Interesting puzzle! It seems that TH, as we know it, is not very effective here, yet still the puzzle can be solved in elegant ways.
My own solution, using uniqueness: Show - Code: Select all
.-------------------.-------------------.-----------------.
| 4 23569 23567 | 3569 3567 3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 2378 | 389 4 378 | 5 6 27 |
:-------------------+-------------------+-----------------:
| 19 19 356 | 356 2 4 | 8 7 356 |
| 7 2356 4 | 1 8 356 | 236 9 2356 |
| 356 8 2356 | 7 356 9 | 14 14 2356 |
:-------------------+-------------------+-----------------:
| 8 356 9 |#24 3567 1 | 3467 #25–4 367 |
| 356 7 1 |#24 9 356 | 346 #25–4 8 |
| 2 4 356 | 3568 3567 35678 | 1367 1–5 9 |
'-------------------'-------------------'-----------------'
UR 24r78c48 => -4r78c18, -5r9c8
- Code: Select all
.-------------------.-------------------.-----------------.
| 4 23569 23567 | 3569 3567 3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 2378 | 389 4 378 |#5 #6 27 |
:-------------------+-------------------+-----------------:
| 19 19 356 | 356 2 4 | 8 7 356 |
| 7 2356 4 | 1 8 356 | 236 9 2356 |
| 356 8 2356 | 7 356 9 | 1 4 2356 |
:-------------------+-------------------+-----------------:
| 8 356 9 |#24 3567 1 |#3467 #25 367 |
| 356 7 1 |#24 9 356 |#346 #25 8 |
| 2 4 356 | 3568 3567 35678 | 1367 1 9 |
'-------------------'-------------------'-----------------'
# in b9 cannot be a 2456 quadruple (note reverse BUG 56) => -6r78c7
The two equivalent solutions:
- Code: Select all
.-------------------.-------------------.-----------------.
| 4 23569 23567 | 3569 3567 3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 2378 | 389 4 378 |#5 #6 27 |
:-------------------+-------------------+-----------------:
| 19 19 356 | 356 2 4 | 8 7 356 |
| 7 2356 4 | 1 8 356 | 236 9 2356 |
| 356 8 2356 | 7 356 9 | 1 4 2356 |
:-------------------+-------------------+-----------------:
| 8 356 9 |#24 3567 1 |#4 #5 367 |
| 356 7 1 |#24 9 356 |#6 #2 8 |
| 2 4 356 | 3568 3567 35678 | 1367 1 9 |
'-------------------'-------------------'-----------------'
- Code: Select all
.-------------------.-------------------.-----------------.
| 4 23569 23567 | 3569 3567 3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 2378 | 389 4 378 |#5 #6 27 |
:-------------------+-------------------+-----------------:
| 19 19 356 | 356 2 4 | 8 7 356 |
| 7 2356 4 | 1 8 356 | 236 9 2356 |
| 356 8 2356 | 7 356 9 | 1 4 2356 |
:-------------------+-------------------+-----------------:
| 8 356 9 |#24 3567 1 |#6 #2 367 |
| 356 7 1 |#24 9 356 |#4 #5 8 |
| 2 4 356 | 3568 3567 35678 | 1367 1 9 |
'-------------------'-------------------'-----------------'
Upon relabeling 56, the rest of the grid is the same in both cases, so there is no way to decide which diagonal should be 24 and which should be 56.
- Code: Select all
.-------------------.-------------------.-----------------.
| 4 23569 23567 | 3569 3567 3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 2378 | 389 4 378 |#5 #6 27 |
:-------------------+-------------------+-----------------:
| 19 19 356 | 356 2 4 | 8 7 356 |
| 7 2356 4 | 1 8 356 | 236 9 2356 |
| 356 8 2356 | 7 356 9 |#1 #4 2356 |
:-------------------+-------------------+-----------------:
| 8 356 9 |#24 3567 1 |#347 #25 367 |
| 356 7 1 |#24 9 356 |#34 #25 8 |
| 2 4 356 | 3568 3567 35678 |#137–6#1 9 |
'-------------------'-------------------'-----------------'
If 6r9c7 then r369c78 is a virtual 45 pair, resolved only by all the givens of 56.
This VP then creates a DP with 245b9, 24b8.
56 in the rest of the grid are unresolvable, ie. contra.
-6r9c7, stte
Xsudo's solution, without using uniqueness:
After a two-string kite (-2r1c3), a short AIC (-2r3c3) and basics:
- Code: Select all
.------------------.-------------------.-----------------.
| 4 23569 #3567 | 3569 #3567 #3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 378 | 389 4 378 | 5 6 27 |
:------------------+-------------------+-----------------:
| 19 19 #356 |#356 2 4 | 8 7 356 |
| 7 #356 4 | 1 8 #356 | 236 9 2356 |
|#356 8 2 | 7 #356 9 | 14 14 356 |
:------------------+-------------------+-----------------:
| 8 #356 9 | 24 7–356 1 | 3467 245 367 |
|#356 7 1 | 24 9 #356 | 346 245 8 |
| 2 4 #356 |#3568 #3567 #35678 | 1367 15 9 |
'------------------'-------------------'-----------------'
Xsudo uses the marked cells as truths.
