Extreme Puzzle 116 Daily Telegraph

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Extreme Puzzle 116 Daily Telegraph

Postby Gee » Sun Dec 14, 2008 2:59 pm

Below is the Daily Telegraph Extreme 116. This is as far as I got with puzzle. It seems like a very difficult puzzle.

At this point I entered the original puzzle into the Scanraid Solver and asked Scanraid to grade it. Much to my surprise it couldn't grade it because, I assume of it's difficulity. Then I ran through the puzzle step-by-step. To my surprise, Scanraid did solve the puzzle. It had the various and sundry AIC's and ALS's but it also had about (5) Bowman's Bingo and (1) Pattern Overlay in the solution. While I have heard of both Bowman’s Bingo and Pattern Overlays , I am not familiar with these techniques and they seem quite complex to use.

My question is, since most of you are familiar with these Extreme Puzzles, did any of you find any solution to this puzzle without resorting to Bowman’s Bingo and the Pattern Overlay? If so, what did techniques did you use to solve the puzzle. What I am asking is, can this puzzle be solved without the Bowman's Bingo and Pattern Overlay?

I would appreciate any comments you may have concerning this puzzle.


Code: Select all

 
 *-----------------------------------------------------------------------------*
 | 34      24      249     | 4569    8       36      | 7       1       259     |
 | 1       6       2489    | 459     49      7       | 2489    2459    3       |
 | 7       348     5       | 1       349     2       | 489     469     689     |
 |-------------------------+-------------------------+-------------------------|
 | 345     1458    148     | 2       14      36      | 1489    4569    7       |
 | 6       9       1478    | 478     5       48      | 1248    3       28      |
 | 2       134578  13478   | 468     1347    9       | 148     456     568     |
 |-------------------------+-------------------------+-------------------------|
 | 9       27      6       | 3       27      1       | 5       8       4       |
 | 8       234     234     | 49      249     5       | 6       7       1       |
 | 45      1457    147     | 78      6       48      | 3       29      29      |
 *-----------------------------------------------------------------------------*
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Postby daj95376 » Sun Dec 14, 2008 4:36 pm

I think you posted the wrong PM.

Code: Select all
         XYZ-Wing [r5c4]/[r5c6]+[r9c4]   <> 8    [r6c4]
 r5  b5  Locked Candidate 1              <> 8    [r5c379]
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Postby storm_norm » Sun Dec 14, 2008 5:26 pm

Gee,

if that PM grid is correct, then you were a hairs breath away from solving it.
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Postby Gee » Mon Dec 15, 2008 12:05 am

Gosh, I don't know what to say! I had a horrible time with this puzzle and I have posted it below as it appeared as Extreme 116. Thank you for your inputs.

Any and all comments appreciated.

Code: Select all

 *-----------*
 |...|.8.|...|
 |16.|..7|..3|
 |7.5|1.2|...|
 |---+---+---|
 |...|2..|..7|
 |.9.|.5.|.3.|
 |2..|..9|...|
 |---+---+---|
 |...|3.1|5.4|
 |8..|...|.71|
 |...|.6.|...|
 *-----------*

 



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Postby m_b_metcalf » Mon Dec 15, 2008 1:50 am

Gee wrote:Gosh, I don't know what to say! I had a horrible time with this puzzle and I have posted it below as it appeared as Extreme 116. Thank you for your inputs.

Any and all comments appreciated.


It starts off with singles, but later becomes very hard, SE 8.4, unusual for a newspaper puzzle.

Regards,

Mike Metcalf
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Postby udosuk » Mon Dec 15, 2008 2:47 am

Gee,

Danny has posted the textbook move (xyz-wing) to advance from your position.

