the use of any stratagy in this thread is up to the player that wants to use them or not.
by deffinition:
a Puzzle: is classified as a puzzle if and only when the problem has a singular solution that is unique.
with a sudoku there is candidate digit arangments that when they are expressed in whole form mutiple solutions.
the simplisted is known as the deadly rectangle.
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ab . . | . ab . | . . .
ab . . | . ab . | . . .
. . . | . . . | . . .
this idea is that the above arangment forms 2 solutions.
either:
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a . . | . b . | . . .
b . . | . a . | . . .
. . . | . . . | . . .
or
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b . . | . a . | . . .
a . . | . b . | . . .
. . . | . . . | . . .
a valid solution may never have a pm state of 2 solutions.
Unique Rectangle 1.1: theory
{ this is also known as avoidable rectangles }
Definition: an a/b/b/a pattern in a solution grid is anything isomorphic to that shown below:Code:
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. . . | .
a . . | b
b . . | a
---------+---
. . . | .
Fact: if a solution grid (not necessarily unique) contains an a/b/b/a pattern on four unclued cells, C, then C=b/a/a/b is also a solution.
Theorem: if a puzzle-in-progress (that does not necessarily have a unique solution) has pencilmarks as shown below on four unclued cells then the bottom right value resolves to '3':Code:
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. . . | .
1 . . | 2
2 . . | 13
---------+---
. . . | .
Proof: suppose to the contrary the bottom right value resolves to '1'. Then (vacuously) the solution grid contains the 1/2/2/1 pattern on four unclued cells, C. So, by the Fact above, C=2/1/1/2 is also a solution. But wait! - the pencilmarks do not allow that other solution - contradiction.
example of it in use:
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.-------.-------.-------.
| . 4 . | . . . | 1 5 . |
| 8 . . | . . 7 | . . 6 |
| . . 1 | . 8 . | . . 4 |
:-------+-------+-------:
| 3 . . | 2 6 . | . . . |
| . . . | 4 . 1 | . . . |
| . . . | . 9 5 | . . 2 |
:-------+-------+-------:
| 1 . . | . 7 . | 4 . . |
| 5 . . | 9 1 . | . . 3 |
| . 9 6 | . . . |. 2 . |
'-------'-------'-------'
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.-------------------.-------------------.-------------------.
| 679 4 379 | 36 2 369 | 1 5 8 |
| 8 25 259 | 1 4 7 | 29 3 6 |
| 269 236 1 | 5 8 369 | 29 7 4 |
:-------------------+-------------------+-------------------:
| 3 17 479 | 2 6 8 | 5 14 79 |
| 279 257 2579 | 4 3 1 |#6 #8 79 |
| 46 168 48 | 7 9 5 | 3 14 2 |
:-------------------+-------------------+-------------------:
| 1 238 238 | 68 7 26 | 4 9 5 |
| 5 278 2478 | 9 1 24 |#7-8 #6 3 |
| 47 9 6 | 38 5 34 | 78 2 1 |
'-------------------'-------------------'-------------------'
the cells with # are none givens digits from the starting grid, knowing this implies they cannot form a ur pattern arangement.
thus R8C7 cannot = 8.
Unique Rectangle type 1:
A UR Type 1 exists, if one of the four cells of a possible UR has additional candidates. If those candidates were eliminated, the UR would exist, causing two solutions. It is therefore absolutely necessary, that one of the additional candidates has to be true. That means, that the UR candidates can be eliminated from the cell with the additional candidates. If only one additional candidate exists, it can be placed in that cell immediately.
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.-------------.---------------.-----------.
| 5 34 2 | 134 134 8 | 9 6 7 |
| 1 3-89@ 89@ | 7 36 69 | 4 5 2 |
| 49 6 7 | 5 24 29 | 3 8 1 |
:-------------+---------------+-----------:
| 2 1 3 | 6 5 7 | 8 4 9 |
| 6 5 4 | 8 9 1 | 2 7 3 |
| 7 89@ 89@ | 23 23 4 | 6 1 5 |
:-------------+---------------+-----------:
| 8 2 1 | 9 67 56 | 57 3 4 |
| 3 47 6 | 124 1247 25 | 157 9 8 |
| 49 479 5 | 14 8 3 | 17 2 6 |
'-------------'---------------'-----------'
In the example the candidates 8 and 9 nearly form an UR in r2c23 and r6c23. The UR is only prevented by the additional candidate 3 in r2c2. 8 and 9 can be eliminated from r2c2 (or 3 can be placed in r2c2).
