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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sun Oct 17, 2010 4:49 am Post subject: Puzzle 10/10/17: C XY |
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| Code: | +-----------------------+
| . . . | . 7 . | . . . |
| . 9 . | . . . | 6 8 . |
| . . . | 6 . . | 7 . 9 |
|-------+-------+-------|
| . . 9 | . . 7 | . . 5 |
| 8 . . | . 6 . | . . . |
| . . . | 1 . 3 | . 6 8 |
|-------+-------+-------|
| . 8 7 | . . . | 4 9 . |
| . 1 . | . . 9 | 8 5 . |
| . . 6 | 8 . 4 | . . 1 |
+-----------------------+
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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: Sun Oct 17, 2010 9:43 am Post subject: |
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this is a solution way out in the left field bleacher seats. enjoy...
| Code: | +----------------------+-------------------+-------------+
| 136-2 346-2 38(2) | 9 7 8(2) | 5 1(2) 34 |
| 7 9 35(2) | 235 345 1 | 6 8 34 |
| 135-2 345-2 358(2) | 6 345 58(2) | 7 1(2) 9 |
+----------------------+-------------------+-------------+
| 236 236 9 | 24 8 7 | 1 34 5 |
| 8 23 1 | 24(5) 6 (25) | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+----------------------+-------------------+-------------+
| 3(5) 8 7 | 3(5) 1 6 | 4 9 2 |
| 234 1 34(2) | 7 23 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+----------------------+-------------------+-------------+ |
this uses an almost x-wing on 2's in r12c68
#1...[x-wing(2)r13c68] = (2-5)r5c6 = (5)r5c4 - (5)r7c4 = (5)r7c1 - (5=2)r9c2 - (2)r8c3 = (2)r123c3; r13c12 <> 2
-------------------
| Code: | +-----------------------+-----------------+-------------+
| 136 346 238 | 9 7 28 | 5 12 34 |
| 7 9 235 | 235 345 1 | 6 8 34 |
| 135 345 2358 | 6 345 258 | 7 12 9 |
+-----------------------+-----------------+-------------+
| 3-6(2) (623) 9 | (24) 8 7 | 1 (34) 5 |
| 8 (23) 1 | (245) 6 25 | 9 (34) 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+-----------------------+-----------------+-------------+
| 3(5) 8 7 | 3(5) 1 6 | 4 9 2 |
| 234 1 34 | 7 23 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+-----------------------+-----------------+-------------+ |
AUR situation here...
AUR on 234 in r45c248 shows that there is a strong link on the 6 in r4c2 and the 5 in r5c4
#2...AUR[(6)r4c2 = (5)r5c4] - (5)r7c4 = (5)r7c1 - (5=2)r9c2 - (2)r45c2 = (2)r4c1; r4c1 <> 6
-------------------------
| Code: | +-----------------+--------------+-----------+
| 6 34 28 | 9 7 28 | 5 1 34 |
| 7 9 235 | 235 345 1 | 6 8 34 |
| 1 345 358 | 6 345 58 | 7 2 9 |
+-----------------+--------------+-----------+
| (23) 6 9 | 24 8 7 | 1 34 5 |
| 8 3-2 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+-----------------+--------------+-----------+
| (35) 8 7 | 35 1 6 | 4 9 2 |
| 34-2 1 34 | 7 23 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+-----------------+--------------+-----------+ |
xy-wing to finish.
#3...(2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2; r5c2 and r8c1 <> 2
------------------
after looking at the xy-wing more carefully, I went back to the initial grid after basics and found this one stepper.
| Code: | +------------------------+-------------------+-------------+
| 123(6) 3-2(46) 238 | 9 7 28 | 5 12 3(4) |
| 7 9 235 | 25(3) 45(3) 1 | 6 8 (34) |
| 1235 2345 2358 | 6 45(3) 258 | 7 12 9 |
+------------------------+-------------------+-------------+
| (236) 36-2 9 | 24 8 7 | 1 34 5 |
| 8 3-2 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+------------------------+-------------------+-------------+
| (35) 8 7 | 35 1 6 | 4 9 2 |
| 34(-2) 1 34(2) | 7 (23) 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+------------------------+-------------------+-------------+ |
notice that if the 6 is false in r4c1 then the xy-wing exists (2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2, this would eliminate the 2's in r45c2 and r8c1...
if the 6 is true, then.
(6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2
so after all that. neither the xy-wing nor the 2 in r9c2 can both be false.
eliminates a number of 2's...
r8c1 <> 2
r45c2 <> 2
r1c2 <> 2
xy-wing[(2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2] = (6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2
(2)r4c1
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(3)r4c1 - (3=5)r7c1 - (5=2)r9c2
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(6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2 |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Oct 17, 2010 4:38 pm Post subject: |
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I had some fun too - but not quite on norm's scale!
