PANFOPCWHTTAPA 3 (11th - 16th Nov) Score Discuss
stuck in this sudoku159 posts • Page 5 of 7 • 1 2 3 4 5 6 7
@ 2015-06-25 12:28 AM (#18495 - in reply to #8611) (#18495) Top

kishy72



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kishy72 posted @ 2015-06-25 12:28 AM

PS : The 34 thing at the bottom was done in the hope that there was a Y-Wing hidden.There seems to be nothing of that sort until someone proves me wrong.......
@ 2015-06-25 4:45 AM (#18496 - in reply to #18495) (#18496) Top

debmohanty




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debmohanty posted @ 2015-06-25 4:45 AM

Rishi told me that this was easy/medium difficulty, so I don't think you should expect a *-wing there.
@ 2015-06-25 1:45 PM (#18501 - in reply to #18496) (#18501) Top

harmeet



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harmeet posted @ 2015-06-25 1:45 PM

If R2C8 is 2 ==> R2C2 is 3 and R2C1 is 2. ==> R6C2 is 4 ==> R6C8 is 3 ===> R7C8 has no valid entry (both 2 and 3 are used up in C8).
Therefore R2C8 is 1. Proceed from here.
@ 2015-06-25 2:07 PM (#18502 - in reply to #18501) (#18502) Top

kishy72



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kishy72 posted @ 2015-06-25 2:07 PM

harmeet - 2015-06-25 1:45 PM

If R2C8 is 2 ==> R2C2 is 3 and R2C1 is 2. ==> R6C2 is 4 ==> R6C8 is 3 ===> R7C8 has no valid entry (both 2 and 3 are used up in C8).
Therefore R2C8 is 1. Proceed from here.


If R2C8 is 2 ,then why can't R2C2 be a 1?????

In fact ,it has to be a 1 as there is no other place for 1 in R2 when you consider the above assumption.
@ 2015-06-25 2:26 PM (#18503 - in reply to #18502) (#18503) Top

harmeet



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harmeet posted @ 2015-06-25 2:26 PM

Because R1C7 will be 3, thereby forcing R2C2 to be 3. I'm only going by the image you posted with the marks. Yet to try and solve the sudoku.
@ 2015-06-25 2:37 PM (#18505 - in reply to #18503) (#18505) Top

kishy72



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kishy72 posted @ 2015-06-25 2:37 PM

harmeet - 2015-06-25 2:26 PM

Because R1C7 will be 3, thereby forcing R2C2 to be 3. I'm only going by the image you posted with the marks. Yet to try and solve the sudoku.


Yes.Now it's clear.I was not able to follow what you said earlier.I don't think we need to go down to boxes 6,9 etc., for proving that contradiction.So the logic is that if there is a 2 in R2C8 ,then 1 and 3 will be cramped for space in Box 1.

Great work and some good spotting!Well done!!

Thanks a lot !

Kishore
@ 2015-06-25 2:37 PM (#18506 - in reply to #18503) (#18506) Top

harmeet



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harmeet posted @ 2015-06-25 2:37 PM

Yeah you said it yourself. If R2C8 is 2 then there cannot be 1 anywhere in R2. Hence R2C8 must be 1.
@ 2015-06-25 2:48 PM (#18508 - in reply to #18506) (#18508) Top

kishy72



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kishy72 posted @ 2015-06-25 2:48 PM

harmeet - 2015-06-25 2:37 PM

Yeah you said it yourself. If R2C8 is 2 then there cannot be 1 anywhere in R2. Hence R2C8 must be 1.


No!!You are getting me wrong.I said that if R2C8 is 2 ,then 1 cannot be anywhere in R2 EXCEPT R2C2. BUT it doesn't work that way!

