1. On the ICP Algorithm Esther Ezra, Micha Sharir Alon Efrat

2. The problem: Pattern Matching Input: A = {a 1, …, a m }, B = {b 1, …, b n } A,B  R d Goal: Translate A by a vector t  R d s.t.  (A+t,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 t i-1 Assign each point a+t i-1  A+t i-1 to its NN b = N B (a+t i-1 ) Compute the relative translation  t i that minimizes  with respect to this fixed (frozen) NN-assignment. Translate the points of A by  t i (to be aligned to B). t i  t i-1 +  t i. Stop when the value of  does not decrease. The overall translation vector previously computed. t 0 = 0. Some of the points of A may acquire new NN in B.

4. The algorithm: Convergence [Besl & Mckay 92]. 1.The value of  =  2 decreases at each iteration. 2.The algorithm monotonically converges to a local minimum. This applies for  =   as well. a1a1 a2a2 a3a3 b1b1 b2b2 b3b3 b4b4 A local minimum of the ICP algorithm. The global minimum is attained when a 1, a 2, a 3 are aligned on top of b 2, b 3, b 4.

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 R 1  (A+t,B) = RMS(t) = Iteration 1: b 1 = 0 b 2 = 4 a 3 = 1a 4 = 3 a 1 = -3-   t 1  1 m=4, n=2. (b 1 +b 2 )/2 = 2 a 2 = -1  (A,B)  3 Solving the equation: RMS’(t)=0

8. An example in R 1 : Cont. Iteration 2: b 1 = 0 b 2 = 4 a 4 +  t 1  t 2 = 1 (b 1 +b 2 )/2 = 2 a 3 +  t 1 a 2 +  t 1 a1+t1a1+t1 Iteration 3: b 1 = 0 b 2 = 4 NN-assignment does not change.  t 3 = 0 (b 1 +b 2 )/2 = 2 a 1 +  t 1 +  t 2 a 2 +  t 1 +  t 2 a 3 +  t 1 +  t 2 a 4 +  t 1 +  t 2  (A+  t 1,B)  2  (A+  t 1 +  t 2,B)  1

9. Number of iterations: An upper bound The value of  is reduced at each iteration, since 1.The relative translation  t minimizes  with respect to the frozen NN-assignment. 2.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 R 1 Critical event: The NN of some point a i is changed. Two “consecutive” NN-assignments differ by one pair (a i,b j ). #NN-assignments  |{(a i,b j ) | a i  A, b j  B}| = nm. b1b1 a2a2 amam a1a1 b2b2 bnbn aiai bjbj b j+1 tt a i crosses into a new Voronoi cell.

11. The bound in R 1 Is the quadratic upper bound tight??? A linear lower bound construction (n=2):  (m) b 1 = 0 a m-1 = 2k-1a m = 2k+1a 1 = -2k-1 (b 1 +b 2 )/2 = 2k Tight! (when n=2). b 2 = 4k a 2 = -2k+1 Using induction on i=1,…, k/2. m=2k+2, n=2

12. Structural properties in R 1 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  t i  0 for each i  0, or  t i  0 for each i  0. Use induction on i.

13. A super-linear lower bound construction in R 1. b 1 = 0 b 2 = 1 m=n The construction (m=n): a 1 = -n -  (n-1), a i = (i-1)/n -1/2 + , b i = i-1, for i=1, …, n. Theorem: The number of iterations of the ICP algorithm under the above construction is  (n log n).  1 = 1/2

14. b 2 = 1 b 3 = 2 b 4 = 3 b 2 = 1 Iteration 2: Iteration 3: At iteration 4, the NN of a 2 remains unchanged (b 3 ): Only n-2 points of A cross into V(b 4 ).

