## Sue De Coq Revisited Again (ASI#1)

Advanced methods and approaches for solving Sudoku puzzles
Mage wrote:After the following AIC loop, you need only a single pair to crack the puzzle.
(sorry for the probably non-accademic AIC notation) :

7r3c9=7r56c9 - (7=1)r4c7 - (1=2)r4c8 - 2r13c8=2r3c9 - loop
==> r3c9<>35, r5c7<>7, r4c14<>1, r6c8<>2

Mage

Very nice!

Slightly different version of the loop:
r3c9 =2= r6c9 -2- r4c8 -1- r4c7 -7- r13c7 =7= r3c9 => r4c14,r5c79,r6c8<>1, r6c8<>2, r3c9<>35, r5c7<>7
hobiwan
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tarek wrote:a SDC puzzle
Code: Select all
`. . .|8 . 2|. . .. . 4|. . .|3 . .. 6 .|. . .|. 8 .-----+-----+-----9 . .|4 . 8|. . 5. . .|. . .|. . .8 . .|2 . 5|. . 4-----+-----+-----. 1 .|. . .|. 2 .. . 6|. . .|1 . .. . .|3 . 7|. . .`
tarek

Okay...so I just promised Don that I wouldn't start posting again...and I'm not going to, but just this once as I had so much fun solving ronk's puzzle and now tarek's puzzle.

First, (after SSTS, of course) I found a couple Sue De Coq's. Unfortunately, all of the one's I found still left an xy-wing to solve the puzzle (at least they do individually). Then I noticed the potential deadly pattern in r46c257[37/36/67]. This gives the easy elimination r6c8<>9. However, this again leaves a couple of xy-wings. On the hand, a larger chain using the same potential deadly pattern, implies r7c9<>9 and this does crack the puzzle in one step.

But then I remembered that not everyone like uniqueness steps and so I tried to see another way to kill that 9. It took all of a click of the pairs button in simple sudoku followed by clicking on the 9 button to see that a transported xyz-wing will do the job. More precisely, the xyz-wing in r2c8/r9c89[569] implies a strong inference (9)r2c8=r9c89. From this, a simple chain is born:
(9)r6c7 = r56c8 - r2c8 = r9c89, => r9c7<>9.

There are more things in there, though. An hxy-loop (forgive me, denis) that jumps out at you if you look in rn-space. Almost certainly, there will be some xyt-chains that will crack it (or almost xy-chains, if you're into that sort of thing).

Awesome puzzle, tarek.
re'born

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re'born wrote:Okay...so I just promised Don that I wouldn't start posting again...

Gosh, that makes it sound as if I was asking for a promise not to post. Let's be perfectly clear- I was more like pl-e-a-d-d-ding for the guy to post.

And an interesting post it is! fwiw: I am a major supporter of looking on the assumption of uniqueness as if it had been added to the sudoku constitution- maybe there's time to add a Proposition (to those abroad & elsewhere- relates to our impending U.S. election ). However, I like the plan B using transport, a method you've championed and which is too easily forgot.
DonM
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DonM wrote:
re'born wrote:Okay...so I just promised Don that I wouldn't start posting again...

Gosh, that makes it sound as if I was asking for a promise not to post. Let's be perfectly clear- I was more like pl-e-a-d-d-ding for the guy to post.

And an interesting post it is! fwiw: I am a major supporter of looking on the assumption of uniqueness as if it had been added to the sudoku constitution- maybe there's time to add a Proposition (to those abroad & elsewhere- relates to our impending U.S. election ). However, I like the plan B using transport, a method you've championed and which is too easily forgot.

Don is correct and I apologize for the insinuation

While I'm here, I might as well mention:
(6=5)r1c7-(5)r9c7=(5-6)r9c8=(6)r6c8, => r46c7<>6, solving the puzzle.

As well,
re'born wrote:Almost certainly, there will be some xyt-chains that will crack it

(9=6)r9c9 - (6)r1c9 = (6)r1c7 - (6=7)r4c7 - (7=9)r6c7/(6)r1c7, => r9c7<>9
(Okay, so this, strictly speaking, isn't an xyt-chain, but it is easier than the legitimate one:
(9=6)r9c9 - (6=7)r1c9 - (7=3)r1c3 - (3=5)r1c5/(7)r1c9 - (5=6)r1c7 - (6=7)r4c7 - (7=9)r1c7/(6)r1c7
that makes the same exclusion.)

