This post details forms of Sue de Coq not revealed in the founding post of October 24, 2005, widening the definition of classic Sue de Coq to all ALS and including a single alternate version. It introduces another form of Sue de Coq with a single alternate that is the basis for a specific, logically defined trial, leading to eliminations or a solution.
As with anything new (accept my blog) there was some stir up when Sue’s post came out. Some contested the logic, or the motivation. Some found, or thought they found, additional SdC’s that Sue missed. Your reader assignment last post was to find all Sue de Coq chutes and victims of the form N(a + b)(c + d), with both alternate bv in the remainders, and to look for a clueless form with 5 chute numbers. These are the forms of Sue de Coq justified by Sue’s 5 number rationale.
Circling clockwise from the NW box, there is:
NWc2 = 8(1+9)(2+3),
with bv 19 in c2 and 23 in NW. It removes 3r2c1, 1r6c2, 1r7c2 and 3r1c3.
NEr2 = (3+9)(5+8)(2),
with bv 39 in r2 and 58 in NE, removes 3r2c1 and 9NWr2c12 via bv 59 in r2, but three 2-candidates see all three 2’s in NEr2, one of which must be true.
Another clueless classic is SWc1 = (1+9)(3+6)(5), with bv 19 in SW and 36 in c1. It removes other 1’s in SW, and 3r2c1. Since the chute must also contain a 5, 5r2c1 and 5r8c3 are eliminated.
Sue’s original post reported these three, but missed the toxic set removals on the clueless ones.
Sue also missed the ALS example:
In NWr1=4(3+9)(5+8) the r3c12 ALS 239 in the box remainder supplies 3 or 9 to the box, removing 39r2c1 and 39r2c2.
Like a 39 bv, the ALS can only give up 3 or 9, not both. But the ALS 239 must therefore contain 2. The other 2 in the NE box is history!
The Hodoku Sue de Coq page (follow the Miscellaneous link) provides an extraordinary example of ALS removal power in the Sue de Coq.
The clueless 5 number chute Ec8 contains 4(3+5)(7+8), removing 35r7c8 and 7 and 8 E candidates outside the ALS 789. But to supply 7 or 8, the ALS must contain 9, and four others are removed.
There is another form of classic Sue de Coq that is seldom seen or discussed. A chute with two clues and one alternative, of form
MN(a + b)
with an ALS containing a and b in a remainder, can be considered a Sue de Coq, even though the chute has only 4 numbers. In Sue’s puzzle, SWr7 = 46(1+2) is an example chute, but it has no remainder ALS containing 1 and 2. I couldn’t find a single case among my review puzzles, but I do remember using it. I’m reserving this blog space for an example. Why don’t you find one for me? I’ll give you credit right here.
Sue de Coq is a one/two punch: The two alternate ALS that mark the SdC restrict the chute to 3 numbers, but the remainders apply this restriction for removals within their cells.
This division of functions prompts me to introduce another form of Sue de Coq, which occurs much more frequently than the classic two alternate form of the original post that is generally recognized. Let’s call it the single alternate Sue de Coq, or maybe the SASdC.
Suppose the 5 number SdC chute contains a clue, but only one remainder ALS is present, say (c + d) in the diagram. The chute cannot contain N,c and d. If it contains c or d, then it also contains a or b. Box removals of c or d outside of the ALS still apply.
But it is possible for c and d to be missing altogether. Then the chute contains N and a and b, most often in a specific order. The trial for the missing alternative chute I call a verification of the single alternate SdC. It is a logically constructed trial that yields either the elimination of the SdC, or a chute solution, but I consider it an open question whether or not to apply it as early as the bv scan, where its use could obscure results obtainable without trial.
A nice example comes from a later review of Paul Stephens puzzles.
But the chute can also be described as NWr2 = 2(3+4)(6+8)+382. The first term describes the chute if 4 or 6 is present, and the second term, if both are missing.
So what do we do? First I establish that the classic first term SdC has a result. In this case, the ALS 689 of r2c59 forces a 6-candidate from the row remainder. Check. Then I put NWr2 = 382 on trial. I look for some immediate contradiction, or in PowerPoint, copy the slide and follow up. In blog analysis I do that with the trial trace described on the trials page(menu bar). This type of trace points to the shortest set of inference paths to the contradiction, if there is one. In this case, NWr2 = 382 fails, because it forces 8r2c9 and two 8’s into c4. I’ve found that arrows to the next positive result give the most readable result.
With this reasonably quick contradiction, I can pretend I saw it, and use the result. The Sysudoku policy on such trials is spelled out on the Order of Battle page(menu bar again). If the contradiction is one that nobody is going to believe I saw, I defer use of this result until I’ve tried everything else. I think this is the right policy for the blog, but the use of logically constructed trials is a personal decision. It hinges on what you take as a “logically constructed” trial. Your standing as a sysudokie doesn’t depend on it. But you are hereby deputized to dispute the use of arbitrary guesses as logically constructed trials.
Having attended this Sysudoku briefing on ALS based, single alternate Sue de Coq, you are now considered to be armed and dangerous for puzzles with Sue de Coq formations of all kinds. Sue de Coq’s “Sudoku carnage” post contains a remarkable number of classic, two alternate Sue de Coq. These are actually very rare, but the single alternate Sue de Coq introduced in this post occurs so frequently that it keeps Sue de Coq at the top on my advanced solving “to do” list.
Next we will look atthe elimination method known as APE. Its full name is Aligned Pair Exclusion, or Aligned Pair Elimination. It is often found in the search for Sue de Coq, because it depends on ALS, especially bv, in the same way.