The New Sudoku Players' Forum

by **Myth Jellies** » Wed Jan 31, 2007 4:18 am

ronk wrote:Would those likely have been examples of the following pattern?...

I did not know how to complete the pattern, so I don't really know.

In some cases, though, I had a color molecule segment something like this...

Code: Select all: ...- A = a - J = j - b = B - ... | | +---- c ----+ || C | . . .

In this case, the colors a, b, and c form a potential hub. One just needs to find a rim to connect it to

by **ronk** » Wed Jan 31, 2007 4:47 am

Myth Jellies wrote:In some cases, though, I had a color molecule segment something like this...

Code: Select all
...- A = a - J = j - b = B - ... | | +---- c ----+ || C | . . .

In this case, the colors a, b, and c form a potential hub/

That link between 'a' and 'b' is an added wrinkle. I guess that would make it a "distributed hub"

and definitely not the diagram I cited.

BTW my prior post now illustrates varations that weren't coded, meaning my test was likely too narrow in scope. Perhaps someone else will try and have better results.

by **Steve K** » Wed Jan 31, 2007 9:11 am

Code: Select all: Puzzle start *-----------* |8..|.4.|...| |.95|..3|...| |...|2..|..6| |---+---+---| |6..|...|2..| |.2.|7.1|.4.| |..3|...|..9| |---+---+---| |7..|..9|...| |...|5..|68.| |...|.2.|..1| *-----------* Puzzle point before pattern exists. A non-trivial enterprise just to get here! *-----------* |8..|.4.|...| |.95|6.3|...| |...|2.8|..6| |---+---+---| |6..|...|2..| |.2.|761|.4.| |..3|..2|.69| |---+---+---| |7..|..9|...| |...|5..|68.| |...|.26|..1| *-----------*

This is a fairly difficult puzzle. After the simple sudoku solving set is appled to the last puzzle position above, one must make the following eliminations to get to the point where I found the pattern that I wish to discuss:

1) r7c4=1 == r1c4=1 -- r1c4=9 == r3c5=9 -- r3c4=5 == r6c5=5 -- r6c5=8 == r7c5=8 forbids r7c5=1 and forbids r7c4=8
2) Y wing style elimination:
r9c3=4 == r9c3=8 -- r9c4=8 == r6c4=8 -- r6c4=4 == r4c6=4 forbids r4c3=4 (called Y wing style as it has a vertex and two endpoints, but in this case one endpoint and one vertex are strong in location, not cells)

The puzzle is now advanced to:

Code: Select all: Puzzle possibility matrix where pattern occurs. *-----------------------------------------------------------------------------* | 8 1367 1267 | 19 4 57 | 13579 123579 2357 | | 124 9 5 | 6 17 3 | 1478 127 2478 | | 134 1347 147 | 2 1579 8 | 134579 13579 6 | |-------------------------+-------------------------+-------------------------| | 6 14578* 178 | 39 39 45& | 2 157 578 | | 59 2 89 | 7 6 1 | 358 4 358 | | 145 1457# 3 | 48 58& 2 | 157 6 9 | |-------------------------+-------------------------+-------------------------| | 7 13468# 12468 | 134 38& 9 | 345 235 2345 | | 12349 134 1249 | 5 137 47 | 6 8 234 | | 345 3458* 48 | 348 2 6 | 79 79 1 | *-----------------------------------------------------------------------------* * = multi hub. # = Spokes & = AIC chain pieces that act like an ALS.

Perhaps this is almost off topic. I have found patterns similar to the hub and spoke idea in puzzles, but often they involve mulitple hubs. In this example, I see an almost hidden pair 58 at r49c2. But for the 5 at r6c2 and the 8 at r7c2, it would be a hidden pair. If there is no hidden pair at r49c2, then r6c5=8, thus r4c6=5, thus r4c2=4. This allows one to eliminate 17 from r4c2 and 4 from r9c2. A AIC chain representation of this step:

{Hidden pair 58 at r49c2} == {r6c2=5 == r7c2=8 -- r7c5=8 == r6c5=8} -- r6c5=5 == r4c6=5 -- r4c6=4 == r4c2=4.
Forbids r4c2 from being anything but 458.
forbids r9c2 from being 4.

