The New Sudoku Players' Forum

[denis_berthier]

To be totally clear, I didn't say the trick was not valid. I said the trick involves some hidden information input. And I can say it now, this hidden information input is indeed the assumption of uniqueness. And the way it appears is very subtle and unusual.

I'm not sure what you might mean by hidden information.

It shouldn't be too difficult to imagine that, if you reduce the complexity of a puzzle (a purely logical complexity measure that is moreover stable under isomorphism) by some magical trick, the trick must involve some information input that was not available in the initial puzzle. As I said above, this information input is the assumption of uniqueness.

Here's how I look at it: The original puzzle had a unique solution. (This argument doesn't rely on uniqueness as a condition for working; it will prove or disprove uniqueness in the process.)

A really strange statement to start with, immediately followed by a denial that it is used in the argument. However, I shall show that you are using uniqueness in a hidden form.

There are 3 (in the case of EF1) digits which only appear once, all in the same house, after basics.
For EF1, then, if we remove those 3 unique digits, what property does the puzzle we are left with have? If it was unique before, then clearly it now has 3! = 6 solutions - the filled digits were serving to disambiguate which permutation we actually had.

Notice that you're mentioning uniqueness again (though you claim not to use it).
But before dealing with this, let me mention some ambiguity in what you are doing. Are you doing your transformations on the initial puzzle or on the PM obtained after Singles?

1) In the first interpretation (which doesn't really make sense), here is what we really get.

Now, for those 6 solutions, what are the values of the empty cells in c4? 124, in some order.

That happens to be true, but not for the reason you're stating. Without assuming uniqueness, we don't know how many solutions there are, but it is totally irrelevant here.
After deleting any 1 2 4 given in column 3 and allowing only 124 to be in r235c4, we have:

Code: Select all

+-------------------------------+-------------------------------+-------------------------------+
! 9         8         123456789 ! 123456789 123456789 123456789 ! 123456789 123456789 123456789 !
! 7         123456789 123456789 ! 124       123456789 6         ! 5         123456789 123456789 !
! 123456789 5         123456789 ! 124       7         123456789 ! 9         6         123456789 !
+-------------------------------+-------------------------------+-------------------------------+
! 123456789 123456789 123456789 ! 123456789 8         123456789 ! 123456789 3         123456789 !
! 123456789 123456789 123456789 ! 124       123456789 7         ! 8         123456789 123456789 !
! 123456789 7         123456789 ! 3         123456789 123456789 ! 123456789 123456789 9         !
+-------------------------------+-------------------------------+-------------------------------+
! 123456789 123456789 123456789 ! 123456789 5         123456789 ! 123456789 123456789 6         !
! 123456789 123456789 123456789 ! 6         123456789 9         ! 3         8         123456789 !
! 123456789 123456789 123456789 ! 123456789 123456789 3         ! 123456789 7         123456789 !
+-------------------------------+-------------------------------+-------------------------------+

What we can know by using only Singles is the following:

Code: Select all

hidden-single-in-a-row ==> r8c3 = 7
hidden-single-in-a-row ==> r7c4 = 7
hidden-single-in-a-column ==> r9c4 = 8
hidden-single-in-a-column ==> r3c6 = 8
hidden-single-in-a-block ==> r2c9 = 8
hidden-single-in-a-column ==> r7c8 = 9
hidden-single-in-a-column ==> r4c4 = 9
hidden-single-in-a-column ==> r1c4 = 5
hidden-single-in-a-column ==> r2c5 = 9
hidden-single-in-a-block ==> r1c5 = 3
hidden-single-in-a-block ==> r3c9 = 3
hidden-single-in-a-row ==> r1c3 = 6

RESOLUTION STATE:
   9         8         6         5         3         124       1247      124       1247     
   7         1234      1234      124       9         6         5         124       8         
   124       5         124       124       7         8         9         6         3         
   12456     1246      1245      9         8         1245      12467     3         12457     
   123456    123469    123459    124       1246      7         8         1245      1245     
   124568    7         12458     3         1246      1245      1246      1245      9         
   12348     1234      12348     7         5         124       124       9         6         
   1245      124       7         6         124       9         3         8         1245     
   12456     12469     12459     8         124       3         124       7         1245
195 candidates, 1345 csp-links and 1345 links. Density = 7.11%

There is a clear loss of information compared to the original puzzle after Singles (and this Sukaku puzzle cannot be solved by any whip of any length).
We do find your 6 possibilities for column c4. But making the 124 to ABC change doesn't allow a solution by Singles.


