On the ICP Algorithm

1
On the ICP Algorithm

• Esther Ezra, Micha Sharir
• Alon Efrat

2
The problem Pattern Matching

• Input
• A a1, , am, B b1, , bn A,B ? Rd
• Goal
• Translate A by a vector t ? Rd s.t. ?(At,B) is
  minimized.
• The (1-directional) cost function

The cost function (measures resemblance).
The nearest neighbor of a in B.
A , B
3
The algorithm Besl Mckay 92

• Use local improvements
• Repeat At each iteration i, with cumulative
  translation ti-1
• Assign each point ati-1 ? Ati-1 to its NN b
  NB(ati-1)
• Compute the relative translation ?ti that
  minimizes ? with respect to this fixed (frozen)
  NN-assignment.Translate the points of A by ?ti
  (to be aligned to B). ti ? ti-1 ?ti .
• Stop when the value of ? does not decrease.

The overall translation vector previously
computed. t0 0 .
Some of the points of A may acquire new NN in B.
4
The algorithm Convergence

• Besl Mckay 92.
• The value of ??2 decreases at each iteration.
• The algorithm monotonically converges to a local
  minimum.
• This applies for ??? as well.

A local minimum of the ICP algorithm.
a1
a2
a3
b1
b2
b3
b4
The global minimum is attained when a1, a2, a3
are aligned on top of b2, b3, b4 .
5
NN and Voronoi diagrams

• Each NN-assignment can be interpreted by the
  Voronoi diagram V(B) of B.
• b is the NN of a
• a is contained in the Voronoi cell V(b) of V(B).

6

• The problem under the RMS measure

7
The algorithm An example in R1

• ?(At,B) RMS(t)
• Iteration 1

Solving the equation RMS(t)0
?(A,B) ? 3
m4, n2.
a3 1
a4 3
a1 -3-?
a2 -1
b1 0
b2 4
?t1 ? 1
(b1b2)/2 2
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An example in R1 Cont.
Iteration 2
?(A?t1,B) ? 2
a4 ?t1
a3 ?t1
a2 ?t1
a1?t1
b1 0
b2 4
?t2 1
(b1b2)/2 2
Iteration 3
?(A?t1?t2,B) ? 1
a1 ?t1 ?t2
a2 ?t1 ?t2
a3 ?t1 ?t2
a4 ?t1 ?t2
b1 0
b2 4
NN-assignment does not change. ?t3 0
(b1b2)/2 2
9
Number of iterations An upper bound

• The value of ? is reduced at each iteration,
  since
• The relative translation ?t minimizes ? with
  respect to the frozen NN-assignment.
• The value of ? can only decrease with respect to
  the new NN-assignment (after translating by ?t).
• No NN-assignment arises more than once!
• iterations NN-assignments (that the algorithm
  reaches)

10
Number of iterations A quadratic upper bound in
R1
am
a2
a1
ai
b1
b2
bn
bj
bj1
?t
ai crosses into a new Voronoi cell.

• Critical event The NN of some point ai is
  changed.
• Two consecutive NN-assignments
  differ by one pair (ai,bj) .
• NN-assignments ? (ai,bj) ai ? A, bj ? B
  nm.

11
The bound in R1

• Is the quadratic upper bound tight???
• A linear lower bound construction (n2) ?(m)

m2k2, n2
a2 -2k1
am-1 2k-1
am 2k1
a1 -2k-1
b1 0
b2 4k
(b1b2)/2 2k
Using induction on i1,, k/2 .
Tight! (when n2) .
12
Structural properties in R1

• Lemma
• At each iteration i?2 of the algorithm
• Corollary (Monotonicity)
• The ICP algorithm always moves A in the same
  (left or right) direction.Either ?ti ? 0 for
  each i?0, or ?ti ? 0 for each i?0.

