Dynamic Delaunay Triangulation

1
Dynamic Delaunay Triangulation

• Theory Lunch / Mar 12, 2003
• Benoît Hudson

2
A Delaunay Mesh

• We have points in the plane
• We want a mesh of triangles
• Delaunay triangulation
• defining feature triangles' circumscribing
  circle is empty
• Generalizes to any Rd

3
So what?

• This means
• the triangulation is well-defined (unique for a
  set of points)
• can compute in ?(n lg n)
• dual of Voronoi diagram
• minimum angle is maximized
• visualization, numerical stability

4
Dynamic Delaunay

• We can compute the Delaunay in O(n lg n)
• New point comes in.
• most of the diagram is unchanged
• want to pay o(n lg n) ideally, O(lg n)
• Old point is removed.
• again, little changes
• Queries check which triangle a point is in.
• or, draw the triangulation

5
2-d is hard let's go shopping!

• 1-d Delaunay
• segments instead of triangles
• circumcircle of a segment is just the segment
• A good structure for this
• fast insert / delete
• fast query
• for now random insertion / deletion order.

6
Binary Search Trees!

• Random inserts just use vanilla insert.

a
a
empty
insert a
insert b
b
a
a
b
-inf
inf
-inf
inf
-inf
inf
7
Deletions

• Deleting p
• Want structure as if p were never inserted
• ease of analysis now the structure is unique
• Answer
• keep track of insertion time.
• rotate p to a leaf, then remove it.
• just like deleting from a treap

8
Delete b
a
a
a
a
b
c
c
c
b
d
d
c
e
d
e
b
e
d
e
b
a
c
d
e
9
Another view

• Leaves are the current intervals
• Nodes have intervalsof when a new nodewas
  inserted
• Search for p look at left, right interval

-inf,inf
-inf,a
a,inf
-inf,b
b,a
c,b
-inf,c
e,a
b,e
c,d
d,b
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Sameness

• It's the same tree
• nodes are the vertices
• label edges by tightest interval so far

• nodes are the intervals
• label nodes by childrens' midpoints

Easy to update labels through rotate.
11
Back to 2-d

• Intervals are now triangles.
• Insertion

p
12
2-d Data Structure

• What's a node?
• Either Node vertex p
• Or Node triangle T

13
Previous work

• Lots of previous work ca. 91-93
• Guibas et al, Melhorn et al, Devillers et al
• All of it has triangles as nodes
• Devillers et al
• insertion is nice easy
• deletion is hell to explain
• 1-d analogue cutting out the node,
• patching everything up nicely
• requires 2 or 3 auxiliary structures

14
Vertices as nodes

• Better analogy to search trees.
• Starting case an infinite-size triangle
• (1-d was infinite-size segment)
• We have a DAG
• nodes vertices
• edges triangles (1-d was intervals)

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Insertion

• Inserting a point
• find where it goes
• 1-d find the segment whose circumcircle
  (segment) it's in
• 2-d find the triangles whose circumcircles it's
  in
• create new node, edges
• 1-d bury the old segment with 2 new ones
• 2-d bury the (many) old triangles with (many)
  new ones

16
Insertion pseudo-code

• Find encroached triangles that are leaves
• those whose circumcircle p lies inside
• Make all of them have p be their burier
• For each triangle,
• for each edge (i,j)
• if another triangle has edge (i,j) skip (both are
  being buried)
• else, create new triangle (p,i,j) with p its
  creator

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Rotation in 2-d?

• We are rotating x to be below y
• Always rotate oldest child up
• Goal new structure is as if x was inserted after
  y, not before.
• Afterwards, need to
• update who x's parents are now pointing to
• move some triangles from x to y
• destroy some triangles
• create some triangles

18
Rotation pseudo-code

• Rules
• y keeps triangles it created, unless x is a
  vertex
• y creates new triangles that x blocked
• y takes parents of x that y encroaches
• x keeps triangles it created, unless y encroaches
• x destroys old triangles that y now blocks
• x takes triangles y created, if x is a vertex

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Rotation in 1-d
xa yb

• Rotate a down.

-inf,inf
a

1. b keeps b,inf
2. b creates -inf,b
3. b takes -inf,inf
4. a keeps -inf,a
5. a destroys a,inf
6. a takes a,b

a,inf
-inf,a
b
b,inf
a,b
a
b
-inf
inf
20

1. b keeps b,inf
2. b creates -inf,b
3. b takes -inf,inf
4. a keeps -inf,a
5. a destroys a,inf
6. a takes a,b

-inf,inf
-inf,inf
a
b
a,inf
-inf,a
b,inf
a
b
a,b
-inf,a
b,inf
a,b
a
b
a
b
-inf
inf
-inf
inf
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Does this work?

• To prove
• Rotation produces the structure that would have
  been produced had x been inserted after y
• Triangles not affected by the rotation should not
  be affected.
• Triangles affected by the rotation are correctly
  moved / created / destroyed.

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A proof (1)

• Triangles not affected by the rotation should
  have creator burier not depend on insertion
  order of x,y
• T created after y
• then must not be encroached by x or y
• T buried before x
• someone else encroaches and is still before y

23
A proof (2)

• Affected triangles are correctly updated.
• Each rule is correct.
• By proof (1), anything not included in those
  rules isn't affected.
• QED.

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How big?

• Clearly, DAG has n nodes (vertices).
• How many edges (triangles) ?
• Forward analysis how many will we create?
• dunno.
• Backward analysis do it, and see what the output
  looks like.

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Backwards

• Look at last ith vertex inserted. How many
  triangles did it create?
• Total triangles in N-node triangulation is at
  most 3N-3
• i real vertices 3 infinite ones 3i3 total
• random vertex has (3i3)/i 33/i triangles.
• Sum 1..n 3nHn O(n).

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Small, but fast?

• Insertions take time proportional to triangles
  encroached on by p.
• Can show that expected triangles is O(lg n).
• big picture count how many triangles there can
  be that have j vertices encroach on them.
• find probability a triangle has p encroach on it
• sum over all triangles Clarkson, Shur

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Corollary

• Depth must be O(lg n) expected.
• So point location queries are fast.
• Also, deletion is fast O(lg n) rotations
• Rotations involve triangles created by or buried
  by x,y. Constant number of these.