Thread - stuck in this sudoku

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

kishy72

Posts: 419

Country : India

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

Country : India

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

Posts: 668

Country : India

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 .

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

Posts: 739

Country : India

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

Posts: 51

Country : India

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

Posts: 739

Country : India

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

Posts: 329

Country : India

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

Posts: 337

Country : Switzerland

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

Posts: 102

Country : United States

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.

Edited by ghirsch 2015-11-15 4:51 AM

@ 2015-11-15 6:13 AM (#19947 - in reply to #19946) (#19947) Top

swaroop2011

Posts: 668

Country : India

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.

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

Posts: 102

Country : United States

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

Posts: 419

Country : India

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 ?

Edited by kishy72 2016-04-05 1:39 PM

@ 2016-04-05 6:00 PM (#21436 - in reply to #21433) (#21436) Top

rajeshk

Posts: 542

Country : India

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 ?

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

Posts: 419

Country : India

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
----------------

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

Posts: 419

Country : India

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

Posts: 6

Country : China

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

Posts: 419

Country : India

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.

@ 2019-11-23 12:36 PM (#27362 - in reply to #8611) (#27362) Top

kishy72

Posts: 419

Country : India

kishy72 posted @ 2019-11-23 12:36 PM

Little Killer from LMD portal by 'Realshaggy'

This sudoku is pure evil.I tried everything that I usually encounter in Little Killer Sudokus from totalling clues,seeing min-max possibilities, clue interaction etc., but this sudoku just yielded nothing.I have been trying for the past week or so to complete this without success.It's kind of demotivating in a way that after so many years of solving sudoku, there are still some that I am unable to be complete. Kindly someone share the break-in for this ' psycho little killer '.

(LK.jpg)

Attachments
----------------

LK.jpg (45KB - 2 downloads)

@ 2019-11-23 6:15 PM (#27363 - in reply to #8611) (#27363) Top

Puzzle_Maestro

Posts: 25

Country : United Kingdom

Puzzle_Maestro posted @ 2019-11-23 6:15 PM

If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42.

Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs.

@ 2019-11-25 8:52 AM (#27364 - in reply to #27363) (#27364) Top

kishy72

Posts: 419

Country : India

kishy72 posted @ 2019-11-25 8:52 AM

Puzzle_Maestro - 2019-11-23 6:15 PM

If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42.

Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs.

Thanks a lot ! I could complete it with your pointer.


	@ 2015-11-13 2:29 AM (#19912 - in reply to #19909) (#19912) Top
kishy72 Posts: 419 Country : India	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 Country : India	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 Posts: 668 Country : India	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 Posts: 739 Country : India	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 Posts: 51 Country : India	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 Posts: 739 Country : India	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 Posts: 329 Country : India	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 Posts: 337 Country : Switzerland	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 Posts: 102 Country : United States	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. Edited by ghirsch 2015-11-15 4:51 AM
	@ 2015-11-15 6:13 AM (#19947 - in reply to #19946) (#19947) Top
swaroop2011 Posts: 668 Country : India	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. 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 Posts: 102 Country : United States	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 Posts: 419 Country : India	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 ? Edited by kishy72 2016-04-05 1:39 PM
	@ 2016-04-05 6:00 PM (#21436 - in reply to #21433) (#21436) Top
rajeshk Posts: 542 Country : India	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 ? 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 Posts: 419 Country : India	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 ---------------- CP.png (61KB - 1 downloads) IB.png (100KB - 0 downloads)
	@ 2019-10-25 1:40 PM (#27344 - in reply to #8611) (#27344) Top
kishy72 Posts: 419 Country : India	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 Posts: 6 Country : China	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 Posts: 419 Country : India	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.
	@ 2019-11-23 12:36 PM (#27362 - in reply to #8611) (#27362) Top
kishy72 Posts: 419 Country : India	kishy72 posted @ 2019-11-23 12:36 PM Little Killer from LMD portal by 'Realshaggy' This sudoku is pure evil.I tried everything that I usually encounter in Little Killer Sudokus from totalling clues,seeing min-max possibilities, clue interaction etc., but this sudoku just yielded nothing.I have been trying for the past week or so to complete this without success.It's kind of demotivating in a way that after so many years of solving sudoku, there are still some that I am unable to be complete. Kindly someone share the break-in for this ' psycho little killer '. (LK.jpg) Attachments ---------------- LK.jpg (45KB - 2 downloads)
	@ 2019-11-23 6:15 PM (#27363 - in reply to #8611) (#27363) Top
Puzzle_Maestro Posts: 25 Country : United Kingdom	Puzzle_Maestro posted @ 2019-11-23 6:15 PM If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42. Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs.
	@ 2019-11-25 8:52 AM (#27364 - in reply to #27363) (#27364) Top
kishy72 Posts: 419 Country : India	kishy72 posted @ 2019-11-25 8:52 AM Puzzle_Maestro - 2019-11-23 6:15 PM If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42. Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs. Thanks a lot ! I could complete it with your pointer.