- Code: Select all
Almost locked sets xz rule
If
1)A,B be almost locked sets.
2)Specific common candidate (lets say x) can appear for A and B only in specific unit (U)
Then Any other common candidate (lets say z) cant appear outside of A and B if it can see all the z candidates in both A and B.
The reason?
If z appear then both A and B are locked but only one can get the x.
Its easy to see that its a generalization of the xyz wing rule (that's the reason for the name).
+----------------+----------------+----------------+
| 7 1 4 | 2 3 9 | 8 6 5 |
| 2 5 9 | 8 6 1 | 4 7 3 |
| 3 6 8 | 57 4 57 | 2 1 9 |
+----------------+----------------+----------------+
| 4 38 7 | 56 1 2568 | 35 9 28 |
| 9 38 6 | 47 58 247 | 35 24 1 |
| 5 2 1 | 3 9 *48 | 7 48 6 |
+----------------+----------------+----------------+
| 8 9 3 | 1 7 45 | 6 25 24 |
| 6 4 2 | 9 58 3 | 1 58 7 |
| 1 7 5 |*46 2 *468 | 9 3 48 |
+----------------+----------------+----------------+
A={R6C6,R9C6}
B={R9C4}
x=6
z=4
+----------------+----------------+----------------+
| 579 5789 12 | 3 4 12 | 589 59 6 |
| 459 6 249 | 7 8 29 | 1 3459 3459 |
| 34 89 1349 | 19 6 5 | 489 7 2 |
+----------------+----------------+----------------+
| 79 4 8 | 5 2 79 | 3 6 1 |
| 6 1 5 | 89 3 489 | 7 2 49 |
| 2 3 79 | 6 1 479 | 459 8 459 |
+----------------+----------------+----------------+
| 1 2 39 | 4 7 38 | 6 359 3589 |
| 3457 57 6 | 128 9 138 |*24 134 3478 |
| 8 *79 *3479 | 12 5 6 |*249 1349*3479 |
+----------------+----------------+----------------+
A={R9C2,R9C3,R9C7,R9C9}
B={R8C7}
x=2
z=4
+-------------+-------------+-------------+
| 24 7 8 | 24 6 5 | 1 9 3 |
| 9 3 24 | 248 1 48 | 56 7 56 |
| 5 1 6 | 7 3 9 | 8 4 2 |
+-------------+-------------+-------------+
| 28 9 23 | 458 7 6 |*35 1 *45 |
|*17 6 5 | 3 9 14 | 2 8 *47 |
| 178 4 *13 | 58 2 18 | 357 6 9 |
+-------------+-------------+-------------+
| 6 5 7 | 1 4 2 | 9 3 8 |
| 14 2 14 | 9 8 3 | 67 5 67 |
| 3 8 9 | 6 5 7 | 4 2 1 |
+-------------+-------------+-------------+
A={R5C1,R6C3}
B={R4C7,R4C9,R5C9}
x=7
z=3
OR
+-------------+-------------+-------------+
| 24 7 8 | 24 6 5 | 1 9 3 |
| 9 3 24 | 248 1 48 | 56 7 56 |
| 5 1 6 | 7 3 9 | 8 4 2 |
+-------------+-------------+-------------+
| 28 9 23 | 458 7 6 | 35 1 45 |
|*17 6 5 | 3 9 14 | 2 8 47 |
|*178 4 13 | 58 2 *18 | 357 6 9 |
+-------------+-------------+-------------+
| 6 5 7 | 1 4 2 | 9 3 8 |
| 14 2 14 | 9 8 3 | 67 5 67 |
| 3 8 9 | 6 5 7 | 4 2 1 |
+-------------+-------------+-------------+
A={R6C1,R6C6}
B={R5C1}
x=7
z=1
Almost locked sets xz rule
31 posts
• Page 1 of 3 • 1, 2, 3
Almost locked sets xz rule
by bennys » Sun Nov 27, 2005 7:07 am
Last edited by bennys on Wed Dec 07, 2005 2:27 am, edited 2 times in total.
