dailysudoku.com Forum Index dailysudoku.com
Discussion of Daily Sudoku puzzles
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Puzzle 10/09/04: A

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Puzzles by daj
View previous topic :: View next topic  
Author Message
daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Sat Sep 04, 2010 5:18 am    Post subject: Puzzle 10/09/04: A Reply with quote

Code:
 +-----------------------+
 | . . . | . . . | . 7 . |
 | . 6 . | 9 . 2 | 8 4 . |
 | . . 8 | 7 . 6 | 9 . 5 |
 |-------+-------+-------|
 | . 8 7 | . . 9 | . . . |
 | . . . | . 1 . | . . . |
 | . 2 4 | 6 . 8 | . . 9 |
 |-------+-------+-------|
 | . 3 1 | . . . | 2 . . |
 | 2 7 . | . . . | . 9 . |
 | . . 9 | . . 1 | . . 8 |
 +-----------------------+

Play this puzzle online at the Daily Sudoku site
Back to top
View user's profile Send private message
peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK

PostPosted: Sat Sep 04, 2010 8:14 am    Post subject: Reply with quote

Initially two moves
Quote:
xyz-wing(46-5) r8c5 ; r8c6<>5
x-wing(5)
Concurrent to these two moves I noticed a shortcut...
Quote:
kraken x-wing(5) r28c35 fin (5)r8c6 - r1c6=r2c5 ; r46c5<>5
Back to top
View user's profile Send private message
JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Sat Sep 04, 2010 10:12 am    Post subject: Reply with quote

Peter, nice shortcut. Even if I noticed its possibility, I didn't take the time to directly examine its consequence ! Instead, I took a longer journey and I had fun with this puzzle. I hope the following is of some interest ...

1st run : 1 step solution (Fishes useless)

Empty Rectangle : (5)R2B7 : =>r9c5<>5
X Wing (7)C69/r57 : => r5c7<>7
    After colouring the bivalues in B2 (GEM), one obtains : 5-SIS AIC : (75)R6C5 (53)R2C5 (35)R2C3 (56)R8C3 6R9 : (7)r6c5=(6)r9c5 : => r9c5<>7
2nd run : smaller depth AIC
    XYZ Wing (4-56), pivot at R8C5 : r8c6<>5
    X Wing (5)R28/c35 : r456c35<>5
3rd run : single digit exclusions on 5

The 2 1st runs are equivalent to r6c5<>5 and furthermore there are a lot of exclusions on 5. So, trying r6c5=5, it appears that there is no place for a 5 in box B8 after a network of inferences on the candidates for 5. Writing the Forbiding Matrix for such a net, one obtains the following single step :
    Mutant Starfish (5)R2B2789/r79c356 with endofin r2c5 or t chain (5)R2B2789 : (5)r2c5=(5)r8c5 : => r469c5<>5
The Starfish may be simplified into the
    Franken Swordfish (5)R28B2/c356 with endofin r2c5 : => r469c5<>5 Very Happy ...
JC
Back to top
View user's profile Send private message
peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK

PostPosted: Sat Sep 04, 2010 11:43 am    Post subject: Reply with quote

JC, I am unclear as to how to go about colouring bivalues - is that what Sudopedia refers to as Medusa colouring? I just follow them around the grid trying to make pincers - which I why I rarely spot moves like your first one where the chain ends on the cell where you make the elimination.
Back to top
View user's profile Send private message
JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Sat Sep 04, 2010 7:02 pm    Post subject: Reply with quote

peterj wrote:
JC, I am unclear as to how to go about colouring bivalues - is that what Sudopedia refers to as Medusa colouring? ....

Yes, indeed. Here are some details, certainly written many times in an equivalent manner elsewhere.

While pencilmarking I simultaneously do Locked (hidden, then naked) Subsets and Single Digit Exclusions (as complete as possible). I also tag the bi-locals (. for bi-locals in a box or locked candidates, - for bi-locals in a row and | for bi-locals in a column). The tagging helps me, in particular, to identify the "hub" cells (cells containing at least 2 candidates belonging to bi-locals). The "hub" cells, the bivalue cells and the X Chain snippets are the essential ingredients of simple AICs (without groups and ALS), as well as the basis of Medusa Colouring. As an AIC may be read in both directions from any node, it may be engineered from any bivalue cell or from 2 candidates outside the "hub" cell (spokes). Colouring the nodes of an AIC is then a convenient way to trace it.

