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Saturday puzzle January 7 - a fishy one!

 
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keith



Joined: 19 Sep 2005
Posts: 3355
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
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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,
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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
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keith



Joined: 19 Sep 2005
Posts: 3355
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
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