Use induction on i.
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A super-linear lower bound construction in R1.
The construction (mn) a1 -n - ?(n-1), ai
(i-1)/n -1/2 ?, bi i-1, for i1, , n.
mn
b1 0
b2 1
?1 1/2
Theorem The number of iterations of the ICP
algorithm under the above construction is ?(n log
n).
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Iteration 2
b2 1
b3 2
Iteration 3
b3 2
b2 1
b4 3
At iteration 4, the NN of a2 remains unchanged
(b3) Only n-2 points of A cross into V(b4).
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The lower bound construction in R1

• Round j n-1 right pts of A assigned to bn-j1
  and bn-j2
• Consists of steps where pts of A cross the
  bisector
• ?n-j1 (bn-j1 bn-j2) / 2
• Then at each such step i,
• pts of A that cross ?n-j1 is exactly j
• (except for the last step).
• At the last such step, pts of A that cross
  ?n-j1 or ?n-j2 is exactly j-1.

steps in the round ? n/j .
At the next round, steps in the round ? n/(j-1)
.
16
The lower bound construction in R1

• The overall number of steps

17
The lower bound construction in R1Proof of the
claim

• Induction on j
• Consider the last step i of round j

llt j pts have not yet crossed ?n-j1
Black pts still remain in V(bn-j2).
bn-j2
bn-j1
bn-j3
The points a2, , al1 cross ?n-j1, and
an-(j-l-2), , an cross ?n-j2 . Overall number
of crossing points l j - l -1 j - 1.
QED
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General structural properties in Rd

• Theorem
• Let A a1, , am, B b1, , bn be two
  point sets in Rd.
• Then the overall number of NN-assignments, over
  all translated copies of A, is O(mdnd).
  Worst-case tight.
• Proof
• Critical event The NN of ait changes from b to
  b
• t lies on the common boundary of V(bai) and
• V(bai)

t
b - ai
b - ai
Voronoi cells of the shifted diagram V(B-ai) .
b - ai
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NN Assignments in Rd

• Critical translations t for a fixed ai
• Cell boundaries in V(B-ai) .
• All critical translations
• Union of all Voronoi edges
• Cell boundaries in the overlay M(A,B) of
  V(B-a1),, V(B-am).
• NN-assignments cells in M(A,B)

Each cell of M(A,B) consists of translations with
a common NN-assignment
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NN Assignments in Rd

• Claim
• The overall number of cells of M(A,B) is O(mdnd).
• Proof
• Charge each cell of M(A,B) to some vertex v of
  it(v is charged at most 2d times).
• v arises as the vertex in the overlay Md of only
  d diagrams.
• The complexity of Md is O(nd) Koltun Sharir
  05.
• There are d-tuples of diagrams
  V(B-a1),, V(B-ad).

QED
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Lower bound construction for M(A,B)
1/m
d2
A
Vertical cells Minkowski sums of vertical cells
of V(B) and (reflected) horizontal points of A .
1
B
M(A,B) contains ?(nm) horizontal cells that
intersect ?(nm) vertical cells ?(n2m2) cells.
Horizontal cells Minkowski sums of horizontal
cells of V(B) and (reflected) vertical points of
A .
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NN Assignments in Rd

• Lower bound construction can be extended to any
  dimension d
• Open problem
• Does the ICP algorithm can step through all
  (many) NN-assignments in a single execution?
• d1
• Upper bound O(n2).
• Lower bound ?(n log n).

Probably no.
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Monotonicity

• d1
• The ICP algorithm always moves A in the same
  (left or right) direction.
• Generalization to d?2
• ? connected path obtained by concatenating the
  ICP relative translations ?t. (? starts at the
  origin)
• Monotonicity ? does not intersect itself.

?
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Monotonicity

• Theorem
• Let ?t be a move of the ICP algorithm from
  translation t0 to t0 ?t . Then RMS(t0? ?t) is
  a strictly decreasing function of ?.
• ? does not intersect itself.

Cost function monotone decreasing along ?t.
?t
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Proof of the theorem
The Voronoi surface, whose minimization diagram
is V(B-a).

• SB-a(t) minb?B at-b2 minb?B (t2
  2t(a-b) a-b2)
• RMS(t)
• SB-a(t) - t2 is the lower envelope of n
  hyperplanes
• SB-a(t) - t2 is the boundary of a
  concave polyhedron.
• Q(t) RMS(t) - t2 is the average of SB-a(t)
  - t2a?A
• Q(t) is the boundary of a concave
  polyhedron.

The average of m Voronoi surfaces S(B-a).
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Proof of the theorem
Replace f(t) by the hyperplane h(t) containing
it. h(t) corresponds to the frozen NN-assignment.
NN-assignment at t0 defines a facet f(t) of Q(t),
which contains the point (t0,Q(t0)).
t0
The value of h(t)t2 along ?t is monotone
decreasing
?t
Since Q(t) is concave, and h(t) is tangent to
Q(t) at t0, the value of Q(t)t2 along ?t
decreases faster than the value of h(t)t2 .
QED
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More structural properties of ?

