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Nightmare, September 20

 
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Marty R.



Joined: 12 Feb 2006
Posts: 5119
Location: Rochester, NY, USA

PostPosted: Wed Oct 11, 2006 11:02 pm    Post subject: Nightmare, September 20 Reply with quote

This is my second attempt and the second is just as big a mess as the first. What am I missing? All I've been able to do is a next-to-useless Finned X-Wing on "6."

Code:
----------------------------------------------------------
|2459  17    245   |357   8     1257  |1247  23479 6     |
|259   3     6     |57    125   4     |127   8     12579 |
|245   17    8     |9     12356 12567 |1247  2347  12457 |
----------------------------------------------------------
|23568 268   235   |1     4     2567  |9     267   278   |
|1     246   7     |8     269   269   |3     5     24    |
|2458  2468  9     |567   256   3     |12478 2467  12478 |
----------------------------------------------------------
|468   5     1     |2     69    689   |478   479   3     |
|7     24689 234   |346   1369  1689  |5     249   2489  |
|2348  2489  234   |345   7     589   |6     1     2489  |
----------------------------------------------------------   
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Thu Oct 12, 2006 3:35 am    Post subject: Go Tigers! Reply with quote

Marty,

Sudoku Susser says you need chains:

5 x Comprehensive Forcing Chains
3 x Simple Forcing Chains
1 x Comprehensive Naked Sets
1 x Unique Rectangles and Loops
3 x Intersection Removal
2 x Simple Hidden Sets
2 x Simple Naked Sets
9 x Pinned Squares

Go Tigers!

Keith
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Thu Oct 12, 2006 8:36 am    Post subject: Reply with quote

Code:
 *-----------*
 |...|.8.|..6|
 |.36|..4|.8.|
 |..8|9..|...|
 |---+---+---|
 |...|14.|9..|
 |1.7|8..|.5.|
 |..9|..3|...|
 |---+---+---|
 |.51|2..|..3|
 |7..|...|5..|
 |...|.7.|61.|
 *-----------*

 *--------------------------------------------------------------------*
 | 2459   17     245    | 357    8      1257   | 1247   23479  6      |
 | 259    3      6      | 57     125    4      | 127    8      12579  |
 | 245    17     8      | 9      12356  12567  | 1247   2347   12457  |
 |----------------------+----------------------+----------------------|
 |-23568 *268    235    |*1      4      2567   | 9     *267    278    |
 | 1     #246    7      | 8      269    269    | 3      5      24     |
 |-24568 *2468   9      |*567    256    3      | 12478 *2467   12478  |
 |----------------------+----------------------+----------------------|
 | 468    5      1      | 2      69     689    | 478    479    3      |
 | 7     *24689  234    |*346    1369   1689   | 5     *249    2489   |
 | 2348   2489   234    | 345    7      589    | 6      1      2489   |
 *--------------------------------------------------------------------*

Marty, you just stopped a tad too soon. The finned swordfish (r468c248 fin in r5c2) on the sixes enables you to solve for the 6 in r7c1. From there it slowly reduces via basics to...
Code:
 *--------------------------------------------------------------------*
 | 59     17     24     | 35     8      127    |-1247   39     6      |
 | 259    3      6      |%57     125    4      |%127   %8     #579    |
 |*245    17     8      | 9      12356  1267   |-1247  -347   *457    |
 |----------------------+----------------------+----------------------|
 | 3      8      5      | 1      4      267    | 9      267    27     |
 |*1     -246    7      | 8      26     9      | 3      5     *24     |
 |#24     246    9      |%567    256    3      |%8     %2467   1      |
 |----------------------+----------------------+----------------------|
 | 6      5      1      |%2      9      8      |%47    %47     3      |
 | 7      249    234    | 346    136    16     | 5      29     8      |
 | 8      249    234    | 34     7      5      | 6      1      29     |
 *--------------------------------------------------------------------*

Then the fours (*) have an r35c19 finned (r6c1) sashimi x-wing.
And the sevens (%) have an r267c478 finned (r2c9) swordfish.

Now it is time for a little multidigit coloring. Start with the bivalue cell in r1c1 and give each digit the conjugate colors, 'B' and 'b'. Extend those conjugate colors as far as possible and you end up with something like this.
Code:
 *----------------------------------------------------------------------*
 | 5B9b    17      24    |*3B5b    8       127   | 124     3b9B    6    |
 | 259B    3       6     |*57      125     4     |*127b    8       579b |
 | 245     17      8     | 9       123b56  1267  | 124     3B4b    57   |
 |-----------------------+-----------------------+----------------------|
 | 3       8       5     | 1       4       267   | 9       267     2B7b |
 | 1       26      7     | 8       26      9     | 3       5       4    |
 | 24      246     9     | 567     256     3     | 8       267     1    |
 |-----------------------+-----------------------+----------------------|
 | 6       5       1     | 2       9       8     | 4b7B    4B7b    3    |
 | 7       249B    234   | 346     13B6    16    | 5       2B9b    8    |
 | 8       249b    234   | 34      7       5     | 6       1       2b9B |
 *----------------------------------------------------------------------*

Note that color 'b' inhibits both the five and the seven from r2c4, therefore color 'b' cannot be true.

One could also find the following related AIC
(7=5)r2c4 - (5=3)r1c4 - (3)r1c8 = (3-4)r3c8 = (4)r7c8 - (4=7)r7c7 =>
chain implies that r2c4 = 7, r7c7 = 7, or both. Since r2c7 sees both r2c4 and r7c7, it cannot contain a 7, and the puzzle is unlocked.

