'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

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

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.

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

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.

Nice.

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 .

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 :) .

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?

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

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

Center has to be 1. In case we put 7 there then there will be no place left to put 7 in 6th Box.