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Not easy - no URs . . .

 
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Mon Jan 21, 2008 9:43 pm    Post subject: Not easy - no URs . . . Reply with quote

. . . or skyscrapers or ERs, etc. in sight in this one, M3620506 (72). Post basics:
Code:

+---------------+----------------+---------------+
| 2457 1   3457 | 6     279  479 | 237  8    379 |
| 6    289 78   | 1     279  3   | 4    79   5   |
| 247  29  347  | 24    8    5   | 2367 679  1   |
+---------------+----------------+---------------+
| 12   7   9    | 258   125  6   | 58   3    4   |
| 24   28  6    | 23458 2359 479 | 1    57   78  |
| 3    5   148  | 48    17   47  | 9    2    6   |
+---------------+----------------+---------------+
| 159  3   15   | 7     4    2   | 568  1569 89  |
| 8    6   57   | 9     35   1   | 357  4    2   |
| 1579 4   2    | 35    6    8   | 357  1579 379 |
+---------------+----------------+---------------+

Play this puzzle online at the Daily Sudoku site

I did quickly spot ALS in R5C9 & C7R489, which gets rid of the 7 in R9C9 but took some time to make any further progress.

I have done it, but wasn't too happy about my method - a certain amount of trial & error to get started, which I then justified as an AIC. And when I'd found the AIC, I discovered that I'd missed fairly powerful ALS. Quite nice puzzle after that. Hope you do it more convincingly than I did!
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Tue Jan 22, 2008 1:59 am    Post subject: Reply with quote

Victor,

Your grid isn't quite simplified: the <7> in r5c6 can be removed due to locked candidates. Since the <7>s in b6 are strongly linked, a bit of coloring not only removes that r9c9 <7>, it makes some determinations:
Code:
+---------------+----------------+---------------+
| 2457 1   3457 | 6     279  479 | 237  8    379 |
| 6    289 78   | 1     279  3   | 4    79   5   |
| 247  29  347  | 24    8    5   | 2367 679  1   |
+---------------+----------------+---------------+
| 12   7   9    | 258   125  6   | 5R8G 3    4   |
| 24   28  6    | 23458 2359 49  | 1    5G7R 7G8R|
| 3    5   148  | 48    17   47  | 9    2    6   |
+---------------+----------------+---------------+
| 159  3   15   | 7     4    2   | 568  1569 89  |
| 8    6   57   | 9     35   1   | 357  4    2   |
| 1579 4   2    | 35    6    8   | 357  1579 379r|
+---------------+----------------+---------------+

I Medusa color (RG) b6. Then, the 5R in r4c7 creates a red {37} pair in r89c7 and means that the <9> in r9c9 would be red (r). This traps that <7>, since it sees 7G. But, it also determines all the values in b6 since it creates 8r in r7c9, which is a wrap with 8R in r5c9: green must be true.

After that, I resorted to some extensive "extended" Medusa that led to another trap and wrap, but did not solve the puzzle. However, there was then a W-Wing followed by an XY-Wing.

Then, more extended Medusa for more trapping and another wrap.

Finally, this resulted in a BUG+1.
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Wed Jan 23, 2008 3:23 am    Post subject: Reply with quote

Here is a Medusa/3D Coloring implementation for finding AICs.

Three of the conjugate strings (aA, bB, dD) are highlighted below

Code:
+-----------------+-----------------+---------------+
| 2457 1     3457 | 6     279  4D79 | 237  8    379 |
| 6    2b8a9 7a8A | 1     2B79 3    | 4    79   5   |
| 247  29    347  | 2D4d  8    5    | 2367 679  1   |
+-----------------+-----------------+---------------+
| 12   7     9    | 258   125  6    | 58   3    4   |
| 24   2a8A  6    | 23458 2359 479  | 1    57   78  |
| 3    5     148a | 4a8A  17   47   | 9    2    6   |
+-----------------+-----------------+---------------+
| 159  3     15   | 7     4    2    | 568  1569 89  |
| 8    6     57   | 9     35   1    | 357  4    2   |
| 1579 4     2    | 35    6    8    | 357  1579 379 |
+-----------------+-----------------+---------------+


a sees b in r2c2, B sees D in box 2, d sees a in col 4

Thus we have the color AIC or color molecule...

