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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Sun Nov 11, 2012 1:01 am Post subject: Still more Eureka |
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This is the Fiendish from Nov. 9 that Dan posted on the other forum. How would my first move be notated? The 9 in A makes B=4 which creates an 89 pair in C which takes out the 9 from D and then 59 from E. I'm fairly confident that what I wrote is incorrect.
| Code: | *-------------------------------------------------------------*
| 7 56 1 | 89 2 56 | 89 3 4 |
| 689 3 4 | 7 89 56 | 2 589 1 |
| 89 2 589 | 1 4 3 | 7 589 6 |
|---------------------+-------------------+-------------------|
| 2 D59 E569 | 39 7 4 | 139 1-69 8 |
| 1 479 -679 | 2389 89 28 | 349 A69 5 |
| 3 C489 C89 | 6 5 1 |B49 2 7 |
|---------------------+-------------------+-------------------|
| 4 678 3 | 5 16 9 | 18 178 2 |
| 689 689 2 | 48 16 7 | 5 148 3 |
| 5 1 78 | 248 3 28 | 6 478 9 |
*-------------------------------------------------------------*
(6=9)r5c8-(9=4)r6c7-(4=NP89)-(9=5)r4c2-(59=6)r4c3-->r5c8, r6c3<>6 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Nov 11, 2012 10:51 am Post subject: Re: Still more Eureka |
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| Marty R. wrote: | This is the Fiendish from Nov. 9 that Dan posted on the other forum. How would my first move be notated? The 9 in A makes B=4 which creates an 89 pair in C which takes out the 9 from D and then 59 from E. I'm fairly confident that what I wrote is incorrect.
| Code: | *-------------------------------------------------------------*
| 7 56 1 | 89 2 56 | 89 3 4 |
| 689 3 4 | 7 89 56 | 2 589 1 |
| 89 2 589 | 1 4 3 | 7 589 6 |
|---------------------+-------------------+-------------------|
| 2 D59 E569 | 39 7 4 | 139 1-69 8 |
| 1 479 -679 | 2389 89 28 | 349 A69 5 |
| 3 C489 C89 | 6 5 1 |B49 2 7 |
|---------------------+-------------------+-------------------|
| 4 678 3 | 5 16 9 | 18 178 2 |
| 689 689 2 | 48 16 7 | 5 148 3 |
| 5 1 78 | 248 3 28 | 6 478 9 |
*-------------------------------------------------------------*
(6=9)r5c8-(9=4)r6c7-(4=NP89)-(9=5)r4c2-(59=6)r4c3-->r5c8, r6c3<>6 |
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Three comments:
1: You forgot to include the cells involved in the NP89 term; otherwise it is a valid statement although it is not necessary to include the notation "NP".
(6=9)r5c8-(9=4)r6c7-(4=89)r6c23-(9=5)r4c2-(59=6)r4c3-->r5c8, r6c3<>6
2: The next two terms dealing with the NP59 are awkward but probably understandable. An alternative is to treat the NP59 in the same manner as the prior NP89.
(6=9)r5c8-(9=4)r6c7-(4=89)r6c23-(59=6)r4c23-->r5c8, r6c3<>6
3: Another alternative for the notation is to treat the four cells in box 4 as a single als(45689).
(6=9)r5c8-(9=4)r6c7-(4=6)r64c23 --> r5c8, r6c3<>6
I have also seen other posts that include the "als" notation when the als is large/involved.
(6=9)r5c8-(9=4)r6c7-als(4=6)r65c23 --> r5c8, r6c3<>6
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 11, 2012 6:03 pm Post subject: |
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Thanks Ted and welcome back.  |
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