Clique homological simplification problem is NP-hard

Transcription

1 Clique homological simplification problem is NP-hard NingNing Peng Department of Mathematics, Wuhan University of Technology Foundations of Mathematics September 9, 2017

2 Outline 1 Homology and Persistent homology 2 The Clique homological simplification 3 CHS is NP-hard

3 Abstarct We consider the problem that input as a pair (G, H) of graphs and asks whether there exists a simplicial complex X which realizes the persistent homology of Cl(G) into Cl(H). We show that this problem is NP-hard.

4 Topological Data Analysis Input: Data (Finite point set S R d ). Assumption: Data was sampled from underlying space X. Goal: recover topology of X (Homology class). Use standard methods, such as Cell complex, Vietoris-Rips complex.. Be creative. Compute Persistent Homology. Software: Javaplex, Matlab, Perseus, GIC, etc. Workflow: Point cloud =) nested complexes =) persistent module =) barcode or diagram.

5 Definition Let V = {v 0, v 1,...,v m } be a non-empty finite set. An (abstract) n-simplex is a subset of V with = n + 1. A family K of subsets of V is an (abstract) simplicial complex if ; 62 K, andforany 2 K and any non-empty,wehave 2 K. The dimension of K is dim(k) =max{n : there is an n-simplex in K}.

6 What topological features does the following data exhibit? Discrete points have trivial topology.

7 What topological features does the following data exhibit? Discrete points have trivial topology.

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14 If the d is too small... Then, we detect noise.

15 If the d is too small... Then, we detect noise.

16 If the d is too small... Then, we detect noise.

17 If the d is too large...

18 If the d is too large...

19 If the d is too large...

20 If the d is too large...

21 Idea: Cell Complex ANSWER: Consider all distances.

22 Example: Creating a simplicial complex Step 0: Start by adding 0-dimensional vertices(0-simplices)

23 Example: Creating a simplicial complex Step 1: add 1-dimensional edges (1-simplices). Note: These edges must connect two vertices. I.e., the boundary of an edge is two vertices.

24 Example: Creating a simplicial complex Step 2: Add 2-dimensional triangles (2-simplices). Boundary of a triangle is a cycle consisting of 3 edges.

25 Example: Creating a simplicial complex Step 3: Add 3-dimensional tetrahedrons (3-simplices). Boundary of a 3-simplex is a cycle consisting of its four 2-dimensional faces.

26 Example: Creating a simplicial complex Step n: Addn-dimensional n-simplices, v 1, v 2,, v n+1.

27 Chain Complex Let K be a simplicial complex. For a ring R, then-chain group C n (K) of K over R is the free R-module whose basis consists of the oriented n-simplicies in K. Definition Let C n (K) bethen-chain group of K (over Z 2 ). The boundary n : C n (K)! C n 1 (K) is a homomorphism defined n ([v 0, v 1,, v n ]) = X i [v 0, v 1,, ˆv i,, v n ], where ˆv i indicates that v i is deleted from the sequence a. a If we want to n over a ring R in general, n n([v 0, v 1,, v n]) = X i ( 1) i [v 0, v 1,, ˆv i,, v n]. In the case of R = Z 2,since1+1=0,wemaynottake( to consider the orientation. 1) i and do not need

28 Chain Complex 0-chain [A] [B] [C]+[D]+[O] any linear comb of 0-simplice 0-chain group C 0.

29 Chain Complex 1-chain [B, C] [C, F ]+[M, N] [C, G]+[G, D]+[D, C] any linear comb of 1-simplice 1-chain group C 1.

30 Chain Complex 2-chain [J, I, K] [D, C, G]+[G, C, F ] [D, C, G]+[G, C, F ]+[F, E, H] any linear comb of 2-simplice 2-chain group C 2.

31 Simplicial Complex K-chain any linear comb of k-simplice K-chain group C k.

32 Cycle and Boundary K-Cycle and K-Boundary 1-cycle [C, E]+[E.F ]+[F, C] [C, F ]+[F, G]+[G, C] 1-boundary [C, F ]+[F, G]+[G, C] 1-cycle group: Z 1 1-boundary group: B 1

33 Homology Group H k = Z k /B k Z k : K-cycle group B k : K-boundary group H k : K-th homology classes. A set of K-th homology classes. Each homology by a non-bounded cycle [F, I ]+[I, J]+[J, G]+[G, F ] [C, E]+[E, F ]+[F, C]

34 Betti number Generators of H k indenpendent homology classes. H 0 s generators: {[A], [B], [O]}. H 1 s generators: {[C, E], [E, F ], [F, C], [F, I ]+ [I, J]+[J, G]+[G, F ]}. K-th Betti number k=rank(h k ) is the cardinality of generatores 0 = 3, cardinality of components 1 = 2, cardinality of holes

35 Filtration Filtration: Holes born and die: the life time of each hole. The persistent homology: hole as homology class, the life time of homoloy class.

