This fairly easy puzzle requires only one step to solve.
..3.7.9............6.521.8...........42.6.35.39..5..14.2..1..3.9..8.5..6..4...8..
puzzle: Show
Good solving.
Robert
+--------------------------+---------------------------+-------------------------+
| f12458 158 3 |ga46 7 ga468 | 9 ga246 125 |
| f124578 1578 1578 | 3469 3489 34689 | 124567 2467 1257 |
| e47 6 9 | 5 2 1 | d47 8 3 |
+--------------------------+---------------------------+-------------------------+
| 15678 1578 15678 | 123479 3489 234789 | d267 2679 2789 |
| 178 4 2 | 179 6 789 | 3 5 789 |
| 3 9 678 | 27 5 278 | d267 1 4 |
+--------------------------+---------------------------+-------------------------+
| 5678 2 5678 | 4679 1 4679 | 457 3 579 |
| 9 137 17 | 8 34 5 | c1247 b247 6 |
| 1567 1357 4 | 23679 39 23679 | 8 79 1579 |
+--------------------------+---------------------------+-------------------------+
Cenoman wrote:(8=642)r1c468 - r8c8 = r8c7 - (2=674)r346c7 - (4)r3c1 = (24)r12c1 - (246=8)r1c468 => +8 r1c6; lclste
eleven wrote:Cenoman wrote:(8=642)r1c468 - r8c8 = r8c7 - (2=674)r346c7 - (4)r3c1 = (24)r12c1 - (246=8)r1c468 => +8 r1c6; lclste
Very nice !
I saw, that a missing 8r1c6 would lead to a 18 pair in r1c12 (r1, c9, b9) and a contradicting 1 in r2c9 then, but could not find this chain with the 24 pair.
Mauriès Robert wrote:Hi Cenoman,
I prefer your resolution with Kraken to the one with AIC because it is like crossing two conjugated tracks P(4r12c1) and P(4r3c1) like this :
P(4r12c1) : 4r12c1->246r1c148->8r1c6
P(4r3c1) : 4r3c1->7r3c7->26r46c7->2r8c8->46r1c48->8r1c6
=> r1c6=8, stte
Which can be spelled, I believe, in the Kraken way:
42r12c1-(246=8)r1c468
4r3c1-(4=7)r3c7-(7=62)r46c7-2r8c7=r8c8-(2=468)r1c468
=> r1c8=8, stte
Robert
(8=642)r1c468 - r8c8 = r8c7 - (2=674)r346c7 - (4)r3c1 = (24)r12c1 - (246=8)r1c468 => +8 r1c6