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hi.i m posting this sudoku taken from fed sudoku.yours truthfully has reached till this point without any problem but was unable to proceed beyond this point even after trying miserably for 20 mins.So people who solve sudokus unbelievably fast and make others blink in amazement!help this poor fellow to see light in this sudoku.

Edited by kishy72 2012-09-21 8:53 PM

Hi Kishore, After uploading the image in http://tinypic.com you will get "HTML for Websites".
Copy that and paste in your post.

"HTML for Websites" will look like : <a href="http://tinypic.com?ref=1zn20so" target="_blank"><img src="http://i47.tinypic.com/1zn20so.png" border="0" alt="Image and video hosting by TinyPic"></a>

You can also copy and paste "Direct Link for Layouts" which looks like : http://i47.tinypic.com/1zn20so.png

i am posting the link...thnks deb.i could not have posted this without your help.

Focus on R7C2 and R9C2 and column 2. You should be able find the next steps.

Spoiler: show

Edited by debmohanty 2012-09-21 10:05 PM

Administrator - 2012-09-21 9:51 PM

Focus on R7C2 and R9C2 and column 2. You should be able find the next steps.

Spoiler: show

hi i m not getting it.How did u lock (3/6)in (R7C2 /R9C2).(3 and 6) eliminated from R2C2,R3C2,R4C2-perfectly ok.But only 3 is getting eliminated from R5C2 .So 6 might be present in R5C2 right?!correct me if i am missing something.

kishy72 - 2012-09-22 11:44 AM

Administrator - 2012-09-21 9:51 PM

Focus on R7C2 and R9C2 and column 2. You should be able find the next steps.

Spoiler: show

In the bottom left box, the only digits that can be there are 3,6,7,8 right? Now in C2, you already have a place in R2 for 7-8, and R8 for 7-8. So 7-8 can't be in any of the rest of the cells in C2. So, R7 and R9 of C2 cannot have 7-8, and the only remaining digits that are possible from the original 4 is 3 and 6.

So basically R9C3 contains either 7 or 8 and therefore......................clear now . will be posting many more stuck sudokus .Thanks guys!

second in the list of sudokus ....where i m unable to cross beyond this point ...fatigue sets in after a certain point...not able to lock any number after this ....

Where do you put 2 in column 3?

Spoiler: show

It should solve pretty smoothly after that.

Administrator - 2012-09-23 8:28 PM

Where do you put 2 in column 3?

Spoiler: show

It should solve pretty smoothly after that.

Beautiful!i completed very fast after putting that number...great spotting!...hope i remember these small intricacies and implement them in future sudokus.thnk u !

third in the series of stuck sudokus...
how do i proceed from here fellas?!

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R1C6 can only be 5-7 due to normal single eliminations of 3,6,2. Similarly, R1C5 can only be 2-5-7. As you have already marked R1C9 as 5-2, that makes a triplet meaning 7 can only be in R1C5/6. It can also only be in R5C5/6, as you have marked. So in C4, you get just one spot to place the 7.

prasanna16391 - 2012-09-25 4:59 PM

R1C6 can only be 5-7 due to normal single eliminations of 3,6,2. Similarly, R1C5 can only be 2-5-7. As you have already marked R1C9 as 5-2, that makes a triplet meaning 7 can only be in R1C5/6. It can also only be in R5C5/6, as you have marked. So in C4, you get just one spot to place the 7.

amazing ..and here was i under the false assumption all these years that all it takes to solve a sudoku is a set of simple rules and logic when the tricks to be learnt in the trade is endless....i know i certainly wouldn't have placed that 7 after making that deduction of locking it in R1C5/R1C6....i m learning new tricks in sudoku ....all the 3 sudokus i failed to solve has some triplets...!!!!!i feel i m improving gradually after joining LMI....thnks prasanna!

Edited by kishy72 2012-09-25 7:50 PM

4th in the series ..help me out on this one fellas....hope u guys are not too busy in prep for wc...i am sure i combed the entire sudoku for triplets and could not notice any

Edited by kishy72 2012-09-27 11:05 PM

I remember how I solved this one(thats a rarity, I generally struggle to re-solve things).

