| View previous topic :: View next topic |
| Author |
Message |
| wapati
| | Joined: 12 Jun 2007 | | Posts: 622 | | : | | Location: Canada |
Items |
|
 Posted: Thu Sep 06, 2007 6:02 pm Post subject: If a BUG is needed, does it have to be the last hard step? |
 |
|
I just posted a puzzle that ends with a BUG, and I thought "of course the BUG is the last "technique".
Now I am wondering. A BUG could set up an x-wing or an xy-wing, or some other pattern that wasn't visible before the BUG.
If a BUG is "needed" does it have to be the last "pattern"?
Anyone have a contrary puzzle? |
|
| Back to top |
|
 |
| re'born
| | Joined: 26 Oct 2007 | | Posts: 4 | | : | |
Items |
|
| Posted: Fri Oct 26, 2007 1:42 am Post subject: |
|
|
wapati,
Presuming you are talking about a BUG+1, then it will be the last non-single step. Let x be the BUG obstruction. Placing x will eliminate all of the other candidates in that cell. But each of these candidates, by definition, shows up only twice in any unit. Hence, we can immediately place those candidates conjugate to the removed candidates. This will cascade until the puzzle is solved, for if it didn't, you would necessarily stop in a BUG grid, an impossibility. |
|
| Back to top |
|
|
| wapati
| | Joined: 12 Jun 2007 | | Posts: 622 | | : | | Location: Canada |
Items |
|
| Posted: Fri Oct 26, 2007 9:04 pm Post subject: |
|
|
Thanks!
Further proof that I don't fully understand uniqueness type methods. 
It should have been obvious, I now suspect.  |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in
Log in to check your private messages
