akash.doulani - 2011-02-14 11:38 AM

...wasnt able to do anything in the last 30 minutes . someone please help me get over this serious problem.

rakesh_rai - 2011-02-14 12:02 PM

One solution could be ...to start the test 30 minutes after you click on "Start"

akash.doulani - 2011-02-14 11:38 AM

it was a great set of sudokus. started with 1st one but dont know how , i was stuck in that for a good 15 minutes and ultimately got it wrong. then did the 4 killers . then did the jigsaw diagonal. aroud 30 minutes were left and i started with disjoint diagonal . and as deb nicely put it "Akash continued his trademark not doing anything in last 30 minutes :-) " i dont know how i got it wrong . so wasnt able to do anything in the last 30 minutes . someone please help me get over this serious problem.

rakesh_rai - 2011-02-14 11:59 AM

Is there a neat and elegant approach for solving the 150 pointer? Can someone share on the forum? I got stuck after a while...and had to resort to trial and error to complete it.

Rohan Rao - 2011-02-14 4:16 PM

First of all, nice set of puzzles Rishi. Got completely mixed up with the variants in the middle of the test, but managed to come out clean :D

The 150 pointer was the Diagonal-Antiknight Sudoku. I started the test with this sudoku. I got a couple of numbers and was then stuck. A little guesswork made me progress with a couple of more numbers. By this time 4 minutes were gone. Then it struck me. Observe the given numbers. All 4s and 8s are symmetrically opposite. All 2s and 5s are symmetrically opposite. Similarly, 1s and 7s, 6s and 9s. Also, the 3s are symmetrically placed. Once you realise this, I think the puzzle is not very difficult because the entire grid will have these pairs in symmetrical opposite cells.
Even after getting a few numbers, you will observe the pattern holds. Hence, I directly placed a 3 in the middle cell (R5C5). Then I solved the puzzle in about 3mins using this logic.
Solving the toughest puzzle in 7 minutes does give you a boost in confidence :-) But I still couldn't complete all 12 :|

If I found it, I'm sure many of the top solvers must have used this logic too :-)

Rohan.

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