How To Spot a Sue De Coq

Advanced methods and approaches for solving Sudoku puzzles

How To Spot a Sue De Coq

Postby Yogi » Mon Sep 26, 2022 9:25 pm

..83.7..926....3.........4.3...521.87.4..8..5.5........8...6..7.....52....1......
Hodoku says that after basics this puzzle can be reduced to Singles with a Sue De Coq elimination
but how do you spot such an animal, and how does it work? You can cut to the chase with
148327..9267594381.3.861742396452178714638925852..9....8.2.6..7.7..852...21..38..77

Code: Select all
+----------------+----------------+----------------+
| 1    4    8    | 3    2    7    | 56   56   9    |
| 2    6    7    | 5    9    4    | 3    8    1    |
| 59   3    59   | 8    6    1    | 7    4    2    |
+----------------+----------------+----------------+
| 3    9    6    | 4    5    2    | 1    7    8    |
| 7    1    4    | 6    3    8    | 9    2    5    |
| 8    5    2    | 17   7    9    | 46   36   346  |
+----------------+----------------+----------------+
| 459  8    359  | 2    1    6    | 45   359  7    |
| 469  7    39   | 9    8    5    | 2    1369 346  |
| 4569 2    1    | 79   47   3    | 8    569  46   |
+----------------+----------------+----------------+


Previously I always ended up solving these puzzles with alternative techniques. This time I think I found the SDC and understood the eliminations, but I’m interested in your comments on the technique for this puzzle. I haven’t seen any recent threads on the basic idea of SDC, only comments about variations, extensions, programming and algorithms, which are a bit off-planet for me. I note that there are some examples in Sue De Coq’s original thread on Disjointed Subsets, which I will work through in time.
Anyway, what I see here is that cells r7c137 and r8c3 form a bent Quad, which allows the elimination of 4 from r7c5, solving it to 1, reducing the puzzle to stte. The logic is fairly simple: If r7c7 is 4, then r7c5 can’t be 4. If r7c7 is 5, then 4 occurs in the intersection and again r7c5 can’t be 4 (this is true for either option in r8c3.)
The same sort of thing happens in the box. If r8c3 is 3, then 9 occurs in the intersection. If r8c3 is 9 then 3 occurs in the intersection, allowing the elimination of 3 and 9 from anywhere else in the box. This true for either option in r7c7.
I see this technique as having characteristics much in common with the XYZ Wing, in that it depends on a cell outside a box and a cell inside a box, which can’t see each other, but both interacting with cells inside the box which they can see, in this case a group of cells called an intersection, rather than a single cell called a pivot. This could help you spot them if you are scanning for both at the same time.
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Re: How To Spot a Sue De Coq

Postby Hajime » Tue Sep 27, 2022 7:51 am

Code: Select all
148327..9267594381.3.861742396452178714638925852..9....8.2.6..7.7..852...21..38..
No 77 at the end.
This can be solved using only several XYZ-Wings and no other basic or advanced methods.
My solver did spot 12 XYZ-Wings but I have no exit in the loop during XYZ-Wing searching. So maybe less will be sufficient.
SukakuExplainer analysis spots 1 x Bidirectional Cycle (7,0), but XYZ-Wing is only SER=4.4 .
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Re: How To Spot a Sue De Coq

Postby jco » Wed Sep 28, 2022 10:11 pm

I always had great difficulty spotting SDCs (but I like it a lot!).
Often I stumble on SDCs as Loops (i.e., I find some loop and then realize that there is a related SDC).
I can spot ALPs (and when I see one, I know that a SDC should be present).
I describe next how I spot ALPs.

Code: Select all
.------------------------------------------------------------.
| 1     4     8     | 3     2     7     | 56    56     9     |
| 2     6     7     | 5     9     4     | 3     8      1     |
| 59    3     59    | 8     6     1     | 7     4      2     |
|-------------------+-------------------+--------------------|
| 3     9     6     | 4     5     2     | 1     7      8     |
| 7     1     4     | 6     3     8     | 9     2      5     |
| 8     5     2     | 17    17    9     | 46    36     346   |
|-------------------+-------------------+--------------------|
| 459   8     359   | 2     14    6     | 5-4   1359   7     |
|(46)-9 7     39    | 19    8     5     | 2    (6)139 (46)3  |
| 4569  2     1     | 79    47    3     | 8     59-6  (46)   |
'------------------------------------------------------------'

Loop (4-6)r8c1 = (6)r8c89 - (6=4)r9c9 - (4)r8c9 = (4-6)r8c1 => -9 r8c1, -6 r9c8, -4 r7c7; ste

*spotting* the ALP:
we cannot have both 4 and 6 out of r8c89 since the only place for both at row 8, would be a r8c1.
So, either 4 or 6 must be at r8c89.
This fact combined with (46)r9c9 already gives the eliminations -4 r7c7 and -6 r9c8 in any case.
Also, we cannot have both 4 and 6 at r8c89 (r9c9 would be without a valid digit), so 1,3 or 9 must be there.
However, due to (39)r8c3, (19)r8c4, in each case we conclude that -9 r8c1. By this I mean:
(1)r8c8 - (1=9)r8c4
||
(3)r8c89 - (3=9)r8c3
||
(9)r8c8
=> -9 r8c1
This reasoning above is only to identify the pattern (ALP) and check the resulting eliminations.
When posting, I prefer to write the move as the above loop.

There is another ALP available (same reasoning, so only the loop is written below).
Code: Select all
.------------------------------------------------------------.
| 1     4     8     | 3     2     7     | 56    56     9     |
| 2     6     7     | 5     9     4     | 3     8      1     |
| 59    3     59    | 8     6     1     | 7     4      2     |
|-------------------+-------------------+--------------------|
| 3     9     6     | 4     5     2     | 1     7      8     |
| 7     1     4     | 6     3     8     | 9     2      5     |
| 8     5     2     | 17    17    9     | 46    36     346   |
|-------------------+-------------------+--------------------|
| 45(9) 8    (39)5  | 2     14    6     | 45   (39)-15 7     |
| 46-9  7    (39)   | 19    8     5     | 2     1369   346   |
| 456-9 2     1     | 79    47    3     | 8     569    46    |
'------------------------------------------------------------'

Loop (3-9)r7c8 = (9)r7c13 - (9=3)r8c3 - (3)r7c3 = (3-9)r7c8 => -15 r7c8, -9 r89c1; ste

For this puzzle, (due to several BVCs) what called my attention first was the following almost xy-chain:
Code: Select all
.-----------------------------------------------------------.
| 1     4     8     | 3     2     7     | 56    56    9     |
| 2     6     7     | 5     9     4     | 3     8     1     |
| 59    3     59    | 8     6     1     | 7     4     2     |
|-------------------+-------------------+-------------------|
| 3     9     6     | 4     5     2     | 1     7     8     |
| 7     1     4     | 6     3     8     | 9     2     5     |
| 8     5     2     | 17    17    9     | 46    36    346   |
|-------------------+-------------------+-------------------|
| 459   8     359   | 2    a14    6     | 5-4   1359  7     |
| 469   7    c39    |b19    8     5     | 2     1369 d36(4) |
| 4569  2     1     | 79    47    3     | 8     569  e46    |
'-----------------------------------------------------------'

[(4=1)r7c5 - (1=9)r8c4 - (9=3)r8c3 - (3=*6)r8c9 - (6=4)r9c9] = (4)r8c9 => -4 r7c7; ste

EDIT: changed "... 4 or 6 must be at row 8" to "... 4 or 6 must be at r8c89" for clarity.
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