- bennys
- Posts: 156
- Joined: 28 September 2005
another example
by bennys » Thu Dec 01, 2005 9:21 pm
- Code: Select all
Puzzle: Menneske.no Very Hard #3272193
+-------+-------+-------+
| . . . | . 9 8 | . . . |
| . . 7 | 5 . . | 1 . . |
| . 5 . | . 3 . | . 6 . |
+-------+-------+-------+
| 4 . . | 9 . . | . 7 2 |
| . 7 3 | . . . | 4 9 . |
| . . 6 | . . 7 | 5 . . |
+-------+-------+-------+
| 7 2 . | . 5 . | . 3 . |
| . . 1 | . . 6 | 7 . . |
| . . . | 8 7 . | . . . |
+-------+-------+-------+
+-------------------+-------------------+-------------------+
| 136 1346 24 | 12467 9 8 | 23 245 3457 |
| 3689 34689 7 | 5 246 24 | 1 248 3489 |
| 18 5 2489 | 1247 3 124 | 289 6 478 |
+-------------------+-------------------+-------------------+
| 4 *18 *58 | 9 *168 35 | 368 7 2 |
| 1258 7 3 |*26 1268 *25 | 4 9 168 |
| 1289 189 6 | 234 1248 7 | 5 18 138 |
+-------------------+-------------------+-------------------+
| 7 2 489 | 14 5 149 | 689 3 1468 |
| 3589 3489 1 | 234 24 6 | 7 2458 4589 |
| 356 346 459 | 8 7 12349 | 29 1245 145 |
+-------------------+-------------------+-------------------+
A={R4C2,R4C3,R4C5}
B=(R5C4,R5C6}
x=6
z=5
and we get r4c6<>5
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- Posts: 156
- Joined: 28 September 2005
by rubylips » Fri Dec 02, 2005 11:46 pm
That's right, though a far trickier chain is required later on.
- Code: Select all
1. Consider the chain r4c3+<5|6>+r4c5~6~r5c4+<6|3>+r4c6.
When the cell r4c6 contains the value 5, some other value must occupy the cell r4c3, which means that the value 3 must occupy the cell r4c6 - a contradiction.
Therefore, the cell r4c6 cannot contain the value 5.
- The move r4c6:=5 has been eliminated.
The value 3 is the only candidate for the cell r4c6.
2. The cell r1c7 is the only candidate for the value 3 in Column 7.
3. The cell r8c4 is the only candidate for the value 3 in Column 4.
4. The cell r4c3 is the only candidate for the value 5 in Row 4.
5. The cell r6c9 is the only candidate for the value 3 in Row 6.
6. The cell r5c6 is the only candidate for the value 5 in Row 5.
7. The values 3 and 6 occupy the cells r9c1 and r9c2 in some order.
- The moves r9c1:=5, r9c1:=9, r9c2:=4 and r9c2:=9 have been eliminated.
The cell r8c1 is the only candidate for the value 5 in Column 1.
8. Consider the chain r5c1-2-r6c1-9-r6c2~9~r8c2-9-r8c9~9~r2c9+<9|6>+r2c5=6=r5c4.
When the cell r5c4 contains the value 2, some other value must occupy the cell r5c1, which means that the value 6 must occupy the cell r5c4 - a contradiction.
Therefore, the cell r5c4 cannot contain the value 2.
- The move r5c4:=2 has been eliminated.
The value 6 is the only candidate for the cell r5c4.
- rubylips
- Posts: 149
- Joined: 01 November 2005
by Bob Hanson » Tue Dec 06, 2005 1:46 am
First, let me say that I'm really impressed by this method, bennys.
Congratulations!
I'd like to suggest an alternative reading that shows that this technique is just scratching the surface of a much wider application. If I write the above as
{18, 58, 18*}--{6*,26,25}
I'm saying that two locked sets are strongly linked via 168*.
The common 5 amounts to a "weak link" that leads to the inconsistency:
{1, 8, *}--{*,6,2}
That is, no digit available in r4c5.
Another way to look at this, though, is simply as a short strong chain weakly associated with two linked sets:
I'm trying to show a weak link there between the 58 and the {18, 18} set.