2) Suppose now you were talking of the PM after Singles (and no other "basics" are needed here).

Now, for those 6 solutions [...]

In this case, your statement is obviously false. Whatever the initial number of solutions can be, it is unchanged and the reason for this is very simple: the first steps of the solution are 3 Singles that re-establish the 3 deleted values. At this point, no information has been gained, no information has been lost:

Code: Select all

(solve-sukaku-grid
   +----------------------+----------------+-----------------+
   ! 9    8     6         ! 5    3     124 ! 1247  124  1247 !
   ! 7    24    3         ! 124  9     6   ! 5     124  8    !
   ! 24   5     123456789 ! 24   7     8   ! 9     6    3    !
   +----------------------+----------------+-----------------+
   ! 12   126   123456789 ! 9    8     5   ! 1267  3    127  !
   ! 123  1236  9         ! 124  1246  7   ! 8     5    124  !
   ! 8    7     5         ! 3    1246  124 ! 1246  124  9    !
   +----------------------+----------------+-----------------+
   ! 134  134   8         ! 7    5     124 ! 124   9    6    !
   ! 5    14    7         ! 6    124   9   ! 3     8    124  !
   ! 6    9     123456789 ! 8    14    3   ! 14    7    5    !
   +----------------------+----------------+-----------------+
)
hidden-single-in-a-row ==> r9c3 = 2
hidden-single-in-a-row ==> r4c3 = 4
naked-single ==> r3c3 = 1
Resolution state:
   9         8         6         5         3         124       1247      124       1247     
   7         24        3         124       9         6         5         124       8         
   24        5         1         24        7         8         9         6         3         
   12        126       4         9         8         5         1267      3         127       
   123       1236      9         124       1246      7         8         5         124       
   8         7         5         3         1246      124       1246      124       9         
   134       134       8         7         5         124       124       9         6         
   5         14        7         6         124       9         3         8         124       
   6         9         2         8         14        3         14        7         5         

After that, the resolution state and the resolution path are exactly the same as for the original puzzle (in Z4).
Notice that, you could also re-add the 1 in r3c4; that wouldn't change anything to the above rules and decided values.

And, in fact, a different order for each of the solutions. We don't care about which is correct, yet. We just know that whatever permutation was in c3, there is a corresponding permutation in c4.

In this second interpretation, this statement becomes meaningless.


Now, back to uniqueness.

Effectively, what we are doing by putting ABC into c4, rather than 124 into c3,[...]

What you are really doing here is assuming that, by some unspecified means, one can reach a resolution state where the 3 values in c4 are decided. But the only general way this can be guaranteed is if the solution is unique. After that, you don't have to know the real values for applying Singles and reach a solution, but you couldn't apply Singles if you didn't make this assumption first.
At this point you are assuming there is a unique solution.
From there, you can do whatever you want, it will be impossible to erase this assumption by any form of circular reasoning.

[mith]

Denis, I don't disagree that the puzzle has to be unique for this to work. (To solve any puzzle uniquely, you need the puzzle to have a unique solution.) The distinction I see here is that using the trick proves uniqueness in the process (unlike, say, a hidden rectangle, which assumes that there is no solution if the grid has that deadly pattern, where some invalid puzzle could actually have 2 solutions to that case and 3 total).