15. The lower bound construction in R 1 Round j: n-1 right pts of A assigned to b n-j+1 and b n-j+2 Consists of steps where pts of A cross the bisector  n-j+1 = (b n-j+1 + b n-j+2 ) / 2 Then at each such step i, # pts of A that cross  n-j+1 is exactly j (except for the last step). At the last such step, # pts of A that cross  n-j+1 or  n-j+2 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 R 1 The overall number of steps:

17. The lower bound construction in R 1 : Proof of the claim Induction on j: Consider the last step i of round j : b n-j+2 b n-j+3 b n-j+1 The points a 2, …, a l+1 cross  n-j+1, and a n-(j-l-2), …, a n cross  n-j+2. Overall number of crossing points = l + j - l -1 = j - 1. Black pts still remain in V(b n-j+2 ). l< j pts have not yet crossed  n-j+1 QED

18. General structural properties in R d Theorem: Let A = {a 1, …, a m }, B = {b 1, …, b n } be two point sets in R d. Then the overall number of NN-assignments, over all translated copies of A, is O(m d n d ). Worst-case tight. Proof: Critical event: The NN of a i +t changes from b to b’ t lies on the common boundary of V(b–a i ) and V(b’–a i ) Voronoi cells of the shifted diagram V(B- a i ). t b - a i b’ - a i b’’ - a i

19. # NN Assignments in R d Critical translations t for a fixed a i = Cell boundaries in V(B-a i ). All critical translations = Union of all Voronoi edges = Cell boundaries in the overlay M(A,B) of V(B-a 1 ),…, V(B-a m ). # NN-assignments = # cells in M(A,B) Each cell of M(A,B) consists of translations with a common NN-assignment

20. # NN Assignments in R d Claim: The overall number of cells of M(A,B) is O(m d n d ). Proof: Charge each cell of M(A,B) to some vertex v of it (v is charged at most 2 d times). v arises as the vertex in the overlay M d of only d diagrams. The complexity of M d is O(n d ) [Koltun & Sharir 05]. There are d-tuples of diagrams V(B-a 1 ),…, V(B-a d ). QED

21. Lower bound construction for M(A,B) B d=2 A 1/m 1 Horizontal cells: Minkowski sums of horizontal cells of V(B) and (reflected) vertical points of A. Vertical cells: Minkowski sums of vertical cells of V(B) and (reflected) horizontal points of A. M(A,B) contains  (nm) horizontal cells that intersect  (nm) vertical cells:  (n 2 m 2 ) cells.

22. # NN Assignments in R d 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? d=1: Upper bound: O(n 2 ). Lower bound:  (n log n). Probably no.

23. Monotonicity d=1: 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. 

24. Monotonicity Theorem: Let  t be a move of the ICP algorithm from translation t 0 to t 0 +  t. Then RMS(t 0 +   t) is a strictly decreasing function of .  does not intersect itself. Cost function monotone decreasing along  t. tt

25. Proof of the theorem S B-a (t) = min b  B ||a+t-b|| 2 = min b  B (||t|| 2 + 2t·(a-b) + ||a-b|| 2 ) RMS(t) = S B-a (t) - ||t|| 2 is the lower envelope of n hyperplanes S B-a (t) - ||t|| 2 is the boundary of a concave polyhedron. Q(t) = RMS(t) - ||t|| 2 is the average of {S B-a (t) - ||t|| 2 } a  A Q(t) is the boundary of a concave polyhedron. The Voronoi surface, whose minimization diagram is V(B-a). The average of m Voronoi surfaces S(B-a).

26. Proof of the theorem NN-assignment at t 0 defines a facet f(t) of Q(t), which contains the point (t 0,Q(t 0 )). Replace f(t) by the hyperplane h(t) containing it. h(t) corresponds to the frozen NN-assignment. The value of h(t)+||t 2 || along  t is monotone decreasing Since Q(t) is concave, and h(t) is tangent to Q(t) at t 0, the value of Q(t)+||t 2 || along  t decreases faster than the value of h(t)+||t 2 ||. QED t0t0 tt

27. More structural properties of  Corollary: The angle between any two consecutive edges of  is obtuse. Proof: Let  t k,  t k+1 be two consecutive edges of . Claim:  t k+1   t k  0  b’ = N B (a+t k ) a b = N B (a+t k-1 ) tktk a must cross the bisector of b, b’. b’ lies further ahead QED

28. Cauchy-Schwarz inequality More structural properties: Potential Lemma: At each iteration i  1 of the algorithm, (i) (ii) Corollary: Let  t 1, …,  t k be the relative translations computed by the algorithm. Then Due to (i) Due to (ii)

29. More structural properties: Potential Corollary: d=1: Trivial d  2: Given that ||  t i ||  , for each i  1, then Since tktk  does not get too close to itself. 