Oh, and I forgot to mention that there is another xyz-transport opportunity that makes a different exclusion which solves the puzzle. A hint is that the xyz-wing pivot is r5c6.
re'born

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re'born wrote:
re'born wrote:Almost certainly, there will be some xyt-chains that will crack it

(9=6)r9c9 - (6)r1c9 = (6)r1c7 - (6=7)r4c7 - (7=9)r6c7/(6)r1c7, => r9c7<>9

(Okay, so this, strictly speaking, isn't an xyt-chain, but it is easier than the legitimate one:
(9=6)r9c9 - (6=7)r1c9 - (7=3)r1c3 - (3=5)r1c5/(7)r1c9 - (5=6)r1c7 - (6=7)r4c7 - (7=9)r1c7/(6)r1c7
that makes the same exclusion.)

True, but your "legitimate one" would look a lot prettier with the concept of grouped "xy-nodes" ...

(9=6)r9c9 - (6=7)r1c9 - (7=5)r1c35 - (5=6)r1c7 - (6=9)r46c7 => r9c7<>9

Now where have I seen that before? Isn't that what most people call ALSs?
ronk
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ronk wrote:True, but your "legitimate one" would look a lot prettier with the concept of grouped "xy-nodes" ...

(9=6)r9c9 - (6=7)r1c9 - (7=5)r1c35 - (5=6)r1c7 - (6=9)r46c7 => r9c7<>9

Now where have I seen that before? Isn't that what most people call ALSs?

Touché! It's nice to hear from you again, Ron.
re'born

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here ronk wrote:Is there a simpler solution path?

Thanks to Mage, hobiwan and Denis for posting alternate moves to my Sue de Coq (SdC) move. Thanks also to Luke451 and re'born for commenting on the puzzle.

Let me summarize the suggested moves this way. There are six strong sets {(a) r4c7, (b) r4c8, (c) 2c9 or (d) 2b3, and (e) 7c9 or (f) 7b3} which yield four different continuous loops. Four of the six strong sets (aka inferences, links) have r3c9 in common. All four continuous loops have the identical ten eliminations.

Code: Select all
` 6      57     9      | 234    2345   2345   | f1347  d1245     8 23     58     23     | 7      458    1      |  469    4569     569 4      578    1      | 69     2358   69     | f37    d25   cdef2357----------------------+----------------------+------------------------ 157    9      8      | 123    6      2357   |a17    b12       4 157    2      46     | 1489   1457   45789  |  16789  3       e1679 17     3      46     | 12489  1247   2479   |  5      12689  ce12679----------------------+----------------------+------------------------ 239    1      23     | 5      34     3468   |  34689  7        369 8      46     57     | 12346  9      23467  |  1346   1456     1356 39     46     57     | 13468  1347   34678  |  2      145689   13569r3c9 =2= r13c8 -2- r4c8 -1- r4c7 -7- r56c9 =7= r3c9 = continuous (Mage's; reversed)r3c9 =2= r6c9 -2- r4c8 -1- r4c7 -7- r13c7 =7= r3c9 = continuous (hobiwan's)r3c9 =2= r6c9 -2- r4c8 -1- r4c7 -7- r67c9 =7= r3c9 = continuous (Denis's; loop closed & rotated)r3c9 =2= r13c8 -2- r4c8 -1- r4c7 -7- r13c7 =7= r3c9 = continuous (a 4th combination)All four loops have the same ten eliminations:r4c14,r5c79,r6c89<>1, r6c8<>2, r3c9<>35, r5c7<>7`

I made modifications to original chains to make them look as similar as possible. My personal preference is the one based on Denis's last nrct-chain, because it only uses five cells. However, some might not like the overlap at r6c9.

ronk also wrote:Can you find the Sue de Coq?

The hidden answer has been added here. hobiwan, re'born, it's not clear to me whether or not you found the SdC.

[edit: "7b3" was incorrectly "b3"]
Last edited by ronk on Thu Oct 30, 2008 6:04 am, edited 1 time in total.
ronk
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ronk wrote:
ronk also wrote:Can you find the Sue de Coq?

The hidden answer has been added here. hobiwan, re'born, it's not clear to me whether or not you found the SdC.

I did. I'm PM'd my solution to Don a couple of days ago before I worked up the courage to post publicly again. Here is the relevant excerpt from that message:

re'born to DonM wrote:
By the way, I found the Sue de Coq in ronk's puzzle at the end of your thread. It's a bit odd, but subset counting makes it brutally clear why it works. The cells are r1c8, r2c789, r3c8 and r4c8. The eliminations are r689c8<>1, r6c8<>2, r1c7<>4, r3c9<>5.
re'born

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ronk wrote:hobiwan, re'born, it's not clear to me whether or not you found the SdC.

