The New Sudoku Players' Forum

[pyrpoliser]

Hi to everybody.

Its been some time now that I am thinking on a new method for solving the extreme sudoku puzzles.I mean those that can not be solved with the basic strategies as described at http://www.scanraid.com/BasicStrategies.htm And I thought of sharing my findings with you hoping that you could possibly help in developing or proving the points of this method.

I started thinking of how the numbers get distributed on the grid on the first place and how this affects the final solution.
Everybody knows that each number has to appear 9 times in rows, 9 times in clolumns and 9 times in blocks.Everybody also knows that each number has to fill positions 1,2,3,4,5,6,7,8,9 of the 9 rows and columns without any exception or repeatition whatsoever.
But I guess not many people have noticed that this is not so in blocks!! When I personally noticed that, and after spending some time on it, I am absolutely convinced that the key in solving all sudokus is hidden in the way the numbers are distributed in the blocks.

If you watch the final solution of any puzzle, no matter how easy or difficult it was, you will notice that most numbers violate their filling up of block positions in at least 2 occurences each.
Heres an example of a puzzle, its final solution and the distribution of each number in the blocks.



In this example you can see that only 2 numbers 7 and 9 actually appear in all block positions 1 to 9. The rest violate the pattern by appearing twice or even 3 times at the same block positions. If we take e.g number 1 it appears twice at Block position 9 and twice at block position 2, hence it never appears at block positions 3 and 8. This affects other numbers in the grid causing a sort of chain reaction, because some other digit has to fill those positions, or at least one of them, but at the same time leaving other positions behind.

I studied about 300 puzzles so far, never found a puzzle to have less than 6 numbers actually VBPN (violating the block positioning norm). In fact I found puzzles in which all 9 numbers were VBPN. The average and quite usual number of VBPN is 7, sometimes you find 6, sometimes 8, but I would say 7 is the average. This means 7 out of 9 numbers or about 80% do VBPN.

Second observation is that no single number can VBPN at just one position.It has to do at least 2 violations, i.e appear twice in at least 2 positions. 3 violations are also common. I never found a case of 4 violations although I found a couple of 3 ½ violations : -) e.g 2 times at BP (block position) 8, 2 times at BP 3 and 3 times at BP 4.

Like I said I tried this method after all basic strategies failed to solve the puzzle, so I am not sure of what use this method can be at earlier stages of puzzle solving.

The third observation is that what really directs the solution of the puzzle is the digit (or one of the digits) that appear mostly as a solved square after all basic strategies have failed.That specific digit has the highest chances to be in VBPN mode . The problem is to find that digit and make right use of it. This is not always possible.

A fourth observation is that the average of 2 numbers that eventually will NOT cause VBPN, appear at very few solved squares after all basic strategies failed to solve the puzzle.


Well how can all these, be of any use? Here are some applications I found so far. There might be more though, please feel free to contribute.

1)Suppose you have a puzzle in which a number is already solved at 7 cells and this number is already verified to be in VBPN at just one position., by appearing at BPs 3,8,6,4,9,2,8 (violation is at block position 8). So you need at least another violation. If the unsolved squares that contain this number allow you only 2 options a) BP 4,1 and B)BP 5,1 then you can tell immediately that the right choice is the first one.

At this point I should say that this method would really be a fire setter to all puzzles if after studying the positioning of numbers in solved squares when all basic strategies have failed, it were possible to tell for sure which numbers will cause a VBPN and which not. In fact I actually found a method, later to discover that it was true simply bacause most extreme puzzles are computer generated, and the damn computers simply follow a specific pattern…Anyway if someone can find a method to do that, that would really be a great breakthrough.

2)So the second use of this method is a sort of "guessing". However in this case your "guessing" has at least 80% chance to be true.

Heres an example.



As you can see the leading number here is 4. It is already solved at all 9 positions and has 2 VBPN. So 4 can be of no use anymore. Heres the complete table

1 appears 6 times with VBPN
2 appears 6 times with VBPN
3 appears 2 times without VBPN
4 appears 9 times with VBPN
5 appears 3 times without VBPN
6 appears 4 times without VBPN
7 appears 4 times without VBPN
8 appears 6 times with VBPN
9 appears 6 times without VBPN


Knowing that the average number of digits that DO NOT cause VBPN in every puzzle is 2, if someone would ask you to pick a group of 4 that would almost certainly contain those 2, wouldn’t you choose 3,5,6 and 7 because of their comparatively low solved squares? Well, solve this puzzle and find out if you were right or wrong.

