The New Sudoku Players' Forum

[coloin]

Code: Select all

+---+---+---+
|74.|.98|5..|
|8.9|..6|...|
|.65|.7.|...|
+---+---+---+
|...|.69|.5.|
|9..|7..|...|
|65.|84.|.97|
+---+---+---+
|...|6.4|8..|
|5..|...|41.|
|4..|...|.3.|
+---+---+---+


Definitely Maybe !

[denis_berthier]

.
Hi coloin

When you propose a puzzle, it would be nice to add its linear form:
74..985..8.9..6....65.7........69.5.9..7.....65.84..97...6.48..5.....41.4......3.

Maybe^2

Code: Select all

Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 7      4      123    ! 123    9      8      ! 5      26     1236   !
   ! 8      123    9      ! 12345  1235   6      ! 1237   247    1234   !
   ! 123    6      5      ! 1234   7      123    ! 1239   248    123489 !
   +----------------------+----------------------+----------------------+
   ! 123    12378  123478 ! 123    6      9      ! 123    5      12348  !
   ! 9      1238   12348  ! 7      1235   1235   ! 1236   2468   123468 !
   ! 6      5      123    ! 8      4      123    ! 123    9      7      !
   +----------------------+----------------------+----------------------+
   ! 123    12379  1237   ! 6      1235   4      ! 8      27     259    !
   ! 5      23789  23678  ! 239    238    237    ! 4      1      269    !
   ! 4      12789  12678  ! 1259   1258   1257   ! 2679   3      2569   !
   +----------------------+----------------------+----------------------+
196 candidates.

naked-triplets-in-a-row: r4{c1 c4 c7}{n3 n2 n1} ==> r4c9≠3, r4c9≠2, r4c9≠1, r4c3≠3, r4c3≠2, r4c3≠1, r4c2≠3, r4c2≠2, r4c2≠1

Code: Select all

Trid-OR3-relation for digits 1, 3 and 2 in blocks:
        b1, with cells (marked #): r1c3, r2c2, r3c1
        b2, with cells (marked #): r1c4, r2c5, r3c6
        b4, with cells (marked #): r6c3, r5c2, r4c1
        b5, with cells (marked #): r6c6, r5c5, r4c4
with 3 guardians (in cells marked @): n5r2c5 n8r5c2 n5r5c5

   +----------------------+----------------------+----------------------+
   ! 7      4      123#   ! 123#   9      8      ! 5      26     1236   !
   ! 8      123#   9      ! 12345  1235#@ 6      ! 1237   247    1234   !
   ! 123#   6      5      ! 1234   7      123#   ! 1239   248    123489 !
   +----------------------+----------------------+----------------------+
   ! 123#   78     478    ! 123#   6      9      ! 123    5      48     !
   ! 9      1238#@ 12348  ! 7      1235#@ 1235   ! 1236   2468   123468 !
   ! 6      5      123#   ! 8      4      123#   ! 123    9      7      !
   +----------------------+----------------------+----------------------+
   ! 123    12379  1237   ! 6      1235   4      ! 8      27     259    !
   ! 5      23789  23678  ! 239    238    237    ! 4      1      269    !
   ! 4      12789  12678  ! 1259   1258   1257   ! 2679   3      2569   !
   +----------------------+----------------------+----------------------+

Trid-OR3-whip[4]: r7n9{c2 c9} - r7n5{c9 c5} - OR3{{n5r5c5 n5r2c5 | n8r5c2}} - r4c2{n8 .} ==> r7c2≠7

***** STARTING ELEVEN_S REPLACEMENT TECHNIQUE *****
RELEVANT DIGIT REPLACEMENTS WILL BE NECESSARY AT THE END, based on the original givens.
Trying in block 1

AFTER APPLYING ELEVEN''S REPLACEMENT METHOD to 3 digits 1, 2 and 3 in 3 cells r3c1, r2c2 and r1c3,
the resolution state is:

Code: Select all

   +----------------------+----------------------+----------------------+
   ! 7      4      3      ! 123    9      8      ! 5      1236   1236   !
   ! 8      2      9      ! 12345  1235   6      ! 1237   12347  1234   !
   ! 1      6      5      ! 1234   7      123    ! 1239   12348  123489 !
   +----------------------+----------------------+----------------------+
   ! 123    78     478    ! 123    6      9      ! 123    5      48     !
   ! 9      1238   12348  ! 7      1235   1235   ! 1236   123468 123468 !
   ! 6      5      123    ! 8      4      123    ! 123    9      7      !
   +----------------------+----------------------+----------------------+
   ! 123    1239   1237   ! 6      1235   4      ! 8      1237   12359  !
   ! 5      123789 123678 ! 1239   1238   1237   ! 4      123    12369  !
   ! 4      123789 123678 ! 12359  12358  12357  ! 123679 123    123569 !
   +----------------------+----------------------+----------------------+

THIS IS THE PUZZLE THAT WILL NOW BE SOLVED.
RELEVANT DIGIT REPLACEMENTS WILL BE NECESSARY AT THE END, based on the original givens.

finned-x-wing-in-rows: n1{r4 r1}{c4 c7} ==> r2c7≠1
biv-chain[4]: r4n1{c7 c4} - r1c4{n1 n2} - r3c6{n2 n3} - r6n3{c6 c7} ==> r6c7≠1, r4c7≠3
biv-chain[5]: c1n2{r7 r4} - r4n3{c1 c4} - r6n3{c6 c7} - r2c7{n3 n7} - b9n7{r9c7 r7c8} ==> r7c8≠2
z-chain[5]: c7n1{r5 r9} - c7n6{r9 r5} - c8n6{r5 r1} - r1n1{c8 c4} - r4n1{c4 .} ==> r5c9≠1
t-whip[5]: r6c3{n1 n2} - r6c7{n2 n3} - r2c7{n3 n7} - c8n7{r2 r7} - r7c3{n7 .} ==> r8c3≠1, r9c3≠1, r5c3≠1
whip[5]: r3c6{n2 n3} - r6c6{n3 n1} - b4n1{r6c3 r5c2} - b4n3{r5c2 r4c1} - r4c4{n3 .} ==> r5c6≠2
whip[6]: r1c4{n2 n1} - r4c4{n1 n3} - b4n3{r4c1 r5c2} - b4n1{r5c2 r6c3} - r6c6{n1 n2} - b2n2{r3c6 .} ==> r9c4≠2
whip[6]: r1c4{n2 n1} - r4c4{n1 n3} - b4n3{r4c1 r5c2} - b4n1{r5c2 r6c3} - r6c6{n1 n2} - b2n2{r3c6 .} ==> r8c4≠2
whip[6]: r3c6{n2 n3} - r6n3{c6 c7} - r6n2{c7 c3} - b4n1{r6c3 r5c2} - r5n3{c2 c5} - c5n2{r5 .} ==> r9c6≠2
t-whip[6]: b8n2{r9c5 r8c6} - r3c6{n2 n3} - r6c6{n3 n1} - r6c3{n1 n2} - r4c1{n2 n3} - r4c4{n3 .} ==> r5c5≠2
whip[1]: c5n2{r9 .} ==> r8c6≠2
t-whip[5]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 .} ==> r9c7≠3
t-whip[5]: r4c7{n1 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 .} ==> r9c7≠1
whip[1]: c7n1{r5 .} ==> r5c8≠1
t-whip[6]: r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 .} ==> r4c7≠2
naked-single ==> r4c7=1
t-whip[6]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 r9c7} - c7n6{r9 .} ==> r5c7≠3
t-whip[6]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 r9c7} - c7n9{r9 .} ==> r3c7≠3
z-chain[4]: c7n3{r6 r2} - r2n7{c7 c8} - c8n4{r2 r3} - c8n8{r3 .} ==> r5c8≠3
whip[6]: r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 n2} - r5n2{c9 .} ==> r9c3≠2
whip[6]: r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 n2} - r5n2{c9 .} ==> r8c3≠2
whip[1]: b7n2{r7c3 .} ==> r7c5≠2, r7c9≠2
hidden-pairs-in-a-block: b8{n2 n8}{r8c5 r9c5} ==> r9c5≠5, r9c5≠3, r9c5≠1, r8c5≠3, r8c5≠1
z-chain[4]: b5n2{r4c4 r6c6} - r3c6{n2 n3} - c5n3{r2 r7} - c1n3{r7 .} ==> r4c4≠3
singles ==> r4c4=2, r1c4=1, r4c1=3, r7c1=2, r3c6=2, r3c7=9
finned-x-wing-in-columns: n1{c5 c2}{r5 r7} ==> r7c3≠1
stte

[coloin]

I understand eleven's clever replacement method...its logical
You could also try all six ways to put the 1-2-3 clues in box 1..and see how it solves [ ] That would be logical too !