I came to know the logic after your second post where you mentioned that if R2C8 is 2 then R1C7 is 3 .Continuing on this line of thinking we have the following :

If R2C8 is 2 ---->R1C7 is 3 ---->R2 (C7/C9) will be 45 and 457 respectively ---->Now looking at R2,there is a triplet 457 ,45 and 457 --->1 and 3 can only be in Box 1 which cannot be the case obviously looking at the image ---->Therefore 2 is not at R2C8 and 1 is the digit.
@ 2015-07-05 7:11 PM (#18689 - in reply to #18508) (#18689) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-07-05 7:11 PM

someone please help me from this step, solving this
I tried finding x, y wing but no success. sorry if numbers are messed up :)

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Edited by gaurav.kjain 2015-07-05 7:12 PM
@ 2015-07-05 9:44 PM (#18691 - in reply to #8611) (#18691) Top

akash.doulani



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akash.doulani posted @ 2015-07-05 9:44 PM

R7c6 will be 5 because of BUG
@ 2015-07-06 8:21 AM (#18692 - in reply to #18691) (#18692) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-07-06 8:21 AM

akash.doulani - 2015-07-05 9:44 PM

R7c6 will be 5 because of BUG



Thanks Akash.

I would want to see more logical opening.
Appreciate anyone help.
@ 2015-07-06 9:42 AM (#18693 - in reply to #8611) (#18693) Top

prasanna16391



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prasanna16391 posted @ 2015-07-06 9:42 AM

If 6 is in R9C2, using 4th box, it can't be in R6C5.
If 5 is in R9C6, using 2nd box, it can't be in R6C5.

Reading both of those together should show that the placements together force no digit in R6C5.
@ 2015-07-06 10:45 AM (#18694 - in reply to #8611) (#18694) Top

vopani



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vopani posted @ 2015-07-06 10:45 AM

@Akash
Seems like you found a bug in this sudoku! ;-)

@Gaurav
Why do you want to see a more logical opening? What is your intention behind it? What is it that you will gain by learning about Swordfish or BUG or any of these extremely advanced techniques?
@ 2015-07-06 10:53 AM (#18695 - in reply to #18694) (#18695) Top

prasanna16391



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prasanna16391 posted @ 2015-07-06 10:53 AM

Rohan Rao - 2015-07-06 10:45 AM

@Gaurav
Why do you want to see a more logical opening? What is your intention behind it? What is it that you will gain by learning about Swordfish or BUG or any of these extremely advanced techniques?


Fair point. In a competition at this stage I would always prefer T&E.
@ 2015-07-06 11:01 PM (#18696 - in reply to #18693) (#18696) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-07-06 11:01 PM

prasanna16391 - 2015-07-06 9:42 AM

If 6 is in R9C2, using 4th box, it can't be in R6C5.
If 5 is in R9C6, using 2nd box, it can't be in R6C5.

Reading both of those together should show that the placements together force no digit in R6C5.


Thanks Prasanna.
@ 2015-09-08 12:03 PM (#19269 - in reply to #8611) (#19269) Top

kishy72



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kishy72 posted @ 2015-09-08 12:03 PM

A Consecutive Sudoku from the US Sudoku Team Qualification round by Wei-Hwa Huang.I greatly enjoyed this very delightful set ! All the sudokus had a unique solve path and were beautifully constructed.I was more or less sprinting through the test when I hit this one and spent around 27-28 mins on this single sudoku.I adopted a very mechanical approach and found that R2C6 was the start in my case.I eliminated digits one by one in R2C6 which was very tiresome.Surely, there must be an alternative easier solve than the one mentioned.If someone has a better start,please let me know.....
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Kishore


Edited by kishy72 2015-09-08 12:03 PM
@ 2015-09-08 4:33 PM (#19275 - in reply to #8611) (#19275) Top

prasanna16391



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prasanna16391 posted @ 2015-09-08 4:33 PM

What I did was,

1. Mark odds/evens. In 2nd box, there are three distinct consecutive pairs, which would take one odd and one even each, meaning two odds and one even go in the series of 3. The series of 3 cannot have 7 and so cannot have 9 either and so must be 123 or 345. Either way, 3 is part of it, and there must be groups of 6-7 and 8-9 in that box elsewhere.

2. Then, note that R3C8 to R3C7 is a series of 6 and that R1C7 = R2C6.

3. Where can 8 be in the 3rd box? Can't be in R1C8 because that is odd (using C8). Can't be in R1C7 or R2C7 because both are 3rd/4th cells of a 6-cell series (considering R1C7=R2C6. 8 cannot be in R2C8 either because that causes a 5-6 pair in 2nd box which isn't allowed going by step 1. So, 8 is at an extreme of the 6-cell series.

