shye wrote:i'll share mine then!
Interesting, my solution path was almost identical before I started looking for a way to skip step two.
totuan wrote:To reduce complexity of the AUR and the number of steps, I try to eliminate 1r4c6 first
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I would say it's easier to use the UR directly to reduce the complexity of the net.
It's even simpler with externals – you need 1 or 5 in r6c38, resulting in a triple with c46 => 9r6c5, 2r6c46.
The 7r5c7 – r7c5 deduction is really cool, there is also an oddagon (Broken Wing) for it:
- Code: Select all
+---------+---------+---------+
| . . . | . . . | . . . |
| . . # | . G . | . # . |
| . # . | . . . | . . . |
+---------+---------+---------+
| . . . | . . . | . . . |
| . G . | . . . | a G . |
| . . . | . . . | . G . |
+---------+---------+---------+
| . . . | . b . | . . . |
| . # . | . G G | . # . |
| . . . | . . . | . . . |
+---------+---------+---------+
In r28c28b1 7 has to appear outside #, but ab see all the guardians => a – b
BTW with the trivalue oddagon, there are usually eliminations in the rest of the grid as well.
I was curious about this one – and it turns out that 6s only have one pattern which doesn't break the 367 template.
This is enough to get btte, the effective elimination is 6r2c4.
Maybe someone can use this to get a one-stepper.
Marek