Vanhegan fiendish November 11, 2012

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Vanhegan fiendish November 11, 2012

Postby ArkieTech » Sun Nov 11, 2012 7:28 am

Code: Select all
 *-----------*
 |...|5.3|.7.|
 |...|6.8|..4|
 |6..|79.|.35|
 |---+---+---|
 |3..|1..|4..|
 |42.|...|.87|
 |..7|..2|..9|
 |---+---+---|
 |91.|.74|..3|
 |2..|9.5|...|
 |.5.|2.6|...|
 *-----------*


Play/Print this puzzle online
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Re: Vanhegan fiendish November 11, 2012

Postby tlanglet » Sun Nov 11, 2012 12:49 pm

Code after basics:
Code: Select all
 *-----------------------------------------------------------*
 | 18    89    1289  | 5     4     3     | 268   7     268   |
 | 57    37    35    | 6     2     8     | 19    19    4     |
 | 6     4     28    | 7     9     1     | 28    3     5     |
 |-------------------+-------------------+-------------------|
 | 3     689   589   | 1     568   7     | 4     256   26    |
 | 4     2     15    | 3     56    9     | 156   8     7     |
 | 158   68    7     | 4     568   2     | 3     156   9     |
 |-------------------+-------------------+-------------------|
 | 9     1     6     | 8     7     4     | 25    25    3     |
 | 2     37    348   | 9     13    5     | 678   46    168   |
 | 78    5     348   | 2     13    6     | 789   49    18    |
 *-----------------------------------------------------------*

My first puzzle after a vacation, and it was difficult for me. I only found a few moves and they were not effective until I found a winner....

7r9c1=(7-9)r9c7=(9-1)r2c7=r5c7-r5c3=r1c3-(1=8=7)r19c1 => 7r9c1
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Re: Vanhegan fiendish November 11, 2012

Postby ArkieTech » Sun Nov 11, 2012 2:03 pm

tlanglet wrote:My first puzzle after a vacation, and it was difficult for me. I only found a few moves and they were not effective until I found a winner....

7r9c1=(7-9)r9c7=(9-1)r2c7=r5c7-r5c3=r1c3-(1=8=7)r19c1 => 7r9c1


Nice! :D another direction same result

Code: Select all
 *-----------------------------------------------------------*
 | 18    89    1289  | 5     4     3     |d268   7     268   |
 | 57   a37   b35    | 6     2     8     | 19    19    4     |
 | 6     4     28    | 7     9     1     |d28    3     5     |
 |-------------------+-------------------+-------------------|
 | 3     689   589   | 1     568   7     | 4     256   26    |
 | 4     2    c15    | 3     56    9     |d156   8     7     |
 | 158   68    7     | 4     568   2     | 3     156   9     |
 |-------------------+-------------------+-------------------|
 | 9     1     6     | 8     7     4     |d25    25    3     |
 | 2     37    348   | 9     13    5     |d678   46    168   |
 | 78    5     348   | 2     13    6     | 789   49    18    |
 *-----------------------------------------------------------*
als aic
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-(1r5c7=7r8c7)als:r13578c7 => -7r8c2; lclste
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Re: Vanhegan fiendish November 11, 2012

Postby daj95376 » Sun Nov 11, 2012 5:21 pm

_
Interpreting an interesting network found by my solver. (-or- Kraken Cell on r9c7)

Code: Select all
 +--------------------------------------------------------------+
 |  18    89    1289  |  5     4     3     |  268   7     268   |
 |  57    37    35    |  6     2     8     |  1-9   19    4     |
 |  6     4    *28    |  7     9     1     | *28    3     5     |
 |--------------------+--------------------+--------------------|
 |  3     689   589   |  1     568   7     |  4     256   26    |
 |  4     2     15    |  3     56    9     |  156   8     7     |
 |  158   68    7     |  4     568   2     |  3     156   9     |
 |--------------------+--------------------+--------------------|
 |  9     1     6     |  8     7     4     |  25    25    3     |
 |  2     37    348   |  9     13    5     |  678   46    168   |
 | *78    5     348   |  2     13    6     | *78+9  49    18    |
 +--------------------------------------------------------------+
 # 55 eliminations remain

 a) (9)r9c7 = (78)r9c17 + (28)r3c37;     -> (b)

 b) Skyscraper on <8> in (*);            -> (c)

 c) (8=1)r1c1 - r6c1 = r6c8 - (1=9)r2c8  =>     r2c7<>9
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Re: Vanhegan fiendish November 11, 2012

