Uniqueness Type 6 - UR meets X-Wing

Advanced methods and approaches for solving Sudoku puzzles

Postby Havard » Tue May 09, 2006 11:06 am

ronk wrote:
RW wrote:It is indeed possible to construct a deadly pattern of 7 cells (or other odd amounts) (...) but these would have at least one trivalue cell.

Yes, most of us are aware of the Forming MUGs from BUG-Lite composites thread.

If Havard found one with only bivalue cells, I'd be very interested to see it.
Me too.


Here you go. This one is from puzzle nr. 1430:

Code: Select all
23    1     69    | 7     4     36    | 5     8     29
5     78    69    | 12*   12*   368   | 37    4     379
4     78    23    | 9     5     38    | 237   6     1
------------------+-------------------+------------------
7     23    23    | 8     6     9     | 1     5     4
6     9     4     | 5     3     1     | 78    2     78
1     5     8     | 24*   7     24*   | 9     3     6
------------------+-------------------+------------------
8     23    7     | 6     9     5     | 4     1     23
23    4     1     | 23    8     7     | 6     9     5
9     6     5     | 14+3* 12*   24*   | 238   7     238


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Postby ronk » Tue May 09, 2006 11:40 am

Havard wrote:Here you go. This one is from puzzle nr. 1430:

Code: Select all
23    1     69    | 7     4     36    | 5     8     29
5     78    69    | 12*   12*   368   | 37    4     379
4     78    23    | 9     5     38    | 237   6     1
------------------+-------------------+------------------
7     23    23    | 8     6     9     | 1     5     4
6     9     4     | 5     3     1     | 78    2     78
1     5     8     | 24*   7     24*   | 9     3     6
------------------+-------------------+------------------
8     23    7     | 6     9     5     | 4     1     23
23    4     1     | 23    8     7     | 6     9     5
9     6     5     | 14+3* 12*   24*   | 238   7     238

Thanks. What technique made the exclusion r9c4<>2 and what was the candidate grid at the time?

TIA, Ron

[edit: It seems strange to find this pattern in a puzzle solvable by Angus' Simple Sudoku.]
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Postby Havard » Tue May 09, 2006 11:53 am

ronk wrote:Thanks. What technique made the exclusion r9c4<>2 and what was the candidate grid at the time?

TIA, Ron


Code: Select all
23    1     69    | 7     24    2346  | 5     8     239
5     78    69    | 1236  12    2368  | 237   4     12379
4     78    23    | 9     125   2358  | 237   6     137
------------------+-------------------+------------------
7     23    23    | 8     6     9     | 1     5     4
6     9     45    | 45    3     1     | 78    2     78
1     45    8     | 245   7     245   | 9     3     6
------------------+-------------------+------------------
8     235   7     | 2356  9     2356  | 4     1     235
23    2345  1     | 2345  8     7     | 6     9     235
9     6     45    | 12345 1245  2345  | 238   7     2358

UR in r2c45 - r9c45. Due to this pattern:
Code: Select all
           
ab     abX
        |
       a|
     a  |
abY-----abZ
eliminate b in abZ


Strong links: r2c4=1=r9c4=1=r9c5

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Postby ronk » Tue May 09, 2006 1:01 pm

Consider the following:
Code: Select all
 .    ab   .    | ab    .    .    | .    .    .   
                |                 |               
 .    .    .    | bc    .    .    | .    bc   .   
                |                 |               
 .    ab   .    | abc+X .    .    | .    bc   .   

for r3c4=X

For a puzzle with a unique solution, X cannot be eliminated from the abc+X cell. Therefore, any one of abc may already have been eliminated and the "BUG-Lite Type 1" would still apply ... with an odd number of cells. Duh!

Indeed, any two of abc may have been eliminated -- and X may be the single digit d -- and we would still have a valid exclusion from a deadly pattern of all bivalues.
Code: Select all
 .    ab   .    | ab    .    .    | .    .    .   
                |                 |               
 .    .    .    | bc    .    .    | .    bc   .   
                |                 |               
 .    ab   .    | a+d   .    .    | .    bc   .   

for r3c4=d


Hmm. Is this factoid already in the Myth Jellies' MUG thread? And since we're effectively "freely inventing" the missing candidates of a MUG pattern, shouldn't we be calling this a MUG-Lite?:D
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Postby Havard » Tue May 09, 2006 1:11 pm

Well, my example with 7 cells is a BUG-Lite+1.
Code: Select all
 .    ab   .    | ab    .    .    | .    .    .   
                |                 |               
 .    .    .    | bc    .    .    | .    bc   .   
                |                 |               
 .    ab   .    | ac+X  .    .    | .    bc   .   


Each candidate represented exactly twice in all Rows, Columns and Boxes.

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Postby RW » Tue May 09, 2006 3:21 pm

Havard wrote:I made a quick search through the top 1465 with BUG's after Locked sets (and UR), but before strong links /fishes.

...

