The New Sudoku Players' Forum

[Myth Jellies]

An interesting uniqueness deduction

I managed to get the current extreme 111 to the following

Code: Select all

 *-----------*
 |...|.96|...|
 |..8|.5.|..3|
 |5.1|...|9.6|
 |---+---+---|
 |..3|2.5|..8|
 |...|...|...|
 |1..|3.9|7..|
 |---+---+---|
 |2.6|...|1.9|
 |8..|.6.|3..|
 |...|12.|...|
 *-----------*

 *-----------------------------------------------------------*
 | 3     2     47    | 478   9     6     | 458   1     57    |
 | 69    69    8     | 47    5     1     | 24    247   3     |
 | 5     47    1     | 478   3     2     | 9     478   6     |
 |-------------------+-------------------+-------------------|
 | 467   467   3     | 2     1     5     | 46    9     8     |
 | 49    489   25    | 6     78+4  78+4  | 25    3     1     |
 | 1     4568  245   | 3     48    9     | 7     2456  245   |
 |-------------------+-------------------+-------------------|
 | 2     347   6     | 5     78+4  78+34 | 1     478   9     |
 | 8     1     457   | 9     6     47    | 3     2457  2457  |
 | 47    35    9     | 1     2     38    | 68    568   47    |
 *-----------------------------------------------------------*

From here one can color the sevens and pretty much solve the puzzle. There is also a uniqueness deduction which didn't go anywhere, but I found interesting because I had never made one much like it before. The AIC looks something like...

(34#1 = 78#1)r7c6 -UR- (78#3 = 4#1)r57c5|r5c6 - (4=8)r6c5 - (8)r7c5 = (8&3)r79c6 => r7c6 <> 7

This UR deduction is pretty well buried, and I was wondering if anyone else noticed a simpler UR deduction that accomplished the same thing.

[ttt]

Hi Myth,
Nice find and interesting! But as AIC, why don’t present above like this - Either (34)r7c6 or (4)r57c5/r5c6:

Code: Select all

(34)r7c6=AUR(78)r57c56=(4)r57c5/r5c6-(4=8)r6c5-(8)r7c5=(3&8)r79c6 => r7c6<>7

BTW, It seems your path to solve this one difference with mine – I start this puzzle from triple 478 at col.8

ttt

[eleven]

With a net it looks simpler:

r7c6=7 -> r8c6=4 -> r5c6=8 -> r6c5=4 -> r5c5=7 -> r7c5=8
Deadly pattern.

[ronk]

With a net it looks simpler:

r7c6=7 -> r8c6=4 -> r5c6=8 -> r6c5=4 -> r5c5=7 -> r7c5=8
Deadly pattern.

OK, asserting r7c6=7 crashes the puzzle by creating a deadly pattern. By what logic do you even try that asssertion

[eleven]

OK, asserting r7c6=7 crashes the puzzle by creating a deadly pattern. By what logic do you even try that asssertion

Very probably i would have missed that. And for sure i would have missed this complicated AIC-logic with 5 extra candidates.

But in potential UR's i use to look for strong links. So r7c6=7 -> r5c5=7 is obvious. From here its not so far to see, that also the 8's are forced in the other corners.

[Glyn]

Here's another arrangement leading to r7c6<>7

Code: Select all

(78)URr57c56
||
(7)r28c6
||
(8)r6c5-(8=47)r7c5|r8c6
||
(8-3)r9c6=(3)r7c6

[Mage]

...
The AIC looks something like...

(34#1 = 78#1)r7c6 -UR- (78#3 = 4#1)r57c5|r5c6 - (4=8)r6c5 - (8)r7c5 = (8&3)r79c6 => r7c6 <> 7

This UR deduction is pretty well buried, and I was wondering if anyone else noticed a simpler UR deduction that accomplished the same thing.

Very probably i would have missed that. And for sure i would have missed this complicated AIC-logic with 5 extra candidates.