A = a - b = B - D = d - a = A

...which means either A or A is true

Code:
+------------------+-------------------+---------------+
| 457   1     3457 | 6     279   4D7E9 | 237  8    39  |
| 6     2b9B  8    | 1     2B79  3     | 4    79   5   |
| 47    2B9b  347  | 2D4d  8     5     | 2367 679  1   |
+------------------+-------------------+---------------+
| 1e2E  7     9    | 2f5F  1E25f 6     | 8    3    4   |
| 2e4E  8     6    | 23f4D 23F9  49    | 1    5    7   |
| 3     5     1E4e | 8     1e7E  4E7e  | 9    2    6   |
+------------------+-------------------+---------------+
| 159   3     1e5E | 7     4     2     | 56   169  8   |
| 8     6     57   | 9     3f5F  1     | 3F57 4    2   |
| 1579  4     2    | 3F5f  6     8     | 357  179  39  |
+------------------+-------------------+---------------+


Tossing in two more conjugate strings. We have E seeing f along r4. Thus we have the color AIC...

e = E - f = F, hence either F or e or both must be true.

The 2 in r5c5 sees both an F and an e, therefore r5c5 <> 2.

Extending the fF string and adding a new conjugate string along c9 gives

Code:
+------------------+--------------------+----------------+
| 457   1     3457 | 6     279   4D7E9f | 237  8    3g9G |
| 6     2b9B  8    | 1     2B79  3      | 4    79   5    |
| 47    2B9b  347  | 2D4d  8     5      | 2367 679  1    |
+------------------+--------------------+----------------+
| 1e2E  7     9    | 2f5F  1E25f 6      | 8    3    4    |
| 2e4E  8     6    | 23f4D 3F9f  4f9F   | 1    5    7    |
| 3     5     1E4e | 8     1e7E  4E7e   | 9    2    6    |
+------------------+--------------------+----------------+
| 159   3     1e5E | 7     4     2      | 56   169  8    |
| 8     6     57   | 9     3f5F  1      | 3F57 4    2    |
| 1579  4     2    | 3F5f  6     8      | 357  179  3G9g |
+------------------+--------------------+----------------+


Now we have f seeing G along r1, and F seeing G along r9, so G must be false

g = G - f = F - G = g

Code:
+------------------+--------------------+-----------------+
| 457   1     4E57 | 6     279F  4D7E9f | 27   8     3    |
| 6     2b9B  8    | 1     2B7b  3      | 4    7B9b  5    |
| 47    2B9b  3    | 2D4d  8     5      | 26e7 6E79B 1    |
+------------------+--------------------+-----------------+
| 1e2E  7     9    | 2f5F  1E25f 6      | 8    3     4    |
| 2e4E  8     6    | 23f4D 3F9f  4f9F   | 1    5     7    |
| 3     5     1E4e | 8     1e7E  4E7e   | 9    2     6    |
+------------------+--------------------+-----------------+
| 9     3     1e5E | 7     4     2      | 5e6E 1E6e  8    |
| 8     6     57   | 9     3f5F  1      | 3F57 4     2    |
| 1E57  4     2    | 3F5f  6     8      | 3f57 1e7E  9    |
+------------------+--------------------+-----------------+


Among other things we have E seeing b along r5 and B in r3c8, so it can't be true,

e = E - b = B - E = e

which gets you to the following BUG+1 grid.

Code:
 *--------------------------------------------------*
 | 45+7 1    57   | 6    79   49   | 2    8    3    |
 | 6    29   8    | 1    27   3    | 4    79   5    |
 | 47   29   3    | 24   8    5    | 6    79   1    |
 |----------------+----------------+----------------|
 | 1    7    9    | 25   25   6    | 8    3    4    |
 | 2    8    6    | 34   39   49   | 1    5    7    |
 | 3    5    4    | 8    1    7    | 9    2    6    |
 |----------------+----------------+----------------|
 | 9    3    1    | 7    4    2    | 5    6    8    |
 | 8    6    57   | 9    35   1    | 37   4    2    |
 | 57   4    2    | 35   6    8    | 37   1    9    |
 *--------------------------------------------------*
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Wed Jan 23, 2008 3:45 am    Post subject: Reply with quote

The preceeding is a good instructive example of the use of Medusa multi-coloring and how it relates to inference logic in general.

However, it also exemplifies for me why it is that I prefer the "extended" form of single cluster Medusa that I've been using lately. First, at least in this case, both approches produced the same overall result (the identical BUG+1). Second, the extended single-cluster Medusa does not require any need to make explicit note of weak link connections between color clusters/strings/chains. Nor does it require one to determine which pair/pairs of various colors are strongly linked and which are not. This makes the "extended" approach much more "user-friendly."

