To start, look at where the 3 can go in R2. It will give you a few placements. To not ruin the rest of the solve, i'll let you try to solve it from this point. If there's another sticking point, just show where you're at after this.

Edited by Para 2013-04-25 9:05 PM

Hi !!Thnks for the nice tip .I was able to proceed from there till here.I see that this stage is ripe for a bifurcation where the numbers will rapidly fall in place after that but i was looking for the logical continuation and the right line of thinking has eluded me for close to 45mins from this point.How do i continue from here?!


Edited by kishy72 2013-04-27 11:26 PM

In the upper right region, the 7 is in Row 2 or row 3. There is another 7 in row 2 or row 3 which is neighbour R3C6. R2C3 can't be a 7.

Edited by caudmont 2013-04-28 12:54 AM

Thnks .But i got stuck again in the same sudoku just after placing the 4s in first 3 rows and 7s in first 2 rows.I am not going to waste everyone's time by posting that sudoku once more.I would greatly appreciate a spoiler though again from that stage.
Time to move on to the next one.I have completed the 108 point killer after a torturous grind.So this little killer would be the last i would be posting from the serb grand prix and i hope that i dont get clueless multiple times just like the previous.Excepting a few deductions i was unable to make much of a start here.Applause to all those who have completed this.



Edited by kishy72 2013-05-02 7:04 PM

A few steps I can add to this. R6C1 being 4 makes the rest 9/8-9/8-9, so that'd make the 40 line next to it a max of 7-8/7-8/7 which leaves a sum of 18 in the 4th box. So, its 5-6. If its 5-6, the 15 sum from the top of the grid comes into play. The 4th box part of it can be a minimum of 8 and maximum of 9 so the 2nd box part of it can be 6 or 7. Now you need to take the 40 sum at the top into consideration, and use the fact that 1 and 2 are already taken in the 2nd box in the 6/7 sum, and the remaining can sum to minimum 8, and maximum can only be 9, to satisfy the long 15 sum from the right. So thats 5/6. Which means the possibilities for the 40 at the top are reduced to doubles. Thats complex in itself, but its as far as I could get by "clean" logic.

After that I couldn't get a really clear path to it. I could solve by thinking of a few chains and cramping around the left of higher numbers, but I think I'll wait for the authors to post a better way than that.

Tripod sudoku from the recently concluded Turkey round .I had no idea on how to go about it during the test.But after completing the test i paid closer attention to the rules and did a bit of scrutiny to see if i could gain anything meaningful from the rules and i did.Since all points where 3 lines meet are given , it hit me that either 2 lines are shaded or only one is shaded in the plus (+) that surrounds each dot.So some slitherlink kind of work let me to this stage from where i could not continue further.Since this Tripod sudoku is present in the Italian round too i am interested to see how it works.So fellas how do i continue and are there any general rules to keep in mind while solving this variant(I dont want to be stuck again)


Edited by kishy72 2013-06-26 9:32 PM

Quick note that might help - Tripod isn't toroidal, so you can basically draw the outer square in it's entirety, which will give you few more "crosses" between cells, using the all tripods marked rule. So, more "threads" across same regions can be made. Also, I used the 7s and thought about how every region can/can't reach the 7s and got the two other 7s without drawing much borders. (you missed a 7 in R1C2 for example that can be deduced without drawing all borders, because you basically have 2 regions in the row, and the right one will have the 7 in it once you draw the outer square). Maybe that helps.

Nice.Thnks a lot!I think i can complete it now.

Pls help me in solving this sudoku.

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.