The New Sudoku Players' Forum

[pyrpoliser]

I would like to clear out that the method when used as a high probability option IS NOT at all something like flipping a coin. In fact the risk can be calculated early enough.

Heres an example from a puzzle already posted in this thread. As you can see after applying basic strategies we are already in VBPN regarding digits 2,5,6 and 9. Like I said in previous messages at this stage those digits are useless.



Digit 4 though is not.. It appears at Block1 at 4,5,6,7,8,9 at Block2 at 4,8 at Block 4 at 8,9 at Block7 at, 1,2,3 and at Block8 at 7,8. It also appears at only 4 solved cells which is comparatively low i.e at positions 3,6,2,4. So unless you can calculate your risk you shouldn’t proceed using this digit. Striking out the solved positions from the unsolved we end up with positions 5,8,9,1 and 7.
This is the ONE and ONLY possibility for digit 4 to eventually be non VBPN. Moreover it is a forced line once 4 is set at Block1 pos 5.

At this stage one can easily check how many the total combinations for digit 4 are. From a quick look I can count 11.(Correct me if I am wrong). So the risk is 1 out of 11 or 9%. Taking up this risk you immediately know that a 4 has to be excluded from Block1 position 5. Doing that in just 2 steps you end up with digit 3 at Block7 position 1 solved.

NB. There are cases when a digit is not in VBPN mode but if you strike out the positions where it was already solved then it proves itself that its going to be in VBPN mode no matter what. That digit is of no use anymore. For example if you check digit 5 at the 4 ½ puzzle of tso you will see that it ends at the following possible BPs: 56,5678,3,35,1345679,57,7,2345,45. After striking out positions 3 and 7 from the unsolved squares you end up with incompatibility so digit 5 will end up in VBPN mode no matter what.


USE OF THE METHOD WITH PROVEN 100% MATHEMATICAL CERTAINTY

I am working on this. For anyone interested to join me in this effort the scope is to prove that a digit is going to end up in VBPN mode. This proof must come from other digits not from the digit iitself. My current line of reasearch is to check what one can do with digits that are already in VBPN mode . Most of them reveal the BPs they left/ or will eventually leave behind.

[tso]

Code: Select all

 *-----------------------------------------------------------*
 | 7     9     238   | 1    -28    36    | 5    [68]   4     |
 | 134   346   2348  | 24    5     36    | 178   9     167   |
 |[14]   456   458   | 7    [48]   9     | 2     3    [16]   |
 |-------------------+-------------------+-------------------|
 | 5     7     6     | 8     9     1     | 3     4     2     |
 | 8     1     9     | 23    23    4     | 6     7     5     |
 | 2     34    34    | 5     6     7     | 9     1     8     |
 |-------------------+-------------------+-------------------|
 | 34    345   345   | 6     17    8     | 17    2     9     |
 | 6     8     7     | 9     13    2     | 4     5     13    |
 | 9     2     1     | 34    347   5     | 78    68    367   |
 *-----------------------------------------------------------*


First of all, at this point there is a 5-cell simple xy-type forcing chain that basically solves the puzzle -- 100% of the time.

r3c5=8 -> r1c5=2
r3c5=4 -> r3c1=1 -> r3c9=6 -> r1c8=8 -> r1c5=2
Therefore, r1c5=2

... but that's besides the point ...

At this stage one can easily check how many the total combinations for digit 4 are. From a quick look I can count 11.(Correct me if I am wrong). So the risk is 1 out of 11 or 9%. Taking up this risk you immediately know that a 4 has to be excluded from Block1 position 5. Doing that in just 2 steps you end up with digit 3 at Block7 position 1 solved.

There are 8 combinations, not 11.

Regardless, you conclusion is unsupported and will give you no advantage over picking at random. Excluding 4 from r2c2 is a guess, and not a particularly educated one. Jeez, you can *always* have at *least* a 8 out of 9 chance of being right when making an exclusion! Since only ONE number will end up in a cell, there are EIGHT digits you can guess at excluding. You'll usually be right. It means nothing. After a few guesses, you'll get one wrong and that's that. Let's try it. Pick a Sudoku. don't show it to me, but hold it up so the rest of the forum can see it. Let me concentrate --- the center cell is NOT an 8! Purty good, huh? The great and powerful TSO has spoken!

Your suppositions are supported by neither logic nor statistical evidence. In this case, your fallacious tactic is more complex than the valid tactic it replaces.

