The New Sudoku Players' Forum

[mturner]

Code: Select all

*-----------------------------------------*
| 38    389 4    | 2   179  17  | 6 18 5  |
| 26    28  7    | 45  45   16  | 3 18 9  |
| 1     69  5    | 3   69   8   | 2 7  4  |
*-----------------------------------------*
| 9     4   2    | 16  17   167 | 5 3  8  |
| 368   1   368  | 9   38   5   | 4 2  7  |
| 7     5   38   | 48  2348 23  | 9 6  1  |
*-----------------------------------------*
| 4     26  1    | 7   26   9   | 8 5  3  |
| 5     38  9    | 16  128  23  | 7 4  26 |
| A2368 7   A368 | 58 -2358 4   | 1 9  A26|
*-----------------------------------------*


In this puzzle, there is a multi-value chain
+3[r9c5] -3[r8c6] +3[r8c2] -8[r8c2] +8[r8c5] -8[r9c45]

The 3 in r8c6 and the two 8's in r9c56 are the ends of the chain and form a conjugate pair in row 9.

There is also an ALS in row 9 in cells r9c1, r9c3, and r9c9 with values 2368 (Marked with an A).

This ALS along with the conjugate pair form a locked set in row 9. Any candidate in row 9 with value 236, or 8 that is not part of the locked set can be eliminated. In this case, the 2 in r9c5.

Is this a new elimination or am I re-plowing old ground?

[Luke]

I never learned Medusa. I'm wondering if you are familiar with als-xz?

[David P Bird]

As an Alternating Inference Chain your elimination looks like this:

(236=8)ALS:r9c139 – (8=3)r8c2 - (3)r8c5 = (35)r9c45 => r9c5 <> 2

Within your cell set there is a < Sue de Coq > pattern. The 4 cells (38)r8c2, (2368)r9c1, (368)r9c3 & (26)r9c9 together hold 4 digits and make a 2 sector locked set that must hold (38) in box 7 and (26) in row 9. Hence (2)r9c5 must be false.

So your discovery isn't new in itself, but how you used Medusa to help you spot it may possibly be a new twist - I'm not sure.

[7b53]

there is a < Sue de Coq > pattern. The 4 cells (38)r8c2, (2368)r9c1, (368)r9c3 & (26)r9c9 together hold 4 digits and make a 2 sector locked set that must hold (38) in box 7 and (26) in row 9. Hence (2)r9c5 must be false.

nice to see you posting again. David

Code: Select all

*-----------------------------------------*
| 38    389 4    | 2   179  17  | 6 18 5  |
| 26    28  7    | 45  45   16  | 3 18 9  |
| 1     69  5    | 3   69   8   | 2 7  4  |
*-----------------------------------------*
| 9     4   2    | 16  17   167 | 5 3  8  |
| 368   1   368  | 9   38   5   | 4 2  7  |
| 7     5   38   | 48  2348 23  | 9 6  1  |
*-----------------------------------------*
| 4     26  1    | 7   26   9   | 8 5  3  |
| 5     38  9    | 16  128  23  | 7 4  26 |
| A2368 7   A368 | 58 -2358 4   | 1 9  A26|
*-----------------------------------------*

I see either r9c1 or r9c3 is a locked set {3,8} with r8c2.
so what is left is either r9c1=26 or r9c3=6. either one will then be a locked set with r9c9 {2,6}. resulting r9c5<>2

is this the correct way to look at it ?

[mturner]

David Bird,
Thanks for the better notation. It's an area I'm weak in.

Although there is in fact a Sue de Coq pattern as you describe, it is only there by coincidence. The chain could just as easily connect the two ends via another path (in some other example).

7b53,
I think you are also looking at it as a Sue de Coq.

The real point here is that the (8)r9c45 and the (3)r9c5 are connected by a 3D Medusa chain somehow and create a conjugate pair. Combine that conjugate pair with an appropriate ALS and we can make some eliminations.

[Luke]

Here's the als-xz I mentioned. It seems a bit more basic, although the SdQ is short and sweet:

Code: Select all

 
 *-----------------------------------------------------------*
 | 38    389   4     | 2     179   17    | 6     18    5     |
 | 26    28    7     | 45    45    16    | 3     18    9     |
 | 1     69    5     | 3     69    8     | 2     7     4     |
 |-------------------+-------------------+-------------------|
 | 9     4     2     | 16    17    167   | 5     3     8     |
 | 368   1     368   | 9     38    5     | 4     2     7     |
 | 7     5     38    | 48    2348  23    | 9     6     1     |
 |-------------------+-------------------+-------------------|
 | 4     26    1     | 7     26    9     | 8     5     3     |
 | 5     38    9     | 16    128   23    | 7     4     26    |
 | 2368  7     368   | 58    2358  4     | 1     9     26    |
 *-----------------------------------------------------------*

(2=38)r8c26-(8=362)r9c139 ==> r9c5<>2

I never learned Medusa, so I have never heard of a "Medusa chain." I have been under the impression that Medusa was a method discredited as trial-and-error. Am I off base in thinking that?