Anyway lets continue.Leaving aside 4 which is already solved we seem to have 4 leader numbers here all with 6 solved squares 3 of which already have VBPN. So 1,2 and 8 can be of no use. (All combinations in their unsolved cells will simply add more VBPN). So our only hope is 9. We cannot tell with absolute certainty that 9 will eventually have a VBPN but combining the fact that 9 is one of the leading numbers plus the fact that any randomely selected number has an 80% chance to have a VBPN, then I would say the chances for 9 having a VBPN is near 95% (an estimate).

Analysing 9s
They appear at BPs 1,2,4,6,7,9
Remaining BPs 3,5,8
Block 5 a 9 in pos 5 fills up BPs 5-8-3 violates nothing
Block 5 a 9 in pos 2 fills up BPs 2-5-9 violates 2-9
Block 5 a 9 in pos 2 fills up BPs 2-8-6 violates 2-6
Block 5 a 9 in pos 8 fills up BPs 8-5-3 violates nothing

Conclussion: we need 9 at Block 5 pos 2. Setting 9 at this position cracks the puzzle!

Notice the astonishing effects here. You are not simply setting 9 at that position! You are actually damping a whole bunch of 3 other digits as well!

Here’s another example.



After applying basic strategies we end up with this pattern

1 appears 1 times without VBPN
2 appears 2 times without VBPN
3 appears 2 times without VBPN
4 appears 3 times without VBPN
5 appears 3 times without VBPN
6 appears 2 times without VBPN
7 appears 7 times without VBPN
8 appears 5 times without VBPN
9 appears 2 times without VBPN

Just for your information the digits that eventually do not cause VBPN in this puzzle are 2 and 9. Not surprising if you notice that their unsolved squares at this stage rank at the bottom of the scale.
Whats interesting in this case is number 7. It is definetely the leader of the puzzle and fortunately for us it is solved at 7 squares without a single VBPN. It is almost 100% certain (like I said this method at this stage is purely based on statistics-no absolute 100% certainty) that 7 is in VBPN mode and that the remaining two 7s must both cause block position violations.If you chose to set Block6 pos 6 to 7, that would not cause a VBPN, and would almost certainly be wrong. So setting Block 6 pos4 and block 9 pos9 to 7 causes two VBPN as expected. If you care to solve the puzzle you will see that these selected positions are correct.

I am really waiting your contributions regarding this method. Please do test it and let us see what we get.

NB. I call this method Pyrpolising = Setting on fire.

[pyrpoliser]

Hmmm, I dont see the images getting loaded.

Anybody knows why?

Isn't this the code for images???
[img] [/img]
Anyway for the moment if anyone is interested please download them separately

[RW]

Hi pyrpoliser, welcome to this forum!

That's some interesting thoughts you have there, haven't seen any similar technique earlier. Some questions first.

1)Suppose you have a puzzle in which a number is already solved at 7 cells and this number is already verified to be in VBPN at just one position., by appearing at BPs 3,8,6,4,9,2,8 (violation is at block position 8). So you need at least another violation. If the unsolved squares that contain this number allow you only 2 options a) BP 4,1 and B)BP 5,1 then you can tell immediately that the right choice is the first one.

A real life example of this would be welcome, right now I don't quite understand how this situation would be possible.

Code: Select all

 *-----------*
 |..1|...|...|
 |...|...|..1|
 |...|.1.|...|
 |---+---+---|
 |...|...|.1.|
 |1..|...|...|
 |...|..1|...|
 |---+---+---|
 |...|...|...|
 |...|...|...|
 |.1.|...|...|
 *-----------*

In this diagram the numbers are placed as you mentioned, and one can clearly see that no BP 5 is available (which is logical as the number already appears in BPs 2,8,8; three times in the middle column of the three boxes). What have I got wrong?

2)So the second use of this method is a sort of "guessing". However in this case your "guessing" has at least 80% chance to be true.