Anyway, see R2 and R8. whichever way you see it, R2/8C6 has to be 1/3. Should be simple to see, say 1 and 3 are in R2C4 and C5, then both have to be in R8C6, and vice versa. So R7C6 has to be 6.

there is a triplet of 458 in r1r3r5/c6. after that u get a 6 in r7c6. its plain solving after that

akash.doulani - 2012-09-28 12:34 PM

there is a triplet of 458 in r1r3r5/c6. after that u get a 6 in r7c6. its plain solving after that

Yes indeed. I could not spot that...

A general query guys....is it advisable to put more than 2 possible numbers in a box? Initially i was not in the habit of putting possible numbers in a box.In fact i hated that since i felt it made the whole sudoku very clustered.But i have realised that it is impossible to solve sudokus of greater difficulty without making a note of the possible numbers that occur in a box.But still is it advisable to write 3-4 numbers in a box and try to solve it if u get stuck, or to just write numbers where you get only 2 after eliminating the rest..This is something which i had always wanted to know from others because even after putting 2, i have got stuck in many a sudoku and the whole thing gets clear only when i put 3 numbers ...want to be sure that sudokus can be solved without writing more than 2 possible numbers in the box....ideas are welcome...

Its about the method that suits you best. Personally, I don't like using pencil marks at all. I do better on sites without pencil marks, and I try but I can't stop myself from using them subconsciously when they are there as a part of the interface(been trying to). I've talked to others though and the normal thing is to note down numbers to keep track. So I suppose you'll find this way easier. The difficult ones need a lot of pencil marks, because to add difficult tricks, the creator would have to use a larger group of numbers in the grid and solving them requires an understanding of that entire group. So there's nothing wrong in entering 4-5 pencil marks in a cell as long as you don't end up confusing yourself(which is my problem and why I'm faster without that).

Edited by prasanna16391 2012-09-28 3:58 PM

@ kishore : from where do you get those sudokus ? they are pretty interesting to solve. can you share the source ?

I take them from fed sudoku...choose puzzle blogs from LMI home screen -->Fedsudoku--->sudoku archive--->classic printable....
They are my daily dosage of tablets for improving sudoku skills.

kishy72 - 2012-09-28 10:56 PM

I take them from fed sudoku...choose puzzle blogs from LMI home screen -->Fedsudoku--->sudoku archive--->classic printable....
They are my daily dosage of tablets for improving sudoku skills.

Akash is a regular there, he's just not been able to do them lately so didn't know I suppose. Anyway, I'm a regular there too. There's other sites as well with daily sudokus like Sudokucup, mindoku. These 3 are my daily routine anyway. Related to your earlier pencil marks question, mindoku's interface doesn't have pencil marks.

the way please in this

Edited by kishy72 2012-10-06 9:57 PM

Just by going the marks you have done (and assuming that they are correct), where do you put 3 in Box2?

However, I think you have gone wrong somewhere. I couldn't resolve the bottom 3 rows so early in the solve.

i have addled the sudoku so much that i could not see that direct spotting...and yes i have gone wrong. I started from the beginning again and i saw where i went wrong ....Thnks deb!

So fellas another of those tough computer generated ones where i feel it has managed to give the bare minimum number of numbers required .... a tough nut to crack for me....tried for 28mins.....totally warped .....shw me the way Lmians!!

Edited by kishy72 2012-10-22 7:42 PM

and if possible post a link where i can see these versions guys
Untouchable sudoku,Disjoint groups ,Symetric unequal , Antiknight
Queens ,Quadro ,Touchy , DuoDoku.... not even heard of some of these variants...planning to take part in sudoku cup want to try atleast one of each of these versions(hopefully) before the test.

Firstly the Classic - I remember this one. What I did while solving was focus in the first box. By process of elimination, you get that R2C1, R2C3 can only contain either of 4-6-9. You've already marked R2C7 containing 4-6. This makes a triplet and eliminates 6 from the 2-6 in R2C5, so you have a 2 there. Should be ok after that.

