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Best of LMI Sudoku tests — December Sudoku Test — 14th-16th December67 posts • Page 3 of 3 • 1 2 3
@ 2013-12-18 10:20 PM (#13879 - in reply to #13874) (#13879) Top

Richard



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Richard posted @ 2013-12-18 10:20 PM

Ours brun - 2013-12-18 3:17 PM

And anyway, in my case it was rather a pleasant surprise to see that I was able to solve the Number X in around five and a half minutes.


I noticed that you submitted 19 solutions first, without 'X'. When you entered that one few minutes later, I thought you must halve solved already half of the grid earlier, only to finish it after the first lot.
Incredible if you did the complete puzzle that fast.
It has a very narrow solving path, with many different solving steps. In my opinion it was the only real hard puzzle, since it has no shortcuts like the renban-puzzle.
So: Well done!
@ 2013-12-19 12:43 PM (#13881 - in reply to #13864) (#13881) Top

Richard



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Richard posted @ 2013-12-19 12:43 PM

As you will know, I started my SVS (Sudoku Variants Series) earlier this month. You can read about it here .
Richard - 2013-12-05 2:10 PM
Recently I have been busy with puzzles for the German Sudoku Championship 2014 and another sudoku project. I noticed once more the incredible number of Sudoku variations and also that there are a lot of types unexplored by me.
Regards, Richard

The quote above is from the SVS-forum thread. The other project were I was talking about was Best Of Sudokus. Going throug all the Sudoku Tests of 2010 and 2011 was very interesting and fun to do. I found a lot of types that where new(ish) to me and it was indeed one of the reasons I started thinking about SVS.
Since there were 18 tests to be taken into account (with two of them having interacting grids), BoS would contain 20 sudokus. Quite a lot for a test of 120 minutes. So fortunately I could create a few grids of 6x6 to keep the total needed time for top solvers well under the limit of two hours.

Some notes to some of the puzzles then:

Chaos Renban
I like it to include a puzzle that can be solved by all puzzlers in a normal time, but it contains a few shortcuts for top solvers. This one was a good example. Law of leftovers requires that a 5 has to be placed in R34C5. Renban groups with 5 or more cells always require a 5 too, so R4C5 = 5. None of the renban groups in the corners can have the 5 in the center cell; it would block all possible positions of the 5 in the renban group in the opposite corner. Noticing one or both of these shortcuts makes the solving much faster. While I was watching the performances of players, it was clear to me that not everyone found these shortcuts.

Sundoku
The intrinsic logic in the sundoku of LMI Spring Sudoku Test was really brilliant! I had to smile when I recognised it myself. For that reason I thought it would be a good idea to place the sun in a corner cell of the middle block in the test and use the center cell in the Leftover Puzzle. When you didn’t find the logic yet: Digits 1,2,3 in the center block have to go in the sun and both ‘Non-ray-cells’. This means that the 4 can only go in the diagonal ray. If it goes in the horizontal or vertical ray, it would block all possibilities in the sun. So the start of this puzzle is placing 1,2,3,4 in the diagonal ray immediately.

Scattered Sudoku
Using grey cells that are locked in in a white irregular shape is always nice from an aesthetic point of view. I like to do that now and then. But maybe the shape of the regions has scared some people off, because it is the least submitted puzzle. Anyway, Law of leftovers is your best friend in this one: R3567C4 must contain the same digits as R1C1, R2C2, R8C2 and R9C1. This works on the left, right, top and bottom of the grid. No other tricks are necessary.

Number X is alive
This type was high on my list to become the hardest puzzle. Here are the first steps in the solving process: adding the sums of both cages in block 2 leads to an even sum, so the remaining cell has to be odd, and therefor must be 9. The sum of the two cages must be 36 so X is either 3 or 8. If you try X=8, R4C1 = 6 and that is not possible. So X = 3. Now R4C159 can be calculated and filled and R6 can be filled for a lot of cells too.
Later in the solving process R9C28 as (possible) sum and R7C37 play an important role in finishing this one.

Missing Arrows
In Crazy Arrows this one was my favorite (after the test!) but it was not solved that often during the test because it was a real tough one. So I wanted to create an easier one of this type.

I think most of the other puzzles don’t require additional info on solving process, but if you have any question regarding one of the test puzzles, or other feed back; don’t hesitate to write it here in the forum!

At this place I would like thank all authors of LMI-sudoku tests for their inspiring work!

Thank you all for participating; hope to see you all again next time.


@ 2013-12-19 9:37 PM (#13884 - in reply to #13702) (#13884) Top

prasanna16391



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prasanna16391 posted @ 2013-12-19 9:37 PM

I had the Number X out pretty fast too. I agree it's narrow though. Just more of a strength/weakness thing. I took longer on the Sundoku where I missed obvious steps repeatedly.

Oh and also the Blackout Sum, which I broke twice and was the main reason for not being able to finish. After the test, a day later, it was really easy to solve as the points suggested
@ 2014-01-19 6:40 PM (#14132 - in reply to #13702) (#14132) Top

sena



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sena posted @ 2014-01-19 6:40 PM

In No X is alive, author has given the clue that the sums of both cages in block 2 must be even. why? How to manipulate that?
Could you please explain? When the remaining cell can be 2,4,6,8 or 9 how do you choose 9?
@ 2014-01-20 1:01 AM (#14135 - in reply to #14132) (#14135) Top

Ours brun




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Ours brun posted @ 2014-01-20 1:01 AM

It is really easy, actually.

1) We know that the values of these two cages end with the same digit; hence they are both even or both odd.
2) The sum of two even numbers is even. The sum of two odd numbers is also even. So, whatever may be the value of X, the sum of the values of these two cages is even.
3) The sum of the digits from 1 to 9 is equal to 45; so, the value of R2C5 is equal to 45 minus an even number; hence, an odd number; hence, 9.

I hope it is clear enough.

Bastien
@ 2014-01-20 1:03 AM (#14136 - in reply to #14132) (#14136) Top

SKnight



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SKnight posted @ 2014-01-20 1:03 AM

sena - 2014-01-19 9:40 AM

In No X is alive, author has given the clue that the sums of both cages in block 2 must be even. why? How to manipulate that?
Could you please explain? When the remaining cell can be 2,4,6,8 or 9 how do you choose 9?


First, the idea that the sum of the two cages in block 2 (top, center 3x3 region) must be even:
Note that it is not that each individual cage total must be even, just that their sum is even. They both
end with the same digit ("X"), so their sum is even.

Those two cages are everything in block 2 except the center square, and the sum of all digits in a block is 45. Thus
the remaining digit must be odd (45 minus some even number).

The remaining digit must be 9 since we know it is odd, and 1,3,5 and 7 are all eliminated by the placement of those
digits in the same row/column of the one we're looking for.

@ 2014-01-20 5:44 PM (#14141 - in reply to #13702) (#14141) Top

sena



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Country : India

sena posted @ 2014-01-20 5:44 PM

It is very clear now. Rest of the things I completed.
Thank you very much for your immediate response.
Best of LMI Sudoku tests — December Sudoku Test — 14th-16th December67 posts • Page 3 of 3 • 1 2 3
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