Postby Marty R. » Mon Nov 12, 2012 2:24 am

Code: Select all
    *-----------------------------------------------------------*
     | 18    89    1289  | 5     4     3     | 268   7     268   |
     | 57    37    35    | 6     2     8     | 19    19    4     |
     | 6     4     28    | 7     9     1     | 28    3     5     |
     |-------------------+-------------------+-------------------|
     | 3     689   589   | 1     568   7     | 4     256   26    |
     | 4     2     15    | 3     56    9     | 156   8     7     |
     | 158   68    7     | 4     568   2     | 3     156   9     |
     |-------------------+-------------------+-------------------|
     | 9     1     6     | 8     7     4     | 25    25    3     |
     | 2     37    348   | 9     13    5     | 678   46    168   |
     | 78    5     348   | 2     13    6     | 789   49    18    |
     *-----------------------------------------------------------*




(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-(9=4)r9c8-(4=6)r8c8-(6=18)r89c9-????

I needed four moves, but just monkeying around practicing notation with what I see here. What's the end of my string? At this point we've established that r8c8=6 and r89c9=18, so r8c7 must be=7. Would a term like this be legal? (168r8c89,r9c9=7)r8c7?
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Re: Vanhegan fiendish November 11, 2012

Postby tlanglet » Mon Nov 12, 2012 4:40 am

Marty R. wrote:
Code: Select all
    *-----------------------------------------------------------*
     | 18    89    1289  | 5     4     3     | 268   7     268   |
     | 57    37    35    | 6     2     8     | 19    19    4     |
     | 6     4     28    | 7     9     1     | 28    3     5     |
     |-------------------+-------------------+-------------------|
     | 3     689   589   | 1     568   7     | 4     256   26    |
     | 4     2     15    | 3     56    9     | 156   8     7     |
     | 158   68    7     | 4     568   2     | 3     156   9     |
     |-------------------+-------------------+-------------------|
     | 9     1     6     | 8     7     4     | 25    25    3     |
     | 2     37    348   | 9     13    5     | 678   46    168   |
     | 78    5     348   | 2     13    6     | 789   49    18    |
     *-----------------------------------------------------------*




(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-(9=4)r9c8-(4=6)r8c8-(6=18)r89c9-????

I needed four moves, but just monkeying around practicing notation with what I see here. What's the end of my string? At this point we've established that r8c8=6 and r89c9=18, so r8c7 must be=7. Would a term like this be legal? (168r8c89,r9c9=7)r8c7?


Marty, the five cells involved in box 9 form an als(146789)r8c789|r9c89 so the AIC could be written as:
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-als(9=7)r8c789|r9c89 => r8c2<>7

Dan has been using another variation of noting an als as follows:
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-(9r9c8=7r8c7)als:r8c789|r9c89 => r8c2<>7

Still another variation is:
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-(9r9c8=7r8c7)als(146789)r8c789|r9c89 => r8c2<>7

Maybe someone else from the Players' Forum could offer a suggestion about the als notation.

Ted
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Re: Vanhegan fiendish November 11, 2012

Postby ronk » Mon Nov 12, 2012 5:56 am

tlanglet wrote:Marty, the five cells involved in box 9 form an als(146789)r8c789|r9c89 so the AIC could be written as:
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-als(9=7)r8c789|r9c89 => r8c2<>7

Dan has been using another variation of noting an als as follows:
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-(9r9c8=7r8c7)als:r8c789|r9c89 => r8c2<>7

When the number of ALS cells is large, it's probably time to consider using the complementary AHS:

(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=r6c8-(1=9)r2c8-r9c8=(9-7)r9c7=7r8c7 => r8c2<>7

... or the slightly shortened ...

(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-r5c7=(1-9)r2c7=(9-7)r9c7=7r8c7 => r8c2<>7
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Re: Vanhegan fiendish November 11, 2012

Postby Marty R. » Mon Nov 12, 2012 6:52 pm

Thank you Ted and Ron.

I don't think I can use Ted's suggestions because I I've looked at ALS in the past and just don't understand them, so I wouldn't know to put them in the notation. Therefore, I'd have to use Ron's example, although I think I'd get stuck upon encountering a similar situation.