For the variants - You can look at past Monthly tests for some. I think I remember off the top of my head an Anti Knight in Sudoclones and a Queen Sudoku in Diagonal Vision. Apart from that, you have Disjoint groups in the Fed Sudoku site itself, in the Alternatives section. So too Quadro I think. There's also this site(Where disjoint groups is called Offset, Symmetric Unequal is I think No Mirror).

Edited by prasanna16391 2012-10-23 1:05 AM

kishy72 - 2012-10-22 9:20 PM

and if possible post a link where i can see these versions guys
Untouchable sudoku,Disjoint groups ,Symetric unequal , Antiknight
Queens ,Quadro ,Touchy , DuoDoku.... not even heard of some of these variants...planning to take part in sudoku cup want to try atleast one of each of these versions(hopefully) before the test.

You can also find some of them in the monthly test authored by Jan Mrozowski 2 years ago, called "Classic Look-Alikes", it's basically the same concept as the 1st round of sudokucup:

http://logicmastersindia.com/lmitests/?test=M201007S

Fred

Thnks prasanna and Fred for the links ...did not spot the 469 combination in R2C1 ......and by the way took this Queens sudoku from Diagonal vision....valued at 85 points i assumed initially that is too much for a simple sudoku (atleast it appeared to me like that with so many numbers) But i soon realised that this sudoku is worth its weight in gold......tried at 3 different times today to give my mind a break and still could not proceed beyond this point.....i could not use the queen constraint at all with just 1 9 locked till this point.

kishy72 - 2012-10-23 6:55 PM

just to make sure everyone knows the rules of queen sudoku....in simple words 9 is assumed to be a chess queen ...no queen(9) can attack other queens(other 9s)...all other sudoku rules apply

Edited by kishy72 2012-10-23 7:12 PM

and in symmetric unequal what does cells 180deg symmetric to each other mean..????totally blank....could not make out head or tail out of it...kindly explain guys..

The queen sudoku of diagonal vision's instructions booklet is a bit too hard... It's my fault.
We discussed about it there: http://logicmastersindia.com/forum/forums/thread-view.asp?tid=451&s...

Fred

kishy72 - 2012-10-23 7:02 PM

and in symmetric unequal what does cells 180deg symmetric to each other mean..????totally blank....could not make out head or tail out of it...kindly explain guys..

R1C1 is symmetric to R9C9, R1C2 is symmetric to R9C8, R2C1 symmetric to R8C9, etc... does it help?

Fred

Fred76 - 2012-10-23 7:22 PM

The queen sudoku of diagonal vision's instructions booklet is a bit too hard... It's my fault.
We discussed about it there: http://logicmastersindia.com/forum/forums/thread-view.asp?tid=451&s...

Fred

So i am not the only one stuck it seems! But Beautiful theme nevertheless....similar to what u do in a chess game.......think ahead and see if it is advisable to go with that continuation...perhaps thats why this sudoku is rightly named queen sudoku....i was looking for a logical continuation....did not look for a contradiction at all....

Fred76 - 2012-10-23 7:25 PM

kishy72 - 2012-10-23 7:02 PM

and in symmetric unequal what does cells 180deg symmetric to each other mean..????totally blank....could not make out head or tail out of it...kindly explain guys..

R1C1 is symmetric to R9C9, R1C2 is symmetric to R9C8, R2C1 symmetric to R8C9, etc... does it help?

Fred

So going by the same logic R5C1 is symmetric to R5C9 ,R5C2 is symmetric to R5C8 etc...and the only cell symmetric to itself in the entire grid is R5C5...?!

kishy72 - 2012-10-23 7:45 PM
So going by the same logic R5C1 is symmetric to R5C9 ,R5C2 is symmetric to R5C8 etc...and the only cell symmetric to itself in the entire grid is R5C5...?!

That is correct. Now all the additional rule says is "If two different cells are 180⁰ symmetric to each other, they must contain different digits".
More examples: mock12 IB & PB - http://logicmastersindia.com/lmitests/?test=mock12

not able to touch the solution in this untouchable sudoku....cells with same digits dont touch diagonally..first sudoku of this variant which i had tried ...where am i missing the trick here?

Edited by kishy72 2012-10-29 11:01 PM

kishy72 - 2012-10-29 11:00 PM

not able to touch the solution in this untouchable sudoku....cells with same digits dont touch diagonally..first sudoku of this variant which i had tried ...where am i missing the trick here?