There is a "relay" aspect to this 58. Thus, setting r4c6=5, which is weakly linked to BOTH the 58 and the 25 has the following effect:
{8}...{18, 18*}--{6*,26,2}
giving
{8}...{1, 1*}--{6*,6,2}
and
{8}...{1, *}--{*,6,2}
and the logical inconsistency then as no value allowed at r4c5.
My point is simply that there might be any number of ways to strongly link two locked sets -- just like strong chains. That means we could have any wild connection here, traversing who-knows-how-many blocks:
{45}...{58}...{18, 18*}--{6*,26,25}...{57}...{74}
Setting a four that is weakly linked to these two ends does the trick.
It seems to me the possibilities include:
{.....*}--{*......}
where the two subsets intersect in a given cell but have no common digit,
{.....*}-4-{*......}
where they intersect and have one (or more) common digits,
{.....x}--{x'......}
where x and x' are conjugate pairs
{.....x}-x'y'--{y'......}
where x/x' and y/y' are conjugate pairs, but there is an intervening 2-value cell...
what else?
{....x} might be an X-wing, not a subset.
lots of potential here, I think.
ps. r4c6#5 is also eliminated by 3D Medusa as an invalid link between weakly linked strong chains. This is a little 3D Medusa path involving several cells, ultimately showing an inconsistency in row 4, eliminating 2 6s:
(not that I saw that myself!)
I like your way better!
This particular puzzle falls quickly to 3D Medusa. Can you show some of this on a puzzle that would require trial and error because all other standard avenues have been exhausted? If it's finding XYZ-wings, it's doing more than 3D Medusa can do.
Congratulations!
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+-------------------+-------------------+-------------------+
| 136 1346 24 | 12467 9 8 | 23 245 3457 |
| 3689 34689 7 | 5 246 24 | 1 248 3489 |
| 18 5 2489 | 1247 3 124 | 289 6 478 |
+-------------------+-------------------+-------------------+
| 4 *18 *58 | 9 *168 35 | 368 7 2 |
| 1258 7 3 |*26 1268 *25 | 4 9 168 |
| 1289 189 6 | 234 1248 7 | 5 18 138 |
+-------------------+-------------------+-------------------+
| 7 2 489 | 14 5 149 | 689 3 1468 |
| 3589 3489 1 | 234 24 6 | 7 2458 4589 |
| 356 346 459 | 8 7 12349 | 29 1245 145 |
+-------------------+-------------------+-------------------+
A={R4C2,R4C3,R4C5}
B=(R5C4,R5C6}
x=6
z=5
and we get r4c6<>5
I'd like to suggest an alternative reading that shows that this technique is just scratching the surface of a much wider application. If I write the above as
{18, 58, 18*}--{6*,26,25}
I'm saying that two locked sets are strongly linked via 168*.
The common 5 amounts to a "weak link" that leads to the inconsistency:
{1, 8, *}--{*,6,2}
That is, no digit available in r4c5.
Another way to look at this, though, is simply as a short strong chain weakly associated with two linked sets:
- Code: Select all
{58}...{18, 18*}--{6*,26,25}
I'm trying to show a weak link there between the 58 and the {18, 18} set.
There is a "relay" aspect to this 58. Thus, setting r4c6=5, which is weakly linked to BOTH the 58 and the 25 has the following effect:
{8}...{18, 18*}--{6*,26,2}
giving
{8}...{1, 1*}--{6*,6,2}
and
{8}...{1, *}--{*,6,2}
and the logical inconsistency then as no value allowed at r4c5.
My point is simply that there might be any number of ways to strongly link two locked sets -- just like strong chains. That means we could have any wild connection here, traversing who-knows-how-many blocks:
{45}...{58}...{18, 18*}--{6*,26,25}...{57}...{74}
Setting a four that is weakly linked to these two ends does the trick.
It seems to me the possibilities include:
{.....*}--{*......}
where the two subsets intersect in a given cell but have no common digit,
{.....*}-4-{*......}
where they intersect and have one (or more) common digits,
{.....x}--{x'......}
where x and x' are conjugate pairs
{.....x}-x'y'--{y'......}
where x/x' and y/y' are conjugate pairs, but there is an intervening 2-value cell...
what else?
