Some math about the Engineyard contest

Re:Some math about the Engineyard contest by jpwagner on Friday July 17, 2009 @11:55AM (#710142)

Rule 5 from the contest:

  "5.  In the case of a tied Hamming distance
      (entirely possible), the winner will be chosen
       by lottery among people with the best distance"

This says: don't take me seriously.

Hopefully for their sake they get 1000s of people running implementations and by chance someone will get 19 and win. They won't even have the best algorithm, but it will test permutations randomly.

I wrote up a solution, but I prefer scratch off tickets.

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Re:Some math about the Engineyard contest by JimmyL on Friday July 17, 2009 @12:33PM (#710216)

What would you suggest they do? Your follow-on indicates that you think ties should be adjudicated on the basis of the "best" algorithm, but how do you objectively define that?

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Re:Some math about the Engineyard contest by sophacles on Friday July 17, 2009 @12:36PM (#710223)

A simple criterion would be "capable of calculating the most results", which can be objective and meets some definitions of "best algorithm".

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Re:Some math about the Engineyard contest by antirez on Friday July 17, 2009 @02:24PM (#710433)

A simpler and less biased way IMHO is that if two teams will end with a result of, for instance, 30, but one team was able to find two or more strings, this is the winner.

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Re:Some math about the Engineyard contest by jpwagner on Friday July 17, 2009 @12:35PM (#710221)

Here are some thoughts:

First one to hit Hamming distance of 5.

Lowest average Hamming distance over top n trials.

Most trials in x hours.

...

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Re:Some math about the Engineyard contest by JimmyL on Friday July 17, 2009 @12:53PM (#710251)

Something like throughput in trials on some standardized machine could be an interesting criteria - it would test both your hashing algorithm and your permutations.

Having said that, I'm betting that the EY people chose this particular problem since highly-optimized algorithms are easily available to all - meaning that the amount of HW one can throw at the problem (and luck) will likely be the determining factor in who wins.

Which HW should someone use? Well, EY has a cloud computing program, and it's right there....

That is, I think, the point of this contest. Make a problem that's hardware-intensive, so that people will try deploying it on the EY cloud platform and get them into the proverbial front door of the product family. They're not interested in novel algorithms; they're interested in getting people onto their platform, and getting them used to it.

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Re:Some math about the Engineyard contest by grandalf on Friday July 17, 2009 @02:46PM (#710480)

Uh I don't think EY offers a cloud solution that would make this easy... so I am doubtful that was their goal.

It seems that their goal (by having defined a specific dictionary) was to have lots of people tweeting those dictionary words to @engineyard.... sort of a way to get better twitter-driven pagerank.

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Re:Some math about the Engineyard contest by JimmyL on Friday July 17, 2009 @03:24PM (#710573)

Interesting...I think you may be right.

I originally though this - http://www.engineyard.com/private-cloud - was a cloud environment a la EC2 based on the title (which would work with my theory), but reading more about it makes it seem like it's still optimized for Rails, and hence not for this.

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Re:Some math about the Engineyard contest by grandalf on Friday July 17, 2009 @10:41PM (#711274)

Yeah, one of EY's services actually is semi-managed EC2 and they advertise it as such.

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Re:Some math about the Engineyard contest by preview on Friday July 17, 2009 @05:20PM (#710821)

That limits the allowed solutions to those that are purely software. What about a hardware/software hybrid or hardware-only solution (e.g. PCI Express card with one or more FPGAs)?

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