30. The problem under the Hausdorf measure

31. The relative translation Lemma: Let D i-1 be the smallest enclosing ball of {a+t i-1 – N B (a+t i -1) | a  A}. Then  t i moves the center of D i-1 to the origin. Proof: tt o a1a1 b1b1 a2a2 b2b2 a3a3 b3b3 a4a4 b4b4 a 1 -b 1 a 2 -b 2 a 4 -b 4 a 3 -b 3 a 1 -b 1 +  t a 2 -b 2 +  t a 4 -b 4 +  t a 3 -b 3 +  t o The radius of D determines the (frozen) cost after the translation. Any infinitesimal move of D increases the (frozen) cost. D D QED

32. No monotonicity ( No monotonicity (d  2) Lemma: In dimension d  2 the cost function does not necessarily decrease along  t. Proof: a1a1 a2a2 a3a3 b b’ tt r c tt a2+ta2+t a3+ta3+t a1+ta1+t b Initially, ||a 1 -b|| = max ||a i -b|| > r, i=1,2,3. Final distances: ||a 2 +  t-b|| = ||a 3 +  t-b|| = r, || a 1 +  t-b’ || < r. The distances of a 2, a 3 from b start increasing Distance of a 1 from its NN always decreases. QED

33. The one-dimensional problem Lemma (Monotonicity): The algorithm always moves A in the same (left or right) direction. Proof: Let |a*-b*|=max a  A |N B (a)-a|. Suppose a* < b*. a*-b* is the left endpoint of D 0 ; its center c < 0.  t 1 > 0, |  t 1 | < |a*-b*|. a*+  t 1 < b*, so b* is still the NN of a*+  t 1. |a*+  t 1 -b*|=max a  A |a+  t 1 -N B (a)| > |a+  t 1 -N B (a+  t 1 )| a*+  t 1 -b* is the left endpoint of D 1. The lemma follows by induction. a*-b* tt D0D0 0c a–N B (a) Initial minimum enclosing “ball”. 0 a*+  t 1 -b*a+  t 1 –N B (a) a*, b* still determine the next relative translation. QED

34. Number of iterations: An upper bound Theorem: Let  B be the spread of B. Then the number of iterations that the ICP algorithm executes is O((m+n) log  B / log n). Corollary: The number of iterations is O(m+n) when  B is polynomial in n Ratio between the diameter of B and the distance between its closest pair.

35. The upper bound: Proof sketch Use: the pair a*, b* that satisfies |a*-b*|=max a  A |N B (a)-a| always determines the left endpoint of D i. Put I k = |b*-(a*+t k )|. Classify each relative translation  t k Short if Long – otherwise. Overall: O(n log  B / log n). Easy

36. The upper bound: Proof sketch Short relative translations: a  A involved in a short relative translation at the k-th iteration, |a+t k-1 – N B (a+t k-1 )| is large. a  A involved in an additional short relative translation a has to be moved further by |a+t k-1 – N B (a+t k-1 )|. I k = b*-(a*+t k ) must decrease significantly. b1b1 bjbj b j+1 a tktk I k-1 a1a1

37. Number of iterations: A lower bound construction At the i-th iteration: 1.  t i = 1/2 i, for i=1,…,n-2. 2.Only a i+1 crosses into the next cell V(b i+2 ). The overall translation length is < 1. Theorem: The number of iterations of the ICP algorithm under the following construction ( m=n) is  (n). a1a1 a2a2 anan b1b1 b2b2 bnbn n n-1 aiai bibi

38. 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…