• Corollary
• The angle between any two consecutive edges of ?
  is obtuse.
• Proof
• Let ?tk, ?tk1 be two consecutive edges of ? .
• Claim
• ?tk1 ? ?tk ? 0

?
b NB(atk)
a
?tk
a must cross the bisector of b, b. b lies
further ahead
b NB(atk-1)
QED
28
More structural properties Potential

• Lemma
• At each iteration i?1 of the algorithm,
• (i)
• (ii)
• Corollary
• Let ?t1, , ?tk be the relative translations
  computed by the algorithm. Then

Due to (i)
Due to (ii)
Cauchy-Schwarz inequality
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More structural properties Potential

• Corollary
• d1
• Trivial
• d?2
• Given that ?ti ? ? , for each i?1, then

tk
?
Since
? does not get too close to itself.
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• The problem under the Hausdorf measure

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The relative translation

• Lemma
• Let Di-1 be the smallest enclosing ball of
  ati-1 NB(ati-1) a ? A. Then ?ti moves
  the center of Di-1 to the origin.
• Proof

Any infinitesimal move of D increases the
(frozen) cost.
The radius of D determines the (frozen) cost
after the translation.
a3
a2
a3-b3
a3-b3?t
a2-b2
a2-b2?t
b2
b3
o
o
D
b1
?t
b4
D
a4-b4?t
a4-b4
a4
a1-b1
QED
a1
a1-b1?t
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No monotonicity (d ? 2)

• Lemma
• In dimension d?2 the cost function does not
  necessarily decrease along ?t .
• Proof

Final distances a2?t-b a3?t-b r
, a1?t-b lt r .
Initially, a1-b max ai-b gt r , i1,2,3
.
The distances of a2, a3 from b start increasing
?t
a2
a2?t
r
b
a3
?t
c
a3?t
b
a1
a1?t
b
Distance of a1 from its NN always decreases.
b
QED
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The one-dimensional problem

• Lemma (Monotonicity)
• The algorithm always moves A in the same (left or
  right) direction.
• Proof
• Let a-bmaxa?A NB(a)-a .
• Suppose a lt b.
• a-b is the left endpoint
• of D0 its center c lt 0.
• ?t1 gt 0, ?t1 lt a-b .
• a?t1 lt b, so b is still the
• NN of a?t1 .
• a?t1-bmaxa?A a?t1-NB(a)gt
  a?t1-NB(a?t1)
• a?t1-b is the left endpoint of D1.
• The lemma follows by induction.

Initial minimum enclosing ball.
?t
D0
0
c
a-b
aNB(a)
a, b still determine the next relative
translation.
0
a?t1-b
a?t1NB(a)
QED
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Number of iterations An upper bound
Ratio between the diameter of B and the distance
between its closest pair.

• Theorem
• Let ?B be the spread of B.
• Then the number of iterations that the ICP
  algorithm executes is O((mn) log ?B / log n).
• Corollary
• The number of iterations is O(mn) when ?B is
• polynomial in n

35
The upper bound Proof sketch

• Use the pair a, b that satisfies
  a-bmaxa?A NB(a)-a always determines the
  left endpoint of Di.
• Put Ik b-(atk) .
• Classify each relative translation ?tk
• Short if
• Long otherwise.Overall O(n log ?B / log n).

Easy
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The upper bound Proof sketch

• Short relative translations
• a ? A involved in a short relative translation at
  the k-th iteration,
• atk-1 NB(atk-1) is large.
• a ? A involved in an additional short relative
  translation
• a has to be moved further by atk-1
  NB(atk-1) .
• Ik b-(atk) must decrease
  significantly.

b1
bj
bj1
Ik-1
a
a1
?tk
37
Number of iterations A lower bound construction
Theorem The number of iterations of the ICP
algorithm under the following construction (mn)
is ?(n).

• At the i-th iteration
• ?ti 1/2i , for i1,,n-2 .
• Only ai1 crosses into the next cell V(bi2).
• The overall translation length is lt 1.

b1
b2
bn
bi
n-1
n
a1
a2
an
ai
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Many open problems

• Bounds on iterations
• How many local minima?
• Other cost functions
• More structural properties
• Analyze / improve running time,
• vs. running time of directly computing
• minimizing translation
• And so on