How would your method do in this puzzle:

Code: Select all

 *-----------*
 |..5|4.3|..1|
 |.1.|.2.|.4.|
 |...|1..|9..|
 |---+---+---|
 |.72|...|...|
 |...|.3.|.7.|
 |...|8..|3.6|
 |---+---+---|
 |..7|..9|..4|
 |.8.|...|16.|
 |6..|2..|..5|
 *-----------*

I studied about 300 puzzles so far, never found a puzzle to have less than 6 numbers actually VBPN (violating the block positioning norm).

Study the solution to the above puzzle and you'll find no numbers VBPN.

The third observation is that what really directs the solution of the puzzle is the digit (or one of the digits) that appear mostly as a solved square after all basic strategies have failed.That specific digit must be in VBPN mode because if it weren’t, the puzzle would have already been solved.

This I don't understand at all, in the grid I just posted above no digit is in VBPN, but the puzzle is certainly not solved...

The biggest problem with your technique as I see it is that you are talking about 80% chance or 95% chance of being correct, all accepted techniques so far are 100% correct. Some more examples, and especially examples that don't involve guessing, would maybe help.

Btw, there's a sudoku variant called "disjoint groups" with the additional constraint of "no numbers VBPN". Check out the links from this thread.

RW

[keith]

http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=592

Keith

[MCC]

pyrpoliser, re images, I notice that you are putting a space before and after the URL if you remove these, that might correct things.

[ravel]

Independantly now, if because of RW's obviously valid objections the percentages have to be corrected, a probabilistic approach has some charm for me, especially for very hard puzzles.

First:
It is true, that a solution here would not be accepted that is based on guessing a number. This is only "allowed" for a wrong number, because when it turns out to be wrong you implicitely have a proof, that it can be eliminated. I think this is common sense here.
Concerning uniqueness methods there is a minority, which does not accept them, because they depend on the pre-condition that the puzzle is unique.
[edit: reformulated] Now one also could say, when a guess leads to a solution, it is proved, when we know, that the sudoku is unique. (But my POV is, that each number has to be proved, when filled in, not afterwards)

Second:
Sometimes, when i try to find a chain to eliminate a number, i realize that it would solve the puzzle and it takes me much more time to "prove" that it must be there (those are puzzles i do not like). In these cases a "reasonable guessing" would be something between. When i know, that there are 20 times more solution grids, when the number is there and i solve it with that guess, maybe i would be satisfied without an additional prove.

Third:
In sudoku competitions guessing is allowed. When you have a method, which allows you to find a probably correct number quickly, this would be a great help there (provided the puzzles are created "randomly distributed" for the method).

[pyrpoliser]

Thanks for the welcoming RW.

In this diagram the numbers are placed as you mentioned, and one can clearly see that no BP 5 is available (which is logical as the number already appears in BPs 2,8,8; three times in the middle column of the three boxes). What have I got wrong?

Nothing wrong. Simply my mentioning of BP 5,1 was arbitrary. The fact still remains that BPs 4, 1 or BPs 1,4 are the correct choices.
So theoretically the rule is correct but it seems there cannot be any "worthy" applications if 7 cells are already solved. Perhaps the rule could have applications when 6 cells are already solved with just 1 VBPN. So we are looking for another 1 or another 3 VBPN in the remaining 3 cells. This excludes the possibility of another 2.

To be honest I did not find even one case were a number having solved squares and already been in VBPN mode, could produce anything useful by using this method. Unless someone can provide an example where the opposite is true, I believe I should set a rule that

a number already in VBPN mode can produce nothing useful, so forget it.

Study the solution to the above puzzle and you'll find no numbers VBPN.

Correct! : - ((

So 0 out of 9 VBPN, like 9 out of 9 is a possibility. So can you tell me what is the possibility for a 0 out of 9 VBPN? I mean if you used a computer program to generate it then the program should be able to tell you how many it can produce with such a pattern. If this is the case could you please also check for less than 6 VBPN?

The biggest problem with your technique as I see it is that you are talking about 80% chance or 95% chance of being correct, all accepted techniques so far are 100% correct.

Yes my friend, I know that.
Just for the record however, I already mentioned in my previous message that this method is just missing something to be 100% correct. In fact the reason I posted my findings publicly here was to ask for peoples help, for this specific point.

I repeat

…. I should say that this method would really be a fire setter to all puzzles if after studying the positioning of numbers in solved squares -after all basic strategies have failed-, it were possible to tell for sure which numbers will cause a VBPN and which not.

