Exact Algorithms for 2-Clustering with Size Constraints in the Euclidean Plane

Transcription

1 Exact Algorithms for 2-Clustering with Size Constraints in the Euclidean Plane Alberto Bertoni, Massimiliano Goldwurm, Jianyi Lin Dipartimento di Informatica Università degli Studi di Milano SOFSEM Pec pod Sněžkou, Czech Republic - January 24 29, 2015, 41st Int. Conf. on Current Trends in Theory and Practice of Computer Science

2 Introduction Clustering Problem Instance X R d, X = n, m N, 1 < m < n Solution {A 1, A 2,..., A m } partition of X optimal (criterion: weight, variance,...)

3 Introduction Application areas: image analysis bioinformatics unsupervised learning pattern recognition data mining, statistical data analysis Traditional approaches: heuristics (e.g. K-Means) approximate solutions unknown or exponential worst-case computation time Goals of our research: exact solutions complexity results (e.g. NP-hardness) efficient algorithms in the easy cases (polynomial time) size-constraints version of the problem

4 Problems Formal definitions (dimension d, l p norm, p 1) given X = {x 1,...,x n } R d, cluster : A X, A centroid : C A = argmin µ R d a µ p p a A weight of A: W p (A) = a A a C A p p weight of a partition {A 1,..., A m } of X W p (A 1,..., A m ) = W p (A 1 )+W p (A 2 )+ +W p (A m )

5 Problems Clustering Problem (p 1) Istance : X = {x 1,..., x n } R d, X = n m N, 1 < m < n Solution : m-partition {A 1,..., A m } of X with minimum W p (A 1,...,A m ) Parameters: p, d, m, n Known results: - NP-hard for m = 2 and arbitrary d (p = 2) [Aloise et al. 09] - NP-hard for d = 2 and arbitrary m (p = 2) [Dasgupta 07, Mahajan-et-al.09] - p = 2 : Minimum Sum of Squares Clustering [Aloise, Hansen 09] - k-means Heuristic [Lloyd 57, MacQueen 67, Vattani 09]

6 Problems Clustering Problems with Constraints Constraint types: must-link, cannot-link, diameter, size [Wagstaff-Cardie 00, Bradley-et-al.00, Zhu-et-al.10] Size-Constrained-Clustering(p) (SCC) Instance : X = {x 1,..., x n } R d, X = n integers k 1, k 2,..., k m, i k i = n Solution : m-partition {A 1,..., A m } of X with A i = k i with i = 1,...,m and minimum W p (A 1,...,A m ) Variants: m-scc : fixed m SCC-d : fixed d m-scc-d : both m and d fixed

7 Previous Results Complexity Results 1) p > 1 2-SCC is NP-hard [BGLS 12] arbitrary d HALF-PARTITION reduction from MINIMUM-BISECTION it holds p > 1 2) p 1 SCC-1 is NP-hard [Sacca 10] arbitrary m reduction from 3-PARTITION Notice: Clustering(d = 1, p = 2) FP 3) SCC-2(p = 2, k i {2, 3}) is NP-hard [Lin 13]

8 Previous Results 4) p Q\N s.t. Centroid localization is difficult Problem p-centroid Localization (p-cl) Instance : X = {x 1,...,x n } N, h N Question : C X > h? Problem SQRT-Sum [Garey-Johnson76, Allender-et-al 06] Instance : a 1,..., a r, b 1,...,b s N Question : a1 + + a r > b b s? SQRT-Sum NP? 3 = SQRT-Sum P -LC [Saccà 10, BGLS12] 2

9 Previous Results Tractable cases Necessary conditions : fixed p N, d e m 1) 2-SCC-1 (uniform unary p N) FP [BGLS12] 2) p N +, d N + 2-SCC-d FP [Lin 13] 3) In R 2 with Manhattan norm (d = 2, p = 1) [BGLP14, Pini 14] 2-SCC-2 Problem with p = 1 on input (X, k), X = n Plain Algorithm T(n, k) = O(n 2 log n) (Simpler) Full Algorithm ( k = 1,...) T(n) = O(n 3 log n)

