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vgr's profile
Venkatesh Rao
Venkatesh Rao
Venkatesh Rao
@vgr

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Venkatesh Rao

@vgr

Conversational account. For work follow @ribbonfarm, @breaking_smart, @artofgig. Tweets are 90% vacuous views, apathetically held. Mediocritopian. IKEA builder.

Los Angeles, CA
venkateshrao.com
Joined August 2007

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    1. Venkatesh Rao‏ @vgr 28 Sep 2019
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      Is there an easy algorithm to a) count the number of equivalent shortest paths on a manhattan grid b) enumerate them systematically? I wrote a dumb random algo and I could run it for a long time to count unique paths but I imagine there's a closed-form solution?pic.twitter.com/Moj8hm9NcM

      7 replies 3 retweets 9 likes
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      Venkatesh Rao‏ @vgr 28 Sep 2019
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      I'm actually interested in a second-order problem: count the fraction of paths that go through a particular segment. If you walked to work on a manhattan grid everyday, making random choices, how long before you walked a specific block? A specific sequence of blocks?

      12:09 PM - 28 Sep 2019
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      • Lindsay Clark https://thisrentaldoesnotexist.com/ Nick Parker James Tinged Words Bushra Farooqui 🎄🌉 Pirates sleep soundly (for Santa) Iris van Rooij
      5 replies 1 retweet 6 likes
        1. New conversation
        2. \u221e‏ @hdevalence 28 Sep 2019
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          Replying to @vgr

          i think this decomposes as the product of two full sub-blocks and reduces to the earlier problem.

          1 reply 0 retweets 2 likes
        3. \u221e‏ @hdevalence 28 Sep 2019
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          Replying to @hdevalence @vgr

          the number of shortest paths through the highlighted segment is the product of the number of shortest paths to its entrance and the number of shortest paths from its exitpic.twitter.com/lNAhxcvFUA

          0 replies 0 retweets 1 like
        4. End of conversation
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        2. Bob Kerns (No PhD, but I know what one is, WSJ)‏ @BobKerns 28 Sep 2019
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          Replying to @vgr

          Subdivide the problem. Path from start to segment start. Path from segment end to end.

          2 replies 0 retweets 2 likes
        3. Venkatesh Rao‏ @vgr 28 Sep 2019
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          Replying to @BobKerns

          makes sense, thanks

          0 replies 0 retweets 0 likes
        4. End of conversation
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        2. Maynard Handley‏ @handleym99 28 Sep 2019
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          Replying to @vgr

          Before you can get any further, you have to define "making random choices"... 90% of the confusion thinking about stochastic problems boils down to not specifying the EXACT measure of interest.

          1 reply 0 retweets 0 likes
        3. Maynard Handley‏ @handleym99 28 Sep 2019
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          Replying to @handleym99 @vgr

          For at least some versions of "making random choices" this will collapse to Gaussians (because everything sane does...), and cylinder sets on Gaussians (ie paths through a value range) are fairly easily calculated.

          0 replies 0 retweets 0 likes
        4. End of conversation
        1. jonathan‏ @jonathanvoelkle 29 Sep 2019
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          Replying to @vgr

          By now you know that there are (M+N)! /(M! N!) paths from A to B. Every path that does not pass the box, will pass exactly one crossed point (see drawing), call it C. You can count the paths from A to B through C using the formula, take the sum over all such point C, et voilàpic.twitter.com/Kajmvyh4ap

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        2. twisted metal vibes‏ @skullpile5 29 Sep 2019
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          Replying to @vgr

          The stationary distribution of an unweighted random walk on a graph is proportional to the degree of each node. Specifically the probability of being in node i is the degree of the node divided by twice the number of edges in the graph.

          1 reply 0 retweets 1 like
        3. twisted metal vibes‏ @skullpile5 29 Sep 2019
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          Replying to @skullpile5 @vgr

          The visitation frequency of a node is exactly the inverse of that.

          1 reply 0 retweets 1 like
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