[JasonLion]

3D Medusa Coloring is an approach to marking cells with two colors in order to find alternating inference chains (AIC). Generally only conjugate pairs and cells with exactly two pencil marks are included as strong links in the chains, though some people extend medusa coloring to include other kinds of strong links.

It is common to use medusa coloring without identifying a specific chain. The pattern of coloring builds up a net of links, and most people do not bother to identify which specific chain within that net was used to make a specific elimination. There will always be a chain that can be identified associated with each elimination, there simply isn't any need to locate it.

[daj95376]

In this puzzle, there is a multi-value chain
+3[r9c5] -3[r8c6] +3[r8c2] -8[r8c2] +8[r8c5] -8[r9c45]

The 3 in r8c6 and the two 8's in r9c56 are the ends of the chain and form a conjugate pair in row 9.

I guess that I'm the only one who has problems with your "chain". First off, it starts with a weak link on <3> and ends with a weak link on <8>. There's nothing to be learned from your chain. Then you switch to discussing <3> in r8c6 and the <8s> in r9c45. This is based on an initial strong link on <3> and a final weak link on <8>. Again, nothing to be learned from your chain. How do you get a conjugate pair???

Your chains using Eureka notation:

Code: Select all

(3)r9c5 - (3)r8c6 = (3-8)r8c2 = (8)r8c5 - (8)r9c45

          (3)r8c6 = (3-8)r8c2 = (8)r8c5 - (8)r9c45


[mturner]

daj95376,
Rather than get caught in notation details, here are the links of the chain:
Strong link on 3 in box 8 (r9c5 to r8c6) (the only 3's in the box) ( + to -)
Strong link on 3 in row 8 (r8c6 to r8C2) (the only 3's in the row) (- to +)
Strong link in r8c2 (3 to 8 ) bi-value cell (+ to -)
Strong link on 8 (r8c2 to r8c5) the only 8's in the row (- to +)
Strong link on 8 (r8c5 to node r9c45)(+ to -) (either the 8 in r8c5 is true or one of the 8's in r9c45 is true)

Notice the starting node is + and the ending node is -. Therefore only one of them is true. That is the conjugate pair.
(8r9c45 and 3r9c5)

If the 3 in r9c5 is false, then after following the strong links noted above, one of the 8's in r9c45 must be true. This forces the 8s in r9c13 to both be false. Combining with r9c9, with have a locked set (r9c139) on 236, forcing the 2 in r9c5 to be false.

If the 3 in r9c5 is true, the 2 in r9c5 is false.

In either case, the 2 in r9c5 is false.

Luke451,
Although you can use an ALS-XZ to do the elimination in this case, it does not prove the method I am showing is invalid or not new. As of now I don't have a better example of what I'm trying to show.

[David P Bird]

I wasn't expecting my response to get so much reaction! When I first became aware of SdCs I actively sought the basic pattern, but nowadays I don't bother because, as this example shows the eliminations will follow on from a scan for Almost sets. When two Almost sets overlap, potentially they can combine to form a two-sector locked set just as here. When a hit is recognised it's then easier to notate the locked set than the AIC. This approach will also identify extended SdC patterns involving further digits.

For a locked naked set any external candidate that sees all instances of itself in one house covered by the set is then false.
For a locked hidden set any internal candidate that isn't locked in the pattern cells in one of houses is false.

I use < GEM > to track AICs which allows weak and strong inferences to be followed. It takes quite a bit of practice to use, but if that investment is made, it can track through different group nodes, ANSs, AHSs etc which I didn't believe that Medusa could do – hence my comment. Unlike Medusa which can follow multiple conjugate segments using various pairs of colours, GEM can only follow the various spurs that fan out from a single starting chain, but these can be followed further.

[JasonLion]

This all reminds me of ALS's in AIC's and the use of ALS subsets.

[daj95376]

daj95376,
Rather than get caught in notation details, here are the links of the chain:
Strong link on 3 in box 8 (r9c5 to r8c6) (the only 3's in the box) ( + to -)
Strong link on 3 in row 8 (r8c6 to r8C2) (the only 3's in the row) (- to +)
Strong link in r8c2 (3 to 8 ) bi-value cell (+ to -)
Strong link on 8 (r8c2 to r8c5) the only 8's in the row (- to +)
Strong link on 8 (r8c5 to node r9c45)(+ to -) (either the 8 in r8c5 is true or one of the 8's in r9c45 is true)

Notice the starting node is + and the ending node is -. Therefore only one of them is true. That is the conjugate pair.
(8r9c45 and 3r9c5)

If the 3 in r9c5 is false, then after following the strong links noted above, one of the 8's in r9c45 must be true. This forces the 8s in r9c13 to both be false. Combining with r9c9, with have a locked set (r9c139) on 236, forcing the 2 in r9c5 to be false.