[edited to fix the swordfish marked with %]
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Marty R.



Joined: 12 Feb 2006
Posts: 5119
Location: Rochester, NY, USA

PostPosted: Thu Oct 12, 2006 4:47 pm    Post subject: Reply with quote

Keith/MJ,

Both of you used some unfamiliar terms which I'll have to research. Thanks for taking the time to help.

Marty
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Thu Oct 12, 2006 7:06 pm    Post subject: Reply with quote

Hi Marty,

suppose you are more familiar with these methods:
Code:
 *--------------------------------------------------------------------*
 | 2459   17     245    | 357    8      1257   | 1247   23479  6      |
 | 259    3      6      | 57     125    4      | 127    8      12579  |
 | 245    17     8      | 9      12356  12567  | 1247   2347   12457  |
 |----------------------+----------------------+----------------------|
 | 23568  268    235    | 1      4      2567   | 9      267    278    |
 | 1      246    7      | 8      269    269    | 3      5      24     |
 | 2458   2468   9      | 567    256    3      | 12478  2467   12478  |
 |----------------------+----------------------+----------------------|
 | 468    5      1      | 2      69     689    | 478    479    3      |
 | 7      24689  234    | 346    1369   1689   | 5      249    2489   |
 | 2348   2489   234    | 345    7      589    | 6      1      2489   |
 *--------------------------------------------------------------------*
With 3 strong links you can eliminate 6 from r4c1:
r7c1=r8c2-r8c4=r6c4-r6c8=r4c8
(if 6 not in r7c1 then in r8c2,r6c4 and r4c8, i.e. either r7c1 or r4c8 is 6 => r4c1<>6)
=> r7c1=6

With quadruple and turbot fish (2 strong links) in 4 you come here:
Code:
 *--------------------------------------------------------------------*
 | 59     17     24     | 35     8      127    | 1247   39     6      |
 | 259    3      6      | 57     125    4      | 127    8      579    |
 | 245    17     8      | 9      12356  1267   | 1247   347    57     |
 |----------------------+----------------------+----------------------|
 | 3      8      5      | 1      4      267    | 9      267    27     |
 | 1      26     7      | 8      26     9      | 3      5      4      |
 | 24     246    9      | 567    256    3      | 8      267    1      |
 |----------------------+----------------------+----------------------|
 | 6      5      1      | 2      9      8      | 47     47     3      |
 | 7      249    234    | 346    136    16     | 5      29     8      |
 | 8      249    234    | 34     7      5      | 6      1      29     |
 *--------------------------------------------------------------------*
r3c1=5 => r1c1=9 => r2c9=9 => r3c9=5
=>r3c9<>5

Then with a pair and triple you get r2c57=12 and a turbot fish in 7, which solves the puzzle.
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Marty R.



Joined: 12 Feb 2006
Posts: 5119
Location: Rochester, NY, USA

PostPosted: Thu Oct 12, 2006 10:52 pm    Post subject: Reply with quote

Quote:
With 3 strong links you can eliminate 6 from r4c1:
r7c1=r8c2-r8c4=r6c4-r6c8=r4c8
(if 6 not in r7c1 then in r8c2,r6c4 and r4c8, i.e. either r7c1 or r4c8 is 6 => r4c1<>6)
=> r7c1=6


Ravel,

I don't understand the notation, but I can follow along and see where r4c8=6 if r7c1 doesn't. However, I don't see where three strong links come into play.

Quote:
With quadruple and turbot fish (2 strong links) in 4 you come here:


I'm afraid I don't understand this either.

Sorry.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Fri Oct 13, 2006 7:35 am    Post subject: Reply with quote

Marty,

see here a good description for strong links.
The notation r7c1=r8c2-r8c4=r6c4-r6c8=r4c8 is my own (therefore i explained the logic), "=" is for a strong link (a conjugated pair for the candidate), "-" for a "weak" one, i.e. the two cells share a unit (row, column, box).
So you have: if 6 ist not in r7c1, then it must be in r8c2, then it cannot be in r8c4, then it must be in r6c4 and so on.
If you have such alternating strong and weak links (which also could be strong ones), starting and ending with a strong one, then you can eliminate the number from cells, that see both the first and the last cell in the chain.

This is nothing else than advanced coloring (if all links are strong it is simple coloring), also called x-chains, i.e. simple forcing chains only using one number, nothing difficult for you, so forgive me, when the notation confused you.
You can write it as:
r7c1=6 => r6c1<>6
r7c1<>6 => r8c2=6 => ....=> r4c8=6 => r6c1<>6

Quote:
With quadruple and turbot fish (2 strong links) in 4 you come here
You can find 2 quads (if i remember right) then and can place the 4 in r5c9 after using 2 strong links in 4.
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Marty R.



Joined: 12 Feb 2006
Posts: 5119
Location: Rochester, NY, USA

PostPosted: Fri Oct 13, 2006 4:33 pm    Post subject: Reply with quote

Quote:
You can write it as:
r7c1=6 => r6c1<>6
r7c1<>6 => r8c2=6 => ....=> r4c8=6 => r6c1<>6


This type of logic is something I'm well familiar with.

"Advanced coloring" is a new term to me, presumably something different from "multi-coloring", a technique that I've had trouble grasping. So advanced is a form of simple coloring except for the alternating weak and strong links. But the basic principle appears to be the same; a single chain with an even number of cells so the starting and ending cells are of opposite polarity, so to speak.

I'll add this to the repertoire. Thanks.
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