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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Mon Oct 06, 2008 3:41 pm Post subject: Hopelessly stuck |
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This is another Outlaw puzzle from Paul's Pages. I haven't made a single move, although a hidden UR will remove the 5 from r8c5. Most if the time Extended Medusa can bail me out, but not here.
Code: |
+--------------+-------------------+---------------+
| 5 178 1467 | 134 134 2 | 678 9 3467 |
| 167 2 1467 | 1349 8 1479 | 567 345 34567 |
| 3 78 9 | 5 6 47 | 1 48 2 |
+--------------+-------------------+---------------+
| 4 17 3 | 8 12 5 | 27 6 9 |
| 17 6 8 | 1249 1249 149 | 257 35 357 |
| 2 9 5 | 6 7 3 | 4 1 8 |
+--------------+-------------------+---------------+
| 8 35 16 | 7 13459 149 | 569 2 1456 |
| 179 35 17 | 12349 123459 6 | 589 458 145 |
| 169 4 2 | 19 159 8 | 3 7 156 |
+--------------+-------------------+---------------+
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Play this puzzle online at the Daily Sudoku site |
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Tue Oct 07, 2008 3:56 am Post subject: |
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marty, don't be too down about this one. according to sudocue this could involve
3 medusa traps
4 medusa bridges
4 medusa complexes |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Oct 07, 2008 7:45 am Post subject: |
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Medusa will help when extended ...
Starting at r3c6, This is where Medusa takes us (red and green background).
Looking only at "red", we see
a) r7c6<>4 (because of r3c6=4)
b) r7c7=9 (because of r8c7=8, sl 9 in box 9), thus r7c6<>9
If r7c6 is neither 4 nor 9, it must be 1
c) r2c4=9 (because of r2c6=7, sl 9 in row 2), thus r9c4<>9
If r9c4<>9 it must be 1.
This is a Medusa wrap, it solves r3c6 and some more cells.
After that (and I did use that UR wsl Marty mentioned), a skyscraper (6) is enough to solve the puzzle. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Oct 07, 2008 11:26 am Post subject: |
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nataraj wrote: | Looking only at "red", we see
a) r7c6<>4 (because of r3c6=4)
b) r7c7=9 (because of r8c7=8, sl 9 in box 9), thus r7c6<>9
If r7c6 is neither 4 nor 9, it must be 1
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We must be working on different puzzles, because my solution has [r7c6]=4. In fact, it's what finally cracks the puzzle for me. |
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ravel
Joined: 21 Apr 2006 Posts: 536
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Posted: Tue Oct 07, 2008 1:08 pm Post subject: |
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Nataraj said: This is a Medusa wrap, which means that this (red) branch cant be true. It would lead to both r7c6=1 (b) and r9c4=1 (c).
Therefore the red starting point r3c6=4 must be wrong and the greens are true.
PS:
Extended Medusa is also known as Red Green Transport (but the wrap - a contradiction in one branch, which you often will get - is not mentioned there).
This is probably the most effective technique to solve harder puzzles on paper, where the normal arsenal of techniques fails. Just try to find a good starting point (e.g. bivalue cells or conjugated numbers), where both branches easily can be extended.
Drawbacks: It is hard to describe the solution to others and if you are (un)lucky, one branch directly will solve the puzzle. |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Tue Oct 07, 2008 3:51 pm Post subject: |
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Extended Medusa usually rescues me on difficult puzzles, but I tried it twice here without any success. Thanks for the responses. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Oct 07, 2008 7:12 pm Post subject: |
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Thanks for the clarification, ravel! Actually what I did was not Medusa per se and not Asellus' extension, but the "graded equivalence marks", or GEM for short. GEM can lead to eliminations (what in Medusa parlance is called "traps") or to a contradiction ("wrap"), or both. I rarely use Medusa/GEM - I've decided for myself that the method is all too dependent on the starting point -, but since Medusa was explicitly mentioned by Marty and storm_norm, I gave it a try.
The results (if there are any) can be described adaequately by AICs, although in many instances one needs grouped or even branched AICs and that does get very messy. therefore I rely mostly on the drawings like I did in this example, here, together with a few immediately obvious conclusions ("if x=a then y<>a then ..." seems to be the most understandable and simple type of explanation. Personally I've never felt comfortable with reasoning involving green and red, although it is extremely helpful to remember that in Medusa colored grids either all reds are true or all greens)
----
edited to correct meaning of GEM = graded equivalence marks. The article on GEM is in sudopedia
Last edited by nataraj on Tue Oct 07, 2008 9:21 pm; edited 1 time in total |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Tue Oct 07, 2008 8:52 pm Post subject: |
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daj95376 wrote: | We must be working on different puzzles, because my solution has [r7c6]=4. |
Not at all. nataraj's statement that r7c6 is neither <4> nor <9> is part of a color "extension" based upon the condition that red is true. As he immediately shows, red cannot possibly be true so that statement has no bearing upon the puzzle's ultimate solution. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Oct 07, 2008 9:22 pm Post subject: |
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Okay, I now understand that red represents a forcing net for [r3c6]=4 that leads to the contradiction [r7c6]=1 and [r9c4]=1.