Perhaps the Medusa multi-coloring technique can reveal more underlying implications than can the "extended" technique. But then again, maybe not. So far, I suspect that it is the latter.
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Wed Jan 23, 2008 4:40 am    Post subject: Reply with quote

The problem with the "extended method" is that it marks up the page based on assuming a particular color is true (or false). It can do basically anything that T&E can do (because assuming a color is true is the same as assuming a set of candidates is true) and its results are dependent on the color/candidate that you start with. The real multi-color Medusa method makes no such assumptions and has no start point dependence. It just marks up each readily apparent individual conjugate string and then you can use a simple who-sees-who to create AIC color molecules. It is not as powerful as the extended method, and still requires a lot of searching to note the deductions; but to my way of thinking it is much more satisfying. It is a way of coming up with AICs without using a single if-then which allows you to put AICs in the same toolbox as singles, naked pairs, and x-wings.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Wed Jan 23, 2008 7:12 am    Post subject: Reply with quote

I felt much the same when I was first exploring the method. However, I now believe that T&E has nothing (or something vanishingly close to nothing) to do with it. For one thing, the extensions work for both colors simultaneously: it isn't just picking one color to assume as true. The truth assumption is a shorthand for the extended marking rules. One could, alternately, construct "extended" marking rules that didn't explicitly mention any truth assumption. It is truly a bifurcation method and has exactly the same issues as any such method (i.e. assuming that one set of candidates or another is true).

On the other hand, the multi-coloring involves selecting weak links to exploit. To my mind, there is more T&E involved in that process. (And I say this as someone who has exploited involved AICs to a considerable extent in solutions and enjoyed doing so. Finding them has a substantial amount of trial and error feel... though, admittedly, an "educated" sort of T&E that improves the odds of success.) The typical advanced grid has a very large number of weak links present. How does one choose? It isn't so easy to see where any particular weak link "bridge" of two clusters will lead. Even if one says, "Well, first color multiple clusters then examine the weak link connection possibilities," this only adds another T&E dimension: In exchange for fewer weak links to consider between the established clusters, one still must choose which clusters to color.

With extended Medusa, I don't have to choose any weak links to exploit or more than one cluster to color in hopes of linking them effectively. Yet, the results seem to be the same! Once I make my initial choice of where to start the Medusa coloring, everything else follows strictly mechanically. Either it will produce eliminations or it won't. Perhaps a different starting choice would produce better results. But that's true of Medusa in general, of whatever sort.

So for me, I'm much more comfortable with extended Medusa, aesthetically, than I am with Medusa multi-coloring... precisely the opposite assessment, and even though the logical underpinnings of the multi-coloring is interesting to someone like me.

As for that question of "assuming something is true," don't ALL sudoku solving methods involve precisely that? And, aren't all such methods (other than forcing) bifurcative? An AIC typically assumes that a victim is true, then shows that this leads to the conclusion that the victim is false. So, we accept the alternate. Even DP solutions assume that either the puzzle is valid or it isn't. Everywhere we look we have alternate assumptions of truth, as we should have (in my humble opinion). In fact, that isn't the interesting question at all.

The interesting question is: which is easier to use?

In answer to that question, I believe it is no contest: "extended" Medusa wins by a mile!
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Wed Jan 23, 2008 5:30 pm    Post subject: Reply with quote

Asellus wrote:
I felt much the same when I was first exploring the method. However, I now believe that T&E has nothing (or something vanishingly close to nothing) to do with it. For one thing, the extensions work for both colors simultaneously: it isn't just picking one color to assume as true. The truth assumption is a shorthand for the extended marking rules. One could, alternately, construct "extended" marking rules that didn't explicitly mention any truth assumption. It is truly a bifurcation method and has exactly the same issues as any such method (i.e. assuming that one set of candidates or another is true).

Other bifurcative methods include X-coloring, and Ariadne's Thread. Your extended Medusa, known elsewhere by other names, is assumptive just like these, and noting that it uses multiple assumptions instead of only one does not improve it in my eyes. There is no reason to make an assumption to find an AIC or use standard Medusa, therefore it would not be proper to lump them into your bifurcative bin.

