## Help with my first very hard please.

Advanced methods and approaches for solving Sudoku puzzles

### Help with my first very hard please.

Hi guys. My first attempt at a very hard from the sudoku program having a more or less 100% sucess rate with the Times puzzles. What a difference!

Starting grid was;

8 x x = x x 1 = x 3 x
x x 1 = 8 x x = x x 4
x x x = 9 x x = x 7 x
================
x 5 x = x x 9 = 6 x x
x 4 x = x x x = x 5 x
x x 9 = 7 x x = x 4 x
================
x 8 x = x x 2 = x x x
3 x x = x x 5 = 9 x x
x 2 x = 1 x x = x x 5

Quite quickly I got to;
8 9 x = 5 x 1 = 2 3 6
x x 1 = 8 x x = 5 9 4
x x x = 9 x x = x 7 x
================
x 5 x = x x 9 = 6 x x
x 4 x = x x x = x 5 9
x x 9 = 7 5 x = x 4 x
================
x 8 x = x x 2 = x x x
3 x x = x x 5 = 9 x x
x 2 x = 1 x x = x x 5

This is the stage I am at now. I have got all my pencil marks in. I even spotted an xwing on th e5's but it doesn;t help me remove any candidates. In fact, I can't see a single candidate I can remove anywhere.

Would one of you kind souls give me a nudge in the right direction please. Coming up to 2 hours staring at this one. PS sorry about the layout, I can't figure out the best way.

Cheers

Paul.
paulf2127

Posts: 26
Joined: 23 May 2005

The key to solve it is column 3. Take a really good look at it :)
Animator

Posts: 469
Joined: 08 April 2005

### Solving this one

Yep.

Column three is certainly the key.

When you are stuck on these types of puzzles, one key strategy to solving them is to look for groups of cells in either the same row, column or 3x3 square that as a group only have as many option numbers as there are cells you are looking at.

Not to put too fine a point on it, this occurs in row 1, 7, 8 and 9 in the puzzle above. This means that the only possibilities for cell 3,3 are 2 and 3. This enables you to place a 5 in R3C1 and R7C3. After this, it snowballs.

Cheers,

Andrew
Guest

### ?

I dont understand what pattern to look for. I am supposed to look on my "pencil marks" and see if things reappears in R,C or 3x3s, then what conclutions do I draw from theese patterns when I find them?
Guest

Column three has been suggested as being the key, but I can't figure out how. What is it that I'm missing?

I suggest starting in column one. Find N common candidates in N cells (as described here) in order to make some eliminations to your pencilmarks. You will then be able to fill in a cell in column two, and from there things will unravel nicely.

updated: fixed spelling
scrose

Posts: 322
Joined: 31 May 2005

aaaah, now i got it, thanx for the link!
Guest

hi guys thanks for the help with the learners like my self - but i have just got to make one very big comment " is this a boys thing?" why cant i see the logic? in the puzzle in question the column 3 was first given as the clue to the solving - then one of the helpers said to eliminate the unwanted candidates in column three you first had to look at all the candidates in column one!!! the logic behind the correct elimination of unneeded candidates continues to evade me ! hence my comment above -
here is column 3 in question (4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467) ok and after looking at this i am supposed to be able to place a (2,3) 1n cell 3 of the column which then in turn "allows" me to place the number 5 in cell 7 of the same column ! will somebody with patience and a kind heart explain to me just how you apply logic to arrive at this position. incidently i arrived at the same state as paulf2127 so my original efforts werent totally useless - its just how to eliminate the uneeded candidates via "logic" that has me tearing my hair out ! cheers
goldie5218

Posts: 37
Joined: 27 May 2005

### Eureka

Goldie,

The trick in column three was to notice that there was a box at the top which had only two candidates. I can't remeber the numbers now because I finished it. But, whichever one it was out of the two was going to leave a pair at the bottom. From memory I think the candidates were 3, 6 and 8. This meant that there was only one place the 5 could go in that column and it snowballed after that.

Obviously with these very hard puzzles there is a single thing that sets it off and it is just trying to spot it.

PaulF
paulf2127

Posts: 26
Joined: 23 May 2005

sorry Goldie,

I meant 4, 5 and 7 not 3, 6 and 8. Should have read your post better.
paulf2127

Posts: 26
Joined: 23 May 2005

OK, I think I have a kind heart and this is much more interesting than my work is today:

Goldie, you stated that column 3 is this

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

Excellent - that's what I got.