However, using nearly the identical set of cells, here is another way to crack it:

Code: Select all
+-------------------------+-------------------------+-------------------------+
| 34      24      249     | 4569    8       36      | 7       1       259     |
| 1       6       2489    | 459     49      7       | 2489    2459    3       |
| 7       348     5       | 1       349     2       | 489     469     689     |
+-------------------------+-------------------------+-------------------------+
| 345     1458    148     | 2       14      36      | 1489    4569    7       |
| 6       9       1478    |-478     5      *48      | 1248    3       28      |
| 2       134578  13478   | 468     1347    9       | 148     456     568     |
+-------------------------+-------------------------+-------------------------+
| 9       27      6       | 3       27      1       | 5       8       4       |
| 8       234     234     | 49      249     5       | 6       7       1       |
| 45      1457    147     |*78      6      *48      | 3       29      29      |
+-------------------------+-------------------------+-------------------------+

r59c4 form strong link of 7 @ c4
r9c46 form strong link of 8 @ r9,b8
r59c6={48} form strong link of 4 @ c6

As a result, r5c4 can't be 4 (because the 3 strong links above will force r5c6=4).

I guess, this is almost like a strong wing, lacking only the 4th strong link to complete the loop (compared to this one).

Also, if you don't oppose to use uniqueness assumption, these 4 cells form a type of UR:

With the strong link of 8 @ r9c46, a 4 @ r5c4 will force r59c46 to be [4884], a deadly pattern.

(Perhaps some other UR experts can tell you which type of UR it is.)

:idea:
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Postby udosuk » Mon Dec 15, 2008 2:56 am

m_b_metcalf wrote:It starts off with singles, but later becomes very hard, SE 8.4, unusual for a newspaper puzzle.

If the puzzle was made symmetrical by inserting a given clue of 9 @ r8c4, the SE rating would become lower, but IMO still a bit too hard for a newspaper puzzle, even with the "Extreme" branding.:!:
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Postby m_b_metcalf » Mon Dec 15, 2008 5:03 am

udosuk wrote:
m_b_metcalf wrote:It starts off with singles, but later becomes very hard, SE 8.4, unusual for a newspaper puzzle.

If the puzzle was made symmetrical by inserting a given clue of 9 @ r8c4, the SE rating would become lower, but IMO still a bit too hard for a newspaper puzzle, even with the "Extreme" branding.:!:

Sorry, but it stays at 8.4.

Regards,

Mike Metcalf
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Postby aran » Mon Dec 15, 2008 5:57 am

udosuk wrote:Gee,

Danny has posted the textbook move (xyz-wing) to advance from your position.

However, using nearly the identical set of cells, here is another way to crack it:

Code: Select all
+-------------------------+-------------------------+-------------------------+
| 34      24      249     | 4569    8       36      | 7       1       259     |
| 1       6       2489    | 459     49      7       | 2489    2459    3       |
| 7       348     5       | 1       349     2       | 489     469     689     |
+-------------------------+-------------------------+-------------------------+
| 345     1458    148     | 2       14      36      | 1489    4569    7       |
| 6       9       1478    |-478     5      *48      | 1248    3       28      |
| 2       134578  13478   | 468     1347    9       | 148     456     568     |
+-------------------------+-------------------------+-------------------------+
| 9       27      6       | 3       27      1       | 5       8       4       |
| 8       234     234     | 49      249     5       | 6       7       1       |
| 45      1457    147     |*78      6      *48      | 3       29      29      |
+-------------------------+-------------------------+-------------------------+

r59c4 form strong link of 7 @ c4
r9c46 form strong link of 8 @ r9,b8
r59c6={48} form strong link of 4 @ c6

As a result, r5c4 can't be 4 (because the 3 strong links above will force r5c6=4).

I guess, this is almost like a strong wing, lacking only the 4th strong link to complete the loop (compared to this one).

Also, if you don't oppose to use uniqueness assumption, these 4 cells form a type of UR:

With the strong link of 8 @ r9c46, a 4 @ r5c4 will force r59c46 to be [4884], a deadly pattern.

(Perhaps some other UR experts can tell you which type of UR it is.)