4. The series has to be 8-7-6-5-4-3 or 3-4-5-6-7-8. Either way, R1C7, R2C7 will form a 5-6 pair, which means 7 cannot be in R2C9 or R3C9 because 6 is taken. So, 7 must be in R2C7 and the series of 6 is done.



(Consecutive.png)



Attachments
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Attachments Consecutive.png (14KB - 6 downloads)
@ 2015-09-08 6:25 PM (#19279 - in reply to #19275) (#19279) Top

kishy72



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kishy72 posted @ 2015-09-08 6:25 PM

prasanna16391 - 2015-09-08 4:33 PM

What I did was,

1. Mark odds/evens.


For reasons unknown,I fail to use this simple yet highly powerful notation.Now,I clearly understand where I am missing the trick in this variant of sudoku.Often it is the consecutive sudoku that tends to trip me up a lot in any test.Just like the in/out rule of an irregular,I am beginning to comprehend that this notation of odd/even is indispensable when it comes to consecutive and possibly a lot other variants.

Can't be in R1C7 or R2C7 because both are 3rd/4th cells of a 6-cell series


I never spotted this 6-cell series and I never would have if it hadn't been for the explanation you gave.This is mainly due to the fact that the series bends back on itself.I was under the impression that it leads to additional possibilities rather than a straight 6-cell consecutive chain.I am curious to know whether you spotted it right away?If yes,what made you sure that it is a series of straight 6 consecutive numbers ?

and thanks a ton for the lovely explanation and for making me a better solver right from this moment !


Kishore



@ 2015-09-08 8:24 PM (#19280 - in reply to #8611) (#19280) Top

prasanna16391



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prasanna16391 posted @ 2015-09-08 8:24 PM

Because of the converse rule, a 'T' shape of consecutive bars will always be a series of 4. The series in question clearly extends on the T since the R2C7 continuation to R2C8 cannot be the same as R2C6. I also notice a series of 8 in the bottom left btw, but didn't use that till later. The series of 8 is a bit more difficult to see, but note that R7C3 is the intersection of two T-shapes, so if it is the same series going outward from there then the same digit will be in R6C2 and R8C3.
@ 2015-09-08 8:52 PM (#19281 - in reply to #19280) (#19281) Top

kishy72



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kishy72 posted @ 2015-09-08 8:52 PM

Thank you.That makes it clear!
@ 2015-09-16 7:43 PM (#19405 - in reply to #8611) (#19405) Top

abhi265645



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abhi265645 posted @ 2015-09-16 7:43 PM

Not been able to move after this... Please help. TIA<a href="http://tinypic.com?ref=32zjnma" target="_blank"><img src="http://i60.tinypic.com/32zjnma.jpg" border="0" alt="Image and video hosting by TinyPic"></a>
@ 2015-09-16 8:32 PM (#19407 - in reply to #8611) (#19407) Top

swaroop2011




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swaroop2011 posted @ 2015-09-16 8:32 PM

I think last row, 1 will come at C3.
@ 2015-09-17 11:58 AM (#19411 - in reply to #19407) (#19411) Top

abhi265645



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abhi265645 posted @ 2015-09-17 11:58 AM

Thank you @Swaroop.
@ 2015-09-24 6:59 PM (#19448 - in reply to #8611) (#19448) Top

kishy72



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kishy72 posted @ 2015-09-24 6:59 PM

'Flower' from Riad's contest.I have managed to complete the other 8 except this.Strangely,most of the puzzles to me seemed to be a contest in itself and took more than an hour and close to a 2.Can someone post the continuation here?

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Edited by kishy72 2015-09-24 7:00 PM
@ 2015-09-24 10:29 PM (#19449 - in reply to #19448) (#19449) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-09-24 10:29 PM

Some cells are divided into GROUP A and B and possible Group positions are marked. 'A' group 2 cells should be same as 'A' middle cell group and 'B' group cells in Box2 should become part of B group cells in marked positions.
B connections are shown, so because of that B group cells as shown connected, with violet colour should be same

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kishy72 - 2015-09-24 6:59 PM

'Flower' from Riad's contest.I have managed to complete the other 8 except this.Strangely,most of the puzzles to me seemed to be a contest in itself and took more than an hour and close to a 2.Can someone post the continuation here?

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