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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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 Posted: Fri May 02, 2008 12:00 am Post subject: Fortnight |
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M5184793 (86)
| Code: |
+----------------+--------------+---------------+
| 28 9 36 | 7 25 58 | 36 1 4 |
| 2478 2456 1 | 9 3 48 | 2567 567 257 |
| 247 2345 345 | 6 124 14 | 2357 8 9 |
+----------------+--------------+---------------+
| 149 1345 3459 | 134 8 7 | 2569 3569 25 |
| 6 34 8 | 5 49 2 | 79 379 1 |
| 129 1235 7 | 13 6 19 | 4 359 8 |
+----------------+--------------+---------------+
| 5 7 49 | 8 149 3 | 19 2 6 |
| 149 146 469 | 2 7 459 | 8 459 3 |
| 3 8 2 | 14 1459 6 | 1579 4579 57 |
+----------------+--------------+---------------+
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Play this puzzle online
I enjoyed doing this puzzle - mix of standard & non-standard moves, including one AIC I couldn't avoid. Not easy, but I did have a fortnight in which to stare at it . . . |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri May 02, 2008 9:27 am Post subject: |
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| Code: | .------------------.------------------.------------------.
| 28 9 36 | 7 25 58 | 36 1 4 |
| 2478 2456 1 | 9 3 48 | 2567 567 257 |
| 247 2345 345 | 6 12 14 | 2357 8 9 |
:------------------+------------------+------------------:
| 149 1345 359 | 134 8 7 | 2569 3569 25 |
| 6 34 8 | 5 49 2 | 79 379 1 |
| 129 125 7 | 13 6 19 | 4 359 8 |
:------------------+------------------+------------------:
| 5 7 49 | 8 149 3 | 19 2 6 |
| 149 16 469 | 2 7 59 | 8 459 3 |
| 3 8 2 | 14 159 6 | 1579 4579 57 |
'------------------'------------------'------------------' |
my stuck point
medusa perhaps? |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Fri May 02, 2008 10:17 am Post subject: |
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Consider how the two 14s are related . . . |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri May 02, 2008 3:48 pm Post subject: |
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| Quote: | | Consider how the two 14s are related . . . |
Victor,
How about continuing the explanation? I can see they're a W-Wing connected by the 1s in c5, but I can't do anything with it, even with pincer coloring. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri May 02, 2008 5:43 pm Post subject: |
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| Quote: | | How about continuing the explanation? I can see they're a W-Wing connected by the 1s in c5, but I can't do anything with it, even with pincer coloring. |
I reached the same point as the grid above but can't get any further. I am not into W-wings, but I agree that a w-wing doesn't seem to work here. Can't see how the <14>s help in any other way but I'm not into colouring either. (Don't wanna be!) |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Fri May 02, 2008 8:18 pm Post subject: |
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Consider the 1s, seeing how they affect the other 14, via the middle box. Say you start with the bottom left 14 cell. If it's 1 then r6c6 must be 1 and so the other 14 must be NOT-1. Alternatively, if it's NOT-1 in the bottom left cell, then it's 1 in the top 14 cell. And the same thing goes if you start with the top right cell: if it's 1, the bottom left cell is 4, if it's not-1 the other cell is 1. So the two 14 cells are conjugate: one must be 1,the other 4, like 3 steps of a colouirng chain in either number. So we can continue colouring in 4s in either direction.
I think that means we can eliminate the 4s in r2c2 and r4c3. That gives a hidden pair 48 in r2.
That's where I need an AIC. Suppose r6c1 is 9. I haven't written down the whole thing, but I saw that it gives a contradiction that means that r6c1<9> r6c6 = 9. This contradiction means that r6c1 <> 9.
Edit. I tried to write out a longer explanation, but made a mess - sorry. Must go now - to be continued.
Last edited by Victor on Fri May 02, 2008 9:27 pm; edited 2 times in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri May 02, 2008 9:19 pm Post subject: |
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| Quote: | Consider the 1s, seeing how they affect the other 14, via the middle box. Say you start with the bottom left 14 cell. If it's 1 then r6c6 must be 1 and so the other 14 must be NOT-1. Alternatively, if it's NOT-1 in the bottom left cell, then it's 1 in the top 14 cell. And the same thing goes if you start with the top right cell: if it's 1, the bottom left cell is 4, if it's not-1 the other cell is 1. So the two 14 cells are conjugate: one must be 1,the other 4, like 3 steps of a colouirng chain in either number. So we can continue colouring in 4s in either direction.
I think that means we can eliminate the 4s in r2c2 and r4c3. That gives a hidden pair 48 in r2. |
If I understand W-Wings correctly (certainly no sure thing), the fact that it's a W-Wing means the 4s are pincers. What I can't follow is how you can color either 4 so as to eliminate something. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Fri May 02, 2008 9:31 pm Post subject: |
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Marty: | Quote: | | If I understand W-Wings correctly |
Point is, this isn't just a W-wing: you'd be right if it were. It's a full-on conjugate link, of the kind described by Keith in his discussion about M-wings etc. It's exactly the same effect as 3 steps of a colouring chain in 4s. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Fri May 02, 2008 10:36 pm Post subject: |
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victor, I like that logic a lot.
