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Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Wed Nov 08, 2006 9:05 pm    Post subject: Nightmare November 8th 2006 (oops sorry not 11th)

I am new to this site and the pencilmark solving techniques. I like new techniques i didn't use before like finned fish and empty rectangles.
I don't really like the medusa technique when it takes massive forms (using a lot of cells). Maybe i dislike it because i can't use it properly. When i try it, it usually takes strange proportions eventually leading to a big contradiction which solves the puzzle completely. When checking the solver it uses smaller steps with multiple medusa colouring techniques which just makes the solving look nicer.
In todays nightmare i managed to bypass the medusa bridge though. And i find the puzzles more enjoyable to solve that way.

After the opening you end up with this situation.

 Code: 1  36  256 478 2467 2678  2578 2357   9     59   8   4  79   27    3   1257   6   127  39   7  26   1    5  2689*  28   234  2348   6  12   9   5   17    4    3     8   127  38  123  7  89   136  689   4    129   5  358  4  15   2   137  789  179   179   6  47  16  168  3    9    5  12678 1247 12478  47   5  138  6   247  27   189   139  138   2   9  36  47    8    1   567  3457  347

Then the solver the solver comes with a medusa bridge to eliminate the 9 in R3C6 using a lot of squares.

 Code: 1  36" 256 478 2467 2678  2578 2357   9     59   8   4  79   27    3   1257   6   127  39*  7  26*  1    5  2689*  28   234  2348   6  12   9   5   17    4    3     8   127  38  123  7  89   136  689   4    129   5  358  4  15   2   137  789  179   179   6  47  16  168  3    9    5  12678 1247 12478  47   5  138  6   247  27   189   139  138   2   9  36  47    8    1   567  3457  347

I found it easier to eliminate digit 6 in R1C2. When you look at row 3 there are only 3 cells (R3C1, R3C3 and R3C6) that can contain either digit 6 or 9. So of those 3 cells 2 should contain either digit 6 or 9.
When digit 6 is placed in R1C2 cells, a R3C3 will be a naked single for digit2 and R3C1 will be a hidden single for digit 3 in box 1. This interferes with what is stated earlier. Thus digit 6 can be eliminated from R1C2.
I hope my explanation is clear enough.

This eventually has the same result as the medusa bridge.
I am not really sure under which technique this falls, but i found it more delicate then the medusa bridge the solver recommended.

greetings

Para

p.s. I'd like to try some nightmares without the medusa colouring technique. If someone could tell me which puzzles to try then, i would appreciate it.

Last edited by Para on Thu Nov 09, 2006 2:36 pm; edited 1 time in total
GreenLantern
Regular

Joined: 30 May 2006
Posts: 14

Posted: Thu Nov 09, 2006 1:12 am    Post subject:

I think this is actually the November 8 Nightmare:
 Code: 1 . . | . . . | . . 9  . . 4 | . . 3 | . 6 .  . 7 . | 1 5 . | . . . -------+-------+------  . . 9 | . . 4 | 3 8 .  . . 7 | . . . | 4 . 5  . 4 . | 2 . . | . . 6 -------+-------+------  . . . | 3 9 . | . . .  . 5 . | 6 . . | . . .  2 . . | . 8 1 | . . .

 Para wrote: I found it easier to eliminate digit 6 in R1C2. When you look at row 3 there are only 3 cells (R3C1, R3C3 and R3C6) that can contain either digit 6 or 9. So of those 3 cells 2 should contain either digit 6 or 9. When digit 6 is placed in R1C2 cells, a R3C3 will be a naked single for digit2 and R3C1 will be a hidden single for digit 3 in box 1. This interferes with what is stated earlier. Thus digit 6 can be eliminated from R1C2. I hope my explanation is clear enough.

 Code: +----------------------+----------------------+----------------------+ | 1      36     256    | 478    2467   2678   | 2578   2357   9      | | 59     8      4      | 79     27     3      | 1257   6      127    | | 39     7      26     | 1      5      2689   | 28     234    2348   | +----------------------+----------------------+----------------------+ | 6      12     9      | 5      17     4      | 3      8      127    | | 38     123    7      | 89     136    689    | 4      129    5      | | 358    4      15     | 2      137    789    | 179    179    6      | +----------------------+----------------------+----------------------+ | 47     16     168    | 3      9      5      | 12678  1247   12478  | | 47     5      138    | 6      247    27     | 189    139    138    | | 2      9      36     | 47     8      1      | 567    3457   347    | +----------------------+----------------------+----------------------+

