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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Tue Dec 18, 2007 5:07 am Post subject: An EXTREMELY difficult Menneske |
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Code: | This is about the highest rating I have seen on the Menneske site.
Maybe we can all chip away at it.
Puzzle: M2190446sh(1323)
+-------+-------+-------+
| 4 7 . | . . . | . . . |
| . 5 6 | . . 3 | . 4 . |
| . 1 . | . 4 . | . 2 . |
+-------+-------+-------+
| . . . | . 2 . | 1 . . |
| . . . | 9 . 7 | . . . |
| . 4 . | . . . | . . 5 |
+-------+-------+-------+
| . . . | . 8 1 | . . . |
| 5 . . | . . . | 3 8 . |
| 3 . . | 7 . 5 | . . 4 |
+-------+-------+-------+ |
Basics get you to here:
Code: | +----------------------+----------------------+----------------------+
| 4 7 2389 | 12568 15 2689 | 5689 13569 13689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 | 568 4 689 | 56789 2 36789 |
+----------------------+----------------------+----------------------+
| 6789 3689 5789 | 4568 2 468 | 1 3679 36789 |
| 1268 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 2789 | 168 13 68 | 26789 3679 5 |
+----------------------+----------------------+----------------------+
| 2679 269 4 | 3 8 1 | 25679 5679 2679 |
| 5 69 17 | 24 69 24 | 3 8 17 |
| 3 2689 1289 | 7 69 5 | 269 169 4 |
+----------------------+----------------------+----------------------+ |
I suggest a UR to start.
Keith |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Dec 18, 2007 7:46 am Post subject: |
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O.K., I'll chip first.
As Keith suggested, I tried a UR (69) in r89c25, which leaves r9c2 with {2,8}
This cell is the fin in a finned x-wing on "2". This is the pattern, the fin is marked @. xwing would be in in c29. Take out 2 from r7c1.
Code: |
+·····+·····+·····+
· *·o *· ·
· /-/ · |· ·
·*-----* |· ·
· · |· ·
· · |· ·
+·····+····|+·····+
· · |· ·
· · |· ·
·o o o· |· *·
· · |· /-/|·
·o o· |·* |·
+·····+····|+····|+
·o o · |·o *·
· · |· ·
· ·#---#· ·
· · · ·
· @ o· ·o ·
+·····+·····+·····+
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Next I did some Medusa coloring starting with the {1,7} in r8c9, which led me to this short AIC:
(7=1)r8c9-(1)r9c8=(1-5)r1c8=(5)r7c8
which removes "7" from r7c8.
From there it is a short step (a few box/line eliminations box 6) to here:
Code: | +--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 12568 15 2689 | 5689 13569 13689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 | 568 4 689 | 56789 2 36789 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 5789 | 4568 2 468 | 1 3679 3689 |
| 1268 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 2789 | 168 13 68 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 679 269 4 | 3 8 1 | 25679 569 2679 |
| 5 69 17 | 24 69 24 | 3 8 17 |
| 3 28 1289 | 7 69 5 | 269 169 4 |
+--------------------------+--------------------------+--------------------------+ |
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ravel
Joined: 21 Apr 2006 Posts: 536
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Posted: Tue Dec 18, 2007 8:37 pm Post subject: |
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Really hard, cant see something nice.
Code: | +--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 12568 #15 2689 | 5689 #13569 13689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 |#568 4 689 |#56789 2 #36789 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 5789 | 4568 2 468 | 1 3679 3689 |
| 1268 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 2789 | 168 13 68 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 679 269 4 | 3 8 1 | 25679 569 2679 |
| 5 69 17 | 24 69 24 | 3 8 #17 |
| 3 28 1289 | 7 69 5 | 269 #169 4 |
+--------------------------+--------------------------+--------------------------+
| There is a cycle
(1)r1c5-(1)r1c8=(1)r9c8-(1=7)r8c9-(7)r3c9=(7-5)r3c7=(5)r3c4-(1)r1c5
making all links strong, so
r1c49<>1 (1 in r1c5 or r1c8), r7c9<>7 (7 in r8c9 or r3c9), r1c4<>5 (5 in r3c4 or r1c5).