Kishy: BE CAREFUL NOT TO POST THE SUDOKUS OF A COMPETITION BEFORE IT IS FINISHED !

The competition is now finished, but it was not the case when you posted this one.

That being said, try to place the 2 in row 5, and then you can place lot of other digits

Fred

Fred76 - 2012-10-30 4:32 AM

Kishy: BE CAREFUL NOT TO POST THE SUDOKUS OF A COMPETITION BEFORE IT IS FINISHED !

The competition is now finished, but it was not the case when you posted this one.

Kishy72, Consider this as your first and last warning. Never post puzzles from LMI / other sites while the competition is running. Always wait for few hours after the competition ends.

Fred76 - 2012-10-30 4:32 AM

Fred

BE CAREFUL NOT TO POST THE SUDOKUS OF A COMPETITION BEFORE IT IS FINISHED

This thought did cross to me immediately after i had posted it but i left it at that thinking that the person had just as much a chance of going wrong as getting it correct by seeing this sudoku but anyway i know that that is not the point.It wont occur again.

Administrator - 2012-10-30 8:48 AM

Fred76 - 2012-10-30 4:32 AM

Kishy: BE CAREFUL NOT TO POST THE SUDOKUS OF A COMPETITION BEFORE IT IS FINISHED !

The competition is now finished, but it was not the case when you posted this one.

Kishy72, Consider this as your first and last warning. Never post puzzles from LMI / other sites while the competition is running. Always wait for few hours after the competition ends.

You could have removed my post instead of admonishing me like this or you could have said to me in private before u posted this.This looks like slapping me in the face and saying 'Dont post sudokus in this forum'.I m not a fugitive under warrant

kishy72 - 2012-10-30 10:33 AM

You could have removed my post instead of admonishing me like this or you could have said to me in private before u posted this.This looks like slapping me in the face and saying 'Dont post sudokus in this forum'.I m not a fugitive under warrant

Well you'll have to be to solve the next test :P Anyway, in all seriousness, what you are doing(posting on forum and asking doubts) is great. In fact we've been encouraging beginners to do that for some time now. Its just that, when something like this happens, we need to make sure it doesn't happen again through anyone, not just you. I think that was the reason for the public warning. Do be careful in the future :)

I just wanted to say a quick Thank You for the tricks I've learned only on these 2 pages. I have never looked for triplets in a puzzle! This is the next way to go! Thank you very much!

http://i46.tinypic.com/14yba10.jpg

I have had trouble seeing a way to solve this puzzle completely avoiding trial and error. Can someone please help me?

I tried to fit it rigth into the post, but it was rather big.

Edited by purzelbaumfan 2013-03-29 4:59 PM

I can proceed further by seeing that 5 in Row 7 put in either place puts 7 in R7C4, which can be seen using row 4 for one placement of 5, and the other is pretty straightforward. There's another sticking point in proving that R9C2 can't be 4 (4-5 in the 7th box means 5 must be in R4C2, where'll 7 go in C2?). Its pretty difficult, but thats the closest I can come to a solid path.

Thank you very much. That was a very decent solving path without trying too much. I appreciate your help.

This is another puzzle I didn't get any further with. Any suggestions?

Edited by purzelbaumfan 2013-03-30 10:08 PM

There's a way to eliminate 4 from R5C6 (It then goes in R7C5 and eliminates both possibilities in column 8!). Also, only 4/8 can be in R9C9 because of R7C5, which means one of 4/8 in Row 7 is in C8/9, so R7C6 is 3/9 only. After that, I could prove that R9C4 cannot be a 3 by just using 3s and 9s and seeing the crowd-out that happens in R1C9. But this last step's bordering on guessing, so I'm not sure if thats the intended path.

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?

Edited by purzelbaumfan 2013-04-03 11:34 PM

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?

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

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?

Edited by kishy72 2013-04-25 8:40 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

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?!

Edited by kishy72 2013-04-27 11:26 PM

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

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.

Edited by kishy72 2013-05-02 7:04 PM

kishy72 - 2013-05-02 7:04 PM
A better pic?!Not able to get the resolution that i want.

