The New Sudoku Players' Forum

[Hajime]

Hmm, yesterday's Valentine was questionable?
Here a diamant

Code: Select all

                     
  . . . . . . . 1 9   
  . . 4 1 . 6 7 . 3   
  . 3 . . 2 . . 4 .   
  7 . . . . . . . 1   
  5 . . . 6 . . . 8   
  . 6 . 3 . . . 7 .   
  . . 8 . . . 5 . .   
  . 7 . 5 . 4 . . .   
  3 . . . 8 . . . .   

.......19..41.67.3.3..2..4.7.......15...6...8.6.3...7...8...5...7.5.4...3...8....


[yzfwsf]

M2-Wing: 6r4c7 = (6-5)r4c8 = r2c8 - (5=6)r3c9 => r13c7<>6; stte

[Hajime]

Nice, one M2 wing has 5 links. =-=-=
I had 2, one turbot crane and one skyscraper, both 3 links, =-=

[P.O.]

r3c9=5 - c8n5{r2 r4} - r4n6{c8 c7} => b3 empty of 6 => r3c9 <> 5 ste.

[Cenoman]

Same logic as P.O.'s
L2-Wing: (5)r2c8 = (5-6)r4c8 = r4c7 - r13c7 = (6)r3c9 => -5 r3c9; ste

@Hajime:
In an AIC, the total number of links is always an odd number (n strong links + n-1 weak links), due to the alternance requirement. So, to give the size of a solution, just sum the strong links: any wing has length 3, turbot or SS have length 2 each. P.O.'s move is a whip(3).

[Hajime]

@Hajime:
In an AIC, the total number of links is always an odd number (n strong links + n-1 weak links), due to the alternance requirement. So, to give the size of a solution, just sum the strong links: any wing has length 3, turbot or SS have length 2 each. P.O.'s move is a whip(3).

Thanks Cenoman,
I will adapt my link-count, also in SiSeSuSo (in due time)

[yzfwsf]

Nice, one M2 wing has 5 links. =-=-=
I had 2, one turbot crane and one skyscraper, both 3 links, =-=

Are you sure you can find the turbot crane in the initial state?

[Hajime]

Are you sure you can find the turbot crane in the initial state?

After another test I cannot reproduce. So I was mistaken. Very strange...
Now I found:
UVWXYZ-Wing Type B on {235689} in c8e5789,c8e2,b3e9 (-6)c8,b3 => (-6)r7c9 (-6)r8c9 (-6)r9c9 as very first method; stte
c8e2 is the pivot, meaning column 8 element 2 = r2c8