Then (with r1c4<>1)
(1)r5c1=(1)r6c1-(1)r6c4)=(1-2)r2c4=(2)r2c1-(2)r5c1
(either r5c1=1 or r6c1=1=>... => r2c1=2) => r5c1<>2
But then ? I tried a Medusa with the 1's (8 strong links), but no success. |
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Tue Dec 18, 2007 9:47 pm Post subject: |
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this has a 13,633 rating on the sudokusolver.co.uk solver
the highest any of the VH's get is 2,000-3,000
and just to give a base comparison, the white belt sudoku book puzzles have a 20-100 rating
the hardest puzzles known to date average 100,000-300,000
and oh by the way, this puzzle was completed by the solver in 13 seconds using its guessing routine.
the "golden nugget" referenced in some other threads as the toughest so far took the solver over 4 minutes to hack away brute force.
norm |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Tue Dec 18, 2007 10:12 pm Post subject: |
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I was playing with Sudoku Susser on this one. It has something called "Trebor's Tables".
For example, R1C5 is <15>.
If R1C5 is <5>, you can solve <5> and <7> in R3, then <1> in C9.
So, for all possible values of R1C5, R1C9 is not <1>.
Interesting, but I think it would take a human an infinite amount of time (and pencils and paper and erasers) to do this. |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 5:48 am Post subject: |
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ravel,
Your nice cycle chain should actually read:
(1)r1c5-(1)r1c8=(1)r9c8-(1=7)r8c9-(7)r3c9=(7-5)r3c7=(5)r3c4-(5=1)r1c5
(At first glance, it looked as if you'd written a discontinous loop.)
And, you missed some resulting eliminations: (7-5)r3c7 also becomes a conjugate link, and hence a bivalue. So, <6>, <8> and <9> are all eliminated from r3c7.
I haven't yet tried to determine whether this is of any help with the puzzle. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Dec 19, 2007 6:49 am Post subject: |
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For people with a visual mind like mine, here's the graphic representation of ravel's nice loop, together with the eliminations it makes. Haven't got time yet to continue but I am sure we'll get this one in a finite amount of time.
Together with ravel's second AIC, this is where we are:
Code: |
+--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 268 15 2689 | 5689 13569 3689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 | 568 4 689 | 57 2 36789 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 5789 | 4568 2 468 | 1 3679 3689 |
| 168 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 2789 | 168 13 68 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 679 269 4 | 3 8 1 | 25679 569 269 |
| 5 69 17 | 24 69 24 | 3 8 17 |
| 3 28 1289 | 7 69 5 | 269 169 4 |
+--------------------------+--------------------------+--------------------------+
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And I agree, with so many strong links in "1", there must be something Medusa can do for us. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Dec 19, 2007 7:22 am Post subject: |
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O.K., here's the Medusa color plot, starting from "1"
There is already one elimination we can make immediately:
r4c3<>7 (sees red 7 in r8c3 and green 5 in its own cell)
happy hunting .... |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Dec 19, 2007 7:32 am Post subject: |
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Some more from Medusa:
the elimination of 7 from r4c3 gives a strong link on 7 in c3 -> color 7 in r6c3 green.
If RED is true, then in col 3, candididate 2 can go ONLY into cell r6c3. (reason:
- red 1 in r2c4 => r2c1=2 => r1c3<>2
- red 5 in r5c3 => r5c3<>2
- red 1 in r9c3 => r9c3<>2)
This allows elimination of 8 and 9 from r6c3. R6c3 is now bi-value and we can color 2 in r6c3 red.
or plain "code":
Code: |
+--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 268 15 2689 | 5689 13569 3689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 | 568 4 689 | 57 2 36789 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 589 | 4568 2 468 | 1 3679 3689 |
| 168 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 27 | 168 13 68 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 679 269 4 | 3 8 1 | 25679 569 269 |
| 5 69 17 | 24 69 24 | 3 8 17 |
| 3 28 1289 | 7 69 5 | 269 169 4 |
+--------------------------+--------------------------+--------------------------+
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 9:11 am Post subject: |
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While nataraj was doing that coloring, I was doing something similar but exploiting some Deadly Patterns (DPs). I hope I haven't made any logical lapses!
Code: | +------------------------+------------------------+------------------------+
| 4 7 2389 | @268 15 @2689 | 5689 13569 3689 |
| 289 5 6 | 128 7 3 | 89 4 189 |
| 89 1 389 |$ 568 4 $ 689 | 57 2 36789 |
+------------------------+------------------------+------------------------+
| 6789 3689 5789 | @4568 2 @468 | 1 3679 3689 |
| 168 2368 258 | 9 135 7 | 4 36 2368 |
| 126789 4 2789 |$@168 13 $@68 | 2689 3679 5 |
+------------------------+------------------------+------------------------+
| 679 269 4 | 3 8 1 | 25679 569 269 |
| 5 69 17 | @24 69 @24 | 3 8 17 |
| 3 28 1289 | 7 69 5 | 269 169 4 |
+------------------------+------------------------+------------------------ |
"$" marks a {68} UR.