To break down the argument (and to be clear, I am talking about the state of the puzzle after singles, where most of 356789 are filled and c3 is complete; that should be clear from the grids anyway, but to make sure there is no confusion):

1. Any solution of the first puzzle (whether the puzzle is unique or not!) must have some permutation of 124 in c4.
2. Any solution of the second puzzle (whether the puzzle is unique or not!) must have some permutation of ABC in c3.
3. If the first puzzle can be solved uniquely (we don't know this yet!), there are exactly 6 solutions of the in-between state (the puzzle without 124 or ABC filled in either column):

Code: Select all

.------------------.----------------.------------------.
| 9     8      6   | 5    3     124 | 1247   124  1247 |
| 7     124    3   | 124  9     6   | 5      124  8    |
| 124   5      124 | 124  7     8   | 9      6    3    |
:------------------+----------------+------------------:
| 124   1246   124 | 9    8     5   | 12467  3    1247 |
| 1234  12346  9   | 124  1246  7   | 8      5    124  |
| 8     7      5   | 3    1246  124 | 1246   124  9    |
:------------------+----------------+------------------:
| 1234  1234   8   | 7    5     124 | 124    9    6    |
| 5     124    7   | 6    124   9   | 3      8    124  |
| 6     9      124 | 8    124   3   | 124    7    5    |
'------------------'----------------'------------------'

or

Code: Select all

.------------------.----------------.------------------.
| 9     8      6   | 5    3     ABC | ABC7   ABC  ABC7 |
| 7     ABC    3   | ABC  9     6   | 5      ABC  8    |
| ABC   5      ABC | ABC  7     8   | 9      6    3    |
:------------------+----------------+------------------:
| ABC   ABC6   ABC | 9    8     5   | ABC67  3    ABC7 |
| AB3C  AB3C6  9   | ABC  ABC6  7   | 8      5    ABC  |
| 8     7      5   | 3    ABC6  ABC | ABC6   ABC  9    |
:------------------+----------------+------------------:
| AB3C  AB3C   8   | 7    5     ABC | ABC    9    6    |
| 5     ABC    7   | 6    ABC   9   | 3      8    ABC  |
| 6     9      ABC | 8    ABC   3   | ABC    7    5    |
'------------------'----------------'------------------'

Each of these solutions must have a different permutation of 124 in c3, ABC in c4... and also ABC in c3 and 124 in c4. (Say there were 2 solutions with the same order of 124 in c4 but have different orders in c3. For the sake of argument, let's say one of these has 142 in order in c3, and the other has 421 in order in c3. We can take the second solution, and permute the digits 4 -> 1 -> 2 -> 4 for the entire grid to arrive at a new solution with 142 in c3 but a different permutation in c4; that is, there are at least 2 solutions to the original puzzle. On the other hand, we could have 2 solutions with the same order of 124 in c4 and in c3, in which case we can permute the digits to each different order and end up with 12 solutions total.)

4. Each of these 6 solutions has one of the 6 different permutations for 124 in c3, 124 in c4, ABC in c3, and ABC in c4; it should be clear, then, that there is a one-to-one correspondence between 124 and ABC. (If for one of these grids 124 and ABC map in c4 one way and in c3 a different way, we can do a similar permutation as in the previous point to find a second solution for the same fill of one of the columns.)
5. Now, the key distinction: If the puzzle with ABC in c4 solves uniquely - and we can see it does by solving it with singles - this proves the original puzzle must also solve uniquely (same argument again - if the original didn't solve uniquely, we would be able to derive another solution for the ABC version), and therefore that the mapping between solutions is valid.

Rather than relabeling, it may be more intuitive to think of as a coloring argument. ABC in c4 are effectively colors, and regardless of what digits are in those cells, we can make deductions about what colors must go in other cells; if we end up concluding that color A is in r4c3, and the puzzle has a 4 in r4c3, then we know color A is 4 for the entire grid.