Here's another arrangement leading to r7c6<>7

Code: Select all

(78)URr57c56
||
(7)r28c6
||
(8)r6c5-(8=47)r7c5|r8c6
||
(8-3)r9c6=(3)r7c6

Besides the UR, this simple AIC seems also easy to find :

(7)r5c6=(7)r5c5 - (4)r5c5=(4)r5c7 - (4=7)r8c7 => r7c6<>7

[eleven]

Here is a similar example.

Puzzle from Patterns Game by JPF.

Code: Select all

 +-------+-------+-------+
 | . . 1 | 2 . . | . . . |
 | . 3 . | . . 4 | 5 . . |
 | 6 . . | . . . | . 2 . |
 +-------+-------+-------+
 | 2 . . | 1 . . | . 6 . |
 | . . . | . . . | . . . |
 | . 5 . | . . 7 | . . 8 |
 +-------+-------+-------+
 | . 4 . | . . . | . . 5 |
 | . . 6 | 9 . . | . 1 . |
 | . . . | . . 3 | 7 . . |
 +-------+-------+-------+

(yes i know it can be solved with wings)
There are several potential UR's, 58 (r8c5<>8), 37 (r5c13<>3, r7c3<>3) and this one:

Code: Select all

+-------------------+-------------------+-------------------+
| 45    78    1     | 2     3578  6     | 3489  37    349   |
| 9     3     2     | 78    1     4     | 5     78    6     |
| 6     78    45    | 35    378   9     | 1348  2     134   |
+-------------------+-------------------+-------------------+
| 2     9    #348   | 1    #3458  58    | 34    6     7     |
| 347   6    #3478  | 34   #348   2     | 19    5     19    |
| 1     5     34    | 6     9     7     | 2     34    8     |
+-------------------+-------------------+-------------------+
| 37    4     379   | 78    2     1     | 6     389   5     |
| 357   2     6     | 9     4578  58    | 348   1     34    |
| 8     1     59    | 45    6     3     | 7     49    2     |
+-------------------+-------------------+-------------------+

Potential UR 38, strong links for 8 in c3, r5, i.e. r5c3=3 -> (r4c3=8 & r5c5=8).
Additionally we have r5c3=3 -> r5c4=4 and r5c5=8 -> r4c6=5, therefore r3c5=3 - deadly pattern.

[DonM]

Since this is this week's Extreme, I can't check this out until Sunday, but it's nice to see you back Myth- thought you'd gone mything or something.

[daj95376]

Code: Select all

*-----------------------------------------------------------*
 | 3     2     47    | 478   9     6     | 458   1     57    |
 | 69    69    8     | 47    5     1     | 24    247   3     |
 | 5     47    1     | 478   3     2     | 9     478   6     |
 |-------------------+-------------------+-------------------|
 | 467   467   3     | 2     1     5     | 46    9     8     |
 | 49    489   25    | 6     78+4  78+4  | 25    3     1     |
 | 1     4568  245   | 3     48    9     | 7     2456  245   |
 |-------------------+-------------------+-------------------|
 | 2     347   6     | 5     78+4  78+34 | 1     478   9     |
 | 8     1     457   | 9     6     47    | 3     2457  2457  |
 | 47    35    9     | 1     2     38    | 68    568   47    |
 *-----------------------------------------------------------*

This UR deduction is pretty well buried, and I was wondering if anyone else noticed a simpler UR deduction that accomplished the same thing.

Code: Select all

[r6c5]=4 => (78) UR Type 1 => [r7c6]<>78
[r6c5]=8 => (47) UR Type 1 => [r7c6]<>47

[r7c6]<>7


[Luke]

With a net it looks simpler:

r7c6=7 -> r8c6=4 -> r5c6=8 -> r6c5=4 -> r5c5=7 -> r7c5=8
Deadly pattern.

OK, asserting r7c6=7 crashes the puzzle by creating a deadly pattern. By what logic do you even try that assertion

This is one of those fine lines separating logic from guessing, right?

