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02/19/06 Daily Sudocue.net Nightmare

 
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Paladin



Joined: 10 Feb 2006
Posts: 15

PostPosted: Wed Feb 22, 2006 7:18 pm    Post subject: 02/19/06 Daily Sudocue.net Nightmare Reply with quote

I could use some help with this puzzle. Although www.sudokusolver.co.uk indicates that this puzzles is only rated at 1252 (which means that it shouldn't be that hard to solve if you can apply an advanced solving technique), I don't seem to be able to get anywhere. I can't seem to find anything that will allow me to solve more than four (4) cells.

The start grid, and my four cell answers, are as follows:

Code:

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

My feeble answers at r3c1, r8c1, r5c4, r7c5:

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


Any help I could be given would be very much appreciated. Thank you.
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Marty R.



Joined: 12 Feb 2006
Posts: 5180
Location: Rochester, NY, USA

PostPosted: Wed Feb 22, 2006 10:01 pm    Post subject: Reply with quote

You're not alone; I solved the same four cells and hit a brick wall. Embarassed
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Ruud



Joined: 18 Jan 2006
Posts: 31

PostPosted: Sun Feb 26, 2006 12:13 am    Post subject: Reply with quote

I am not always in the mood to provide clues for my daily nightmares, but this one deserves a little extra attention.

A series of new techniques has been developed by a couple of guys at the sudoku players forum. These techniques were way over our simple heads until Myth Jellies came up with the Filet-O-Fish theory. This was both brilliant and simple. From this, a couple of techniques have been developed, and one of those I implemented in SudoCue.

This is it:

Finned-X-Wing.

Try to find a situation in the sudoku (there is one in the nightmare, right after you placed the 4th digit) that looks like an X-Wing, but it has one additional candidate in the defining rows (or columns) in the same box as one of the 4 cells that form the X-Wing.

When there are candidates in that same box, that will be eliminated by the X-Wing, they will also be eliminated by the extra candidate that must be true if the X-Wing does not exist.

Try to find it. This is a one-time-only hint from me. The finned X-Wing will not have it's last appearance in the daily nightmares series, so get used to it!

In case you wonder: there are 2 other techniques that I recently implemented. Those are BUG and XYZ-Wing. Better learn about them too.

Ruud.

P.S. Has anyone of you solved the clueless sudoku yet?
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David Bryant



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

PostPosted: Sun Feb 26, 2006 1:37 am    Post subject: Here's how I worked it out Reply with quote

Here -- I'll give more of a hint than Ruud did!

This puzzle isn't particularly elegant -- it's just a hard grind all the way through. Here's a (general) description of the way I solved it.

1. Resolved top center 3x3 box into {2, 7}, {3, 5}, and {1, 4} pairs.
2. Found {1, 4, 9} triplet in row 3, setting r3c1 = 6.
3. Found hidden pair {1, 3} in r7c1 & r9c1, setting r8c1 = 8.
4. The "4" in column 1 lies in middle left 3x3 box.
5. r5c4 = 8 (unique horizontal); r7c5 = 8 (unique vertical)
6. Coloring on "5" reveals pair {7, 9} in r2c1.
7. Coloring on "6" reveals {2, 4, 9} in r8c9.
8. XY-Wing from r2c1 eliminates "4" at r6c8; r6c1 = 4 (unique horizontal).
9. The "7" in row 6 lies in middle center 3x3 box.
10. {5, 9} pair revealed in column 4 -- r9c4 = 7.
11. {3, 5} pair revealed in column 5 -- r7c5 = 8.
12. r8c7 = 5 (unique horizontal), revealing {1, 2} pair in r7c7 & r9c7.
13. Triplet {4, 7, 9} lies in row 2.

After these steps and the associated eliminations the puzzle looks like this.

