2014
DOI: 10.1017/s0963548314000200
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An Optimal Algorithm for Finding Frieze–Kannan Regular Partitions

Abstract: In this paper we prove that two local conditions involving the degrees and co-degrees in a graph can be used to determine whether a given vertex partition is Frieze–Kannan regular. With a more refined version of these two local conditions we provide a deterministic algorithm that obtains a Frieze–Kannan regular partition of any graphGin timeO(|V(G)|2).

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Cited by 10 publications
(23 citation statements)
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“…Recall that ω < 2.373 is the matrix multiplication exponent. In [9] they gave a different algorithm which improved the dependence of the running time on n from O ε (n ω+o (1) ) to O ε (n 2 ), while sacrificing the dependence of ε. Namely, it was shown that there is a deterministic algorithm that finds, in…”
Section: Algorithmic Weak Regularitymentioning
confidence: 99%
See 2 more Smart Citations
“…Recall that ω < 2.373 is the matrix multiplication exponent. In [9] they gave a different algorithm which improved the dependence of the running time on n from O ε (n ω+o (1) ) to O ε (n 2 ), while sacrificing the dependence of ε. Namely, it was shown that there is a deterministic algorithm that finds, in…”
Section: Algorithmic Weak Regularitymentioning
confidence: 99%
“…In [9], an alternative theorem is given for finding an irregular pair. The statement below is a consequence of [9, Theorem 8.1].…”
Section: Theorem 32mentioning
confidence: 99%
See 1 more Smart Citation
“…We apply Theorem 4 to G with approximation parameter ε, and obtain an equitable partition P : V = V 1 ∪ · · · ∪ V K on our vertex set with the desired properties such that 1/ε < K < 2 ε −c = 2 √ log log n . This can be done deterministically in n 2+o(1) -time using the algorithm of Dellamonica et al [15]. 2.…”
Section: Proof Of Theoremmentioning
confidence: 99%
“…either (1) correctly states that d (G, G ) ε, or (2) outputs sets S and T such that |e G (S, T ) − e G (S, T )| > ε −O (1) . In particular, the fact that G may not have bounded weights presents a challenge in applying results from [1,2].…”
mentioning
confidence: 99%