Asellus wrote:
On the other hand, the multi-coloring involves selecting weak links to exploit. To my mind, there is more T&E involved in that process. (And I say this as someone who has exploited involved AICs to a considerable extent in solutions and enjoyed doing so. Finding them has a substantial amount of trial and error feel... though, admittedly, an "educated" sort of T&E that improves the odds of success.)


There is no T&E involved in noting that two conjugate strings see each other. The conjugate strings are your markers--your building blocks--and it takes no T&E to either find them or determine if they are linked.

Asellus wrote:
The typical advanced grid has a very large number of weak links present. How does one choose? It isn't so easy to see where any particular weak link "bridge" of two clusters will lead. Even if one says, "Well, first color multiple clusters then examine the weak link connection possibilities," this only adds another T&E dimension: In exchange for fewer weak links to consider between the established clusters, one still must choose which clusters to color.


I repeat, there is no T&E involved in noting that two conjugate strings see each other. No assumptions are needed to determine if two conjugate strings see each other regardless of how many potential weak links there may be in a puzzle. Any conjugate string or "cluster" may be considered as a building block for an AIC just as any bivalued cell may be considered a building block for a naked pair. Having more potential building blocks has nothing to do with adding a T&E dimension.

Asellus wrote:
With extended Medusa, I don't have to choose any weak links to exploit or more than one cluster to color in hopes of linking them effectively. Yet, the results seem to be the same! Once I make my initial choice of where to start the Medusa coloring, everything else follows strictly mechanically. Either it will produce eliminations or it won't. Perhaps a different starting choice would produce better results. But that's true of Medusa in general, of whatever sort.

Your statement is false, real Medusa is start point independent. You get the same relationships between the clusters and the same potential eliminations regardless of what cluster you may initially consider. The question of whether you will consider/discover the combinations that result in deductions early or late in your process of consideration is moot. In some cases a color molecule diagram can aid you in that discovery. Whether or not an AIC string of clusters results in a reduction, the OR condition between the two endpoint colors remains a viable condition throughout the solve process. Color labels can remain on the grid throughout the solve process as well, and may be useful at a later time.

Asellus wrote:
As for that question of "assuming something is true," don't ALL sudoku solving methods involve precisely that? And, aren't all such methods (other than forcing) bifurcative? An AIC typically assumes that a victim is true, then shows that this leads to the conclusion that the victim is false. So, we accept the alternate. Even DP solutions assume that either the puzzle is valid or it isn't. Everywhere we look we have alternate assumptions of truth, as we should have (in my humble opinion). In fact, that isn't the interesting question at all.

The interesting question is: which is easier to use?

In answer to that question, I believe it is no contest: "extended" Medusa wins by a mile!

NO basic methods require that you assume something is true to find them. If you are making such assumptions then you are not really using them. Basic methods are either extremely basic (singles) or involve building a pattern from smaller building blocks. A naked pair uses two bivalue cells as strong link building blocks that are arranged just so. A hidden pair involves two bilocation blocks arranged just so. An x-wing involves two bilocation blocks arranged in a different configuration. AICs and Medusa also use strong link building blocks arranged in a particular configuration. I have just provided an example of a way to use Medusa to find AICs without ever assuming that any candidate or color is either true or false. Avoiding these assumptions is the interesting point because if we allow them then we might as well just use Ariadne's Thread to flush out the solution once we exhaust the basics and be done with it. Ariadne's Thread is just as "easy" as extended Medusa--for me "ease" does not translate directly into "interesting".
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Asellus



Joined: 05 Jun 2007
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PostPosted: Wed Jan 23, 2008 6:24 pm    Post subject: Reply with quote

Well, we certainly see this world differently. I can't say I'm pursuaded.

When I said "ALL sudoku methods," I should have been clear I was speaking of all advanced methods. I did not mean to include basic methods such as naked locked sets.
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Thu Jan 24, 2008 8:45 pm    Post subject: Reply with quote

Thanks guys, found that very interesting and helpful. I'm not qualified to comment on whether or not T & E is involved in extended Medusa. I'll just remark, as a solver, that different styles of thought / approach suit different people. I happen to be comfortable thinking in chains, tho' I acknowledge of course that you have to find them first - guess I'll have to try Medusa.
I did find the first elimination, of the 8 in R2C2, using a certain amount of T & E (albeit of an 'educated' kind). But I did then see that R5C5 <> 2 without T & E, by looking at how conjugate links interact, and that felt good.
Anyway, thanks
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