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.

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

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?

Fair point. In a competition at this stage I would always prefer T&E.

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.





Attachments
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Consecutive.png (14KB - 6 downloads)

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.



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.