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stuck on a very hard puzzle

 
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debbie



Joined: 27 Mar 2006
Posts: 7

PostPosted: Fri Apr 07, 2006 5:14 pm    Post subject: stuck on a very hard puzzle Reply with quote

This was in the newspaper last Saturday. I put it down for a week and picked it up again and still can't figure out the next move:

26x 9xx x48
4xx 38x 2x6
7x8 642 15x

8xx 4x9 6x1
3x9 168 5x4
614 xxx xxx

586 793 412
9x7 x14 x6x
14x xx6 xxx
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keith



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

PostPosted: Fri Apr 07, 2006 5:40 pm    Post subject: DB Puzzle? Reply with quote

Debbie,

Look in the "Other Puzzles" forum:

http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=590

Keith
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debbie



Joined: 27 Mar 2006
Posts: 7

PostPosted: Fri Apr 07, 2006 7:27 pm    Post subject: Reply with quote

Thanks Keith. Same puzzle, but the "explanation" made absolutely no sense to me (as I am not familiar with alpha stars)... I can not follow the logic at all.
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David Bryant



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

PostPosted: Sat Apr 08, 2006 12:20 am    Post subject: Here's a translation ... Reply with quote

Hi, Debbie! Let me translate someone_somewhere's jargon into everyday terminology.

At the point where you got stuck the matrix, including candidate lists, looks like this.
Code:
  2     6    135    9    57    157   37     4     8
  4    59    15     3     8    157    2    79     6
  7    39     8     6     4     2     1     5    39
  8    257   25     4    237    9     6    237    1
  3    27     9     1     6     8     5    27     4
  6     1     4    25   2357   57   3789  23789  379
  5     8     6     7     9     3     4     1     2
  9    23     7    258    1     4    38     6    35
  1     4    23    258   25     6   3789  3789  3579

First we will prove that there cannot be a "3" at r9c9.

-- r1c7 either contains a "3" or a "7" (this is the "alpha star").
-- If r1c7 = 3 then r9c3 = 3 (because there are only two places for a "3" in column 3).
-- If r1c7 = 7 then r3c9 = 3 (because there are only two places for a "3" in the top right 3x3 box).
-- Either way there cannot possibly be a "3" at r9c9.

Next we will prove that there cannot be a "3" at r8c7.

-- r3c2 either contains a "3" or a "9" (this is the second "alpha star").
-- If r3c2 = 3 then r1c7 = 3 (because there are only two places for a "3" in the top right 3x3 box).
-- If r3c2 = 9 then r8c2 = 3 (because there are only two places for a "3" in column 2).
-- Either way there cannot possibly be a "3" at r8c7, so r8c7 = 8.

You can place a couple more "8"s after setting r8c7 = 8. Now the matrix looks like this.
Code:
  2     6    135    9    57    157   37     4     8
  4    59    15     3     8    157    2    79     6
  7    39     8     6     4     2     1     5    39
  8    257   25     4    237    9     6    237    1
  3    27     9     1     6     8     5    27     4
  6     1     4    25   2357   57    379    8    379
  5     8     6     7     9     3     4     1     2
  9    23     7    25     1     4     8     6    35
  1     4    23     8    25     6    379   379   579


-- Since the "2" in row 6 must lie in the middle center 3x3 box there cannot be a "2" at r4c5.

-- There cannot be a "7" at r6c7:
If r1c7 = 7 then r6c7 <> 7;
If r1c7 = 3 then r3c9 = 9 and r6c7 = 9 <> 7 (only two places for a "9" in row 6).

-- There cannot be a "3" at r6c9:
If r3c9 = 3 then r6c9 <> 3;
If r3c9 = 9 then r2c8 = 7, r5c8 = 2, r4c8 = 3, and r6c9 <> 3.

Making these slight changes to the matrix lets us finish this puzzle off.
Code:
  2     6    135    9    57    157   37     4     8
  4    59    15     3     8    157    2    79     6
  7    39     8     6     4     2     1     5    39
  8    257   25     4    37     9     6    237    1
  3    27     9     1     6     8     5    27     4
  6     1     4    25   2357   57    39     8    79
  5     8     6     7     9     3     4     1     2
  9    23     7    25     1     4     8     6    35
  1     4    23     8    25     6    379   379   579

We see that r1c5 cannot be a "7", because if it were a "7" both r1c7 and r4c5 would have to be "3", and that would make it impossible to fit a "3" in row 6. Therefore r1c5 = 5, and the rest is very easy. dcb

PS There are undoubtedly other ways to skin this cat. I just translated what someone_somewhere had to say about this puzzle into language that is more easily understood.
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mastertsai



