There is an xy-wing in box4 and box7. In simple terms, it means you can rule out one of the two possibilities at R4C1.

Note that you have some unmarked pencil marks at R7C3 and R8C1.

Thank u Deb. Now I could complete the sudoku. But before that I have to be familiar with so many techniques like XY wing and all that.
I need a lot of practice.

I was actually about to post this just before Deb posted the XY-Wing thing.

Anyway, here's how I would've solved it (and something I'm able to spot more easily) - A 3 in R3C2 means it has to be in R7C3. A 2 in R3C1 means it has to be in R7C3. So both of these can't be true, so 4's in R3C2. I generally don't know names of the techniques I use, I just use them

Dynasty sudoku from the US sudoku Team Qualification created by Wei Hwa hwang aka onigame.I have always had difficulty solving sudokus with a puzzle element attached to them .I find these kind of sudokus to be sort of a 'speed breaker' since they greatly reduce my solving pace.....So where is the continuation in here?!(Rules : 1-7 and 2 shaded cells in each row,column and Box.Shaded cells don't touch orthogonally.White cells have to be connected)



Edited by kishy72 2014-05-13 8:19 PM

After giving a look to it,
i could only find next step as r9c3 as should be black and cannot be 4. Little complicated logic:
if its 4 then it will give black at r1c3 and which immediately give black at r5c4 and r3c4 ( only place for that column ) sp this makes r4c3 as black and thus r4c2 becomes 4. Also there is 5 because of all these at r1c4. Now notice that r1c2 nothing can come because black can't come being adjacent and also 4 and 5 is elliminated in that row and column.
Thus r9c3 has to be black. And it gives r9c2 is 4. Immediately r1c3 cannot be black for kind of similar reason as above ( if it's black then r1c4 will be 5, again r1c2 nothing can fit as 4 and 5 already present in either row or column and black being adjacent)
Thus giving r1c2 is black and thus r3c1 is black.
After this it gets solve quickly :)

Let me know if i am wrong. May be this method was not good, but thats the way i figured out. Would be interesting to know if anyone else come up with some alternative way. :)

Hiii Swaroop!Thanks a lot!Pretty sound reasoning indeed!But don't you think that the deduction is a bit too lengthy for comfort and borders a bit more on guess work.I doubt whether someone could visualize that far without putting pen to paper.Well a few maybe?!Surely there must be some cutthroat logic from where I got stuck to proceed further.

To illustrate my point take a look at this Triomino sudoku from the Japanese Grand Prix round.
Rules: Place a digit from 1 to 6 into each empty cell or blacken the cell so that each digit appears exactly once in every row, column, and outlined 3x3 region along with three black cells. Each black cell should be part of an orthogonally connected group of three blackened cells (a triomino). No two triominoes can share an edge.

As I see ,you can proceed further by taking a guess ,lets say at R4C4 assuming it to be a shaded cell which leads to the following ---->1 at R4C9,1 at R6C6----->Shaded cells at R5C9,R6C9,R6C8---->6 at R6C7---> shaded cell at R8C7--->2 in R8C6(you can't have a shaded cell because if you do you won't get 3 shaded cells in Box 9 (which are part of a triomino))--->2 in R5C4,5 in R5C5 --->Shaded cells in R1C6,R7C6 and R9C6(only cells in that column)---->shaded cells in R9C5 and R9C4(to get a triomino) and at last a contradiction since you get 4 shaded cells in Box 8.Phew!So you can't have a shaded cell at R4C4 and you proceed beyond after that.
But do you think there is a better way without going the earlier path that I stated.There is a much stronger contination indeed!Look at cells R7C1 and R7C2 .These cells must contain atleast one shaded cell.So there cannot be a shaded cell in R9C2 because if there is, then you would have 4 shaded cells in Box 7 which violates the rule.So R9C2 must be a number and cannot be anything except a 5 and from where the sudoku proceeds like knife through butter!!This is where I feel ,the top solvers differentiate themselves from the rest.It was not before spending a lot of time that I managed to get a 5 at R9C2.So it is something like this that I would be mighty interested to know in the earlier Dynasty sudoku.
----Kishore----


Edited by kishy72 2014-05-15 9:00 PM

Hi,
I know as i said mine not be good way, because it has long way deducing. But after that thing it gets solve very easily. I am still trying to find if any other way (surely there must be). And yes in Trimino Sudoku that was the key point, luckily i got it quickly during competition (though overall performance was bad).
My case is exactly opposite to yours - Classics slow me down drastically, Variants i am comfortable with. :)

Edited by swaroop2011 2014-05-15 9:50 PM

Just look at row 5. (R5C3 and R5C4) have to be ( a 5 and a black cell ). From this we can easily deduce that R6C7 is 5. Is this enough to proceed further?

oh yes correct hm :)
awesome. Didn't thought about it :)

Could you please help me in proceeding with this 6x6 sudoku.
The numbers in black are givens and the ones in green is what I have entered. I am unable to proceed further.