"@" marks an 8-cell {2468} DP. To be clear: r1 has {268}; r4 {468}; r6 {68}; r8 {24}.
Note that the <5>s in c4 are strongly linked, as are the <9>s in c6. Also, if r6c4 is <1> then r3c4 must be <5>, and if r6c4 is NOT <1> then r4c4 must be <5>, both consequences independent of the DPs.
Now... note that <5>r4c4 and <9>r1c6 cannot both be true or else the $ UR results; thus, they are weakly linked. Note also that <5>r3c4 and <9>r4c6 cannot both be false since this would require r6c1 to be <1>, in order to prevent $, making r3c4 <5>, which is a contradiction; thus, they have a strong inferential link.
Medusa coloring from {15}r1c5 gets us here (I didn't notice the possible <7> elimination in r4c3 pointed out by nataraj):
Code: | +--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 268 1G5R 2689 | 5689 1R3569 3689 |
| 289 5 6 | 1R28 7 3 | 89 4 1G89 |
| 89 1 389 | 568 4 689 | 5R7G 2 367R89 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 5G789 | 45R68 2 468 | 1 3679 3689 |
| 1268 2368 25R8 | 9 135G 7 | 4 36 2368 |
| 126789 4 2789 | 1G68 13 68 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 67G9 269 4 | 3 8 1 | 2567R9 569 269 |
| 5 69 1G7R | 24 69 24 | 3 8 1R7G |
| 3 28 1R289 | 7 69 5 | 269 1G69 4 |
+--------------------------+--------------------------+--------------------------+ |
Next, IF RED IS TRUE then <9> in r3c6 must also be true, based on the links described above. Marking this <9> "r", I next mark other resulting determinations (i.e. those from other "r" and "R" values) "r" as well. These can be used for color trapping against "G":
Code: | +--------------------------+--------------------------+--------------------------+
| 4 7 2389r | 2r68 1G5R 268r9 | 5689 1R3569 3689 |
| 2r89 5 6 | 1R28 7 3 | 89 4 1G89 |
| 8r9 1 3r89 |#5G6r-8 4 689r | 5R7G 2 367R89 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 #5G-78-9 | 45R68 2 4r68 | 1 3679 3689 |
| 1268 2368 25R8 | 9 135G 7 | 4 36 2368 |
| 126789 4 2789 |#1G-68r 13 6r8 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 67G9 269 4 | 3 8 1 | 2567R9 569 269 |
| 5 69 1G7R | 24r 69 2r4 | 3 8 1R7G |
| 3 28 1R289 | 7 69 5 | 269 1G69 4 |
+--------------------------+--------------------------+--------------------------+ |
Now: <7> and <9> are removed from r4c3 (due to 5G in r4c3 plus 9r in r1c3 and 7R in r8c3); <8> is removed from r3c4 (due to G and r values in the same cell); and <6> is removed from r6c4 (similarly).
The coloring can now be modified and continued, including the <8> and <9> elimination in r6c3 already noted by nataraj:
Code: | +--------------------------+--------------------------+--------------------------+
| 4 7 2389r | 2r68 1G5R 268r9 | 5689 1R3569 3689 |
| 2r89 5 6 | 1R28 7 3 | 89 4 1G89 |
| 8r9 1 3r89 | 5G6R 4 689r | 5R7G 2 367R89 |
+--------------------------+--------------------------+--------------------------+
| 6789 3689 5G8R | 45R68 2 4r68 | 1 3679 3689 |
| 1268 2368 25R8 | 9 135G 7 | 4 36 2r368r |
| 126789 4 2R7G | 1G8R 13 6r8 | 2689 3679 5 |
+--------------------------+--------------------------+--------------------------+
| 67G9 2r69 4 | 3 8 1 | 2567R9 569 269 |
| 5 69 1G7R | 24r 69 2r4 | 3 8 1R7G |
| 3 28r 1R289 | 7 69 5 | 2r69 1G69 4 |
+--------------------------+--------------------------+--------------------------+ |
We now find two "r" values in cell r5c9, which constitutes a "modified" Medusa "Wrap": All "R" values are false and all "G" values true; but we don't know the status of the "r" values because they were based on the assumption that "R" was true.