It has several cells in common with this SIN:
Code: | SIN: 4r3c6 7r2c6 9r2c4 1r9c4 9r7c6 9r8c7 8r1c7 [r3c8]=EMPTY
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Oct 08, 2008 7:23 am Post subject: |
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Having sort of started this "extended" Medusa thing on this board, I have come to avoid any "extensions" that are not obvious without markings (not that I'm saying the extensions above are not obvious... that is up to each person to decide) so as to avoid anything too elaborate that threatens to become forcing (or extended color markings that then have to be erased to continue). Instead, I prefer to opt for Medusa multi-coloring.
GEM is probably a more efficient approach to this puzzle. But, GEM is complex and not so easy to learn. Medusa multi-coloring is more straightforward than GEM, if not as powerful. In the case of this puzzle, it is only moderately efficient. An example of a grid template that can be used for pencil and paper Medusa multi-coloring is here.
I started in r4c7 with an "Aa" cluster, then added a "Bb" cluster with weak link bridge on the <7>s in c2 to give an "ab" strong pair.
Code: | +------------------+--------------------+------------------+
| 5 1A78B 1467 | 134 134 2 | 6-78b 9 3467 |
| 167 2 1467 | 1349 8 147B9 | 567 345 34567 |
| 3 7B8b 9 | 5 6 4B7b | 1 4b8B 2 |
+------------------+--------------------+------------------+
| 4 1a7A 3 | 8 1A2a 5 | 2A7a 6 9 |
| 1A7a 6 8 | 1249 1249 149 | 2a57 35 357 |
| 2 9 5 | 6 7 3 | 4 1 8 |
+------------------+--------------------+------------------+
| 8 35 16 | 7 13459 149 | 569 2 1456 |
| 179 35 17 | 12349 123459 6 | 58B9 458b 145 |
| 169 4 2 | 19 159 8 | 3 7 156 |
+------------------+--------------------+------------------+ |
This quickly traps <7> in r1c7, though this is not immediately helpful.
I next added 6 more clusters, "Cc" through "Hh":
Code: | +------------------+---------------------+------------------+
| 5 1A78B 1467 | 134 134 2 | 6B8b 9 3467 |
| 16h7 2 1467 | 1349c 8 147B9C | 567 345 34567 |
| 3 7B8b 9 | 5 6 4B7b | 1 4b8B 2 |
+------------------+---------------------+------------------+
| 4 1a7A 3 | 8 1A2a 5 | 2A7a 6 9 |
| 1A7a 6 8 | 1249 1249 149 | 2a57 35 357 |
| 2 9 5 | 6 7 3 | 4 1 8 |
+------------------+---------------------+------------------+
| 8 35 1H6h | 7 13459 149 | 5-69d 2 1456 |
| 17e9f 35 1e7E | 12349 123459 6 | 58B9D 458b 145 |
| 16H9F 4 2 | 1G9g 159 8 | 3 7 156h |
+------------------+---------------------+------------------+ |
There are several new weak link bridges, such as "BC" in cell r2c6 which creates a "bc" strong pair. The strong pairs resulting from such weak link bridges are:
bc, bd, AE, EF, CG, fG, dF, fh, Eh
For those not familiar with the technique, wherever two strong pairs share a conjugate color pair, a strong pair is induced. Here, for instance, we have "bc" and "CG" which induce the additional strong pair "bG" due to the "Cc" conjugates. The strong pairs induced in this way at this point are:
bE, aF, bG, EG, dG, dh
Despite all those strong pairs, I find only one trap: the "dh" trap of <6> in r7c7 (shown above). From this elimination there are simplifications that result in 6b in r2c7, trapping <7> in that cell and leading to more simplifications. It also gives 6b in r1c3 and a "BH" strong pair via r2 or b1. This in turn produces several more strong pairs: aH, cH, dH, EH, BE, Bf, Bd, GH, and others, but these are enough. The eliminations shown below (<7>s in r12c3 with a "BE" trap; <1>s in r7c56 and r9c1 with a "GH" trap; <9> in r8c7 with a "Bd" trap) solve the puzzle.
Code: | +---------------------+---------------------+-------------------+
| 5 1A78B 146b-7 | 134 134 2 | 6B8b 9 347 |
| 16h7 2 146-7 | 1349c 8 147B9C | 5B6b 345 3457 |
| 3 7B8b 9 | 5 6 4B7b | 1 4b8B 2 |
+---------------------+---------------------+-------------------+
| 4 1a7A 3 | 8 1A2a 5 | 2A7a 6 9 |
| 1A7a 6 8 | 1249 1249 149 | 2a7A 35 35 |
| 2 9 5 | 6 7 3 | 4 1 8 |
+---------------------+---------------------+-------------------+
| 8 35 1H6h | 7 -13459 -149 | 5D9d 2 1456H |
| 17e9f 35 1e7E | 12349 123459 6 | 58B-9D 458b 145 |
| -16H9F 4 2 | 1G9g 159 8 | 3 7 156h |
+---------------------+---------------------+-------------------+ |
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