Now, look at brackets (i.e. rows) 1, 7, 8 and 9 for this column. There are only four possibilities for the numbers in these four cells. All four of these brackets only contain the numbers 4, 5, 6 and 7. There are four numbers and four possible cells.

This means that the numbers 4, 5, 6 and 7 must go in those four cells. There are no possible other numbers in those cells and if you placed any one of those numbers in a different cell in that column, there would not be enough possible numbers left to go into the four cells in question.

Therefore, we can eliminate the possibilities that 4, 5, 6 or 7 go in any of the other cells in this column.

This leaves the possibilities for the column as:

(4,7) (1) (23) (238) (238) (9) (4567) (467) (467)

Now, this automatically gives us a 5 in row 7 (it is the only 5 left). It also gives us a 5 in R3C1 as this is now the only possibility for a 5 in that row. We have eliminated the only other possibility of a 5 in that row - (cell R3C3)

Cheers,

Andrew
abailes

Posts: 27
Joined: 02 June 2005

hi andrew thnks very very much for your taking the trouble to help me, your explanation obviously makes sense to you and others with the 'logic' key implanted in the mind, but plse bear with me, why the decision in getting rid of 4,5,6 in cell three and not 4,5,6 in cell seven?"
cheers
goldie5218

Posts: 37
Joined: 27 May 2005

goldie5218,

there are two questions you can ask yourself: where can the 2, 3 and 8 go? and another one is, where can the 4, 5, 6 and 7 go?

Here is the current list of candidates:

r1c3: 4, 7
r3c3: 2, 3, 4, 5, 6
r3c4: 2, 3, 7, 8
r3c5: 2, 3, 6, 7, 8
r3c7: 4, 5, 6, 7
r3c8: 4, 6, 7
r3c9: 4, 6, 7

In these you can see two groups of numbers, since it might not be very clear, I'll post another list of candidates (some which can be removed due to other constraints!)

Let's assume for a second the list looks like this:

r1c3: 4, 5, 6, 7
r3c3: 2, 3, 4, 5, 6, 8
r3c4: 2, 3, 7, 8
r3c5: 2, 3, 6, 7, 8
r3c7: 4, 5, 6, 7
r3c8: 4, 5, 6, 7
r3c9: 4, 5, 6, 7

Do you see the two groups in that list? If yes, then you should be able to find the same group in the original list of candidates aswell...
Animator

Posts: 469
Joined: 08 April 2005

Here is my column one solution. The candidates for column one are {8}, {267}, {2456}, {127}, {1267}, {126}, {145679}, {3}, {4679}. Notice that the candidates 4, 5, and 9 only occur in three cells. Therefore, the other candidates can be eliminated from those three cells, leaving {8}, {267}, {45}, {127}, {1267}, {126}, {459}, {3}, {49}. In particular, with the elimination of candidate 1 from r7c1, there is only one candidate 1 remaining in lower-left box.
scrose

Posts: 322
Joined: 31 May 2005

Goldie,

the simple answer to your question is that cell three has a 2 and a 3 in it as well. It therefore has other options left open to it other than the 4,5,6,7 group we have found.

The key paragraph in my post above is:

This means that the numbers 4, 5, 6 and 7 must go in those four cells. There are no possible other numbers in those cells and if you placed any one of those numbers in a different cell in that column, there would not be enough possible numbers left to go into the four cells in question.

You cannot apply this logic to a group with cell three in it, because of the other options (2 or 3) that you have in this cell.

Does that make sense?

Andrew

P.S. as Animator alludes to, there is also a group of 2, 3 and 8. in rows 3, 4 and 5. This works in much the same way - three cells are the only possible answers for these three numbers. This is just a little harder to spot (because all cells have other options as well) and when going through this logic for the first time, noting this may confuse more than it helps.
abailes

Posts: 27
Joined: 02 June 2005

ok andrew thnks again for your patience - so let me test my (hazy) understanding - when looking for a "pattern" in a column/row/block i must try and identify a group of candidates which will all appear in the collumn/row/block and will occupy the number of cells equal to the pattern spotted ie three diff candidates for three diff cells - now the trick as you have mentioned, is that these same cells may contain other candidates as well as the pattern candidates? but these "extra" candidates can be eliminated for the purpose of identifying a single cell that can only hold a single candidate?? am i getting on the same page at last? cheers
goldie5218

Posts: 37
Joined: 27 May 2005

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