On Testing Convexity and Submodularity

On Testing Convexity and Submodularity Michal Parnas Dana Ron Ronitt Rubinfeld

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

Examples • The object can be a graph (represented by its adjacency matrix), and the property can be 3-colorabilty. • The object can be a string and the property can be membership in given regular language. • The object can be a function and the property can be linearity.

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 offunctions, graphs, strings, sets of points ... Many algorithms with complexity that is sub-linear in (or even independent of) size of object.

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

Submodular and Supermodular Functions Let I = I1I2  …  Id, d2 be a product space where Iq . In particular consider Iq={0,…,nq}. For x,y Idefine the join of x and y as:(x1,…,xd)  (y1,…,yd) = (min(x1,y1),…,min(xd,yd )and the meet as:(x1,…,xd) (y1,…,yd) = (max(x1,y1),…,max(xd,yd ) Definition:A function f : I is submodular iffor every x,y I we have f(x y) + f(x  y)  f(x)+f(y) . If inequality holds in opposite direction then f is supermodular.

vi’,j vi’,j’ Vi+1,j Vi+1,j+1 Vi,j Vi,j+1 vi’,j - vi,j  vi’,j’ - vi,j’ vi,j vi,j’ Monge Matrices - (Important)Special Case of Submodular Functions: d=2 When d=2 represent function by 2-dimensional matrix V={vi,j} , i,j = 0 to n. Matrix is a Monge matrix if for every 0  i < i’  n , 0  j < j’  n , have vi,j+vi’,j’  vi,j’+vi’,j.If inequality holds in opposite direction then Inverse Monge. Suffice that inequality holds for every i,j and i’=i+1, j’=j+1.

Why are we interested in such functions (matrices)? Submodular functions and specifically Monge matrices, play an important role in combinatorial optimization. For many optimization problems (e.g., transportation problems, searching and selecting, TSP) if the input is a matrix having the Monge (submodular) property, then they can be solved more efficiently.

Testing Inverse Monge Matrices • Testing algorithm is given distance parameter and query access to matrix V (that is, can probe matrix). • If V is Inverse Monge then should accept (w.p. 1). • If V is -far from any Inverse Monge matrix (that is should modify more than an -fraction of its entries to become Monge) then should reject w.p. at least 2/3. We give an algorithm whose query and time complexity is O(log2n/) .

Characterization of I-Monge Matrices Given matrix V, define matrix CV’ = {ci,j} , 0  i,j  n : CV’ V vi,j= kimj ck,m ci,j= (vi,j-vi-1,j) - (vi,j-1-vi-1,j-1) ci,0=vi,0-vi-1,0 c0,0=v0,0 c0,j=v0,j-v0,j-1 Claim1: For every 0  i,j  n , vi,j= kimj ck,m. Claim2: Let CV = {ci,j} , i,j = 1 to n. The matrix V is an I-Monge matrix i.f.f. CV is non-negative.

CV (i’,j) (i’,j’) (i,j) (i,j’) (Bad) Candidate Testing Algorithm Given characterization, natural suggestion is to sample from CV and reject if see negative entry. Doesn’t quite work:V may be far from I-Monge but CV contains only single negative entry. However, in this case there are many sub-matrices of CV with negative sum of elements. Observe: for any sub-matrix of CV , can compute sum of elements given access to V in constant time. Sum of elements:vi’,j’-vi-1,j’ – vi’,j-1 +vi-1,j-1

Testing Algorithm - Definitions Definition1: A sub-row of CV is called legal if it results from bisecting its row in a recursive manner. A legalsub-column is defined analogously. A sub-matrix is legal if both its rows and its columns are legal.

(i’,j) (i’,j’) Sum of elements in M:vi’,j’-vi-1,j’ – vi’,j-1 +vi-1,j-1 M (i,j) (i,j’) Testing Algorithm - Definitions Cont’ Definition2:A sub-matrixM of CV is calledgood if the sum of all its elements is non-negative. Otherwise it is bad. Definition3:We say that (i,j) is goodif all legal sub-matrices containing it are good. Otherwise it is bad.

