The New Sudoku Players' Forum

[zhaoqian6318]

Code: Select all

*--------------------------------------------------------------------------*
| 4579    79      34579  | 27      8       1      | 234569  2347    4679   |
| 2       1       379    | 4       6       5      | 39      8       79     |
| 457     6       8      | 3       27      9      | 245     1247    147    |
|------------------------+------------------------+------------------------|
| 3       79      79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
|------------------------+------------------------+------------------------|
| 467     28      147    | 9       27      3      | 468     5       1467   |
| 4579    3       4579   | 6       1       8      | 49      47      2      |
| 679     28      179    | 27      5       4      | 3689    137     1679   |
*--------------------------------------------------------------------------*

??????

[re'born]

Code: Select all

*--------------------------------------------------------------------------*
| 4579    79      34579  | 27      8       1      | 234569  2347    4679   |
| 2       1       379    | 4       6       5      | 39      8       79     |
| 457     6       8      | 3       27      9      | 245     1247    147    |
|------------------------+------------------------+------------------------|
| 3       79      79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
|------------------------+------------------------+------------------------|
| 467     28      147    | 9       27      3      | 468     5       1467   |
| 4579    3       4579   | 6       1       8      | 49      47      2      |
| 679     28      179    | 27      5       4      | 3689    137     1679   |
*--------------------------------------------------------------------------*

??????

Well, there is a UR deduction r14c23<79> => r1c3<>7,9...then it gets trickier.

[RW]

There's also another uniqueness reduction saying r13c7<>4, 10 points to anyone who can find out why. But then it gets trickier...

RW

[re'born]

There's also another uniqueness reduction saying r13c7<>4, 10 points to anyone who can find out why. But then it gets trickier...

RW

I see a [Reverse BUG on the digits 4 and 5 that will (with some work) do the job)]. Is this what you had in mind?

[ronk]

I saw the same, but with digits 1 and 4.

[RW]

I see a ... Is this what you had in mind?

Very good! Actually, it wasn't exactly what I had in mind. I noticed the eliminations as a normal BUG-lite, but the reverse of this same pattern works very well also!

[zhaoqian6318]

Sorry. I don't understand.

???????

[wintder]

I am certain I don't understand uniqueness and bugs, although I am learning, I hope. What I have done here does solve the puzzle quickly, I just don't know if either move is correct. I think the first is a revers bug and the second a bug-lite. Do I have any of it properly done?

Code: Select all

.------------------------.------------------------.------------------------.
| 45-7-9  79     *345    | 27      8       1      |*234569  2347    4679   |
| 2       1       379    | 4       6       5      | 39      8       79     |
|*457     6       8      | 3       27      9      |*245     1247    147    |
:------------------------+------------------------+------------------------:
| 3       79      79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
:------------------------+------------------------+------------------------:
| 467     28      147    | 9       27      3      | 468     5       1467   |
|*4579    3      *4579   | 6       1       8      | 49      47      2      |
| 679     28      179    | 27      5       4      | 3689    137     1679   |
'------------------------'------------------------'------------------------'

The digits 45 marked * form a potential deadly pattern. r1c1 is an
external 45 that fixes the pattern. r1c1 must be 4 or 5.

Now locked candidates in c1 on 9s gives us this.

Code: Select all

 
.------------------------.------------------------.------------------------.
| 45     *79      345    | 27      8       1      | 234569  2347   *79+46  |
| 2       1      *79+3   | 4       6       5      | 39      8      *79     |
| 457     6       8      | 3       27      9      | 245     1247    147    |
:------------------------+------------------------+------------------------:
| 3      *79     *79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
:------------------------+------------------------+------------------------:
| 467     28      147    | 9       27      3      | 468     5       1467   |
| 4579    3       457    | 6       1       8      | 49      47      2      |
| 679     28      17     | 27      5       4      | 3689    137     1679   |
'------------------------'------------------------'------------------------'

The marked cells on 79 would be deadly unless r2c3=3  or r1c9=4 or r1c9=6
There are only 2 3s in r2 and only 2 6s in r1. There are lots of 4s in r1
so I set r1c9 = 4. The rest is routine stuff.

[ravel]

Sorry. I don't understand.
???????