My first solution was..
| Code: | x-chain(5) ; (5)r5c6=r3c5 - r3c2=r9c2 - r9c5=r7c4 ; r5c3<>5
xy-wing(35-4) r7c1 ; r6c3<>4, r8c1<>4
ADP(234)r45c248 ; r4c2=6 |
Then inspired by Norm's efforts I found a different one-stepper ...
| Code: | Consider r3 which contains an AHP(34)r3c25 which exists if r3c13<>3.
If it doesn't exist then all other 3 in r3 are eliminated.
If it does exist then 5 is eliminated from r3c2 (and r3c5) and a chain eliminates the same 3s in row 3.
*-----------------------------------------------------------------*
| 1236 2346 238 | 9 7 28 | 5 12 34 |
| 7 9 235 | 235 345 1 | 6 8 34 |
| 12(3)5 2(-34)(5) 2(3)58| 6 (-34)5 258 | 7 12 9 |
|------------------------+--------------------+-------------------|
| 2(3)6 2(3)6 9 | 24 8 7 | 1 34 5 |
| 8 2(3) 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
|------------------------+--------------------+-------------------|
| (3)5 8 7 | 35 1 6 | 4 9 2 |
| 2(3)4 1 2(3)4 | 7 (23) 9 | 8 5 6 |
| 9 2(5) 6 | 8 (25) 4 | 3 7 1 |
*-----------------------------------------------------------------*
(3)r3c13=AHP(34)r3c25:[(34-5)r3c2]=r9c2 - (5=2)r9c5 - (2=3)r8c5* - (3)r8c13=r7c1 - r4c1=r45c2 ; *r3c5<>3, r3c2<>3 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Oct 17, 2010 5:36 pm Post subject: |
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| storm_norm wrote: | after looking at the xy-wing more carefully, I went back to the initial grid after basics and found this one stepper.
| Code: | +------------------------+-------------------+-------------+
| 123(6) 3-2(46) 238 | 9 7 28 | 5 12 3(4) |
| 7 9 235 | 25(3) 45(3) 1 | 6 8 (34) |
| 1235 2345 2358 | 6 45(3) 258 | 7 12 9 |
+------------------------+-------------------+-------------+
| (236) 36-2 9 | 24 8 7 | 1 34 5 |
| 8 3-2 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+------------------------+-------------------+-------------+
| (35) 8 7 | 35 1 6 | 4 9 2 |
| 34(-2) 1 34(2) | 7 (23) 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+------------------------+-------------------+-------------+ |
notice that if the 6 is false in r4c1 then the xy-wing exists (2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2, this would eliminate the 2's in r45c2 and r8c1...
if the 6 is true, then.
(6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2
so after all that. neither the xy-wing nor the 2 in r9c2 can both be false.
eliminates a number of 2's...
r8c1 <> 2
r45c2 <> 2
r1c2 <> 2
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I think you have one more elimination than you listed for the XY-Wing. This defeats a one-stepper solution.
===== ===== ===== ===== ===== =====
Peter: On notation, I can see ...
(3)r3c13=hp(34)r3c25 - (5)r3c2=r9c2 ...
I'm not sure how I'd write it as an AHP statement. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Oct 17, 2010 6:08 pm Post subject: |
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| peterj wrote: | I had some fun too - but not quite on norm's scale!
My first solution was..
| Code: | x-chain(5) ; (5)r5c6=r3c5 - r3c2=r9c2 - r9c5=r7c4 ; r5c3<>5
xy-wing(35-4) r7c1 ; r6c3<>4, r8c1<>4
ADP(234)r45c248 ; r4c2=6 |
Then inspired by Norm's efforts I found a different one-stepper ...
| Code: | Consider r3 which contains an AHP(34)r3c25 which exists if r3c13<>3.
If it doesn't exist then all other 3 in r3 are eliminated.
If it does exist then 5 is eliminated from r3c2 (and r3c5) and a chain eliminates the same 3s in row 3.