But seriously..... Definitely no tridagon insertion in r5c5 !!!!! But Maybe there are some elsewhere in box 2 and box 4.

I thought the pattern unusual and couldnt advance on it .....I wonder what others can do with it ?

[denis_berthier]

I understand eleven's clever replacement method...its logical
You could also try all six ways to put the 1-2-3 clues in box 1..and see how it solves [ ] That would be logical too !

That's exactly what eleven's method amounts to. It's an educated form of T&E. Long ago, I've explained it here: http://forum.enjoysudoku.com/eleven-s-variable-replacement-method-and-its-complexity-t39277.html.
It's particularly efficient in case a tridagon is present.

I thought the pattern unusual and couldnt advance on it .....I wonder what others can do with it ?

As for me, I don't see anything unusual in the pattern itself.
.

[marek stefanik]

Code: Select all

,-------------------,-------------------,--------------------,
| 7    4     #123   |#123    9     8    | 5     26    1236   |
| 8   *123    9     | 12345 *1235  6    | 1237  247   1234   |
|#123  6      5     | 1234   7    #123  | 1239  248   123489 |
:-------------------+-------------------+--------------------:
|#123  78     478   |#123    6     9    | 123   5     48     |
| 9   *1238   12348 | 7     *1235  1235 | 1236  2468  123468 |
| 6    5     #123   | 8      4    #123  | 123   9     7      |
:-------------------+-------------------+--------------------:
|B123  12379 B7–123 | 6      1235  4    | 8     27    259    |
| 5    23789  23678 | 239   T238   237  | 4     1     269    |
| 4    12789  12678 | 1259  T128–5 1257 | 2679  3     2569   |
'-------------------'-------------------'--------------------'

TH 8-loop 123# => each of 123 has to appear in b15# and in b24# and thus can only appear once in the corresponding 4-loop (*)
Proof: Suppose that WLOG 1 does not appear in b15#.
The 23 pairs in b15# break either b2# (if r1c3 = r6c6), or b4# (if r1c3 = r4c4), i.e. contra.

GE: Any of 123 in r7c13 (B) is in c2 forced into * and in c5 into r89c5 (T) with 8. –123r7c3, –5r9c5, 9.2 skfr

Code: Select all

,------------------,-------------------,----------------,
| 7    4    #123   |#123    9     8    | 5    6   123   |
| 8  y*13–2  9     | 12345 *1235  6    |x13–2 7   1234  |
|#123  6     5     | 1234   7    #123  | 9    48  12348 |
:------------------+-------------------+----------------:
|#123  7     48    |#123    6     9    | 123  5   48    |
| 9   *8–213 12348 | 7     *25–13 25–13| 6    48 x13–2  |
| 6    5    #123   | 8      4    #123  | 123  9   7     |
:------------------+-------------------+----------------:
|B13   139   7     | 6      135   4    | 8    2   59    |
| 5    2389  2368  | 239   T38–2  7    | 4    1   69    |
| 4    1289  1268  | 1259  T18–2  125  | 7    3   569   |
'------------------'-------------------'----------------'

By the above GE, since r7c1 cannot be a 2 anymore, –2r89c5. 2c5\* => –2r25c2
The digit in r2c7 (x) takes b6p6. It then cannot be 2 (c5), so it forms a 13 pair with r2c2 (y). –13r2c459, –13r5c2
Recall that since y does not appear on b1#, it must appear on b5#. Since xy are a 13 pair, we get –13r5c56. 7.7 skfr

Code: Select all

,--------------,--------------,--------------,
| 7    4 af13–2| 123   9   8  | 5    6  123  |
| 8   y13  9   | 245   25  6  |x13   7  24   |
|b123  6   5   | 1234  7  c13 | 9    8  1234 |
:--------------+--------------+--------------:
| 123  7   4   | 13    6   9  | 123  5  8    |
| 9    8  y13  | 7     25  25 | 6    4 x13   |
| 6    5  e123 | 8     4  d13 | 123  9  7    |
:--------------+--------------+--------------:
| 13   9   7   | 6     13  4  | 8    2  5    |
| 5    23  68  | 29    38  7  | 4    1  69   |
| 4    12  68  | 259   18  25 | 7    3  69   |
'--------------'--------------'--------------'

