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 

Saturday puzzle January 7 - a fishy one!

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Other puzzles
View previous topic :: View next topic  
Author Message
keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sat Jan 07, 2006 3:11 pm    Post subject: Saturday puzzle January 7 - a fishy one! Reply with quote

Another good and challenging one:

Code:


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



Best wishes

Keith
Back to top
View user's profile Send private message
someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Sun Jan 08, 2006 9:58 am    Post subject: Reply with quote

Hi,

After applying standard techniques and finding:

Hidden Pair 4 6 in r8c1 and r8c3
Nacked Pair 3 5 in r2c1 and r7c1

I get to:
Code:
1    6    48   3    7    2    48   9    5     

35   2358 258  4    6    9    1278 17   278     
aB   bBbb ...  .    .    .    .... ..   ...     
..   ...b ...  .    .    .    .... ..   ...
     
49   29   7    1    8    5    3    6    24     

8    129  29   6    139  37   1247 5    247     
     
7    15   3    2    15   4    9    8    6     
     
69   4    2569 58   159  78   127  17   3     
     
35   7    1    58   4    38   6    2    9     
Aa   .    .    ..   .    ..   .    .    .     
aA   .    .    ..   .    ..   .    .    .
     
46   58   46   9    2    1    578  3    78     
..   ..   ..   .    .    .    ...  .    ..     
..   aB   ..   .    .    .    ...  .    ..
     
2    3589 589  7    35   6    58   4    1   

1. path: r7c1 = 3, r2c1 <> 3, r2c2 = 3, r2c2 <> 8
2. path: r7c1 = 5, r8c2 <> 5, r8c2 = 8, r2c2 <> 8
and the 2 pathes lead us to the conclusion:
exclude 8 from r2c2

8 not in r9c3, it is in r8c2 or r9c2
Code:
1    6    48   3    7    2    48   9    5
     
35   235  258  4    6    9    1278 17   278
     
49   29   7    1    8    5    3    6    24
     
8    129  29   6    139  37   1247 5    247     
.    ...  aA   .    ..a  ..   .... .    ...
                             
7    15   3    2    15   4    9    8    6     
.    .C   .    .    .C   .    .    .    .
                           
69   4    2569 58   159  78   127  17   3     
..   .    .b.. ..   .bB  ..   ...  ..   .

35   7    1    58   4    38   6    2    9
     
46   58   46   9    2    1    578  3    78
     
2    3589 59   7    35   6    58   4    1     
.    .... Ba   .    .b   .    ..   .    .

1. path: r4c3 = 9, r9c3 <> 9, r9c3 = 5, r5c3 <> 5 (and r9c5 <> 5), r5c2 = 5
2. path: r4c3 = 9, r4c5 <> 9, r6c5 = 9, r6c5 <> 5 and r9c5 <> 5, r5c5 = 5and the two pathes lead us to a contradiction. This means we can:
excluded 9 from r4c3

2 in r4c3 - Sole Candidate
2 in r6c7 - Unique Horizontal
Code:
1    6    48   3    7    2    48   9    5
     
35   235  58   4    6    9    178  17   278     
..   ...  ..   .    .    .    a..  Aa   ...

49   29   7    1    8    5    3    6    24
     
8    19   2    6    139  37   147  5    47     
.    bC   .    .    ..C  ..   B..  .    ..
     
7    15   3    2    15   4    9    8    6
     
69   4    569  58   159  78   2    17   3     
..   .    ...  ..   B.b  ..   .    a.   .

35   7    1    58   4    38   6    2    9     
     
46   58   46   9    2    1    578  3    78     
     
2    3589 59   7    35   6    58   4    1     

1. path: r2c8 = 1, r2c7 <> 1, r4c7 = 1, r4c2 <> 1, r4c2 = 9
2. path: r2c8 = 1, r6c8 <> 1, r6c5 = 1, r6c5 <> 9, r4c5 = 9
and the two pathes lead us to a contradiction. This means we can:
excluded 1 from r2c8
and the rest is just an exercise, up to the solution:

Code:
  1 6 4 3 7 2 8 9 5
  5 3 8 4 6 9 1 7 2
  9 2 7 1 8 5 3 6 4
  8 9 2 6 1 3 4 5 7
  7 1 3 2 5 4 9 8 6
  6 4 5 8 9 7 2 1 3
  3 7 1 5 4 8 6 2 9
  4 5 6 9 2 1 7 3 8
  2 8 9 7 3 6 5 4 1



see u,
Back to top
View user's profile Send private message
David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sun Jan 08, 2006 8:32 pm    Post subject: Another way to solve it Reply with quote

This puzzle is very tough. I did find a way to break through the logjam by considering the forcing chains originating in r6c5, but I wonder if there might not be a shorter path to the solution.