This is the grid now:
Code: | +------------------+---------------+------------------+
| 4 7 2389 | 268 1 2689 | 5689 356 3689 |
| 289 5 6 | 28 7 3 | 89 4 1 |
| 89 1 389 | 5 4 689 | 7 2 3689 |
+------------------+---------------+------------------+
| 689 389 5 | 468 2 468 | 1 7 38 |
| 1 238 28 | 9 5 7 | 4 36 2368 |
| 268 4 7 | 1 3 68 | 28 9 5 |
+------------------+---------------+------------------+
| 7 269 4 | 3 8 1 | 2569 56 269 |
| 5 69 1 | 24 69 24 | 3 8 7 |
| 3 28 289 | 7 69 5 | 269 1 4 |
+------------------+---------------+------------------+ |
[Edit: Final grid was not fully reduced.] |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Dec 19, 2007 9:31 am Post subject: |
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Nice going, Asellus.
I went another way, but I,too arrived at r6c1=7. (Must go over my notes again to post here).
In the meantime, taking "the grid now", there are a few box/line eliminations that can be made:
Code: |
+--------------------------+--------------------------+--------------------------+
| 4 7 2389 | 268 1 2689 | 5689 356 3689 |
| 289 5 6 | 28 7 3 | 89 4 1 |
| 89 1 389 | 5 4 689 | 7 2 3689 |
+--------------------------+--------------------------+--------------------------+
| 689 389 5 | 468 2 468 | 1 7 38 |
| 1 238 28 | 9 5 7 | 4 36 2368 |
| 268 4 7 | 1 3 68 | 28 9 5 |
+--------------------------+--------------------------+--------------------------+
| 7 269 4 | 3 8 1 | 2569 56 269 |
| 5 69 1 | 24 69 24 | 3 8 7 |
| 3 2689 289 | 7 69 5 | 269 1 4 |
+--------------------------+--------------------------+--------------------------+ |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 9:58 am Post subject: |
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I caught that Box-Line thing and editted while you were working on it. (Aren't overlapping posts fun?)
I've made more progress. First, there is a Kite pivoting in Box 1 that removes <9> from r9c7. This leads to some more fancy coloring based on a DP:
Code: | +------------------+------------------+------------------+
| 4 7 2389 | 268 1 2689 | 5689 3r56 3689 |
| 289 5 6 | 28 7 3 | 89 4 1 |
| 89 1 3r89 | 5 4 689 | 7 2 3689 |
+------------------+------------------+------------------+
| 689 3G89B 5 | 468 2 468 | 1 7 3R8G |
| 1 @23R8 @28 | 9 5 7 | 4 36 2368 |
| 268 4 7 | 1 3 68 | 28 9 5 |
+------------------+------------------+------------------+
| 7 2B6P 4 | 3 8 1 | 2569 56 269 |
| 5 6B9P 1 | 24 6P9B 24 | 3 8 7 |
| 3 @28 @289B | 7 6B9P 5 | 2B6P 1 4 |
+------------------+------------------+------------------+ |
A {28} UR is marked "@". r5c2 is <3> and/or r9c3 is <9> (strong inference). Coloring these Red-Blue, then coloring from these Red-Green and Blue-Pink, we can eliminate Red-Blue traps.
Once again, I had to extend the coloring by assuming Red to be true and coloring with "r" for "r-B" traps. This eliminates <9> from r3c3. Maybe it'll do some more... I'll keep looking. |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 10:03 am Post subject: |
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The <3> in r4c2 is toast, also, due to the R-B coloring. That helps a bit. |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 10:49 am Post subject: |
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Next, there was a {68} Type 4 UR in r46c16 (<6>s linked in c1) that removed <8> from r4c1. That gets things here:
Code: | +------------+-------------+------------+
| 4 7 289 | 268 1 2689 | 5 3 689 |
| 289 5 6 | 28 7 3 | 89 4 1 |
| 89 1 3 | 5 4 689 | 7 2 689 |
+------------+-------------+------------+
| 69 89 5 | 468 2 468 | 1 7 3 |
| 1 3 28 | 9 5 7 | 4 6 28 |
| 268 4 7 | 1 3 68 | 28 9 5 |
+------------+-------------+------------+
| 7 26 4 | 3 8 1 | 269 5 29 |
| 5 69 1 | 24 69 24 | 3 8 7 |
| 3 28 89 | 7 69 5 | 26 1 4 |
+------------+-------------+------------+ |
And,
(9)r4c1-(9=8)r4c2-(8=2)r5c3-(2=8)r5c9-(8)r6c7=(8-9)r2c7=(9)r2c1-(9)r4c1; r4c1<>9
This solves the puzzle. |
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ravel
Joined: 21 Apr 2006 Posts: 536
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Posted: Wed Dec 19, 2007 11:51 am Post subject: |
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Asellus wrote: | Your nice cycle chain should actually read:
(1)r1c5-(1)r1c8=(1)r9c8-(1=7)r8c9-(7)r3c9=(7-5)r3c7=(5)r3c4-(5=1)r1c5
...
