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Re:Researchers achieve ‘absurdly fast’ algorithm for network... by tromp on Thursday June 09, 2022 @04:08AM (#31678724)
It means m^(1+eps) for arbitrarily small positive constant epsilon. So you can make it run in time O(m^1.1) but also in time O(m^1.00000001), where the latter will presumably have a larger constant implied by the O(). The technical reason you can make eps arbitrarily small is that it represent a function of m that tends to 0 as m tends to infinity.
Re:Researchers achieve ‘absurdly fast’ algorithm for network... by plopilop on Thursday June 09, 2022 @07:55AM (#31680061)
Mhh, while what you are saying is not wrong (though I'm not convinced about the constant thing, given everything is asymptotic), it's a bit imprecise.

A more exact formulation would be m^(1+o(1)) is equal to m^(1 + eps(m)) for some eps where eps(m) -> 0 for m -> infty