Here you go. This one is from puzzle nr. 1430:

Code: Select all
23    1     69    | 7     4     36    | 5     8     29
5     78    69    | 12*   12*   368   | 37    4     379
4     78    23    | 9     5     38    | 237   6     1
------------------+-------------------+------------------
7     23    23    | 8     6     9     | 1     5     4
6     9     4     | 5     3     1     | 78    2     78
1     5     8     | 24*   7     24*   | 9     3     6
------------------+-------------------+------------------
8     23    7     | 6     9     5     | 4     1     23
23    4     1     | 23    8     7     | 6     9     5
9     6     5     | 14+3* 12*   24*   | 238   7     238



How could you identify this as a BUG-lite, when there is two UR-type 1 making the same reduction... You said you looked for URs first. It seems your solver doesn't handle URs with missing candidates (the missing 2 in r9c4 made both URs invisible). I dare to suggest that any 7 cell BUG-lite+1 can also be solved as two separate URs type 1 (or 1 UR type 1, depending on where the extra candidate is). That's why I made my example earlier in this thread a +2, makes the BUG-lite pattern useful.

It is very likely that the 9 cell BUG-lite+1 also can be solved in smaller parts, if it can't I'd like to see it.:)

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Postby Havard » Tue May 09, 2006 3:45 pm

RW wrote:How could you identify this as a BUG-lite, when there is two UR-type 1 making the same reduction... You said you looked for URs first. It seems your solver doesn't handle URs with missing candidates (the missing 2 in r9c4 made both URs invisible). I dare to suggest that any 7 cell BUG-lite+1 can also be solved as two separate URs type 1 (or 1 UR type 1, depending on where the extra candidate is). That's why I made my example earlier in this thread a +2, makes the BUG-lite pattern useful.

It is very likely that the 9 cell BUG-lite+1 also can be solved in smaller parts, if it can't I'd like to see it.:)

RW


I see your point. My solver does not yet handle UR 's with missing candidates, because I am still not quite sure what to make of them myself, let alone how to program them. Maybe you could give me a short "UR's with missing candidates-for-dummies"-lecture?:) (And for the record: if you take away one of the UR candidates it can no longer by definition be called a UR. I like the idea of extending a principle, but you would have to accept that it is perhaps only you and a few more people that would concider a UR with missing candidates a UR...):)

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Postby ronk » Tue May 09, 2006 3:52 pm

RW wrote:I dare to suggest that any 7 cell BUG-lite+1 can also be solved as two separate URs type 1 (or 1 UR type 1, depending on where the extra candidate is).
I came to the same conclusion while out walking to clear my head.

RW wrote:It is very likely that the 9 cell BUG-lite+1 also can be solved in smaller parts, if it can't I'd like to see it.:)
Me too.:)
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Postby RW » Tue May 09, 2006 4:06 pm

Havard wrote:Maybe you could give me a short "UR's with missing candidates-for-dummies"-lecture?


I've explained the logic behind URs with missing candidates here. You will find that they are not only valid, but valid even in multiple solution puzzles (unless you removed the missing candidate with uniqueness technique, which you wouldn't have done in a multiple solution puzzle:) ).

Havard wrote:And for the record: if you take away one of the UR candidates it can no longer by definition be called a UR.


I think this:

Code: Select all
ab . . |ab . .
X  . . |ab . .

X<>ab


is definitely a unique rectangle, no matter if candidates ab exist in X or not. I guess this is yet another difference in solving without pms and solving with them. Only if you first write down the candidates, you can physically remove them. If you don't write them down, you would not have any problem making the reduction above without considering if the candidates still are there or not.

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Postby ronk » Tue May 09, 2006 4:15 pm

RW wrote:It seems your solver doesn't handle URs with missing candidates ...

My solver didn't either. I just added that capability to UR Type 1 only ... and was disappointed to find no additional exclusions in the top1465.:( Either weren't any, or they were destroyed by other techniques.

I'll try other types anyway.
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Postby tarek » Tue May 09, 2006 5:11 pm

It shouldn't be difficult to program........

As long as each Roof cells has at least one candidate from the floor cells.

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Postby RW » Tue May 16, 2006 5:36 am

I was suprised that I couldn't find this common pattern yet in this "crazy UR"-thread (sorry if it's been discussed somewhere else already):

[edit: removed a diagram and changed the layout to make this post clearer]

Code: Select all
ab      abX
-----------
abY C   ab
bD  bE  abF
bG  bH  abK


where if abX=a => abY=a, without any strong links.