{....x} might be an X-wing, not a subset.
lots of potential here, I think.
ps. r4c6#5 is also eliminated by 3D Medusa as an invalid link between weakly linked strong chains. This is a little 3D Medusa path involving several cells, ultimately showing an inconsistency in row 4, eliminating 2 6s:
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||---c4--|---c5--|---c6--||---c7--|
--------------------------------------
r1 || 12467 | 9 | 8 || 23 |
|| c | | || |
---||-------+-------+-------||-------+
r2 || 5 | 246 | 24 || 1 |
|| | D | || |
---||-------+-------+-------||-------|
r3 || 1247 | 3 | 124 || 289 |
|| | | || |
===||=======================||========
r4 || 9 | 168 | 35 || 368 |
|| | e | aA || Bb |
---||-------+-------+-------||-------+
r5 || 26 | 1268 | 25 || 4 |
|| bC | | Ba || |
---||-------+-------+-------||-------+
(not that I saw that myself!)
I like your way better!
This particular puzzle falls quickly to 3D Medusa. Can you show some of this on a puzzle that would require trial and error because all other standard avenues have been exhausted? If it's finding XYZ-wings, it's doing more than 3D Medusa can do.
- Bob Hanson
- Posts: 75
- Joined: 04 December 2005
by Jeff » Tue Dec 06, 2005 3:46 am
Bob Hanson wrote:This particular puzzle falls quickly to 3D Medusa. Can you show some of this on a puzzle that would require trial and error because all other standard avenues have been exhausted? If it's finding XYZ-wings, it's doing more than 3D Medusa can do.
Hi, Bob. Firstly, welcome to this forum.
Correct me if I am wrong, 3D medusa is similar to a technique called advanced colouring table in this forum. If that is the case, it identifies double implication chains only. xyz-wings are triple implication chains. Bennys' xz rule can be of poly-implications. It can surely make deductions which are beyond 3D medusa. The extension of the xz-rule is full of potential indeed.
- Jeff
- Posts: 708
- Joined: 01 August 2005
by Bob Hanson » Tue Dec 06, 2005 4:50 am
Thank you. Yes, the Medusa idea is pretty much just advanced coloring, as I understand it. But there's no need for "coloring" and derives its name from the 3D representation I introduced at http://www.stolaf.edu/people/hansonr/sudoku.
And, yes, it only does straight-line implications, not this interesting subset business. All it is really is the synthesis of "X-cycles" and "Y-cycles" into one whole piece. I describe it at http://www.stolaf.edu/people/hansonr/sudoku/explain.htm and http://www.stolaf.edu/people/hansonr/sudoku/top95-analysis.htm .
The way I look at it, subset analysis and grid analysis pushes beyond the current "trial and error" level, providing some depth. Simple trial and error just uses singles; even simple subset analysis actually does more than that.
So when I saw this combination of n-tuples, it really intrigued me. Most of it is handled just fine by 3D Medusa, but not xyz-wings, for the reason you mention.
In general, subset analysis isn't necessary -- I've only found one puzzle (among those that CAN be solved with 3D Medusa) that couldn't be solved if I turned off subset analysis. So what that indicates is that it is a rare subset that contains no chains and no alternative solving mechanism.
But I am definitely intrigued by this associating of subsets. Some interesting new rules in there for sure.
And, yes, it only does straight-line implications, not this interesting subset business. All it is really is the synthesis of "X-cycles" and "Y-cycles" into one whole piece. I describe it at http://www.stolaf.edu/people/hansonr/sudoku/explain.htm and http://www.stolaf.edu/people/hansonr/sudoku/top95-analysis.htm .
The way I look at it, subset analysis and grid analysis pushes beyond the current "trial and error" level, providing some depth. Simple trial and error just uses singles; even simple subset analysis actually does more than that.
So when I saw this combination of n-tuples, it really intrigued me. Most of it is handled just fine by 3D Medusa, but not xyz-wings, for the reason you mention.
In general, subset analysis isn't necessary -- I've only found one puzzle (among those that CAN be solved with 3D Medusa) that couldn't be solved if I turned off subset analysis. So what that indicates is that it is a rare subset that contains no chains and no alternative solving mechanism.
But I am definitely intrigued by this associating of subsets. Some interesting new rules in there for sure.
