1. Testing Metric Properties Michal Parnas and Dana Ron

2. Property Testing (Informal Definition) For a fixed property P and any object O, determine whether O has property P, or whether O is far from having property P (i.e., far from any other object having P ). Task should be performed by querying the object (in as few places as possible). ?? ? ? ?

3. Property Testing - Background Initially defined by Rubinfeld and Sudan in the context of Program Testing (of algebraic functions). Goldreich Goldwasser and Ron initiated study of testing properties of (undirected) graphs. Growing body of work deals with properties of functions, graphs, strings, sets of points... Many algorithms with complexity that is sub-linear in (or even independent of) size of object.

4. Motivation Computational: Design testing algorithms that are (much) more efficient than exact decision algorithms for properties. Combinatorial: Gain new understanding about tested property.

5. Testing Metric Properties P - Metric property ; M - n x n rational-valued matrix;   - Distance/approximation parameter; M is said to be -far from property P if must modify more than  fraction of n 2 entries so that M obtains P. Otherwise say that it is -close. Testing algorithm can query M on entries M[i,j]. If M has property P, should accept; If M is  -far from property P, should reject w.p.  2/3.

6. Tree Metrics and Ultametrics An n x n matrix M is a tree metric (additive metric) if exists a tree T with positive weights on edges, such that: There exists a mapping  from [n] into nodes of T; For every i,j[n]={1,…,n}, T(  (i),  (j))=M[i,j]; All nodes to which no i[n] is mapped to, have degree greater than 2. If: T is rooted,  maps only to leaves of T, and distance of all leaves to root is the same, then M is an ultrametric.

7. 7 5 3 2 5 3 5 4 1 53 67 4 2 M[1,2]=8; M[1,3]=12; M[1,4]=10; M[1,5]=15;... Tree Metric 123456 444 3 3 2 2 11 M[1,2]=M[1,3]=M[2,3]=8; M[1,4]=M[1,5]=M[1,6]=12; M[4,5]=M[4,6]=6; M[5,6]=2;... Ultrametric

8. Our Results Can test ultrametrics with |S|= O(log(1/  )/   ). Can test general tree metrics with |S|=O(log(1/  )/   ). Can extend result for ultrametrics to approximate ultrametrics. Can test d-dimensional Euclidean metrics with |S|=O(d log d/  ). Our algorithms all work by taking uniformly selected sample S  [n] and querying M[i,j] for i,j  S. Size of sample is always poly(1/  ) and independent of n. Specifically:

9. Our Results (continued) Testing algorithms can be used to solve relaxed versions of corresponding search problems in time linear in n (and polynomial in 1/  ). That is, can construct tree that agrees with M on all but at most  -fraction of entries. (Note that running time is sub-linear in size of matrix M.)

10. Constructing an Ultrametric Tree Suppose M is an ultrametric. We can construct an ultrametric tree that agrees with M on given subset {1,…,s} in following manner: Initialization: Position points 1 and 2 at equal distance M[1,2]/2 from root node. Iterations: For each point j = 3,…,s add point j to current tree by adding new branch that emits from j’s unique point of departure from tree. This point is determined by closest point in tree.

11. M[1,2]=8; M[1,3]=M[1,4]=M[1,5]=10; M[2,3]=M[2,4]=M[2,5]=10; M[3,4]=2; M[3,5]=6; M[4,5]=6; 12 44 3 1 5 2 3 4 5 11 1

12. Consistency of points with tree For U  [n], let T U denote tree with leaf-set U, that agrees with M on U (if exists, such tree is unique). Def: Say that j  [n] \ U is consistent with T U if adding j to T U as described in construction procedure, results in tree that agrees with M on U+j. Denote set of points consistent with U by  U.

13. The “Scaffold Partition” For U  [n], let T U denote tree with leaf-set U, that agrees with M on U. We refer to tree as scaffold. Def: Let P U be following partition of  U, induced by T U : Points i and j are in same class i.f.f have same point of departure from T U.

14. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2 The scaffold partition

15. Violating Pairs If M is an ultrametric, then for every subset U, and for every two points i,j that belong to different classes in P U, value of M[i,j] is exactly determined by corresponding (different) departure points in T U. Def: Say that i,j   U that belong to different classes in P U are a violating pair w.r.t. T U if distance between them according to scaffold T U differs from M[i,j].

16. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2 If M is ultrametric, must have M[i,j]=8. ji 32

17. Two types of “witnesses” Suppose have scaffold tree T U that agrees with M on U. (If can’t construct such tree, clearly M not ultrametric.) It follows that: If obtain point j that is inconsistent with T U then have witness that M not ultrametric. If obtain pair of points i,j that are violating w.r.t. T U then have witness that M not ultrametric.