These eliminations consider the following strong sets:

1) 5's in column 2: {r9c2, r6c2, r4c2} = 5
2) 8's in column 2: {r9c2, r7c2, r4c2} = 8
3) 8's in column 5: {r7c5, r6c5} = 8
4) 5's in box 5: {r6c5, r4c6} = 5
5) 4's in row 4: {r4c6, r4c2} = 4

Logic:
a) Strong sets 1) and 2) form an Almost Almost Hidden pair 58.
b) If hidden pair 58 at r49c2, then r49c2 is nothing but candidates 58.
c) If not hidden pair 58 at r49c2, then at least one of the following is true:
{r7c2=8, r6c2=5}
d) Clearly, at this point, we can combine b,c to write:
{hidden pair 58 at r49c2} =={r7c2=8 == r6c2=5}
e) But, because of strong set 3): no matter what, if the second part of d be true, we can have no more 5 at r6c5. In fact, r6c5 would be 8, but this is really not material to the chain, rather the fact tht r6c5 is not 5 is material.
f) Because of strong set 4), e) => r4c6=5 =>r4c6<>4.
g)Because of strong set 5), f) => r4c2=4.
i)Conclusion: b) == g): {Hidden pair 58 at r49c2} == {r4c2=4} forbids:
r4c2=17, r9c2=4.
j)Intermediate conclusion: {r4c6=5} == {r4c6=5}, but this forbids nothing as we already have that. This intermediate conclusion is perhaps the most difficult to understand. A whole bit of theory about partial loop closure would have to be introduced as to why this weak link in the chain must be strong, but not some of the other ones. Careful study of the bracketing that I use in the forbidding chain representation of the step shows that although the weak link:
{r6c2=5 == .... == r6c5=8} -- r6c5=5
is indeed strong, it does not directly forbid anything, (ever, IMO).

How does this fit the pattern?
1) The Almost Hidden Pair is the hub(or double hub).
2) The neccessary and sufficient conditions for not having the hidden pair are the spokes(58 in c2 outside of the hub)
3) A short conditional AIC chain shows if not hidden pair 58, then r4c2=4 - this acts like an ALS.

Alternate view:
1) cell r4c6 is the hub, with the weak link between 4 & 5.
2) strong 4's and 5's from that hub are the spokes.
3) Almost Almost Locked set Hidden pair 58 is the rim.
4) A short AIC chain connects the dots, reducing the Almost Almost Locked Hidden pair in fact to an Almost Hidden Pair, conditionally.

The alternate view is best seen reading the chain backwards. Naturally, a properly developed chain has no direction, so reading this chain backwards should make the same sense as forwards.

Code: Select all: Alternate viewpoint of the hub *-----------------------------------------------------------------------------* | 8 1367 1267 | 19 4 57 | 13579 123579 2357 | | 124 9 5 | 6 17 3 | 1478 127 2478 | | 134 1347 147 | 2 1579 8 | 134579 13579 6 | |-------------------------+-------------------------+-------------------------| | 6 14578S* 178 | 39 39 45 HUB | 2 157 578 | | 59 2 89 | 7 6 1 | 358 4 358 | | 145 1457* 3 | 48 58S^ 2 | 157 6 9 | |-------------------------+-------------------------+-------------------------| | 7 13468* 12468 | 134 38^ 9 | 345 235 2345 | | 12349 134 1249 | 5 137 47 | 6 8 234 | | 345 3458* 48 | 348 2 6 | 79 79 1 | *-----------------------------------------------------------------------------* HUB = hub. S = Spokes endpoints * = Almost Almost Locked Set (HIDDEN) ^ = Spoke extension

To understand this alternate viewpoint, the tempation would be to use the existing strong link at the HUB. Suppose, for a moment that I do not do that:
a) If r4c2=4 then clearly (r4c2<>17 and r9c2<>4).
b) If r4c2<>4, then r4c6=4 => r6c5=5 => r7c5 = 8.
c} (r6c5=5 and r7c5=8) => Hidden pair 58 at r49c2, which also forbids:
r4c2=17 and r9c2=4.

Just thought I might introduce some further complications into this idea.

by **Steve K** » Thu Feb 01, 2007 5:26 pm

The tough puzzle for February 2,2007 (they are a few hours ahead) at sudoku.com.au can be solved in many ways. Amongst them, and of equivalent complexity to the others I found, is a modified hub and spoke pattern. By modifed, I mean it again forms only a semiclosed loop, as the previous example above.