{8/9 in r2c23,r6c23 => r2c2<>8, r2c2<>9}
Unique Rectangle Type 2
If in a possible UR two non diagonal cells have only one extra candidate, and that candidate is the same in both cells, all candidates seeing both extra candidates can be eliminated.
The logic is simple: To avoid the UR one of the extra candidates must be true, thus any cell seeing both candidates cannot have the same value placed.
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.---------------.-----------------------.-------------------.
| 14 7 9 | 38 23458 235 | 135 18 6 |
| 3 28 5 | 6 2789 1 | 4 789 29 |
| 146 248 12 | 3789 2345789 23579 | 13579 1789 1259 |
:---------------+-----------------------+-------------------:
| 5 6 4 | 379 379 8 | 19 2 19 |
| 2 9 8 | 5 1 4 | 6 3 7 |
| 7 1 3 | 2 69@ 69@ | 8 5 4 |
:---------------+-----------------------+-------------------:
| 149 234 127 | 1379 2359-7 2359-7| 1579 6 8 |
| 8 5 1-7 | 4 679@ 679@ | 2 19-7 3 |
| 19 23 6 | 1389-7 23589-7 2359-7 | 1579 4 159 |
'---------------'-----------------------'-------------------'
An UR Type 2 can eliminate candidates in more than one house: The additional 7s in r8c56 eliminate all other 7s from both row 8 and block 8.
6/9 in r6c56,r8c56 => r7c456,r8c38,r9c456<>7
Unique Rectangle Type 3
This is a combination of UR with Naked/Locked Subsets. We look for two non diagonal cells in a possible UR that have extra candidates. Since one of those candidates has to be set to avoid the UR, we can treat both cells as one virtual cell containing only the extra candidates and try to build a Naked Subset (all additional cells have to see both UR cells with extra candidates). If such a UR/Naked Subset is found, all subset candidates can be eliminated from all cells outside the subset (but in the same house of course).
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.---------------------.------------------.------------------.
| 5 4679 8 | 3467 3469 347 | 1 2 34 |
| 1369 1469 1369 | 3468 2 1348 | 35 3459 7 |
| 2 1479 1379 | 347 1349 5 | 8 349 6 |
:---------------------+------------------+------------------:
| 4 5679-1 5679-13| 357 13 2 | 356 8 135 |
| 6-13@ 8 2567-13 | 9 134 1347 | 356 345 12345 |
| 13 125@ 1235@ | 458-3 48-13 6 | 9 7 245-13|
:---------------------+------------------+------------------:
| 1689 1569 1569 | 2 3468 348 | 7 35 35 |
| 68 25@ 25@ | 368 7 38 | 4 1 9 |
| 7 3 4 | 1 5 9 | 2 6 8 |
'---------------------'------------------'------------------'
The example shows an UR Type 3 with a Locked Pair: The additional candidates 1 and 3 in r6c23 form the Locked Pair with r6c1 (locked in row 6 and block 4).
this one has two examples:
2/5 in r6c23,r8c23 => r4c23,r5c13,r6c59<>1, r45c3,r5c1,r6c459<>3
2/5 in r6c23,r8c23 => r4c23,r5c3<>1, r45c3<>3, r4c23,r5c3<>6
Unique Rectangle Type 4
We look for additional candidates in two non diagonal cells again, but this time we ignore those extra candidates and concentrate on the UR candidates themselves: If one of the UR candidates is not possible anymore in any other cell of a house that contains both cells with the extra candidates, the other UR candidate can be eliminated from those UR cells.