Note that f the puzzle were not unique, the relabeling trick could still be used... and it would show one of two things: either that there is not a unique mapping between 124 and ABC, or that there is still a unique mapping but there is some other deadly pattern elsewhere in the puzzle (that is, we were able to color c3, but not the entire grid). Either way, using the trick will prove it is not unique (and also which case is correct - in the latter case, we will still be able to map 124 to ABC and thus uniquely solve c4, just not the entire puzzle), though of course there is no guarantee that any particular relabeling is going to make the puzzle easier (I could just as easily have given a puzzle with SER 7.8 which only gets harder if you relabel).

[denis_berthier]

using the trick proves uniqueness in the process

That's where we disagree.

to be clear, I am talking about the state of the puzzle after singles,

This is the state after Singles:

Code: Select all

   9         8         6         5         3         124       1247      124       1247     
   7         24        3         124       9         6         5         124       8         
   24        5         1         24        7         8         9         6         3         
   12        126       4         9         8         5         1267      3         127       
   123       1236      9         124       1246      7         8         5         124       
   8         7         5         3         1246      124       1246      124       9         
   134       134       8         7         5         124       124       9         6         
   5         14        7         6         124       9         3         8         124       
   6         9         2         8         14        3         14        7         5

But you write:

3. If the first puzzle can be solved uniquely (we don't know this yet!), there are exactly 6 solutions of the in-between state (the puzzle without 124 or ABC filled in either column):

Code: Select all

.------------------.----------------.------------------.
| 9     8      6   | 5    3     124 | 1247   124  1247 |
| 7     124    3   | 124  9     6   | 5      124  8    |
| 124   5      124 | 124  7     8   | 9      6    3    |
:------------------+----------------+------------------:
| 124   1246   124 | 9    8     5   | 12467  3    1247 |
| 1234  12346  9   | 124  1246  7   | 8      5    124  |
| 8     7      5   | 3    1246  124 | 1246   124  9    |
:------------------+----------------+------------------:
| 1234  1234   8   | 7    5     124 | 124    9    6    |
| 5     124    7   | 6    124   9   | 3      8    124  |
| 6     9      124 | 8    124   3   | 124    7    5    |
'------------------'----------------'------------------'

What is this in-between state? E.g., where does the 1 in r1c2 come from? Are you now replacing all the cells containing 1 and/or 2 and/or 4 by 124? Obviously, we aren't yet talking of the same thing.

Anyway, whichever state you start from, when you replace the three 124 in c4 by 3 supposedly decided (but unknown) values A, B, C, you fall into the hidden trap mentioned in my previous post.

I notice you didn't propose any explanation to the magical change of complexity level. How can a mere relabelling (because this is how you present this approach) decrease the complexity level so as to make it equal to Singles? What are the hidden resolution rules that replace those needed for the original puzzle?

[mith]

Denis, the in-between state is this hypothetical puzzle, where we don't know anything about the placement of 124 in c3:

Code: Select all

+-------+-------+-------+
| 9 8 6 | 5 3 . | . . . |
| 7 . 3 | . 9 6 | 5 . 8 |
| . 5 . | . 7 8 | 9 6 3 |
+-------+-------+-------+
| . . . | 9 8 5 | . 3 . |
| . . 9 | . . 7 | 8 5 . |
| 8 7 5 | 3 . . | . . 9 |
+-------+-------+-------+
| . . 8 | 7 5 . | . 9 6 |
| 5 . 7 | 6 . 9 | 3 8 . |
| 6 9 . | 8 . 3 | . 7 5 |
+-------+-------+-------+
98653....7.3.965.8.5..78963...985.3...9..785.8753....9..875..965.76.938.69.8.3.75

The placement of the digits in c3 is serving to (uniquely, if the puzzle has a unique solution) determine which of the solutions to the in-between state is true. Likewise, the placement of ABC in c4 is also serving to (uniquely, if the puzzle has a unique solution) determine which of the solutions to the in-between state is true.

We don't know at this point how many solutions there are, either to the original puzzle, the in-between state, or the c4-filled puzzle; but we do know how they relate to each other - the in-between state has 6 solutions for each solution to the original puzzle (or c4-filled puzzle), one for each permutation of 124 (or ABC). We also know that if the original puzzle is unique, the c4-filled puzzle is unique, and conversely if the c4-puzzle is unique, the original puzzle is unique. The singles solve of the c4-filled puzzle thus proves the uniqueness of the original puzzle.