After a series of fairly routine moves I arrived at this position.
Code:
   1      6     48      3      7      2     48      9      5
  35    2358    258     4      6      9    1278    17     278
  49     29      7      1      8      5      3      6     24
   8     129    29      6     139    37    1247     5     247
   7     15      3      2     15      4      9      8      6
  69      4    2569    58     159    78     127    17      3
  35      7      1     58      4     38      6      2      9
  46     58     46      9      2      1     578     3     78
   2    3589    589     7     35      6     58      4      1

Now consider the consequences of placing the various possible values at r6c5, which could be {1, 5, 9}.

r6c5 = 9 ==> r6c1 = 6
r6c5 = 1 ==> r5c5 = 5 ==> r5c2 = 1 ==> {2, 9} pair in r4c2, r4c3 ==> r6c1 = 6

Both of these chains are fairly short and direct. There's more work to be done to deal with the third case.

r6c5 = 5 ==> r6c4 = 8 ==> r7c4 = 5 ==> r7c1 = 3 ==> r2c1 = 5 (first chain)

r6c5 = 5 ==> r9c5 = 3 ==> r7c6 = 8 ==> r6c6 = 7 ==> r4c6 = 3
Also, r6c5 = 5 ==> r5c5 = 1; r5c5 = 1 & r4c6 = 3 ==> r4c5 = 9 ==> r4c3 = 2 (second chain)

Combining these two we get what we need:
r2c1 = 5 & r4c3 = 2 ==> r2c3 = 8 ==> r1c3 = 4 ==> r3c1 = 9 ==> r6c1 = 6

So no matter what value is placed at r6c5 we must have r6c1 = 6; with this additional value placed the rest of the puzzle falls apart rather easily. dcb
Back to top
View user's profile Send private message Send e-mail Visit poster's website
keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sun Jan 15, 2006 4:28 pm    Post subject: BUG me three times! Reply with quote

Because of the forcing chains, I think it is unlikely that two people will solve this puzzle the same way. Here is a solution that involves a Swordfish, and at the end, another BUG.

I don't know how common this BUG pattern is, but it is curious that three of the most recent puzzles in this discussion topic have them.

(This is not how I solved the puzzle. It is a solution generated by a computer program that mimics how a human might solve the puzzle.)

R1C4 must be <3>.
R3C6 must be <5>.
R1C5 must be <7>.
R3C4 must be <1>.
R4C4 must be <6>.
R2C5 must be <6>.
R8C5 is the only square in column 5 that can be <2>.
R9C6 is the only square in column 6 that can be <6>.
R5C9 is the only square in column 9 that can be <6>.
R5C8 is the only square in row 5 that can be <8>.
R1C8 must be <9>.
R5C1 is the only square in row 5 that can be <7>.
R7C9 is the only square in column 9 that can be <9>.
R7C8 is the only square in row 7 that can be <2>.
R8C8 is the only square in column 8 that can be <3>.

Squares R2C1 and R7C1 in column 1 form a simple naked pair. These 2 squares both contain the 2 possibilities <35>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R6C1 - removing <5> from <569> leaving <69>.
R8C1 - removing <5> from <456> leaving <46>.
A set of 2 squares form a simple hidden pair. R8C1 and R8C3 all contain the 2 possibilities <46>. No other squares in row 8 have those possibilities. Since the 2 squares are the only possible locations for 2 possible values, any additional possibilities these squares have (if any) can be eliminated. These squares now become a simple naked pair.
R8C3 - removing <58> from <4568> leaving <46>.

Found a 4-link Comprehensive Forcing Chain. If we assume that square R2C2 is <8> then we can make the following chain of conclusions:
R8C2 must be <5> (force), which means that
R7C1 must be <3> (force), which means that
R2C1 must be <5> (force), which means that
R2C2 must be <3> (R2 pin).
Since this is logically inconsistent, R2C2 cannot be <8>.
Intersection of column 2 with block 7. The value <8> only appears in one or more of squares R7C2, R8C2 and R9C2 of column 2. These squares are the ones that intersect with block 7. Thus, the other (non-intersecting) squares of block 7 cannot contain this value.
R9C3 - removing <8> from <589> leaving <59>.

Found a 5-link Comprehensive Forcing Chain. If we assume that square R9C2 is <9> then we can make the following chain of conclusions:
R3C2 must be <2> (force), which means that
R3C9 must be <4> (force), which means that
R1C7 must be <8> (force), which means that
R9C7 must be <5> (force), which means that
R9C2 must be <8> (R9 pin).
Since this is logically inconsistent, R9C2 cannot be <9>.
R9C3 is the only square in row 9 that can be <9>.
R4C3 must be <2>.
R6C7 is the only square in row 6 that can be <2>.