18. Testing Algorithm for Ultrametrics 1. Uniformly select s=O(log(1/  )/  3 ) points from [n]. Denote set by U. 2. Construct tree T U that agrees with M on U. If fail, reject. 3. Uniformly select m=O(1/  ) pairs of points from [n]. 4. If any of these 2m points is inconsistent with T U, or any of the m pairs is violating w.r.t. T U, then reject. 5. If no step cause rejection then accept.

19. Analysis of Algorithm  If M is ultrametric -- Algorithm always accepts. (No inconsistent points and no violating pairs.)  From now on assume M is  -far from ultrametric. Will show that algorithm rejects w.h.p. Specifically: Either can’t construct T U that agrees with M; or many inconsistent points w.r.t. T U ; or many violating pairs w.r.t. T U ;

20. Special Case (for M  -far from ultrametric) Suppose T U agrees with M, and all but at most (  /3)n 2 pairs of points in  U belong to different classes in P U (are separated). (In particular is the case if all classes of size O(  n).) Claim: Either have > (  /3)n inconsistent points w.r.t. T U or have > (  /3)n 2 violating pairs w.r.t T U. Subject to claim, if M is  -far from ultrametric, then rejected w.h.p. as required.

21. Proof of Claim for special case Assume, contrary to claim, that have  (  /3)n inconsistent points, and  (  /3)n 2 violating pairs. Will show that  ultrametric tree T that agrees with M on all but at most  n 2 entries, in contradiction to assumption on M. Tree T builds on scaffold T U : For every class C in P U create star-shaped sub-tree with leaf set C that is rooted at point of departure of C from T U. Inconsistent points are added arbitrarily. By premise of lemma and (counter) assumptions, num of disagreements  (  /3)n. n + (  /3)n 2 + (  /3)n 2 =  n 2. incon. pts viol. Pairs unsep. pairs

22. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2

23. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2

24. General Case By special case: Gain from separating points to diff classes. Def: Say that point kU is effective separator w.r.t. T U if adding k to U causes  (  n/12) 2 pairs of points to be separated into different classes. k C1C1 C4C4 C3C3 C2C2 C 1,2 C 1, 1

25. General Case By special case: Gain from separating points to diff classes. Def: Say that point kU is effective separator w.r.t. T U if adding k to U causes  (  n/12) 2 pairs of points to be separated into different classes. k C4C4 C3C3 C2C2 C 1,2 C 1, 1

26. General Case (continued) In analysis, view sample U as being selected in phases. In each phase, if  many effective separators then one selected w.h.p. After sufficient num of phases, either have special case (few non-separated pairs), or U s.t. have few effective separators w.r.t. T U. In latter case can show that  class C in P U,  tree T C s.t. for almost all pairs i,jC, M[i,j]= T C (i,j). (Tree is star-shaped/broom-shaped.)

27. General Case (continued) Claim: Either have > (  /4)n inconsistent points w.r.t. T U or have > (  /4)n 2 violating pairs w.r.t T U. Subject to claim, if M is  -far from ultrametric, then rejected w.h.p. as required. Proof of Claim is similar to that in special case: Assume few inconsistent points and violating pairs, show that  tree close to M (contradicting M being  -far from ultrametric).

28. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2

29. 1 3 1 1 1 3 2 1 1 2 2 C1C1 C4C4 C3C3 C2C2

30. Solving Relaxed version of Search Problem Analysis implies that testing algorithm can be used to solve relaxed version of corresponding search problem. That is, if M is ultrametric then, w.h.p. can construct tree that agrees with M on all but at most  -fraction of entries in time linear in n and polynomial in 1/  : Construct scaffold T U on uniformly selected sample U; Partition all points in [n]\U into classes of P U according to distances to points in U; For each class C construct star/broom-shaped tree T C.

31. Testing Approximate Ultrametrics Def: For a given approximation parameter , we say that matrix M is a  -approximate ultrametric if exists ultrametric M’ s.t. for every i,j [n], |M[i,j]-M’[i,j]|  . We describe an algorithm, that for every  and , if M is a  –approximate ultrametric then algorithm accepts M, and if M is  –far from being a c  –approximate ultrametric then algorithm rejects M w.h.p. (c is a fixed constant).

32. Conclusions and Further Research Presented algorithm for testing whether matrix is an ultrametric or far from being an ultrametric. Analysis implies fast solution for relaxed search problem. Mentioned similar results for approximate ultrametrics, general tree metrics and Euclidean metrics. We suspect that results can be improved in terms of dependence on 1/ . We conjecture that can extend result for general tree metrics to approximate variant. Testing other natural metric properties?