Code:
 57   2457  1237   14     9    35     6    48   12478
 79    25    123    6    35     8    47    49    12
  6     8    19    14    27    27     3     5    149
  2    579    6    59    35     4     8     1    37
 579    3    79     8     1    569   47     2    467
  4     1     8     2    67    367    9    36     5
 13     6     4    59     8    259   12     7    39
  8    279   279    3    26     1     5    469   49
 13    29     5     7     4    269   12   3689  3689


And now we find a very nice "5-star constellation" in r2c1, r3c3, r5c3, r5c7, and r2c7:

A. r5c7 = 7 ==> r5c3 = 9 ==> r3c3 = 1
B. r5c7 = 4 ==> r2c7 = 7 ==> r2c1 = 9 ==> r3c3 = 1

Setting r3c3 = 1 is enough to crack the rest of the puzzle wide open. dcb
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TKiel



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

PostPosted: Sun Feb 26, 2006 6:03 am    Post subject: Reply with quote

After doings the 13 items in Davids list my candidate grid looked like this
Code:
 
 *-----------------------------------------------------------------------*
 | 57      2457    1237    | 14   9    35      | 6       48      12478   |
 | 79      25      123     | 6    35   8       | 47      49      12      |
 | 6       8       19      | 14   27   27      | 3       5       149     |
 |-------------------------+-------------------+-------------------------|
 | 2       579     6       | 59   35   4       | 8       1       37      |
 | 579     3       79      | 8    1    569     | 47      2       467     |
 | 4       1       8       | 2    67   367     | 9       36      5       |
 |-------------------------+-------------------+-------------------------|
 | 13      6       4       | 59   8    259     | 12      7       1239    |
 | 8       279     279     | 3    26   1       | 5       469     249     |
 | 13      29      5       | 7    4    269     | 12      3689    123689  |
 *-----------------------------------------------------------------------*

which left r1c2 as a singleton 4, which lead to the placement of four other singles and left the grid looking like this
Code:

 *-----------------------------------------------------------*
 | 57    4     237   | 1     9     35    | 6     8     27    |
 | 79    25    123   | 6     35    8     | 47    49    12    |
 | 6     8     19    | 4     27    27    | 3     5     19    |
 |-------------------+-------------------+-------------------|
 | 2     579   6     | 59    35    4     | 8     1     37    |
 | 579   3     79    | 8     1     569   | 47    2     467   |
 | 4     1     8     | 2     67    367   | 9     36    5     |
 |-------------------+-------------------+-------------------|
 | 13    6     4     | 59    8     259   | 12    7     1239  |
 | 8     279   279   | 3     26    1     | 5     469   249   |
 | 13    29    5     | 7     4     269   | 12    369   8     |
 *-----------------------------------------------------------*

There is a swordfish that excludes 6 from r5c6, leaving r5c9 as a single 6 and that leads to singles the rest of the way.

I think I saw the Finned X-wing but the placement of one of the singles (after Davids list) made the exclusion moot. If someone wanted to send a PM, so as not to ruin it for others, with the exclusion I would appreciate it.
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Feb 26, 2006 7:15 pm    Post subject: Please explain "coloring" Reply with quote

After the position given by David, nothing exotic is needed to solve the puzzle.

What I do not understand, and I have been staring at this thing for hours: What is the "coloring" logic that says R2C1 is not <5>, and that R8C9 is not <6>?

In other words, how do you proceed from the following position without using chains?

Code:


+----------------------+----------------------+----------------------+
| 57     2457   1237   | 14     9      35     | 6      48     12478  |
| 579    24579  12379  | 6      35     8      | 1247   49     12479  |
| 6      8      19     | 14     27     27     | 3      5      149    |
+----------------------+----------------------+----------------------+
| 2      579    6      | 579    357    4      | 8      1      37     |
| 4579   3      79     | 8      1      5679   | 47     2      467    |
| 47     1      8      | 2      367    367    | 9      346    5      |
+----------------------+----------------------+----------------------+
| 13     6      4      | 59     8      259    | 125    7      1239   |
| 8      279    279    | 3      2567   1      | 245    469    2469   |
| 13     279    5      | 79     4      2679   | 12     3689   123689 |
+----------------------+----------------------+----------------------+




Thank you,

Keith


Here is a solution, starting from the position posted above by David:

