The New Sudoku Players' Forum

[QBasicMac]

Working on Today's Puzzle
http://www.dailysudoku.co.uk/sudoku/index.shtml
Fri 22 Jul 2006 (HARD!)

Code: Select all

Original Puzzle
--- 3-9 -7-
--- -2- -61
7-1 --- 3--
4-- 2-- ---
-89 4-5 23-
--- --1 --8
--6 --- 4-5
89- -7- ---
-1- 6-2 ---

Solution So Far
--- 319 -7-
9-- -2- -61
7-1 --6 3--
4-- 2-- ---
189 465 237
--- --1 -48
-76 --- 4-5
89- -7- ---
-1- 6-2 7--

Pencilmarks So Far
256       2456      2458      58        458       48        58        258       24       
35        345       3458      578       458       478       58        58        4       
25        245       2458      58        458       48        589       2589      249     
356       356       357       789       389       378       1569      159       69       
-         -         -         -         -         -         -         -         -       
2356      2356      2357      79        39        37        569       59        69       
23        23        23        189       389       38        189       1289      239     
235       2345      2345      15        345       34        16        12        236     
35        345       345       589       34589     348       89        89        39   


Well I solved it using the following logic, but damn!, isn't there an easier way?

Mac

Solution:
Because of the pencilmark 58's at r1c7 and r2c7, one must be a 5 and the other an 8. Thus the pencilmark 2589 at r3c8 becomes 29. r3c8 must contain a 2 or a 9.

Code: Select all

Looking at the pencilmarks in box 9:
4* 1289  5*
16 12    236
7* 89    39

If 2 then
4* 89   5*
6  1    2
7* 89   3

If 9 then
4* 12   5*
6  12   3
7* 8    9


Hence there is an 8 in c8 in box 9.

Therefore r3c8=9

[SteveF]

You are certainly on the right lines with the 5,8 that you mention, but using ordinary eliminations you should find a cell in column 8 that has only one candidate, r4c8.

[QBasicMac]

Hi, SteveF,

Thanks or the reply. Would you mind further explaining in great detail (similar to my argument above) exactly how I would do that by "ordinary means". I obviously cannot figure it out.

Mac

[SteveF]

The 5,8 pair in r1c7, r2c7 means you can remove other 5's and 8's in box 3 (and as a bonus other 5's in column 7).

In particular you can remove 5 and 8 from r3c8. this leaves only one cell for the candidate 5 in column 8, namely r4c8.

[QBasicMac]

Ouch! How in the world did I miss that?

Thanks!

Mac

[QBasicMac]

It was worse than I thought. My final conclusion that r3c8=9 was wrong. All I showed with all that argument was that r3c8 was either 2 or 9, which I already knew at step 1.

It turned out that 9 was just a lucky correct guess. I would have been better served if 2 was the correct value so I could pay the price for that logic failure.

Mac