Another "Mild" Puzzle that stumps me

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Another "Mild" Puzzle that stumps me

Postby Kit_ISIS_Kat » Fri Jun 09, 2006 3:01 am

In the post from a few weeks ago, it seems that there was a clue or two missing from the puzzle itself. I wonder if this is another example of the same. Any help would be greatly appreciated. More detailed comments are at the bottom of this email. Thanks!

Original Puzzle

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


Partial solution
5..|..8|3..
...|51.|..7
..2|36.|54.
---+---+---
.7.|..6|1..
9..|.7.|..5
6.8|4..|72.


---+---+---
.6.|..5|8..
3..|68.|...
.84|23.|956


{5} {149} {1679} {79} {249} {8} {3} {169} {129}
{48} {349} {369} {5} {1} {249} {26} {689} {7}
{178} {19} {2} {3} {6} {79} {5} {4} {189}

{24} {7} {35} {89} {259} {6} {1} {389} {3489}
{9} {1234} {13} {18} {7} {123} {46} {368} {5}
{6} {135} {8} {4} {59} {139} {7} {2} {39}

{127} {6} {179} {179} {49} {5} {8} {137} {1234}
{3} {1259} {1579} {6} {8} {1479} {24} {17} {124}
{17} {8} {4} {2} {3} {17} {9} {5} {6}


Question relates to 2nd mini-grid
{79} {249} {8}
{5} {1} {249}
{3} {6} {79}

The hint I was given was "Reduce based on Naked pairs". This would leave me with:
{79} {24}** {8}
{5} {1} {24}*
{3} {6} {79}

The program allowed me to place a 2 under the 8* but not a 4. What I want to know is what clue is there to tell me to eliminate the 4 from that cell or to tell me it is the 2 that belongs there?

Or from a different perspective, what clue is there to tell me that the 4 and not the two belongs in the cell** above where the 1 is located?
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Postby tarek » Fri Jun 09, 2006 6:20 am

Hi there, what you are looking at is a HIDDEN pair rather than a naked pair.......The logic behind keeping only 24 in both cells is that if u look in that box......these 2 cells are the only cells which can hold 2 & 4 so one of them must be 2 the other must be 4.you can eliminate the rest....

now you cant go any further with hidden pairs regarding these 2 cells....

the puzzle is solved with advanced methods like colouring, this step is how i see it
Code: Select all
*--------------------------------------------------------*
| 5     149   1679 | 79    24    8    | 3     169   129  |
|*48    3489  369  | 5     1    *24   | 26    689   7    |
| 178   189   2    | 3     6     79   | 5     4     189  |
|------------------+------------------+------------------|
|#24    7     35   | 89    259   6    | 1     389   3489 |
|*9    -24    13   | 18    7     123  |*46    368   5    |
| 6     135   8    | 4     59    139  | 7     2     39   |
|------------------+------------------+------------------|
| 12    6     179  | 179   49    5    | 8     137   1234 |
| 3     1259  1579 | 6     8    *1479 |*24    157   14   |
| 178   158   4    | 2     3     17   | 9     157   6    |
*--------------------------------------------------------*
*--------------------------------------------------------*
| 5    *149   1679 | 79   *24    8    | 3     169   129  |
| 48    3489  369  | 5     1     24   | 26    689   7    |
| 178   189   2    | 3     6     79   | 5     4     189  |
|------------------+------------------+------------------|
|#24   *7     35   | 89    259   6    | 1     389  *3489 |
| 9    -24    13   | 18    7     123  | 46    368   5    |
| 6     135   8    | 4     59    139  | 7     2     39   |
|------------------+------------------+------------------|
| 12    6     179  | 179  *49    5    | 8     137  *1234 |
| 3     1259  1579 | 6     8     1479 | 24    157   14   |
| 178   158   4    | 2     3     17   | 9     157   6    |
*--------------------------------------------------------*
Eliminating 4 From r5c2 (Eyeless Finned Swordfish in Columns 167 with 1 fin in Box 4, or by constructing the Finned Swordfish in rows 147 with 1 fin in Box 4)


& the puzzle is solved

tarek
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Another Solution

Postby Carcul » Sun Jun 11, 2006 10:21 am

Code: Select all
 *-----------------------------------------------------------*
 | 5     149   1679  | 79    24    8     | 3     169   129   |
 | 48    3489  369   | 5     1     24    | 26    689   7     |
 | 178   189   2     | 3     6     79    | 5     4     189   |
 |-------------------+-------------------+-------------------|
 | 24    7     35    | 89    259   6     | 1     389   3489  |
 | 9     1234  13    | 18    7     123   | 46    368   5     |
 | 6     135   8     | 4     59    139   | 7     2     39    |
 |-------------------+-------------------+-------------------|
 | 127   6     179   | 179   49    5     | 8     137   1234  |
 | 3     1259  1579  | 6     8     1479  | 24    157   124   |
 | 178   158   4     | 2     3     17    | 9     157   6     |
 *-----------------------------------------------------------*

[r2c3]=3=[r2c2]={ATILA(4): r2c1|r4c1|r5c2|r5c7|r8c7|(r8c9)|r8c6|r2c6}
=4=[r8c9]-4-[r8c7]-2-[r2c7]-6-[r2c3],

which implies r2c3<>6 and the puzzle is solved.
For an explanation about the notation, check here.
"ATILA" stands for Almost Two Incompatible Loops Argument: check
this post.