Thanks,
Bharath



Edited by ka_bharath 2014-07-02 7:43 PM

its quite easy observe naked pair of 3,4 at R3C6,R6C6 and naked single i.e 4 at R2C6

here is the solution

416532
235164
124356
653421
342615
561243

Just to correct it, I think you meant naked pair of 3,6

hoo no again typo

Thank You.

Sudoku Surprise - What an amazing test with a great set of sudokus.One of those tests which I felt had that rare 'reunion'of sorts between sudokus and puzzles.Rather than reflecting on my performance which was sub-optimal,I thought it would be prudent to get down to brass tacks and try understanding the beautiful intricacies hidden in each puzzle.

So,I came across this Secret Code sudoku which I attempted during the test and spent close to an hour!! on this single puzzle and still could not get to the solution (logically).I guessed the answer key knowing that it had to contain one of these strings 2316,2361,3216,3261 in Column 3 .Once again ,it turned out to be a lucky guess .Hence to be candid, it's only 470 points that I could muster in reality.
The reason that I am posting this here is that 80% of the solvers should have reached the following stage after which they would have hit a dead-end.The Sudoku could not have become more stuck.
So fellas,if someone did find that devastating piece of logic from this point which brings the curtains down on this one ,I would be glad to know.

----Kishore-----


Edited by kishy72 2014-11-26 4:40 PM

I guess everyone was stuck for sometime at this point.
Firstly if you see the arrows which are least pointing directions:eg r6c1 or r1c3 they can point at most 3 directions.
Now the logic i used is if you see 2-3 pair, for the solution to become unique atleast one of 2 or 3 has to be part of secret code. So this restricts further the direction of r6c1 and it has to be pointing right. (This way i got one direction - so atleast now i know that secret code belongs to order which will follow row 6 and it has atleast one of 2 or 3 and it can't be 3rd digit of the code)
This immediately fixes the direction of r3c9 which has to be towards r4c8. Also this implies that secret code cannot contain 4,5,8.
Now this gives that r2c4 arrow has to be towards right or towards r3c5. But now if its towards r3c5 then it has to be 376 due to obvious reasons. but this is not possible immediately you can check that with known arrow directions i.e row 6. So r2c4 arrow is pointing right. which implies code is 371 (as 4 is not part of it and 2 cannot be as it is last).

Hope this helps. I don't know if above is right way or not. But would still love to know if any other shorter way is there :)

*sorry small edit in above r6c1 can be 4 directions but anyways that doesn't change the analysis :)

Excellent!I think that you have provided the precise logical way of solving this sudoku.I seriously doubt if there is a shorter way than the above mentioned one.
Right at step 1 ,I faltered.I mean,I failed to deduce that, atleast one of 2 or 3 has to be part of the code as it did not hit me that the puzzle would fail to be unique without that.Tremendous!!!
Also, I understand that the most important strategy to adopt in this puzzle is to concatenate clues and see if they work in conjunction with other clues.These are the grey areas that needs to worked on.
Thanks a lot for the explanation.Well done !!

Hi all!!I was randomly solving some sudokus from the IB of Swiss Team Qualification round when I came across this Arrow-thermo sudoku .I fail to see a logical continuation from the following point....Can someone spot the same and share .....



Edited by kishy72 2015-06-02 6:07 PM

R5C3 and R7C3 contain (3,5). So, R6C3 = 7, and R1C3 = 6 ......
From here, it can be solved forward.

Advised only seeing just plainly, not gone through whole.. Maybe wrong also

There is also possibility of 3 in R3C3/R3C6.So your deduction is not correct.

7 can only come at R3C9 or R5C9

if you put 7 at R5C9, then 9 is forced at R6C8
which will force 9 at R5C2, this will force 9 at R3C1.

9 at R3C1 will force 9 at R2C8, which is wrong as R6C8 already contains 9

so 7 should come at R3C9.
It should be easy solve from there on.

Somewhat similar to gaurav but a 2 step smaller,

Consider 1,3,4,6 boxes. Each has only 2 or 3 places of placing 9. So visually its easy to see that placing 9 at r3c7 leaves no room to place 9 in box 4. After this its easy i suppose.

Interesting ways of looking at the logic for the same outcome.The first appears to be a bit of a hit and trial and the second approach is shorter but much more difficult and global/broader thinking is required.Nevertheless the latter seems the ideal way for dealing with the issue aptly.So then I assume that there is no other shorter logical path than the ones mentioned.Thanks a lot for sharing the logic ....Gaurav and Swaroop

Kishore