Code: Select all

*--------------------------------------------------------------------------*
| 4579   #79      34579  | 27      8       1      | 234569  2347    4679   |
| 2       1      #379    | 4       6       5      |#39      8       79     |
| 457     6       8      | 3       27      9      | 245     1247    147    |
|------------------------+------------------------+------------------------|
| 3       79      79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
|------------------------+------------------------+------------------------|
| 467     278     147    | 9       27      3      | 468     5       1467   |
| 4579    3       4579   | 6       1       8      |#49      47      2      |
| 679     2789    179    | 27      5       4      | 3689    137     1679   |
*--------------------------------------------------------------------------*

Hm, dont know, what they mean either. All i can see is, [edit - its late] that if you use the marked cells after assuming a 4 in r13c7, with the strong link for 5 in column 7 you get a deadly pattern iin r13c17.
But this neither is what i thought to be a reverse bug (using givens) nor is it simpler than chains to solve the puzzle.
So please explain, guys.

wintder, your deduction is not valid, there are a lot of other possibilities (extra candidates) to "fix" the pattern.

[wintder]

wintder, your deduction is not valid, there are a lot of other possibilities (extra candidates) to "fix" the pattern.

I had thought in using only duos, 45 in this case, that what I did was valid.

I am well aware that I am learning. <sigh>

[RW]

zhaoqian, to understand the basics of the techniques we are discussing, see here. Ruud's first post is a good introduction, follow the links at the bottom of his post for further information and examples.

I had thought in using only duos, 45 in this case, that what I did was valid.

No, you have to focus on all cells containing any of the candidates in the deadly pattern, in all units that the pattern belong to. You got the deadly pattern almost right, it is the 45 pairs in those boxes, but r1c1 is also part of the deadly pattern (see here). It's a subpattern of the third example. If we would have r13c7=45 and r8c13=45, then r1c1 couldn't save us anymore, because assigning the value 4 or 5 to this cell would cause a UR in one direction and a cell without any possible candidates in the other direction.

To avoid the deadly pattern we must make sure that in either box 3 or box 7 there is a digit 4 or 5 outside the deadly pattern. Digit 5 cannot go in any other cell in either of the boxes, so it has to be digit 4 => r7c13 or r13c89 must be 4. Now notice that in column 9, we only have candidate 4 in box 3 or row 7. So, here's the short chain:

if r13c7=4 => r13c89<>4 => r7c9=4 => r7c13<>4
=> the deadly pattern is completed => r13c7<>4

Code: Select all

*--------------------------------------------------------------------------*
|*4579    79     *34579  | 27      8       1      |*234569 #2347   #4679   |
| 2       1       379    | 4       6       5      | 39      8       79     |
|*457     6       8      | 3       27      9      |*245    #1247   #147    |
|------------------------+------------------------+------------------------|
| 3       79      79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
|------------------------+------------------------+------------------------|
|#467     28     #147    | 9       27      3      | 468     5      %1467   |
|*4579    3      *4579   | 6       1       8      | 49      47      2      |
| 679     28      179    | 27      5       4      | 3689    137     1679   |
*--------------------------------------------------------------------------*

Code: Select all

.------------------------.------------------------.------------------------.
| 45     *79      345    | 27      8       1      | 234569  2347   *79+46  |
| 2       1      *79+3   | 4       6       5      | 39      8      *79     |
| 457     6       8      | 3       27      9      | 245     1247    147    |
:------------------------+------------------------+------------------------:
| 3      *79     *79     | 5       4       2      | 1       6       8      |
| 1       4       2      | 8       3       6      | 7       9       5      |
| 8       5       6      | 1       9       7      | 24      24      3      |
:------------------------+------------------------+------------------------:
| 467     28      147    | 9       27      3      | 468     5       1467   |
| 4579    3       457    | 6       1       8      | 49      47      2      |
| 679     28      17     | 27      5       4      | 3689    137     1679   |
'------------------------'------------------------'------------------------'
The marked cells on 79 would be deadly unless r2c3=3  or r1c9=4 or r1c9=6
There are only 2 3s in r2 and only 2 6s in r1. There are lots of 4s in r1
so I set r1c9 = 4. The rest is routine stuff.

Your deadly pattern is correct, but your deduction is wrong. A deadly pattern can never be used to eliminate any other candidate from the pattern cells than those that are part of the pattern itself. If you set r1c9=4, then you also somehow eliminate candidate 6, which cannot be done using a deadly pattern on digits 79. With that pattern you may eliminate 7 and 9 from r1c9, because of the strong interference in row 2 (if r1c9=79 => r2c3=79), but you may not make any further eliminations.

