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stuck in this sudoku
   LMI Essentials -> Solving Techniques159 posts • Page 5 of 7 • 1 2 3 4 5 6 7
akash.doulani
Subject: Re: stuck in this sudoku @ 2015-07-05 9:44 PM (#18691 - in reply to #8611) (#18691) Top


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R7c6 will be 5 because of BUG
gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-07-06 8:21 AM (#18692 - in reply to #18691) (#18692) Top


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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.
prasanna16391
Subject: Re: stuck in this sudoku @ 2015-07-06 9:42 AM (#18693 - in reply to #8611) (#18693) Top


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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.
vopani
Subject: Re: stuck in this sudoku @ 2015-07-06 10:45 AM (#18694 - in reply to #8611) (#18694) Top


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@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?
prasanna16391
Subject: Re: stuck in this sudoku @ 2015-07-06 10:53 AM (#18695 - in reply to #18694) (#18695) Top


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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.
gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-07-06 11:01 PM (#18696 - in reply to #18693) (#18696) Top


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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.
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-08 12:03 PM (#19269 - in reply to #8611) (#19269) Top


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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
prasanna16391
Subject: Re: stuck in this sudoku @ 2015-09-08 4:33 PM (#19275 - in reply to #8611) (#19275) Top


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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)
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-08 6:25 PM (#19279 - in reply to #19275) (#19279) Top


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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



prasanna16391
Subject: Re: stuck in this sudoku @ 2015-09-08 8:24 PM (#19280 - in reply to #8611) (#19280) Top


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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.
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-08 8:52 PM (#19281 - in reply to #19280) (#19281) Top


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Thank you.That makes it clear!
abhi265645
Subject: Re: stuck in this sudoku @ 2015-09-16 7:43 PM (#19405 - in reply to #8611) (#19405) Top




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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>
swaroop2011
Subject: Re: stuck in this sudoku @ 2015-09-16 8:32 PM (#19407 - in reply to #8611) (#19407) Top



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I think last row, 1 will come at C3.
abhi265645
Subject: Re: stuck in this sudoku @ 2015-09-17 11:58 AM (#19411 - in reply to #19407) (#19411) Top




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Thank you @Swaroop.
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-24 6:59 PM (#19448 - in reply to #8611) (#19448) Top


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'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
gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-09-24 10:29 PM (#19449 - in reply to #19448) (#19449) Top


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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?

Image and video hosting by TinyPic
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-24 10:51 PM (#19450 - in reply to #19449) (#19450) Top


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Thanks!!Was difficult to spot.I could get only the A group .Now I see what I missed.

Kishore
gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-09-25 11:39 AM (#19451 - in reply to #19450) (#19451) Top


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Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied.
Someone please help.

cross0002
imag


kishy72
Subject: Re: stuck in this sudoku @ 2015-09-25 12:32 PM (#19452 - in reply to #19451) (#19452) Top


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gaurav.kjain - 2015-09-25 11:39 AM

Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied.
Someone please help.


This is how the crossword goes.See where you have gone wrong.
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gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-09-25 12:42 PM (#19453 - in reply to #19452) (#19453) Top


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Ohh this R6c6/7 and R7C6/7 creating 2x2 which I was thinking no 2x2 is possible in cross number, This was my confusion and this was prohibiting me from putting any more markings. Thanks Kishore

Please give me some starting steps for wall sudoku.

kishy72 - 2015-09-25 12:32 PM

gaurav.kjain - 2015-09-25 11:39 AM

Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied.
Someone please help.


This is how the crossword goes.See where you have gone wrong.
Image and video hosting by TinyPic

kishy72
Subject: Re: stuck in this sudoku @ 2015-09-25 1:31 PM (#19454 - in reply to #19453) (#19454) Top


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Please give me some starting steps for wall sudoku.



This one is hard to explain.I myself kind of stumbled to the solution after a long time.I will run you though the basic starting steps.

Remember that in this sudoku,there are 2 "golden" rules.

Rule 1 :: If there is no wall segment between 2 cells, those 2 cells are consecutive in value.
Rule 2 :: Each digit(unless it is a 1 or 12) will have connection to 2 digits which is consecutive to it .This means that if you get connection to 2 cells ,you can draw wall segments on the other 2 borders and vice versa. Digits 1 and 12 will have wall segments on 3 border grid lines.

* The zero clue indicates that there can be no wall segments on that line .Hence I marked 'x''s on that line .This means cells R23/C1,R23/C2,R23/C3.......R23/C12 are consecutive in value.

*Also I have marked red lines on some borders .These are obtained by deduction from the given clues.For instance,reason out how the (37)clue at the top can be satisfied.

*Look at R4C4 .It cannot be a 1 or 12.There are 2 red wall segments that are obtained by deduction from the clues.So from what I stated above in Rule 2 ,you can draw connections to 2 cells.

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Give it a start and get back if you are stuck !

Kishore

Edited by kishy72 2015-09-25 1:33 PM
gaurav.kjain
Subject: Re: stuck in this sudoku @ 2015-09-26 2:42 PM (#19465 - in reply to #19454) (#19465) Top


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Finally I am done with THE Maze Monster. Kishore your initial steps and explanation helped a lot. Thanks for wonderful explanation.

maze20002



But I have not got some initial markings, which you have shown by deducing from the number clues, Eventually, I got them while solving. Can you give me reason of markings highlighted in BLUE.

maze1





Edited by gaurav.kjain 2015-09-26 2:50 PM
kishy72
Subject: Re: stuck in this sudoku @ 2015-09-26 6:38 PM (#19468 - in reply to #19465) (#19468) Top


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But I have not got some initial markings, which you have shown by deducing from the number clues, Eventually, I got them while solving.



Yes .Some of the deductions are obtained as the solve proceeds.
kishy72
Subject: Re: stuck in this sudoku @ 2015-11-12 9:15 PM (#19907 - in reply to #8611) (#19907) Top


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After an indefinite hiatus,I am posting here in the forum again.At this instance, a Group sum sudoku from the IB of SM - 4

I tried this one a lot and got a lot of pencil marks going but not one concrete digit.I think it would be futile to post the one I had pencil marks on as it yielded nothing.Hence I am posting the image as is.....

Can someone please tell me how to put one digit without much effort ?

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Edited by kishy72 2015-11-12 9:16 PM
vopani
Subject: Re: stuck in this sudoku @ 2015-11-12 10:51 PM (#19909 - in reply to #8611) (#19909) Top


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Not sure if this is the best start, but here's what I did. It took me a while though.

R5C1 & R6C1 = 5 & 6
R5C2 + R6C2 = 5
R5C3 + R6C3 = 5

So, R2C3 + R3C3 = 5 OR 6 OR 7
It cannot be 7 because there is a 7-sum-pair below it (R3C3 + R4C3).
If it is 6, then R2C2 & R3C2 = 4 & 6 and R5C2 & R6C2 = 1 & 4 which is not possible (Basically, you can't fill up the 16-sum and 10-sum of column2-column3 together).

So, R2C3 + R3C3 = 5
R2C2 + R3C2 = 11 = 5 & 6
R2C1 + R3C1 = 3 = 1 & 2
R1C3 + R4C3 = 11 = 5 & 6
R3C1 & R3C3 = 1 & 2
R4C1 & R4C2 = 3 & 4

So,
R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Hence,
R5C2 & R6C2 = 2 & 3

R4C2 = 4 !

From there, it should get solved.
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