## The Almost Sue De Coq (ASI#2)

Advanced methods and approaches for solving Sudoku puzzles

### The Almost Sue De Coq (ASI#2)

The concept of an Almost Sue De Coq pattern is not very complicated once one understands Sue De Coq itself. And at this point in advanced manual solving when we are all searching for new sources of strong links, the concept is extremely attractive especially when all of the many potential weak link ‘targets’ are considered. Also, since according to Ruud’s study in his ‘Sue De Coq revisited’ thread, Sue De Coq itself occurs more frequently than patterns such as jellyfish, naked quads and hidden triples, it stands to reason that the ‘almost Sue De Coq’ will occur much more frequently, particularly since there are at least 4 different places in the pattern where the digit responsible for the ‘almost’ situation can occur. All of which makes it rather surprising that, as far as I can tell, this pattern has never been discussed or used in examples on the forums though I'm sure others have thought about the concept.

The example graphics pretty much tell the story so not much further explanation is necessary except to point out that if one extra digit, beyond that necessary for a valid Sue De Coq, is present in either the core 2 or 3 cells (ie. the aals or aaals within the same box), the bivalue cell A in the box, the bivalue cell B in the row or column or the additional cell C (if present) then that digit can be used as the basis for the strong link.

But for the 3 in r4c1, the basic Sue De Coq pattern is present: the core blue cells (an aals with 1,5,6,7,9 in 3 cells), the brown cell A with 5,7 and the green cell C with 1,9. The logic is straightforward: If not 3 in r4c1 then the Sue De Coq is true and any of the circled green cells are valid weak link ‘targets’. In this case a relatively simple chain eliminates the 9 in r4c1:

aic:
(3)r4c1 = suedecoq[(15679)r4c123/r4c5(19)/r5c3(57)] - (6)r4c7 = r6c7-(6=8)r6c3 - (8=9)r4c2 => r4c1<>9
nice loop:
r4c1 -9- r4c1 =3= suedecoq[(15679)r4c123/r4c5(19)/r5c3(57)] -6- r4c7 =6= r6c7 -6- r6c3 -8- r5c2 -9- r4c1 => r4c1<>9

Here, the situation appears a little more complex, but the principle is the same. Once again, but for one digit, the 8 in r3c9, there would be a valid Sue De Coq. In this case, the core is an aaals (3,4,5,7,8,9 in 3 cells), the brown cell A contains 5,8, the blue cell B contains 3,7 and there is a gold cell C present containing the extra digit 4 necessary because of the aaals of the core. However, cell C also contains one extra digit 8 that makes the Sue De Coq invalid. The logic is similar to the previous example: If not 8 in r3c9 then the Sue De Coq is true and any of the circled green cells are valid weak link ‘targets’. This example shows that not only are there several of the latter available, but also that occasionally a group may be available. The chain above takes advantage of the group of 9s in r3c456.

aic:
(8)r3c9 = sdc[(345789)r3c456/r1c4(58)/r3c7(37)/r3c9(347)] – grp(9)r2c456 – r2c2 – (9=8)r4c2 – r6c3 = (8)r6c5
nice loop:
r3c5 -8- r3c9 =8= sdc[(345789)r3c456/r1c4(58)/r3c7(37)/r3c9(347)] -9- r2c456 =9= r2c2 -9- r5c2 -8- r6c3 =8= r6c5 -8- r3c5

The question might be raised as to whether most or all chains using an almost-sdc could be subsumed by other chains. However, given all the possibilities using the almost-sdc, I rather doubt it. There is, in fact, a separate chain that can be used to achieve the same elimination in the first example above. There may be one also for the second example; I didn’t look very hard. Even if it were possible to find alternate chains for most or all of the possible almost-sdc chains (again, which I doubt), finding an almost-sdc pattern allows one to see many more possibilities at once rather than having to search them out individually. Also is the fact that, in a given overall solution, a chain using an almost-sdc pattern may be possible without requiring a previous move that might have been necessary for an alternate route to achieve the same elimination making the overall solution shorter (more elegant?). Not to mention that the Sue De Coq is a ‘what is’ pattern. There is no additional assumptivity compared to other basic ‘almost’ patterns that makes it less attractive for use.

