Glyn wrote:If we made an additional constraint based on an observation that the givens 1,2,3 do not simultaneously share a row and box and apply that to the pencilmarks we get
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.------------------.------------------.------------------.
| 1 57 57 | 259 256 4 | 2679 8 3679 |
| 2478 9 3478 | 1 68 68 | 267 246 5 |
| 248 245 6 | 2589 7 3589 | 1 49 49 |
:------------------+------------------+------------------:
| 6789 3 789 | 5789 58 2 | 4 1569 6789 |
| 5 1467 4789 | 4789 3 789 | 6789 69 2 |
| 4789 47 2 | 6 1458 5789 | 5789 3 789 |
:------------------+------------------+------------------:
| 24679 8 4579 | 3 456 567 | 569 4569 1 |
| 467 4567 1 | 24578 9 5678 | 3 456 468 |
| 3 456 459 | 458 4568 1 | 25689 7 4689 |
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solvable by singles. However, is there a good reasons to lock down these special candidates? Apart from that I've only been able to pick away at it slowly.
Great discovery!
However you haven't got the trick 100% accurate. Let me try to clarify things for you...
For this very puzzle, with the digit mapping [123456789]->[123564897], each min-row must either have the whole of {123}|{456}|{789}, or one each digit from these 3 sets (e.g. 147, 158, 349).
Proof:Firstly we need to see that the 3 digits {456} must either exists as a whole mini-row or be distributed on 3 different mini-rows within a block.
Suppose this is not the case, i.e. two digits out of {456} are on the same mini-row and the remaining one is on a different mini-row. Then this pattern must also be duplicated in the other 2 blocks of the same band. So within each block we have one mini-row containing two of {456} (i.e. being {45},{56},{46} respectively), one mini-row containing one of {456} (i.e. 6,4,5 respectively), one mini-row without any of {456}.
Now, each (whole) row of this band must also consist of 3 mini-rows, and these must also have two, one, none of {456} respectively. However, the mini-row with one of {456} will always clash with the mini-row with two of {456} (i.e. that sole value will always be the same as one of the two values). To see this fact, just look at the following example:
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456 . 456 | 456 . . | . . .
. . . | 456 . 456 | 456 . .
456 . . | . . . | 456 . 456
No matter how you assign values you'll always run into a contradiction.
Therefore, within a block, each of {456} must either occupy the same mini-row or occupy 3 different mini-rows.
Using the same reasoning, each of {789} must also behave the same.
As a result, each of {123} must also behave in the same manner!
(Explicit proof: Suppose {456} occupy the same mini-row. Then {789} can't occupy 3 different mini-rows, so must also occupy another single mini-row. Then {123} are forced to occupy the remaining single mini-row.
Suppose {456} occupy 3 different mini-rows. Then {789} can't occupy a single mini-row, so must also occupy 3 different mini-rows. Then {123}, with only one cell remaining available on each mini-row, are forced to also occupy 3 different mini-rows.)
(End of proof.)With this trick, it's very easy to crack this puzzle.
(Initially the top band can be both ways but in the middle and bottom band you must have each mini-row containing exactly one from {123},{456},{789} respectively. After a series of singles you can also confirm this property also holds true for the top band. And then you can complete the puzzles using singles only.
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(And also for "version 2" this trick is also applicable: first notice r7c1 must be 3 because r7c123 can't be {789}. So r8c2=7 (hidden single). Now seeing r6c123 can't be {789}, r6c1 must be 4, hence r4c4=5 (symmetry). And the rest are all naked singles. Way cool!
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