MATLAB Programming Contest Blog

November 11th, 2010

The polls are closed and the results are in: contest veteran Nick Howe (who has ruled the Dark more than once) is our newest Prince of Darkness. His code “Equatorial” came in at half the score of his nearest competitor. Congratulations Nick, and well done!

I wanted to point out a change about timing in the rules. We listed the start and stop times for the contest as being 16:00 UTC… but we were off by one. The times are keyed to noon here in Natick, and that is in fact 17:00 UTC. Nick Howe would have won the Darkness prize with either a 16:00 UTC or a 17:00 UTC cut-off, so it had no effect today, but please note the change going forward.

By Ned Gulley

Ned is part of the MATLAB Central team.

17:23 UTC | Posted in Sailing Home | Permalink | 1 Comment »

You can follow any responses to this entry through the RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed.

One Response to “Nick Howe is our Prince of Darkness”

1. Alan Chalker replied on November 11th, 2010 at 19:09 UTC :
   

   I have been able to figure out the scoring formula and am posting it here as I traditionally do. As usual, it’s very similar to the recent contests:

   score = k1*result + k2*e(k3*runtime) + k4*max(complexity-10,0) + k5*nodes

   Where:

   k1 = 0.2
   k2 = 0.01
   k3 = 1/11 (0.090909)
   k4 = 1
   k5 = 0.001

   It’s a little premature to do a leading entry analysis, since we are still in Twilight, but I’ll give it a shot:

   The current leading entry has a time of 25s, result of 8085, cyc of 40, and nodes of 2446. Here’s a breakdown of the current tradeoffs:

   -cyc and score are a 1:1 ratio (i.e. each point shaved off cyc is a point shaved off the score)
   -time and score are a 1:0.1 ratio
   -result and score are a 1:0.2 ratio
   -node and score are a 1:0.001 ratio

   The entry isn’t anywhere close to the ‘knee’ of the time exponential curve, which is rather flat until about ~60s.

Leave a Reply

<pre class="code">
a = magic(3);
sum(a)
</pre>

These postings are the author's and don't necessarily represent the opinions of The MathWorks.