@ghirsch
How did you mark the bottom 3-polyomino?

Why can't it be a 1?



You're right, it could be a star.



I'll give that sort of notation a try, thanks for the advice and starting point.

The reasoning goes this way :

i)R1 and R2 contain a total of 4 stars.
ii)Each 2X2 square can contain only a maximum of 1 star.Therefore cells R1C9,R1C10,R2C9 and R2C10 can contain a maximum of one star.Similarly Cells R1C6,R1C7,R2C6 and R2C7 can have a maximum of 1 star.
iii)To make the total 4 stars,the only place is R1C1.It has to be a star.

Thanks everyone, I figured out the rest of it now.

A NEWS Puzzle from forsmarts which I have no clue on how to go about.There is no particular reason for posting this puzzle now .I found it lying in a corner of my room along with many other unsolved puzzles.While I could do some of the rest,this caused me trouble .I had a look at the solution and still could not figure out the start.So if someone can share the starting steps ,it would be helpful.



Edited by kishy72 2015-08-20 7:34 PM

A few starting thoughts (though no major break-ins):



1) You can eliminate some spaces (the red slashes) because of the existing directions in regions (e.g. R1C6 is north of the N in the region).
2) You can eliminate some spaces (the green slashes) because the only possible entry is already in that row/column (e.g. R1C3 is northwest of the S, but the column already has an N and a W).
3) There are 22 regions (so 44 letters) and 12 rows/columns. Each type of letter can only go in 11 different rows/columns (e.g. there can't be an S in the top column), so every letter must be in 11 of 12 rows/columns.
4) This isn't included in the picture, but the region containing R5C7 can't have an N or S in it, so there must be a W in R5C7 (I'm sure this can be used elsewhere too).

Edited by ghirsch 2015-08-20 11:31 PM

Hi ghirsch. Thanks a lot for the starters.However,I am still not clear.How did you shade R2C5/C6/C7 and lots of other squares too.For instance ,why can't be a E present in R2C5 (East of South) and a W present in R1C6 and R2C6/7 (West of North)

The region containing W(R9C3) has E in R12C5 as per solution.How is it even possible?? E has to be right opposite W right?I mean in line with W.How's it located to the lower right of it or in words South East of it?

Hi Kishy & ghirsch,
I believe you have not interpreted rules correctly.
The alphabets has to just follow the sense of direction.
What I meant by this is, E just has to be on right hand side of W, it doesn't need to be in line with W. it can be up or down but right side.
E.g. Consider the region R8C5 "W" clue region. E can be in R8C6, R7C6 or R9C6 but it cannot be at R12C5.

Similarly for N, W , S as well.
Another example S has to be on right side and at same time downside of W.

Check this given example for clarity:
http://logicmastersindia.com/lmitests/dl.asp?attachmentid=147&view=...

Because then S won't be South of E. Look at it both ways, simply.

Thanks Swaroop and Prasanna ! I could proceed till here after which it's too difficult .Where am I missing the trick?



Edited by kishy72 2015-08-22 12:11 AM

The upside down L tetromino must have an N in the top row. Then you get an NW-pair in row 3.

I don't see how the former leads to the latter?! Where is the NW pair that you refer in row 3?How did you deduce that?

Hi kishy,
I haven't solved completely but looking at your image, I can tell you next few steps immediately:
1) R5C9 and R5C10 --> N will come in that region no matter what and in one of these places. (you can see what happens if N will not come)
2) Because of this R5C12 will be cross.
3) Now this gives N-W pair in Row 3, N or W at R3 -C5C6 AND N OR W at R3 C11-12.
4) So this eliminates N at R3 C1C2.
5) So this gives "S" at R5C1 and "E" at R3C2. So that will give "S" at r8c3.
I believe it should go easy after this.

I can't thank you enough Swaroop.I have done and dusted the puzzle finally and that's a tremendous relief after breaking my head for close to 2 days.The way you explained about that NW Pair in point 3 made me see clearly what I failed to observe....

A Numberlink puzzle.I realized that this puzzle has many ways of completing itself.Where am I going wrong?


Edited by kishy72 2015-08-30 2:50 PM

In addition,in the above puzzle,I had to use the constraint that all cells need to be visited by paths without which I found 'n' number of ways of completing this puzzle.Is there something critical that I am missing here?

This was a broken puzzle (I suppose you have taken this from one old LMI test)


Talking about Numberlink in general, this write up (Numberlink Primer) by Palmer will be an eye-opener for most.

Ah! I see.....It was taken from Nikoli Selection 2010 LMI monthly test.



Thanks a lot Deb!I will look through this primer.

I ran into some problems in the following Hitori puzzle with just a few steps away from completion.Despite being so close,I fail to see a logical continuation here. Can someone post the solution ?



Edited by kishy72 2015-08-31 7:10 PM

one quick next step: R1C2 will be shaded.
Reason: Either of R8C1 or R8C2 will be shaded so that means R7C2 is unshaded in any case. Hence above step.

Thank you ! I failed to see the 7 getting trapped.Done with the puzzle.

A very difficult Hitori Puzzle ! Can someone please post the continuation here where I am stuck?


Edited by kishy72 2015-09-02 1:43 AM

hi,
can you share the original puzzle link ?
I am confused whether what you have circled or not.
E.g. Is R6C5 "4" clue circled or not ?

Sorry.You are right.It's not something which I deduced logically.I forgot to erase it.

R6c5 "4:" clue should be unshaded only.