Still, with this pattern you have two strong links meeting in a "elbow." It's often a good idea to plug in the other UR candidate (the one not involved with the strong links) into the elbow cell or its diagonal. When there's a contradiction, it's likely to leap out at you as it does here.

eleven's observation seems to be born from experience. It doesn't seem like heresy to me, because one could always go ahead and construct an elaborate mathematical chain to prove it.

[eleven]

Nice catch, daj95376.

Seen this way its an easy to find pattern (though very rare, i suppose).

Code: Select all

 123 123   .
  .   .    .
 12   .    .
-------------
 123 12-3X .
  .   .    .
  .   .    .

This is one of those fine lines separating logic from guessing, right?
...
eleven's observation seems to be born from experience.

For me this is not a question of logic or guessing, but tactics.
What to do, if no obvious patterns can be found anymore ? My tactic is to check the grid for patterns and chains with special properties, which can be verified in short time, if they are there or not. Maybe this is not more effective than e.g. try to find a forcing chain or AIC from each bivalue cell or bilocation pair, but this (programs) tactic is just boring for me.

[ronk]

With a net it looks simpler:

r7c6=7 -> r8c6=4 -> r5c6=8 -> r6c5=4 -> r5c5=7 -> r7c5=8
Deadly pattern.

OK, asserting r7c6=7 crashes the puzzle by creating a deadly pattern. By what logic do you even try that assertion

This is one of those fine lines separating logic from guessing, right?

I was hinting that using the contrapositive was more direct. With a BUG-Lite+4 (aka UR+4) pattern, we know at least one of the non-UR candidates in the UR cells must be true. Rather than "crash the puzzle", why not find a common outcome of individual assertion of the non-UR candidates

[Myth Jellies]

Answer a few question brought up...

...But as AIC, why don’t present above like this - Either (34)r7c6 or (4)r57c5/r5c6:

Code: Select all

(34)r7c6=AUR(78)r57c56=(4)r57c5/r5c6-(4=8)r6c5-(8)r7c5=(3&8)r79c6 => r7c6<>7

I use that style when I show the strong links vertically like a Kraken net

(78#4)r57c56:UR
||
(4)r57c5|r5c6 - (4=8)r6c5 - (8)r7c5 = (8&3)r79c6
||
(34)r7c6

But when I present it as a linear string, I think the weak UR link more precisely shows what is happening....

(34#1 = 78#1)r7c6 -UR- (78#3 = 4#1)r57c5|r5c6 - (4=8)r6c5 - (8)r7c5 = (8&3)r79c6 => r7c6 <> 7

The avoidance of a rectangle of cells containing 78 is more accurately represented as weak link because it really is just like avoiding having two 7s in the same house. So I think the weak representation gets deeper into the guts of what is going on.

Glyn's net representation for possible escape vectors for the 7's and 8's is an equivalent alternative to the net rep for alternate digits within the deadly pattern cells

Following daj's idea is interesting. Essentially something like...

(34)r7c6 = (78)r7c6 -UR- (78#3)r57c5|r5c6 = (8-4)r6c5 = (47#3)r57c5|r5c6 -UR- (47)r7c6 = (38)r7c6

The helper cell in its dual locked triplet position is the key. It forces a strong link between 3/4ths of the two AURs. That was the simplification I was looking for. Good catch daj.

Another way of looking at it is to consider the following pattern

Code: Select all

123   123
---------
123   123

In order for the above pattern to not degenerate into a UR, all digits must be represented at least once, and one digit must be represented twice, so it occupies one corner pair and the other two occupy the other corner pair. Now if you add a helper cell that forms locked triples you get

Code: Select all

 123  *123
#123
 ---------
*123  #123

because of the locked triple the starred cells must be equal to each other and hence the hashed cells must also be equal to each other. Thus if you remove a digit from the hashed helper cell you have to remove it from the other hashed cell. This remains true even if you add other options in the hashed corner cell.

Code: Select all

 123  *123
#123
 ---------
*123  #123X

[Myth Jellies]

A similar deduction slightly extended

Code: Select all

#234
 123  *123
#234
 ---------
*123  #123X

Can remove 1 from the hashed corner cell