The New Sudoku Players' Forum

[paulcoz]

I'd really appreciate some input from the Sudoku gurus here.

I just solved this 'hard' puzzle by trial and error (eg. in the row where there is only a 5 and a 9 to be added, I tried 5/9, from top to bottom, and when that didn't work out, back-tracked and substituted 9/5).

I would like to know if there is a more logical way of proceeding with this (semi-completed) puzzle other than guessing numbers? I know of a few strategies, but I'm by no means a Sudoko 'master'. It looks to me like all the numbers left in this puzzle are dependent on combos (?).

What do you think? Is there a strategy that I'm maybe unaware of that I can use to solve this puzzle by logic alone?

.85.4.621
4..1.6.8.
1...8...4
..391475.
...63214.
.148753..
3.1.984.5
...4.12..
.49...81.

The dots are the blanks in the 9x9 grid. Sorry for the poor format.

Paul.

[Shazbot]

you haven't given your candidate list at this stage, so some of these moves you may have completed:

Code: Select all

 *--------------------------------------------------------------------*
 | 79     8      5      | 37     4      379    | 6      2      1      |
 | 4      2379   27     | 1      25     6      | 59     8      379    |
 | 1      23679  267    | 2357   8      379    | 59     379    4      |
 |----------------------+----------------------+----------------------|
 | 268    26     3      | 9      1      4      | 7      5      268    |
 | 5789   579    78     | 6      3      2      | 1      4      89     |
 | 269    1      4      | 8      7      5      | 3      69     269    |
 |----------------------+----------------------+----------------------|
 | 3      267    1      | 27     9      8      | 4      67     5      |
 | 5678   567    678    | 4      56     1      | 2      3679   3679   |
 | 2567   4      9      | 2357   256    37     | 8      1      367    |
 *--------------------------------------------------------------------*


Locked candidates: 2s in column 3 and 3s in box 8, will allow some eliminations
Naked pair: 5/9 in box 3, 6/7 in box 9
Locked candidates: 7s in row 8
Naked pair: 2/6 in column 2
leads to a Naked Triple: 3/7/9 in row 3 (edit: sorry - it's 379, not 579)

You should now have a placement in r3c7, which leads to the rest of the puzzle being solved easily with naked singles.

[paulcoz]

Hi Shazbot,

Thanks for the reply.

I'm not familiar with your Sudoku terminology. Could you by any chance explain what a locked candidate is (I understand naked pair) and also the row / column coordinate system? Are the rows horizontal (top to bottom) and columns vertical (left to right) as in MS Excel?

I'm not sure of which square coordinate r3c7 refers and the basis of the elimination.

Regards,
Paul.

[Shazbot]

Hi Paul,

yes - rows and columns are the same as in Excel - columns going up/down, rows going left/right. r3c7 is row 3, column 7.

For explanations of the terminology used here as well as more advanced techniques, visit Simple Sudoku (I prefer this one for layout and explanation) and SadMan Sudoku (this one has a few more techniques explained)

[Cec]

Hi Shazbot,
"...I'm not familiar with your Sudoku terminology. Could you by any chance explain what a locked candidate is ..

Hi Paul,
To add to Shazbot's reply, you could look at this thread for Pappocom's recommended terminology.

To understand "Locked Candidates"(1) look at Box8 (this is the lower middle 3X3 grid). Notice that within this Box the candidate 3's are restricted solely to either two cells in row9. As each row, column or Box must only contain digits 1 to 9 then any other candidate 3's in this row and outside this Box can be safely excluded. Therefore candidate 3 can be excluded from r9c9.

I couldn't improve any better on the above Simple Sudoku link which explains how "Locked Candidates(2) is identified.

Hope this helps if not let us know.

Cec

[paulcoz]

Hi, thankyou very much for your help.

I still can't see how to deduce another number, sorry. The eliminations you've described make sense (and limit the possibilities a lot) but even that doesn't help me conclusively determine the value of another square (eg. in box 9 - it could be 6/7 in r7c8 or r9c9, despite the elimination of the 3 in r9c9).

"leads to a Naked Triple: 5/7/9 in row 3"

How did you reach this step? Can you spell it out for me? (I now understand your other comments about the naked pairs/locked candidates etc..)

Thanks again,
Paul.

[QBasicMac]

I now understand your other comments about the naked pairs/locked candidates etc..)

Got that down pat, eh? Congratulations!

