Approximating the MST Weight in Sublinear Time Bernard Chazelle (Princeton) Ronitt Rubinfeld (NEC) Luca Trevisan (U.C. Berkeley)

Sublinear Time Algorithms Make sense for problems on very large data sets Go contrary to common intuition that “an algorithm must be given at least enough time to read all the input” Must be probabilistic Must be approximate

Approximation For decision problems: the output is the correct answer either for the given input, or at least for some other input “close” to it. (Property Testing) For optimization problems: the output is a number that is close to the cost of the optimal solution for the given input. (There is not enough time to construct a solution)

Previous Examples The cost of the max cut in a graph with n nodes and cn 2 edges can be approximated to within a factor  in time 2 poly(1/  c). (Goldreich, Goldwasser, Ron) Other results for “dense” instances of optimization problems, for low-rank approximation of matrices,... No results (that we know of) for problems on bounded-degree graphs.

Our Result Given a connected weighted graph G, with maximum degree d and with weights in the range {1,..., w}, we can compute the weight of the minimum spanning tree of G to within a factor of  in time O(dw  -2 log w/  ); we also prove that it is necessary to look at  dw  -2 ) entries in the representation of G. (We assume that G is represented using adjacency lists)

Main Intuition Suppose all weights are 1 or 2 Then the MST weight is equal to n – 2 + # of conn. comp. induced by weight-1 edges weight 1 weight 2 connected components Induced by weight-1 edges MST

Algorithm

Algorithm for weights in {1,2} To approximate the MST weight to within a multiplicative factor (1+  ) it’s enough to approximate c1 to within an additive factor  n (c1:= # of connected components induced by weight-1 edges) To approximate c1 we use ideas from Goldreich-Ron (property testing of connectivity) The algorithm runs in time O(d  -2 log  -1 )

Approximating # of connected components Given a graph G of max degree d with n nodes we want to compute c, the number of connected components of G up to an additive error  n. For every vertex u, define n u := 1 / size of component of u Then c =  u n u And if we call a u := max {n u,  } Then c =  u a u  n

Wrapping up the analysis Can estimate summation of a u using sampling Once we pick a vertex u at random, the value a u can be computed in time O(d/  ) We need to pick O(1/   ) vertices, so we get running time O(d/   )

Algorithm CC-APPROX(  ) Repeat O(1/  2 ) times pick a random vertex v do a BFS from v, stopping after 2/  steps b:= 1 / number of visited vertices return (average of the values b) * n

Improved Algorithm CC-APPROX( , W) Repeat O(1/  2 ) times pick a random vertex v do first step of a BSF from v b:=0; t:=1 (*) flip a coin If heads, and visited <W nodes so far t:=2*t continue BSF until ends or t nodes are visited if BSF ends, b:= 2 #random coins / nodes visited else go to (*) return (average of the values b) * n Inner procedure takes average O(dlog W) time

Analysis Main idea: if v is in a component of size c<W, then b is zero with prob. ~(1 – 1/c) and ~1 with probability ~1/c. The average of b is 1/c. Setting W:=2/  we get –each time, the average of b is within  /2 from the average over v of n v (that is, (# conn. comp.)/n) –Repeating O(1/  2 ) times, the probability of deviating by another factor  /2 is bounded by a constant –The average running time is O(d  -2 logW), that is O(d  -2 log  -1 ).

General Weights Generalize argument for weight 1 and 2. Let c i = # of connected components induced by edges of weight at most i Then the MST weight is n – w +  i=1,..., w-1 c i

Final Algorithm For j=1,..., w-1, call CC-APPROX( ,2w/  ) on the subgraph of G obtained by removing edges of cost >j Get a i, an approximation of c i Return n – w +  i=1,..., w-1 a i Average answer is within  n/2 from cost of MST, and variance is bounded Total running time O(dw  -2 log w/  )

Extensions Low average degree Non-integer weights

Lower Bound

Abstract sampling problem Fix p,  Define two binary distributions A,B Pr[A=1] = p, Pr[A=0]=1-p Pr[B=1] = p+  p, Pr[B=0]=1-p-  p Distinguishing A from B with constant probability requires  (1/p  2 ) samples

Reduction Fix p = 1/w We consider two distributions of weights over a cycle of length n In distribution G, for each edge we sample from A; if A=0 the edge gets weight 1, otherwise it gets weight w In distribution H, same with B H and G are likely to have MST costs that differ by about  n To distinguish them we need to look at  (w/  2 ) edge weights

Higher Degree Sample from G or H as before, –also add d-1 forward edges of weight w+1 from each vertex –randomly permute names of vertices Now, on average, reading t edge weights gives us t/d samples from A or B, so t=  (dw/  2 )

Conclusions A plausibility result showing that approximation for a standard graph problem in bounded degree (and sparse) graphs can be achieved in time independent of number of vertices Use of approximate cost without solution? More problems? –The trivial Max SAT approximation algorithms can be implemented in constant time, and give (an implicit representation of) a solution –Non-trivial Max SAT approximation? (say, 3/4) –Something really useful?