Joined: 08 Apr 2006
Posts: 2

PostPosted: Sat Apr 08, 2006 6:02 am    Post subject: Fortune Angel Sudoku Slover Reply with quote

261 957 348
495 381 276
738 642 159

852 479 631
379 168 524
614 235 987

586 793 412
927 514 863
143 826 795

http://www.fortuneangel.com/99/sudoku.asp
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TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Sat Apr 08, 2006 12:00 pm    Post subject: Reply with quote

Mastertsai,

Everyone is welcome to contribute, but most people want the logic behind the solving steps, not just the final answer. There are hundreds of programs that will solve a Sudoku, the better ones also list the techniques used for each step in the process.
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debbie



Joined: 27 Mar 2006
Posts: 7

PostPosted: Sat Apr 08, 2006 1:33 pm    Post subject: Reply with quote

Thank you so much Tkiel.
But what is the logic for looking at the 3 to start with? What makes this the alpha star? How is this different than process of elimination? I think this is what someone on the other thread referred to as a "forced" move, but I'd like to know how to spot these.
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jabejochke



Joined: 16 Mar 2006
Posts: 21
Location: Reading

PostPosted: Sat Apr 08, 2006 3:29 pm    Post subject: Reply with quote

Debbie,

Great question. I too want to learn from this insight. Of the 15+ two-digit cells on the board, why the 37 at r1c7? Could I bypass some or all of the other 15?

Thanks to all for these discussions.

Jack
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David Bryant



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

PostPosted: Sat Apr 08, 2006 4:39 pm    Post subject: Looking for double-implication chains Reply with quote

Jack wrote:
Of the 15+ two-digit cells on the board, why the 37 at r1c7? Could I bypass some or all of the other 15?

I'm not as good at finding the DIC's as our friend someone_somewhere is, but I have had some luck with the technique. Here's how I go about it.

1. I don't try to use this technique until I'm pretty thoroughly stuck.

2. I start by examining each cell that has been reduced to a pair of possibilities.

3. Usually there's not much to gain from "paired pairs" -- so if there's say {6, 8} in two adjacent cells, those aren't good candidates. You want to concentrate on the "isolated pairs" first.

4. If one chain leading away from a cell depends on the values only (say 36 - 56 - 59) while the other depends on positional clues (only two places for a "3" in this row, or column, or box) then the chains are more likely to be useful than if both chains depend on values only.

5. Short chains are easier to follow than longer ones. So the first time I go around the board looking at pairs I only search to a depth of 2 or 3 moves for each candidate. If that doesn't turn anything up I start chasing longer chains.

I don't know if any of that will help or not, Jack -- I hope it does. dcb
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keith



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

PostPosted: Sat Apr 08, 2006 7:54 pm    Post subject: I don' know how! Reply with quote

Debbie and Jack,

This is the starting point, posted by David:

Code:


+-------------------+-------------------+-------------------+
| 2     6     135   | 9     57    157   | 37    4     8     |
| 4     59    15    | 3     8     157   | 2     79    6     |
| 7     39    8     | 6     4     2     | 1     5     39    |
+-------------------+-------------------+-------------------+
| 8     257   25    | 4     2357  9     | 6     237   1     |
| 3     27    9     | 1     6     8     | 5     27    4     |
| 6     1     4     | 25    2357  57    | 3789  23789 379   |
+-------------------+-------------------+-------------------+
| 5     8     6     | 7     9     3     | 4     1     2     |
| 9     23    7     | 258   1     4     | 38    6     35    |
| 1     4     23    | 258   25    6     | 3789  3789  3579  |
+-------------------+-------------------+-------------------+



You can make some progress by coloring on <3>, etc. I get to here:

Code:


+----------------+----------------+----------------+
| 2    6    135  | 9    57   157  | 37   4    8    |
| 4    59   15   | 3    8    157  | 2    79   6    |
| 7    39   8    | 6    4    2    | 1    5    39   |
+----------------+----------------+----------------+
| 8    257  25   | 4    37   9    | 6    237  1    |
| 3    27   9    | 1    6    8    | 5    27   4    |
| 6    1    4    | 25   2357 57   | 39   8    79   |
+----------------+----------------+----------------+
| 5    8    6    | 7    9    3    | 4    1    2    |
| 9    23   7    | 25   1    4    | 8    6    35   |
| 1    4    23   | 8    25   6    | 379  379  579  |
+----------------+----------------+----------------+



I don't see any way forward, except to use a chain. One that immediately solves the puzzle is present in the original position.

If R1C3 is <3>, then you can quickly see R1C5 = <5> (via R1C7) and R9C5=<5> (via R9C3). This is a contradiction in C5, so R1C3 is not <3>, and the puzzle is solved.