13r5c39 => yr5c3
xr1c3 = r3c1 – r3c6 = r6c6 – r6c3 = xr1c3 => xr1c3, –2r1c3

Code: Select all

,------------,-------------,------------,
| 7   4   13 | 123  9   8  | 5   6  123 |
| 8   13  9  | 245  25  6  |#13  7  24  |
| 2   6   5  | 134  7  #13 | 9   8  4–13|
:------------+-------------+------------:
| 13  7   4  | 13   6   9  | 2   5  8   |
| 9   8   13 | 7    25  25 | 6   4  13  |
| 6   5   2  | 8    4  #13 |#13  9  7   |
:------------+-------------+------------:
| 13  9   7  | 6    13  4  | 8   2  5   |
| 5   23  68 | 29   38  7  | 4   1  69  |
| 4   12  68 | 259  18  25 | 7   3  69  |
'------------'-------------'------------'

RP 13# => –13r3c9, stte

[denis_berthier]

.
I expected coloin to make the same objection as before.
This is a variant of eleven's method. As soon as you consider the value in a cell as a variable, you're implicitly doing T&E.
.

[marek stefanik]

The word 'implicitly' is putting in a lot of work here.

Following a T&E procedure, it is easy to express every step with resolution.
I could therefore say that whenever you use T&E, you are implicitly doing resolution.
You looking at a solution using eleven's method and saying: "Actually, that's T&E(11)." would then be like someone looking at a whip and counting how many steps it would require in resolution (which might line up with a metric you have been very vocal against).

Now, clearly using T&E is easier than using resolution, otherwise people would use resolution instead. Same with eleven's method being easier than T&E.
Therefore, it makes no sense to use one as a complexity metric for solutions based on the others.

But the difference between eleven's method and T&E is much greater.
The whips in your path after relabeling don't even translate to eliminations in the original puzzle.
Maybe that's why coloin hasn't raised the same objection?

[denis_berthier]

Following a T&E procedure, it is easy to express every step with resolution.
I could therefore say that whenever you use T&E, you are implicitly doing resolution.

I've proved long ago that any T&E(1) step can be replaced by a braid. For T&E(2 or 3), this requires much more complex T-braids.
So, the word "easy" doesn't reflect any reality.

You looking at a solution using eleven's method and saying: "Actually, that's T&E(11)."

What I've said precisely is "when understood in terms of T&E-depth, eleven's method is depth 3, which explains why it can solve puzzles at T&E-depth 3". But it seems you have a problem with nuances.

The whips in your path after relabeling don't even translate to eliminations in the original puzzle.

They don't have to. That's the whole point of relabelling: you solve a different puzzle.
For the rest, I'll let coloin speak for himself.
But when you write "Proof: Suppose that WLOG 1 does not appear in b15#...." difficult not to think of a T&E variant.
.

[coloin]

I dont know now what to think about solving any puzzle with assumptions assertions and logical T&E !
There was countless arguments on " Eureka" board way back. But there is no trace of it now !

I chose the puzzle because you couldnt legitimately assert that r5c5 was a 5 ... {because its not}