Finally, I mentioned that the almost Sue De Coq pattern should be found even more frequently than Sue De Coq itself. If you haven’t noticed already, note that the two examples above are not only from the same puzzle, but are at exactly the same point in the puzzle!

(The above puzzle is the UK forum Weekly Extreme #109 ER=8.4. ASI stands for Advanced Sudoku Illustrated)
Last edited by DonM on Mon Oct 27, 2008 11:07 am, edited 2 times in total.
DonM
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### Re: The Almost Sue De Coq (ASI#2)

DonM wrote:at this point in advanced manual solving when we are all searching for new sources of strong links

You've got one, but you fail to recognise it: the z and t extensions.

A pattern occurs more or less frequently according to the patterns after which you apply it. In the present case, I wouldn't use Sue-de-Coq as there is a solution based on much simpler patterns.

interaction row r8 with block b8 for number 8 ==> r7c6 <> 8, r7c5 <> 8, r7c4 <> 8
interaction column c3 with block b4 for number 7 ==> r5c2 <> 7, r4c2 <> 7, r4c1 <> 7
interaction block b5 with column c5 for number 1 ==> r9c5 <> 1, r7c5 <> 1, r2c5 <> 1
interaction block b1 with column c1 for number 5 ==> r9c1 <> 5, r8c1 <> 5, r7c1 <> 5, r4c1 <> 5
nrc-chain[3] n9{r7c9 r8c8} - {n9 n2}r6c8 - n2{r6c7 r7c7} ==> r7c9 <> 2
nrc-chain[3] {n9 n1}r4c5 - n1{r6c5 r6c1} - n3{r6c1 r4c1} ==> r4c1 <> 9
nrc-chain[3] n9{r3c1 r6c1} - n1{r6c1 r6c5} - {n1 n9}r4c5 ==> r3c5 <> 9
nrc-chain[4] n5{r4c9 r5c8} - n4{r5c8 r5c7} - {n4 n2}r7c7 - {n2 n5}r9c9 ==> r7c9 <> 5
nrc-chain[4] n4{r5c7 r5c8} - n5{r5c8 r4c9} - {n5 n2}r9c9 - {n2 n4}r7c7 ==> r3c7 <> 4
nrc-chain[3] n4{r3c5 r3c9} - n8{r3c9 r1c9} - {n8 n5}r1c4 ==> r3c5 <> 5
nrc-chain[4] n4{r5c8 r5c7} - {n4 n2}r7c7 - {n2 n5}r9c9 - n5{r8c8 r5c8} ==> r5c8 <> 9
nrc-chain[2] n9{r5c6 r5c2} - n9{r6c1 r3c1} ==> r3c6 <> 9
hidden-pairs-in-a-row {n5 n9}r3{c1 c4} ==> r3c4 <> 8, r3c4 <> 3, r3c1 <> 7
hidden-pairs-in-a-column {n1 n3}{r2 r7}c4 ==> r7c4 <> 9, r7c4 <> 5, r2c4 <> 9
swordfish-in-columns n9{r6 r3 r8}{c1 c4 c8} ==> r8c6 <> 9, r8c5 <> 9, r6c9 <> 9, r6c5 <> 9
nrc-chain[2] n9{r5c6 r4c5} - n9{r4c9 r7c9} ==> r7c6 <> 9
xyz-chain[3] {n3 n2}r6c9 - {n2 n7}r2c9 - {n7 n3}r3c7 ==> r3c9 <> 3
nrc-chain[3] n9{r6c8 r6c1} - n1{r6c1 r6c5} - {n1 n9}r4c5 ==> r4c9 <> 9
...singles...
GRID 0 SOLVED. LEVEL = L4, MOST COMPLEX RULE = NRC4
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denis_berthier
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Don, interesting work!