Edited by kishy72 2013-05-02 8:08 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.

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)

Edited by kishy72 2013-06-26 9:32 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.

prasanna16391 - 2013-06-26 9:44 PM

Nice.Thnks a lot!I think i can complete it now.

Pls help me in solving this sudoku.

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.

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.

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

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)

Edited by kishy72 2014-05-13 8:19 PM

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. :)

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----

Edited by kishy72 2014-05-15 9:00 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

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?

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 :)

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

Edited by ka_bharath 2014-07-02 7:43 PM

its quite easy observe naked pair of 3,4 at R3C6,R6C6 and naked single i.e 4 at R2C6

here is the solution

416532
235164
124356
653421
342615
561243

Just to correct it, I think you meant naked pair of 3,6

dp_94 - 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

Sudoku Surprise - What an amazing test with a great set of sudokus.One of those tests which I felt had that rare 'reunion'of sorts between sudokus and puzzles.Rather than reflecting on my performance which was sub-optimal,I thought it would be prudent to get down to brass tacks and try understanding the beautiful intricacies hidden in each puzzle.

So,I came across this Secret Code sudoku which I attempted during the test and spent close to an hour!! on this single puzzle and still could not get to the solution (logically).I guessed the answer key knowing that it had to contain one of these strings 2316,2361,3216,3261 in Column 3 .Once again ,it turned out to be a lucky guess .Hence to be candid, it's only 470 points that I could muster in reality.
The reason that I am posting this here is that 80% of the solvers should have reached the following stage after which they would have hit a dead-end.The Sudoku could not have become more stuck.
So fellas,if someone did find that devastating piece of logic from this point which brings the curtains down on this one ,I would be glad to know.

----Kishore-----

Edited by kishy72 2014-11-26 4:40 PM

I guess everyone was stuck for sometime at this point.
Firstly if you see the arrows which are least pointing directions:eg r6c1 or r1c3 they can point at most 3 directions.
Now the logic i used is if you see 2-3 pair, for the solution to become unique atleast one of 2 or 3 has to be part of secret code. So this restricts further the direction of r6c1 and it has to be pointing right. (This way i got one direction - so atleast now i know that secret code belongs to order which will follow row 6 and it has atleast one of 2 or 3 and it can't be 3rd digit of the code)
This immediately fixes the direction of r3c9 which has to be towards r4c8. Also this implies that secret code cannot contain 4,5,8.
Now this gives that r2c4 arrow has to be towards right or towards r3c5. But now if its towards r3c5 then it has to be 376 due to obvious reasons. but this is not possible immediately you can check that with known arrow directions i.e row 6. So r2c4 arrow is pointing right. which implies code is 371 (as 4 is not part of it and 2 cannot be as it is last).

Hope this helps. I don't know if above is right way or not. But would still love to know if any other shorter way is there :)

*sorry small edit in above r6c1 can be 4 directions but anyways that doesn't change the analysis :)

Excellent!I think that you have provided the precise logical way of solving this sudoku.I seriously doubt if there is a shorter way than the above mentioned one.
Right at step 1 ,I faltered.I mean,I failed to deduce that, atleast one of 2 or 3 has to be part of the code as it did not hit me that the puzzle would fail to be unique without that.Tremendous!!!
Also, I understand that the most important strategy to adopt in this puzzle is to concatenate clues and see if they work in conjunction with other clues.These are the grey areas that needs to worked on.
Thanks a lot for the explanation.Well done !!

Hi all!!I was randomly solving some sudokus from the IB of Swiss Team Qualification round when I came across this Arrow-thermo sudoku .I fail to see a logical continuation from the following point....Can someone spot the same and share .....

Edited by kishy72 2015-06-02 6:07 PM

R5C3 and R7C3 contain (3,5). So, R6C3 = 7, and R1C3 = 6 ......
From here, it can be solved forward.

Advised only seeing just plainly, not gone through whole.. Maybe wrong also

Yajendra - 2015-06-02 7:40 PM

R5C3 and R7C3 contain (3,5). So, R6C3 = 7, and R1C3 = 6 ......
From here, it can be solved forward.

