Property Testing n Bounded Degree Graphs

Testing K-Edge-Connectivity in Bounded Degree Graphs

Lecture by: Aviad Lipiner 1Property Testing in Bounded Degree Graphswritten by: Oded Goldreich, Dana RonGiven a predetermined property and a graph we want to distinguish between the 2 cases:1)The graph has the property2) The graph is far from having the property2CorrectnessComplexityOur algorithms work in time polynomial in

Accuracy Algorithm always accepts the graph when it has the tested property.Rejects with high probability ( ) if the graph is - far from having the property

3ModelWe will view bounded-degree graphs as functions defined over pairs (v,i), where v is a vertex and i is a positive integer , .

Meaning the value of (v,i) is the neighbour of v (with a special symbol indicating non-existance of such a neighbour).

4DefinitionsGraphs In this specific case we are discussing undirected graphs of bounded degree d.G=(V,E)V={1,,|V|} , and that for every vertex the edges incident to v have distinct labels in {1,,d}. This labeling may be arbitrary and need not be consistent among neighboring vertices. We associate a function f to the Graph.

That is f(v,i)=u if (u,v) is the edge to v. f(v,i)=0 if there is no such edge.

5Distance Our measure of distance between graphs depends on their degree bound (d).

Meaning the distance between 2 graphs G1 and G2 with degree bound d, Where V(G1) = V(G2) = [N], is:

6So that for every 2 graphs . This notation of distance can naturally be extended to a set C, of N-vertex graphs with degree bound d.

We let represent the class of graphs with N vertices and degree bound d which have property . In case is empty we define dist(G, ) to be 1 for every G.

7 K-Edge-ConnectivityLet be an integer. A graph is said to be k-edge-connected if there are k edge-disjoined paths between every pair of vertices in the graph.

Equivalent definition The sub-graph resulting by omitting any k-1 edges from the graph , is connected.

Assume that

8Testing connectivity (k=1)Here we will assume ( since otherwise except for the graph cannot be connected).

Lemma 2 Let - If a graph G is from the class of N-vertex connected graphs with maximum degree d, then it has more than connected components.

9Proof Assume contrary to the claim that G has no more than connected components.Let be the connected components of G.We will differ between the following 2 cases:1) sum of degrees in each is as most .2)For some connected component , , the sum of degrees is greater than .

10Case 1 Each contains at least 2 vertices of degree d-1 or at least one vertex of degree d-2 at most.If both have a degree of d-1 (or and their degree is d-2).We connect toHere we made the graph connected by adding edges.

This contradicts that G is from the class of connected graphs with degree d.

11Case 2- We cannot add edges without violating the degree bound

. Let Ti be an arbitrary spanning tree of Ci. Ti contains at least 2 vertices so it has at least 2 leaves. At most one of Cis vertices has a degree less than d.Thus the tree Ti has a leaf which has degree in G.This leaf has an edge in Ci which is not in Ti.We can remove this edge from G without disconnecting Ci, now we have 2 vertices with degree less than d.Do the same as in Case 1. We have removed/added at most edges.

We again reached a contradictions and the claim follows.

12Corrollary 3 If a graph from the class of N-vertex connected graphs of degree bound , then G has at least connected components each containing less than vertices. Proof: According to lemma 2, G has at least connected components. The number of connected componenst with at least vertices is at most . So the remaining ones are at least in number.

13From Lemma 2 for every , every graph is - close to the class of connected graphs with degree bound d.So we may assume that .Since every connected component contains at least one vertex we conclude that if G is -far from the class of connected graphs that the probability that a uniformly selected vertex belongs to a connected component with less than vertices is at least .

If we uniformly select vertices, the probability that no selected vertex belongs to a component of size less than is bounded above by

14

Algorithm 4 (input: d, )Here we assume that (otherwise step 2 will reject).

1) Uniformly and independently select vertices in the graph2) For each vertex s selected perform a BFS starting from s until vertices have been reached or no more new vertices can be reached (a small component has been found)3) If any of the above searches finds a small connected component then output REJECT, otherwise output ACCEPT

15The algorithm never rejects a connected graph since a connected graph consists of a single component.By corollary 3 , if a graph is from connected then it is rejected with probability at least .

The query complexity and running time of the algorithm are .

Note that the choice to perform a BFS is quite arbitrary, any other linear-time searching method will do.

16Testing Connectivity for K>1CombinatoricsA subset is said to be k-edge-connected if there are k-disjoint paths between each pair of vertices in S. If , these paths may go through vertices not in S.A subset containing a singe vertex is defined as k-edge-connected.k-edge connected classes - of graph G are maximal subsets of V which are k-edge connected and each vertex in V resides in exactly one such class.

17We will show tests according to the following

Assuming that the graphs we test for are (k-1)-connected.Without any prior assumptions18An auxiliary tree of a (k-1) connected graphIn Y.Dinitz and J. Westbrooks article Maintaining the classes of 4-edge-connecticity in a graph on-line they state the following Fact that we will use without proving

Fact A.1 Let k>1 be an integer and G be a (k-1)-connected graph. Then there exists an auxiliary graph , TG, which is a tree that:

Each k-connected class in G corresponds to a unique node in TG.