Carcul
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Postby daj95376 » Sun Jun 11, 2006 4:53 pm

Technically speaking -- of which I'm not very good it seems -- the partial solution originally presented does have a Naked Pair <79> in block 2.

These steps will get your original puzzle to the following state.

Code: Select all
r6c7    =  7     Hidden Single
r6c1    =  6     Hidden Single
r8c4    =  6     Hidden Single
    b2  -  79    Naked  Pair
r2c4    =  5     Naked  Single
r3c5    =  6     Naked  Single
r3c7    =  5     Hidden Single
r9c7    =  9     Naked  Single
r9c5    =  3     Naked  Single
    b4  -  135   Naked  Triple
r147    -  2     Swordfish

 
 *-----------------------------------------------------------*
 | 5     149   1679  | 79    24    8     | 3     169   129   |
 | 48    3489  369   | 5     1     24    | 26    689   7     |
 | 178   189   2     | 3     6     79    | 5     4     189   |
 |-------------------+-------------------+-------------------|
 | 24    7     35    | 89    259   6     | 1     389   3489  |
 | 9     24    13    | 18    7     123   | 46    368   5     |
 | 6     135   8     | 4     59    139   | 7     2     39    |
 |-------------------+-------------------+-------------------|
 | 127   6     179   | 179   49    5     | 8     137   1234  |
 | 3     1259  1579  | 6     8     1479  | 24    157   14    |
 | 178   158   4     | 2     3     17    | 9     157   6     |
 *-----------------------------------------------------------*


As to your question about why r2c6 can not be 4, the answer is that it forces the elimination of all 4's in block 1 and results in an invalid puzzle.

Code: Select all
r2c6=4 => r5c6=2 => r5c2=4 => r1c2<>4
r2c6=4                     => r2c1<>4 & r2c2<>4
Last edited by daj95376 on Mon Jun 12, 2006 7:17 am, edited 1 time in total.
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Re: Another Solution

Postby ronk » Sun Jun 11, 2006 8:47 pm

Carcul wrote:
Code: Select all
 *-----------------------------------------------------------*
 | 5     149   1679  | 79    24    8     | 3     169   129   |
 | 48    3489  369   | 5     1     24    | 26    689   7     |
 | 178   189   2     | 3     6     79    | 5     4     189   |
 |-------------------+-------------------+-------------------|
 | 24    7     35    | 89    259   6     | 1     389   3489  |
 | 9     1234  13    | 18    7     123   | 46    368   5     |
 | 6     135   8     | 4     59    139   | 7     2     39    |
 |-------------------+-------------------+-------------------|
 | 127   6     179   | 179   49    5     | 8     137   1234  |
 | 3     1259  1579  | 6     8     1479  | 24    157   124   |
 | 178   158   4     | 2     3     17    | 9     157   6     |
 *-----------------------------------------------------------*

[r2c3]=3=[r2c2]={ATILA(4): r2c1|r4c1|r5c2|r5c7|r8c7|(r8c9)|r8c6|r2c6}
=4=[r8c9]-4-[r8c7]-2-[r2c7]-6-[r2c3],

which implies r2c3<>6 and the puzzle is solved.

Just coloring the two conjugate chains in digit 4 is easier to understand.
Code: Select all
 . B . | . b . | . . .
 A 4 . | . . B | . . .
 . . . | . . . | . . .
-------+-------+-------
 a . . | . . . | . . A
 . A . | . . . | a . .
 . . . | . . . | . . .
-------+-------+-------
 . . . | . B . | . . b
 . . . | . . b | A . 4
 . . . | . . . | . . .

The two chains are colored 'Aa' and 'Bb'. In row 2 and box 1 color 'B' (representing digit 4 as) true excludes 'A' true. In row 8, 'b' true also excludes 'A' from being true. Since either 'B' or 'b' must be true, digit 4 may be eliminated from all cells colored 'A'.

One possible nice loop expression: r2c1-4-r4c1=4=r5c2-4-r5c7=4=r8c7-4-r8c6=4=r2c6-4-r2c1,

which implies r2c1<>4 and the puzzle is solved with cascading singles.
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Postby Carcul » Mon Jun 12, 2006 8:24 am

Ronk wrote:Just coloring the two conjugate chains in digit 4 is easier to understand.


Yes, I know. I also spoted that simple solution. But sometimes I think it's interesting (and, who knows, educational) to present/try to find a solution using an alternative way of thought (even if a simpler solution is available, as in this case), instead of keep solving puzzles using always the same way. If Kit_ISIS_Kat don't like my solution, then he/she will simply ignore it, with no problem. But I believe that it's always good to have several options available.

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