purifire - 2011-02-14 6:14 PM

This was my interpretation of this puzzle and the reason I gave it such a high value.

But in a competition the end justifies the means so you dont get marks for methods used but only for the correct solution.

Hope this helps.

motris - 2011-02-14 10:40 PM

I already sent Rishi my approach to the last puzzle. On things like Diagonals I often do minor coloring/trial steps to make eliminations. I certainly took longer than "7 minutes" but it was maybe 15 to do this:

While I didn't instantly notice it, I caught on quickly that the puzzle had rotational digit symmetry (the 2/5 would be placed in symmetric spots, the 1/7, 6/9, 4/8 as well - 3 is self symmetric).

Most of the solve involved considering the most limited digits (first 2s and 5s) and how certain forcing placements cause contradictions. I'll describe steps for "2" but same applies for "5" on other end

After placing the 3's, 2 can be in three positions in box 4, but if it is in R5C2 it leaves no spot for a 2 in column 3, so a 2 is in R45C3, and in C2 in box 7. If the 2 is in R7C2 (pushing the resulting 2's), there is no possible place for a 2 on the LL-UR diagonal, so R8C2 = 2.

Next, either there are two 2s at R4C3 and R3C4 or three 2s at R5C3, R4C4, R3C5. The latter case (more tempting to try) led to an immediate contradiction in the lower-right box as you can't put in 2's to avoid knight constraints or diagonal constraints. So 2 = R4C3 and R3C4.

Then, 2 is in R6C56 but if in R6C5, then no way to put a 2 on the UL-LR diagonal. So 2 in R6C6.

Then (describing 4, but 8 is symmetric), there must be a 4 somewhere in column 6. You know the upper-right has a 4 in either R3C7 or R1C9 and these form (from the anti-knight constraint) a perfect elimination of a 4 from R1C6. This is my favorite step. Anyway, other digits eliminate all but R9C6 for that 4. This gives several more 4's, and all the 2's as a result.

Last necessary step comes from thinking about R5C7 spot. It contains a 7 or a 9. Whichever digit is in this cell must also be in R6C6 because of the eliminations in the center. But a 7 is already forced in C2 in R45, so you can't have the two 7s in the other spots, so this is a 9.

From there it is just entering digits and copying the symmetric logic, but very interesting and challenging puzzle. I think the Anti-Knight Non Consecutive was my favorite, but this was a close second.

rakesh_rai - 2011-02-14 7:31 PM

Paper solvers had it simplified this time with a simpler answer entry system (by and large, with a few exceptions). Of course, there can be further improvements which can be discussed and sorted out.

I am in the minority who are solving online. And there is one thing which I would like to mention, from an improvement perspective in future tests - the representation of the diagonal lines on the grid. This time we had five grids which had a diagonal constraint. The example on the site also had diagonal lines and was represented using gray lines. But in the test four out of the five grids had black (dashed) lines running diagonally. I am attaching an image which brings out the differences clearly. The pencilmarks in the gray one are clear, but the other one requires a lot of eye exercise while solving the sudoku - especially the numbers 1,3,5,7 and 9 - each of which interfere with one or more of the diagonal lines.

We can surely avoid this by choosing slightly light colors for these lines.

pafcio123 - 2011-02-21 12:11 AM

Hi, Is there anywhere password to get the PB?

regards