In addition to nodes corresponding to k-connected classes, there are two types of nodes: empty nodes, and cycle nodes.

All leaves of the auxiliary tree TG correspond to k-connected classes of G. Furthermore, there are exactly k-1 edges (in G) going out from each of these classes.19TG Auxiliary graphIf G is a (k-1) connected graph. Then we can define an auxiliary graph TG , which is a tree, such that for every k-class in G there is a corresponding node in TG.TG might have addition auxiliary nodes but they are not leaves and will not interest us here. If G is k-connected TG has a singe node.Otherwise TG has at least 2 leaves (leaves play a key role in our algorithm)Each leaf corresponds to a k-class C of G which is separated from the rest of the graph by a cut of size k-1.We will later show that for every leaf class C, given a vertex we can efficiently identify that v belongs to a leaf class.

20Lemma A.4In Y.Dinitz and J. Westbrooks article Maintaining the classes of 4-edge-connecticity in a graph on-line they lay the foundations for the following lemma which we will not prove:

Let G be a (k-1)-connected graph, whose auxiliary graph, TG, has L leaves. Then by removing and adding at most 4L edges to G we can transform it into a k-connected graph G

If the maximum degree of G is d then the maximum degree of G is upper bounded by max{d,k}. 21Lemma 7 Let G be a (k-1) connected graph that is from the class of k-connected graphs with maximum degree . Then TG has at least leaves.

Proof:Assume towards contradiction that TG has leaves. Then by Lemma A.4 G can be transformed into a k-connected graph G by removing and adding at most edges. The maximum degree of G is max(k,d)=d. This contradicts that G is from the class of k-connected graphs with maximum degree d.

22Corollary 8 Let G be a (k-1) connected graph that is from the class of k-connected graphs with maximum degree . Then TG has at least leaves each containing at most vertices.

23Algorithm 9 (input: d, ) We will assume that the number of vertices N in G is greater than .1) Uniformly and independently select vertices. 2) For each vertex s selected , check whether s belongs to a k-class leaf which has at most vertices.

3) If any leaf class is discovered then output REJECT, otherwise output ACCEPT.

24 The procedure for checking whether a given vertex belongs to a small k-class leaf always return the correct answer if the vertex does not belong to such a leaf.

If the vertex belongs to such a leaf a correct answer is returned with probability at least

25If a graph is -far from being k-connected there are 2 sources for probability of being wrongly accepted:1) By corrollary 8 The probability that no vertex - s belonging to a small k-class leaf is selected in step 1 is at most

2) The probability that the procedure for identifying a k-class leaf fails given such a vertex

Total at most error probability

26Identifying a k-class leafThe following algorithm indentifies k-class leaves 27Algorithm 15 (input: s, n)Given a vertex s and size bound n1) Do times or until a cut of size less than k is found:

- Random search process: Starting with {s}, maintain the set (S) of vertices visited. In each step as long as |S|k.

2) For every vertex u such that deg(u)>d, all of us neighbours have degree k.48Let G be a (k-1)-connected graph, and TG be its auxiliary tree. If we augment G by an edge with endpoints in the k-connected classes C1 and C2.

Then the classes on the simple path between C1 and C2 in TG form a k-connected class.Lemma A.249Let G be a (k-1)-connected graph, and TG be its auxiliary tree. C1 and C2 two k-connected classes.

If we omit a single edge from each Ci and add 2 edges. If (u1,v1) and (u2,v2) were omitted from C1 and C2 we add (u1,u2) and (v1,v2)

Then the classes on the simple path between C1 and C2 in TG form a k-connected class.Lemma A.350Case 1. We remove the edge (u,v) from the graph. If the graph remains k-connected, no additional modification. Else by Lemma A.2- The auxiliary tree of the graph consists of a simple path, with u belonging to one k-class leaf and v to the other. v now has degree k and is not a singleton leaf. u is not a singleton leaf which has degree at least Now we can apply Lemma A.3 on the 2 leaf-k-classes. This will give a k-connected graph at the cost of 5 edge modifications.

51Here we must consider 2 sub-cases

Case 2.a There exist 2 vertices u1 and u2 , so that their degrees are larger than d and all their neighbors have degree k.Then there must be 2 vertices such that v1 is a neighbor of u1 and v2 is a neighbor of u2.We add an edge between v1 and v2 , increasing their degree to k+1.We then apply Case 1 twice : to the edges (u1,v1), (u2,v2).

We have decreased the excess of the graph by 2 at a cost of 11 edge modifications. Case 2.

52Case 2.b There is a single vertex , u , with a degree greater than d .

If deg(u)>d+1 we will remove at least 2 edges beside u. Let v1 and v2 be 2 neighbors of u.we now add an edge between v1 and v2 and then apply case 1 to (u,v1) and then (u,v2). We have decreased the excess of the graph by 2 making 11 edge modificationsIf deg(u)=d+1 the we will choose v to be any neighbor of u (we will call this neighbor w), deg(w) < d ** . We ass an edge between v and w raising the degree of v to k+1.We apply Case 1 to (u,v)

Cost of 6 edge modifications. 53Now we slightly modify Algorithm 9 so that in step 2 we look for a small set of vertices which is separated from the rest of the graph by a cut of size j (j