Advised only seeing just plainly, not gone through whole.. Maybe wrong also

There is also possibility of 3 in R3C3/R3C6.So your deduction is not correct.

7 can only come at R3C9 or R5C9

if you put 7 at R5C9, then 9 is forced at R6C8
which will force 9 at R5C2, this will force 9 at R3C1.

9 at R3C1 will force 9 at R2C8, which is wrong as R6C8 already contains 9

so 7 should come at R3C9.
It should be easy solve from there on.

Somewhat similar to gaurav but a 2 step smaller,

Consider 1,3,4,6 boxes. Each has only 2 or 3 places of placing 9. So visually its easy to see that placing 9 at r3c7 leaves no room to place 9 in box 4. After this its easy i suppose.

gaurav.kjain - 2015-06-03 7:33 AM

7 can only come at R3C9 or R5C9

if you put 7 at R5C9, then 9 is forced at R6C8
which will force 9 at R5C2, this will force 9 at R3C1.

9 at R3C1 will force 9 at R2C8, which is wrong as R6C8 already contains 9

so 7 should come at R3C9.
It should be easy solve from there on.

Consider 1,3,4,6 boxes. Each has only 2 or 3 places of placing 9. So visually its easy to see that placing 9 at r3c7 leaves no room to place 9 in box 4

Interesting ways of looking at the logic for the same outcome.The first appears to be a bit of a hit and trial and the second approach is shorter but much more difficult and global/broader thinking is required.Nevertheless the latter seems the ideal way for dealing with the issue aptly.So then I assume that there is no other shorter logical path than the ones mentioned.Thanks a lot for sharing the logic ....Gaurav and Swaroop

Kishore

Stuck in this combo variant.Anti-knight + Renban.Same numbers are not chess knight-move connected and all groups of connected cells form consecutive numbers.I know it is not one of those regular sudokus that we get to solve most of the time.But still,it is that long chain of logical thinking and visualization required before hitting a contradiction in these kind of variants that fascinates me and induced me to solve these......

Source and Author Credits : Logic masters deutschland / Zhergan

Edited by kishy72 2015-06-11 7:11 PM

7 will be in r7c5 because in box 5 , 7 can only come at r4c6, r6c6, r6c4 eliminating 7 from r7c6. After this even I am stuck

akash.doulani - 2015-06-11 11:24 PM

7 will be in r7c5 because in box 5 , 7 can only come at r4c6, r6c6, r6c4 eliminating 7 from r7c6. After this even I am stuck

I am afraid but I don't see how you eliminated 7 from R6C5?!If you consider that case 7 may in fact be in R7C6

I'm not sure if the cross you made for R9C4 is to denote that 8 cannot be there, but I assume so (because it is true that it can't be there, you'll reach a contradiction placing an 8 in C3). This means there's a 7-8 pair in C4 in R2 and R6, which gives a 9 in R4C4.

prasanna16391 - 2015-06-12 1:18 AM

I'm not sure if the cross you made for R9C4 is to denote that 8 cannot be there, but I assume so (because it is true that it can't be there, you'll reach a contradiction placing an 8 in C3). This means there's a 7-8 pair in C4 in R2 and R6, which gives a 9 in R4C4.

Indeed!Thanks a lot!!It was close to impossible finding that 78 pair amidst that clutter.It looks almost invisible to the human eye.I got to add here that this sudoku is in no way trivial and tremendously sticky even after using this deduction .I would advise anyone interested to have a go at it seriously and give themselves a pat after completion.I completed it just before and the solving experience and the satisfaction of completion was like nothing before.

Thanks again prasanna for setting the direction in the right path in the above sudoku!

Seriously deadly sudoku, even after placing that 9 i am not sure if there is any simplified steps.

I was trying to solve this Search 9 Sudoku on Fed-Sudoku and am stuck at this point. I don't know how to proceed further. Can anyone give a hint? Have I made any mistake?

Edited by aashay 2015-06-21 11:44 PM

aashay - 2015-06-21 11:44 PM

I was trying to solve this Search 9 Sudoku on Fed-Sudoku and am stuck at this point. I don't know how to proceed further. Can anyone give a hint? Have I made any mistake?

