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

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

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.

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