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.-----------------.-----------.----------------.
| 3 145 24 | 7 6 14 | 25 9 8 |
| 79 8 6 | 29 3 5 | 4 27 1 |
| 1579 1457 249 | 29 8 14 |256-7@ 3 56-7@|
:-----------------+-----------+----------------:
| 17 147 48 | 6 5 2 | 3 178 9 |
| 169 16 89 | 4 7 3 | 1568 158 2 |
| 2 3 5 | 1 9 8 | 67@ 4 67@ |
:-----------------+-----------+----------------:
| 4 9 3 | 8 2 6 | 157 157 57 |
| 8 2 1 | 5 4 7 | 9 6 3 |
| 56 56 7 | 3 1 9 | 258 258 4 |
'-----------------'-----------'----------------'
To see how this works, let us take a look at the example: Possible UR on 6/7 in r3c79 and r6c79. r3c7 and r3c9 have additional candidates (meaning: those are the only cells of the UR that could possible contain neither a 6 nor a 7). There are two houses that see both cells r3c7 and r3c9: Row 3 and block 3. Row 3 has candidate 6 only in the UR cells (so does block 3, but this is not necessary - one house is sufficient), so 6 must go in either r3c7 or r3c9. But that means that 7 can be neither in r3c7 nor in r3c9 or we would have two solutions. We can therefore eliminate 7 from r3c79
Unique Rectangle Type 5
UR Type 5 is a variation of UR Type 2, but now the additional candidate can be in diagonal cells as well. Suppose we have a possible UR with one extra candidate in either two diagonal cells or in three cells. All occurences of that candidate can be eliminated in all cells that see all UR cells containing the additional candidate. The logic behind that move is the same as in a UR Type 2.
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.----------------.-------------.-------------.
| 7 5 46 | 9 8 2 | 46 3 1 |
| 2 14 9 | 6 14 3 | 5 8 7 |
| 13 1346 8 | 14 5 7 | 9 246 24 |
:----------------+-------------+-------------:
| 8 12 12 | 5 9 4 | 3 7 6 |
| 6 47 47 | 3 2 1 | 8 9 5 |
| 5 9 3 | 8 7 6 | 14 124 24 |
:----------------+-------------+-------------:
| 13 1367 5 | 47-1 146@ 9 | 2 146@ 8 |
| 9 8 16 | 2 146@ 5 | 7 146@ 3 |
| 4 1267 1267 | 17 3 8 | 16 5 9 |
'----------------'-------------'-------------'
This example is the only published example for a UR Type 5 with only two additional candidates so far: 1 must be set in either r7c8 or r8c5 to avoid the UR. Cell r7c4 sees both UR cells and cannot be 1 itself.
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.--------------.---------------.--------------.
| 34 9 17 | 8 146 36 | 2 5 467 |
| 8 47 57 | 2 469 3569 | 369 1 4679 |
| 234 6 125 | 7 149 359 | 39 8 49 |
:--------------+---------------+--------------:
| 9 8 4 | 5 2 1 | 7 6 3 |
| 7 2 6 | 4 3 8 | 5 9 1 |
| 5 1 3 | 69 69 679 | 4 2 8 |
:--------------+---------------+--------------:
| 26 3 29 | 69 8 4 | 1 7 5 |
| 1 5 8 | 3 7 269@ | 9-6 4 269@ |
| 246 47 279 | 1 5 269@ | 8 3 269@ |
'--------------'---------------'--------------'
The second UR Type 5 has three additional candidates: 6 must be in r8c6, r8c9, or r9c9. It can be eliminated from r8c7.
Unique Rectangle Type 6
UR Type 6 is a variation of UR Type 4. It is not very common, but highly effective, because it always leads to two placements: Suppose we have a possible UR with two diagonal cells containing extra candidates. If one of the UR candidates is nowhere else in the two rows and two columns building the UR, it can be eliminated from the cells with the extra candidates.
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.--------------.-------------.----------.
| 5 4 1 | 26 3 28 | 7 68 9 |
| 89 39 36 | 16 7 18 | 2 4 5 |
| 78 27 26 | 9 4 5 | 1 68 3 |
:--------------+-------------+----------:
| 47 237 23-5@| 25@ 6 24 | 8 9 1 |
| 49 29 25@ | 12-5@8 124 | 3 7 6 |
| 6 1 8 | 3 9 7 | 4 5 2 |
:--------------+-------------+----------:
| 2 6 7 | 4 1 9 | 5 3 8 |
| 1 8 9 | 7 5 3 | 6 2 4 |
| 3 5 4 | 8 2 6 | 9 1 7 |
'--------------'-------------'----------'
Looking at the example: In rows 4 and 5 and columns 3 and 4 candidate 5 appears only within the UR: It forms an X-Wing. If 5 was placed in one of the cells r4c3/r5c4 that contain the extra candidates, it had to be placed in the other cell too, thus forcing the UR and causing two solutions. It is therefore safe to assume that it cannot be in either of those cells.