Example 1 : puzzle 10/08/04 A, before exclusions by Fishes (1st run) [tagging not included]

Code:
+-----------------------------------------------------------------+
| 459    459    2     | 1      8      4a5A  | 36     7      36    |
| 7      6      35A   | 9      3A5a   2     | 8      4      1     |
| 34     1      8     | 7      3a4A   6     | 9      2      5     |
+-----------------------------------------------------------------+
| 1356   8      7     | 2345   245    9     | 3456   1356   236   |
| 3569   59     356   | 2345   1      3457  | 34567  8      2367  |
| 135    2      4     | 6      57a    8     | 357    135    9     |
+-------------------   -------------------------------------------+
| 8      3      1     | 45     9      457   | 2      56     467   |
| 2      7      56A   | 8      456    3A-45 | 1      9      34    |
| 456    45     9     | 235    256A*7 1     | 357    35     8     |
+-----------------------------------------------------------------+
    Colouring the bivalues in B2 (only 2 possible states in B2) : a=A, from (45)R1C6 for example, then going along C5 or C3, one obtains (7a)r6c5=(6A)r9c5 : => r9c5<>7, r6c5<>6 or (3A)r8c6=(4a)r1c6 : => r1c6<>3, r8c6<>4, through the NP(45)B6 "coloured" in A (AIC with ALS). Knowing the endpoints of the 2 AICs, the details of them may be easily devised.

    Note : in this way, one is turning twice around the XYZ Wing (4-56) without noticing it ! To detect the XYZ, one has either to look for the special pattern or start an AIC from one of the 2 associated bivalue cells!
Example 2 : puzzle 10/08/31 C, after SSTS [tagging not included]

Code:
+--------------------------------------------------------------+
| *3a7A  7a9A  2A9a  |  8     5     234   |  6     1     234   |
|  3A6a  4     5     |  2a6   1     9     |  23    7     8     |
|  8     1     2a6A  |  7     34    2346  |  9     235   2345  |
+--------------------------------------------------------------+
|  5     39    4     |  1     2     7     |  8     36    369   |
|  1     8     37    |  46a   9     4a6   |  237   235   2357  |
|  67    2     69    |  3     8     5     |  1     4     79    |
+--------------------------------------------------------------+
|  2     37    37    |  5     6     8     |  4     9     1     |
|  9     6     18    |  24a   347   1234  |  5     238   237   |
|  4     5     18    |  9     37    123   |  237   2368  2367  |
+--------------------------------------------------------------+
    Colouring the bivalues in B1 (only 2 possible states in B1) : a=A, from R1C1 for example, then going along C4 and R6, yields R1C6 empty in colour a : => a is False ! It remains to build an interpretation of the elimination of a in B1!
Example 3 : puzzle 10/09/03 A, after SSTS [with partial tags]

Code:
+-------------------------------------------------------------------+
| 1.a5*9- 8       6     | 7      2      3     | 159-   15     4     |
| 2       1.5|79  579   | 489    6      489   | 15789  3      179   |
| 3       79      4     | 5      1      89    | 789    6      2     |
+-------------------------------------------------------------------+
| 46      2       1     | 468    7      468   | 3      9      5     |
| 57      3       57    | 1      9      2     | 6      4      8     |
| 469-A   6|A9-   8     | 46     3      5     | 17     2      17    |
+-------------------------------------------------------------------+
| 1569    15|A6|9 59    | 2      8      169   | 4      7      3     |
| 16789   4       3     | 69     5      1679  | 2      18     19    |
| 1789    179     2     | 3      4      179   | 159    158    6     |
+-------------------------------------------------------------------+
    Starting from the "hub" cell R2C2, (1a)r1c1=(5A)r7c2, one obtains (1a)r1c1=(9A)r6c1. The associated AIC is : 4-SIS AIC : 1B1 5C2 6C2 9R6 : (1)r1c1=(9)r6c1 : => r1c1<>9.
JC
Back to top
View user's profile Send private message
peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK

PostPosted: Sun Sep 05, 2010 8:20 am    Post subject: Reply with quote

JC, thanks for taking the time to post this. My approach is much more hit and miss at the moment - though I do very much look for "hub" cells, as you call them, when doing my initial scan and then check for bivalues that can see them. My approach to xy-chains varies from serendipitous discovery (!) to intentional hunts for pincers that can kill a promising looking elimination.
Back to top
View user's profile Send private message
Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sun Sep 05, 2010 3:33 pm    Post subject: Reply with quote

I had an unnecessary step (what else is new?).

XYZ-Wing (345)
M-Wing (34), flightless with pincer transport
X-Wing (5)
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Puzzles by daj All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group