The New Sudoku Players' Forum

[Leren]

Code: Select all

*-----------*
|6.8|...|...|
|9.4|.3.|8..|
|.3.|8..|..5|
|---+---+---|
|...|7..|..9|
|...|.26|57.|
|...|958|24.|
|---+---+---|
|1.7|...|...|
|3.5|.6.|9..|
|.2.|3..|..6|
*-----------*
6.8......9.4.3.8...3.8....5...7....9....2657....95824.1.7......3.5.6.9...2.3....6

[SpAce]

Code: Select all

.-------------.------------------------.----------------------.
| 6    57   8 | b125#4  1479    124579 | c1347  1239  c1247   |
| 9    57   4 |  6      3       1257   |  8     12     127    |
| 2    3    1 |  8      479     479    | d467   69     5      |
:-------------+------------------------+----------------------:
| 5    48   2 |  7      14      3      |  16    168    9      |
| 48*  9    3 |  14*    2       6      |  5     7      18     |
| 7    1    6 |  9      5       8      |  2     4      3      |
:-------------+------------------------+----------------------:
| 1    6    7 | b25#4   489     2459   | e4-3   2358   248    |
| 3    48*  5 |  124*   6     a#4-127  |  9     128   b1278#4 |
| 48*  2    9 |  3      1478    1457   |  147   158    6      |
'-------------'------------------------'----------------------'

Oddagon (4)r58,c14,b7 with four #guardians:

Code: Select all

(4)r1c4 - r1c79 = r3c7 - r7c7 =singles= !
||                     /
(4)r7c4 --------------
||                   /
(4)r8c9 ------------
||
(4)r8c6

=> +4 r7c7,r8c6; stte

[SpAce]

Interesting. The easiest "solution" is to remove the clue 9r6c4 and then try again. While at it, 8r1c3 and 3r8c1 can be removed too, though they make little difference.

Code: Select all

6.8......9.4.3.8...3.8....5...7....9....2657....95824.1.7......3.5.6.9...2.3....6
6........9.4.3.8...3.8....5...7....9....2657.....5824.1.7........5.6.9...2.3....6

toomuch.jpg (18.38 KiB) Viewed 629 times

[eleven]

The easiest "solution" is to remove the clue 9r6c4 and then try again.

Which is guessing like trying r4c5=1. You can't know in forward, if it (the UR or the number) would lead to a solution or if it is wrong.

[SpAce]

The easiest "solution" is to remove the clue 9r6c4 and then try again.

Which is guessing like trying r4c5=1. You can't know in forward, if it (the UR or the number) would lead to a solution or if it is wrong.

Of course; hence the quotation marks. There's no way for a manual solver to know which clues are redundant (if any), so none can be removed (or ignored) to reveal any URs. If that information were somehow available, then the UR would be fair game. Otherwise not.

This is a perfect example of a redundant clue actually destroying information and making the puzzle much harder. It's not imaginary or just making the UR harder to see. It's real. The UR inferences are truly gone when there's a given in one of its corners, or when a given of those UR digits sees any (two) of its corners.

A solved cell wouldn't destroy the UR, though, as long as there's a possibility that the UR could have existed at some point without any givens interfering with it. In this case, if we had the same pencil marks as in my earlier solution but 9r6c4 were somehow a solved digit (or a known redundant given), we'd have a perfectly valid AR(19)r56c24 => +4r5c4; stte.