eleven wrote:- Code: Select all
-  *---------------------------------------------------------------*
 |  5  6     1    |  3     24      7      |  8     24     9      |
 |  2  79   #79   |  1     456     8      |  46    45     3      |
 |  8  3     4    |  59-6  2569    2569   |  16    1257   2567   |
 |----------------+-----------------------+----------------------|
 |  7 #59    2    | #569   1356-9  4      |  16-9  1358   568    |
 |  3  4    #59   |  7     8       1569   |  2     15     56     |
 |  6  1     8    |  2     359     59     |  49    3457   57     |
 |----------------+-----------------------+----------------------|
 |  4  2     6    |  8     7       3      |  5     9      1      |
 |  1  58    3    |  4     259     259    |  7     6      28     |
 |  9  578  #57   | #56    1256    1256   |  3     28     4      |
 *---------------------------------------------------------------*
 5679 in 6 cells r4c2,r259c3,r49c9, 59 must be twice => -9r4c57, -6r3c4; stte
 
Very Nice!! I really like the power of such argument!
Even knowing that the logic of this move is not based on chains,
the role of (79)r2c3 puzzled me for a while. I couldn't help but think in
a chaining way. First I tried to see it as a kraken (double) as follows
- Code: Select all
-                     (6=5)r9c4-(5)r4c4-(5=9)r4c2
 ||
 -(6)r4c4-(6=5)r9c4-r9c3=r5c3-(5=9)r4c2
 ||
 (6=5)r9c4-r9c3=r5c3-(5=9)r4c2-(9)r4c4-
Not pretty! But cell r3c2 is not used. The best I could come up with (r3c2 reappears) was
- Code: Select all
- (9)r3c4=r4c4-(9=5)r4c2-(5=97)r25c2-(7=56)r9c34-(6=59)r4c24 => -9 r4c57, -6 r3c4; ste
 '------------''-------------------------'                 '
 '--------------------------------------------'
[i.e., if 9 is not at r4c2, part of the chain shows that (6)r9c4 and, its continuation, that (59) is locked at r4c24. The chain also shows that (9)r4c2 implies (9)r3c4 (6 false there) and trivially -9 r4c57] 
Anyway, it does not come close to that very nice move!
Edit: corrected typo in the chain and added explanation.