Yajilin Yacht (28th Jul - 17th Aug) has started Discuss
stuck in this sudoku159 posts • Page 6 of 7 • 1 2 3 4 5 6 7
@ 2015-09-25 12:32 PM (#19452 - in reply to #19451) (#19452) Top

kishy72



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kishy72 posted @ 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.
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@ 2015-09-25 12:42 PM (#19453 - in reply to #19452) (#19453) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-09-25 12:42 PM

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

@ 2015-09-25 1:31 PM (#19454 - in reply to #19453) (#19454) Top

kishy72



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kishy72 posted @ 2015-09-25 1:31 PM



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
@ 2015-09-26 2:42 PM (#19465 - in reply to #19454) (#19465) Top

gaurav.kjain



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gaurav.kjain posted @ 2015-09-26 2:42 PM

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
@ 2015-09-26 6:38 PM (#19468 - in reply to #19465) (#19468) Top

kishy72



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kishy72 posted @ 2015-09-26 6:38 PM



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.
@ 2015-11-12 9:15 PM (#19907 - in reply to #8611) (#19907) Top

kishy72



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kishy72 posted @ 2015-11-12 9:15 PM

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
@ 2015-11-12 10:51 PM (#19909 - in reply to #8611) (#19909) Top

vopani



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vopani posted @ 2015-11-12 10:51 PM

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.
@ 2015-11-13 12:26 AM (#19911 - in reply to #19909) (#19911) Top

swaroop2011




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swaroop2011 posted @ 2015-11-13 12:26 AM

Nice.
@ 2015-11-13 2:29 AM (#19912 - in reply to #19909) (#19912) Top

kishy72



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kishy72 posted @ 2015-11-13 2:29 AM

Rohan Rao - 2015-11-12 10:51 PM

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.


Thanks Rohan ! Excellent chain of logical thinking .But seriously,were all those meant to deduced during a test and if anyone would wait for logic to put in a digit ? I am wondering the source of the example .
@ 2015-11-13 5:26 AM (#19913 - in reply to #19912) (#19913) Top

debmohanty




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debmohanty posted @ 2015-11-13 5:26 AM

kishy72 - 2015-11-13 2:29 AM
I am wondering the source of the example .

I picked it from my IPC2011 folder - so it had to be bit puzzle-y in nature.
However, I'm not able to see this puzzle in any IPC2011 booklet, so this may be one of the rejects because "too-hard-to-start"
@ 2015-11-13 7:43 AM (#19914 - in reply to #19913) (#19914) Top

swaroop2011




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swaroop2011 posted @ 2015-11-13 7:43 AM

debmohanty - 2015-11-13 5:26 AM

kishy72 - 2015-11-13 2:29 AM
I am wondering the source of the example .

I picked it from my IPC2011 folder - so it had to be bit puzzle-y in nature.
However, I'm not able to see this puzzle in any IPC2011 booklet, so this may be one of the rejects because "too-hard-to-start"


wow ! not ISC but IPC ?? Well anyways 2011 bangalore offline event was real fun. My first National championship :) .
@ 2015-11-13 9:21 AM (#19915 - in reply to #19912) (#19915) Top

vopani



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vopani posted @ 2015-11-13 9:21 AM

kishy72 - 2015-11-13 2:29 AM

Thanks Rohan ! Excellent chain of logical thinking .But seriously,were all those meant to deduced during a test and if anyone would wait for logic to put in a digit ? I am wondering the source of the example .

Nope. I would have guessed after 30-secs.

It took me ~5-6mins to come up with this logic. Instead, I could've solved this by guessing multiple times in 5-6mins :-)
@ 2015-11-13 2:26 PM (#19917 - in reply to #19909) (#19917) Top

RameshLMI



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RameshLMI posted @ 2015-11-13 2:26 PM

Thanks Rohan.

But I am unable to decipher this logical deduction,

R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Can someone make this clear?
@ 2015-11-13 2:33 PM (#19918 - in reply to #19917) (#19918) Top

vopani



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vopani posted @ 2015-11-13 2:33 PM

RameshLMI - 2015-11-13 2:26 PM

Thanks Rohan.

But I am unable to decipher this logical deduction,

R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Can someone make this clear?