R1C2 (pin) <4>.
R1C4 (force) <1>.
R1C8 (force) <8>.
R3C4 (force) <4>.
R9C9 (pin) <8>.
R5C9 (pin) <6>.
R6C8 (force) <3>.
R4C9 (force) <7>.
R1C9 (force) <2>.
R5C7 (force) <4>.
R2C7 (force) <7>.
R2C9 (force) <1>.
R2C1 (force) <9>.
R3C9 (force) <9>.
R3C3 (force) <1>.
R7C9 (force) <3>.
R8C9 (force) <4>.
R2C8 (force) <4>.
R7C1 (force) <1>.
R7C7 (force) <2>.
R9C1 (force) <3>.
R9C7 (force) <1>.
R4C5 (pin) <3>.
R2C5 (force) <5>.
R2C2 (force) <2>.
R1C6 (force) <3>.
R1C3 (force) <7>.
R2C3 (force) <3>.
R9C2 (force) <9>.
R9C8 (force) <6>.
R4C2 (force) <5>.
R8C2 (force) <7>.
R9C6 (force) <2>.
R8C8 (force) <9>.
R1C1 (force) <5>.
R5C3 (force) <9>.
R8C3 (force) <2>.
R4C4 (force) <9>.
R5C1 (force) <7>.
R7C4 (force) <5>.
R5C6 (force) <5>.
R7C6 (force) <9>.
R8C5 (force) <6>.
R6C5 (force) <7>.
R3C6 (force) <7>.
R3C5 (force) <2>.
R6C6 (force) <6>.
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TKiel



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

PostPosted: Sun Feb 26, 2006 10:13 pm    Post subject: Reply with quote

R5c1(A) is conjugate with r5c6(a) which is conjugate with r1c6(A) which is conjugate with r2c5(a). R2c1 shares a row with r2c5(a) and a column with r5c1(A), one of which must be the value 5. Therefore it can't be.

R6c5(A) is conjugate with r8c5(a). R5c6(B) is conjugate with r5c9(b). A & B share a group so both can't be true, therefore either/both a & b are true. R8c9 shares a row with 'a' and a column with 'b', therefore can't be 6.


Last edited by TKiel on Mon Feb 27, 2006 11:32 am; edited 1 time in total
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David Bryant



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

PostPosted: Sun Feb 26, 2006 11:01 pm    Post subject: How the "coloring" works Reply with quote

Keith wrote:
What I do not understand, and I have been staring at this thing for hours: What is the "coloring" logic that says R2C1 is not <5>, and that R8C9 is not <6>?

In other words, how do you proceed from the following position without using chains?

Tracy has already explained this ... I'll try explaining it with different words.
Code:
+----------------------+----------------------+----------------------+
| 57     2457   1237   | 14     9      35+    | 6      48     12478  |
| 579    24579  12379  | 6      35-    8      | 1247   49     12479  |
| 6      8      19     | 14     27     27     | 3      5      149    |
+----------------------+----------------------+----------------------+
| 2      579    6      | 579    357    4      | 8      1      37     |
| 45~79  3      79     | 8      1      5=6-79 | 47     2      46+7   |
| 47     1      8      | 2      36=7   367    | 9      346    5      |
+----------------------+----------------------+----------------------+
| 13     6      4      | 59     8      259    | 125    7      1239   |
| 8      279    279    | 3      256~7  1      | 245    469    2469   |
| 13     279    5      | 79     4      2679   | 12     3689   123689 |
+----------------------+----------------------+----------------------+

First for the "5"s. There are only two ways to place a "5" in the top center 3x3 box. And there are only two ways to place a "5" in row 5.

A. r1c6 = 5 ==> r5c6 <> 5 ==> r5c1 = 5 ==> r2c1 <>5
B. r1c6 <> 5 ==> r2c5 = 5 ==> r2c1 <>5

So if the "5" in the top center 3x3 box is not in line with cell r2c1 then we must have a "5" at r5c1 ... a "5" at r2c1 is just not possible.

The logic for the "6" is substantially similar. Notice that there are two ways to place a "6" in row 5, and two ways to fit a "6" in column 5 -- the two chains are connected via the middle center 3x3 box.

C. r5c9 = 6 ==> r8c9 <> 6
D. r5c9 <> 6 ==> r5c6 = 6 ==> r6c5 <> 6 ==> r8c5 = 6 ==> r8c9 <> 6

Either the "6" in row 5 lies in column 9, or else the "6" in column 5 lies in row 8. Either way a "6" at r8c9 is impossible.

