescribed area2. max( lt , lb ) < fλ

3. If xN2 concave: lt < (f - fE)λ

4. If xS1 concave: lb < (f – fE)λ

5. If xN2 convex: lt ≥ fE λ

6. If xS1 convex: lb ≥ fE λ

7. If xN2 and If xS2 concave: lb – lt ≥ fE λ

8. If xN1 and If xS1 concave: lt – lb ≥ fE λf = total nr of inner faces

fE = nr of faces inside current face which have an edge on the east boundary

)(minmin

GfA

WA

fH

A

Correctness Theorem

The algorithm finds an octagonal drawing

The initial rectangle is a feasible faceNext slides: If the face corresponding to an

internal node of the slicing tree is feasible we can draw the corresponding slicing path such that both new faces are feasible as well.

Vertical Slicing

North and south boundary ether face or outer cycle: no interfering vertices!

We can draw a straight vertical line such that the area’s are correct.

New faces feasible? Because of condition 2 of the original

face, the new faces are also feasible

Vertical Slicing is Feasible?

Condition 2 + formula for λ : max( lt , lb ) H < fλH = Amin

Horizontal Slicing

We will only show this for shape (a) East node of slicing path fixed Two cases - horizontal line through

fixed eastern nodes intersects: West boundary – three subcases Line segment between xS1 and xS2

Horizontal Slicing – case 1a

Case 1a: Straight line gives correct areasNorth face feasible:2. max( lNt, lNb ) = max( lt, 0 )

≤ max( lt, lb ) < fλ

3. lNt = lt < (f - fE)λ < (f – fNE)λ

South face feasible:2. max( lSt, lSb ) = max( 0, lb )

≤ max( lt, lb ) < fλ

6. lSb = lb ≥ fEλ > fS

Eλ

Horizontal Slicing – case 1b

Case 1b: North face is too largeFix x’S1 such that lNb

= lt + fNEλ

And x’S2 such that the areas are correct

The slicing path always is below the top boundary

Since 7: lb ≥ fEλ + ltmax(lb, lt) ≥ fEλ + lt >

fNEλ + lt = lNb

Thus: lNbH < max(lb,lt)H

Since 2: lNbH < fλH = Amin

Horizontal – case 1b feasibility

Case 1b: North face is too largeNorth face feasible:Since: lNb

= lt + fNEλ (and 7)

lb > lNb > lt2. max( lNt, lNb ) <

max( lt, lb ) < fλ

3. lNt = lt ≤ (f – fE)λ < (f – fNE)λ

Etcetera…

Horizontal Slicing – case 1c

Horizontal Slicing – case 2

Case 2: Horizontal line through fixed vertex intersect south boundary

Fix x’S1 such that lNb = lt + fN

Eλ and x’S2 by required area

Possible since 7:lNb = lt + fN

Eλ < lt + fEλ

≤ lbAnd:max(lt , lb)H < fλH = Amin

What have we shown?

If the face corresponding to an internal node of the slicing tree is feasible we can draw the corresponding slicing path such that both new faces are feasible as well.

Proves the Correctness Theorem

Time Complexity

The Algorithm works in linear time Bottom up computation of areas Each slicing path can be imbedded in

constant time

How To Obtain the Good Slicing Tree?

Rahman et al.If a 2-3 plane graph has a rectangular drawing and every proper inner cycle has at least five legs then it is a good slicing graph and a good slicing tree can be found in linear time!

Proper inner cycle: contains no outer edges

Questions?