The question is: Is this possible?? And if yes can someone tell us how?
There is no doubt that yes this is absolutely possible, and one should be able to cumpute it without having to solve the puzzle. So until this is achieved the method can only be used with risk.

NB. Try this method on the zero VBPN puzzle you provided, assuming this condition was known in advance. You will be surprised of how fast it gets solved!


*********************

Thanks Keith. My images are already hosted somewhere, so I dont think its a matter of hosting. I will ask the moderator.

[tso]

I place 100 coins on the table and tell you that they are all heads up. I cover one with my hand at random and ask you what you know about it. You say, with 100% accuracy, "It is heads up".

I flip 100 coins onto the table and cover one with my hand at random and ask you what you know about it. You look at the rest of the coins, knowing that on the average, half will be face up, half will be face down. You notice that 50 (or 60 or 99) of the coins on the table are face up. Still, all you can say is: "It's 50/50 that it is heads up".


Solution methods depending on characteristics of a 'random' must define 'random' Different puzzles with different characteristics result from different generation methods. (Contrary to what you suggest, different software can and does use very different paths to create a puzzle.) Also, the puzzle designer, compiler and editor all have the choice to select only the puzzles that thwart any specific approach based on 'random' puzzles. The designer may do so unintentionally.

Comparing the relative position of the digits in blocks the way you've chosen to do seems at first blush to be reasonable, but is arbitrary. You could divide the puzzle into ANY set of nine 9-cell regions, disjoint or not, and make similar findings that will be similarly useless. One can transpose any two rows or columns passing through the same three boxes without changing ANY of the charactoristics of the puzzle -- but suddenly a puzzle with no digits VBPN becomes one with several.

As has been mentioned, disjoint group sudoku contain the digits 1-9 in all rows, columns boxes as well as in the block positions. In those puzzles, you need this information to solve them. However, a puzzle designer, carbon or silicon based, if composing a puzzle by starting from a finished grid and subtracting clues -- is free to chose ANY grid -- one in which none, all or some of the digits appear multiply in their block positions -- and as long as she doesn't share this information with the solver, the solver simply cannot take advantage of this.

A composer is free to give you this additional information -- but then you've got a puzzle of a different color. If the composer tells you that none of the digits appear in the same block position twice, then you've got a disjoint group sudoku. But she could also tell you that all the disjoint groups contain at least one repetition, the puzzle being unsolvable without this extra bit of info.

We cannot make judgements about "random" sudoku without specifying the the source or method of selection. Puzzles created by Simple Sudoku and those created by Pappocom Sudoku are NOT similar, even when judged to be a similar difficulty. Only if we had a database of all possible sudoku would we be able to choose a puzzle 'at random' and make some sort of judgement. Think of this -- if you *did* choose randomly from that database, your chance of picking a symmetrical puzzle would be very nearly zero, as the assymmetrical puzzles exist in astronomically larger numbers. Yet if I were to make a judgement based on all the puzzles I see in papers and generated from programs, I might decide they're ALL symmetrical.

[tso]

The second of these two puzzles was created by transposing some of the rows and columns of the first. They have the same solution path, both needing at least a 5-cell forcing chain to solve -- they are essentially the same puzzle. Note how their solution grids differ, only the first having different digits in all similar block locations.

Code: Select all

+-------+-------+-------+
| . . . | . 5 . | . . . |
| . . . | 7 . . | . 2 3 |
| . . 9 | 1 . . | 4 . . |
+-------+-------+-------+
| 2 . . | 5 . . | . . 1 |
| . . . | . 9 . | . . . |
| 8 . . | . . 4 | . 6 7 |
+-------+-------+-------+
| . . . | 6 . 8 | 9 . . |
| . 7 . | 9 . 2 | . 4 5 |
| . 1 . | . . . | . . . |
+-------+-------+-------+

Code: Select all

+-------+-------+-------+
| 1 2 3 | 4 5 6 | 7 8 9 |
| 4 5 6 | 7 8 9 | 1 2 3 |
| 7 8 9 | 1 2 3 | 4 5 6 |
+-------+-------+-------+
| 2 3 4 | 5 6 7 | 8 9 1 |
| 5 6 7 | 8 9 1 | 2 3 4 |
| 8 9 1 | 2 3 4 | 5 6 7 |
+-------+-------+-------+
| 3 4 5 | 6 7 8 | 9 1 2 |
| 6 7 8 | 9 1 2 | 3 4 5 |
| 9 1 2 | 3 4 5 | 6 7 8 |
+-------+-------+-------+

