The New Sudoku Players' Forum

[RobO]

I had a pattern similar to the one below in a puzzle I was doing over the weekend. Where does it fit in to existing solving methods?

(Again, similar to my other post, I'd be surprised if it hadn't been spotted before.)

Just rows 1 to 6. Logic is that due to the unique loop on the 12s in rows 4-6, either r4c1 must be a 5, or r6c9 must be a 4 (or both). So one or both of r2c1 and r1c9 must be a 9. Hence the 9 in r1c2 can be eliminated.

Code: Select all

 .   69 . | .  . . | . . 49         
 59  .  . | .  . . | . . .         
 .   .  . | .  . . | . . .         
---------------------------
 125 . .  | .  . . | . . 12             
 12  . .  | 12 . . | . . .         
 .   . .  | 12 . . | . . 124       



[StrmCkr]

anyway to show the entire puzzle?
that would help alot more..

[RobO]

Can do though it'll be a couple of days before I can access the information. For info, it wasn't on a particularly challenging puzzle - just a 'super hard' rated one from www.brainbashers.com - so I'm sure that there will other easier techniques that I could have used at that point.

[StrmCkr]

without the whole puzzle im not seeing what you are doing.

a guess would be.
A: is valid.

Code: Select all

 .   6 . | .  . . | . . 9         
 9  .  . | .  . . | . . .         
 .  .  . | .  . . | . . .         
---------------------------
 5  . .  | .  . . | . . 12             
 12 . .  | 12 . . | . . .         
 .  . .  | 12 . . | . . 4     

or
B: bug lite (.. i belive...but could be wrong..)

Code: Select all

 .   9 . | .  . . | . . 4         
 5  .  . | .  . . | . . .         
 .  .  . | .  . . | . . .         
---------------------------
 12 . .  | .  . . | . . 12             
 12 . .  | 12 . . | . . .         
 .  . .  | 12 . . | . . 12   

if its a bug
then A must be true...

[RobO]

Not quite what I meant.

r4c1 and r6c9 can't both be 1/2 as that would leave an invalid puzzle (as in your example B). So either r4c1 is a 5, or r6c9 is a 4, or both. Hence one, other or both of r2c1 and r1c9 must be a 9.

[StrmCkr]

if its a bug then you can elliminate all cells from the chain that would
trigger the bug. (ie. to avoid hitting the muti solution state)

thus only A is true.

you would then mark alot more eliminations then just the 9.

all three points in example

in example.

B:
5(r2C1)-9(r1C2)-4(R1C9) trigger it
=> R2C1,<>5,R1C2<>9,R1C9 <>4
.

[RobO]

I don't get it. How can you eliminate, for example, the 5 from r2c1? That could be there with a 6 in r1c2 and a 9 in r1c9, leaving the 4 in r6c9 to break up the multi-solution state.

[RobO]

I also don't understand what you mean by BUG and BUG-lite. Do you have links to explain those?

[ronk]

I had a pattern similar to the one below in a puzzle I was doing over the weekend. Where does it fit in to existing solving methods?

It's a nice loop with a BUG-Lite, as StrmCkr noted, for one of the strong inferences:

r1c2 -9- r2c1 -5- BUG-Lite(r456c149):(r4c1 =5|4= r6c9) -4- r1c9 -9- r1c2 ==> r1c2<>9

[StrmCkr]

http://forum.enjoysudoku.com/viewtopic.php?t=2352&start=0&postdays=0&postorder=asc&highlight=bivalue+universal+grave

this is the thread on
bivavle universal graves

basically the "B.U.G" is any state where muti solutions is left.

and
bug-lites

http://forum.enjoysudoku.com/viewtopic.php?t=3056&start=0&postdays=0&postorder=asc&highlight=bivalue+universal+grave

bug lites are a cross between Unique rectangles and a bug

the diffrenece is there is only a 2 soltution state

instead of 3+ which is usually the charactesitic of a bug

as ronk pointed out:

you have a loop of strong inferences on 9: that trigers the bug-lite state. => R1C2<>9

question ronk:
isnt the ellimiantions:

(all of these from the strong infernece on 9 based on band 1)

=> R2C1,<>5,R1C2<>9,R1C9 <>4

as this is a mutipoint implication loop:
5<=> 9 <=>4 =>> bug.

[daj95376]

I had a pattern similar to the one below in a puzzle I was doing over the weekend. Where does it fit in to existing solving methods?

(Again, similar to my other post, I'd be surprised if it hadn't been spotted before.)

Just rows 1 to 6. Logic is that due to the unique loop on the 12s in rows 4-6, either r4c1 must be a 5, or r6c9 must be a 4 (or both). So one or both of r2c1 and r1c9 must be a 9. Hence the 9 in r1c2 can be eliminated.

Code: Select all

 .   69 . | .  . . | . . 49         
 59  .  . | .  . . | . . .         
 .   .  . | .  . . | . . .         
---------------------------
 125 . .  | .  . . | . . 12             
 12  . .  | 12 . . | . . .         
 .   . .  | 12 . . | . . 124       



I'm just a beginner at Deadly Patterns, but it appears that you have a potential DP in your second band. That said, then either <125> is <5> or else <124> is <4>, In either case, you get <69> is not <9>.

[ronk]

question ronk:
isnt the ellimiantions:

(all of these from the strong infernece on 9 based on band 1)

=> R2C1<>5,R1C2<>9,R1C9 <>4

No, there is nothing to indicate 9s are conjugate in r1 and b1.

[StrmCkr]

thanks ronk

i was using it as a forcing chain based on the bilocal of the 9
(box1 and row1) ...

... basicailly there was three ways of activating the bug...

as placeing:
5 in R2C1 = (9)R1C2 = (4)R1C9 => bug
or:

9 in R1C2 = (5)R2C1 = (4)R1C9 => bug
or:
4 in R1C9) = (9)R1C2) = (5)R2C1 => bug

i noticed that the allocation of digits didnt change when the bug triggers.
but im probably missing information to use the forcing chain... ie the whole grid..

i know the bug works off the 4|5 stoping the bug lite.
using band two..
either application: => r1C2<>9

which is said by daj above and reafermed by ronk again below..

daj95376 wrote:
I'm just a beginner at Deadly Patterns, but it appears that you have a potential DP in your second band. That said, then either <125> is <5> or else <124> is <4>, In either case, you get <69> is not <9>.

[ronk]

I'm just a beginner at Deadly Patterns, but it appears that you have a potential DP in your second band. That said, then either <125> is <5> or else <124> is <4>, In either case, you get <69> is not <9>.

Correct.