As you may remember, I studied almost Sue de Coqs a little while ago and came to the conclusion that they would all be subsumed by AICs with embedded ALS and AHTs. However you have come up with a case which I overlooked in your second example! It's not an almost-almost pattern at all but a straightforward almost disjoint locked set.

I would write it like this:
(8)r3c9 = (345789)DLS:r3c45679,r1c3[(3479)r3,(5)c4,(8)b2] – (9)r2c456 = (9)r2c2 – (9=8)r4c2 – (8)r6c3 = (8)r6c5 => r3c5 <> 8

To expand this notation:
(345789)DLS:r3c45679,r1c3 = we have 6 digits and 6 cells in a disjoint locked set
[(3479)r3, (5)c4,(8)b2] = 4 digits are constrained to single instances in row3, one in column4 and one in box2
Thus each digit will occur just once.

I had always been adding extra digits to the cells involved to create the almost condition, but your second case does it by adding an extra copy of one of the digits in a second intersecting house which I missed.

BTW there are typos in your first chain which I'd would write as:
(3)r4c1 = (15679)DLS:r4c1235,r5c4[(169)r4,(7)c3,(5)b4] - (6)r4c7 = (6)r6c7 - (6=8)r6c3 - (8=9)r4c2 => r4c1 <>9

ie I'd still call it a Disjoint Locked Set even though it’s a almost Sue de Coq!

For the almost cases I was exploring, I came to the conclusion that there were so many ways to add the extra digit it was too time consuming to check the all the possible permutations, as I'd get them anyway with my chains. Now I'll need to check if that’s true for your almost condition too.

I hope you didn't reach that position via a kraken node deduction - such a pity if you did!
David P Bird
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Good to hear from you David.

David P Bird wrote:Don, interesting work!

As you may remember, I studied almost Sue de Coqs a little while ago and came to the conclusion that they would all be subsumed by AICs with embedded ALS and AHTs. However you have come up with a case which I overlooked in your second example! It's not an almost-almost pattern at all but a straightforward almost disjoint locked set.

No I didn't see that, but would be interested- can you give me a link?

I had always been adding extra digits to the cells involved to create the almost condition, but your second case does it by adding an extra copy of one of the digits in a second intersecting house which I missed.

For the almost cases I was exploring, I came to the conclusion that there were so many ways to add the extra digit it was too time consuming to check the all the possible permutations, as I'd get them anyway with my chains. Now I'll need to check if that’s true for your almost condition too.

The significance of the A-SDC remains to be seen. It may well be that often the same eliminations can be accomplished in other ways. I would like to think that that isn't always true because any new possible source of a strong link would be welcome. My hunch is that there will be a few A-SDC moves that couldn't be easily replicated with other traditional means. But even if it ended up that wasn't true, I look at it this way: I am always looking for SDC now because it will accomplish so many eliminations in one move. If I happen to find an A-SDC move along the way, I'll use it instead of looking for an alternate way to make the move- after all, in my book, it is little different than an ALS when it comes to overall assumptivity. In other words, an A-SDC, if nothing else, may be something like a virtual menu of possibilities with no further search necessary!

BTW there are typos in your first chain which I'd would write as:
(3)r4c1 = (15679)DLS:r4c1235,r5c4[(169)r4,(7)c3,(5)b4] - (6)r4c7 = (6)r6c7 - (6=8)r6c3 - (8=9)r4c2 => r4c1 <>9

Thanks for picking that up. Last night, I was in typo heaven between the 2 posts- I don't know what was the matter with me. Thanks to Luke, I removed a number of them- but apparently not all.

I hope you didn't reach that position via a kraken node deduction - such a pity if you did!