There is a mistake in your solve.Look at the 3 clue in R5C3.You have a 9 in R6C4 which violates that.You have gone wrong somewhere in between....

Edited by kishy72 2015-06-22 1:14 AM

kishy72 - 2015-06-22 1:13 AM

There is a mistake in your solve.Look at the 3 clue in R5C3.You have a 9 in R6C4 which violates that.You have gone wrong somewhere in between....

It may not be wrong - the rules are different on the Fed series. All possible arrows are given where a digit points towards a 9 at that distance away from it, so it is very different from the ISC one.

I also thought the same, so I reset it and solved again from scratch, but reached the same point.

aashay - 2015-06-22 8:40 AM

I also thought the same, so I reset it and solved again from scratch, but reached the same point.

What you are missing is the "all possible arrows are marked" of this puzzle. What you have solved so far mostly isn't wrong but there is an extra rule here which isn't in the ISC one - if there is say a 3 in a cell without an arrow, there cannot be a 9 three cells away from that 3 in any direction (I repeat, this is NOT a rule for the ISC one)

Please Classics vs. Innovatives IB & PB for extra practice material for Search 9 with exact rules.

aashay - 2015-06-22 8:40 AM

I also thought the same, so I reset it and solved again from scratch, but reached the same point.

Here's the solution to the sudoku that you had posted with the rule(all possible arrows are marked) that apparently you seem to have missed.

I will add a few differences between the search 9 in Fed sudoku and the one in the ISC so that you don't get confused.

**First and foremost a Search 9 clue in FED doesn't necessarily mean the nearest 9 in the direction of the arrow.This obviously means that there may or may not be multiple 9s in the path of the arrow.However the one in ISC indicates the distance to the NEAREST 9 .

**Also since all possible arrow clues are given in FED,you can't have a digit in cell which equals the distance to 9 in any of its eight directions.This need not necessarily be the case in ISC search 9 which means there is no restriction on the cells which don't have an arrow clue.

I hope it's clear.

Kishore

Diminutive and simple by its appearance yet very far from that........

True or False Sudoku from Sudoku Mania by Rishi Puri aka purifire(Rules :: In the cells with clues,the actual digit is one higher or one lower or same as the clue )

The following is the tipping point of the sudoku, where solvers can decide whether to take a guess and get it done and dusted or to wait and hunt for that beautiful elusive logic which seems to be hidden somewhere deep inside.

I prefer the latter and hence would greatly appreciate some form of help at this point///.I spent close to 50 mins at this stage when I started getting desperate and decided that the logic is simply out of reach.

If someone could see the logical continuation from here,please do share .....

Kishore

Edited by kishy72 2015-06-25 12:27 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.......

Rishi told me that this was easy/medium difficulty, so I don't think you should expect a *-wing there.

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.

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.

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.

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

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

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.

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

Edited by gaurav.kjain 2015-07-05 7:12 PM

R7c6 will be 5 because of BUG

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.

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.

@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?

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.

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.

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.....

Kishore

Edited by kishy72 2015-09-08 12:03 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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Consecutive.png (14KB - 6 downloads)

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

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.

Thank you.That makes it clear!

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>

I think last row, 1 will come at C3.

'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?

Edited by kishy72 2015-09-24 7:00 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

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?

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

Kishore

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.

imag

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.

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

This is how the crossword goes.See where you have gone wrong.

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.

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

Kishore

Edited by kishy72 2015-09-25 1:33 PM

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

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.

Edited by gaurav.kjain 2015-09-26 2:50 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.

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 ?

Edited by kishy72 2015-11-12 9:16 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.

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 .

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"

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 :) .

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 :-)

Thanks Rohan.

But I am unable to decipher this logical deduction,

R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Can someone make this clear?

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.

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.... :)

I thought that you created 2 group sum sudoku 6*6 and this one was too easy for the tournament

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

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

Thanks Swaroop, that did the trick. It's a pretty tough deduction to find though, I guess I still need more practice.

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

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.

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
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CP.png (61KB - 1 downloads)

IB.png (100KB - 0 downloads)

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.

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

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.

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
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LK.jpg (45KB - 2 downloads)

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.

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.