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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Mon Oct 29, 2007 7:37 am Post subject: help please |
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This is from daily-sudoku.com, "very hard" of Oct 21.
(original puzzle:http://www.daily-sudoku.com/save_money.php?id=1581)
After basics, some coloring and an xy-wing I got stuck here:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 278 2468 2468 | 2589 12578 1279 | 3 146789 679 |
| 1 38 5 | 6 4 379 | 89 789 2 |
| 2378 23468 9 | 238 12378 1237 | 148 14678 5 |
+--------------------------+--------------------------+--------------------------+
| 6 249 3 | 2789 278 279 | 15 15 49 |
| 89 7 48 | 1 36 5 | 49 2 36 |
| 29 5 1 | 2349 236 23469 | 7 36 8 |
+--------------------------+--------------------------+--------------------------+
| 4 12389 28 | 2357 12357 123 | 6 35789 379 |
| 5 236 26 | 2347 9 8 | 24 347 1 |
| 2389 12389 7 | 2345 12356 1246 | 24589 34589 349 |
+--------------------------+--------------------------+--------------------------+
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Any hints appreciated. |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Mon Oct 29, 2007 11:14 am Post subject: |
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Tricky one. The easiest thing I can spot to start is a hidden pair in column 6 <46>. After that, here is what I did. While looking for xyz-wings, I spotted the following almost xy-ring (or what Denis Berthier would probably call an xyt-ring):
| Code: | .------------------------.------------------------.------------------------.
| 278 2468 2468 | 2589 12578 1279 | 3 146789 679 |
| 1 38* 5 | 6 4 379 | 89* 79-8 2 |
| 2378 2468-3 9 | 238 12378 1237 | 18-4 14678 5 |
:------------------------+------------------------+------------------------:
| 6 249 3 | 2789 278 279 | 15 15 49 |
| 89 7 48 | 1 36 5 | 49* 2 36 |
| 29 5 1 | 2349 236 46 | 7 36 8 |
:------------------------+------------------------+------------------------:
| 4 1289-3 28 | 2357 12357 123 | 6 35789 379 |
| 5 236* 26* | 347-2 9 8 | 24* 347 1 |
| 2389 1289-3 7 | 2345 12356 46 | 258-49 34589 349 |
'------------------------'------------------------'------------------------' |
The idea is that if it wasn't for the 2 in r8c2, you would have an xy-ring and be able to eliminate:
2 from r8c4,
3 from r379c2,
4 from r39c7,
8 from r2c8,
9 from r9c7.
But given that the 2 in r8c2 doesn't change the max multiplicity of 2 in the set, we can still make these deductions (or Denis would say that the 2 in r8c2 sees the right linking candidate 2 in r8c7).
After clearing up some 4's (locked candidates) in column 8, we get to:
| Code: | .------------------------.------------------------.------------------------.
| 278 2468 2468 | 2589 12578 1279 | 3 146789 679 |
| 1 38* 5 | 6 4 379 | 89* 79* 2 |
| 2378 2468 9 | 238 12378 1237 | 18 14678 5 |
:------------------------+------------------------+------------------------:
| 6 249 3 | 2789 278 279 | 15 15 49 |
| 89 7 48 | 1 36 5 | 49 2 36 |
| 29 5 1 | 2349 236 46 | 7 36 8 |
:------------------------+------------------------+------------------------:
| 4 1289 28 | 2357 12357 123 | 6 35789 379 |
| 5 26-3 26 | 347 9 8 | 24 37* 1 |
| 2389 1289 7 | 2345 12356 46 | 258 3589 349 |
'------------------------'------------------------'------------------------' |
where there is a legitimate xy-chain:
r8c2 -3- {r2c2 -8- r2c7 -9- r2c8 -7- r8c8} -3- r8c2, => r8c2 <> 3. This move solves the puzzle. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Oct 29, 2007 2:46 pm Post subject: |
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great post, re'born!
Thx for the help - I did hope to make something of that almost-ring but could not get over the r8c2 cell. Let me see if I understand the reasoning behind your statement:
| Quote: | | given that the 2 in r8c2 doesn't change the max multiplicity of 2 in the set, we can still make these deductions |
if r8c2=2 then r8c4<>2 PLUS the ring works in the same way (counterclockwise) as if r8c3 had been "2"
if r8c2<>2 we have a standard xy-ring which works in both directions
In both cases, all the candidates that constitute the (weak) links in the ring are removed from all cells outside the ring that "see" the link (i.e. 3 from c2, 8 from r2, 4 from c7)
I hope I got this correctly.
Thanks again! |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Mon Oct 29, 2007 3:47 pm Post subject: |
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| nataraj wrote: | great post, re'born!
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Thank you! 
| nataraj wrote: |
Let me see if I understand the reasoning behind your statement:
| Quote: | | given that the 2 in r8c2 doesn't change the max multiplicity of 2 in the set, we can still make these deductions |
if r8c2=2 then r8c4<>2 PLUS the ring works in the same way (counterclockwise) as if r8c3 had been "2"
if r8c2<>2 we have a standard xy-ring which works in both directions
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You've got it!
The way I think of it is using a process called subset counting (see my post here). The basic idea is to count how many times a number could show up in some set of cells in the solution. Take the set of 6 cells *'d in my post. For each of the digits in the chain, how many times could they possibly show up in a solution? If you look hard, you'll see that each digit only shows up in one unit, and hence can only occur once in a solution. But there are only 6 digits and 6 cells to fill, so each digit must occur exactly once in a solution. Thus, for instance, any cell in row 8 outside of our pattern cannot be a 2 since otherwise it will kill all of the 2's in our pattern, an impossibility.
In general, this means that an xy-chain can have extra candidates in some of the cells, as long as they are placed 'appropriately'. I call these almost xy-chains.
Denis Berthier has a variation on this technique called xyt-chains, a concept that overlaps with almost xy-chains in many situations. For a comparison of the two concepts, see my post almost xy-chains vs. xyt-chains. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Oct 29, 2007 8:37 pm Post subject: |
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| Or, you could just say that it is an ALS Chain Loop. One "node" happens to be a 2-cell ALS rather than a one cell ALS. The XY Chain Loop is just the simplest form. |
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