Code: Select all: Puzzle at start. Hidden pair 46 at r3c37, or something equivalent, is required for this demonstration. *-----------* |...|..6|.2.| |8..|..4|...| |.2.|...|.17| |---+---+---| |4..|1..|.35| |...|...|...| |67.|..2|..8| |---+---+---| |19.|...|.4.| |...|5..|..2| |.6.|4..|...| *-----------* Puzzle before Simple Sudoku solving set, plus a finned X wing on 6's. *-----------* |...|..6|82.| |8..|..4|...| |.2.|...|.17| |---+---+---| |48.|1..|.35| |...|...|...| |67.|3.2|.98| |---+---+---| |19.|...|.4.| |...|5..|..2| |.6.|4..|...| *-----------* BTW, none of the simple sudoku steps, except the hp noted above, are actually required. Single hub, two spokes, an ALS, and a conditional coloring AIC on 6's: *-----------------------------------------------------------------------------* | 3579 1345 134579 | 79 1379 6 | 8 2 349S1 | | 8 13 13679 | 279 12379 4 | 3569HUB 56S2 369S1 | | 39 2 46 | 89 3589 3589 | 46 1 7 | |-------------------------+-------------------------+-------------------------| | 4 8 29 | 1 679 79 | 267 3 5 | | 2359 135 12359 | 6789* 456789 5789 | 12467 67* 146 | | 6 7 15 | 3 45 2 | 14 9 8 | |-------------------------+-------------------------+-------------------------| | 1 9 23578 | 2678* 23678 378 | 357 4 36ALS | | 37 34 3478 | 5 36789 13789 | 13679 678* 2 | | 2357 6 23578 | 4 23789 13789 | 13579 578 139 | *-----------------------------------------------------------------------------* Key: HUB = hub S1 = strong spoke 1, - the 3's in box 3. S2 = strong spoke 2 - the 5's in box 3, or row 2, which ever you prefer. * = Conditonal coloring on 6's. ALS = Almost Locked Set

The hub and spoke pattern, modified with partially closed loop, as a AIC, or forbidding chain:
r12c9=3 == r2c7=3 -- r2c7=5 == r2c8=5 -- r2c8=6 =={coloring on 6's. as AIC on 6's:r8c8 == r5c8 -- r5c4 == r7c4} -- r7c9=6 == r7c9=3
forbids: r9c9=3 and forbids r2c7=69.

Of the weak links proven strong, only the chain endpoints, and those at the hub, forbid anything.

After this point, locked canidates eliminate 9 from r9c9, two more cells solve, and I had to resort to another short chain to solve: (there are lots of alternate routes)

r5c7=1 == r6c7=1 -- r6c3=1 == r6c3=5 -- r79c3=5 == r9c1=5 -- r9c1=2 == r5c1=2 forbidding r5c7=2 and unlocking the puzzle with singletons in some container to the end.

by **Steve K** » Sat Feb 03, 2007 2:52 pm

Since Myth has introduced this idea, I have mused about it a bit. It is interesting to note that although the pattern seems rare in the form as Myth has presented it, the pattern is not rare, IMHO, if one expands the concept to include:
A hub whose spokes interact, fish-like, with the puzzle.
A Hub and Axle, whose spokes interact with an ALS.

My search for this pattern has been by hand, so the sampling is tiny. Nevertheless, I have found an instance of this pattern in about 1/3 of the truly tough puzzles that I have tried to find it on.

The following is an example of a Hub and Axle pattern. One might see it in this case as a Hidden Sue de Cox crossed with a Naked Sue de Cox. Moreover, there is a standard depth 5 AIC that accomplishes the same thing. Neverthess, since it is very common for techniques to overlap, and it is very common for the same eliminations to be justified by alternate means, I do not view this as a negative.

Code: Select all: Puzzle at start *-----------* |8..|...|...| |7..|3.5|6..| |..4|.9.|...| |---+---+---| |..1|..9|.2.| |.5.|.4.|.3.| |.3.|7..|8..| |---+---+---| |...|.2.|5..| |..3|4.1|..7| |...|...|..2| *-----------* Puzzle after SS solving set *-----------* |8.5|..4|...| |7..|3.5|6..| |3.4|.9.|...| |---+---+---| |..1|.39|.2.| |.5.|.4.|.3.| |.3.|7..|8..| |---+---+---| |...|.2.|5..| |..3|4.1|9.7| |...|...|..2| *-----------* Puzzle where pattern exists Hub and Axel with Spokes. *--------------------------------------------------------------------* | 8 169 5 | 126 167 4 | 123 179 139 | | 7 129 29 | 3 18 5 | 6 1489 1489 | | 3 16 4 | 1268 9 2678 | 12 1578 158 | |----------------------+----------------------+----------------------| | 46 78 1 | 568 3 9 | 47 2 456 | | 269H 5 78 | 1268sS 4 268sS | 17 3 169H | | 2469 3 269 | 7 156 26ALS | 8 14569 14569 | |----------------------+----------------------+----------------------| | 169 4789 6789 | 689 2 3678 | 5 1468 13468 | | 256 28 3 | 4 568 1 | 9 68 7 | | 1569 4789 6789 | 5689 5678 3678 | 134 1468 2 | *--------------------------------------------------------------------* Key: r5c19 forms the Hub, considering the 2's, 6's, 9's. The strong set: {r5c19=9} is the Axle for the Hub. r5c46=2, r5c46=6 are the spokes. r6c6=26 is the ALS.