Your explanation is absolutely clear. Since I'm not familiar yet with the
Medusa technique, I'll just take your word that your method is simpler.
Another way of arriving at the r3c6<>9 deduction is with a simple Nice Loop:

 Code: [r3c6]=6=[r3c3]-6-[r1c2]-3-[r3c1]-9-[r3c6] => r3c6<>9

For an explanation of the notation used in Nice Loops, see http://www.sudoku.com/forums/viewtopic.php?t=3628 for more information.
Ron Moore

Joined: 13 Aug 2006
Posts: 72
Location: New Mexico

Posted: Wed Nov 15, 2006 9:33 pm    Post subject: UR "Unique Quad" and Other Comments

Para,

The Sudocue program allows you to enable or disable the various solving techniques it can use in its solution. When a new puzzle is loaded into Sudocue, it will report to you if it is unable to solve the puzzle using only the currently enabled techniques. Thus, if you disable the Medusa techniques and Sudocue does not report an inability to solve the puzzle, then you can be sure they are not needed. To change the set of enabled techniques, from the menu choose Tools, then Options, then Solver. I don't remember exactly when they began, but I feel safe in saying that none of the puzzles in the Archive before Oct 2006 require Medusa techniques.

From the position you posted, I used an ALS XZ rule elimination followed by a 4 cell XY chain to complete the solution, without need of any of the Medusa techniques. (In case you're not yet familiar with the ALS XZ technique, refer to the Solving Guide on this site.) I had not run the solver on this puzzle, but after seeing your post, I ran it with the ALS XZ technique enabled and the Medusa techniques disabled. In your posted position (shown below), it found the elimination of 6 from r1c2 which you mentioned, using two fairly large ALS's -- r1c345678 and r3c3789 -- with 3 as the "restricted common" digit.

 Code: .------------------.------------------.------------------. | 1    B36    256  | 478   2467  2678 | 2578  2-357  9   | | 59    8     4    | 79    27    3    | 1257  6     127  | | 39    7    B26   | 1     5     2689 |A28   A234  A2348 | :------------------+------------------+------------------: | 6     12    9    | 5     17    4    | 3     8     127  | | 38    123   7    | 89    136   689  | 4     129   5    | | 358   4     15   | 2     137   789  | 179   179   6    | :------------------+------------------+------------------: | 47    16    168  | 3     9     5    | 12678 1247  12478| | 47    5     138  | 6     247   27   | 189   139   138  | | 2     9     36   | 47    8     1    | 567   3457  347  | '------------------'------------------'------------------'

To my human eye, the ALS in r3c789 -- marked with "A" in the diagram, was more readily apparent. It can be used with the ALS {r1c2,r3c3} -- marked with "B" -- to eliminate 3 from r1c8, using 2 as the "restricted common" digit. (Since r1c2 and r1c8 are strongly linked for digit 3, this immediately implies r1c2=3, as does your and Sudocue's elimination of 6 from r1c2).

This position is reached after either elimination is followed up with basic techniques:

 Code: .------------------.------------------.------------------. | 1     3     26   | 47    2467  8    | 5     27    9    | | 5     8     4    | 9    A27    3    | 127   6     1-2-7| | 9     7     26   | 1     5    B26   | 8     34   #34   | :------------------+------------------+------------------: | 6     12    9    | 5     1-7   4    | 3     8    #27   | | 3     12    7    | 8     16   C69   | 4     29    5    | | 8     4     5    | 2     3    D79   | 179   179   6    | :------------------+------------------+------------------: | 47    6    *18   | 3     9     5    | 127   1247 *12478| | 47    5    *18   | 6     247   27   | 19    139  *138  | | 2     9     3    | 47    8     1    | 6     5    #47   | '------------------'------------------'------------------'

Here there is a short XY chain, in cells A (r2c5), B, C, D in the diagram, which eliminates candidate 7 from r4c5:

(7=2)r2c5 - (2=6)r3c6 - (6=9)r5c6 - (9=7)r6c6 => r4c5 <> 7

The solution is easily completed after this elimination. This is really not the reason for my response, however.

The solution path found by the Sudocue solver is more complex, but it does find a very interesting uniqueness elimination in the position above. Cells r78c39 (marked with "*" in the diagram) form the potential non-unique rectangle, with surplus candidates 2,3,4,7 on one side of the rectangle, in column 9. These combine with cells r349c9 (marked with "#" in the diagram) to form, in effect, a naked quad in column 9, which eliminates 2 and 7 from r2c9. This is termed a "unique subset" pattern in the Solving Guide, in this case a "unique quad." Alongside this example, my discovery of a "unique triple" described in this post looks rather pale.