No, I didn't, but I am doing a little renovation of the end of that puzzle's solution (well after the above point).
DonM
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### Re: The Almost Sue De Coq (ASI#2)

DonM wrote:aic:
(8)r3c9 = sdc[(345789)r3c456/r1c4(58)/r3c7(37)/r3c9(347)] – grp(9)r2c456 – r2c2 – (9=8)r4c2 – r6c3 = (8)r6c5

nice loop:
r3c5 -8- r3c9 =8= sdc[(345789)r3c456/r1c4(58)/r3c7(37)/r3c9(347)] -9- r2c456 =9= r2c2 -9- r5c2 -8- r6c3 =8= r6c5 -8- r3c5

David P Bird wrote:It's not an almost-almost pattern at all but a straightforward almost disjoint locked set.

I would write it like this:
(8)r3c9 = (345789)DLS:r3c45679,r1c3[(3479)r3,(5)c4,(8)b2] – (9)r2c456 = (9)r2c2 – (9=8)r4c2 – (8)r6c3 = (8)r6c5 => r3c5 <> 8

It looks like both of your chains depend upon the absence of candidates 5 and 9 in r3c79. In that case, one ALS is sufficient.

r3c5 -8- r1c4 -5- als:r3c45679 -9- r2c456 =9= r2c2 -9- r5c2 -8- r6c3 =8= r6c5 -8- r3c5 ==> r3c5<>8

[edit: The verbose version of als:r3c45679 is als:(r3c456,r3c79 =5|3478|9= r3c456) ]
Last edited by ronk on Mon Oct 27, 2008 12:44 pm, edited 2 times in total.
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great work dom. i have found similar findings as you have nice to see a good explination on it.
Some do, some teach, the rest look it up.

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### Re: The Almost Sue De Coq (ASI#2)

ronk wrote:It looks like both of your chains depend upon the absence of candidates 5 and 9 in r3c79. In that case, one ALS is sufficient.

r3c5 -8- r1c4 -5- als:r3c45679 -9- r2c456 =9= r2c2 -9- r5c2 -8- r6c3 =8= r6c5 -8- r3c5 ==> r3c5<>8
An interesting point Ronk. The DLS only depends on 5 being absent from r3c9 as it must be restricted to the DLS cells in column 4, but as written, the continuation chain requires 9 to be a restricted common in box 2. However this isn’t necessary if we go the other way using the row instead of the box:

(8)r3c9 = (345789)DLS:r3c45679,r1c3[(3479)r3,(5)c4,(8)b2] – (9)r3c1 = (9)r2c2 – (9=8)r4c2 – (8)r6c3 = (8)r6c5 => r3c5 <> 8

This is why I like to show the containing houses for each digit in the notation.

In fact the elimination doesn't even need an ALS:
(8)r3c9 = (8)r1c9 - (8=5)r1c4 - (5)r3c4 = (5-9)r3c1 = (9)r2c2 - (9)r5c2 = (9-8)r5c6 = (8)r6c5 => r3c5 <> 8

I think that generally there will be some route or other around the DLS linking the necessary internal candidates - in this case from (8)r3c9 to (8)r1c4

Don I only mentioned my efforts in passing in my post of 26th Sept here http://www.sudoku.org.uk/SudokuThread.asp?fid=4&sid=10274&p1=0&p2=0 As I remember, I was finding more ADLS than I could use because of the requirements for the restricted common candidate to lead to a strong link. Where that strong link existed, the ADLS could be used, but I found I could always get there using a simpler chain. My trials weren't exhaustive though and there may be exceptions. However I'm beginning to think that this may still hold even with the case you found and I missed as demonstrated above.
David P Bird
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### Re: The Almost Sue De Coq (ASI#2)

David P Bird wrote:However this isn’t necessary if we go the other way using the row instead of the box:

(8)r3c9 = (345789)DLS:r3c45679,r1c3[(3479)r3,(5)c4,(8)b2] – (9)r3c1 = (9)r2c2 – (9=8)r4c2 – (8)r6c3 = (8)r6c5 => r3c5 <> 8

This is why I like to show the containing houses for each digit in the notation.