:idea:


Another way of removing 4r5c4 :
459r128c4=6r1c4-(6=3)r1c6-(3=6)r4c6-(6=48)r6c4 (=>pair48 r5c6+r6c4) => <4>r5c4
More generally this approach (hidden triple reasoning) will reduce the puzzle fairly quickly eg
A.349r238c5=2r8c5-(2=7)r7c5-(7=2)r7c2-(2=4)r1c2-(4=3)r1c1
B.349r238c5=2r8c5-(2=7)r7c5-(7=8)r9c4-(8=4)r9c6-(4=5)r9c1
A+B=>r1c4=4
=> <4> r4c5 (placement 1 etc)
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Postby Luke » Mon Dec 15, 2008 7:21 am

udosuk wrote:
Code: Select all
+-------------------------+-------------------------+-------------------------+
| 34      24      249     | 4569    8       36      | 7       1       259     |
| 1       6       2489    | 459     49      7       | 2489    2459    3       |
| 7       348     5       | 1       349     2       | 489     469     689     |
+-------------------------+-------------------------+-------------------------+
| 345     1458    148     | 2       14      36      | 1489    4569    7       |
| 6       9       1478    |-478     5      *48      | 1248    3       28      |
| 2       134578  13478   | 468     1347    9       | 148     456     568     |
+-------------------------+-------------------------+-------------------------+
| 9       27      6       | 3       27      1       | 5       8       4       |
| 8       234     234     | 49      249     5       | 6       7       1       |
| 45      1457    147     |*78      6      *48      | 3       29      29      |
+-------------------------+-------------------------+-------------------------+

r59c4 form strong link of 7 @ c4
r9c46 form strong link of 8 @ r9,b8
r59c6={48} form strong link of 4 @ c6

As a result, r5c4 can't be 4 (because the 3 strong links above will force r5c6=4).

I guess, this is almost like a strong wing, lacking only the 4th strong link to complete the loop (compared to this one).

Also, if you don't oppose to use uniqueness assumption, these 4 cells form a type of UR:

With the strong link of 8 @ r9c46, a 4 @ r5c4 will force r59c46 to be [4884], a deadly pattern.

(Perhaps some other UR experts can tell you which type of UR it is.)

Since I'm no expert so I'll have to reference a few: Here's Dan from a "Fiendish Puzzle" thread:
daj95376 wrote:...a candidate can be missing from a PM and still be used by a UR provided it was previously present and no value has been assigned to the cell.

If that concept applies here, then the "missing" 4 in r9c4 would leave you with a type 2 UR, conjugate pairs with one strong link. ("UR + 2B/ 1SL.")
The strong link on 7 would not be required if this approach is legal.
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Postby daj95376 » Mon Dec 15, 2008 8:01 am

Okay, let's back this puppy up to just after basics where Mike Metcalf's SE 8.4 is still applicable. Now, what do you find?

Code: Select all
   c16   X-Wing                          <> 3    [r1c23],[r4c235]
 r5c5    2-String Kite                   <> 7    [r7c3]

 +--------------------------------------------------------------------------------+
 |  349     24      249     |  4569    8       3456    |  7       1       259     |
 |  1       6       2489    |  459     49      7       |  2489    2459    3       |
 |  7       348     5       |  1       349     2       |  489     469     689     |
 |--------------------------+--------------------------+--------------------------|
 |  3456    1458    1468    |  2       14      3468    |  1489    4569    7       |
 |  46      9       14678   |  4678    5       468     |  1248    3       268     |
 |  2       134578  134678  |  4678    1347    9       |  148     456     568     |
 |--------------------------+--------------------------+--------------------------|
 |  69      27      269     |  3       279     1       |  5       8       4       |
 |  8       2345    2349    |  459     249     45      |  6       7       1       |
 |  45      1457    147     |  4578    6       458     |  3       29      29      |
 +--------------------------------------------------------------------------------+
 # 121 eliminations remain
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Postby ronk » Mon Dec 15, 2008 9:18 am

daj95376 wrote:Okay, let's back this puppy up to just after basics where Mike Metcalf's SE 8.4 is still applicable. Now, what do you find?

