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.

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

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.

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

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

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.

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.

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.

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.

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.

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.