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.