*-----------------------------------------------------------------*
| 1236 2346 238 | 9 7 28 | 5 12 34 |
| 7 9 235 | 235 345 1 | 6 8 34 |
| 12(3)5 2(-34)(5) 2(3)58| 6 (-34)5 258 | 7 12 9 |
|------------------------+--------------------+-------------------|
| 2(3)6 2(3)6 9 | 24 8 7 | 1 34 5 |
| 8 2(3) 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
|------------------------+--------------------+-------------------|
| (3)5 8 7 | 35 1 6 | 4 9 2 |
| 2(3)4 1 2(3)4 | 7 (23) 9 | 8 5 6 |
| 9 2(5) 6 | 8 (25) 4 | 3 7 1 |
*-----------------------------------------------------------------*
(3)r3c13=AHP(34)r3c25:[(34-5)r3c2]=r9c2 - (5=2)r9c5 - (2=3)r8c5* - (3)r8c13=r7c1 - r4c1=r45c2 ; *r3c5<>3, r3c2<>3 |
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peterj, very nice.
just one point to make. the 3's in r3c13 act like a group since they are contained within box 1.
since they are strongly linked to the cell (34) in r3c2, then you can also eliminate the 3's in r1c123 and r2c3 since they would also be a contradiction.
and
since you went through and included the group of 3's in box 7 and the group of 3's in box 4, you can now eliminate the 5 in r3c3 because it now is weakly linked to the 3 in r3c3 and the 5 in r3c2 and becomes a contradiction.
and one more...
you start your chain with
(3)r3c13
then on down the chain you have (3)r78c13
so this is true (3)r13c3 = (3)r78c13...
this forms a x-wing within columns 1 and 3
like this...
| Code: | +-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| 3 . 3 | . . . | . . . |
+-------+-------+-------+
|-3 . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
|(3). . | . . . | . . . |
|(3). 3 | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+ |
and the 3 in r4c1 can be eliminated also. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Oct 17, 2010 6:34 pm Post subject: |
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Danny,
yeah, its hard to see but this part of my chain eliminates the 2 in r1c2
(4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2 ; r1c2 <> 2
if that makes it clearer
xy-wing[(2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2] = (6)r4c1 - (6)r1c1 = (6)r1c2 - (4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Oct 17, 2010 7:03 pm Post subject: |
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| Norm wrote: | notice that if the 6 is false in r4c1 then the xy-wing exists (2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2, this would eliminate the 2's in r45c2 and r8c1
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Norm,
Let me restate my position. There's no way you can deduce more than these three eliminations!
Regards, Danny |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Oct 17, 2010 7:04 pm Post subject: |
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| daj95376 wrote: | Peter: On notation, I can see ...
(3)r3c13=hp(34)r3c25 - (5)r3c2=r9c2 ...
I'm not sure how I'd write it as an AHP statement. |
Danny, yes, I can see that's clearer - thanks! I always struggle with writing HP's in AIC. I also recall a heated discussion about the use of the term AHP as opposed to HP.
Norm, I will study your comments. But I don't initially understand the first one - I see that (3)r3c13=r45c2 but not (3)r3c13=r3c2? You seem to be describing a loop but I only see the discontinuity in r3c2. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Oct 17, 2010 7:43 pm Post subject: |
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While reviewing Peter's single-stepper, and cross-checking it with a single stepper SIN from my solver, I came to the conclusion that everything hinges on <5> in [c2].
| Code: | (5-4)r3c2 = (4)r3c5 => r3c25<>3
(5 )r9c2 - (5=3)r7c1 - r4c1 = ( 3)r45c2 => r3c2<>3
|| - (5=2)r9c5 - (2=3)r8c5 => r3c5<>3
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Oct 18, 2010 12:10 am Post subject: |
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Looks like a busy puzzle. Here is my two step offering of BUG-Lites. Cheers 
BUG-Lite+2 (234)r45c248 mixed SIS: r4c1=2,r5c6=2,r13c2=3; r4c2=6
(2)r4c1-(23=6)als:r45c2; r4c2=6
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(2)r5c6-(23=6)als:r45c2; r4c2=6
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(3)r13c2-(23=6)als:r45c2; r4c2=6
BUG-Lite+4 (34)r123c259 with two analysis: internal SIS and external SIS
Internal SIS: r1c2=2,r3c2=5,r23c5=5; r3c6=8
(2)r1c2-r9c2=(2-5)r9c5=r7c4-r5c4=r5c6-(5=8)r3c6; r3c6=8
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(5)r3c2-(5=8)r3c6; r3c6=8
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(5)r23c5-(5=8)r3c6; r3c6=8
External SIS: r5c2=3,r8c5=3; r4c1=2
Bug-Lite+4 (34)r123c259[(3)r5c2=(3)r8c5]-r7c4=r7c1; r4c1-3=2
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Oct 18, 2010 12:45 am Post subject: |
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Potential DP 23-24-34, r4c2=6 or r5c4=5. R5c4<>5 due to invalidity, thus, r4c2=6. Then XY-Wing (354), r6c3<>4.