It's a lot easier executing the moves than writing them down. :?
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Re: Vanhegan fiendish November 11, 2012

Postby 7b53 » Mon Nov 12, 2012 7:29 pm

can one actually solved these type of puzzles without pencil marks ?
if yes, what techniques are suitable for non pencil markers ? :idea: :)
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Re: Vanhegan fiendish November 11, 2012

Postby ArkieTech » Mon Nov 12, 2012 9:15 pm

Marty R. wrote: I've looked at ALS in the past and just don't understand them, so I wouldn't know to put them in the notation.


7b63 wrote:what techniques are suitable for non pencil markers ?


Code: Select all
 *-----------------------------------------------------------*
 | 18    89    1289  | 5     4     3     |d268   7     268   |
 | 57   a37   b35    | 6     2     8     | 19    19    4     |
 | 6     4     28    | 7     9     1     |d28    3     5     |
 |-------------------+-------------------+-------------------|
 | 3     689   589   | 1     568   7     | 4     256   26    |
 | 4     2    c15    | 3     56    9     |d156   8     7     |
 | 158   68    7     | 4     568   2     | 3     156   9     |
 |-------------------+-------------------+-------------------|
 | 9     1     6     | 8     7     4     |d25    25    3     |
 | 2     3-7   348   | 9     13    5     |d678   46    168   |
 | 78    5     348   | 2     13    6     | 789   49    18    |
 *-----------------------------------------------------------*
als aic
(7=3)r2c2-(3=5)r2c3-(5=1)r5c3-(1r5c7=7r8c7)als:r13578c7 => -7r8c2; lclste


if r5c3 were 1 r5c7 would not be 1 then r2c7 would be 1 and r9c7 would be 9
this would make r8c7 a 7.

So if r2c2 was not a 7 then r8c7 would be a 7 therefore r8c2 could not be 7

I know there must be a simpler way to notate this with Eureka but I know know how. :?
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Re: Vanhegan fiendish November 11, 2012

Postby 7b53 » Wed Nov 14, 2012 4:32 am

hey hey, since when there is a "7b63" ? :D
is there a number missing on the keyboard ? :lol:
ah.............. a little joke .
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Re: Vanhegan fiendish November 11, 2012

Postby denis_berthier » Wed Nov 14, 2012 10:18 am

I don't know how you count the length of the patterns proposed in some of the above solutions, but it seems the following uses shorter chains (largest pattern length = 5).

***** SudoRules 16.2 based on CSP-Rules 1.2, config: W *****
singles ==> r7c4 = 8, r7c3 = 6, r5c6 = 9, r4c6 = 7, r5c4 = 3, r6c4 = 4, r3c6 = 1, r2c5 = 2, r1c5 = 4, r6c7 = 3
biv-chain[2]: c2n7{r2 r8} - c2n3{r8 r2} ==> r2c2 <> 9
biv-chain[2]: r2c8{n1 n9} - r2c7{n9 n1} ==> r2c3 <> 1, r2c1 <> 1
whip[1]: r2n1{c8 .} ==> r1c7 <> 1, r1c9 <> 1
whip[1]: c9n1{r9 .} ==> r9c8 <> 1, r9c7 <> 1, r8c8 <> 1, r8c7 <> 1
whip[2]: r2c7{n9 n1} - r2c8{n1 .} ==> r2c3 <> 9
whip[1]: r2n9{c8 .} ==> r1c7 <> 9
whip[2]: c2n3{r8 r2} - c2n7{r2 .} ==> r8c2 <> 4
hidden-single-in-a-column ==> r3c2 = 4
whip[2]: c2n7{r8 r2} - c2n3{r2 .} ==> r8c2 <> 8
whip[2]: r3n8{c7 c3} - b7n8{r9c3 .} ==> r9c7 <> 8
whip[4]: c3n4{r9 r8} - b7n8{r8c3 r9c1} - r9c9{n8 n1} - r9c5{n1 .} ==> r9c3 <> 3
singles ==> r9c5 = 3, r8c5 = 1, r9c9 = 1
whip[1]: b9n8{r8c7 .} ==> r8c3 <> 8
biv-chain[5]: b9n9{r9c7 r9c8} - r2c8{n9 n1} - r6n1{c8 c1} - c1n5{r6 r2} - c1n7{r2 r9} ==> r9c7 <> 7
singles to the end
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