The New Sudoku Players' Forum

[ronk]

for puzzle #2, it seems somebody (read: not me) ought to be able to use the fact that both r3c9 and r7c1 can't be 1 as that would make a rectangle of death or rectangle of non-uniqueness

Perhaps what you are trying to say is that r3c9 and r7c1 cannot be "1" at the same time .....

Obviously, that's what Tracy meant. Indeed, I think that's what was said.

Ron

[tarek]

In response to your question about the next step for puzzle #2, it seems somebody (read: not me) aught to be able to use the fact that both r3c9 and r7c1 can't be 1 as that would make a rectangle of death or rectangle of non-uniqueness or whatever it's called in r3c4,6 and r7c4,6 (they would all have 2,3 as their only values).

You mean that they form a conjugate........ adding an extra bit of information when doing colouring for 1, the same then could be said about r2c6 & r8c4. Anyway, would that help ????

[TKiel]

Carcul,

As Tarek pointed out, that is what I meant and indeed what I thought I said. What I should have said was that either could be 1 or neither could be 1 but both could not be. Apparently, it wasn't a useful bit of info anyway, but it was about the only thing I could deduce from the candidate listing at that time. My hope was that someone who maybe hadn't noticed could use that to find a forcing chain or something to advance the puzzle.

Tracy

[ronk]

My hope was that someone who maybe hadn't noticed could use that to find a forcing chain or something to advance the puzzle.

I had noticed and could only find ...

[r6c8]-2-[r1c8]-5-[r2c8]-3-[r3c9]-1-(AUR:[r3c46]=1=[r7c46])-1-[r7c1]-2-[r8c2]-7-[r6c2]-2-[r6c8] implying r6c8<>2

... which lead no further.

[TKiel]

Actually, with the info Tarek provided, I think we may be able to form two conjugate chains that end up making an exclusion.

Cells r3c9 & r1c7 are related in that if the first is 1, then the second is not and if the second 1 is the first is not. (I'm not sure they can actually be called conjugate, since that implies both if one is true the other is false and if one if false the other is true, whereas in this case the relationship does not imply the second part.) Cell r2c7 does have a conjugate relationship with r3c9.

Cells r2c6 & r8c4 are related in the same manner and for the same reason (both being true allows the formation of a non-unique rectangle).

Labeling those two chains with A-a & B-b

Code: Select all

   *--------------------------------------------------------------*
 | 3      1      9    | 6      578    48   | 247    25     457  |
 | 6      8      2    | 1357   357    134A | 147B   35     9    |
 | 4      5      7    | 123    9      123  | 8      6      13b  |
 |----------------------+----------------------+----------------|
 | 1278   4      138  | 9      2378   5    | 1267   238    137  |
 | 2578   6      358  | 2378   1      238  | 247    9      3457 |
 | 9      27     1358 | 4      2378   6    | 127    2358   1357 |
 |----------------------+----------------------+----------------|
 | 12B    9      4    | 123    6      123  | 5      7      8    |
 | 12578  27     158  | 1258a  258    9    | 3      4      6    |
 | 58     3      568  | 58     4      7    | 9      1      2    |
 *--------------------------------------------------------------*


A & B share a group, thus we can exclude 1 from r3c4 which shares a group with both a & b, one of which must be true.

I hope this is actually logical, even though it doesn't seem to advance the puzzle much.

Tracy

edit--I'm retracting this as I now don't believe it is correct. I'm going to leave it here in case someone can prove me wrong or they can make use of it in a manner I did not.

[vidarino]

From where tarek left off:

Code: Select all

      3      1      9 |       6    578     48 |     247     25    457
      6      8      2 |     157A   357A   134-|     147     35A     9
      4      5      7 |     123B     9    123B|       8      6     13
----------------------+-----------------------+----------------------
   1278      4    138 |       9   2378      5 |       6    238    137
   2578      6    358 |    2378      1    238 |     247      9   3457
      9     27   1358 |       4   2378      6 |     127   2358   1357
----------------------+-----------------------+----------------------
     12      9      4 |     123      6    123 |       5      7      8
  12578     27    158 |    1258    258      9 |       3      4      6
     58      3      6 |      58      4      7 |       9      1      2


If I have understood things right, there is an Almost Locked Sets xz rule that can be applied here. It doesn't seem to have a big impact, though, but still:

A=R2C458, forming a quad 1357
B=R3C46, forming a triple 123
x=1, only one of the sets can contain a 1
z=3, boths sets also contain a 3
R2C6 is visible to both A and B; ergo R2C6<>3

Vidar

[tarek]

From where tarek left off:

Code: Select all

A=R2C458, forming a quad 1357
B=R3C46, forming a triple 123
x=1, only one of the sets can contain a 1
z=3, boths sets also contain a 3
R2C6 is visible to both A and B; ergo R2C6<>3

Nice

i think what you are saying is that x can be found only in box 2

Tarek

[Wolfgang]

puzzle #1 ... is currently one of only two known 2-constrained up to simple coloring sudokus
2-constrained means the backdoors / magic cells come in pairs

Does it mean that i did hit a pair of backdoors?

What is this technique called, I know that the base is a bifurcation that leads to contradiction elimination, but is there a specific name for it.....

I dont have a specific name. i would call it a forcing net that leads to a contradiction.

[gsf]

puzzle #1 ... is currently one of only two known 2-constrained up to simple coloring sudokus
2-constrained means the backdoors / magic cells come in pairs

Does it mean that i did hit a pair of backdoors?

yes
this puzzle has 201 backdoor pairs
your analysis hit [2,9]=7 [8,8]=4

[Wolfgang]

Today i tried to solve puzzle 2, but i wasnt that lucky as with the first one. I found 2 solutions, the one by showing that both r1c9 and r2c8 <> 5 (=> r1c8=5), but especially the latter needed a long chain including 2 URs and an xy-wing, the other by eliminating 1,2 and 8 from r8c4 (=> r8c4 =5), but here i also got very long chains including an x-wing.
Thats the reason why i stopped to try to solve tough puzzles for a long time. I was sitting before them for hours and ended up with either no solution or one that did not satisfy me. Nevertheless i would be interested in a more elegant solution (if one exists).

[gsf]

Nevertheless i would be interested in a more elegant solution (if one exists).

the puzzle has 6 singleton backdoors

Code: Select all

[1,5]=7[2,4]=1[2,7]=7[5,6]=2[8,4]=5[8,5]=2

so chain(s) leading to the solution of any one of them will crack the puzzle
and might have a chance at being short/elegant

in this case though the candidate count for the backdoors after easy methods
have been applied is >= 3, so the standard "make a guess" or "start analysis" with
"cell(s) with the least number of candidates first" leads the solution (and backtrackers) astray

Code: Select all

  3     1     9   |  6    578    48  | 247    25   457
  6     8     2   | 1357  357   134  | 147    35    9
  4     5     7   | 123    9    123  |  8     6     13
------------------+------------------+------------------
 1278   4    138  |  9    2378   5   |  6    238   137
 2578   6    358  | 2378   1    238  | 247    9    3457
  9     27   1358 |  4    2378   6   | 127   2358  1357
------------------+------------------+------------------
  12    9     4   | 123    6    123  |  5     7     8
12578   27   158  | 1258  258    9   |  3     4     6
  58    3     6   |  58    4     7   |  9     1     2