The eliminations in r7c5 reduce the rating to 6.6.
Can anyone explain this pattern by human means?
16 Truths = {68N1 57N2 149N3 49N4 169N5 1589N6}
51 Links = {3r1456789 5r1456789 6r1456789 7r1 3c12356 5c12356 6c12356 356b4 356b5 356b7 35678b8}
Another puzzle which may require some research is this one, posted by mith
here on June 24, 2021:
- Code: Select all
.............12.34..1.3..25..672.5...17..6...52..81.67.75.68...18.2....66.217..5. ED=10.4/1.2/1.2
It is also solvable by Xsudo, but I can't make a lot of sense of its findings.
Edit: We can explain it using the parity argument:
Suppose 356r7c5.
- Code: Select all
.------------------.-------------------.-----------------.
| 4 23569xy3567 | 3569 z3567 z3567 | 279 8 1 |
| 569 569 5678 | 5689 1 2 | 79 3 4 |
| 139 1239 378 | 389 4 378 | 5 6 27 |
:------------------+-------------------+-----------------:
| 19 19 #356 |#356 2 4 | 8 7 356 |
| 7 #356 4 | 1 8 x356 | 236 9 2356 |
|#356 8 2 | 7 y356 9 | 14 14 356 |
:------------------+-------------------+-----------------:
| 8 #356 9 | 24 x356 1 | 3467 245 367 |
|#356 7 1 | 24 9 y356 | 346 245 8 |
| 2 4 #356 |z3568 z3567 z35678 | 1367 15 9 |
'------------------'-------------------'-----------------'
In b8, each minirow contains one of 356. We can fix its vertical permutation as in the diagram.
Chaining through antidiagonal boxes 745, the horizontal parity in b5 is opposite to it, therefore the minicolumn containing x is directly to the right of the one containing y. There is only one option.
r1c56 form a 7z naked pair, restricting r1c3 to xy.
Whichever digit appears in r1c3 is completely eliminated from either (y) c1 or (x) c2, ie. contra.
Therefore -356r7c5.
Edit2: Got some understanding of the other one. It is far from solving the puzzle, but it should reduce it to T&E(1).
The breakthrough found by Xsudo: Show - Code: Select all
.-------------------.------------------.-------------------.
| 2 34569 3489 | 45689 459 4579 | 16789 1789 189 |
| 79 569 89 | 5689 1 2 | 6789 3 4 |
| 479 469 1 | 4689 3 479 | 6789 2 5 |
:-------------------+------------------+-------------------:
|c3489Aa349 6 | 7 2 #349 | 5 1489 1389 |
|c3489 1 7 | 3459 459 6 | 23489 489 2389 |
| 5 2 Bb349 |#349 8 1 | 349 6 7 |
:-------------------+------------------+-------------------:
|#349 7 5 |Ab349 6 8 | 12349 149 1239 |
| 1 8 #349 | 2 c459 c3459 | 3479 479 6 |
| 6 349 2 | 1 7 Ba349 | 3489 5 389 |
'-------------------'------------------'-------------------'
b4p29 and b8p19 are not the same two digits, either b5p37 or b7p16 would have to contain the same digit.
Consequently, each of 349 must appear in them at least once.
- Code: Select all
.-------------------.------------------.-------------------.
| 2 34569 3489 | 45689 459 4579 | 16789 1789 189 |
| 79 569 89 | 5689 1 2 | 6789 3 4 |
| 479 469 1 | 4689 3 479 | 6789 2 5 |
:-------------------+------------------+-------------------:
| 3489 #349 6 | 7 2 349 | 5 1489 1389 |
| 3489 1 7 | 3459 459 6 | 23489 489 2389 |
| 5 2 #349 | 349 8 1 | 349 6 7 |
:-------------------+------------------+-------------------:
| 349 7 5 |#349 6 8 | 12349 149 1239 |
| 1 8 349 | 2 459 3459 | 3479 479 6 |
| 6 49–3 2 | 1 7 #349 | 3489 5 389 |
'-------------------'------------------'-------------------'
3r1116b9##\r799c22233347 => -3r9c2
This abomination of a fish is probably simpler as a contradiction net: if 3r9c2 then in r1 3c3, in # 3r7c4, in r6 3c7, no 3 left in b9, ie. contra.
This only results in a marginal change of rating though (skfr 10.2 -> 10.1).
Now suppose 4r9c7. ER 4r6b8\r8c34 => -4r8c3, then contra. with singles. Therefore -4r9c7.
15 Truths = {3R1 9C5 239C9 4N26 6N347 7N14 8N3 9N26}
37 Links = {3r679 4r4679 9r14679 3c2346 4c3467 9c2346 57n9 3b567 4b478 9b45678}
skfr 9.0, solvable by YZF_Sudoku (even with uniqueness-based techniques disabled).
Marek