Some people call this technique "multi-coloring" ... to me it just looks like two linked binary chains, so I think of it as "coloring." dcb
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Feb 26, 2006 11:08 pm    Post subject: I still don't get it Reply with quote

Tracy,

Sorry, but I still don't see the first one.


ON <5> in R2C1:

>> R5c1(A) is conjugate with r5c6(a)
I agree

>> which is conjugate with r1c6(A)
no, there is a possible <5> in r7c6

>> which is conjugate with r2c5(a)
I agree

>> R2c1 shares a row with r2c5(a) and a column with r5c1(A),
>> one of which must be the value 5. Therefore it can't be.
I would agree, if there were only two possible 5's in c6.


ON <6> in R8C9:

>> R6c5(A) is conjugate with r8c5(a)
I agree

>> R5c6(B) is conjugate with r6c9(b)
I think you mean r5c9(b). If so, I agree.

>> A & B share a group so both can't be true
I agree. But, both can be false, because r6c6 is possibly 6

>> therefore either/both a & b are true. R8c9 shares a row
>> with 'a' and a column with 'b', therefore can't be 6.
I agree. You have taught me something new!


I missed the <6> because I was looking for a single conjugate chain. I still do not understand the <5>.

Thank you,

Keith
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TKiel



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

PostPosted: Sun Feb 26, 2006 11:59 pm    Post subject: Reply with quote

Keith,

I made an error in my post explaining the conjugate 5's. It's actually exactly like the 6's. R5c1(A) and r5c6(a) are conjugates. R1c6(B) and r2c5(b) are conjugates. A & B share a group and can't both be true, so either/both a & b have to be true. R2c1 shares a row with r2c5(b) and a column with r5c1(a), so it can't be true either.
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Mon Feb 27, 2006 1:00 am    Post subject: Coloring Reply with quote

Tracy and David,

Thank you. These colors, chains and cycles are coming into focus for me.

Keith
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Orsi_AC
Guest





PostPosted: Tue Feb 28, 2006 10:34 pm    Post subject: Re: 02/19/06 Daily Sudocue.net Nightmare Reply with quote

I've solved this puzzle using a single technique:
After you solve the fourth cell you will find in the row 6 cols 5 and 6 (3,6,7). As (6,7) appears together in the same box you can eliminate the number 3 from these cells and the numbers 6 and 7 from the others cells in the same row.
Good look.


Paladin wrote:
I could use some help with this puzzle. Although www.sudokusolver.co.uk indicates that this puzzles is only rated at 1252 (which means that it shouldn't be that hard to solve if you can apply an advanced solving technique), I don't seem to be able to get anywhere. I can't seem to find anything that will allow me to solve more than four (4) cells.

The start grid, and my four cell answers, are as follows:

Code:

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

My feeble answers at r3c1, r8c1, r5c4, r7c5:

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


Any help I could be given would be very much appreciated. Thank you.
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TKiel



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

PostPosted: Thu Mar 02, 2006 11:42 am    Post subject: Reply with quote

Orsi AC,

I'm not aware of a technique that allows one to make those exclusions based on that logic. It sounds like you are combining a hidden pair (two values that appear in only two cells) and a naked pair (two cells containing only the same two values) but 6 & 7 both appear in numerous cells in both the row and box in which they are grouped. A hidden pair allows the exclusion of all non pair values from those two cells and a naked pair allows the exclusion of the paired values from all other cells in the group (row, column, box). You've made exclusions both within & outside the two cells.
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guest
Guest





PostPosted: Fri Mar 03, 2006 11:40 am    Post subject: Nightmare-19th Feb, 2006 Reply with quote

I found this puzzle to be very straightforward without having to go 'fishing' or 'painting'. Basic logic only was needed.
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David Bryant



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

PostPosted: Fri Mar 03, 2006 3:41 pm    Post subject: Show me the money Reply with quote

guest wrote:
I found this puzzle to be very straightforward without having to go 'fishing' or 'painting'. Basic logic only was needed.

Hi, guest! That's a bold assertion.

I have studied this puzzle pretty carefully, and I still don't see how to get through it without XY-Wings, or coloring, or some other "advanced" technique.

Please provide a demonstration of your "basic" logic. I would really like to learn how you did it.

Thanks a bunch! dcb
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