[RW]

First of all, at this point there is a 5-cell simple xy-type forcing chain that basically solves the puzzle -- 100% of the time.

And a BUG-lite (with shorter 3-cell chain) to make the same elimination:

Code: Select all

 *-----------------------------------------------------------*
 | 7     9    #238   | 1    -28    36    | 5     68    4     |
 |*134  *346  *2348  | 24    5     36    | 178   9     167   |
 |#14   #456  #458   | 7     48    9     | 2     3     16    |
 |-------------------+-------------------+-------------------|
 | 5     7     6     | 8     9     1     | 3     4     2     |
 | 8     1     9     | 23    23    4     | 6     7     5     |
 | 2    *34   *34    | 5     6     7     | 9     1     8     |
 |-------------------+-------------------+-------------------|
 |*34   *345  *345   | 6     17    8     | 17    2     9     |
 | 6     8     7     | 9     13    2     | 4     5     13    |
 | 9     2     1     | 34    347   5     | 78    68    367   |
 *-----------------------------------------------------------*
BUG-lite in r267c123: r1c3=3 OR r3c123=4 has to be true

If r1c5=8 => r1c3=2 => (r1c3<>3)
          => r3c5=4 (r3c123<>4)
   =>r1c5<>8

RW

[Carcul]

And a BUG-lite (with shorter 3-cell chain) to make the same elimination:

And a shorter "1-cell chain": r1c3 must be "8" and the puzzle is solved.

Carcul

[ronk]

Code: Select all

 *-----------------------------------------------------------*
 | 7     9    #238   | 1    -28    36    | 5     68    4     |
 |*134  *346  *2348  | 24    5     36    | 178   9     167   |
 |#14   #456  #458   | 7     48    9     | 2     3     16    |
 |-------------------+-------------------+-------------------|
 | 5     7     6     | 8     9     1     | 3     4     2     |
 | 8     1     9     | 23    23    4     | 6     7     5     |
 | 2    *34   *34    | 5     6     7     | 9     1     8     |
 |-------------------+-------------------+-------------------|
 |*34   *345  *345   | 6     17    8     | 17    2     9     |
 | 6     8     7     | 9     13    2     | 4     5     13    |
 | 9     2     1     | 34    347   5     | 78    68    367   |
 *-----------------------------------------------------------*


BUG-lite in r267c123: r1c3=3 OR r3c123=4 has to be true

That seems like a confusing way to illustrate both of the following two possible BUG-Lites ...

Code: Select all

 *-----------------------------------------------------------*
 | 7     9     238   | 1     28    36    | 5     68    4     |
 |*134   346  *2348  | 24    5     36    | 178   9     167   |
 | 14    456   458   | 7     48    9     | 2     3     16    |
 |-------------------+-------------------+-------------------|
 | 5     7     6     | 8     9     1     | 3     4     2     |
 | 8     1     9     | 23    23    4     | 6     7     5     |
 | 2    *34   *34    | 5     6     7     | 9     1     8     |
 |-------------------+-------------------+-------------------|
 |*34   *345   345   | 6     17    8     | 17    2     9     |
 | 6     8     7     | 9     13    2     | 4     5     13    |
 | 9     2     1     | 34    347   5     | 78    68    367   |
 *-----------------------------------------------------------*
... and ...
 *-----------------------------------------------------------*
 | 7     9     238   | 1     28    36    | 5     68    4     |
 |*134  *346   2348  | 24    5     36    | 178   9     167   |
 | 14    456   458   | 7     48    9     | 2     3     16    |
 |-------------------+-------------------+-------------------|
 | 5     7     6     | 8     9     1     | 3     4     2     |
 | 8     1     9     | 23    23    4     | 6     7     5     |
 | 2    *34   *34    | 5     6     7     | 9     1     8     |
 |-------------------+-------------------+-------------------|
 |*34    345  *345   | 6     17    8     | 17    2     9     |
 | 6     8     7     | 9     13    2     | 4     5     13    |
 | 9     2     1     | 34    347   5     | 78    68    367   |
 *-----------------------------------------------------------*

... or am I missing something?

[pyrpoliser]

Regardless, you conclusion is unsupported and will give you no advantage over picking at random. Excluding 4 from r2c2 is a guess, and not a particularly educated one. Jeez, you can *always* have at *least* a 8 out of 9 chance of being right when making an exclusion!