If the 3 in r9c5 is true, the 2 in r9c5 is false.

In either case, the 2 in r9c5 is false.

I did not realize that you were using +/- as a form of multi-digit coloring for strong links. Eureka equivalent of your entire logic:

Code: Select all

(3)r9c5 = r8c6 - (3=8)r8c2 - r8c5 = (8)r9c45 - (8=236)r9c139  =>  r9c5<>2


[mturner]

JasonLion,

This all reminds me of ALS's in AIC's and the use of ALS subsets.

Thank you for your most interesting suggestion. I believe this is the applicable example from the link. Please correct me if I am wrong.

Postby Myth Jellies » Sat May 27, 2006 5:53 pm
AIC candidate premises can involve cells that span multiple houses as long as you are careful. Take the following hypothetical snippet...

Code: Select all

Code: Select all

     *--------------------------------------------------------------------*
     | .      .      .      | .     D12347  .      | .     A37'    .      |
     | .      .      .      |C125    .      .      | .      .      .      |
     | .      .      .      |C125    .      .      | .      .      .      |
     +----------------------+----------------------+----------------------+
     | .      .      .      | .     C345    .      | .      .      .      |
     | .      .      .      | .     C345    .      | .      .      .      |
     | .      .      .      |B56     .      .      | .     A367    .      |
     +----------------------+----------------------+----------------------+
     | .      .      .      | .      .      .      | .      .      .      |
     | .      .      .      | .      .      .      | .      .      .      |
     | .      .      .      | .      .      .      | .      .      .      |
    *--------------------------------------------------------------------*


A3 = A(6&7') - B6 = B5 - C5 = C(1&2&3&4) - D(1234) = D7 - A(6&7') = A3
Thus we show that the A cells must contain 3 (or cannot be 6&7), therefore we can remove all other 3's from that column.

In Myth Jellies's example, there is a chain between (7)r1c8 and the (6)r6c8, and they also form a conjugate pair. In addition, r16c8 form an ALS. Because of this, Myth Jellies is able to make conclusions about the 3 in that ALS and therefore about 3's in c8. I would say this is an additional (and perhaps forgotten and overlooked) elimination that is available to 3D Medusa chains.

In my example, the two nodes that make up the conjugate pair are not part of an ALS. I use an additional ALS to make conclusions about candidates outside of the chain and outside the ALS. I think that is an important difference.

[David P Bird]

This is the AIC notation for Myth's elimination:

(37=6)ALS:r168 - (6=5)r6c4 - (12345=7)ANS:r145c5,r34c4 - (7=3)r1c8 => r23458789c8 <> 3

In this case there are two overlapping AANSs in b2 & c5 that combine to give the 5 cell 2-sector ANS. If (7)r1c5 was missing they would form a 2-sector Naked Set composed of overlapping ANSs as I described before. Note that in this case the pattern wouldn't be classed as a SdC.

I conject that when Myths pattern exists there will usually be simpler routes using external cells to achieve the same effect. This would make the cases when they must be identified to crack a puzzle rare. Consequently people don't generally look for them so making the chances of cases being reported even rarer!

mturner, Medusa colours conjugate candidates to guarantee the links can be followed in either direction and remain compatible throughout a colour set. I therefore have trouble understanding what you describe as a Medusa chain when it contains group nodes. I can see various options to extend the colouring that present different banana skins, but I can't second guess the way you work. How for instance would you colour the candidates in either your or Myth's example?

As an observation, the links in an ANS are only conjugate when all but 2 of the candidates are locked in the cells (when it also qualifies as being an AHS).This seems to limit the cases you will be able to find using colouring alone That is unless there are external cells that provide another route to take - which is very similar to my earlier point.

[JasonLion]

In my example, the two nodes that make up the conjugate pair are not part of an ALS. I use an additional ALS to make conclusions about candidates outside of the chain and outside the ALS. I think that is an important difference.

From my perspective now in 2013, Myth showed that ALS could be used as links in AIC, which is essentially the same thing you have done.

However, it is true that isn't exactly what he said, and might not be the way his post would have been understood in 2006. He used an ALS as one of the end points of the chain, not a link. The development of the ER technique briefly passed through a similar period, when ER were initially only used at the end of a chain, but later became used as links at any point. That process, starting with "unusual" links at the ends of chains and then generalizing slightly to allow them as links at any point in the chain has since become a common progression, generally now with no real time between the steps, just the slight simplification that it is easier and clearer to find examples with the "new" element at the end of the chain.

There is however still a fair amount of room for uncertainty/confusion. You are using medusa coloring instead of AIC. Medusa coloring is equivalent to AIC in my book, but it is hardly ever used these days and not everyone agrees on that. It also isn't obvious that my generalization of Myth's post is valid as "obvious" to current practitioners or not.