Just to give something concrete for how the coloring approach looks (without removing or changing any pencilmarks or digits):

Code: Select all

.--------------.----------------.-----------------.
| 9    8     6 | 5    3     124 | 1247  124  1247 |
| 7    24    3 | 124  9     6   | 5     124  8    |
| 24   5     1 | 24   7     8   | 9     6    3    |
:--------------+----------------+-----------------:
| 12   126   4 | 9    8     5   | 1267  3    127  |
| 123  1236  9 | 124  1246  7   | 8     5    124  |
| 8    7     5 | 3    1246  124 | 1246  124  9    |
:--------------+----------------+-----------------:
| 134  134   8 | 7    5     124 | 124   9    6    |
| 5    14    7 | 6    124   9   | 3     8    124  |
| 6    9     2 | 8    14    3   | 14    7    5    |
'--------------'----------------'-----------------'

Whatever is in r5c4 can't be in r23c4, so it must be in r1c6.
Then it can't be in r1c789, so it must be in r2c8.
Then it can't be in r6c8, along with r6c56 (r5c4), so it must be in r6c7.
Then it can't be in r78c7, so it must be in r8c9.
Then it can't be in r8c5, along with r7c6 (r1c6), so it must be in r9c5.
Since r5c4 can't be 6, r6c7 is not 6 either, resolving the 6s in r4c7, r5c2 and r6c5.
Since r4c7 is 6, 7 is resolved in r1c7 and r4c9.
Since r5c2 is 6, 3 is resolved in r5c1 and r7c2.
Since r5c4 can't be 3, and whatever is r5c4 is not in r8c2 (r8c9) or r9c3 (r9c5), it must be in r7c1.
Then it can't be in r3c1, along with r2c2 (r2c8), so it must be in r3c3.
But we know r3c3 is 1, so r5c4 must also be 1 (along with every other cell which has the same value). And so on for the other cells/digits.

What we are doing is restating the problem not in terms of which digits are in which cells, but rather which cells must have the same (or different) values. The apparent complexity disparity here is due to re-framing the problem in this way: we're effectively transforming this from a problem of placing the digits 124 into r235c4 (along with the rest of the grid) to a problem of placing the cells r235c4 (whatever their values) onto the digits in r349c3. Viewed as a contradiction chain starting with some candidate choice in r5c4, it is actually more complex that what is needed to solve the original puzzle without the trick:

Code: Select all

From YZF's solver:

Dynamic Contradiction Chain: If r5c4=2 Then r4c2=2 And r4c2<>2 simultaneously,so r5c4<>2
Chain 2:r5c4=2?r3c4<>2?r3c1=2?r2c2<>2
Chain 1:r5c4=2?r5c2<>2
Chain 0:(r5c2<>2+r2c2<>2) ? r4c2=2
Chain 10:r5c4=2?r3c4<>2
Chain 9:r5c4=2?r2c4<>2
Chain 8:r5c4=2?r6c5<>2
Chain 7:r5c4=2?r5c5<>2
Chain 6:(r5c5<>2+r6c5<>2) ? r8c5=2?r8c9<>2
Chain 5:(r2c4<>2+r3c4<>2) ? r1c6=2?r1c9<>2
Chain 4:r5c4=2?r5c9<>2
Chain 3:(r5c9<>2+r1c9<>2+r8c9<>2) ? r4c9=2?r4c2<>2