What a wise observation!! What you will actually end up with is 8 pieces, each one of which will have 1:9 chance, out of which you will not have the slightest clue which one to choose. So unless you have a criterion with which to pick up your group of 8, your group is totally useless. This criterion is what will enable you utilise your chance of 8:9 collectively and NOT as tiny pieces of 1:9. In fact if you notice I did not turn down any of the remaining possibilities. I left them there and utilised their common criterion which is a 4 missing from Block1 pos5

The great and powerful TSO has spoken!

The greatest…

[pyrpoliser]

USE OF METHOD WITH 100% MATHEMATICAL CERTAINTY

I will use the same puzzle that I used in my previous message
The table below shows the Block positions after applying basic strategies.
The target is to prove that digit 4 cannot hold the line of Block positions 5,8,3,9,6,2,1,7,4 which is the one and only possible way for digit 4 avoid falling into VBPN mode.



Setting those positions in the table and striking out numbers we end up at the results shown on the side table.
Looking at digit 8: Either it takes block position 2 or 4 in block 3 it is going to end to one VBPN. We need another one.The only possible is at position 3.Doing so however will end up with Block1 pos 3 holding an 8 and at the same time Block2 pos 2 also holding an 8 which is impossible as they belong to the same row.

Conclussion: Digit 4 cannot be in non VBPN mode.The rest is as per my previous message.


NB. As it is obvious that this forum is dominated by great people, this is my last word here. Good bye.

[RW]

That seems like a confusing way to illustrate both of the following two possible BUG-Lites ...

Can there be anything less confusing than this?

Code: Select all

ab .  .  |ab .  .  |ab .  . 
ab .  .  |ab .  .  |ab .  . 
ab .  .  |ab .  .  |ab .  . 
= Double solution

I just added some better illustrated examples of the pattern here.

pyrpoliser, so you proved that 4 has to be in VBPN mode, but that's not the real reason why you can make the exclusion. Your real reasoning is:

Code: Select all

If digit '4' isn't in VBPN mode, then there has to be two digits '8' in row 1
If r2c2=4, then digit '4' is not in VBPN mode
   => if r2c2=4, then there has to be two digits '8' in row 1
=> r2c2<>4

Expressed as an error net:

Code: Select all

If r2c2=4 => r3c2=6 => r3c3=5
          => r3c5=4 => r1c5=8 => r2c7=8 => r1c3=8


If you prefer to look for this kind of eliminations in a table of block positions instead of a sudoku grid, then go ahead and do so. But still doesn't mean that VBPN modes has anything to do with solving the puzzle, implications that lead to contradictions are what your really looking for.

If you only know that "if digit 'a' is not in VBPN mode, then there is a contradiction" = useless information
If you only know that "if cell C=a, then digit 'a' isn't in VBPN" = useless information
Combine the two and you have useful information which gives C<>a. But in terms of finding this elimination the VBPN part is useless, you can also skip right to the fact that "if C=a, then there is a contradiction".

But as you seem to like your BP table, there's also another way to find contradictions in it; in the table on the right there is no 7, 8 or 9 for digit '8' in B123 => no 8 in row 3. Lets you make the elimination even before you start thinking about if '8' is in VBPN mode or not.

RW

[pyrpoliser]

RW,

just out of politeness and due respect for the couple of conversations we had on this topic, I will reply to your message, but this is going to be my last post here for the reasons stated above.
I agree with almost everything you said.
The VBPN however was the condition that let me spot the exact cell and the exact digit where I should look for contradiction. The alternative would be to check each digit in each cell separately. So this application of the method may prove beneficial in more complicated puzzles having many cells unsolved with many digits each.

I dont know if there are other applications of the method. I asked for help, and all I received so far was just a lot of push backs from some people riding one high horse after another.

So long, and thank you for your constructive comments.

[al-b]

i am sorry dude, but i am having a hard time trying to figure out what u mean by block positions. and i am having a hard time understanding the grid with b1, b2, b3... going across the top. could you please tell me exactly what a block position is? i find myself to be a lot better than most people i know that play sudoku. i have it on my sprint phone and i am now a black sudoku (even tho i play in brown bcuz it's easier to c), and i play on the insane level regularly. i pretty much have my technique down, but i am looking for some insight on some techniques that i am not aware of. i just learned a new one last week, but it usually only works after u have over 50% of the grid filled already. but yea, i need to know exactly what a block position is. that grid that u have with the yellow going thru it is hard to understand, but that's just bcuz i'm ignorant to the fact of what a block position actually is. can anybody help me out on this?