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Testing the Diameter of Graphs Michal Parnas Dana Ron
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Property Testing of Graphs Let G = (V,E) be an undirected graph. A Testing Algorithm of property P: The algorithm can query on the incidence relations of vertices in G. If G has property P: Accept. If G is “far” from P: Reject.
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Previous Representations of Graphs Adjacency Matrix [GGR]. Queries: Is (u,v) E. -far: n 2 edges should be modified. Dense graphs. Incidence Lists of bounded length d [GR]. Queries: Who is i’th neighbor of v? -far: d n edges should be modified. Sparse bounded degree graphs. 1 1 1 10 0 00 0 1 n 1n1n 1 2 … d 1 n
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Our Model G = (V,E) an undirected graph, |V| = n, |E| m. Representation: Incidence lists of varying length. Queries: Who is i’th neighbor of v? -far: m edge modifications. 1 n
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A Testing Algorithm A testing algorithm for a parameterized property P s is given: Distance parameter 0 < < 1. Query access to a graph G having at most m edges. Boundary function ( ). The algorithm: Should accept with probability at least 2/3, if G has property P s. Should reject with probability at least 2/3, if G is -far from property. PsPs Accept Reject
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The Diameter Problem Question: Is the diameter of G at most D or is it -far from diameter (D)? The algorithms differ in: The boundary function ( ). The query and time complexities. The feasible values of .
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Our Results Time and Query Complexity: 1. 2.
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Related Work Testing Algebraic Properties (Linearity and Low degree) Program Testing: Blum & Luby & Rubinfeld, Rubinfeld, Rubinfeld & Sudan... PCP: Babai & Fortnow & Lund, Babai & Fortnow & Lund & Szegedy, Feige & Goldwasser, Lovasz & Safra & Szegedy, Arora & Lund & Safra, Arora & Safra... Testing Graph Properties (Colorability, Connectivity, Properties defined by first order formula) Goldreich & Goldwasser & Ron, Goldreich & Ron, Alon & Fischer & Krivelevich & Szegedy, Alon & Krivelevich. Testing Other Properties (Monotonicity, Regular languanges) Goldreich & Goldwasser & Lehman & Ron, Dodis & Goldreich & Lehman & Raskhodnikova & Ron & Samorodnitsky, Ergun & Kannan & Kumar & Rubinfeld & Viswanathan, Kearns & Ron, Alon & Krivelevich & Newman & Szegedy.
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Algorithm Input: D, n, m, . Parameters: C, k, . Set Uniformly select starting vertices. For each starting vertex - perform a BFS to distance at most C until k vertices are reached. If at most S starting vertices reach < k vertices then accept, otherwise reject. Time and Query Complexity: O(k 2 S)= O(k 2 / n,m )
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Illustration of the Algorithm 1 S C 2 3
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Proof of Correctness Good Vertex: If C-neighborhood contains k vertices. Bad Vertex: If C-neighborhood contains < k vertices. C We Show: Diameter D Almost (all) vertices are good. Diameter > (D) Many vertices are bad.
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Lemma 1: If at least (1-1/k)n of the vertices are good, then the graph can be transformed into a graph with diameter at most 4C+2 by adding at most 2n/k edges. Proof: c c c c Good Bad Reducing the Diameter
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Lemma 2: If at least (1- /2) n of the vertices are good, where k = (4/ )ln(4/ ), then the graph can be transformed into a graph with diameter at most 2C+2, by adding at most n edges. Proof: Select centers in a greedy manner and connect them. Balls may overlap. Corollary: If G is -far from diameter 2C+2, then there exist more than n n,m /2 bad vertices, where k= (4 / n,m ) ln(4 / n,m ). Proof: Set = n,m in Lemma 2.
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Proof of Item 1 Parameters: C = D = 0 Diameter D All vertices are good. Diameter > (D) = 2D+2 bad vertices. Lemma 2
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Proof of Item 2 Parameters: Diameter > 2C+2 = (D) = Lemma 2 bad vertices (fraction of bad 2 ).
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Item 2 - Continued Diameter D Fraction of bad vertices /2. Lemma 3: Let Diameter D number of bad vertices k i+1. Lemma 3
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Proof of Lemma 3 Assume there are more than k i+1 bad vertices. Diameter D D/2 - neighborhoods intersect. D/2 u1u1 utut u2u2 v u2u2 v utut u1u1 u 1,…,u t bad vertices, t = k i.
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Tree Lemma (Special Case) Let T be a tree of height h and size t. There exists a leaf in T whose 4h/3-neighborhood contains at least vertices. v i = 2, C = 2D/3. By Tree Lemma, there exists a leaf u j whose C-neighborhood contains at least vertices. u j is not bad. Proof: (Proof of Lemma 3)
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Tree Lemma Let T be a tree of height h and size t, and let a < h. There exists a leaf in T whose (h+a)-neighborhood contains at least vertices. Proof: Define - Solve recursively -
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-far For any fixed parameterized property P s, any 0 0, a graph G having at most m edges is -far from property P s if the number of edges that need to be added and/or removed from G in order to obtain a graph having the property, is greater than m. Otherwise, G is -close to P s.
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Range of Corollary: Every connected graph with n vertices and m edges is -close to having diameter D for every Theorem: Every connected graph on n vertices can be transformed into a graph of diameter at most D by adding at most 2n/(D-1) edges. Set:
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Reducing the Diameter of a Graph Lemma: If the C-neighborhood of each vertex contains k vertices, then the graph can be transformed into a graph with diameter at most 4C+2 by adding at most n/k edges. c c c c
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Our Results For (D) = 2D+2, and every , a one-sided error algorithm. For every (D) = 2 i log (D/2 + 1) a two-sided error algorithm, where = For Example: i = 2, (D) = 4D/3 + 2, = i = log(D/2 + 1), (D) = D + 4 Time and Query Complexity:
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