finned-x-wing-in-rows: n1{r4 r1}{c4 c7} ==> r2c7≠1
biv-chain[4]: r4n1{c7 c4} - r1c4{n1 n2} - r3c6{n2 n3} - r6n3{c6 c7} ==> r6c7≠1, r4c7≠3
biv-chain[5]: c1n2{r7 r4} - r4n3{c1 c4} - r6n3{c6 c7} - r2c7{n3 n7} - b9n7{r9c7 r7c8} ==> r7c8≠2
z-chain[5]: c7n1{r5 r9} - c7n6{r9 r5} - c8n6{r5 r1} - r1n1{c8 c4} - r4n1{c4 .} ==> r5c9≠1
t-whip[5]: r6c3{n1 n2} - r6c7{n2 n3} - r2c7{n3 n7} - c8n7{r2 r7} - r7c3{n7 .} ==> r8c3≠1, r9c3≠1, r5c3≠1
whip[5]: r3c6{n2 n3} - r6c6{n3 n1} - b4n1{r6c3 r5c2} - b4n3{r5c2 r4c1} - r4c4{n3 .} ==> r5c6≠2
whip[6]: r1c4{n2 n1} - r4c4{n1 n3} - b4n3{r4c1 r5c2} - b4n1{r5c2 r6c3} - r6c6{n1 n2} - b2n2{r3c6 .} ==> r9c4≠2
whip[6]: r1c4{n2 n1} - r4c4{n1 n3} - b4n3{r4c1 r5c2} - b4n1{r5c2 r6c3} - r6c6{n1 n2} - b2n2{r3c6 .} ==> r8c4≠2
whip[6]: r3c6{n2 n3} - r6n3{c6 c7} - r6n2{c7 c3} - b4n1{r6c3 r5c2} - r5n3{c2 c5} - c5n2{r5 .} ==> r9c6≠2
t-whip[6]: b8n2{r9c5 r8c6} - r3c6{n2 n3} - r6c6{n3 n1} - r6c3{n1 n2} - r4c1{n2 n3} - r4c4{n3 .} ==> r5c5≠2
whip[1]: c5n2{r9 .} ==> r8c6≠2
t-whip[5]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 .} ==> r9c7≠3
t-whip[5]: r4c7{n1 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 .} ==> r9c7≠1
whip[1]: c7n1{r5 .} ==> r5c8≠1
t-whip[6]: r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 .} ==> r4c7≠2
naked-single ==> r4c7=1
t-whip[6]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 r9c7} - c7n6{r9 .} ==> r5c7≠3
t-whip[6]: r6c7{n3 n2} - r5n2{c9 c3} - r6c3{n2 n1} - r7c3{n1 n7} - b9n7{r7c8 r9c7} - c7n9{r9 .} ==> r3c7≠3
z-chain[4]: c7n3{r6 r2} - r2n7{c7 c8} - c8n4{r2 r3} - c8n8{r3 .} ==> r5c8≠3
whip[6]: r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 n2} - r5n2{c9 .} ==> r9c3≠2
whip[6]: r6c3{n2 n1} - r7c3{n1 n7} - c8n7{r7 r2} - r2c7{n7 n3} - r6c7{n3 n2} - r5n2{c9 .} ==> r8c3≠2
whip[1]: b7n2{r7c3 .} ==> r7c5≠2, r7c9≠2
hidden-pairs-in-a-block: b8{n2 n8}{r8c5 r9c5} ==> r9c5≠5, r9c5≠3, r9c5≠1, r8c5≠3, r8c5≠1
z-chain[4]: b5n2{r4c4 r6c6} - r3c6{n2 n3} - c5n3{r2 r7} - c1n3{r7 .} ==> r4c4≠3
singles ==> r4c4=2, r1c4=1, r4c1=3, r7c1=2, r3c6=2, r3c7=9

so there is rows of negative statements of where the 1,2,3 cant go .. then we get the statement that r3c7 is a 9

When do you go back and remove the 123 in box 1 ? [ as they are 5/6 times wrong clues] .. or do you relabel the puzzle at the end based on the true box 9 ?

I also chose the puzzle because after assuming [ somehow] r5c5 isnt a 5 then you know that one or both the 5 @r2c5 or the 8@r5c2 are correct.. but no easy way [ for eleven !] to say which.

I admit now that these type of puzzles are all of little interest if solving them is going to be academic !!

Maybe the hardest puzzles have the ugliest solving path and this one is not endearing !!!

Thanks for both your assessments though.

[denis_berthier]

. or do you relabel the puzzle at the end based on the true box 9 ?

yes, that's how the method works. The original puzzle is transformed before being solved (or after being partly solved) and it is easily transformed back at the end (you know you only have to do some permutation of the 3 digits and you have the original clues to see which permutation works.

.I also chose the puzzle because after assuming [ somehow] r5c5 isnt a 5 then you know that one or both the 5 @r2c5 or the 8@r5c2 are correct.. but no easy way [ for eleven !] to say which.

You can know it at the end. But it is of no interest.

[eleven]

Anyway, nice solution by Marek (who already had introduced the TH-8-loop) !

[RichardGoodrich]

Pay no attention to those who want linear. If you gave it linear they would want grid. Is there something special here?

[eleven]

For some yes, but i am sure, not for you.