The idea here is prove a Hidden pair at the Hub no matter what. In this case, one can prove the Hub to be either Hidden pair 29 or Hidden Pair 69. The spokes are nicely on top of each other in box 5, as noted. There is no requirement, though that these spokes share the same cells. Rather, it is only required that the spokes both "look" into box 5 along r5. The ALS 26 completes the loop, as it prevents both spokes from looking into box 5 at the same time. If they did, then clearly cell r6c6 would be empty.

As AIC (forbidding chain):
Hidden pair 69|r5c19 = 6|r5c46 - (6=2)|r6c6 - 2|r5c46 = Hidden pair 29|r5c19
=> r5c9<>1 & r6c5,r4c4<>6.

The Hidden pair 29 at r5c19 does not have r5c9=2 as a possibility. This, as with naked ALS, makes no difference. A hidden pair can be defined as limiting two candidates to two locations within a large container. For example, even if the two canidates are each limited, in fact, to just one location each, it is still a hidden pair.

(note: modified some terms used to clarify pattern interpretation. Changed, to the best of my ability, language form to forum conventions.)

by **Mike Barker** » Sat Feb 03, 2007 3:44 pm

Once the problem is turned around so that the rim is considered to be an A*LS, then the problem can be reformulated as a Kraken A*LS. The idea is that if M+1 digits link to a candidate elimination cell with the same digit then that digit can be eliminated from the CEC. Another way of describing a Kraken A*LS is if a candidate in a cell links to M+1 of the digits of the A*LS either directly (they share a unit), via strong links (as is shown here), grouped strong links, bivalues, ALS, or pick your favorite link, then it can be eliminated since placing the candidate will result in N-1 candidates in N cells. In the triple spoke pattern:

Code: Select all: +----------------+-------------------+----------------+ | . 1789 . | . 2789 3789 | 789 . . | | . . . | . . . | . . . | | . . . | . . . | . . . | +----------------+-------------------+----------------+ | . 1A* . | 1a2b3c* . . | . . . | | . . . | . 2B* . | . . . | | . . . | . . 3C* | . . . | +----------------+-------------------+----------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | +----------------+-------------------+----------------+

consider a "1" in column 2 except for rows 1 and 4. This candidate directly links to the "1" in r1c2, to the "2" via r1c5-2-r5c5=2=r4c4=1=r4c2-1-r_c2, and to the "3" via r1c6-3-r6c6=3=r4c4=1=r4c2-1-r_c2. Thus {123}=M+1 digits of the A2LS links to "1" in column 2 and thus all "1"s except for rows 1 and 4 can be eliminated.

This expands the idea of a wheel to include ALS in the links, but probably doesn't help finding the beasties. Also from the wheel perspective, its probably easier to see more of the eliminations with a wheel than with a Kraken A*LS so even though the idea can be reformulated as a Kraken A*LS doesn't mean it should be.

Here's a real life example from the Sudoku Unification Model thread which was used to help solve Emily's Friends Puzzle. Here there are four spokes or tentacles, the CEC is r6c4, and the M+1 digits are {289}. This was the last thing that was tried in order to solve the puzzle so it is complicated. Clearly there can be much simpler examples as well.