Dynamic Contradiction Chain: If r5c4=4 Then r4c7=6 And r4c7<>6 simultaneously,so r5c4<>4
Chain 9:r5c4=4?r3c4<>4
Chain 8:r5c4=4?r2c4<>4
Chain 7:(r2c4<>4+r3c4<>4) ? r1c6=4
Chain 6:r1c6=4?r1c9<>4
Chain 5:r1c6=4?r1c7<>4
Chain 4:r1c6=4?r1c8<>4
Chain 3:(r1c8<>4+r1c7<>4+r1c9<>4) ? r2c8=4?r6c8<>4
Chain 2:r5c4=4?r6c5<>4
Chain 1:r5c4=4?r6c6<>4
Chain 0:(r6c6<>4+r6c5<>4+r6c8<>4) ? r6c7=4?r6c7<>6?r4c7=6
Chain 16:r5c4=4?r2c4<>4
Chain 15:r5c4=4?r3c4<>4
Chain 14:(r2c4<>4+r3c4<>4) ? r1c6=4?r1c9<>4
Chain 13:r5c4=4?r5c9<>4
Chain 12:(r5c9<>4+r1c9<>4) ? r8c9=4?r8c2<>4?r8c2=1?r4c2<>1
Chain 11:r3c4<>4?r3c1=4?r2c2<>4?r2c2=2?r4c2<>2
Chain 10:(r4c2<>2+r4c2<>1) ? r4c2=6?r4c7<>6

But, when re-framed as a coloring or relabeling problem, it amounts to solving the second (c4-filled) grid with singles, and then mapping back to match the original puzzle's permutation in c3.

[denis_berthier]

Effectively, what we are doing by putting ABC into c4, rather than 124 into c3,[...]

What you are really doing here is assuming that, by some unspecified means, one can reach a resolution state where the 3 values in c4 are decided.

I still agree with my above statement. However, the following jump to uniqueness was too quick:

But the only general way this can be guaranteed is if the solution is unique.

There is indeed another general way this can be done, but not guaranteed to work: T&E(1) or T&E(2) (one can't say in advance where one will have to stop). Try successively all the candidates or all the pairs of compatible candidates in r235c4. This seems to be more in line with the explanations via bijections.
I must say this is still less enticing to me than merely assuming uniqueness, because it involves unrestricted complexity - even if hidden in terms of bijections.

[denis_berthier]

Viewed as a contradiction chain starting with some candidate choice in r5c4, it is actually more complex that what is needed to solve the original puzzle without the trick

Now, we agree.

[mith]

Try successively all the candidates or all the pairs of compatible candidates in r235c4. This seems to be more in line with the explanations via bijections.

You can certainly look at it that way. The distinction I would draw here is that we are not actually trying candidates at all - we're making deductions about the values of those cells independent of their actual values.

An alternative way of looking at it: we're trying all permutations of candidates simultaneously - like a quantum computer, the waveform collapses when we make a measurement (that is, when our coloring intersects with a known value).

(one can't say in advance where one will have to stop)

One can say exactly where one will have to stop here: you stop when either the cell-equivalence deductions reach the known digits in c3, or when you can't make any more deductions and have to try something else... though even here, if there are deductions made they remain true and potentially useful.

There is no particular reason why one would even need to focus on a particular house to "fill" with colors or letters or whatever. One could simply look at the relationships between cells and track how those interact around the grid. Candidate marking will tell me that r2c2 can only be 2 or 4. It will also tell me that r3c4 can only be 2 or 4. But I can see from the relationship between the cells that they in fact have to contain the same digit, whether it is 2 or 4. The relabeling trick (or coloring) is simply a more intuitive way to conceptualize it for a manual solver, IMO. (Whether that is true for computer solvers, or whether it would even be useful to check for something like this given the relative rarity of a valid setup, I have no idea.)

[yzfwsf]

Algebraic method, I think it is a general trial filling method, rather than some specific numbers.

[denis_berthier]

Try successively all the candidates or all the pairs of compatible candidates in r235c4. This seems to be more in line with the explanations via bijections.

You can certainly look at it that way. The distinction I would draw here is that we are not actually trying candidates at all - we're making deductions about the values of those cells independent of their actual values.

That's partly true. And only because of very special conditions (all the other candidates decided). Otherwise, these deductions wouldn't lead very far.