Another way to view the above is to note that r78c9 are strongly linked for digit 8; therefore neither cell can be 1 since that would force the other to 8, forming the non-unique rectangle. This leaves only r2c9 available for candidate 1 in column 9. However, this would miss the unique quad pattern.

To GreenLantern: I haven't broken down to using the Medusa techniques, but I think "Medusa wrap" is an aid to locating "nice loops" in the grid, though perhaps not called by that name. In the example in the Solving Guide, Ruud gives an equivalent loop for the conflict revealed by the Medusa wrap. I seldom use formal notation for loops, and I suppose it's time I began. Looking at your nice loop (shown below), shouldn't the link between r1c2 and r3c1 (via candidate 3) be a strong link? Otherwise I don't see the alternation of strong and weak links and I don't see how any inference could be drawn.

 Code: [r3c6]=6=[r3c3]-6-[r1c2]-3-[r3c1]-9-[r3c6] => r3c6<>9

To Ruud: On the page describing Eureka notation, in the portion under "Strong Inferences" shown below, I believe the second line is in error, if my interpretation of your symbology is correct. I interpret the "∨" symbol to mean logical "or." What's needed is logical "exclusive or." To avoid more possibly confusing symbology, perhaps you could use "<>." as you explain in the following text.

 Code: (P=false => Q=true) & (Q=false => P=true) (P ∨ Q) P=Q

A nit: In the introductory paragraph, next to last sentence, there is a "reversible" error.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Wed Nov 15, 2006 10:46 pm    Post subject: Hi Yes i know about enabling and disabling. I already turned off unique patterns and BUG +1, mostly because i preferably try to avoid them even when i see them. I don't know if there are any puzzles in which you are forced to use them. I tried disabling medusa. But it got annoying to hear every time that the solver settings weren't correct to solve the puzzle so i just kept it on. I just like to watch sudocue solve the puzzle and comparing it to how i solved the puzzle. I was just surprised about how sudocue solved the puzzle this time. I read the ALS XZ section but i don't really understand how to use it. Of those 3 techniques mentioned in that section (XYZ-wing, APE and ALS XZ) i really only get how to use and spot the XYZ-wing. When i read ALS XZ and see examples of it, i don't really know what happened and how to recognize it. I was mostly curious under what technique my step to eliminate the 6 in r1c2 falls because i couldn't quite place it. I used it in other puzzles as well since i first did it. Just a pattern i see easily i guess. After that elimination i was able to solve it pretty much without any problems. I'll check of some of the earlier puzzles with medusa turned off. Thank you. Para
GreenLantern
Regular

Joined: 30 May 2006
Posts: 14

Posted: Thu Nov 16, 2006 1:43 am    Post subject:

 Ron Moore wrote: I seldom use formal notation for loops, and I suppose it's time I began. Looking at your nice loop (shown below), shouldn't the link between r1c2 and r3c1 (via candidate 3) be a strong link? Otherwise I don't see the alternation of strong and weak links and I don't see how any inference could be drawn.

Nice Loops can be one of four principal types: 1)Strong Inference Loops
3) Alternating Inference Loops (eg x-cycles and x-wings), and 4) Mixed
Inference Loops. The Discontinuous Nice Loop I wrote is in the fourth
category. So long as the loop strictly follows Nice Loop propagation
rules, the inferences it makes are valid.

http://www.sudoku.com/forums/viewtopic.php?t=2143
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

Posted: Fri Nov 17, 2006 9:02 am    Post subject:

 Code: .------------------.------------------.------------------. | 1    D36    256  | 478   2467  2678 | 2578  2357   9   | | 59    8     4    | 79    27    3    | 1257  6     127  | |C39    7    A26   | 1     5     2689 |B28   B234  B2348 | :------------------+------------------+------------------: | 6     12    9    | 5     17    4    | 3     8     127  | | 38    123   7    | 89    136   689  | 4     129   5    | | 358   4     15   | 2     137   789  | 179   179   6    | :------------------+------------------+------------------: | 47    16    168  | 3     9     5    | 12678 1247  12478| | 47    5     138  | 6     247   27   | 189   139   138  | | 2     9     36   | 47    8     1    | 567   3457  347  | '------------------'------------------'------------------'

Another way of looking at it is using the Almost Locked Set in r3c789 in an AIC:

(6=2)A - (2=4&8&3)B - (3)C = (3)D which implies A (r3c3) is 6 or D (r1c2) is 3. In either case, r1c2 cannot be 6.