Example (top1465 #461, also discussed here):

Code: Select all
 19    3     256   | 2457  89    47    | 4568  45678 1678
 8     7     25    | 6     24    1     | 9     45    3
 19    4     56    | 357   89    37    | 568   2     1678
-------------------+-------------------+------------------
 2     15    134   | 8     37    6     | 345   4579  79
 35    68    9     | 2347  2347  347   | 1     57    678
*37    68   *347   | 1     5     9     | 34    68    2
-------------------+-------------------+------------------
 4     19    137   | 37    367   2     | 68    689   5
*357   59   *37    | 347   3467  8     | 2     1     69
 6     2     8     | 9     1     5     | 7     3     4


The diagonal pair + 2SL eliminates 7 from r6c3 and r8c1, but there is also:

If r6c3=3 => r8c1=3 => r6c3<>3

Later in the same puzzle:

Code: Select all
 19    3     256   | 2457  89    47    | 4568  45678 1678
 8     7     25    | 6     24    1     | 9     45    3
 19    4     56    | 357   89    37    | 568   2     1678
-------------------+-------------------+------------------
 2     15    13    | 8     37    6     | 45    4579  79
 35    68    9     | 2347  2347  347   | 1     57    68
 7     68    4     | 1     5     9     | 3     68    2
-------------------+-------------------+------------------
 4     19   *137   |*37    367   2     | 68    689   5
 35    59   *37    |*347   467   8     | 2     1     69
 6     2     8     | 9     1     5     | 7     3     4


The diagonal pair + 1SL eliminates 7 from r8c4. The following eliminations can also be made without SLs:

If r7c3=3 => r8c4=3
If r8c4=3 => r7c3=3

=> r7c3<>3, r8c4<>3

XY-wing then solves the puzzle.

RW
Last edited by RW on Tue May 16, 2006 2:55 pm, edited 1 time in total.
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Postby ronk » Tue May 16, 2006 11:48 am

RW wrote:
Code: Select all
 19    3     256   | 2457  89    47    | 4568  45678 1678
 8     7     25    | 6     24    1     | 9     45    3
 19    4     56    | 357   89    37    | 568   2     1678
-------------------+-------------------+------------------
 2     15    134   | 8     37    6     | 345   4579  79
 35    68    9     | 2347  2347  347   | 1     57    678
*37    68   *347   | 1     5     9     | 34    68    2
-------------------+-------------------+------------------
 4     19    137   | 37    367   2     | 68    689   5
*357   59   *37    | 347   3467  8     | 2     1     69
 6     2     8     | 9     1     5     | 7     3     4


The diagonal pair + 2SL eliminates 7 from r6c3 and r8c1, but there is also:

If r6c3=3 => r8c1=3 => r6c3<>3

For this example, I see that

r6c3=3 -> r7c3<>3
r6c3=3 -> r8c3=7 -> r7c3<>7
r6c3=3 -> r6c3<>4 -> r8c1=5 -> r8c2=9 ->r7c2=1 -> r7c3<>1

which is contradictory and therefore r6c3<>3. But I don't see how -- in general -- you can say ...
RW wrote:
Code: Select all
ab      abX


abY     ab

where if abX=a => abY=a, without any strong links.

Why isn't the following a valid outcome?
Code: Select all
b      a


Y      b

I'm obviously not getting the connection to this:
RW wrote:Mostly appears as:

Code: Select all
ab      abX
-----------
abY C   ab
bD  bE  abF
bG  bH  abK

Would you please clarify?
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Postby ravel » Tue May 16, 2006 2:29 pm

With this pattern its the last remaining a in the box. Nice short AUR-chain:)
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Postby RW » Tue May 16, 2006 5:54 pm

ronk wrote:But I don't see how -- in general -- you can say ...
RW wrote:
Code: Select all
ab      abX


abY     ab

where if abX=a => abY=a, without any strong links.

Why isn't the following a valid outcome?
Code: Select all
b      a


Y      b


[edit: edited the original post to clarify, the diagram you quoted does not include the conditions that imply if abX=a => abY=a. I added the diagram to show the four UR cells, the next line added the condition abX=a => abY=a and the second diagram showed what this would look like in a grid.]

It is not valid because of the definition of the pattern I gave you. If abX=a => abY=a. If you put 'a' in abX and 'y' in abY, there would be a box without an 'a'. In this pattern (the 3 rows below the line are supposed to represent a box):

Code: Select all
ab      abX
-----------
abY bC  ab
bD  bE  abF
bG  bH  abK
(C, D, E, F, G, H and K <> a)

candidate a appears only in column 3 and the cells that are part of the deadly pattern. If you put 'a' in abX, the only cell left for 'a' in the lower box is in abY.

ronk wrote:For this example, I see that

r6c3=3 -> r7c3<>3
r6c3=3 -> r8c3=7 -> r7c3<>7
r6c3=3 -> r6c3<>4 -> r8c1=5 -> r8c2=9 ->r7c2=1 -> r7c3<>1


To clarify my chain:
if r6c3=3 => r7c3<>3 => r8c3<>3 => r8c1=3 (=> r6c1=7 => r8c3=7 => UR: r6c3<>3)

As I said, this pattern is really common (allowed 5 eliminations in the mentioned puzzle) and it's also easy to spot. I would call it a UR because I can make the elimination without taking any steps outside the pattern (excluding the elimination steps, which I don't count as steps anyway).

Another version of this that I frequently use is:

Code: Select all
ab      ab
-----------
abY bC  abX
bD  bE  abF
bG  bH  abK

eliminate b from abX
(if abX=b => ab(above)=a =>abY=a)


RW
Last edited by RW on Tue May 16, 2006 3:11 pm, edited 1 time in total.
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