Almost Locked Set XZ-Rule: A=r2c789 - {4569}, B=r134c8 - {1245}, X=4,5, Z=4,5 => r1c7<>4, r3c9<>5

It missed the eliminations for 1 and two in c8.
hobiwan
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ronk wrote:After the Simple Sudoku Technique Set (SSTS), this puzzle can be solved with a Sue de Coq, a naked triple, a naked pair and singles. Can you find the Sue de Coq Is there a simpler solution path
Code: Select all
`top1465 #136.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2..After SSTS: 6      57     9      | 234    2345   2345   | 1347   1245   8 23     58     23     | 7      458    1      | 469    4569   569 4      578    1      | 69     2358   69     | 37     25     2357----------------------+----------------------+--------------------- 157    9      8      | 123    6      2357   | 17     12     4 157    2      46     | 1489   1457   45789  | 16789  3      1679 17     3      46     | 12489  1247   2479   | 5      12689  12679----------------------+----------------------+--------------------- 239    1      23     | 5      34     3468   | 34689  7      369 8      46     57     | 12346  9      23467  | 1346   1456   1356 39     46     57     | 13468  1347   34678  | 2      145689 13569`

hidden answer (triple click text area to see) wrote:Sets: A = {r13c8} = {1245}; B = {r2c789} = {4569}; C = {r4c8} = {12}

Sets A,B share digits 4,5; sets A,C share digits 1,2

Elims: r1c7<>4, r3c9<>5, r6c8<>12, r89c8<>1
Thanks for the reveal, Ronk. I was beginning to think you'd forgotten abt this conundrum. Earlier I figured the solution here "must be a wrinkle outside the 'core-a-b' pattern mentioned above." Now I see what the wrinkle is (other than the one you've put on me forehead . )

I had never seen a SdC where B (or C) wasn't simply a single bivalue cell. Now I have. It makes sense, really. It's still N cells with N candidates, with B and C sharing core candidates but not sharing digits with each other.

Knowing this and actually finding one are two far different things. Still, thanks for the learning experience.

Luke
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Yes, thanks ronk for that example. That and other contributions here have been appreciated & are just what I hoped this thread to accomplished. I may put this up as a graphic with the initial post since it is a good example of a SDC that doesn't jump out at you- there may be others like that lurking in other puzzles that we'd like to be able to find.
DonM
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Code: Select all
`Tarek's SDC. . .|8 . 2|. . .. . 4|. . .|3 . .. 6 .|. . .|. 8 .-----+-----+-----9 . .|4 . 8|. . 5. . .|. . .|. . .8 . .|2 . 5|. . 4-----+-----+-----. 1 .|. . .|. 2 .. . 6|. . .|1 . .. . .|3 . 7|. . .`

