The New Sudoku Players' Forum

[tlanglet]

Here's a nice one, using externals:

AUR(56)r12c29[6r1c1=6r7c2] => -6r78c1,r1c2; stte

Blue,

The possible external inferences for this AUR are:
Digit 5:
5r2c68 which is a row perspective,
5r9c9 which is a column perspective, or
r23c8 which is a block perspective

Digit 6:
6r1c1 which is a row and block perspective, or
6r7c2 which is a column perspective

The initial strong inference is formed by selecting any one external inference for each digit. Thus possibilities include 5r9c9=6r1c1, 5r9c9=6r7c2, 5r23c8=6r7c2, etc.

Your set of external inferences is not valid; they are redundant and do not consider digit 5.

Ted

[blue]

I'm on my way out the door.
Just a couple of quick comments ...

Code: Select all

 6r1c1 = { HP(56-8)r1c29 + HP(56-9)r12c2 } ; DP  =>  =6 r1c1

That doesn't appear to work for the r12c2 HP (with 6r7c2 present). Am I missing something ?

Your set of external inferences is not valid; they are redundant and do not consider digit 5.

Did you review this part ?

If 6r1c1 is false, there's a <56> locked pair in r1c29, that eliminates 8r1c9.
If 6r7c2 is false, there's a <56> locked pair in r12c2, that eliminates 9r2c2.
If both are false, the combination produces the bare UR pattern.

[tlanglet]

Your set of external inferences is not valid; they are redundant and do not consider digit 5.

Did you review this part ?

If 6r1c1 is false, there's a <56> locked pair in r1c29, that eliminates 8r1c9.
If 6r7c2 is false, there's a <56> locked pair in r12c2, that eliminates 9r2c2.
If both are false, the combination produces the bare UR pattern.

Blue,

The basic logic of this puzzle dictates that 6r1c1=6r7c2; both are true or both are false. You correctly observed that if they are false, then a deadly pattern will result, thus r1c1, r7c2 must be false.

Your process did not use an AUR() as a starting point to determine if conclusions could be derived. More specifically, your notation, AUR(56)r12c29[6r1c1=6r7c2], is not based on the fundamental requirements for using external inferences derived from an AUR().

Your observation and conclusion is correct; the notation is not.

Ted

[blue]

The basic logic of this puzzle dictates that 6r1c1=6r7c2; both are true or both are false.

I'll be darn, your right !
The '=' was meant to indicate a strong link relationship, though -- at least one must be true.

You correctly observed that if they are false, then a deadly pattern will result, thus r1c1, r7c2 must be false.

Both true though, not both false, or else the deadly pattern would result.
[ At least one true, to avoid the deadly pattern. Then as you say, one true implies both true. ]

Your process did not use an AUR() as a starting point to determine if conclusions could be derived.

The bit above, should address that concern.

[daj95376]

I'm on my way out the door.
Just a couple of quick comments ...

Code: Select all

 6r1c1 = { HP(56-8)r1c29 + HP(56-9)r12c2 } ; DP  =>  =6 r1c1

That doesn't appear to work for the r12c2 HP (with 6r7c2 present). Am I missing something ?

You missed the part where I said, "Essentially, 6r1c1 works as an external in [b1] and [r1] for the DP."

[blue]

You missed the part where I said, "Essentially, 6r1c1 works as an external in [b1] and [r1] for the DP."

Right ... hidden pair in the box when it's false.
Very nice

Added: Ted, I should apologize for not seeing this bit, a few posts back:

Digit 6:
6r1c1 which is a row and block perspective, or

Until my "aha moment" here, I had never considered a "block perspective" -- still learning.
I was confused about your "redundant" comment too, which seems now, to have more to do with the fact
that r1c1<>6 also leaves the hidden box pair, than that 6r1c1 and 6r7c2 would be true or false together.

[daj95376]

[Withdrawn: All I did was muddy up the discussion with a non-related example.]

[David P Bird]

Sorry guys, I wasn't following this thread.

The weak inference I referred before to is between two opposite sides holding the same HP.

In this case you can select the sides to be in (r1 & r2) or (c2 and c9) but you can't use one from each.
(Some of the posted chains have assumed (5) is locked in the 4 cells but it isn't.)