I hope this image tells the tale of the elimination of the 4 in r2c2.
notice the 1's in boxes 2 and 8, and how column 5 connects them. if you say that the 1 in r9c4 is true, then you can follow my links to see how the 1 in r3c6 is not true. and you start in r3c6 and go the other way. this says that either of the 4's in r9c4 and r3c6 is true.
then follow the red/green coloring from those fours to eliminate the 4 in r2c2.
continuing the coloring after that elimination gives you the elimination of the 4 in r8c1...
then another four in r4c2, but only after some box/line interactions that solves r3c1 as 7.
edit: my mistake, thanks Earl, good eye.
Last edited by storm_norm on Sat May 03, 2008 1:22 am; edited 4 times in total |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Sat May 03, 2008 12:11 am Post subject: Fortnite |
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Norm,
Your analysis is impressive. But I don't understand how the 4 in R4C1 is "red" if there is no strong pair in either its row, column, or box?
Earl |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat May 03, 2008 12:58 am Post subject: |
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| that is a good eye Earl, its my mistake, i will edit it. r3c1 should be solved as a 7 after my second image. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sat May 03, 2008 8:23 am Post subject: |
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Thanks Norm, for a clear explanation! - you've done my job for me. I was going to try to explain better what I meant (hadn't much time yesterday).
I actually saw this stuff via the middle box
| Code: | .------------------.------------------.------------------.
| 28 9 36 | 7 25 58 | 36 1 4 |
| 2478 2456 1 | 9 3 48 | 2567 567 257 |
| 247 2345 345 | 6 12 14D | 2357 8 9 |
:------------------+------------------+------------------:
| 149 1345 359 | 134B 8 7 | 2569 3569 25 |
| 6 34 8 | 5 49 2 | 79 379 1 |
| 129 125 7 | 13B 6 19C | 4 359 8 |
:------------------+------------------+------------------:
| 5 7 49 | 8 149 3 | 19 2 6 |
| 149 16 469 | 2 7 59 | 8 459 3 |
| 3 8 2 | 14A 159 6 | 1579 4579 57 |
'------------------'------------------'------------------' |
If you think of the Bs as a sort of grouped 1, then it's just a colouring chain A, B, C, D in 1s. So the 1s in A & D are conjugate - 3 steps in a colouring chain, and therefore so are the 4s, and then you can continue colouring as Norm has explained.
-----------------------------------------------------------------------------------
Now for phase 3, which is actually an AIC, but just one step from being an XY-chain, and thus easy to see. | Code: | .------------------.------------------.------------------.
| 28* 9 36 | 7 25* 58* | 36 1 4 |
| 48 256 1 | 9 3 48* | 2567 567 257 |
| 7 2345 35 | 6 12 14* | 2357 8 9 |
:------------------+------------------+------------------:
| 149 1345 359 | 134 8 7 | 2569 3569 25 |
| 6 34 8 | 5 49 2 | 79 379 1 |
| 129# 125 7 | 13 6 19* | 4 359 8 |
:------------------+------------------+------------------:
| 5 7 49 | 8 149 3 | 19 2 6 |
| 19 16 469 | 2 7 59 | 8 459 3 |
| 3 8 2 | 14 159 6 | 1579 4579 57 |
'------------------'------------------'------------------' |
Suppose # is 9. Then r1c1 is 2, and the next * is 5 , ..... , the 19* must be 9. This is a contradiction: so we can remove the 9 in #. (I do know how to write this out as a 'proper' AIC, but why bother? I don't know how other people feel, but I personally find a long AIC written out in full very off-putting.)
That gives us two 12s, in # and in r3c5, which are connected in a 3-step colouring chain in 2s. So the 1s must also be in a 3-step colouring chain and we can continue colouring in 1s. (If you're unsure about this, read 'Chapter 2' in Keith's explanation: http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=2143)
You may find yet another pair linked like this - two 25s.
(Apologies again for a lousy explanation yesterday - one of those bad days: wife packing up to visit grandchild, pesky unannounced visitors, etc., etc.) |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sat May 03, 2008 9:49 pm Post subject: |
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| Marty R. wrote: | | I can see they're a W-Wing connected by the 1s in c5, but I can't do anything with it, even with pincer coloring. |
Actually, if you see those {14}s as a W-Wing with <4> pincers, the <4> at r9c4 can be transported to r7c3, via b9 and the grouped link in b7, eliminating <4> in r3c3.
Of course, the Medusa coloring with grouped links described above is more powerful. In any case, it shows the utility of grouped links. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sat May 03, 2008 11:30 pm Post subject: |
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Great solutions, thats fun  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun May 04, 2008 12:15 am Post subject: |
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| Quote: | | Actually, if you see those {14}s as a W-Wing with <4> pincers, the <4> at r9c4 can be transported to r7c3, via b9 and the grouped link in b7, eliminating <4> in r3c3. |
Yes, it's plain as day, now that I know about it.  |
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