Note that every UR Type 6 is accompanied by a pair of Hidden Rectangles, each of them leading to the same placements.
Hidden Rectangle
Hidden Rectangles are very versatile, because they can be used in possible URs that contain up to three cells with arbitrary additional candidates (the UR is hidden under a clutter of additional candidates - not to be confused with Almost Unique Rectangles).
We need a possible UR with two or three cells containing additional candidates (with only one cell the UR Type 1 should be used). Now take one UR cell without additional candidates and check the row and the column containing the opposite corner of the UR. If one UR candidate is nowhere outside the UR in those two houses, the other UR candidate can be eliminated from the opposite corner.
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.----------.----------------.-------------.
| 3 9 4 | 6 8 2 | 1 5 7 |
| 6 1 8 | 7 3 5 | 4 2 9 |
| 5 2 7 | 4 19 19 | 3 6 8 |
:----------+----------------+-------------:
| 1 5 3 | 29 4 7 | 289 89 6 |
| 4 7 29 | 5 6 8 | 29 3 1 |
| 8 6 29 | 1239 129 139 | 7 4 5 |
:----------+----------------+-------------:
| 7 3 6 | 8 159@ 4 | 59@ 19 2 |
| 9 4 5 | 12 127 16 | 68 178 3 |
| 2 8 1 | 39 57-9@ 369 | 569@ 79 4 |
'----------'----------------'-------------'
To see how this works, let us look at the example: We have only one cell without additional candidates: r7c7, which becomes our starting point. The opposite corner is cell r9c5 contained in row 9 and column 5. Candidate 5 is nowhere outside the UR in those two houses, so 9 can be eliminated from r9c5. Why? If we look at column 5, we see that candidate 5 has to be placed in r7c5 or in r9c5. If it is placed in r9c5, that cell cannot be 9. If it is placed in r7c5, it cannot be in r9c5, and therefore has to be placed in r9c7 too. This forces r7c7 to be 9, and that means that r9c5 cannot be 9 as well or we would have two solutions. Since both possible placements of candidate 5 in column 5 lead to r9c7<>9, 9 can be eliminated from that cell.
Obscured Unique Rectangle: Types 1- 6
{these are combination of naked/hidden sets of pairs} [my idea and concept]
' obsuce rectangles are naked and hidden subsets in full or combinations found on a grid can be used to form a unique rectangles with out actually implying the deductions of the hidden subsets first.'
these rectangles can be any of the basic 6 types and are a varation of the hidden class.
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.-------------------.------------------.---------------------.
| 26 1236 124 | 7 89-16@ 5 | 2389 249 4-89@ |
| 2567 9 8 | 1246 16 1236 | 2357 2457 457 |
| 257 2357 2457 | 24 89@ 23 | 6 1 89-457@|
:-------------------+------------------+---------------------:
| 9 2578 6 | 125 157 127 | 12578 2457 3 |
| 1 2578 257 | 3 567 4 | 25789 25679 56789 |
| 257 4 3 | 8 1567 9 | 1257 2567 567 |
:-------------------+------------------+---------------------:
| 8 567 57 | 9 2 67 | 4 3 1 |
| 3 1567 9 | 156 4 167 | 57 8 2 |
| 4 12567 1257 | 156 3 8 | 579 5679 5679 |
'-------------------'------------------'---------------------'
R13C5(89) hidden pair + R3C9(89) hidden pair + R1C9(89)
=> R1C9<>89
as well as the normal hidden eliminations
R1C5<>16, R3C9 <>457
Almost Unique Rectangle:{aur}
is any pattern of 2 digits that is almost forming a unique rectangle arangment:
to use this stratagy
a chain or sequence of deductions/events that forces the pattern to complete as a unique rectangle cannot occur.
there is mutiple ways to form an AUR:
{i need a simplistic example to demenstrate it.}
More info on AURs
