KScore = Q/(Q-1), where Q=log10(1-K)
0.010.0043458
0.10.043755
0.20.088348
0.50.23138
0.90.5
0.990.66667
0.9990.75
0.99990.8
0.999990.83333
0.99999999990.90909
0.999999999999999999990.95238
0.9999999999999999999999999999990.96774
1-10-nn/(n+1)
1-10-990.99
• What are Kurchan squares ?
http://primepuzzles.net/puzzles/puzz_252.htm

• Can I submit known solutions ?
Yes ! It will be interesting to compare known solutions to the ones found by the contestants.

• Do we have to submit 51 solutions ?
No, since when submit a NxN grid, the scorer will compute the score for all 3 parts.
Thus, with only 17 submissions, you can get scores for all the problems.

• What does log10 mean ?
It's the logarithm in base 10.
For example, we have log(1)=0 since 100=1 and log(10)=1 since 101=10
In most programming languages, you can compute this as log(A) or log(A)/log(10) if log(10) doesn't return 1.

• Are you aware that it is possible to calculate the value of the top score from the information given when you improve one of your results ?
Yes, and this is on purpose ! It's fine as long as you keep this for yourself.
Please, do not publish any result, information or solution.
Note that the best results are improved regularly, so you may only know the best result of a given N at a given moment.

• Why is the scoring function so complicated ?
We did some preliminary tests, and the best scores on part 1 and 2 needed a precision of 20+ digits to discriminate between a good result and the top record.
For the Kurchan's part, there is no such constraint.

• I have submitted my first entry for a problem (e.g. n=20). Why do I get no points for it in one category ?
Someone else already has submitted a much better entry for the problem category where you got 0 points.
Due to the non-linear scoring function your score contribution from this problem is less than 0.00005, which leaves your total score unchanged.

• How can I compute the sum of products for the broken diagonals?
The following pseudocode will compute the sum of the products for diagonals, under the assumption that the n^2 matrix elements are stored in A[i,j], 1<=i<=n, 1<=j<=n:

sum = 0
for x = 0 to n-1 step 1
product = 1
for y = 0 to n-1 step 1
m = (x+y) modulo n
product = product*A[m+1, y+1]
next y
sum = sum+product
next x

for x=0 to n-1 step 1
product = 1
for y=0 to n-1 step 1
m = n-1-((x+y) modulo n)
product = product*A[m+1, y+1]
next y
sum = sum+product
next x

• I still don't understand broken diagonals, help !
http://en.wikipedia.org/wiki/Broken_diagonal