The New Sudoku Players' Forum

[daj95376]

I'm slightly confused by one of the UR's which appears not to be there.

First, I suspect that eleven evaluated his first three URs before performing the eliminations. This is perfectly acceptable.

Second, a candidate can be missing from a PM and still be used by a UR provided it was previously present and no value has been assigned to the cell. This occurs in eleven's fourth UR.

Now, don't take what I say as being correct until ronk, Pat, udosuk, and a few others have time to review it.

As Draco would say, ... Cheers

[udosuk]

Now, don't take what I say as being correct until ronk, Pat, udosuk, and a few others have time to review it.

Okay I'll do it.

The UR 37 in r26c89 can't happen as a 1 is mandatory in r2c89. So no eliminations there.

Personally I think that move is a valid UR:

r6c9=3 => r2c9=7 (c9) & r6c8=7 (r6) & r2c8=3 (c8) => r26c89={37} => deadly pattern
Therefore r6c9<>3.

The fact that 1 @ r2 is locked @ r2c89 isn't relevant at all, as the whole point of the move is a placing of 3 @ r6c9 leads to a deadly pattern (which is as good as a contradiction when we're assured of uniqueness). For a puzzle with a unique solution whenever we see a deadly pattern we know something must be contradictory in that state. Just that the contradictory point (no 1 @ r2) is more obvious than what the other URs lead to.

Alternatively, you can achieve the elimination with a contradiction chain:

r6c9=3 => r2c9=7 (c9) & r2c8=3 (c8) => no 1 @ r2 => contradiction
Therefore r6c9<>3.

Or more elegantly:

1 @ r2 locked @ r2c89.
If r2c8=1 => 3 @ c8 locked @ r46c8 => r6c9 <>3
If r2c9=1 => r6c9=7 (c9) => r6c9<>3
Therefore r6c9<>3.

The existence of these deductions doesn't invalidate the UR move above. They just make it look clumsy and redundant.

[Glyn]

Thanks daj95376 and udosuk.
I assumed that if r6c9=3 then it would define the status of the 3's and 7's in the deadly pattern thus stopping it being deadly.
What you're saying Matt is that any scenario which would leave just the {37} combo in these cells, regardless of whether their orientation is defined, means that the pattern applies. (Given the setter's assurance of uniqueness).
With that in mind eleven's 4th UR is ok too.
I will have to 'mug' up on uniqueness I am obviously missing some of it's potential.

[Luke]

Thanks daj95376 and udosuk.
I assumed that if r6c9=3 then it would define the status of the 3's and 7's in the deadly pattern thus stopping it being deadly.
What you're saying Matt is that any scenario which would leave just the {37} combo in these cells, regardless of whether their orientation is defined, means that the pattern applies. (Given the setter's assurance of uniqueness).
With that in mind eleven's 4th UR is ok too.

Another way to look at it would be to say if r6c9=3, then both 1 and 3 would have to be in r2c8. That seems to take uniqueness out of consideration.

[ronk]

In order to prevent the (37)r26c89 deadly pattern, at least one of r4c8=3, r9c9=3 and r1c8=7 must be true.

r4c8=3 -> r6c9<>3
r9c9=3 -> r6c9<>3
r1c8=7 -> r6c8<>7 -> r6c9=7 -> r6c9<>3

[daj95376]

I see how ronk's three assignments would "prevent the (37)r26c89 deadly pattern". Very nice!

What I don't see is "at least one of ... must be true". That's okay, though.

Going back to how I best understand URs: one of [r26c89]=1 must be true to prevent (37)UR.

Code: Select all

 +-----------------------------------------------------------------------+
 |  15     12     125    |  678    78     3      |  9      67     4      |
 |  349    6      349    |  457    2      45     |  8      137    137    |
 |  8      7      34     |  46     9      1      |  236    5      26     |
 |-----------------------+-----------------------+-----------------------|
 |  134    1348   7      |  48     5      2      |  16     136    9      |
 |  59     59     6      |  1      3      7      |  4      2      8      |
 |  2      48     13     |  9      6      48     |  5      137    137    |
 |-----------------------+-----------------------+-----------------------|
 |  147    124    124    |  3      178    68     |  1267   9      5      |
 |  13579  123    12359  |  57     4      569    |  1267   8      26     |
 |  6      59     8      |  2      17     59     |  137    4      13     |
 +-----------------------------------------------------------------------+

1=[r2c8]=3=[r46c8]-    3- [r6c9]
  [r2c9]=1 => [r9c9]=3 => [r6c9]<>3
  [r6c8]=1 => [r6c3]=3 => [r6c9]<>3
  [r6c9]=1             => [r6c9]<>3


also: one of [r26c89]=3 must be true to prevent (17)UR.
also: one of [r26c89]=7 must be true to prevent (13)UR.

So, the question is: Which of the three values is duplicated?

[ronk]

I see how ronk's three assignments would "prevent the (37)r26c89 deadly pattern". Very nice!

What I don't see is "at least one of ... must be true".

If one of the UR cells must contain a non-UR digit, then a non-UR cell must contain a UR digit.

[daj95376]

I see how ronk's three assignments would "prevent the (37)r26c89 deadly pattern". Very nice!

What I don't see is "at least one of ... must be true".

If one of the UR cells must contain a non-UR digit, then a non-UR cell must contain a UR digit.

... and all of the non-UR cell possibilities reduce to the three entries you listed. Okay, and thanks for the information!