Testing Algorithm Test I-Monge * Uniformly and independently select 8/ entries in CV;* for each entry selected, check that it is good. * If all are good then accept, o.w., reject. Running time of Alg is O(log2n/) (since each entry is containedin log2n legal sub-matrices). If V is I-Monge then Alg always accepts (1-sided error). We show that if V is accepted w.p. > 1/3 then it is -close to I-Monge.

Analysis of Testing Algorithm If V accepted w.p. > 1/3 then C=CV contains less than/4 fraction of bad entries. We show that can modify C to obtain non-negative matrix C’={c’i,j} for which the following holds: The corresponding I-Monge matrix V’={v’i,j} (that is v’i,j= kimjc’k,m), is -close to V.

Analysis of Testing Algorithm Cont’ R is union of good sub-matrices that cover all bad ones. R is at most e-fraction of matrix. CV v’i,j= vi,j Can modify at most all entries of R inC=CVto non-negative entries obtaining C’ so that corresponding V’ does not differ from V outside of R. R

Conclusions and Further Research • Give polylog algorithm for testing Monge Matrices. • Give logarithmic time algorithm for testing convexity in 1-d. • Lower Bounds? • Extending result to higher dimensions (that is d>2)? (possibly can be done very similarly in special case of distribution matrices.) • What happens to optimization algorithms if matrix is only close to Monge?

Analysis of Testing Algorithm I If V accepted w.p. > 1/3 then C=CV contains less than/4 fraction of bad entries. Can modify C to obtain non-negative matrix C’={c’i,j} for which the following holds: The corresponding I-Monge matrix V’={v’i,j} (that is v’i,j= i’ij’jc’i’,j’), is -close to V. For each bad entry (i,j), consider maximal bad legal sub-matrix M, that contains (i,j). Due to maximality, M is contained in good (legal square) submatrix M’ that is 4 times larger. Consider union Rof all such good sub-matrices. R“covers” all bad entries in C and its total area is at most an -fraction of C. Our goal: Modify only entries of C that belong to R in order to obtain non-negative C’ s.t. corresponding V’ differs from V only in entries (i,j) in R.

Analysis of Testing Algorithm I Cont’ Region R is union of disjoint square submatrices where each is good. That is, for each of these sub-matrices M, every row and every column is non-negative. We shall “re-fill” M so that all its entries become non-negativeand the sum of entries in each of its rows and columns remains the same. It follows that for every (i,j) outside of R, the value v’i,j (which is the sum of entries c’i’,j’ “below’’ it) is the same as the original value vi,j . Hence, analysis reduced to following task: given non-negative values a1,…,ar and b1,…,bs such that ai= bj, fill r  s matrix with non-negative entries so that the sum of i’th row is ai and the sum of the j’th column is bj . Task can be easily performed by a greedy algorithm. 

Polylog Testing Algorithm: Sketch of Analysis As in analysis of Alg I, if V accepted w.p. > 1/3 then C=CV contains less than/4 bad’ entries. Here too consider unionR of good’ sub-matrices containing maximal bad’ legal sub-matrices. R as bounded size as before. Main difference, R is no longer nice union of disjoint square sub-matrices with non-negative rows and columns. Rather, R is some general “creature” with less stringent constraints on rows and columns. Still, can show how to fillR with non-negative entries so that for every (i,j) not in R, sum of new entries below it is the same as the sum of old entries. Do so in an iterative manner, each time “breaking” off a “block” from R, filling it, and modifying constraints on sums of rows and columns in what remains of R.

Our Starting Point: Combinatorial Auctions and Submodular Set Functions In Combinatorial Auctions several different items are sold concurrently, and bidders are interested in purchasing subsets of items. Each bidder specifies a valuation functionv which assigns each subset of items S a price v(S). Typically assumed that v obeys certain properties. Specifically, that v is a submodularset function. That is: for every two subsets S,T :v(S T) + v(ST)  v(S) + v(T) .

On Testing Convexity and Submodularity