Look at Row4 and Row5. Two sums of 18 and 9 are given (which used 8 cells out of 12).
So, the remaining 4 cells: R4C1 + R4C2 + R5C1 + R5C2 = 42 - 9 - 18 = 15 (to make the total of the two rows 42)

Since R4C1 and R4C2 were found to be 3 & 4, we get R5C1 + R5C2 = 15 - 7 = 8.
And using the 16-sum, R6C1 + R6C2 = 8.
@ 2015-11-13 11:29 PM (#19919 - in reply to #8611) (#19919) Top

neerajmehrotra



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neerajmehrotra posted @ 2015-11-13 11:29 PM

I was lucky to have guessed a better start...could solve it easily then.... Its almost impossible for me decipher Rohan's Logic during a competition.... :)
@ 2015-11-13 11:57 PM (#19920 - in reply to #8611) (#19920) Top

Fred76




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Fred76 posted @ 2015-11-13 11:57 PM

I thought that you created 2 group sum sudoku 6*6 and this one was too easy for the tournament
@ 2015-11-15 4:47 AM (#19946 - in reply to #8611) (#19946) Top

ghirsch



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ghirsch posted @ 2015-11-15 4:47 AM

I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do.

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Edited by ghirsch 2015-11-15 4:51 AM
@ 2015-11-15 6:13 AM (#19947 - in reply to #19946) (#19947) Top

swaroop2011




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swaroop2011 posted @ 2015-11-15 6:13 AM

ghirsch - 2015-11-15 4:47 AM

I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do.

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9 at r7c7 due to 34 pair at r1c7 and r6c7

Edited by swaroop2011 2015-11-15 6:13 AM
@ 2015-11-15 6:39 AM (#19948 - in reply to #8611) (#19948) Top

ghirsch



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ghirsch posted @ 2015-11-15 6:39 AM

Thanks Swaroop, that did the trick. It's a pretty tough deduction to find though, I guess I still need more practice.
@ 2016-04-05 1:38 PM (#21433 - in reply to #8611) (#21433) Top

kishy72



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kishy72 posted @ 2016-04-05 1:38 PM

Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ?

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Edited by kishy72 2016-04-05 1:39 PM
@ 2016-04-05 6:00 PM (#21436 - in reply to #21433) (#21436) Top

rajeshk



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rajeshk posted @ 2016-04-05 6:00 PM

kishy72 - 2016-04-05 1:38 PM

Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ?

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Center has to be 1. In case we put 7 there then there will be no place left to put 7 in 6th Box.
@ 2019-10-25 1:25 PM (#27343 - in reply to #8611) (#27343) Top

kishy72



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kishy72 posted @ 2019-10-25 1:25 PM

The following sudoku is from CSOC 63 PB.I couldn't understand the English version of the rules given in the example image.I assumed it to be a clone sudoku and started solving like that in contest.However, it quickly broke after that.


Can someone clarify what the rules imply ?



(CP.png)



(IB.png)



Attachments
----------------
Attachments CP.png (61KB - 1 downloads)
Attachments IB.png (100KB - 0 downloads)
@ 2019-10-25 1:40 PM (#27344 - in reply to #8611) (#27344) Top

kishy72



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kishy72 posted @ 2019-10-25 1:40 PM

Also, can the admin kindly specify alternate websites for uploading images here ? Earlier, I used to upload from tinypic which seems to have shut down now.
The images posted above are screenshots and are too big and uncomfortable looking.
@ 2019-10-26 7:23 PM (#27348 - in reply to #8611) (#27348) Top

xiao01wei



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xiao01wei posted @ 2019-10-26 7:23 PM

CSOC 63 P13's rule:
if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes:
A        C
B        D
and
if A>B, and then the result is C>D at the same time
if A<B, and then the result is C<D at the same time
if C>D, and then the result is A>B at the same time
if C<D, and then the result is A<B at the same time
the same inequality between the four numbers in two shapes with corresponding position

Edited by xiao01wei 2019-10-26 7:30 PM
@ 2019-10-26 11:38 PM (#27350 - in reply to #27348) (#27350) Top

kishy72



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kishy72 posted @ 2019-10-26 11:38 PM

xiao01wei - 2019-10-26 7:23 PM

CSOC 63 P13's rule:
if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes:
A        C
B        D
and
if A>B, and then the result is C>D at the same time
if A if C>D, and then the result is A>B at the same time
if C the same inequality between the four numbers in two shapes with corresponding position


Thanks a lot Xiao Wei for the clarification ! I finished the sudoku now.It is certainly an interesting variant and I will look forward to seeing more of it in future.
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