36 Persistent Homology Recall H k : k-th homology classes i.e. the set of K-th homology classes Define H i,j k K 0 K 1 K 2 K n. A set of K-th homology classes such that They born at or before K i They are still alive at K j : Persistent Homology group for filtration

37 Let a filtered complex K have a filtration K: K 0 K 1... K M = K. For the i-th complex K i has associated boundary n i and groups C n (K i ), Z n (K i ), B n (K i )andh n (K i ). For i, j, n 2 N, thej-persistent n-th homology group Hn i,j (K) of K i is the image of the canonical induced homomorphism from H n (K i )toh n (K i+j ), that is, Hn i,j (K) ' Z n (K i )/(B n (K i+j ) \ Z n (K i i,j )). n = rank Hn i,j (K) isthe j-persistent nth Betti number of K i.

38 Persistent Homology Group

39 Barcode

40 The Clique homological simplification Homological simplification Let G =(V, E) be a simple graph, that is, a graph without loops and multiple edges, where V is the set of vertexes, and E is the set of edges. Here we call an abstract simplicial complex by a simplicial complex, for short. Definition 1 The clique complex C(G) of a graph G(V, E) isthesimplicial complex (G) on the vertex set V whose simplies are all cliques V,thatis, C(G) :={ V : 8x, y 2 (x 6= y! {x, y} 2 E)} 2 The independence complex I (G) is the simplicial complex whose faces are all the independent sets in G. WehaveCl(G) =I (Ḡ).

41 The Clique homological simplification Definition Let L K be two simplicial omplexes. The simplicial complex X is said to be a homological simplification of the pair (K, L) ifl X K and the maps j : H p (L)! H p (X ) and i : H p (X )! H p (K) induced by inclusions are respectively surjective and injective for all integers p > 0. We only consider homology groups over a field F (especially, set F = Z 2 ). Then, X is a homological simplification of the pair (K, L) if and only if X realizes the persistent homology of L into K.

42 CHS is NP-hard The problem: CHS Definition The problem CHS (short for clique homological simplification) takes as input a pair (G, H) of graphs and asks whether there exists a simplicial complex X which realizes the persistent homology of Cl(G) into Cl(H). The size of Cl(G) may not be polynomial of the size of G. ThenCHSmay not be NP.

43 CHS is NP-hard The problem is NP-hard Theorem The problem CHS and IHS are NP-hard. Since independent sets of G are the same as the cliques of Ḡ, weseethat I (G) is isomorphic to Cl(Ḡ) as an abstract simplicial complex. Then, we only show the CHS is NP-hard.

44 CHS is NP-hard We describe a reduction algorithm that transforms in linear time any instance of the 3SAT problem into an instance (Cl(G), Cl(H)) of the homological simplification problem in such a way that (Cl(G), Cl(H)) has a homological simplification if and only if E has a satisfying assignment.

45 CHS is NP-hard Lemma For every finite 2-dimensional simplicial complex K, we can find a graph G such that H (Cl(G)) ' H (K) in polynomial-time computation. Proof. Let G be a graph whose vertices are nonempty simplices of K and apair( 1, 2) of vertices is an edge if 1 2 or 2 1.DefineH as the same way for L. We can compute G and H in polynomial time of the size of input, because K and L are 2-dimensional. Then, since Cl(G) =K 0,wehaveH (Cl(G)) ' H (K).

46 CHS is NP-hard Proof of Main Theorem Outline of proof: reduction from 3-SAT 1 Given any instance of 3SAT, construct a simplicial pair (K, L) with trivial persistent homology group H (L! K). 2 Reconstruction: X with trivial homology and L X K. 3 The pair (K, L) has a homological simplification if and only if the formula is satisfiable.

47 CHS is NP-hard By the lemma, since the above K and L are at most 2-dimensional, we can compute G and H in polynomial time of the size of input such that H (Cl(G)) ' H (K) andh (Cl(H)) ' H (L). Then, Cl(G) andcl(h) have a homological simplification if and only if is satisfible.

48 CHS is NP-hard Afra Zomorodian and Gunnar Carlsson, Computing Persistent Homology Discrete and Computational Geometry, Volume 33, Issue 2, pp , Chen C, Freedman D. Hardness results for homology localization. Discrete and Computational Geometry, 2011, 45(3): Adamaszek M, Stacho J. Complexity of simplicial homology and independence complexes of chordal graphs. Computational Geometry, 2016, 57: Attali D, Lieutier A. Optimal reconstruction might be hard. Discrete and Computational Geometry, 2013, 49(2):

49 CHS is NP-hard Thank you very much!