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avibryant's profile
Avi Bryant
Avi Bryant
Avi Bryant
@avibryant

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Avi Bryant

@avibryant

Happiest when working on Random Forests from random beaches.

Galiano Island
avibryant.com
Joined November 2006

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    1. Michael G. Noll‏ @miguno 14 Oct 2014

      @posco Is it known CMS behavior that ++ w.r.t. heavy hitters is kinda "broken" because for heavy hitters ++ is not associative?

      1 reply 0 retweets 0 likes
    2. P. Oscar Boykin‏ @posco 14 Oct 2014
      Replying to @miguno

      @miguno SketchMap's approach is not strictly associative, but the fraction approach (in CMS) should be. It is "approximately associative" :/

      3 replies 1 retweet 1 like
    3. Michael G. Noll‏ @miguno 14 Oct 2014
      Replying to @posco

      @posco For instance, if the 1st CMS counted German tweets, then reducing across all countries' CMS will return German-biased results (AFAIU)

      1 reply 0 retweets 0 likes
      Avi Bryant‏ @avibryant 14 Oct 2014
      Replying to @miguno

      @miguno @posco that seems wrong. It should be true that HH(A + B) is a strict subset of HH(A) + HH(B), right?

      10:48 AM - 14 Oct 2014
      5 replies 0 retweets 0 likes
        1. New conversation
        2. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @avibryant

          @avibryant @posco True. AFAIU the intersection is the problem as it's not associative. It introduces "left-biased" monoculture when reducing

          1 reply 0 retweets 0 likes
        3. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @miguno

          @avibryant @posco s/intersection/subset/

          0 replies 0 retweets 0 likes
        4. End of conversation
        1. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @avibryant

          @avibryant @posco CMS-"local" HH on "right" CMS's are less likely to become global HH when reducing.

          0 replies 0 retweets 0 likes
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        1. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @avibryant

          @avibryant @posco The problem is similar to micro vs macro average.

          0 replies 0 retweets 0 likes
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        1. New conversation
        2. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @avibryant

          @avibryant @posco For special case of 2 CMS you can trick by unioning both HHs, re-querying estimates from /combined/ CMS, then dropping HH

          1 reply 0 retweets 0 likes
        3. Avi Bryant‏ @avibryant 14 Oct 2014
          Replying to @miguno

          @miguno @posco right, that special case is exactly what I was imagining. Can't you just always use that trick?

          1 reply 0 retweets 0 likes
        4. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @avibryant

          @avibryant @posco Not in general, given the way the monad magic adds 1 CMS at a time. For the case of N CMS you must union all N HHs at once

          1 reply 0 retweets 0 likes
        5. Michael G. Noll‏ @miguno 14 Oct 2014
          Replying to @miguno

          @avibryant @posco And even then the problem is not yet solved. I'll try to summarize in a GH issue. Still hope I'm mistaken. ;-)

          1 reply 0 retweets 0 likes
        6. Avi Bryant‏ @avibryant 14 Oct 2014
          Replying to @miguno

          @miguno @posco looking forward to the issue, because so far I'm not getting it :)

          1 reply 0 retweets 0 likes
        7. Michael G. Noll‏ @miguno 15 Oct 2014
          Replying to @avibryant

          @avibryant @posco Summarized at https://github.com/twitter/algebird/issues/353 …

          1 reply 0 retweets 2 likes
        8. Avi Bryant‏ @avibryant 15 Oct 2014
          Replying to @miguno

          @miguno @posco aha, yep. The fundamental issue with topN was known, the bug in the top% implementation was not. Thanks for all the detail!

          2 replies 0 retweets 0 likes
        9. Michael G. Noll‏ @miguno 15 Oct 2014
          Replying to @avibryant

          @avibryant @posco Sure, np. Given this limitation of topN CMS, I begin to feel it should not be included in AB or only with a big warning :)

          0 replies 0 retweets 0 likes
        10. End of conversation
        1. Avi Bryant‏ @avibryant 14 Oct 2014
          Replying to @avibryant

          @miguno @posco s/strict//. But point is: for HH > N%, you should always be able to find HH(A+B) by just evaluating all of HH(A) and HH(B).

          0 replies 0 retweets 0 likes
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