The New Sudoku Players' Forum

[tarek]

Code: Select all

+-------+-------+-------+
| . 2 . | 5 . 7 | . 1 . |
| . 1 . | . . . | . . 5 |
| . . 7 | . . 8 | 4 . . |
+-------+-------+-------+
| . . 2 | 8 . . | 6 . 1 |
| 3 . . | . . . | . . . |
| . . . | . . 5 | . . 7 |
+-------+-------+-------+
| 2 . . | . . 4 | 7 . . |
| . 8 . | . . . | . 4 6 |
| . . 9 | . 1 . | . . . |
+-------+-------+-------+
.2.5.7.1..1......5..7..84....28..6.13.............5..72....47...8.....46..9.1....


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[SpAce]

Code: Select all

.-----------.----------.---------------------.
| 89+  2  6 | 5  4   7 | 89+3  1   ab3(8)[9] |
| 89+  1  4 | 3  29  6 | 89+2  7     5       |
| 5    3  7 | 1  29  8 | 4     6    a2[9]    |
:-----------+----------+---------------------:
| 4    7  2 | 8  3   9 | 6     5     1       |
| 3    6  5 | 2  7   1 | 89    89    4       |
| 1    9  8 | 4  6   5 | 23    23    7       |
:-----------+----------+---------------------:
| 2    5  1 | 6  8   4 | 7     39  b(3)-9    |
| 7    8  3 | 9  5   2 | 1     4     6       |
| 6    4  9 | 7  1   3 | 5     28    28      |
'-----------'----------'---------------------'

UR (89)r12c17 using externals

(9)r13c9 =UR= (83)r17c9 => -9 r7c9; stte

[pjb]

Code: Select all

 89      2       6      | 5      4      7      |c389    1      89-3   
 89      1       4      | 3      29     6      |c289    7      5     
 5       3       7      | 1      29     8      | 4      6     b29     
------------------------+----------------------+---------------------
 4       7       2      | 8      3      9      | 6      5      1     
 3       6       5      | 2      7      1      |c89     89     4     
 1       9       8      | 4      6      5      | 23     23     7     
------------------------+----------------------+---------------------
 2       5       1      | 6      8      4      | 7      39    a39     
 7       8       3      | 9      5      2      | 1      4      6     
 6       4       9      | 7      1      3      | 5      28     28     


(3=9)r7c9 - (9=2)r3c9 - (2=3)r125c7 => -3 r1c9; stte
or very similar
(8=2)r9c9 - (2=9)r3c9 - (9=8)r126c7 => -8 r1c9; stte

Phil

[Ngisa]

Code: Select all

+----------------+----------------+--------------------+
| 89     2     6 | 5     4      7 | 389     1      389 |
| 89     1     4 | 3    b29     6 |c289     7      5   |
| 5      3     7 | 1    a29     8 | 4       6      2-9 |
+----------------+----------------+--------------------+
| 4      7     2 | 8     3      9 | 6       5      1   |
| 3      6     5 | 2     7      1 | 89      89     4   |
| 1      9     8 | 4     6      5 |d23     e23     7   |
+----------------+----------------+--------------------+
| 2      5     1 | 6     8      4 | 7      f39    g39  |
| 7      8     3 | 9     5      2 | 1       4      6   |
| 6      4     9 | 7     1      3 | 5       28     28  |
+----------------+----------------+--------------------+


(9=2)r3c5 - r2c5 = r2c7 - r6c7 = (2-3)r6c8 = (3-9)r7c8 = (9)r7c9 => - 9r3c9; stte


Clement

[Cenoman]

Code: Select all

 +-----------------+-----------------+-------------------+
 |  89   2    6    |  5    4    7    | b39+8  1  ca38+9  |
 |  89   1    4    |  3    29   6    | a28+9  7    5     |
 |  5    3    7    |  1    29   8    |  4     6    2-9   |
 +-----------------+-----------------+-------------------+
 |  4    7    2    |  8    3    9    |  6     5    1     |
 |  3    6    5    |  2    7    1    |  89    89   4     |
 |  1    9    8    |  4    6    5    |  23    23   7     |
 +-----------------+-----------------+-------------------+
 |  2    5    1    |  6    8    4    |  7     39  c39    |
 |  7    8    3    |  9    5    2    |  1     4    6     |
 |  6    4    9    |  7    1    3    |  5     28   28    |
 +-----------------+-----------------+-------------------+

BUG+3
(9)b3p34 == (8)r1c7 - (8=39)r17c9 => -9 r3c9; ste

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 |c89   2    6    | 5    4    7    |b389  1   b389  |
 | 9-8  1    4    | 3    29   6    |a289  7    5    |
 | 5    3    7    | 1    29   8    | 4    6    29   |
 *----------------+----------------+----------------|
 | 4    7    2    | 8    3    9    | 6    5    1    |
 | 3    6    5    | 2    7    1    | 89   89   4    |
 | 1    9    8    | 4    6    5    | 23   23   7    |
 *----------------+----------------+----------------|
 | 2    5    1    | 6    8    4    | 7    39   39   |
 | 7    8    3    | 9    5    2    | 1    4    6    |
 | 6    4    9    | 7    1    3    | 5    28   28   |
 *--------------------------------------------------*


