Vanhegan Fiendish March 23, 2013

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Vanhegan Fiendish March 23, 2013

Postby ArkieTech » Sat Mar 23, 2013 2:12 am

Code: Select all
 *-----------*
 |9.5|..1|..8|
 |.42|...|.9.|
 |...|9..|.23|
 |---+---+---|
 |5..|476|8..|
 |...|2.8|...|
 |..7|539|..4|
 |---+---+---|
 |65.|..3|...|
 |.7.|...|34.|
 |3..|8..|7.1|
 *-----------*


Play/Print this puzzle online
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Re: Vanhegan Fiendish March 23, 2013

Postby 7b53 » Sat Mar 23, 2013 3:16 am

how about solving this one without PMs....

Code: Select all
 *-----------------*
 |9 3 5|6 2 1|4 7 8|
 |. 4 2|3 . 7|. 9 .|
 |7 . .|9 . 4|. 2 3|
 |-----+-----+-----|
 |5 . .|4 7 6|8 . .|
 |4 . .|2 1 8|. . 7|
 |. . 7|5 3 9|. . 4|
 |-----+-----+-----|
 |6 5 1|7 4 3|. 8 .|
 |. 7 .|1 . .|3 4 .|
 |3 . 4|8 . .|7 . 1|
 *-----------------*
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Re: Vanhegan Fiendish March 23, 2013

Postby JC Van Hay » Sat Mar 23, 2013 7:54 am

1. After Singles, further Single Digit Eliminations will only be done by Locked Candidates.
2. There are Pairs in B289. For other N-tuples ..., wait and see.
3. Even with PMs, I am unable to spot in no time the best next step, as I need to carefully analyze at least the B/B-Plot.
4. However, R89 only contains bivalues without partition :=> the puzzle will be solved by first testing their candidates.
5. So, in no time : r8c1=2->solution, while r8c1=8->contradiction [For example : R89 + C8 -> r4c3 is empty].
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Re: Vanhegan Fiendish March 23, 2013

Postby ArkieTech » Sat Mar 23, 2013 10:35 am

Code: Select all
 *--------------------------------------------------*
 | 9    3    5    | 6    2    1    | 4    7    8    |
 | 18   4    2    | 3    58   7    | 156  9    56   |
 | 7    168  68   | 9    58   4    | 15   2    3    |
 |----------------+----------------+----------------|
 | 5    12   39   | 4    7    6    | 8    13   29   |
 | 4    69   369  | 2    1    8    | 59   35   7    |
 | 128  128  7    | 5    3    9    | 26   16   4    |
 |----------------+----------------+----------------|
 | 6    5    1    | 7    4    3    | 29   8    29   |
 | 28   7    89   | 1    69   25   | 3    4    56   |
 | 3    29   4    | 8    69   25   | 7    56   1    |
 *--------------------------------------------------*
als xy-wing
(9=2)r9c2-(2=1)r4c2-(1=9)r4c38 => -9r5c2,r8c3; ste
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Re: Vanhegan Fiendish March 23, 2013

Postby Leren » Sat Mar 23, 2013 10:55 am

Code: Select all
*-----------------------------------------------------*
| 9    3    5     | 6    2    1     | 4    7    8     |
| 18   4    2     | 3    58   7     | 156  9    56    |
| 7    168  68    | 9    58   4     | 15   2    3     |
|-----------------+-----------------+-----------------|
| 5   a12  c39    | 4    7    6     | 8    13  b29    |
| 4   d69   369   | 2    1    8     | 59   35   7     |
| 128  128  7     | 5    3    9     | 26   16   4     |
|-----------------+-----------------+-----------------|
| 6    5    1     | 7    4    3     | 29   8    29    |
| 28   7    89    | 1    69   25    | 3    4    56    |
| 3   e9-2  4     | 8    69   25    | 7    56   1     |
*-----------------------------------------------------*

-2 r4c2 = (2-9) r4c9 = r4c3 - r5c2 = r9c2 => -2 r9c2; stte

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Re: Vanhegan Fiendish March 23, 2013

Postby Marty R. » Sat Mar 23, 2013 3:47 pm

Code: Select all
*-----------------------------------------------------*
| 9    3    5     | 6    2    1     | 4    7    8     |
| 18   4    2     | 3    58   7     | 156  9    56    |
| 7    168  68    | 9    58   4     | 15   2    3     |
|-----------------+-----------------+-----------------|
| 5    12   39    | 4    7    6     | 8    13   29    |
| 4    69   369   | 2    1    8     | 59   35   7     |
| 128  128  7     | 5    3    9     | 26   16   4     |
|-----------------+-----------------+-----------------|
| 6    5    1     | 7    4    3     | 29   8    29    |
| 28   7    89    | 1    69   25    | 3    4    56    |
| 3    29   4     | 8    69   25    | 7    56   1     |
*-----------------------------------------------------*


M-Wing (9=2)r9c2-r4c2=(2-9)r4c9=r5c7=>r5c2<>9
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Re: Vanhegan Fiendish March 23, 2013

Postby 7b53 » Sun Mar 24, 2013 2:17 am

JC Van Hay wrote:the puzzle will be solved by first testing their candidates.
r8c1=2->solution, while r8c1=8->contradiction [For example : R89 + C8 -> r4c3 is empty].

interesting... but isn't this kind of like trial and error ?
I found the two end cells of digit 9 (r4c3 and r9c2). same eliminations as ArkieTech.

A solver most likely will call for an xy-chain. there are couple of them in fact.
question is can one spot them without PMs ? it's very limited without pencilmarks, isn't it ?
and of course there's always exception.
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Re: Vanhegan Fiendish March 23, 2013

Postby eleven » Sun Mar 24, 2013 1:56 pm

Yes, it is not one of the puzzles, which can be solved easier without pencilmarks, but there are many ways how to do it.
Code: Select all
 *-----------------*
 |9 3 5|6 2 1|4 7 8|
 |. 4 2|3 . 7|. 9 .|
 |7 . .|9 . 4|. 2 3|
 |-----+-----+-----|
 |5 x .|4 7 6|8 . .|
 |4 - .|2 1 8|- . 7|
 |x x 7|5 3 9|o . 4|
 |-----+-----+-----|
 |6 5 1|7 4 3|# 8 .|
 |. 7 .|1 . .|3 4 .|
 |3 # 4|8 . .|7 . 1|
 *-----------------*

Just one example:
You should have noticed the 128 triple in box 4 (with the 8 locked to r6c12).
You should know that r7c7 and r9c2 only can be 29.
You should have noticed that one of them must be 9 (2 strong links in c27).
Then there is a simple chain r7c7=9->r6c7=2->r4c2=2->r9c2=9.
So r9c2 must be 9.
I know, that this sounds complicated for people who are used to pencilmarks. But how much time did you invest for practicing to solve without pencilmarks ?
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Re: Vanhegan Fiendish March 23, 2013

Postby 7b53 » Sun Mar 24, 2013 11:18 pm

eleven wrote:You should have noticed the 128 triple in box 4 (with the 8 locked to r6c12).
You should know that r7c7 and r9c2 only can be 29.
You should have noticed that one of them must be 9 (2 strong links in c27).
Then there is a simple chain r7c7=9->r6c7=2->r4c2=2->r9c2=9.
So r9c2 must be 9.

nice to see another path.
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