Now what Para did was the little known trick of using the complement of the Almost Locked Set, which I've called the Weak ALS. Note that the remaining cells of a group that are not part of the ALS form the Weak ALS. The digits that are not in the ALS must be in the Weak ALS. If there are N cells in the Weak ALS, then there must be N-1 non-ALS digits in the Weak ALS. What this means is that there is room for only one ALS digit in the Weak ALS cells. Thus there is an exclusionary weak link between every ALS digit candidate in the Weak ALS cells (that is why I call it a weak ALS). You can use these weak links in an AIC.

 Code: .------------------.------------------.------------------. | 1     36    256  | 478   2467  2678 | 2578  2357   9   | | 59    8     4    | 79    27    3    | 1257  6     127  | |A39    7    A26   | 1     5    A2689 |B28   B234  B2348 | :------------------+------------------+------------------: | 6     12    9    | 5     17    4    | 3     8     127  | | 38    123   7    | 89    136   689  | 4     129   5    | | 358   4     15   | 2     137   789  | 179   179   6    | :------------------+------------------+------------------: | 47    16    168  | 3     9     5    | 12678 1247  12478| | 47    5     138  | 6     247   27   | 189   139   138  | | 2     9     36   | 47    8     1    | 567   3457  347  | '------------------'------------------'------------------'

The Weak ALS represented by cells marked with an A. The non-ALS digits are 6 & 9, and they must be included in A. The weak ALS rule tells you there is a weak link between the 3 in r3c1 and the 2 in r3c3. Thus you have...

(3)r1c2 = [3(r3c1)&6&9]A - [6&9&2(r3c3)]A = (6)r3c3 => r1c2 <> 3

...which is precisely the deduction that Para came up with. There is more on Weak ALS at the following link.
http://www.sudoku.com/forums/viewtopic.php?p=34826&highlight=#34826
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

Posted: Fri Nov 24, 2006 6:50 am    Post subject:

Myth Jellies wrote:
 Code: .------------------.------------------.------------------. | 1    D36    256  | 478   2467  2678 | 2578  2357   9   | | 59    8     4    | 79    27    3    | 1257  6     127  | |C39    7    A26   | 1     5     2689 |B28   B234  B2348 | :------------------+------------------+------------------: | 6     12    9    | 5     17    4    | 3     8     127  | | 38    123   7    | 89    136   689  | 4     129   5    | | 358   4     15   | 2     137   789  | 179   179   6    | :------------------+------------------+------------------: | 47    16    168  | 3     9     5    | 12678 1247  12478| | 47    5     138  | 6     247   27   | 189   139   138  | | 2     9     36   | 47    8     1    | 567   3457  347  | '------------------'------------------'------------------'

Another way of looking at it is using the Almost Locked Set in r3c789 in an AIC:

(6=2)A - (2=4&8&3)B - (3)C = (3)D which implies A (r3c3) is 6 or D (r1c2) is 3. In either case, r1c2 cannot be 6.

Now what Para did was the little known trick of using the complement of the Almost Locked Set, which I've called the Weak ALS. Note that the remaining cells of a group that are not part of the ALS form the Weak ALS. The digits that are not in the ALS must be in the Weak ALS. If there are N cells in the Weak ALS, then there must be N-1 non-ALS digits in the Weak ALS. What this means is that there is room for only one ALS digit in the Weak ALS cells. Thus there is an exclusionary weak link between every ALS digit candidate in the Weak ALS cells (that is why I call it a weak ALS). You can use these weak links in an AIC.

 Code: .------------------.------------------.------------------. | 1     36    256  | 478   2467  2678 | 2578  2357   9   | | 59    8     4    | 79    27    3    | 1257  6     127  | |A39    7    A26   | 1     5    A2689 |B28   B234  B2348 | :------------------+------------------+------------------: | 6     12    9    | 5     17    4    | 3     8     127  | | 38    123   7    | 89    136   689  | 4     129   5    | | 358   4     15   | 2     137   789  | 179   179   6    | :------------------+------------------+------------------: | 47    16    168  | 3     9     5    | 12678 1247  12478| | 47    5     138  | 6     247   27   | 189   139   138  | | 2     9     36   | 47    8     1    | 567   3457  347  | '------------------'------------------'------------------'

The Weak ALS represented by cells marked with an A. The non-ALS digits are 6 & 9, and they must be included in A. The weak ALS rule tells you there is a weak link between the 3 in r3c1 and the 2 in r3c3. Thus you have...

(3)r1c2 = [3(r3c1)&6&9]A - [6&9&2(r3c3)]A = (6)r3c3 => r1c2 <> 6

...which is precisely the deduction that Para came up with. There is more on Weak ALS at the following link.
http://www.sudoku.com/forums/viewtopic.php?p=34826&highlight=#34826
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