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colmmacc's profile
Colm MacCárthaigh
Colm MacCárthaigh
Colm MacCárthaigh
@colmmacc

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Colm MacCárthaigh

@colmmacc

AWS, Apache, Crypto, Irish Music, Haiku, Photography

Seattle
notesfromthesound.com
Joined April 2008

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    1. Colm MacCárthaigh‏ @colmmacc 3 Jul 2019
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      In distributed systems, consensus protocols etc, it's easy to miss how varied network partitions can be. They don't have to be symmetrical, so A can see B, but B can't see A. In a 4 node setup, there are over 1200 potential partitions, and a flaky network can reproduce them all.pic.twitter.com/qZUiI6Vb2D

      4 replies 78 retweets 219 likes
    2. Hillel‏ @hillelogram 3 Jul 2019
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      Replying to @colmmacc

      Is it 2^{num edges}? I'm getting ~4000 partitions and wonder what I'm doing wrong :/

      2 replies 0 retweets 1 like
      Colm MacCárthaigh‏ @colmmacc 3 Jul 2019
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      Replying to @hillelogram

      I stopped counting at 1,296, but it can be up to 2,400. From node A's perspective - it can have visibility of BCD, or just BC, BD, CD, B, C, D - which is 7 partitions. The same can simultaneously happen (or not) from B, C, and D's perspective, so that gives us 7^4.

      10:06 AM - 3 Jul 2019
      • 4 Likes
      • Neil Eriksson Diego Magalhães ☁ Kieran Tully Hillel
      2 replies 0 retweets 4 likes
        1. New conversation
        2. Colm MacCárthaigh‏ @colmmacc 3 Jul 2019
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          Replying to @colmmacc @hillelogram

          I actually excluded the "node can see all the other nodes" cases, and so used 6^4, because that can be debatable, and so said "over 1200" for safety. Either way it's way more than people think of.

          2 replies 0 retweets 4 likes
        3. Balaji Raghavan‏ @BalajiRaghava13 3 Jul 2019
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          Replying to @colmmacc @hillelogram

          Shouldn't both node can see all other nodes and node can see none of the nodes be counted as well? Since even if A can see all the other nodes, B could potentially not see some of the nodes which we would want to count as a partition configuration.

          1 reply 0 retweets 2 likes
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        1. Kieran Tully‏ @KieranTully 3 Jul 2019
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          Replying to @colmmacc @hillelogram

          Thanks for posting this, I wasn't finding it simple until you laid out. However, did you forget the case where a node is fully isolated & can't see any other nodes? Making 8 possible partitions for each of 4 nodes, or 4096 total in the worst case?

          0 replies 0 retweets 1 like
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