Now the hard part: actually seeing them. LOL

Locked candidate 3 in box 8
Locked candidate 3 in row 1
Locked candidate 2 in column 3
Locked candidate 9 in column 7
Naked pair 67 in box 9
Locked candidate 7 in row 8
Naked pair 26 in column 2

If you do the eliminations hinted at above, the puzzle solves with no triples.

Mac

[QBasicMac]

If you do the eliminations hinted at above, the puzzle solves with no triples.

I should have added that you should arrive at the position shown below. If you have trouble, let us know.

Mac

Code: Select all

-85  -4-  621
4--  1-6  -8-
1--  -8-  --4

--3  914  75-
---  632  14-
-14  875  3--

3-1  -98  4-5
---  4-1  2--
-49  ---  81-


Code: Select all

+----------------+----------------+-------------+
| 79    8    5   | 37    4    379 | 6   2   1   |
| 4     379  27  | 1     25   6   | 59  8   37  |
| 1     379  267 | 257   8    79  | 59  37  4   |
+----------------+----------------+-------------+
| 268   26   3   | 9     1    4   | 7   5   268 |
| 5789  579  78  | 6     3    2   | 1   4   89  |
| 269   1    4   | 8     7    5   | 3   69  269 |
+----------------+----------------+-------------+
| 3     26   1   | 27    9    8   | 4   67  5   |
| 5678  57   678 | 4     56   1   | 2   39  39  |
| 256   4    9   | 2357  256  37  | 8   1   67  |
+----------------+----------------+-------------+


[Cec]

"leads to a Naked Triple: 5/7/9 in row 3"

"..How did you reach this step?.."

Hi Paul,
The above help from QBasicMac should enable you to solve this puzzle. Just thought I'd mention I couldn't spot the 579 Triple in r3 either.
Cec

[emm]

That's because it's a 379!

[Cec]

Thanks em - We'll have to nick-name you "Eagle-eye". Thinking what Santa is wrapping for me is affecting my concentration.
Cec

[Shazbot]

oops - sorry. Butterfingers strikes again!

[paulcoz]

Per QBasicMac's table, I don't understand how you can eliminate the following numbers:

6 in r3c2
3 in r3c6
7 in r7c2
6 in r8c2
7 in r9c1

Also, if these numbers are eliminated, I don't see how you can rule out one or another of the remaining choices.

Please advise (and sorry for being such a dope!),
Paul.

[QBasicMac]

sorry for being such a dope!

Been there - Done that!

The locked candidate 3 in box 8 erases r9c9(3), I see you got that

The next is the locked candidate 3 in row 1. Here is where I get confused. How did you erase 3 from r3c4 without also erasing the one in r3c6? Anyway, the pattern is that the only place a 3 can go in box 2 is in row 1. So ALL other 3's in box 2 can be erased.

The locked candidate 2 in col 3 you got (erased 2 from r2c2 and r3c2)

The locked candidate 9 in col 7 You got (erased 9 from r2c9 and r3c8)

Due to the naked pair 67 in box 9 you erased 6 and 7 from other box 9 cells.

The locked candidate 7 in row 8? Ok, the only place that a 7 appears on row 8 is in row 8 of box 7. So 7 can be erased from the other two rows in box 7, that's r7c2 and r9c1, right?

The naked pair 26 in column 2 accounts for the last two, erase 6 from r3c2 and r8c2.

Start over and do the above in that order and see if you don't reach the pencilmark display I posted.

Starting from that display the rest are singles. Here are some, in order:

r3c3 has the only 6 in box 1. (Place it)

r2c3 has the only 2 (same box) (Place it)

That leaves r3c4 with the only 2 in box 2.

etc. Continue until you can't find any single anywhere and post your pencilmarks. They are there somewhere - guaranteed

Mac

[Cec]

Per QBasicMac's table, I don't understand how you can eliminate the following numbers:
6 in r3c2
3 in r3c6
7 in r7c2
6 in r8c2
7 in r9c1

Explanations for these five cases would be: x=exclude
(a) Naked pair 26 in c2 = x other 6's in c2
(b) 3's locked in r1Box2= x other 3's in Box2
(c) Naked pair 76 in Box9 = x other 7's in Box9 = 7's locked in r8Box7
=x other 7's in Box7
(d) Naked pair 26 in c2=x other 6's in c2
(e) 7's locked in r8Box7=x 7 in r9c1

Hope I haven't made mistakes. Now well after midnight - too tired. This might help you answer your other queries - hopefully will again look at it myself but Santa needs me to select pressies for my grand children and he has a tight schedule.

Cheers Cec