Improved Approximation Bounds for
Planar Point Pattern Matching
(under rigid motions)
Minkyoung Cho
Department of Computer Science
University of Maryland
Joint work with David M. Mount
Example
Problem Definition
Point Pattern Matching : Given a pattern set P (size m) and a
background set Q (size n), compute the transformation T that
minimizes some distance measure from T(P) to Q.
•
Transformation :
a. Translation
b. Translation + Rotation (Rigid Transformation)
c. Translation + Rotation + Scale ….
•
Distance Measure:
a. Mean Squared Error
b. (Bidirectional, Directed) Hausdorff distance
c. Absolute distance
d. Hamming distance …
Directed Hausdorff distance
P
h(P, Q)  max min d(p, q)
pP
Q
qQ
Def: maximum distance of a
set(P) to the nearest point in
the other set(Q)
q1
q3
p3
i.e.
h(P, Q) = d(p2 , q2 )
h(Q, P) = d(p3, q4 )
p2
q2
Property: Not symmetric
p1
q4
Previous Result
Running time
Approximation
Bound
Chew, et. al
[CGH+97]
O(m3 n2 log2mn)
optimal
Goodrich, Mitchell, Orletsky
[GMO94]
O(n2 m logn)
4
Indyk, Motwani,
Venkatasubramanian [IMV99]
O(n4/3 Δ 1/3 logn)
(1 + ε)σ
Cardoze, Schulman [CS98]
O(n2 logn + logO(1) Δ)
(1 + ε)σ
Recall |P| = m, |Q| = n
Δ : the ratio of the distances between the farthest and closest pairs of
points
σ : an upper bound on the Hausdorff distance given by user or computed by
using binary search
Our result
• Improved Alignment-Based Algorithm of GMO.
• Approximation factor is always ≤ 3.13,
• Approximation factor ≤ 3 + 1/(√3ρ).
ρ = ½ diam(P)/hopt
where diam(P) denote the diametric distance of P and
hopt denote the optimal Hausdorff distance.
• Lower bound ≥3 + 1/(10ρ2)
– we present an example
Talk Overview
- Serial alignment algorithm (GMO’94)
- Symmetric alignment algorithm (ours)
- Analysis of the approximation factor for symmetric alignment
- translation
- rotation
- Lower bound
- Future work & conclusion
Serial Alignment Algorithm [GMO94]
1. Pick a diametrical pair (p1, p2) in P
2. For all possible pairs (qi, qj) in Q,
translate p1 to qi
rotate to align p1p2 with qiqj
compute Hausdorff distance
3. Return the transformation with minimum Hausdorff distance
qj
p2
p2
qi
p1
p1
Simple Example
For unique transformation between two planar point sets, we need
at least two points from each set.
p1
p2
q1
p1
q1
q2
p2
q2
optimal
p1
p2
q1
2 x optimal
q2
Symmetric alignment Algorithm
1. Pick a diametric pair (p1, p2) in P
2. For all possible pairs (qi, qj) in Q,
translate the midpoint of p1 & p2 to the midpoint of qi & qj
rotate to align p1p2 with qiqj
compute Hausdorff distance
3. Return the transformation with minimum Hausdorff distance
qj
p2
qi
p1
p2
p1
Comparison
q3
q3
p3
q1
<
p2
p1
q2
q1
Serial alignment
p2
p1
q2
Symmetric alignment
q3
q3
p3
q1
p3
p1
p3
>
p2
q2
q1
p1
p2
q2
Main Theorem
Theorem. Consider two planar point sets P and Q whose optimal
Hausdorff distance under rigid transformations is hopt. Recall that
ρ = ½ diam(P)/hopt, where diam(P) denotes the diameter of P. Then the
for all ρ > 0, the approximation ratio of symmetric alignment satisfies:


1
Asym (ρ)  min  3 
, 4ρ 2  2ρ  1 
3ρ


Talk Overview
- Serial alignment algorithm (GMO’94)
- Symmetric alignment algorithm (ours)
- Analysis of the approximation factor for symmetric alignment
- translation
- rotation
- Lower bound
- Future work & conclusion
Outline of the Proof
1. Suppose that we know the optimal transformation T* between P
and Q. ( i.e. h(T*(P), Q) = hopt and
each point in P has initial displacement distance ≤ hopt)
2. Apply Symmetric alignment algorithm (translation + rotation) to
the optimal solution
- compute the upper bound of translation displacement distance
- compute the upper bound of rotation displacement distance
3. Add these three displacement distances. It will become the
approximation factor.
Illustration of displacement distance
1. Initial Displacement
2. Translation Displacement
3. Rotation Displacement
disp
Appx 
hopt
disp
hopt
t
r
q1
mq
p1
hopt = 1
mp
p2
q2
Why can we assume an optimal placement?
Optimal
Arbitrary
Algorithm’s result is independent of initial placement
Basic Set-Up
Our approximation factor is sensitive to a geometric parameter ρ.
hopt :The optimal Hausdorff distance
ρ : half ratio of the diametric distance of P and hopt
α : the acute angle between line segment p1p2 and q1q2
Assume hopt = 1 and ρ > hopt
mq
q1
p1
hopt = 1
ρ
mp
α
p2
q2
Translation Displacement
q1
s1
ρ cos α
mq
α
p1
ρ
hopt = 1
q1  ( ρ cos α  s1, h)
q2  ( ρ cos α  s2 , h)
s1  s2 

mq  
,h
 2

h
mpp  (0,0)
m
q2
s2
p2
| s1 |  1  (ρ sin α  h) 2
| s2 |  1  (ρ sin α  h) 2
Translation Displacement (Con’t)
s  s2 
mq   1
,h
 2

| s1 |  1  (ρ sin α  h) 2
| s2 |  1  (ρ sin α  h) 2
2
2
2
2

  s1  s2 
s

s
s

s


2
1
2
1
2
2
  
T  mpmq  
  h  
  h2
2
 2 

  2 
 s1 2  s2 2 
  h2  1  ρ2 sin 2 α
 
2


q
2
s1
1
s1 q1
p1
mq
ρ
hopt = 1
α
mq
p2
mp
q2
s2
s2
q2
Rotation Displacement
A rotation displacement distance depends on angle(α) and
distance(x) from center of rotation.
And, the maximum rotational distance will be | R | ≤ 2√3ρ sin(α/2).
α
q1
mq
p1
ρ
hopt
mp
p2
q2
Rotation and Distance from Center of Rotation
α

2 x sin 
2

ρ
p1

m
3ρ
p
x
p2
p1
The distances from
the rotation center
are at most √3ρ
α
x

mp
p2
The rotation distance with
side length x and angle α
is 2x sinα/2
.
Approximation factor: Putting it Together
Approximation factor = Translation + Rotation + Initial Displacement
= |T + R| + 1 ≤ |T| + |R| + 1
α
1
 1  ρ sin α  2 3ρ sin  1  3 
2
3ρ
2
2
Recall… hopt = 1
Due to the restriction of time & space, we just show the case, ρ -> ∞
lim 1  ρ 2 sin 2 α  2 3ρ sin α  1
ρ
2
α

 1  ρ α  2 3ρ   1
2
2
2
3
,
 1  x  3x  1 x  ρα  ρ sin α 
2
2
appx  3
Talk Overview
- Serial alignment algorithm (GMO’94)
- Symmetric alignment algorithm (ours)
- Analysis of the approximation factor for symmetric alignment
- translation
- rotation
- Lower bound
- Future work & conclusion
Lower Bound Example
1
 3
10ρ2
3ρ
q1 α  arcsin 3 


 2ρ 
p1
hopt = 1
p2
ρ
mp
mq
q2
Future Work
•
•
Does there exist a factor 3 approximation based on simple point
alignments
Improve running time & robustness?
Thanks
Result
Plot of approximation factor as function of ρ
ρ<1
3ρ
q1
hopt = 1
p1
ρ
p2
(1  ρ) 2  ( 3ρ) 2  1  2ρ  4ρ 2  3, ρ  1.xx
q2
One fixing
We assumed that the diameter pair has a corresponding pair. (Even
though it can be a point, not a pair)
If the minimum Hausdorff distance from symmetric alignment is
bigger than ρ, we return any transformation which the midpoint
mp matches with any point in Q.
This is quite unrealistic case since the transformation is not
uniquely decided and any point in Q can be matched with all
points in P.( meaningless )
q1
p1
q1
p1
p2
p2
q2
Minor error for proof of GMO’94
q1
p2
p1
q2
For High Dimension
GMO algorithm works for all dimension. How about symmetric
alignment? If we match d-points symmetrically, it’s unbounded.
However, if we follow GMO algorithm except matching midpoint
rather than match one of the points, then our algorithm can be
extend to all dimension with better approximation factor.