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mrdmnd

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Matthew Redmond - Data Engineer, Linear Algebraicist, and Computational Geometer. Executor all sorts of shenanigans at Google X.

Views expressed in HN comments are my own.

mttrdmnd@gmail.com

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Yeah, I think this is the goal - remember; there are some problems that only need to be solved correctly once! Imagine something like a millennium problem - you'd be willing to wait a pretty long time for a proof of the RH!

Location: San Francisco, CA

Remote: Yes

Willing to relocate: No

Technologies: C++, Rust, Python, Java, Scala, Lua, PostGIS, SQL, all the standard geospatial data science stuff

Résumé/CV: https://drive.google.com/file/d/1-Zsa5C7fQKJmj7YgUJlRa4b_V_9...

Email: mttrdmnd@gmail.com

--

Mid-career jack-of-all trades problem solver with a decade of software engineering, data science, and analytics experience. Very experienced working with large geospatial datasets, planning, simulation, and control policy work. Looking for roles involving physical science, chemistry, energy, or other fun science and engineering problems.

Location: San Francisco Bay Area

Remote: Yes, flexible.

Willing to relocate: no

Technologies: C, C++, Python, Scala, Mathematica, Matlab

Resume: Contact me directly

Email: mttrdmnd@gmail.com

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Interested in research grade problems. MIT-trained computational geometer. Current interests include earth science prediction problems, robotics, and geometric optimization.

Most recently worked at Google X as balloon systems simulation engineer. Contact me for some wild stories about airspace ;)

As someone who just did this internally:

Do it. It's worth your time. Very well paced exercises, and it walks you through the flow quite nicely.

Interesting that the optimal solution is not "slide it along the space diagonal" but rather "slide it parallel to a face".

I remember a classic MIT problem set in computer science where we reduced the problem of finding a "profitable arbitrage loop" in a directed graph to a shortest path problem: imagine a graph where nodes are currencies and edges are exchange rates: we're seeking to find some loop of nodes U,V,W,...U such that the cumulative product of the edge weights is greater than unity.

The trick is to transform the graph by taking the negative log of the edge weights, which turns problem of finding a cumulative product > 1 into one of finding a negative sum loop. Then, you can just run the Bellman Ford algorithm and if it detects a negative cost cycle in the transformed graph, this corresponds to a positive arbitrage cycle in the original graph.

I always thought that was a neat application.