- Bob Hanson
- Posts: 75
- Joined: 04 December 2005
Re: another example
by ronk » Tue Dec 06, 2005 5:21 am
bennys wrote:Puzzle: Menneske.no Very Hard #3272193
- Code: Select all
+-------------------+-------------------+-------------------+
| 136 1346 24 | 12467 9 8 | 23 245 3457 |
| 3689 34689 7 | 5 246 24 | 1 248 3489 |
| 18 5 2489 | 1247 3 124 | 289 6 478 |
+-------------------+-------------------+-------------------+
| 4 *18 *58 | 9 *168 &35 | 368 7 2 |
| 1258 7 3 |*26 1268 *25 | 4 9 168 |
| 1289 189 6 | 234 1248 7 | 5 18 138 |
+-------------------+-------------------+-------------------+
| 7 2 489 | 14 5 149 | 689 3 1468 |
| 3589 3489 1 | 234 24 6 | 7 2458 4589 |
| 356 346 459 | 8 7 12349 | 29 1245 145 |
+-------------------+-------------------+-------------------+
A={R4C2,R4C3,R4C5}
B=(R5C4,R5C6}
x=6
z=5
and we get r4c6<>5
It's interesting that when the elimination cell r4c6 (tagged &) is added to each of A and B, both new sets are also "almost locked sets" ...
C={r4c2,r4c3,r4c5,r4c6} = {13568}
D ={r4c6,r5c4,r5c6} = ( {2356}
It's probably just coincidence, but it's also interesting that the 3 occurs once only ... in the intersection of C and D.
Last edited by ronk on Tue Dec 06, 2005 1:29 am, edited 2 times in total.
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- Joined: 02 November 2005
- Location: Southeastern USA
by Jeff » Tue Dec 06, 2005 5:25 am
Bob Hanson wrote:In general, subset analysis isn't necessary -- I've only found one puzzle (among those that CAN be solved with 3D Medusa) that couldn't be solved if I turned off subset analysis. So what that indicates is that it is a rare subset that contains no chains and no alternative solving mechanism.
In terms of manual solving, which is what I am interested in, subset analyses such as the xz rule, xyz-wings, empty cell contradiction and implied locked set should be executed before 3D medusa or bilocation/bivalue plot, simply because they are much much more user friendly. To this effect, the statement could well be "In general, 3D medusa and bilocation/bivalue plot aren't necessary if subset analyses are to be conducted on every grid in the first place".
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- Posts: 708
- Joined: 01 August 2005
by bennys » Tue Dec 06, 2005 6:34 am
Bob
I am not sure I understand exactly what are you looking for but the way that I see it almost locked sets behave like "directional cells" so for example
I am not sure I understand exactly what are you looking for but the way that I see it almost locked sets behave like "directional cells" so for example
- Code: Select all
+----------------------+----------------------+----------------------+
| . . . | . . . | . . . |
| . . . | 127 . . | . . . |
| 593 593 5936 | 527 125 . | . . . |
+----------------------+----------------------+----------------------+
| . . 67 | x . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+----------------------+----------------------+----------------------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+----------------------+----------------------+----------------------+
x<>7
xy wing for almost locked sets
+----------------------+----------------------+----------------------+
| . . . | . . . | . . . |
| . . . | 127 . . | 396 . . |
| . . . | 527 125 . | 3596 965 . |
+----------------------+----------------------+----------------------+
| . . . | . . . | . . . |
| . . . | 784 . . | . . . |
| . . . | 3784 384 . | x . . |
+----------------------+----------------------+----------------------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+----------------------+----------------------+----------------------+
x<>3
Last edited by bennys on Tue Dec 06, 2005 3:21 am, edited 1 time in total.
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- Posts: 156
- Joined: 28 September 2005
by Myth Jellies » Tue Dec 06, 2005 10:37 am
I'm not sure if it was made clear, but it looks like the z-value has to be in both A and B.
In addition, there are some interesting results if you combine an almost locked sets xz-pattern with either the Pattern Overlay Method (POM) or with partial-POM. The Almost Locked Sets XZ-Rule tells you that any solution pattern (or valid rookery) for z has to be part of either A or B (or both). Thus any solution pattern that is not part of either A or B can be eliminated. The POM part of things lays out all of your solution patterns so that partial eliminations can be made, even if there was no handy candidate spanning all of the z-values in A and B.