Using abbreviated notation it's easy to get into a tangle with AURs, so for checking purposes I find it best to expand it into its fundamental components.

eg
(8)r1c9 = (56)r1c29 -[UR]- (56)r2c29 = (9)r2c2 - (9=6)r8c2 - (6)r2c2 = (6)r2c9 => r1c9 <> 6

Note how the cells on either side of the -[UR]- add up to the 4 required and also how easy it is to check that every link is sound.
This doesn't kill the puzzle though.

This does:

Code: Select all

 *--------------------*--------------------*--------------------*
 | 67    56f   4      | <2>   <1>   <9>    | 78    <3>   568e   |
 | <3>   569fg 17     | <4>   <8>   57     | <2>   1579  56e    |
 | 179   <8>   2      | 57    <6>   <3>    | 79a   1579  4      |
 *--------------------*--------------------*--------------------*
 | <2>   7     <5>    | 9     4     6      | 1     <8>   <3>    |
 | <8>   <1>   <6>    | 57    <3>   57     | <4>   <2>   <9>    |
 | <4>   <3>   9      | 1     2     8      | <5>   6     <7>    |
 *--------------------*--------------------*--------------------*
 | 679   69    37     | <8>   <5>   1      | 3679b <4>   2      |
 | 569   2     <8>    | 3     <7>   <4>    | 69b   59c   <1>    |
 | 157   <4>   137    | <6>   <9>   <2>    | 378   57    58d    |
 *--------------------*--------------------*--------------------*

(9)r3c7 = (9)r78c7 - (9=5)r8c8 - (5)r9c9 = (56)r12c9 -[UR]- (56)r12c2 = (9)r2c2 => r2c1 <> 9 ste

[tlanglet]

The initial strong inference is formed by selecting any one external inference for each digit. Thus possibilities include 5r9c9=6r1c1, 5r9c9=6r7c2, 5r23c8=6r7c2, etc.

Your set of external inferences is not valid; they are redundant and do not consider digit 5.

I understand that you want the strong inference to be between candidates for both DP values. But, that can't always be the case. Consider this UR Type 4 that I "borrowed" from an old collection posted by Mike Barker. There is an X-Wing on <9> and the only external candidates are: 7r5c7, 7r6c1, and 7r6c2.

Danny.

Note that I said all AUR digits need to be considered. In your example with the x-wing overlay, we find that no external inferences exist for digit 9 because of the x-wing overlay so your external inferences on digit 7 are proper. In the puzzle of this thread, both digits have external inferences but were not included.

Ted

[eleven]

My 2 cents:
DP 56r12c29 => r1c2<>6, r2c9<>5, stte

[David P Bird]

My 2 cents:
DP 56r12c29 => r1c2<>6, r2c9<>5, stte

Eleven, are you Uruguayan by any chance?

(For non-football fans, they don't feel bound by any rules but will call foul if others don't comply.)

[daj95376]

_

General Constraint:

If the absence of one or more external candidates reduces the UR to a DP, then at least one of these candidates must be true.

Strong-Link Constraint:

In order to form a strong link between external candidates, no subset of these candidates can reduce the UR to a DP.



blue's solution met the first constraint, but it doesn't pass the second constraint. However, my scenario can now be expressed as:

Code: Select all

 UR(56)r12c29[6r1c1=6r1c1]


_

[Sudtyro2]

Hey folks, slow down just a little...I'm learning entirely too fast!

[Luke]

Code: Select all

*-----------------------------------------------------------*
 | 67   *56    4     | 2     1     9     | 78    3    *56+8  |
 | 3    *56#9  17    | 4     8     57    | 2    -9157 *56    |
 | 179   8     2     | 57    6     3     | 79    1579  4     |
 |-------------------+-------------------+-------------------|
 | 2     7     5     | 9     4     6     | 1     8     3     |
 | 8     1     6     | 57    3     57    | 4     2     9     |
 | 4     3     9     | 1     2     8     | 5     6     7     |
 |-------------------+-------------------+-------------------|
 | 679   69    37    | 8     5     1     | 3679  4     2     |
 | 569   2     8     | 3     7     4     | 69   #59    1     |
 | 157   4     137   | 6     9     2     | 378   57   #58    |
 *-----------------------------------------------------------*

What, nobody likes the w-wing?

(9)r2c2=(5)r9c9-(5=9)r8c8 ==>r2c8<>9

[blue]

(withdrawn)