Yajilin Yacht (28th Jul - 17th Aug) has started Discuss
stuck in this sudoku159 posts • Page 3 of 7 • 1 2 3 4 5 6 7
@ 2013-04-03 11:33 PM (#10511 - in reply to #8611) (#10511) Top

purzelbaumfan



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purzelbaumfan posted @ 2013-04-03 11:33 PM

Thank you. I have managed to solve some other puzzles of the "extreme" level. Now I tried a "hardcore" one and I can't get any further. I don't even know where to start a chain! What do you say?

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Edited by purzelbaumfan 2013-04-03 11:34 PM
@ 2013-04-04 5:32 AM (#10517 - in reply to #10511) (#10517) Top

motris



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motris posted @ 2013-04-04 5:32 AM

I'd wager I've never met a person who can solve that puzzle logically or even deduce 1 digit before guessing. I'd even say I've yet to even find a computer solver that can do it -- One solver completely stopped without any digits and the other required a lot of Bowman's Bingos which are essentially guesses.

You can probably check some of these puzzles out this way too. I recommend plugging them into sudokuwiki.org/sudoku.htm which can quickly evaluate the next step at a point when you get stuck.

Can I ask what book this is that is publishing such crazy puzzles?
@ 2013-04-04 8:09 PM (#10526 - in reply to #8611) (#10526) Top

purzelbaumfan



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purzelbaumfan posted @ 2013-04-04 8:09 PM

I sometimes put sudokus into the solver you recommended. But it still is a computer and some of the strategies are just way over the top and too funky. The last times I posted puzzles in here it was the same way: I first put them into the sudoku solver to see what it had to say, and then I put it in here. I found that the "human solutions" were always better to understand and there were still some nice tricks about it.

Unfortunately, this time I didn't bother to put it into the solver because I was sure it would give me extraordinary strategies I'd never use again.

The book is Nr. 31 from Stefan Heine: http://www.ps-heine.de/archives/1037
@ 2013-04-25 8:39 PM (#10814 - in reply to #10526) (#10814) Top

kishy72



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kishy72 posted @ 2013-04-25 8:39 PM

With the Sudoku Caravan moving next to the UK and then the US in a short notice i thought it would be appropriate now to get the unworked worked out.So after a bit of an exile from this stuck forum (which was more self imposed) i m back to being stuck again .So this is going to be the first of the three 'Big Fish sudokus' that i m going to post.This movable digits sudoku looked like a fun puzzle to solve until the point i got stuck from where it became suddenly 'immovable' and made me to call it quits.So fellas can you find the continuation and shw the way to me?
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Edited by kishy72 2013-04-25 8:40 PM
@ 2013-04-25 9:01 PM (#10816 - in reply to #10814) (#10816) Top

Para



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Para posted @ 2013-04-25 9:01 PM

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
@ 2013-04-27 11:25 PM (#10849 - in reply to #10816) (#10849) Top

kishy72



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kishy72 posted @ 2013-04-27 11:25 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?!
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Edited by kishy72 2013-04-27 11:26 PM
@ 2013-04-28 12:53 AM (#10850 - in reply to #10849) (#10850) Top

caudmont



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caudmont posted @ 2013-04-28 12:53 AM

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
@ 2013-05-02 7:04 PM (#10914 - in reply to #10850) (#10914) Top

kishy72



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kishy72 posted @ 2013-05-02 7:04 PM

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.
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Edited by kishy72 2013-05-02 7:04 PM
@ 2013-05-02 8:02 PM (#10915 - in reply to #10914) (#10915) Top

kishy72



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kishy72 posted @ 2013-05-02 8:02 PM

kishy72 - 2013-05-02 7:04 PM
A better pic?!Not able to get the resolution that i want.
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Edited by kishy72 2013-05-02 8:08 PM
@ 2013-05-02 8:26 PM (#10916 - in reply to #10914) (#10916) Top

prasanna16391



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prasanna16391 posted @ 2013-05-02 8:26 PM

kishy72 - 2013-05-02 7:04 PM

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.



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.
@ 2013-06-26 9:32 PM (#11398 - in reply to #10916) (#11398) Top

kishy72



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kishy72 posted @ 2013-06-26 9:32 PM

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)
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Edited by kishy72 2013-06-26 9:32 PM
@ 2013-06-26 9:44 PM (#11400 - in reply to #11398) (#11400) Top

prasanna16391



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prasanna16391 posted @ 2013-06-26 9:44 PM

kishy72 - 2013-06-26 9:32 PM

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)


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.
@ 2013-06-26 10:09 PM (#11401 - in reply to #11400) (#11401) Top

kishy72



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kishy72 posted @ 2013-06-26 10:09 PM

prasanna16391 - 2013-06-26 9:44 PM

kishy72 - 2013-06-26 9:32 PM

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)


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.


Nice.Thnks a lot!I think i can complete it now.
@ 2013-09-05 7:29 PM (#12588 - in reply to #8611) (#12588) Top

deepika m



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Country : India

deepika m posted @ 2013-09-05 7:29 PM




Pls help me in solving this sudoku.
@ 2013-09-05 11:51 PM (#12595 - in reply to #8611) (#12595) Top

debmohanty




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debmohanty posted @ 2013-09-05 11:51 PM

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.


@ 2013-09-06 2:04 PM (#12602 - in reply to #8611) (#12602) Top

deepika m



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Country : India

deepika m posted @ 2013-09-06 2:04 PM

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.
@ 2013-09-06 3:04 PM (#12603 - in reply to #8611) (#12603) Top

prasanna16391



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prasanna16391 posted @ 2013-09-06 3:04 PM

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
@ 2014-05-13 8:19 PM (#15273 - in reply to #8611) (#15273) Top

kishy72



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kishy72 posted @ 2014-05-13 8:19 PM

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)

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Edited by kishy72 2014-05-13 8:19 PM
@ 2014-05-14 2:39 AM (#15276 - in reply to #15273) (#15276) Top

swaroop2011




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swaroop2011 posted @ 2014-05-14 2:39 AM

kishy72 - 2014-05-13 8:19 PM

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)


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. :)
@ 2014-05-15 9:00 PM (#15292 - in reply to #15276) (#15292) Top

kishy72



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kishy72 posted @ 2014-05-15 9:00 PM

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----
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Edited by kishy72 2014-05-15 9:00 PM
@ 2014-05-15 9:33 PM (#15293 - in reply to #8611) (#15293) Top

swaroop2011




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swaroop2011 posted @ 2014-05-15 9:33 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
@ 2014-05-15 10:54 PM (#15296 - in reply to #15293) (#15296) Top

rakesh_rai




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rakesh_rai posted @ 2014-05-15 10:54 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?
@ 2014-05-15 11:17 PM (#15297 - in reply to #15296) (#15297) Top

swaroop2011




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swaroop2011 posted @ 2014-05-15 11:17 PM

rakesh_rai - 2014-05-15 10:54 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 :)
@ 2014-07-02 7:38 PM (#15886 - in reply to #8611) (#15886) Top

ka_bharath



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ka_bharath posted @ 2014-07-02 7:38 PM

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

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Edited by ka_bharath 2014-07-02 7:43 PM
@ 2014-07-02 8:01 PM (#15887 - in reply to #8611) (#15887) Top

dp_94



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dp_94 posted @ 2014-07-02 8:01 PM

its quite easy observe naked pair of 3,4 at R3C6,R6C6 and naked single i.e 4 at R2C6
stuck in this sudoku159 posts • Page 3 of 7 • 1 2 3 4 5 6 7
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