Coupling the xz-pattern with a partial-POM analysis of rows 7-9 of
A={R9C2,R9C3,R9C7,R9C9}; B={R8C7}; x=2; z=4;
Patterns for (z=4) in A and B include 4acdef. We can eliminate any z-pattern that is not in either A or B. Thus we can eliminate pattern 4b.
Coupling the xz-pattern with a partial-POM analysis of rows 4-6 of
A={R4C2,R4C3,R4C5}; B=(R5C4,R5C6}; x=6; z=5;
Patterns for (z=5) in A and B include only 5a. (Thus 5a is the solution for candidate 5's.) We can eliminate any z-pattern that is not in either A or B.
Thus we can eliminate pattern 5b.
But if we didn't see that, we could also do
A={R4C2,R4C5,R4C7}; B={R6C8,R6C9}; x=3; z=8;
A and B includes 8abcdeghijklm; thus we can eliminate 8f.
And
A={R4C2,R4C5,R4C7}; B={R6C8,R5C9}; x=6; z=8;
to similarly eliminate 8e.
Theoretically, the equations from each partial-POM cell should lead to these results, but this is a really quick and cool shortcut.
In addition, there are some interesting results if you combine an almost locked sets xz-pattern with either the Pattern Overlay Method (POM) or with partial-POM. The Almost Locked Sets XZ-Rule tells you that any solution pattern (or valid rookery) for z has to be part of either A or B (or both). Thus any solution pattern that is not part of either A or B can be eliminated. The POM part of things lays out all of your solution patterns so that partial eliminations can be made, even if there was no handy candidate spanning all of the z-values in A and B.
Coupling the xz-pattern with a partial-POM analysis of rows 7-9 of
- Code: Select all
+----------------+----------------+----------------+
| 579 5789 12 | 3 4 12 | 589 59 6 |
| 459 6 249 | 7 8 29 | 1 3459 3459 |
| 34 89 1349 | 19 6 5 | 489 7 2 |
+----------------+----------------+----------------+
| 79 4 8 | 5 2 79 | 3 6 1 |
| 6 1 5 | 89 3 489 | 7 2 49 |
| 2 3 79 | 6 1 479 | 459 8 459 |
+----------------+----------------+----------------+
| 1 2 39 | 4 7 38 | 6 359 3589 |
| 3457 57 6 | 128 9 138 |*24 134 3478 |
| 8 *79 *3479 | 12 5 6 |*249 1349*3479 |
+----------------+----------------+----------------+
gives
+-------------------+-------------------+-------------------+
| 1 2 3ab | 4 7 3cdef | 6 3g 3h |
| 9abc | 8a | 5ab 5cd |
| | | 9de 8bc |
| | | 9fg |
| | | |
| 3cd 5bd 6 | 1a 9 1b |*2b 1c 3f |
| 4abc 7b | 2a 3abgh |*4d 3e 4f |
| 5ac | 8b 8c | 4e 7cd |
| 7a | | 8a |
| | | |
| 8 *7c *3efgh | 1 5 6 |*2a 1ab *3bd |
| *9df *4def | 2 |*4a 3ac *4c |
| *7d | |*9a 4b *7ab |
| *9eg | | 9b *9c |
+-------------------+-------------------+-------------------+
A={R9C2,R9C3,R9C7,R9C9}; B={R8C7}; x=2; z=4;
Patterns for (z=4) in A and B include 4acdef. We can eliminate any z-pattern that is not in either A or B. Thus we can eliminate pattern 4b.