(one can't say in advance where one will have to stop)

What I said is:

T&E(1) or T&E(2) (one can't say in advance where one will have to stop)

You misunderstood the "where"; I meant at which level of T&E.

...something like this given the relative rarity of a valid setup

It was interesting to bring this back to light for a while. But right, not anything I'd care to add to my solver or to mention in a book.

[Cenoman]

The second puzzle simplifies with the same trick, though not to singles.

I was troubled when I tried, my first attempt was unsuccessful. Therefore, I tried to understand the blockage.

Code: Select all

 +-----------------------+-------------------+-----------------------+
 |  1234   8      6      |  123   7    9     |  1234   1234   5      |
 |  1234   1234   9      |  8     5    123   |  7      6      1234   |
 |  7      5      123    |  123   4    6     |  9      8      123    |
 +-----------------------+-------------------+-----------------------+
 |  124    9      124    |  5     3    124   |  8      7      6      |
 |  5      1234   7      |  6     8    124   |  1234   1234   9      |
 |  8      6      1234   |  124   9    7     |  5      1234   1234   |
 +-----------------------+-------------------+-----------------------+
 |  9      234    8      |  7     1    34    |  6      5      234    |
 |  6      134    5      |  9     2    8     |  134    134    7      |
 |  1234   7      1234   |  34    6    5     |  1234   9      8      |
 +-----------------------+-------------------+-----------------------+

The intermediate puzzle, without the knowledge of the arrangement of 1,2,3,4 in c5

Code: Select all

 +-----------------------+-----------------------+-----------------------+
 |  1234   8      6      |  1234   7      9      |  1234   1234   5      |
 |  1234   1234   9      |  8      5      1234   |  7      6      1234   |
 |  7      5      1234   |  1234   1234   6      |  9      8      1234   |
 +-----------------------+-----------------------+-----------------------+
 |  1234   9      1234   |  5      1234   1234   |  8      7      6      |
 |  5      1234   7      |  6      8      1234   |  1234   1234   9      |
 |  8      6      1234   |  1234   9      7      |  5      1234   1234   |
 +-----------------------+-----------------------+-----------------------+
 |  9      1234   8      |  7      1234   1234   |  6      5      1234   |
 |  6      1234   5      |  9      1234   8      |  1234   1234   7      |
 |  1234   7      1234   |  1234   6      5      |  1234   9      8      |
 +-----------------------+-----------------------+-----------------------+

This puzzle has 24 solutions (Andrew's Stuart site), and noticeable, 1, 2, 3, 4 are candidates in 36 cells alone, forming naked quads in all 27 sectors (no interaction with other digits, all placed in every sector).

Code: Select all

 +-----------------------+-------------------+-----------------------+
 |  abcd   8      6      |  abcd  7    9     |  abcd   abcd   5      |
 |  abcd   abcd   9      |  8     5    abcd  |  7      6      abcd   |
 |  7      5      abcd   |  abcd  abcd 6     |  9      8      abcd   |
 +-----------------------+-------------------+-----------------------+
 |  abcd   9      abcd   |  5     abcd abcd  |  8      7      6      |
 |  5      abcd   7      |  6     8    abcd  |  abcd   abcd   9      |
 |  8      6      abcd   |  abcd  9    7     |  5      abcd   abcd   |
 +-----------------------+-------------------+-----------------------+
 |  9      abcd   8      |  7     abcd abcd  |  6      5      abcd   |
 |  6      abcd   5      |  9     abcd 8     |  abcd   abcd   7      |
 |  abcd   7      abcd   |  abcd  6    5     |  abcd   9      8      |
 +-----------------------+-------------------+-----------------------+

This is exactly the same puzzle, written using a different notation.

Algebraic method, I think it is a general trial filling method, rather than some specific numbers.

Fully in line with yzfwsf

Now, following mith's idea, we can raise the questions:
- does the knowledge of the arrangement a, b, c, d in any of the 27 sectors yields a unique solution of the puzzle ?
- for sectors answsering "yes", are there some, "simplifying" the puzzle solution ? (using mith's wording)
- and last, but not least, how to spot these specifically ?