Is this a guess? I don't think so, though there is no systematic way to identify squares to check. I check squares with both two and with three possibilities, but I only follow the implications for at most three steps and only through squares that have two possibilities.

Best wishes,

Keith
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jabejochke



Joined: 16 Mar 2006
Posts: 21
Location: Reading

PostPosted: Sun Apr 09, 2006 1:05 pm    Post subject: Reply with quote

David and Keith,

Thanks for the responses. I haven't yet digested them -- but just wanted to recognize how quickly your responses followed our expressed need.

Jack
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Havard
Guest





PostPosted: Sun Apr 09, 2006 3:09 pm    Post subject: Reply with quote

Hi.

There is an Empty Rectangle that will crack the puzzle from this position:

Code:
+----------------+----------------+----------------+
| 2    6    135  | 9    57   157  | 37   4    8    |
| 4    59   15   | 3    8    157  | 2    79   6    |
| 7    39   8    | 6    4    2    | 1    5    39   |
+----------------+----------------+----------------+
| 8    257  25   | 4    37   9    | 6    237  1    |
| 3    27   9    | 1    6    8    | 5    27   4    |
| 6    1    4    | 25   2357 57   | 39   8    79   |
+----------------+----------------+----------------+
| 5    8    6    | 7    9    3    | 4    1    2    |
| 9    23   7    | 25   1    4    | 8    6    35   |
| 1    4    23   | 8    25   6    | 379  379  579  |
+----------------+----------------+----------------+


You can read about empty rectangles here:
http://www.sudoku.com/forums/viewtopic.php?t=3251

It is hiding in the 7's! :)

Havard
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keith



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

PostPosted: Sun Apr 09, 2006 9:40 pm    Post subject: More Details, Please? Reply with quote

Havard,

This old dog has a hard time learning new tricks.

Can you please explain how an Empty Rectangle applies in this example?

Thank you,

Keith
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ravel
Guest





PostPosted: Mon Apr 10, 2006 9:38 am    Post subject: Reply with quote

I also cannot see an elimination with 1 ER, but Havard also uses combinations of ER's here

This is sample 3:

Code:
X X . | . . . | X . X
X X . | . . . | X . X
. . + | . . . | . + .
------+-------+------
. . a | . . . | . . .
. / . | . . . | . . .
b . . | . . . | . * .
------+-------+------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .

You can apply it in 2 ways (marked left and right) to eliminate 2 7's:

Code:
+----------------+----------------+----------------+
| 2    6    135  | 9   *57   157+ |b37b  4    8    |
| 4    59   15   | 3    8    157  | 2   a79a  6    |
| 7    39   8    | 6    4    2    | 1    5    39   |
+----------------+----------------+----------------+
| 8    257  25   | 4    37   9    | 6    237  1    |
| 3    27   9    | 1    6    8    | 5    27   4    |
| 6    1    4    | 25  +2357 57*  | 39  +8+   79   |
+----------------+----------------+----------------+
| 5    8    6    | 7    9    3    | 4    1    2    |
| 9    23   7    | 25   1    4    | 8    6    35   |
| 1    4    23   | 8    25   6    | 379  379  579  |
+----------------+----------------+----------------+
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ravel
Guest





PostPosted: Mon Apr 10, 2006 11:05 am    Post subject: Reply with quote

PS: This is all you need for the puzzle, because it can be applied also earlier, where one first gets stuck.
Code:
+----------------+----------------+----------------+
| 2    6    135  | 9   *57   157  | 37   4     8    |
| 4    59   15   | 3    8    157  | 2    79    6    |
| 7    39   8    | 6    4    2    | 1    5     39   |
+----------------+----------------+----------------+
| 8    257  25   | 4    37   9    | 6    237   1    |
| 3    27   9    | 1    6    8    | 5    27    4    |
| 6    1    4    | 25   2357 57*  | 3789 23789 379  |
+----------------+----------------+----------------+
| 5    8    6    | 7    9    3    | 4    1     2    |
| 9    23   7    | 258  1    4    | 38   6     35   |
| 1    4    23   | 258  25   6    | 3789 3789  5789 |
+----------------+----------------+----------------+

Each of the 5's you get, solves the puzzle.
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Mon Apr 10, 2006 11:22 am    Post subject: Hard puzzle Reply with quote

I believe the empty rectangle to be in box 6: the 7 contained in the box must be placed in row 6 or column 8.

The conjugate cells for 7 in row 2, r2c68, interact withthe column and exclude 7 from r6c6.

Steve
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ravel
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PostPosted: Mon Apr 10, 2006 2:24 pm    Post subject: Reply with quote

Yes, of course, how could i miss this conjugated pair ? I keep to be blind sometimes.
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