On the ICP Algorithm

1
On the ICP Algorithm

• Esther Ezra Micha Sharir
• Alon Efrat

2
The algorithm Besl Mckay 92

• 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

• The algorithm is based on local improvements.
• Repeat At each iterarion i
• 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.
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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 .
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NN and Voronoi diagrams

• Each NN-assignment can be visualized 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).

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• The problem under the RMS measure

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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 Continue
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
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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)

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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.

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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) .
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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.

• Lemma
• Suppose that the n-1 rightmost points of A are
  assigned
• to bn-j1 and bn-j2 , j1, , n. Consider the
  round j of all iterations in
• which some of the point of A cross ?n-j1
  (bn-j1 bn-j2) / 2 .
• Then
• At each such iteration i, .
• The overall number of points of A that cross
  ?n-j1 is exactly j. (except for the last
  iteration).
• At the last such iteration, the overall number of
  points of A that cross either ?n-j1 or ?n-j2 is
  exactly j-1.

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

• The overall number of iterations

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The lower bound construction in R1Proof of the
lemma

• Using induction on j, properties (i), (ii)
  (iii)
• Consider the last iteration i of round j

llt j points has not crossed ?n-j1 yet.
The black points still remain at 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).
• Proof
• Critical event The NN of ait is changed from
  bj to bj1
• The translation t lies on the common boundary
• of V(bjai) and V(bj1ai) .

t
bj1 - ai
bj - ai
Voronoi cells of the shifted diagram V(B-ai) .
bk - ai
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The upper bound in Rd

• t critical events for (a fixed) ai
• Boundaries of all cells in V(B-ai) .
• t critical event for all ai, i1,,m
• Boundaries of all cells 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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The upper bound in Rd

• Claim
• The overall number of cells of M(A,B) is O(mdnd).
• Proof
• We 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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The upper bound in RdLower 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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The upper bound in Rd

• The lower bound construction can be generalized
  to any dimension d.
• Open problem
• Does the ICP algorithm can step through all
  (many) NN-assignment 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 d?1
• Let ? be the 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.

The cost function is 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.
The NN-assignment corresponding to t0 defines a
facet f(t) of Q(t), which contains the point
(t0,Q(t0)).
t0
The image of ?t along h(t)t2 is a monotone
decreasing function.
?t
Since Q(t) is concave, and h(t) is tangent to
Q(t) at t0, the image of ?t along Q(t)t2
decreases more strongly than the image along
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

?
a must cross the bisector of b, b.
b NB(atk)
a
?tk
b NB(atk-1)
QED
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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-Schwartz 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 Hausdorff 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 decreases 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 increase.
?t
a2
a2?t
r
b
a3
?t
c
a3?t
b
a1
a1?t
b
The 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, w.l.o.g., a lt b.
• a-b is the left endpoint
• of D0, and 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 using 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
The 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 (the constant of proportionality
  is linear in the degree of that polynomial bound).

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The upper bound Proof sketch

• Based on the fact that 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 involves a short relative translation at
  the k-th iteration,
• atk-1 NB(atk-1) is large.
• a ? A involves in an additional short relative
  translation
• a has to be translated further by
  atk-1 NB(atk-1) .
• Ik b-(atk) must decrease
  significantly.

b1
bj
bj1
Ik-1
a
a1
?tk
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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