Since r3c1=9 doesn't remove digit 8 from the "DLS", how does one read that from right-to-left
ronk
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### Re: The Almost Sue De Coq (ASI#2)

denis_berthier wrote:
DonM wrote:at this point in advanced manual solving when we are all searching for new sources of strong links

You've got one, but you fail to recognise it: the z and t extensions.

A pattern occurs more or less frequently according to the patterns after which you apply it. In the present case, I wouldn't use Sue-de-Coq as there is a solution based on much simpler patterns.

Denis, thankyou so much for taking the time to contribute. Having never seen you solve a puzzle, I'll have to take your word for the fact that, manually, you would have found those simpler patterns. fwiw: I already mentioned that I did find an alternate chain for the 1st almost-SDC above. And actually, I didn't use the other one in the solution at all though I do suggest above that there may be an alternate for it as well. All of this is moot however, since the point of this thread was not to present this as the only possible solution to this puzzle (something which, if you read closely, is explicity clear), but to point out a pattern that has never been addressed and which might be of some use in manual solving.

However, I will complement the overall solution found by your SudoRules computer solver and the rapidity with which it was found. I continue to be humbled by my inability to match computer solvers.
DonM
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### Re: The Almost Sue De Coq (ASI#2)

DonM,
Due to your allergy to computers , you're missing the point.
I'm not criticising the fact that you're illustrating your pattern with an example that may have other solutions.
What I'm criticising is your indirect claim (via a reference to Ruud) that Sue de Coq may be a more frequent pattern than it seems.
For such a claim, you should at least provide an example that can't be solved with elementary chains that anyone, human or computer, would have found before a Sue-de-Coq.
Having good graphics is a great thing and you're certainly good at that, but good graphics are not enough to make a good example.
denis_berthier
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### Re: The Almost Sue De Coq (ASI#2)

ronk wrote:
David P Bird wrote:However this isn’t necessary if we go the other way using the row instead of the box:

(8)r3c9 = (345789)DLS:r3c45679,r1c3[(3479)r3,(5)c4,(8)b2] – (9)r3c1 = (9)r2c2 – (9=8)r4c2 – (8)r6c3 = (8)r6c5 => r3c5 <> 8

This is why I like to show the containing houses for each digit in the notation.

Since r3c1=9 doesn't remove digit 8 from the "DLS", how does one read that from right-to-left

You've gone back to your old ways of thinking in Nice Loops again it seems, whereas thinking strictly in weak and strong inferences will set you straight! The DLS spec indicates that the 8 must be located in box 2. Coming in from the right, there is a weak inference between the (9)r3c1 = true and the DLS = true, which is all that is needed. If r3c1<>9 and r3c9=8 both arguments are false, but the inference is still good.
David P Bird
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### Re: The Almost Sue De Coq (ASI#2)

David P Bird wrote:The DLS spec indicates that the 8 must be located in box 2. Coming in from the right, there is a weak inference between the (9)r3c1 = true and the DLS = true, which is all that is needed. If r3c1<>9 and r3c9=8 both arguments are false, but the inference is still good.

Thanks, but that's too opaque for me, so I'll stick with an ALS. Besides, I've already got more 3-letter acronyms than I can handle.
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### Re: The Almost Sue De Coq (ASI#2)

denis_berthier wrote:DonM,
Due to your allergy to computers , you're missing the point. I'm not criticising the fact that you're illustrating your pattern with an example that may have other solutions.