And, you missed some resulting eliminations: (7-5)r3c7 also becomes a conjugate link, and hence a bivalue. So, <6>, <8> and <9> are all eliminated from r3c7. | Thanks for correcting the chain, i am still learning. And yes, i totally missed to use the link inside the cell.
Big step then.
From your last grid after that step it can be solved with uniqueness also. First there is a type 6 UR 56 in r17c78, which takes out 6 from r1c7. Then you have a unique loop 69.
Code: | *---------------------------------------------------*
| 4 7 2389 | 268 1 2689 | 5 36 3689 |
| 289 5 6 | 28 7 3 | 89 4 1 |
| 89 1 389 | 5 4 689 | 7 2 3689 |
|-----------------+----------------+----------------|
| 689 389 5 | 468 2 468 | 1 7 38 |
| 1 238 28 | 9 5 7 | 4 36 2368 |
| 268 4 7 | 1 3 68 | 28 9 5 |
|-----------------+----------------+----------------|
| 7 #269 4 | 3 8 1 |#269 5 29 |
| 5 #69 1 | 24 #69 24 | 3 8 7 |
| 3 28 289 | 7 #69 5 |#269 1 4 |
*---------------------------------------------------*
| One of the 2's in the loop cells must be true => r7c9<>2. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Dec 19, 2007 12:37 pm Post subject: |
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Asellus wrote: | First, there is a Kite pivoting in Box 1 that removes <9> from r9c7 |
I tried, I did. But can't seem to be able to see a kite or any coloring on "9". Where did the 9 in r7c2 go? In "the grid now" it is still there, in my derived position (after box/line) it is still there, in the "fancy coloring based on a DP:" grid it is gone. Of course, that would be a consequence of r9c7<>9, but with that 9 in r7c2 I see no way to remove the 9 from r9c7.
The "9"s in my grid form this pattern:
Code: |
+·····+·····+·····+
· o· *·o o·
· · |· ·
·*-----------# ·
· · |· ·
·@ o· *· o·
+·····+·····+·····+
·*-* · · ·
· · · ·
· · · ·
· · · ·
· · · ·
+·····+·····+·····+
· o · ·o o·
· · · ·
· #-----# · ·
· · | · ·
· o· # ·o ·
+·····+·····+·····+
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Wed Dec 19, 2007 2:35 pm Post subject: |
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Quote: | One of the 2's in the loop cells must be true => r7c9<>2. |
Which solves the puzzle! Congratulations! |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Dec 19, 2007 10:02 pm Post subject: |
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nataraj,
You're not seeing the Kite because you aren't considering strong links with a "grouped end." The "string" of the Kite in r2 is obvious. The other string is in c3: pincer end at r9c3 is strongly linked with the 2 <2>s grouped in Box 1, r13c3. Or...
(9)r9c7-(9)r2c7=(9)r2c1-(9)r13c3=(9)r9c3-(9)r9c7
ravel,
Nice DP work at the end! (I wish I had noticed that Type 6!) It is interesting how much of the solution to this puzzle depended upon various DPs. I guess that's not surprising since DPs induce links that are otherwise not at all evident.
Just so it's clear... the grid from which you last worked with those DPs is the grid that resulted after that involved coloring step utilizing the two DPs in the central tower (not the grid after your earlier AIC loop). |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Thu Dec 20, 2007 5:52 am Post subject: |
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[quote="Asellus"]nataraj,
You're not seeing the Kite because you aren't considering strong links with a "grouped end." The "string" of the Kite in r2 is obvious. The other string is in c3: pincer end at r9c3 is strongly linked with the 2 <2>s grouped in Box 1, r13c3. Or...
(9)r9c7-(9)r2c7=(9)r2c1-(9)r13c3=(9)r9c3-(9)r9c7
Thanks for the explanation, Asellus! Indeed, I did not look at that r13c3 group.
@ravel & Asellus: Congratulations! |
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