RW

[udosuk]

I have no idea about all these uniqueness techniques you guys are talking about.
I just try to solve it without assuming uniqueness, albeit less elegantly ().

First, from this position:

Code: Select all

 *-----------------------------------------------------------------------------*
 | 4579    79      34579   | 27      8       1       | 234569  2347   #4679    |
 | 2       1      @379     | 4       6       5       |*39      8      #79      |
 | 457     6       8       | 3       27      9       | 245     1247   #147     |
 |-------------------------+-------------------------+-------------------------|
 | 3       79     @79      | 5       4       2       | 1       6       8       |
 | 1       4       2       | 8       3       6       | 7       9       5       |
 | 8       5       6       | 1       9       7       | 24      24      3       |
 |-------------------------+-------------------------+-------------------------|
 | 467     28      147     | 9       27      3       | 468     5      #1467    |
 | 4579    3       4579    | 6       1       8       | 49      47      2       |
 | 679     28     @179     | 27      5       4       | 3689    137    -1679    |
 *-----------------------------------------------------------------------------*

ALS-xy-wing (don't worry, I'll explain the logic in detail below):
ALS A: r249c3={1379}
ALS B: r1237c9={14679}
ALS C: r2c7={39}
restricted common between A & C: 3
restricted common between B & C: 9
common between A & B: r9c9=1

Therefore r9c9 can't be 1.

Logic:
r2c7 must be 3 or 9.
If r2c7=3, then r249c3 becomes a naked triple of {179} with 1 fixed at r9c3 => r9c9 can't be 1.
If r2c7=9, then r1237c9 becomes a naked quad of {1467} => r9c9 can't be 1.
Therefore r9c9 can't be 1 no matter what.

Code: Select all

 *-----------------------------------------------------------------------------*
 |*4579   *79     *34579   |*27      8       1       |*234569 -2347   *4679    |
 | 2       1       379     | 4       6       5       |@39      8       79      |
 | 457     6       8       | 3       27      9       | 245     1247    147     |
 |-------------------------+-------------------------+-------------------------|
 | 3       79      79      | 5       4       2       | 1       6       8       |
 | 1       4       2       | 8       3       6       | 7       9       5       |
 | 8       5       6       | 1       9       7       |@24      24      3       |
 |-------------------------+-------------------------+-------------------------|
 | 467     28      147     | 9       27      3       | 468     5       1467    |
 | 4579    3       4579    | 6       1       8       |@49      47      2       |
 |#679     28     #179     |#27      5       4       | 3689   #137    #679     |
 *-----------------------------------------------------------------------------*

ALS-xyz-wing (or something like that)
ALS A: r268c7={2349}
ALS B: r9c13489={123679}
ALS C: r1c123479={2345679}
semi-restricted common between A & C: 2
semi-restricted common between B & C: 2
common among A, B & C: r1c8=3

Therefore r1c8 can't be 3.

Logic:
On r1, 1 of the 3 cells r1c478 must be 2.
If r1c4=2, then r9c13489 becomes a naked quint of {13679} with 3 fixed at r9c8 => r1c8 can't be 3.
If r1c7=2, then r268c7 becomes a naked triple of {349} with 3 fixed at r2c7 => r1c8 can't be 3.
If r1c8=2, then r1c8 can't be 3.
Therefore r1c8 can't be 3 no matter what.

The puzzle can then be solved with a turbot fish on 7, a box-line elimination on 4 and singles.

[ravel]

Finally a nice solution to this hard graft puzzle. As usual its easier for me to find and see the ALS's as chains.
E.g. the first one:
r9c9 = 1 => r1|2c9=9 = r2c7=3 => r24c3=79 => r9c3=1 - contradiction

I have no idea about all these uniqueness techniques you guys are talking about.:

To summarize, why i didn't get the 10 points:
You have to
- see that with the strong link for 5 in column 7, r1|3c7=4 => r13c7=45
- see (know) the combined uniqueness pattern as marked by RW
- see that no 5 can be outside the pattern in all 3 boxes. and r7c13 are the only cells where 4 can be outside the pattern
- see that r13c9 <> 4 => r7c9=4 => r7c13<>4
Since i didn't do my homework, i missed to consider point 2. In fact this is point 1 for uniqueness hunters.

[zhaoqian6318]

understand
Thank you!

[wintder]

Thanks for your input, ravel and RW.

RW, it took me a while to see why you weren't concerned with the 4s in r678c1. I have learned quite a bit from this example, and the related links.