After SSTS, the resultant PM shows 4 possible distributed disjoint subsets - 2 DDS & 2 ADDS per http://forum.enjoysudoku.com/viewtopic.php?t=5423
Any one of which is sufficient to solve the puzzle afterwards with SSTS. Sue de Coq's can be described as distributed disjoint subsets, but I'm not sure if ADDS (the distributed disjoint asubset below) qualifies as a SDC.
Code: Select all
`After SSTS.------------------.------------------.------------------.| 1     9     37   | 8     357   2    |+56    4     67   || 2     8     4    | 6     579   19   | 3    +59    179  || 5     6     37   | 17    4     139  | 2     8     179  |:------------------+------------------+------------------:| 9     37    2    | 4     36    8    |+67    1     5    || 6     4     5    | 17    379   139  | 8     3-9   2    || 8     37    1    | 2     369   5    |+679   36-9  4    |:------------------+------------------+------------------:| 7     1     9    | 5     8     6    | 4     2     3    || 3     5     6    | 9     2     4    | 1     7     8    || 4     2     8    | 3     1     7    | 569   569   69   |'------------------'------------------'------------------'distributed disjoint asubset[3x4] c7b36.<5679> at r1c7 r2c8 r4c7 r6c7r5c8 <> 9 r6c8 <> 9.------------------.------------------.------------------.| 1     9     37   | 8     35-7  2    | 56    4     67   || 2     8     4    | 6    +579  +19   | 3    +59    17-9 || 5     6     37   |+17    4     39-1 | 2     8     179  |:------------------+------------------+------------------:| 9     37    2    | 4     36    8    | 67    1     5    || 6     4     5    | 17    379   139  | 8     3     2    || 8     37    1    | 2     369   5    | 679   36    4    |:------------------+------------------+------------------:| 7     1     9    | 5     8     6    | 4     2     3    || 3     5     6    | 9     2     4    | 1     7     8    || 4     2     8    | 3     1     7    | 569   569   69   |'------------------'------------------'------------------'distributed disjoint subset[2x4] r2b2.<1579> at r2c5 r2c6 r2c8 r3c4 r1c5 <> 7 r2c9 <> 9 r3c6 <> 1.------------------.------------------.------------------.| 1     9     37   | 8     35    2    | 56    4     67   || 2     8     4    | 6     57-9 +19   | 3     59    17   || 5     6    +37   |+17    4    +39   | 2     8     19-7 |:------------------+------------------+------------------:| 9     37    2    | 4     36    8    | 67    1     5    || 6     4     5    | 17    379   13-9 | 8     39    2    || 8     37    1    | 2     369   5    | 679   369   4    |:------------------+------------------+------------------:| 7     1     9    | 5     8     6    | 4     2     3    || 3     5     6    | 9     2     4    | 1     7     8    || 4     2     8    | 3     1     7    | 569   569   69   |'------------------'------------------'------------------'distributed disjoint subset[3x4] r3c6b2.<1379> at r2c6 r3c3 r3c4 r3c6r2c5 <> 9 r3c9 <> 7 r5c6 <> 9.------------------.------------------.------------------.| 1     9     37   | 8     35    2    |+56    4     67   || 2     8     4    | 6     57   +19   | 3    +59    17   || 5     6     37   | 17    4     39   | 2     8     19   |:------------------+------------------+------------------:| 9     37    2    | 4    +36    8    | 7-6   1     5    || 6     4     5    | 17    379  +13   | 8     39    2    || 8     37    1    | 2     369   5    | 679   369   4    |:------------------+------------------+------------------:| 7     1     9    | 5     8     6    | 4     2     3    || 3     5     6    | 9     2     4    | 1     7     8    || 4     2     8    | 3     1     7    | 569   569   69   |'------------------'------------------'------------------'distributed disjoint asubset[4x5] r2c6b35.<13569> at r1c7 r2c6 r2c8 r4c5 r5c6r4c7 <> 6`
PIsaacson

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PIsaacson wrote:distributed disjoint asubset[3x4] c7b36.<5679> at r1c7 r2c8 r4c7 r6c7
r5c8 <> 9 r6c8 <> 9

I don't think you can call this a SDC (at least not according to SDC's original definition), but it is a perfectly valid ALS-XZ.
PIsaacson wrote:distributed disjoint subset[2x4] r2b2.<1579> at r2c5 r2c6 r2c8 r3c4
r1c5 <> 7 r2c9 <> 9 r3c6 <> 1

Definitely a SDC.
PIsaacson wrote:distributed disjoint subset[3x4] r3c6b2.<1379> at r2c6 r3c3 r3c4 r3c6
r2c5 <> 9 r3c9 <> 7 r5c6 <> 9

Definitely a SDC (I would have missed r5c6<>9!)
PIsaacson wrote:distributed disjoint asubset[4x5] r2c6b35.<13569> at r1c7 r2c6 r2c8 r4c5 r5c6
r4c7 <> 6

Not an SDC (too many boxes involved), but a valid ALS-XY.
hobiwan
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hobiwan,

Thanks for your analysis, but I'm still confused by the definition of a Sue de Coq based on the following from http://www.sudopedia.org/wiki/Sue_de_Coq:
The Sue de Coq technique uses two intersecting sets A and B, where A is a set of N cells with N candidates in a line, B is a set of N cells with N candidates in a box, and the sets A - B and B - A have no common candidates. Candidates can be eliminated from the cells in the line that are not in A, and the cells in the box that are not in B.

This technique was introduced by a member using the alias Sue de Coq in the Sudoku Players' forum, who named it Two-Sector Disjoint Subsets, but a remark that this was a true Sue de Coq gave the technique its present name.

The reason I'm confused is that all of my examples contain (at least) an intersecting A ALS and a B ALS. Shouldn't the wiki be corrected to state that the A/B sets are N cells with N+1 candidates?
PIsaacson

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Joined: 02 July 2008

yeah i am also have the same question as paul.
Some do, some teach, the rest look it up.

StrmCkr

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