Coupling the xz-pattern with a partial-POM analysis of rows 4-6 of
- Code: Select all
+-------------------+-------------------+-------------------+
| 136 1346 24 | 12467 9 8 | 23 245 3457 |
| 3689 34689 7 | 5 246 24 | 1 248 3489 |
| 18 5 2489 | 1247 3 124 | 289 6 478 |
+-------------------+-------------------+-------------------+
| 4 *18 *58 | 9 *168 35 | 368 7 2 |
| 1258 7 3 |*26 1268 *25 | 4 9 168 |
| 1289 189 6 | 234 1248 7 | 5 18 138 |
+-------------------+-------------------+-------------------+
| 7 2 489 | 14 5 149 | 689 3 1468 |
| 3589 3489 1 | 234 24 6 | 7 2458 4589 |
| 356 346 459 | 8 7 12349 | 29 1245 145 |
+-------------------+-------------------+-------------------+
gives
+-------------------+-------------------+-------------------+
| 4 #*1abc *5a | 9 #*1defg 3a |#3b 7 2 |
| #*8abc *8def | #*6a 5b |#6bc |
| | #*8ghij |#8klm |
| | | |
| 1de 7 3 |*2c 1ab *2e | 4 9 1cfg |
| 2ab |*6b 2d *5a | 6a |
| 5b | 6c | 8cfij |
| 8ghk | 8abdelm | |
| | | |
| 1f 1g 6 | 2a 1c 7 | 5 #1ad #1be |
| 2cde 8jm | 3b 2b | #8adg #3a |
| 8il 9b | 4a 4b | #8beh |
| 9a | 8cfk | |
+-------------------+-------------------+-------------------+
A={R4C2,R4C3,R4C5}; B=(R5C4,R5C6}; x=6; z=5;
Patterns for (z=5) in A and B include only 5a. (Thus 5a is the solution for candidate 5's.) We can eliminate any z-pattern that is not in either A or B.
Thus we can eliminate pattern 5b.
But if we didn't see that, we could also do
A={R4C2,R4C5,R4C7}; B={R6C8,R6C9}; x=3; z=8;
A and B includes 8abcdeghijklm; thus we can eliminate 8f.
And
A={R4C2,R4C5,R4C7}; B={R6C8,R5C9}; x=6; z=8;
to similarly eliminate 8e.
Theoretically, the equations from each partial-POM cell should lead to these results, but this is a really quick and cool shortcut.
- Myth Jellies
- Posts: 593
- Joined: 19 September 2005
by ronk » Tue Dec 06, 2005 1:20 pm
Here is my stab at a generalized description of bennys' Almost Locked Set xz-rule
Now the first paragraph above is a very long set of requirements, so it's easy to jump to the conclusion that this "pattern" will rarely occur ... and that may indeed be the case. But there is a lot of "depth" available in this technique from ...
P.S. This probably looks pretty sad to a set theorist, but I've never taken a set theory course and never attempted a "set description" like this before, so please cut me some slack folks and be gentle.
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Given value sets G = {uvwxz} and H = {qrsxz} such that ...
the (m + 1) values of G are assignable to m cells of cell set A,
the (n + 1) values of H are assignable to n cells of cell set B,
cell sets A and B are disjoint (non-overlapping),
cell set A exists entirely in unit (house, group, domain) J,
cell set B exists entirely in unit (house, group, domain) K,
units J and K intersect,
member {x} of G is assignable to only one cell of A,
member {x} of H is assignable to only one cell of B,
members {x} of G and {x} of H are assignable only in unit J.
Since {x} may exist only once in unit J ...
either {x} of G will be eliminated,
or {x} of H will be eliminated,
or both {x} of G and {x} of H will be eliminated.
If {x} of G is eliminated, G is no longer almost-locked but locked with G = {uvwz}.
If {x} of H is eliminated, H is no longer almost-locked but locked with H = {qrsz}.
At least one locked set is then a naked quad (or naked pair or naked triple) containing {z}. Therefore, for cells in the intersection of units J and K but disjoint from A and B, {z} may be eliminated.
Now the first paragraph above is a very long set of requirements, so it's easy to jump to the conclusion that this "pattern" will rarely occur ... and that may indeed be the case. But there is a lot of "depth" available in this technique from ...
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1) any one or more of {uv} in G and/or {qr} in H need not exist,
2) if any member of {uv} is the same as any member of {qr}, then that value is effectively another {z} and may also be eliminated from the same intersection of J and K.
P.S. This probably looks pretty sad to a set theorist, but I've never taken a set theory course and never attempted a "set description" like this before, so please cut me some slack folks and be gentle.
Last edited by ronk on Tue Dec 06, 2005 9:49 am, edited 3 times in total.
- ronk
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