Alternatively:
Finned X-Wing (5), r3c6<>5
To kill DP 23-24-34, r4c2=6 is only killer
XY-Wing (354), r6c3<>4 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Oct 18, 2010 1:33 am Post subject: |
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| storm_norm wrote: | this is a solution way out in the left field bleacher seats. enjoy...
after looking at the xy-wing more carefully, I went back to the initial grid after basics and found this one stepper.
| Code: | +------------------------+-------------------+-------------+
| 123(6) 3-2(46) 238 | 9 7 28 | 5 12 3(4) |
| 7 9 235 | 25(3) 45(3) 1 | 6 8 (34) |
| 1235 2345 2358 | 6 45(3) 258 | 7 12 9 |
+------------------------+-------------------+-------------+
| (236) 36-2 9 | 24 8 7 | 1 34 5 |
| 8 3-2 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+------------------------+-------------------+-------------+
| (35) 8 7 | 35 1 6 | 4 9 2 |
| 34(-2) 1 34(2) | 7 (23) 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+------------------------+-------------------+-------------+ |
notice that if the 6 is false in r4c1 then the xy-wing exists (2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2, this would eliminate the 2's in r45c2 and r8c1...
if the 6 is true, then.
(6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2
so after all that. neither the xy-wing nor the 2 in r9c2 can both be false.
eliminates a number of 2's...
r8c1 <> 2
r45c2 <> 2
r1c2 <> 2
xy-wing[(2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2] = (6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2 |
Norm, I love those crazy, unique moves; they are what makes sudoku fun for me.
I also do not see how you get r1c2<>2 with the xy-wing. The chain for the fin (6)r4c1 makes the deletion but not the xy-wing.
[Edited to remove invalid suggestion]
Ted
Last edited by tlanglet on Mon Oct 18, 2010 5:19 am; edited 1 time in total |
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ttt
Joined: 06 Dec 2008
Posts: 42
Location: vietnam
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| Posted: Mon Oct 18, 2010 4:59 am Post subject: |
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| tlanglet wrote: | | storm_norm wrote: | | Code: | +------------------------+-------------------+-------------+
| 123(6) 3-2(46) 238 | 9 7 28 | 5 12 3(4) |
| 7 9 235 | 25(3) 45(3) 1 | 6 8 (34) |
| 1235 2345 2358 | 6 45(3) 258 | 7 12 9 |
+------------------------+-------------------+-------------+
| (236) 36-2 9 | 24 8 7 | 1 34 5 |
| 8 3-2 1 | 245 6 25 | 9 34 7 |
| 45 7 45 | 1 9 3 | 2 6 8 |
+------------------------+-------------------+-------------+
| (35) 8 7 | 35 1 6 | 4 9 2 |
| 34(-2) 1 34(2) | 7 (23) 9 | 8 5 6 |
| 9 (25) 6 | 8 25 4 | 3 7 1 |
+------------------------+-------------------+-------------+ |
..........
eliminates a number of 2's...
r8c1 <> 2
r45c2 <> 2
r1c2 <> 2
xy-wing[(2=3)r4c1 - (3=5)r7c1 - (5=2)r9c2] = (6)r4c1 - (6)r1c1 = (6-4)r1c2 = (4)r1c9 - (4=3)r2c9 - (3)r2c45 = (3)r3c5 - (3=2)r8c5 - (2)r8c13 = (2)r9c2 |
I also do not see how you get r1c2<>2 with the xy-wing. The chain for the fin (6)r4c1 makes the deletion but not the xy-wing.
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I think that we can see Norm’s deductions as extending of chain and I like to present as:
(2)r9c2=(2)r8c13-(2=3)r8c5-(3)r23c5=(3)r2c4-(4=3)r2c9-(4)r1c9=(4)r1c2-(6)r1c2=(6)r1c1-(6)r4c1=(XY-wing:235)r47c1/r9c2
Elimination r1c2=2 at (4)r1c2 and extend chain to eliminate the rest at the end of chain.
ttt |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Oct 18, 2010 5:15 am Post subject: |
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Norm & ttt,
I now understand r1c2<>2. The strong link on 4 in r1c29 does the deed. That was particularly insightful Norm.
Ted |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Oct 18, 2010 5:26 am Post subject: |
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remember when you see this...
A=B-C=D-E=F
it also means that A=D
and A=F
C=F
of course A=B
so in my chain the (4)r1c2 is strongly linked to the 2 in r9c2
(4)r1c2 = (2)r9c2; that is the basis for the elimination. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Oct 18, 2010 7:13 pm Post subject: |
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Thanks ttt. Now it makes sense:
| Code: | (2)r9c2 = ... = (4-2|6)r1c2=(6)r1c1-(6)r4c1 = (2)r4c1
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(3)r4c1 - (3=5)r7c1 - (5=2)r9c2
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