Squares R5C2 and R5C5 in row 5, R8C2 and R8C7 in row 8 and R9C2, R9C5 and R9C7 in row 9 form a Swordfish pattern on possibility <5>. All other instances of this possibility in columns 2, 5 and 7 can be removed.
R2C2 - removing <5> from <235> leaving <23>.
R6C5 - removing <5> from <159> leaving <19>.
Found a 7-link Simple Forcing Loop. If we assume that square R5C2 is <5> then we can make the following chain of conclusions:
R8C2 must be <8>, which means that
R8C9 must be <7>, which means that
R4C9 must be <4>, which means that
R3C9 must be <2>, which means that
R3C2 must be <9>, which means that
R4C2 must be <1>, which means that
R5C2 must be <5>.
This forms a inherently bi-directional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R5C2 and R8C2 must be <5>.
One of R8C2 and R8C9 must be <8>.
One of R8C9 and R4C9 must be <7>.
One of R4C9 and R3C9 must be <4>.
One of R3C9 and R3C2 must be <2>.
One of R3C2 and R4C2 must be <9>.
One of R4C2 and R5C2 must be <1>.
Thus we can deduce that:
R9C2 - cannot contain <5> because of R5C2 and R8C2.
R8C7 - cannot contain <8> because of R8C2 and R8C9.
R2C9 - cannot contain <7> because of R8C9 and R4C9.
A set of 2 squares form a simple hidden pair. R2C7 and R2C8 all contain the 2 possibilities <17>. No other squares in row 2 have those possibilities. Since the 2 squares are the only possible locations for 2 possible values, any additional possibilities these squares have (if any) can be eliminated. These squares now become a simple naked pair.
R2C7 - removing <8> from <178> leaving <17>.
Found a 8-link Simple Forcing Loop. If we assume that square R6C8 is <7> then we can make the following chain of conclusions:
R4C9 must be <4>, which means that
R3C9 must be <2>, which means that
R2C9 must be <8>, which means that
R2C3 must be <5>, which means that
R6C3 must be <6>, which means that
R6C1 must be <9>, which means that
R6C5 must be <1>, which means that
R6C8 must be <7>.
This forms a inherently bi-directional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R6C8 and R4C9 must be <7>.
One of R4C9 and R3C9 must be <4>.
One of R3C9 and R2C9 must be <2>.
One of R2C9 and R2C3 must be <8>.
One of R2C3 and R6C3 must be <5>.
One of R6C3 and R6C1 must be <6>.
One of R6C1 and R6C5 must be <9>.
One of R6C5 and R6C8 must be <1>.
Thus we can deduce that:
R4C7 - cannot contain <7> because of R6C8 and R4C9.

The puzzle is now:

Code:


  1   6   48    3   7   2     48  9   5   
  35  23  58    4   6   9     17  17  28   
  49  29  7     1   8   5     3   6   24   

  8   19  2     6   139 37    14  5   47   
  7   15  3     2   15  4     9   8   6   
  69  4   56    58  19  78    2   17  3   

  35  7   1     58  4   38    6   2   9   
  46  58  46    9   2   1     57  3   78   
  2   38  9     7   35  6     58  4   1   



The puzzle can be reduced to a Bivalue Universal Grave (BUG) pattern, by making this reduction:
R4C5=<39>.
These are called the BUG possibilities. In a BUG pattern, in each row, column and block, each unsolved possibility appears exactly twice. Such a pattern either has 0 or 2 solutions, so it cannot be part of a valid Sudoku.
When a puzzle contains a BUG, and only one square in the puzzle has more than 2 possibilities, the only way to kill the BUG is to remove both of the BUG possibilities from the square, thus solving it.
R4C5 - removing <39> from <139> leaving <1>.
From this deduction, the following moves are immediately forced:
R4C2 must be <9>.
R4C7 must be <4>.
R5C5 must be <5>.
R6C5 must be <9>.
R4C9 must be <7>.
R1C7 must be <8>.
R4C6 must be <3>.
R8C9 must be <8>.
R6C8 must be <1>.
R5C2 must be <1>.
R9C5 must be <3>.
R6C4 must be <8>.
R6C6 must be <7>.
R7C4 must be <5>.
R6C1 must be <6>.
R2C8 must be <7>.
R7C1 must be <3>.
R8C2 must be <5>.
R2C9 must be <2>.
R9C7 must be <5>.
R9C2 must be <8>.
R7C6 must be <8>.
R8C7 must be <7>.
R1C3 must be <4>.
R2C7 must be <1>.
R2C2 must be <3>.
R3C9 must be <4>.
R3C1 must be <9>.
R3C2 must be <2>.
R6C3 must be <5>.
R8C1 must be <4>.
R2C3 must be <8>.
R2C1 must be <5>.
R8C3 must be <6>.

Enjoy!

Keith
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Other puzzles 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