Code: Select all
 c16   X-Wing                          <> 3    [r1c23],[r4c235]
 r5c5    2-String Kite                 <> 7    [r7c3]

 349    24     249    |  4569   8      3456   |  7      1      259
 1      6      2489   |  459    49     7      |  2489   2459   3
 7      348    5      |  1      349    2      |  489    469    689
----------------------+-----------------------+----------------------
 3456   1458   1468   |  2      14     3468   |  1489   4569   7
 46     9      14678  |  4678   5      468    |  1248   3      268
 2      134578 134678 |  4678   1347   9      |  148    456    568
----------------------+-----------------------+----------------------
 69     27     269    |  3      279    1      |  5      8      4
 8      2345   2349   |  459    249    45     |  6      7      1
 45     1457   147    |  4578   6      458    |  3      29     29

 # 121 eliminations remain

Suspect your question is directed to Gee, but I see an ALS nice loop and one net (due to a kraken cell) leading to 5 placements ...

01) r8c2 -4- r1c2 -2- r7c2 -7- als:r9c123 -4- r8c2 ==> r8c2<>4

02) kraken cell: either r8c2=5 or r8c2=3 or r8c2=2
r8c2 -5- r9c1 =5= r4c1
r8c2 -3- r8c3 =3= r6c3 -3- r4c1
r8c2 -2- r8c5 =2= r7c5 =7= r6c5 =3= r6c23 -3- r4c1 ==> r4c1<>3

[edit: removed whitespace from daj's pencilmarks]
Last edited by ronk on Mon Dec 15, 2008 5:51 am, edited 1 time in total.
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Postby hobiwan » Mon Dec 15, 2008 9:43 am

daj95376 wrote:Okay, let's back this puppy up to just after basics where Mike Metcalf's SE 8.4 is still applicable. Now, what do you find?

Nothing really good (and nothing without Forcing chains). One possibility (of many):
Code: Select all
Forcing Chain Verity => r1c1=3
  r8c2=2 r7c5=2 r6c5=7 r3c5=3 r1c1=3
  r8c2=3 r1c1=3
  r8c2=4 r1c2=2 r7c2=7 r6c5=7 r3c5=3 r1c1=3
  r8c2=5 r4c1=5 r1c1=3

Singles to
.--------------------.----------------.-----------------.
| 3    24      249   | 4569  8    456 | 7     1     259 |
| 1    6       2489  | 459   49   7   | 2489  2459  3   |
| 7    48      5     | 1     3    2   | 489   469   689 |
:--------------------+----------------+-----------------:
| 456  1458    148   | 2     14   3   | 1489  4569  7   |
| 46   9       1478  | 4678  5    468 | 1248  3     268 |
| 2    134578  13478 | 4678  147  9   | 148   456   568 |
:--------------------+----------------+-----------------:
| 9    27      6     | 3     27   1   | 5     8     4   |
| 8    2345    234   | 459   249  45  | 6     7     1   |
| 45   1457    147   | 4578  6    458 | 3     29    29  |
'--------------------'----------------'-----------------'
Forcing Chain Contradiction in r9c8 => r2c5<>4
  r2c5=4 r8c5=9 r7c5=2 r6c5=7 r5c3=7 r5c7=1 r2c7=2 r2c3=8 r3c2=4 r1c23<>4 r1c46=4 r2c5<>4