Code: Select all

+-------+-------+-------+
| . 9 . | 1 . . | . . 4 |
| . . . | . 5 . | . . . |
| . . . | 7 . . | 2 3 . |
+-------+-------+-------+
| . . . | . 9 . | . . . |
| 8 . . | . . 4 | 6 7 . |
| 2 . . | 5 . . | . 1 . |
+-------+-------+-------+
| . . . | 6 . 8 | . . 9 |
| . . 7 | 9 . 2 | 4 5 . |
| . . 1 | . . . | . . . |
+-------+-------+-------+

Code: Select all

+-------+-------+-------+
| 7 9 8 | 1 2 3 | 5 6 4 |
| 1 3 2 | 4 5 6 | 8 9 7 |
| 4 6 5 | 7 8 9 | 2 3 1 |
+-------+-------+-------+
| 5 7 6 | 8 9 1 | 3 4 2 |
| 8 1 9 | 2 3 4 | 6 7 5 |
| 2 4 3 | 5 6 7 | 9 1 8 |
+-------+-------+-------+
| 3 5 4 | 6 7 8 | 1 2 9 |
| 6 8 7 | 9 1 2 | 4 5 3 |
| 9 2 1 | 3 4 5 | 7 8 6 |
+-------+-------+-------+

We can do this with any puzzle you are solving. If your premise were true, you would have to account for how I can magically change the odds of one choice or another by swapping rows or columns.

[ravel]

Good point, tso.

A probabilistic method i thought about of course had to be invariant for equivalent puzzles and this one is not. So this method only can be of interest for 100% correct choices. But even if some would be possible, it could not be more mighty than nishio (or grouped coloring), because - if i did not miss something - the conclusions are done from the positions of a single number only.

[RW]

So can you tell me what is the possibility for a 0 out of 9 VBPN? I mean if you used a computer program to generate it then the program should be able to tell you how many it can produce with such a pattern.

I didn't use a computer, I chose the solution of a disjoint group puzzle and manually removed clues to make a tough valid normal sudoku. I'm not at all suprised that it's easy when the disjoint group characteristic is known, but as tso pointed out, you cannot know this unless someone tells you. I have no idea of how many disjoint grids there are, they might be quite hard to count as equivalent grids in different permutations apparently should be counted.

I agree with ravel that anything that could help in finding "educated guesses" in extreme puzzles could be of value, question is how educated the guesses based on this technique are. Would be interesting to know how much this can be altered by permutating the grid. Is it possible that all grids could be turned into 0 or 9 VBPNs? With 1679616 possible permutations (without swapping the different digits) this doesn't sound at all impossible. Unfortunately swapping the boxrows/columns doesn't affect this so we are left with 46656 possibilities for each grid, which I don't think is enough...

As for the pictures, have you tried saving them in a different format? I don't know which formats are accepted by the forum.

RW

[pyrpoliser]

Edited

[pyrpoliser]

Oh by the way I changed the format of one image from .bmp to .gif and it works fine now.
I will change the format of the rest and upload them again.

[tso]

I don't think there is anything here. You seem to have discovered a way to turn lead into gold, a perpetual motion machine. It *seems* like something's there only because you're not looking closely.

1)Suppose you have a puzzle in which a number is already solved at 7 cells and this number is already verified to be in VBPN at just one position., by appearing at BPs 3,8,6,4,9,2,8 (violation is at block position 8). So you need at least another violation. If the unsolved squares that contain this number allow you only 2 options a) BP 4,1 and B)BP 5,1 then you can tell immediately that the right choice is the first one.

You have divided by zero.

Here is your example:

Code: Select all

+-------+-------+-------+
| . . 7 | . . . | . . . |
| . . . | . . . | . . 7 |
| . . . | . 7 . | . . . |
+-------+-------+-------+
| . . . | . . . | . 7 . |
| 7 . . | . . . | . . . |
| . . . | . . 7 | . . . |
+-------+-------+-------+
| . . . | x . . | x . . |
| . . . | x . . | x . . |
| . 7 . | . . . | . . . |
+-------+-------+-------+


The digit 7 appears in BP 3,8,6,4,9,2,8. There are only two cells left that can take a 7, marked with 'x'. The four form a rectangle -- they will *always* form a rectangle -- and therefore will be in two OR four block positions, not three. The choice you suggest is impossible. Either of the two final placements of the 7 will lead to a total of 2 VBPN.

