The codes will be submitted to a tester, who then uses a range of sudokus (not known by the contestants beforehand) which he tests for each code. The winner is primarily determined by the number of solved sudokus, if there is a tie then by speed.
so, would it make sense, if your solver rejects the hard ones or not ?
I mean, you can't know in advance whether you will have enough time
to finish all puzzles , can you ?
So pick the easy ones, while putting the hard ones on some stack,
only to be considered again once you still have time after
finishing the easy ones, etc.
The tester will get lets say 50 sudokus. Then he will test thouse 50 with one contestant, he gets how many was solved, and the total time. Then he tests the SAME 50 sudokus on the next contestant, he writes down how many was solved, and the speed, and so on...
The winner will be that has the MOST Sudokus solved, if there's a tie (i.e. someone has the same number of sudokus solved as someone else) then the winner is the one that has the fastest time.
I'm telling you now, that my solver will solve ALL the sudokus, how fast it will solve, that's another story... I'm sure it won't be faster than Merri's solver, and I'm sure that Merri's solver will solve ALL the sudokus also...
So if you decide so "pick" the ones that you want to solve, then you will lose the contest FOR SURE.
Ok. Here's some more new ones.
Again, if any of you find I screwed up, by finding more than one solution, please tell me so I can address this porblem.
No real big changes to my old code, but it can now solve the hardest set in about 2780 ms. NotLKH's newest set solved in about 260 ms. I've a yet-another-code in the works which is supposed to be able to solve all by pure logic... however, I'm not sure if the logic I've thought for it is valid. I also don't know if I have the time to complete it before the contest deadline, because I started up a new site a week ago and it is still under construction, yet have become the most popular site I administrate.
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