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Menneske Super Hard (56)

 
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Fri Jul 24, 2009 4:19 pm    Post subject: Menneske Super Hard (56) Reply with quote

I found a one step solution for this puzzle but otherwise it took me seven steps. I don't always solve these puzzles at this level.
Code:

  *-----------*
 |.4.|..2|..7|
 |6.2|.3.|...|
 |.5.|8.4|.2.|
 |---+---+---|
 |.9.|...|2..|
 |...|489|...|
 |..7|...|.9.|
 |---+---+---|
 |...|5.8|.3.|
 |..1|.9.|7.8|
 |2..|3..|.4.|
 *-----------*

Enjoy..............

Ted
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Fri Jul 24, 2009 8:32 pm    Post subject: Reply with quote

this might be a little off the wall...
notice that you can create a pair {13} in row 3 with the 13 in r3c1 and the pseudocell {13} created by the type 3 UR69r39c79... all this is true IF the pair {79} is false in r37c1

in other words the {79} pair r37c1 and the type 3 UR69[(13)r3c179] cannot both be false.
Code:
.------------------.------------------.------------------.
|13-9   4     8    | 169   156   2    | 3569  16    7    |
| 6     17    2    | 179   3     157  | 48    18    1459 |
|*1379  5    *39   | 8     167   4    |U369   2    U1369 |
:------------------+------------------+------------------:
| 1358  9     4    | 167   1567  1357 | 2     168   1356 |
| 135   2     36   | 4     8     9    | 356   7     1356 |
| 358   16    7    | 16    2     35   | 48    9     345  |
:------------------+------------------+------------------:
|*79    67    69   | 5     4     8    | 1     3     2    |
| 4     3     1    | 2     9     6    | 7     5     8    |
| 2     8     5    | 3     17    17   |U69    4    U69   |
'------------------'------------------'------------------'


(79)r37c1 = {type 3 UR69[(13)r179]} - (3=9)r3c3; r1c1 <> 9

for a one stepper.

picture form



Last edited by storm_norm on Sat Jul 25, 2009 12:01 am; edited 2 times in total
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Fri Jul 24, 2009 9:35 pm    Post subject: Reply with quote

Norm, your analysis is extraordinary!

My one step solution also used the UR 69 in r39c79, but I just followed the implications that r3c7=3 or r3c9=13.
(3)r3c79 - (3)r3c3 = (3)r5c3;
(1)r3c9 - (1)r45c9 = (1)r4c8 - (1)r4c456 = (1)r6c4 - (1=6)r6c2 - (6=3)r5c3;

Thus, r5c3=3 if the UR constraints are satisfied.

Ted
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ttt



Joined: 06 Dec 2008
Posts: 42
Location: vietnam

PostPosted: Sat Jul 25, 2009 4:14 am    Post subject: Reply with quote

For AUR(69)r39c79, I did view on other way: at least one of [(9)r3c13, (6)r3c5] must be true
(9)r3c13=(6-7)r3c5=(7)r3c1-(7=9)r7c1 => r1c1<>9

ttt
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sat Jul 25, 2009 6:33 am    Post subject: Reply with quote

ttt, ted,

very nice
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Sat Jul 25, 2009 8:18 am    Post subject: Reply with quote

Not to take anything away from Norm's solution path, which was genius, but here is a little hint I have found that may help the more mortal among us.

When evaluating a UR, if you find yourself considering locked sets with the extra UR digits, take a peek instead at where the base digits have to land when they get moved out of the UR. Oftentimes you can replace an error-prone deduction (with the quantum "cell") with a much simpler equivalent. For example here we have a limited number of escapes for the six and/or nine isolated to r3
Code:
.------------------.------------------.------------------.
|13-9   4     8    | 169   156   2    | 3569  16    7    |
| 6     17    2    | 179   3     157  | 48    18    1459 |
|*1379# 5    *39#  | 8     167#  4    |U369   2    U1369 |
:------------------+------------------+------------------:
| 1358  9     4    | 167   1567  1357 | 2     168   1356 |
| 135   2     36   | 4     8     9    | 356   7     1356 |
| 358   16    7    | 16    2     35   | 48    9     345  |
:------------------+------------------+------------------:
|*79    67    69   | 5     4     8    | 1     3     2    |
| 4     3     1    | 2     9     6    | 7     5     8    |
| 2     8     5    | 3     17    17   |U69    4    U69   |
'------------------'------------------'------------------'

a quick and easy AIC is available using mostly the same cells

(9)r3c12 = (9#2)r39c79 -UR- (6#2)r39c79 = (6-7)r3c5 = (7)r3c1 - (7=9)r7c1 => r1c1 <> 9

notation: (9#2)r39c79 means 9 shows up twice in those cells
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Sat Jul 25, 2009 5:38 pm    Post subject: Reply with quote

I arrived late to the discussion, but decided to see what I'd have done with the UR. Turns out, my approach is essentially identical to Ted's, but with a different elimination.

Code:
 [r3c9]<>1 => <69> UR Type 2 in r37c79 => [r3c3]<>3
 -or-
 (1)r3c9 - r5c9 = r5c1 - (1=6)r6c2 - (6=3)r5c3 - (3)r3c3
 +--------------------------------------------------------------+
 |  139   4     8     |  169   156   2     |  3569  16    7     |
 |  6     17    2     |  179   3     157   |  48    18    1459  |
 |  1379  5     9-3   |  8     167   4     | *369   2    *369+1 |
 |--------------------+--------------------+--------------------|
 |  1358  9     4     |  167   1567  1357  |  2     168   1356  |
 |  135   2     36    |  4     8     9     |  356   7     1356  |
 |  358   16    7     |  16    2     35    |  48    9     345   |
 |--------------------+--------------------+--------------------|
 |  79    67    69    |  5     4     8     |  1     3     2     |
 |  4     3     1     |  2     9     6     |  7     5     8     |
 |  2     8     5     |  3     17    17    | *69    4    *69    |
 +--------------------------------------------------------------+
 # 70 eliminations remain
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