With 27 trials, the answer to those questions are:
- yes for all sectors, to the first,
- yes for 10 sectors (r2346, c346, b125), to the second.
The answer to this second question, is subjective. The quality "simplifying the puzzle solution" is just my own appreciation: I have selected the sectors yielding a solution with a few short chains (2-fishes, X-chains, and possibly wings).
Among the 10 sectors, three are notably simplifying the puzzle solution: r6, c4, b5, with three X-chain solution, two X-chain solution, three X-chain solution resp.

- the answer to the third question is: I don't know...

Here is the solution for c4:

Code: Select all

 +-----------------------+-------------------+-----------------------+
 |  bcd*^  8      6      |  a     7    9     |  bcd^   bcd^   5      |
 |  abd-c* ab-c   9      |  8     5    cd^   |  7      6      abcd^  |
 |  7      5      ac-d   |  b     cd*  6     |  9      8      acd^   |
 +-----------------------+-------------------+-----------------------+
 |  abcd*  9      abcd   |  5     abd* abd   |  8      7      6      |
 |  5      ab     7      |  6     8    abd   |  abcd   abcd   9      |
 |  8      6      abd    |  c     9    7     |  5      abd    bd     |
 +-----------------------+-------------------+-----------------------+
 |  9      abcd   8      |  7     abc  abc^  |  6      5      bcd^   |
 |  6      abcd   5      |  9     abc  8     |  abcd   abcd   7      |
 |  abc    7      abc    |  d     6    5     |  abc    9      8      |
 +-----------------------+-------------------+-----------------------+

Skyscraper(*) (d)r12c1 = r4c1 - r4c5 = r3c5 => -d r3c3
X-chain(^) (c)r1c1 = r1c78 - r23c9 = r7c9 - r7c6 = r2c6 => -c r2c12; lclste
[NP(ab)r25c2 => -ab r78c2; NP(cd)r78c2 => -c r9c13; ste]
Solution:

Code: Select all

 +-----------------------+-------------------+-----------------------+
 |  c      8      6      |  a     7    9     |  b      d      5      |
 |  d      b      9      |  8     5    c     |  7      6      a      |
 |  7      5      a      |  b     d    6     |  9      8      c      |
 +-----------------------+-------------------+-----------------------+
 |  b      9      c      |  5     a    d     |  8      7      6      |
 |  5      a      7      |  6     8    b     |  d      c      9      |
 |  8      6      d      |  c     9    7     |  5      a      b      |
 +-----------------------+-------------------+-----------------------+
 |  9      c      8      |  7     b    a     |  6      5      d      |
 |  6      d      5      |  9     c    8     |  a      b      7      |
 |  a      7      b      |  d     6    5     |  c      9      8      |
 +-----------------------+-------------------+-----------------------+

Comparing to the givens in c5: d=4, a=3, b=1, c=2 (selecting 1 out of 24 ways to affect 1, 2, 3, 4 to a, b, c, d)
the simplification is there, but...

Note: the same study on the first puzzle gives similar results: ste for r23, c468, b2 [i.e. not every sector having NT(124)]

Any other suggestion than such tedious T&E ?

[mith]

It’s a good question, and one I haven’t investigated fully. I suspect the answer is no, but it’s certainly possible we could find some property of these grids that allows for making a better guess at which houses might be “weak”.

Personally, I find it no more tedious to look for cell equivalence truths than it is to look for complex chains, but obviously YMMV. My approach on getting stuck in a puzzle like this would likely be to look for individual deductions and see if the lead anywhere - at least in the 3 digit case, it’s quite easy to spot a few at a glance (any pair of empty cells sharing both a row/column and a box is weak - the third cell in each house must be equivalent). Really, that may be one such property worth investigating - r23c4 is one such pair, and c4, b2, and even r23 are on the stte list (though it doesn’t apply to c6, for example).

[eleven]

I used this trick here for the first time.
No uniqueness assumption needed.