Singles to
.--------------------.----------------.-----------------.
| 3    24      9     | 456   8    456 | 7     1     25  |
| 1    6       248   | 45    9    7   | 248   245   3   |
| 7    48      5     | 1     3    2   | 489   469   689 |
:--------------------+----------------+-----------------:
| 456  1458    148   | 2     14   3   | 1489  4569  7   |
| 46   9       1478  | 4678  5    468 | 1248  3     268 |
| 2    134578  13478 | 4678  147  9   | 148   456   568 |
:--------------------+----------------+-----------------:
| 9    27      6     | 3     27   1   | 5     8     4   |
| 8    2345    234   | 9     24   45  | 6     7     1   |
| 45   1457    147   | 4578  6    458 | 3     29    29  |
'--------------------'----------------'-----------------'
Forcing Chain Verity => r9c6=8
  r6c2=8 r3c2=4 r1c2=2 r7c2=7 r9c4=7 r9c6=8
  r6c3=8 r4c23<>8 r4c7=8 r2c3=8 r8c3=2 r7c5=2 r9c4=7 r9c6=8
  r6c4=8 r9c6=8
  r6c7=8 r2c3=8 r8c3=2 r7c5=2 r9c4=7 r9c6=8
  r6c9=8 r1c9=5 r1c2=2 r7c2=7 r9c4=7 r9c6=8

2 W-Wings, 1 XY-Chain


I found a solution in two steps, but the second step is a monster (Forcing Chain: r1c1=3, Forcing Net: r5c7=1, Singles).
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Postby udosuk » Mon Dec 15, 2008 5:16 pm

aran wrote:Another way of removing 4r5c4 :
459r128c4=6r1c4-(6=3)r1c6-(3=6)r4c6-(6=48)r6c4 (=>pair48 r5c6+r6c4) => <4>r5c4

Not a huge fans of chains like that, but I wonder why did you include the cells r14c6 when you can have it shorter in the opposite direction:

Code: Select all
+-------------------------+-------------------------+-------------------------+
| 34      24      249     |#4569    8       36      | 7       1       259     |
| 1       6       2489    |#459     49      7       | 2489    2459    3       |
| 7       348     5       | 1       349     2       | 489     469     689     |
+-------------------------+-------------------------+-------------------------+
| 345     1458    148     | 2       14      36      | 1489    4569    7       |
| 6       9       1478    |-478     5      *48      | 1248    3       28      |
| 2       134578  13478   |*468     1347    9       | 148     456     568     |
+-------------------------+-------------------------+-------------------------+
| 9       27      6       | 3       27      1       | 5       8       4       |
| 8       234     234     |#49      249     5       | 6       7       1       |
| 45      1457    147     | 78      6       48      | 3       29      29      |
+-------------------------+-------------------------+-------------------------+

4r5c4-(4=8)r5c6-(48=6)r6c4-(6=459)r128c4-4r5c4 => <4>r5c4

In terms of ALS-xz:
ALS A: r128c4={4569}
ALS B: r5c6+r6c4={468}
restricted common: x=6 (@ r16c4)
common: z=4 (to be eliminated @ r5c4)

aran wrote:More generally this approach (hidden triple reasoning) will reduce the puzzle fairly quickly eg
A.349r238c5=2r8c5-(2=7)r7c5-(7=2)r7c2-(2=4)r1c2-(4=3)r1c1
B.349r238c5=2r8c5-(2=7)r7c5-(7=8)r9c4-(8=4)r9c6-(4=5)r9c1
A+B=>r4c1=4
=> <4> r4c5 (placement 1 etc)

Not sure why you say this move is useful, as the previous elimination (r5c4<>4) is enough to solve the puzzle, while this elimination (r4c5<>4) doesn't help much at all.:?:



Luke451, thank you very much for the elaborations on the UR.:)
udosuk
 
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Postby Draco » Mon Dec 15, 2008 6:58 pm

Another solution (from Gee's PMs) is a contradiction chain, which is I think is just a variation of udosuk's solution (resulting in assignment of a different cell):

r5c9=8 r5c7=2 r5c3=1 r5c4=7 r9c4=8 r9c6=4 r5c6=0 ==> r5c9=2

Singles to solve from there.

Cheers...

- drac
Draco
 
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