8r2c7 =BUG+2= 9r1c79 - (9=8)r1c1 => -8 r2c1 ; stte

[SpAce]

Hi Steve,

8r2c7 =BUG+2= 9r1c79 - (9=8)r1c1 => -8 r2c1 ; stte

Nice! When I considered whether to use the UR or the BUG I didn't see that possibility!

For the daily notational detail, I'd like to draw attention to the '=BUG+2=' part. Should it be that or '=BUG+3='?

It's not an easy question, and I've struggled with it myself. I know I've used both options at various times. In fact, these days I usually write it without the +N to avoid the dilemma altogether (though it's not optimal, as it leaves the burden of figuring it out to the reader). If I did include the +N, I'd probably use +3 in this case. Here's my current reasoning:

To me the 'a =PATTERN= b' notation should be read as: 'a OR b because of PATTERN'. In other words, the PATTERN (whatever it is) acts as a catalyst that provides the derived strong link between a and b. What is that PATTERN in this case? Is it BUG+2 or BUG+3? I would say the latter. As further support for that point of view, there's a total of three candidates on both sides of the strong link.

When would I use BUG+2 instead? That would be with a normal strong link: '8r2c7 = BUG+2', which is read as: 'either 8r2c7 OR BUG+2'. That's a valid strong link, but it makes the chain hard to continue with the normal AIC notation, because without any further qualifications 'BUG+2' doesn't have any weak links. Thus, the catalyst notation seems like the best option giving us the strong link to the 9s as we want.

A couple more ways to write it (just to point out some possibilities, not to recommend them):

8r2c7 = [BUG = 9r1c79 - (9=8)r1c1] => -8 r2c1

BUG = [8r2c7 == 9r1c79 - (9=8)r1c1] => -8 r2c1

(8=9)r1c1 - r1c79 = [BUG = 8r2c7] => -8 r2c1

In those samples, the strongly-linked 'BUG' is really a 'BUG+0', i.e. a DP (contradiction, FALSE).

[SteveG48]

I agree. It should be BUG+3

[SpAce]

I agree. It should be BUG+3

Great! I think the real difficulty is if one wants to use that notation within nested chains. In this case it doesn't make any sense, but just to make a point:

8r2c7 =BUG+3= [9r1c7 =BUG+2= 9r1c9] - (9=8)r1c1 => -8 r2c1

Does that compute? I think so, but I'm not 100% sure. What do you think? Should they both be '=BUG+3='? My rationale for the latter ='BUG+2=' is the previously mentioned '8r2c7 = BUG+2' strong link.

(Normally I just use '==' in those cases to keep it simpler and shorter, but I still want to know what the correct explicit form is.)

[rjamil]

Hi,

Well, it's not to be perfect but,I see the latest yzfwsf's program YZF_Sudoku finds BUG+3 move as per RW's logic as follows:
BUG + 3 : => r1c7<>9; (but the move not produces STTE solution. Similarly, its counter Bivalue Universal Grave + 3 not detected.)

However, other similar elimination wise simple moves detected are:
W-Wing: 89 in r1c1,r5c7 connected by 8 in r2c17 => r1c7<>9
XYZ-Wing: 289 in r2c7 r5c7 r3c9 => r1c7 <> 9

Note: no further discussion needed as I have difficulty in identifying the BUG+candidates in this case, especially 8 @ r1c7.

R. Jamil

[SteveG48]

Great! I think the real difficulty is if one wants to use that notation within nested chains. In this case it doesn't make any sense, but just to make a point:

8r2c7 =BUG+3= [9r1c7 =BUG+2= 9r1c9] - (9=8)r1c1 => -8 r2c1

Does that compute?

Yes, I like it. It was recognition of the fact that eliminating 8r2c7 changed things from an almost BUG with 3 guardians to an almost BUG with 2 guardians that led me to write BUG+2 initially, but I didn't have a way to say it. Your way serves nicely.

I could also have simply written 8r2c7 = (8,9)r21c1 - 9r1c79 = BUG => +8 r2c7 . Would that have been better?

Of course I prefer Cenoman's solution, but that's another story.

[SpAce]

Hi Steve,

Just a quick downer until we can get back to the notation issues. rjamil's comment about the BUG+ candidates made me look at yours again. It seems that you have the 8 and 9 switched places in r12c7, which breaks the BUG logic. It's quite funny that we both missed that! The correct guardians are: 8r1c7, 9r1c9, 9r2c7 as in Cenoman's solution.