10 Algorithms for 2-SCC-2 with p = 2 2-SCC-2 Problem (p = 2) Instance: X = {x 1,..., x n } R 2, X = n k {1,..., n/2 } Solution: 2-partition {A, B} of X with A = k s.t. W 2 (A, B) = a A a C A b B b C B 2 2 is minimum Main results : - Plain Algorithm in T(n, k) = O(n 3 k log 2 n) time - Full Algorithm ( k = 1,..., n/2 ) in T(n) = O(n 2 log n) time (both working in O(n) space)

11 Ingredients : 1) separation result (straight line) [BGLS12] 2) number of k-sets in the plane O(n 3 k) [Erdös-et-al73, Dey98] 3) dynamic data structure for convex hull insert, delete operations O(log 2 n) [Overmars-van Leeuwen81]

12 Separation Result [BGLS12] Hypotheses: p > 1 {A, B} is optimal solution of 2-SCC-d with A = k Thesis: c R s.t. u R d : u A implies u C A p p u C B p p < c u B implies u C A p p u C B p p > c = x C A p p x C B p p = c (x R d ) is a hypersurface separating A and B. { the hypersurface is a straight line = If d = 2 = p then A is a k-set 2x(C Bx C Ax )+2y(C By C Ay ) = c + C B 2 C A 2

13 Idea of the plain algorithm: 1) build an initial k-set A, compute W 2 (A, Ā), the convex-hulls Conv(A) and Conv(Ā), a bitangent(a, b), 2) S := A, W := W 2 (A, Ā) Conv(Ā) (a, b) b a Conv(A)

14 Conv(Ā) Conv(A) (a, b) a b Conv(Ā ) Conv(A ) (a, b ) (a, b) a b

15 NextBitangent iteration (0) Conv(Ā ) v = a v u u = b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

16 NextBitangent iteration (1) Conv(Ā ) v = a v u u b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

17 NextBitangent iteration (2) Conv(Ā ) a v v u u b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

18 NextBitangent iteration (3) Conv(Ā ) a v v u u b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

19 NextBitangent iteration (4) Conv(Ā ) a v v u u b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

20 NextBitangent iteration (5) Conv(Ā ) a v u u v b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

21 NextBitangent iteration (6) Conv(Ā ) a v u u v b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

22 NextBitangent iteration (7) Conv(Ā ) a (a, b ) v = b u = a b Conv(A ) Procedure NextBitangent(a, b) A := Insert(Delete(A, a),b) Ā := Insert(Delete(Ā, b),a) u := b; u := Succ(u) v := a; v := Succ(v) repeat if (v, v )Left(u, v) then u := u ; u :=Succ(u) elseif (u, u )Right(u, v) elseif (u, u )Right(v, v ) else u := u ; u :=Succ(u) until (u, u )Right(u, v) (v, v )Left(u, v) return (u, v)

23 Loop-exit condition: E = (u, u )Right(u, v) (v, v )Left(u, v) Three possible enter cases ( E) : (u, v) (u, v) v 1) (v, v )Left(u, v) = Move(u) u u v 2) (u, u )Right(u, v) = Move(v) u v v u

24 3) (u, u )Left(u, v) (v, v )Right(u, v) v (u, v) v (u, v) v v u u u u 3a) (u, u )Right(v, v ) = Move(v) 3b) (u, u )Left(v, v ) = Move(u)

25 Full Algorithm (idea): 1) for k = 1, 2,...,n/2 do { sk := w k = + 2) (q 1,..., q n ) := Sort y (X) (w.r.t. y-coordinate < y ) 3) for i = 1, 2,..., n do A := {q X : q y q i }; k := A W(A, Ā) := weight of (A, Ā) if W(A, Ā) < w k then Update(s k, w k ) { sk := (A, Ā) w k = W(A, Ā) l := horizontal line through q i repeat turn l counter-clockwise around q i till next point q in X update A (insert or delete q) update k = A and W(A, Ā) if W(A, Ā) < w k then Update(s k, w k ) until A= initial cluster 4) return s k for all k

26 Conclusions Further current work - Manhattan norm: 2-SCC-2 under l 1 (p = 1) solvable in O(n 2 log n) time [BGLP14] - Extension to higher dimensions for all d N + 2-SCC-d FP ( p N + ) [Lin 13] - Extension to larger number of clusters (m > 2) - Relaxing number of clusters M-RCC-d = Clustering Problem in dimension d, arbitrary partition size m, size of clusters in M = {2, 3}-RCC-2 is NP-hard (p = 2) [Lin 13] THANKS FOR YOUR ATTENTION