Code: Select all: 3-celled/2-link Kraken AALS (r4c2=3=r5c1=6=r6c123-6-, r4c8-2-r5c478-6-, r4c4-8-r1235c4-6-, r4c8-9-r7c8=9=r7c5-9-r6c123569-6-): r4c248=289 => r6c4<>6 +--------------------------+-----------------------+--------------------+ | 56 26 246 | 1458$ 7 9 | 3 1246 128 | | 1 9 2467 | 458$ 3 48 | 24578 2467 248 | | 357 37 8 | 145$ 2 6 | 457 147 9 | +--------------------------+-----------------------+--------------------+ | 4 2378*@ 1 | 78* 5 2378 | 89 239* 6 | | 368@#$ 5 9 | 468#$ 1468 1238 | 248# 24# 7 | | 678@#$% 2678@#$% 267@#$% | 4789-6 4689% 23478% | 1 5 348% | +--------------------------+-----------------------+--------------------+ | 2 178 5 | 3 489% 1478 | 6 479% 14 | | 6789 1678 367 | 2 14689 1478 | 479 13479 5 | | 679 4 367 | 679 169 5 | 279 8 123 | +--------------------------+-----------------------+--------------------+

by **ronk** » Sat Feb 03, 2007 4:47 pm

Mike Barker wrote:Once the problem is turned around so that the rim is considered to be an A*LS, then the problem can be reformulated as a Kraken A*LS. The idea is that if M+1 digits link to a candidate elimination cell with the same digit then that digit can be eliminated from the CEC. Another way of describing a Kraken A*LS is if a candidate in a cell links to M+1 of the digits of the A*LS either directly (they share a unit), via strong links (as is shown here), grouped strong links, bivalues, ALS, or pick your favorite link, then it can be eliminated since placing the candidate will result in N-1 candidates in N cells.

The relationship between M and N is not entirely clear. Is it N+M candidates for the N cells of the A*LS?

Also A*LS mathematically implies (A^M)LS or (A**M)LS, so how about we use -- for specific sizes in specific situations, ALS, A2LS, A3LS, etc -- in other words, AmLS? An omitted value of 'm' would be understood to mean M=1.

It's probably going to take me a while to decipher your "Kraken A*LS."

by **Mike Barker** » Sat Feb 03, 2007 5:14 pm

Yes there are N+M digits in N cells for an AmLS. A*LS was popular back when I wrote the Kraken A*LS post to imply ALS, AALS, AAALS or as you say A1LS=ALS, A2LS, A3LS = A^mLS. Note that Death Blossom or Kraken Blossom is a special form of this where N=1 and thus all candidates must be linked to the candidate elimination cell since N+M=M+1.

by **ronk** » Sat Feb 03, 2007 8:57 pm

[edit: "fuzzy logic" post deleted]

by **Mike Barker** » Sun Feb 04, 2007 12:29 am

Overkill's chained APE is a good example. r3c2 is the stem with 3 candidates (an AALS or A2LS). The elimination at r2c2 requires all three links in order to achieve a contradiction (in this case no valid placements in r3c2) and thus the elimination. Each link/chain is independent so I don't believe the weak inferences can add anything

. For a multicell A*LS things are basically the same with a requirement for M+1 digits linked in the A*LS (this could actually entail more than M+1 links/chains). This leaves N-1 digits for N cells as the contradiction condition. For example, an ALS requires two digits to be linked. In the xz-rule these are the restricted common and the eliminated candidate. In the xz-mer rule they are the two restriced commons. Of course I'd love to be wrong - I desperately would like to see some new simpler method to solve these harder puzzles.

Code: Select all: r3c2-3-r3c7-4-r3c5-9-r2c6-4-r2c1-1- or with ALS r3c2-3-ALS:r3c57-9-ALS:r2c16-1- r3c2-4-r2c1-1- r3c2-9-r1c2-4-r2c1-1- or with ALS r3c2-9-ALS:r2c1|r1c2-1- +----------------+----------------+----------------+ | 7 {49} 6 | 5 1 3 | 48 489 2 | | {14}[1349] 2 | 8 6 {49} | 7 5 349 | | 5 [349] 8 | 7 {49} 2 | {34} 1 6 | +----------------+----------------+----------------+ | 3 6 7 | 9 2 5 | 148 48 14 | | 9 2 5 | 4 8 1 | 36 367 37 | | 48 48 1 | 6 3 7 | 9 2 5 | +----------------+----------------+----------------+ | 6 7 39 | 1 5 49 | 2 34 8 | | 2 5 4 | 3 7 8 | 16 69 19 | | 18 18 39 | 2 49 6 | 5 347 347 | +----------------+----------------+----------------+

by **ronk** » Sun Feb 04, 2007 1:24 pm

Mike Barker wrote:Of course I'd love to be wrong - I desperately would like to see some new simpler method to solve these harder puzzles.
r3c2-3-r3c7-4-r3c5-9-r2c6-4-r2c1-1-
r3c2-4-r2c1-1-
r3c2-9-r1c2-4-r2c1-1- or with ALS r3c2-9-ALS:r2c1|r1c2-1-

No, you're right. Without realizing it, I was writing about this situation ...

