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Testing the Diameter of Graphs
Michal Parnas Dana Ron
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.
Previous Representations of Graphs
• Adjacency Matrix [GGR].
Queries: Is (u,v) E.
-far: n2 edges should be modified.
Dense graphs.
• Incidence Lists of bounded length d [GR].
Queries: Who is i’th neighbor of v?
-far: dn edges should be modified.
Sparse bounded degree graphs.
1
1
1
10
0
00
0
1 n
1
n
1 2 … d
1 2 … d
1
n
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
A Testing AlgorithmA testing algorithm for a parameterized property Ps 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 Ps.
• Should reject with probability at least 2/3, if G is -far from property . β(s)P
Ps
β(s)P
Accept
Reject
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 .
Our Results ( D ) R e m a r k s
2 D + 2 A n y O n e S i d e dE r r o r
2D12
11
i
m)2i(
nlogn 2i
11
T w o S i d e dE r r o r
4 D / 3 + 2 4/1n~ i = 2
D + 4 )n(logpoly/1f o r D = p o l y ( l o g n )
i = l o g ( D / 2 + 1 )
3ε
1O~
Time and Query Complexity:
1.
2.
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.
AlgorithmInput: 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.
mn,ε/1S
Time and Query Complexity: O(k2S)= O(k2/n,m)
εn
mε mn,
Illustration of the Algorithm
1
S
C2
3
Proof of Correctness
Good Vertex: If C-neighborhood contains k vertices.
Bad Vertex: If C-neighborhood contains < k vertices.
CWe Show:
• Diameter D Almost (all) vertices are good.
• Diameter > (D) Many vertices are bad.
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
cGood
Good
Good
GoodBad
Bad
BadBad
Reducing the Diameter
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 nn,m/2 bad vertices, where k= (4 /n,m)ln(4 /n,m).
Proof: Set = n,m in Lemma 2.
Proof of Item 1
Parameters: C = D = 0
• Diameter D All vertices are good.
• Diameter > (D) = 2D+2 bad vertices.
mn,mn, εε
4ln
4k
mn,ε
4S
n2
mn,εLemma 2
Proof of Item 2
Parameters:
22
D
2
DC
1i
mn,mn, εε
4ln
4k
4
εα mn,
mn,ε
48S
• Diameter > 2C+2 = (D) =
2D12
11
i
n2
mn,εLemma 2bad vertices (fraction of bad 2).
Item 2 - Continued
• Diameter D Fraction of bad vertices /2.
Lemma 3: Let
Diameter D number of bad vertices ki+1.
22
D
2
DC
1i
Lemma 3
Proof of Lemma 3
Assume there are more than ki+1 bad vertices.
Diameter D D/2 - neighborhoods intersect.
D/2
u1
ut
u2
v
u2
v
utu1
u1,…,ut bad vertices, t = ki.
2
Dh
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.tv
3
h
h3
2
i = 2, C = 2D/3. By Tree Lemma, there exists a leaf uj whose C-neighborhood containsat least vertices. uj is not bad.kkt i
Proof:
(Proof of Lemma 3)
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.1)a/)ahlog((
1
t
|)u(|maxmin)a,h,t(f ahleafuT
a2,ah,1b
1tf1bmin)a,h,t(f 1b
Proof: Define -
Solve recursively -
-farFor any fixed parameterized property Ps, any 0 < < 1, and any integer m > 0,a graph G having at most m edges is -far from property Ps
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 Ps.
Range of
Corollary:Every connected graph with n vertices and m edges
is -close to having diameter D for every Dm
n2ε
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.
εn
mε mn, Set:
Reducing the Diameter of a GraphLemma: If the C-neighborhood of each vertex contains k vertices, then the graph can be transformed into a graph withdiameter at most 4C+2 by adding at most n/k edges.
c
c
c
c
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
4/1n~
2D12
11
i
3
1O~
εTime and Query Complexity:
m)2i(
nlogn 2i
11