The only other possibility after 7 of one digit are placed is:

The digit 7 appears in BP 3,8,6,4,9,1,8.

Code: Select all

+-------+-------+-------+
| . . 7 | . . . | . . . |
| . . . | . . . | . . 7 |
| . . . | . 7 . | . . . |
+-------+-------+-------+
| . . . | . . . | 7 . . |
| 7 . . | . . . | . . . |
| . . . | . . 7 | . . . |
+-------+-------+-------+
| . . . | x . . | . x . |
| . . . | x . . | . x . |
| . 7 . | . . . | . . . |
+-------+-------+-------+


In this example, the block positions to be taken will be (1 and 5) OR (2 and 4). Either way, the VBPN will be 2.

You'll *never* have a choice of (1 and 4) OR (1 and 5).

[tso]

Second observation is that no single number can VBPN at just one position.It has to do at least 2 violations, i.e appear twice in at least 2 positions. 3 violations are also common. I never found a case of 4 violations although I found a couple of 3 ½ violations : -) e.g 2 times at BP (block position) 8, 2 times at BP 3 and 3 times at BP 4.

You're not looking very hard. Check the position of the 3's in this puzzle -- what you would call a case of "4 ½ violations" -- the maximum possible.

Code: Select all

+-------+-------+-------+
| 3 . 2 | . . . | . 7 . |
| . . . | . 7 . | 5 . 3 |
| . . . | . 3 . | 8 6 4 |
+-------+-------+-------+
| 8 1 . | 3 . . | 7 . 6 |
| . . 3 | 9 . . | . . . |
| 4 7 . | . 5 . | . . . |
+-------+-------+-------+
| 1 . . | . . . | 3 . . |
| 7 9 . | . 8 3 | . . . |
| . 3 . | 5 . . | . . . |
+-------+-------+-------+

Code: Select all

+-------+-------+-------+
| 3 8 2 | 6 4 5 | 9 7 1 |
| 6 4 1 | 8 7 9 | 5 2 3 |
| 9 5 7 | 2 3 1 | 8 6 4 |
+-------+-------+-------+
| 8 1 9 | 3 2 4 | 7 5 6 |
| 5 2 3 | 9 6 7 | 1 4 8 |
| 4 7 6 | 1 5 8 | 2 3 9 |
+-------+-------+-------+
| 1 6 4 | 7 9 2 | 3 8 5 |
| 7 9 5 | 4 8 3 | 6 1 2 |
| 2 3 8 | 5 1 6 | 4 9 7 |
+-------+-------+-------+

(Violating what? Since there is no requirement that the digits do not repeat in the same block positions, what is it that they are violating?)

Here's a puzzle that is *very* difficult -- though it has a unique solution, most people would have to give up or guess.

Code: Select all

+-------+-------+-------+
| . . . | . . . | 6 . 5 |
| . . . | 3 . . | . 9 . |
| . 8 . | . . 4 | . . 1 |
+-------+-------+-------+
| . 4 . | . 2 . | 9 7 . |
| . . . | . . . | . . . |
| . 3 1 | . 8 . | . 6 . |
+-------+-------+-------+
| 9 . . | 6 . . | . 2 . |
| . 1 . | . . 7 | . . . |
| 5 . 4 | . . . | . . . |
+-------+-------+-------+


Of course, if I were to tell you that ALL NINE DIGITS had "4 1/2 VBPN", then it wouldn't take but a minute to find this:

Code: Select all

+-------+-------+-------+
| 1 2 3 | 9 7 8 | 6 4 5 |
| 4 5 6 | 3 1 2 | 8 9 7 |
| 7 8 9 | 5 6 4 | 2 3 1 |
+-------+-------+-------+
| 6 4 5 | 1 2 3 | 9 7 8 |
| 8 9 7 | 4 5 6 | 3 1 2 |
| 2 3 1 | 7 8 9 | 5 6 4 |
+-------+-------+-------+
| 9 7 8 | 6 4 5 | 1 2 3 |
| 3 1 2 | 8 9 7 | 4 5 6 |
| 5 6 4 | 2 3 1 | 7 8 9 |
+-------+-------+-------+