## Previously Solved...but now I'm stuck!

Advanced methods and approaches for solving Sudoku puzzles

### Previously Solved...but now I'm stuck!

I have previously solved this 'very hard' puzzle but have tried to re-create to explain to a friend, and now I'm stuck!! Any suggestions??

4** *2* ***
5*2 3*8 649
*6* *4* *72

72* *** *61
**6 2** 4**
94* *** 728

28* 763 *14
6*4 5*2 **7
*** *9* 2*6
Guest

Posts: 312
Joined: 25 November 2005

There are only two places for a 5 in box 2 [edit: "box 3" was a typo], both in column 6, so you can remove candidate 5 from the rest of the row.

This makes a naked pair [16][16] in the center box.

Exclusions from this lead to a naked triple in column 4 [169][19][16].

Exclusions from the triple leave only two places for a 9 in the center box, both in column 6, so you can remove candidate 9 frin the rest of the column.

There is an x-wing in 1's in r28c25.
Last edited by tso on Tue Sep 13, 2005 4:06 pm, edited 1 time in total.
tso

Posts: 798
Joined: 22 June 2005

Sorry, don't understand... surely, column 6 is in boxes 2, 5 and 8 - not in box 3 ?
Guest

Posts: 312
Joined: 25 November 2005

Yeah, that's how I see it. You can place all the 5's starting with boxes 2,5.

PS : Don't read the rest of the hints until you've tried to finish it yourself.
emm

Posts: 987
Joined: 02 July 2005

In box 2, I can see that 5 will be in either r1c6 or r3c6. In box 5, 5 will be in r4c5, r5c5 or r6c5 but I haven't been able to narrow it down further. Am I missing something obvious?
Guest

Posts: 312
Joined: 25 November 2005

One thing here is to recognise that other people’s advice can be wrong. Think of it as a challenge, like orienteering, which can be more interesting when they turn the signposts around.

I have got the same problem as you with this puzzle, I can't for the life of me see what I did.

Personally, I'd stick with tso - although he did make one mistake, it is box 2 not box 3 - no-one's perfect!
emm

Posts: 987
Joined: 02 July 2005

Hi pch,

I just solved it by using my newly developed Execl program.
Not a easy one.
It took me three times of trial and error.
But believe it or not. It took me 10 minutes altogether.

437 926 185
512 378 649
869 145 372

728 439 561
156 287 493
943 651 728

285 763 914
694 512 837
371 894 256

Uncle Wong.
uncle_wong_hk

Posts: 8
Joined: 12 September 2005

Uncle Wong, I think you're new here so can I make a suggestion? It's OK to give people clues, it's even OK to give them false leads (by mistake of course), but it's not OK to give them the whole answer served up like Peking Duck on a platter. You're so right - we just loss the fun!
emm

Posts: 987
Joined: 02 July 2005

Sorry.
I am really the newbie for it.
uncle_wong_hk

Posts: 8
Joined: 12 September 2005

I forgive you - don't do it again!
emm

Posts: 987
Joined: 02 July 2005

Maybe my original question was not clear: I, too, have the solution because I solved it a few days ago. My problem is trying to recreate it and I became stuck at the point in the original post. If the solution can only be determined by trial and error, then I will accept the fact and sign off but I was trying to understand if there was a logic that could be applied.
Guest

Posts: 312
Joined: 25 November 2005

I totally understand your original question - it's just that I'm easily distracted.

Can you follow tso's advice? It was sound (apart from the wrong box - but anyone can do that). If we're lucky, he might give us another clue.
emm

Posts: 987
Joined: 02 July 2005

Here is a workable method for your reference.

1. List out the possible numbers for each empty cell.
4(1379)(13789)(169)2(15679)(1358)(358)(35)
5(17)23(17)8649
(138)6(1389)(19)4(159)(1358)72
72(358)(489)(358)(459)(359)61
(138)(135)62(13578)(1579)4(359)(35)
94(135)(16)(135)(156)728
28(59)763(59)14
6(139)45(18)2(389)(389)7
(13)(1357)(1357)(148)9(14)2(358)6
2. See if there are any empty cell that get only one possible number.
3. If the answer is yes, fill the cell with this number.
4. If the answer is no (for this case, the answer is no),
try to fill in a empty cell with a possible number (by guess).
Select the empty cell that when you fill in the number,
many other empty cells will be effected.
5. List out the possible numbers for each empty cell again.
6. Repeat the above steps again and again.
If you get the right answer, that is fine.
But if you cannot get the right answer, that means the guess work
in step 4 is wrong. You have to redo the works starting at this point.

Hope that it will help.
uncle_wong_hk

Posts: 8
Joined: 12 September 2005

Uncle Wong, thanks for your input again, but pch was asking for help in finding a way to solve the puzzle without making a guess. That is what he meant by "without trial and error." Any sudoku (if it is a valid one) can be solved by guessing eventually, but most of us here are looking for ways to solve without needing to guess. pch wants to know if that can be done in this particular example. (I suspect he suspects that maybe he solved it by accident the first time, used faulty logic that by luck gave a correct answer.)
Doyle

Posts: 61
Joined: 11 July 2005

Thanks. Followed tso's advice regarding the naked triple and x-wing. These are both valid and have enabled me to progress (although I have not yet completed the puzzle).
Guest

Posts: 312
Joined: 25 November 2005

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