=X1=r5c5=X2=

... where two negations that have a common outcome are sufficient for that outcome to be true. I was incorrectly trying to use two assertions instead.

(That fuzzy thinking post is now deleted.)

by **Steve K** » Fri Mar 02, 2007 3:11 am

Here is an interesting example of this idea. It is not, in my opinion, the most efficient way to solve this particular puzzle, but nevertheless, it is interesting:

Code: Select all: Puzzle at start *-----------* |9..|2..|..6| |...|..4|...| |5.2|.3.|.7.| |---+---+---| |..3|9..|.2.| |...|.5.|...| |.4.|..7|1..| |---+---+---| |.1.|.8.|2.3| |...|5..|...| |8..|..1|..9| *-----------* puzzle after SS solving set *-----------* |9.4|2.5|..6| |...|..4|..2| |5.2|.3.|.7.| |---+---+---| |..3|9..|.2.| |...|.5.|...| |.4.|..7|1..| |---+---+---| |.1.|.8.|2.3| |...|5..|...| |8..|..1|..9| *-----------* possibility matrix where pattern occurs *--------------------------------------------------------------------* | 9 378 4 | 2 17 5 | 38 138 6 | | 1367 3678 1678 | 678 679 4 | 3589 58 2 | | 5 68 2 | 168 3 689 | 489 7 148 | |----------------------+----------------------+----------------------| | 167 5*678 3 | 9 14 68 | 4678 2 4578 | | 1267 6789* 16789 | 14 5 23 | 34678 34689 478 | | 26 4 58#9 | 38# 26 7 | 1 389 58 | |----------------------+----------------------+----------------------| | 467 1 5679 | 467 8 69 | 2 456 3 | | 3467 23679H 679 | 5 24679 23 | 4678 1468 1478 | | 8 23567H 567 | 3*467 2467 1 | 4567 456 9 | *--------------------------------------------------------------------* There are two hubs, marked H1 & H2. There are three spokes coming out of the bridged hub(2's act as the bridge) The three spokes are 9s, 5s, and 3s. The 8's marked # are a proven strong set as a result of the pattern.

Here is the resultant puzzle after making all the eliminations from this pattern:

Code: Select all: *--------------------------------------------------------------------* | 9 378 4 | 2 17 5 | 38 138 6 | | 1367 3678 1678 | 678 679 4 | 3589 58 2 | | 5 68 2 | 168 3 689 | 489 7 148 | |----------------------+----------------------+----------------------| | 167 5678 3 | 9 14 68 | 4678 2 4578 | | 1267 6789 1678 | 14 5 23 | 34678 34689 478 | | 26 4 589 | 38 26 7 | 1 39 5 | |----------------------+----------------------+----------------------| | 467 1 5679 | 467 8 69 | 2 456 3 | | 3467 29 679 | 5 24679 23 | 4678 1468 1478 | | 8 235 567 | 3467 2467 1 | 4567 456 9 | *--------------------------------------------------------------------* Of course, now with the newly solved 5, the puzzle advances a bit further.

Hopefully the pattern is clear. It considers exactly 6 native strong sets:
3's in row 9, 5's in column 2, 9's in column 2, 2's in column 2, cell r6c4=38, cell r6c3=589.
The newly proven strong sets then are:
8's in row 6 at c34, cell r9c2=235, cell r8c2=29, 9s in box 4 limited to r5c2 and r6c3.

The standard puzzle mark-up that I use clearly indicates that this pattern will exist, as cell r9c2 gets tripled circled, and cell r8c2 gets double circled.

by **daj95376** » Fri Mar 02, 2007 8:09 am

Some of your eliminations correspond to eliminations in six (equivalent) Forcing Chains/Networks. Interesting.

Code: Select all: r2c1 <> 67 Forcing Chain/Net on [r4c6]|[r5c6]|[r6c1]|[r6c4]|[r6c5]|[r8c6] r2c2 <> 367 r2c5 <> 6 r4c2 = 5 r4c7 <> 8 r4c9 <> 5 r5c2 <> 67 r5c3 <> 89 (matches your -9) r6c3 <> 5 r6c8 <> 8 (matches your -8) r6c9 = 5 (matches your -8) r8c2 <> 367 (matches your -367) r8c5 <> 26 r9c2 <> 567 (matches your -67) r9c5 <> 47

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