Edit: On second thought, I think it was only I who missed that. Looks like your original intention was correct after all, since it wasn't really based on the BUG+3 (unlike I assumed). I'll get back to this.

[SpAce]

Hi rjamil,

Well, it's not to be perfect but,I see the latest yzfwsf's program YZF_Sudoku finds BUG+3 move as per RW's logic as follows:
BUG + 3 : => r1c7<>9; (but the move not produces STTE solution.

Yes, RW's logic works here. In fact, its elimination could be included (due to a subchain) in Cenoman's solution, though it makes no difference:

(9)b3p34 == (8)r1c7 - (8=39)r17c9 => -9 r1c7,r3c9; ste

Similarly, its counter Bivalue Universal Grave + 3 not detected.)

I don't know what you mean by that. What is a "counter BUG+3"?

Note: no further discussion needed as I have difficulty in identifying the BUG+candidates in this case, especially 8 @ r1c7.

I would rather think it specifically warrants further discussion! Or don't you want to learn?

You're right that the guardian cell r1c7 is the trickiest of the three. The reason is that it sees at least one other guardian cell in each of its houses (row, col, box) and shares digits with them. The process is easier if you start with the other two cells with three candidates, i.e. r1c9 and r2c7, and identify their guardians first. They're easier because they both have a house in which they're the sole guardian cells.

First an optional step. If you notice that all three (potential) guardian cells (r1c7, r1c9, r2c7) are in box 3, you can easily count what the guardian digits must be, without yet worrying about their exact cells. In box 3 there are exactly two 2s and two 3s, so neither of those digits can be a guardian. On the other hand, there are three 8s and four 9s. That tells us directly that if we have a BUG+ situation (which we don't yet know) the guardians must be two 9s and one 8, because that would leave us with two of every digit in box 3 (if the guardians were removed).

Next you should figure out which guardian digits go to which cell. There are three possible configurations with three cells and two digits, but only one is correct (if any) -- it's the one that leaves exactly two instances of each digit in each house if the guardians were removed. If you did the previous step, you could simply figure out where the 8 must be, and the 9s would necessarily go to the other two cells. However, that's actually a bit harder route here. Like I said, it's easier to start with r1c9 or r2c7. Let's start with the latter, because it's the simplest of them all.

If you look at r2c7 you should notice that it's the only guardian cell in its row, as the other two are bivalue cells. Next you should just check which digit exists three times in that row. The answer is 9, so it must be the guardian. If 9r2c7 were removed, there would be exactly two of each digit (2,8,9) in that row, as per BUG requirements. Thus 9r2c7 can be locked as a guardian. Note that it also removes the possibility that 8r2c7 could be a guardian.

Similarly, if you look at r1c9 you should notice that it's the only guardian cell in its column. Again, if you count the digit instances in that column you notice that 9 is the only one that exists three times, so it must be the guardian digit. Thus 9r1c9 can be locked as a guardian. It also removes the possibility that 8r1c9 is a guardian.

Whether you did the optional step or not, you now know that the third guardian must be 8r1c7 because r1c7 is the only trivalue cell left, and 8 is the only digit in box 3 that has three candidates left. There are also three 8s in every house r1c7 sees: r1,c7,b4. Thus 8r1c7 can be locked as a guardian.

The last step is to double-check that if those three guardians were removed you'd actually have a BUG+3 (all bivalue cells, every digit exactly twice in every house). It holds, so we have a BUG+3 with guardians 8r1c7, 9r1c9, 9r2c7.

--
Edit 1. Fixed a typo (box 4 -> box 3).
Edit 2. bivalue -> trivalue

[SpAce]

I could also have simply written 8r2c7 = (8,9)r21c1 - 9r1c79 = BUG => +8 r2c7 . Would that have been better?

At the very least I think I now understand your original intention. In that light I have to reconsider what I wrote earlier. I'll get back to this after I've thought about it a bit more!

Of course I prefer Cenoman's solution, but that's another story.

Well, Cenoman's is a very clean BUG+3 solution, and also nice with the (non-degenerate) naked-pair ending. Yours, however, used a totally different angle, which makes it interesting (especially since I didn't see it at first). Like I said, I now have to look at it again from the correct angle before making further comments!

[rjamil]

Hi SpAce,

First of all, many thanks for detail guidance with patience. I will try to digest it slowly. (Since, I have already lost my control once over there.)

I don't know what you mean by that. What is a "counter BUG+3"?

What I understand from yzfwsf's post is that, there are two different scenarios as per elimination point of view. I assumed BUG+n for elimination(s) from BUG+candidates of BUG+cells. Whereas, "counter" Bivalue Universal Grave + n for elimination(s) from other cells that are based on BUG+cells/guardian candidates (like Cenoman's BUG+3 move. But, it needs confirmation from yzfwsf.)

R. Jamil