Live by the sword, die by the sword: If the point here is that my alleged limitation is somehow related to a computers, I would suggest that to blast the grid through your solver and then conclude what you would do based on that output (ie. I wouldn't use Sue-de-Coq as there is a solution based on much simpler patterns.) indicates someone so dependent on computers that inappropriate conclusions about human sudoku solving in general are based on them. After all, that output doesn't a) prove that for all puzzles, there would always be a solution based on simpler patterns or b) that even if so, you would be able to find them unassisted by computer. The latter is a critical point: Can one determine from computer output, what is always the best course of action for manual solving? Perhaps, but certainly you haven't proved it yet and having spent a lot of time pattern-solving, I have serious doubts.

What I'm criticising is your indirect claim (via a reference to Ruud) that Sue de Coq may be a more frequent pattern than it seems. For such a claim, you should at least provide an example that can't be solved with elementary chains that anyone, human or computer, would have found before a Sue-de-Coq. Having good graphics is a great thing and you're certainly good at that, but good graphics are not enough to make a good example.

What you are saying would have value except for one critical point: Yes, I am inferring that 'Sue de Coq may be a more frequent pattern than it seems.', but I never said that it would solve some puzzles that simpler methods could not which would demand some sort of support if presented as an absolute. In any event, the one point doesn't infer the other point. Without practice, it is difficult to identify a SDC. Likewise, it takes some time to understand and apply the almost-SDC. Finding examples that prove any point with any certainty will take some time. Which is why the whole tone of the article above is that of a suggestion with 2 examples that are intended to raise interest in the possibilities, if any, of a possible new pathway that can't be replicated by other means, or if not, then the possibility of something that will still be of use to manual solvers in other ways.

But speaking of the graphics and related to our previous enjoyable discussion elsewhere: they prove all that I intended to prove- that the method proposed can be used in an actual puzzle. A foundation (in the other thread) is also provided to assist in understanding the premise. You might benefit from applying this principle yourself. You don't need all the pretty pictures; even just a blow-by-blow explanation of how you manually solved a puzzle using text-based graphics would suffice. Darn, I wish you had done that in your book...or somewhere.

Still, in Ruud's study, Sue De Coq came up more frequently than a number of other patterns that manual solvers often look for routinely. If true, that makes the SDC even more worth looking for due to the several possible eliminations. However, since that would also mean that the almost SDC would come up even more frequently there may be even more of an incentive to look for the pattern. I'm raising questions about a possible benefit. The tone of your reply implies a certainty of no benefit. Any studies you may care to do/submit to prove the point would be interesting.

(Edited for clarity.)
Last edited by DonM on Tue Oct 28, 2008 10:33 am, edited 2 times in total.
DonM
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### Re: The Almost Sue De Coq (ASI#2)

DonM wrote:Live by the sword, die by the sword: If the point here ...

I think the point here is you consider any critic as a personal attack and, as a good Eureka immigrant, you can only counter attack with unrelated stuff and slanderous allegations instead of providing a honest answer.

You're saying SdC "may" be more frequent than it seems. If you want to raise interest in it, it's your job to prove this or at least to give a more convincing example.

DonM wrote:The tone of your reply implies a certainty of no benefit.

There was no tone of my reply but in your imagination. Just a fact: your example doens't prove what you only suggested "may" be true.
denis_berthier
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### Re: The Almost Sue De Coq (ASI#2)

denis_berthier wrote:
DonM wrote:Live by the sword, die by the sword: If the point here ...

I think the point here is you consider any critic as a personal attack and, as a good Eureka immigrant, you can only counter attack with unrelated stuff and slanderous allegations instead of providing a honest answer.

I'll leave it to others to determine if there is anything personal or slanderous in my comments above or whether you yourself have always responded to others in a manner that couldn't be taken personally in your threads. Certainly, I don't see anything more 'pointed' in my comments, than the implications of your 'allergy to computers' comment or the reason for your first post for that matter. I am actually surprised that given your mathematical professorial world you would be so sensitive to anything here as being personal; in my professional world it's nothing more than a spirited debate. One man's honest answer is apparently another man's personal attack, depending on who's saying it, I guess. However, always happy to hear from you.
DonM
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