Helen

(Sometimes I think that this procedure is part of the contest rules.)

As of midnight Tuesday, the leading entry has the following stats:
Result: 134617
Time: 40.0336
Cyc: 21
Nodes: 7289
Score: 13508.84

Because of the way the scoring formula works, the ONLY way to beat 13000 is to lower the result. Even if the time, cyc and nodes were reduced to 1 each, the score would only improve by 45 points to 13464. In order to reach 13000, the result needs to improve to ~129500, or by approximately 5117 (which is a 3.8% improvement).

That might not seem like much, but it actually is. Based upon the null move solvers submitted at the start of the contest, we know that there are a total of 359141 points in the original test suite. The current result is only 37% of the max value of the test boards.

However, keep in mind that there is a non-zero lower limit to the score for a ‘perfect solver’. While it’s impossible for us to know what that is without examining the test suite, we can make some good guesses based upon the best results statistics. The best result so far is 133225, almost 1400 points better than the current leader (although it took 148 seconds to run), but still 3700 points away from 13000, assuming it could run in less than 1/3 the time it currently does.

Another way of looking at it is the % improvement column on the stats webpage. The score has only improved by ~4% since Sunday morning until right now. Thus in order to break the goal, in the next 12 hours entries need to outperform ~72 hours worth of steady improvements.

Examining the current leading solver against random individual boards in the sample test suite, I found that there might be room for slight improvement, however overall it’s doing a really good job, as one might expect.

Thus, while I’m sorry if this bursts anyone’s bubbles, hopefully this will allow some competitors to refocus on just getting a top entry instead of trying to beat an arbitrary score. Good luck everyone for the remainder of the contest!

The code is entry 47616 viewable at:
http://www.mathworks.com/contest/wiring.cgi/view_submission.html?id=47616

Hope this helps!

[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 23 9 21 35 0 0 0 0 0
0 0 0 0 0 9 0 0 0 0 9 0 23 27 76 0 0 0 0
0 0 0 0 0 0 27 0 0 0 44 0 0 23 0 49 49 58 0
0 0 0 0 23 0 0 23 0 0 27 23 35 23 0 0 0 0 0
0 0 0 0 0 0 0 76 0 0 0 0 58 49 0 0 0 0 0
0 0 0 0 0 0 0 0 0 35 0 0 9 0 0 0 0 0 0
0 0 0 0 0 0 0 35 0 0 0 21 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 49 0 23 58 0 0 0 0 0
0 0 0 0 0 49 35 21 0 23 0 0 23 0 0 0 0 0 0
0 0 0 0 0 49 0 23 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 49 21 18 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]

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

Where:

k1 = 0.1
k2 = 2
k3 = 2/30 (0.06666666…)
k4 = 1
k5 = 0.001

The current leading entry has a time of 56s, result of 141891, cyc of 25, and nodes of 4196. Here’s a breakdown of the current tradoffs:

-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 1:5.7 ratio
-result and score are a 1:0.1 ratio
-node and score are a 1:0.001 ratio

David Jones entries have already settled in just at the ‘knee’ of the time exponential curve, which is rather flat until about ~50s. However, because of the new time exponent constant, we are going to see much more payoff in this content in focusing on reducing the execution time versus other scoring elements, probably down to the ~10 second range.

Unfortunately that probably also means that people are going to end up taking the lead due to ‘luck of the draw’ in the minor variations we always see in execution times, since they will be amplified more in the total score compared to in the past.

Hope this helps everyone!

Remove the connectors for (49) from [6,4] to [11,7] and reconnect the [11,7] (49) straight down until it T’s into the existing 49-connectors. Reuses the one bridge across the (35,35) pair to (23,23) pair but via few connectors.

How practical is this challenge problem? Wouldn’t there be a previous step that asks are the parts on the circuit layed out in such a manner that the minimum cost wiring is a consideration?

:-)
Sam I Am

1. Bridges can only be used for crossovers, they can’t be
used to have two 90 angles at the same coordinate.
2. You can not bridge pins, if I am not wrong you’ll pay
for them but they’ll be ignored.
3. The example that described the cost of the bridge should
give a net score of 25+3-(11*2) = 6
4. The size of the boards (as well as other parameters in
the testsuite) are sampled from a distribution, therefore
theoretically speaking there is no limit, but in practice
you could expect that most likely the distributions of the
board sizes wil resemble those in the sample testsuite.
5. “T” connections and “+” connections are possible and you
do not pay extra but the number of wires used (3 and 4
respectively)
6. Darkness does not mean “do not answer questions”, I’ll be around for a while, let me know if you have questions.

Good luck,

P.S. As orginizer I get the chance to peek at some of your
entries :), we have started receiving some interesting
pieces of code.

Lucio
The MathWorks Contest Team

Is the net scorce not 6? (-22+28)

Thanks Andy!
Helen

The segments for the connector between the 8 pins could be written like so.
w = [ 2 3 2 4 ]
[ 2 4 3 4 ]
[ 3 4 3 5 ]
[ 3 5 4 5 ]
[ 4 5 5 6 ]

Emmanual - Great to hear that you are getting ready for the contest! An important part of the contest is the time constraints and the different phases (darkness, twilight, daylight). To practice, look at some of the past contests and practice you MATLAB skills there. Alan Chalker has posted a mid-contest code analysis midway of most contests. These are a truly great read!