The New Sudoku Players' Forum

[tabber]

Hi All,

I am stuck on the moderate killer from Aug 31st (Available at http://www.timesonline.co.uk/article/0,,7-1757275,00.html ).

I have thought of a notation to paste these in the forum using letters to denote groups. Hope it makes sense.

Here is the start grid:

Code: Select all

ABCCDDEEF
ABGGGDDHF
IJKKKKKHL
IJMMNOOHL
PJMNNOQQR
PSTTUVQQR
WSTUUVXXR
WYZZZZ$%%
WY@@$$$%%


A-7 B-5 C-11 D-28 E-10 F-12 G-16 H-10 I-13 J-16 K-15 L-7 M-22 N-17 O-11 P-10 Q-27 R-16 S-12 T-6 U-15 V-11 W-15 X-12 Y-12 Z-16 $-16 %-22 @-15

Started off with K(15) which has to be 1+2+3+4+5, from there, L(7) had to be 6-1. Had some tricky moments, but have now got completely stuck here:-

Code: Select all

5174**283
246738519
983***476
46***7321
32***1***
7*13*****
**2******
*********
*********


I did think of putting the letters in this one too, instead of the *, but hopefully this is clear enough. They are so hard to write down like this, I hope you can understand where i'm coming from. In your answer please refer to groups as things like G(16) or just G, and use the other standard terms as with normal su doku.

Thanks in advance,

Tabber

[Pi]

The link appeares to be dead, sorry

[tabber]

Sorry, thats just because of the bracket and fullstop at the end of the URL. Try this instead (or remove it in the browser)

http://www.timesonline.co.uk/article/0,,7-1757275,00.html

cheers.

(EDIT: I have fixed the URL in the original post by adding a space before the bracket)

[Paranoia]

I'm stuck on the same one, but you're a fairway further than me. If any one could look at the image and help me it'd be most appreciated.

http://paranoia.uni.cc/paranoia/killersudoku_mod.JPG (56k warning!)

If you could tell me where to look, rather than what to do, if that makes sense

[tabber]

Hi, well I think I can remember the next number i got after that. I was there for a while myself.

Think about what numbers make-up 28, and what the maximum number you have left is bearing in mind the candidates in box 3 for that 28. I'll give you another clue if you don't get it yet.

[Paranoia]

Cheers, that sorted it. I'm kicking myself for not spotting earlier.

The 28-4, leaves 24, which can only be made using 9+8+7, or the 28-5, leaves 23, which can only be made using 6+8+9. Either way, an eight is required, and can only go in once space.

Also, a 9 is required, so I can eliminate the 9 from the top right cell, giving me a 3 there, and a 9 in the cell bellow it.

Cheers mate, hopefully progress will be a bit smoother now

[tabber]

Yep, that's it!

That was a very gentle nudge I gave you too....

I'm hoping i'll make some headway on the way home tonight, but i've been stuck for at least half an hour at my current position. If anyone has any subtle clues, let me know.

I think I like these Killer ones. A real challenge.

[Paranoia]

I think you should be able to get the S(12) complete.

What candidates have you got in the $(16) in the bottom right hand square?

[Paranoia]

After your tip, I've gone on to solve it with no further tricky spots.

I'll do my best to return the favour when you get back to me mate

[tabber]

Cheers Paranoia - Got it sorted on the way home. I hadn't heard of the '45 rule' until yesterday after I posted that, when I read about it in another thread. I was a bit sceptical, but it actually really helps... Using that, managed to get a couple of numbers that got it going again.

I am curious how you got S(12) so quickly though, I had to wait until about halfway through before I got that. If I open your original image and pretend that isn't filled in, I can see possibles of 9-3 or 7-5. How did you eliminate the 9-3? I assume it was something to do with %(22) in the bottom right. Can't see it myself though.

I'm on the tricky from the same day now, which is going reasonably well.

[Paranoia]

I am curious how you got S(12) so quickly though, I had to wait until about halfway through before I got that. If I open your original image and pretend that isn't filled in, I can see possibles of 9-3 or 7-5. How did you eliminate the 9-3? I assume it was something to do with %(22) in the bottom right. Can't see it myself though.

Eh? I'm not following. For S(12) I had 9-3. 7-5 was elimated a while back, because of the 7-5 in X(12).

[tabber]

Sorry, I meant X(12). How did you eliminate 9-3 so early on?

[Paranoia]

Ok, IIRC, for bottom cell of R(16), I had 3 candidates: 4,7,8.

For the bottom right square, I had a total value (aiming for 45), of 34 (22+12) + the bottom cell of R(16) + the right two cells of $(16).

If BCR(16) (bottom cell), was 8, then the RC$(16) (right cells) must equal 3, as the total square must equal 45. Because of the 2 in E(10), this was not possible.

If BCR(16) was 7, then RC$(16) must equal 4, and due to the 3 in O(11), that is not possible.

BCR(16) must therefore equal 4, and RC$(16) must total 7. With 2,3,4,5 all elimated from cells higher in the collumn, this makes a pair of 1-6 6-1, eliminating those candidates from the rest of the bottom right square.

In F(12), I have a pair of candiates: 3-9, 9-3 (please tell me you had these too :p), so neither 3 or 9 can go in the right side of the bottom right square. so 2-8, 8-2 must go in the right two cells of %(22).

The left two cells of %(22) must now total 12. 4-8, 8-4 is not possible because of the BCR(16) being a 4, and 5-7, 7-5 is not possible because of the 7 in H(10). This leaves only 3-9, 9-3. A pair, and thus these candidates can be elimated from other cells in the bottom right square, leaving only 7-5 for X(12).

I *think* that's how I did it. I can't remember totally, so I tried to remember and make it up as I go along. There might be some breakdown in my logic somewhere, but I hope thats correct and you can follow it.

[tabber]

Wow, yes I was able to follow that, thanks. Very impressive. That gives me a good angle for future puzzles.