Hi
rjamil,
Here are some BUG examples like I promised. First a two-stepper with a reduced BUG+3 and then a one-stepper using the full BUG+6 directly.
- Code: Select all
.---------------.------------------.-----------.
| 3 9 2 | 7 4 8 | 5 1 6 |
| 8 4 7 | 56 16 15 | 2 9 3 |
| 6 1 5 | 2 9 3 | 4 8 7 |
:---------------+------------------+-----------:
| 9 56 38 | 68+5 23+6 7 | 1 4 25 |
| 2 57 48+3 | 58 13 14+5 | 6 37 9 |
| 1 67+5 34 | 9 26+3 45 | 8 37 25 |
:---------------+------------------+-----------:
| 4 8 6 | 3 5 9 | 7 2 1 |
| 7 2 9 | 1 8 6 | 3 5 4 |
| 5 3 1 | 4 7 2 | 9 6 8 |
'---------------'------------------'-----------'
So, we have six guardians in six cells to begin with. Finding a common elimination for them is not trivial in this case, but fortunately we can get rid of three of the guardians with a simple loop:
- Code: Select all
.----------------.------------------.------------.
| 3 9 2 | 7 4 8 | 5 1 6 |
| 8 4 7 | 56 16 15 | 2 9 3 |
| 6 1 5 | 2 9 3 | 4 8 7 |
:----------------+------------------+------------:
| 9 56 a38 | 568 b23-6 7 | 1 4 c25 |
| 2 57 48-3 | 58 13 145 | 6 37 9 |
| 1 67-5 f34 | 9 236 e45 | 8 37 d25 |
:----------------+------------------+------------:
| 4 8 6 | 3 5 9 | 7 2 1 |
| 7 2 9 | 1 8 6 | 3 5 4 |
| 5 3 1 | 4 7 2 | 9 6 8 |
'----------------'------------------'------------'
Step 1:
(3)r4c3 = (3-2)r4c5 = r4c9 - (2=5)r6c9 - (5=4)r6c6 - (4=3)r6c3 - loop => -3 r5c3, -5 r6c2, -6 r4c5
(3)r4c3 = (32)r4c59 - (2=54'3)r6c963 - loop => -3 r5c3, -5 r6c2, -6 r4c5
- Code: Select all
3r4c3 3r4c5
2r4c5 2r4c9
2r6c9 5r6c9
5r6c6 4r6c6
3r6c3 4r6c3
-------------------------------
-3r5c4 -6r4c5 -5r6c2
5x5 (Rank 0): {23R4 6N369 \ 45r6 2c9 3c3 4n5} => -3 r5c3, -5 r6c2, -6 r4c5
What's left is a much simpler BUG+3:
- Code: Select all
.------------.------------------------.-----------.
| 3 9 2 | 7 4 8 | 5 1 6 |
| 8 4 7 | 56 16 15 | 2 9 3 |
| 6 1 5 | 2 9 3 | 4 8 7 |
:------------+------------------------+-----------:
| 9 56 38 | a68[+5] 23 7 | 1 4 25 |
| 2 57 48 | 8-5 13 b14[+5] | 6 37 9 |
| 1 67 c34 | 9 c26+3 c4(5) | 8 37 25 |
:------------+------------------------+-----------:
| 4 8 6 | 3 5 9 | 7 2 1 |
| 7 2 9 | 1 8 6 | 3 5 4 |
| 5 3 1 | 4 7 2 | 9 6 8 |
'------------'------------------------'-----------'
Step 2:
- Code: Select all
a: (5)r4c4
||
b: (5)r5c6
||
c: (3)r6c5 - (3=4)r6c3 - (4=5)r6c6
=> -5 r5c4; stte
In other words, 2/3 guardians (5r4c4 and 5r5c6) see the elimination directly. The third 3r6c5 needs a short chain (which is also a locked set, actually) to imply 5r6c6 which also sees the elimination. Thus all 3/3 have proved that they could eliminate it, so it can be safely eliminated.
As an AIC:
(5)b5p16 =BUG= (3)r6c5 - (3=4)r6c3 - (4=5)r6c6 => -5 r5c4; stte
(5)b5p16 == (345)r6c536 => -5 r5c4; stte
Of course we can also do it in one step with the BUG+6 directly:
- Code: Select all
.-------------------.--------------------------.-------------.
| 3 9 2 | 7 4 8 | 5 1 6 |
| 8 4 7 | 56 16 15 | 2 9 3 |
| 6 1 5 | 2 9 3 | 4 8 7 |
:-------------------+--------------------------+-------------:
| 9 56 38 | a68(+5) cd23+6 7 | 1 4 25 |
| 2 cd(5)7 b48+3 | 8-5 cd13 a14(+5) | 6 cd37 9 |
| 1 c67+5 b34 | 9 bcd26+3 b4(5) | 8 37 25 |
:-------------------+--------------------------+-------------:
| 4 8 6 | 3 5 9 | 7 2 1 |
| 7 2 9 | 1 8 6 | 3 5 4 |
| 5 3 1 | 4 7 2 | 9 6 8 |
'-------------------'--------------------------'-------------'
- Code: Select all
a: (5)b5p16
||
b: (3)r5c3,r6c5 - (3=4)r6c3 - (4=5)r6c6
||
c: (5-6)r6c2 = (62-3)r64c5 = r5c5 - (3=7)r5c8 - (7=5)r5c2
||
d: (62-3)r46c5 = r5c5 - (3=7)r5c8 - (7=5)r5c2
=> -5 r5c4; stte
Again, the same two guardians (5r4c4 and 5r5c6) see the elimination directly, but now they're only 2/6, so we need chains for the other four.
As a (compressed) AIC:
(5)b5p16 == [(54,3)r6c63,r5c3,r6c5 == (5-6)r6c2 = (62,3)r645c5 - (3=75)r5c82] => -5 r5c4; stte
- Code: Select all
5r5c2 7r5c2
7r5c8 3r5c8
3r5c5 3r4c5 3r6c5
2r4c5 2r6c5
5r6c6 4r6c6
4r6c3 3r6c3
4r6c3 3r6c3
6r6c5 6r6c2
5b5p16 6r4c5 3r5c3 3r6c5 5r6c2
-------------------------------------------------------
-5r5c4