The New Sudoku Players' Forum

by **Jeff** » Wed Mar 01, 2006 5:19 pm

Code: Select all: . . . | . 6 . | . 6 6 . . . | . . 6 | . . 6 . . 6 | . . . | . . . ------+-------+------ . 6 . | 6 . 6 | . .-6 . 6 . | . 6 . | 6 6 . . . . | 6 . . | 6 . . ------+-------+------ 6 . . | . 6 . | 6 . 6 6 . . | 6 . . | . . . 6 . . | . . 6 | . 6 .

Here is a possible grouped x-cycle with multiple inference to deduce r4c9<>6

[r4c9]-6-[r56c7](=6=[r6c4]-6-[r5c5])=6=[r7c7]-6-[r7c5]=6=[r1c5]-6-[r2c6]=6=[r2c9]-6-[r4c9] => r4c9<>6

by **Ruud** » Wed Mar 01, 2006 6:06 pm

Havard wrote:I was wondering if anyone could see a pattern here:

Code: Select all
. . . | . 6 . | . 6 6 . . . | . . 6 | . . 6 . . 6 | . . . | . . . ------+-------+------ . 6 . | 6 . 6 | . .-6 . 6 . | . 6 . | 6 6 . . . . | 6 . . | 6 . . ------+-------+------ 6 . . | . 6 . | 6 . 6 6 . . | 6 . . | . . . 6 . . | . . 6 | . 6 .

This may be an alternative:

There are 3 candidates for 6 in box 9.

R7C9 kills R4C9 with a direct hit.
R9C8 forces R12C9 which kills R4C9.
R7C7 follows a longer route: it forces R6C4 -> R8C1 -> R9C6 -> R1C5 -> R2C9, killing R4C9

All 3 candidates of box 9 eliminate R4C9, so it has to go.

Ruud.

by **Havard** » Wed Mar 01, 2006 9:33 pm

Thanks for your replies!

Here is another one:

Code: Select all: . . . | . 4 4 | 4 . . . . 4 |-4 . 4 | . . . . . 4 | . . . | . 4 . ------+-------+------ . . . | 4 4 4 | . 4 . . . . | 4 . . | 4 . 4 4 . . | . . . | . . . ------+-------+------ . . . | 4 4 . | 4 . . . 4 . | . . . | . . . . . . | . 4 . | 4 . 4

one way of looking at this one is using this rule:

Code: Select all: . . . | . . C | . . . . . . | . . C'| . . . . . . | * * C"| . A . ------+-----|-+---|-- . . . | . . D | . B . . . . | . . . | . . . . . . | . . . | . . . ------+-------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . .

fillet o' skyscraper I belive it ended up being called

and then using the fact that we have one more strong link:

Code: Select all: . . . | . . C | . . . . . E | * * C'| . . . . . F | * * C"| . A . ------+-----|-+---|-- . . . | . . D | . B . . . . | . . . | . . . . . . | . . . | . . . ------+-------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . .

and in this puzzle this translates to:

Code: Select all: . . . | . . 4 | . . . . . 4 |-4 . 4 | . . . . . 4 | . . | | . 4 . ------+-----|-+---|-- . . . | . . 4 | . 4 . . . . | . . . | . . . . . . | . . . | . . . ------+-------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . .

Now this is from this puzzle:

Code: Select all: . . 3 | 6 . . | . 9 7 1 . . | . 9 . | 6 2 3 6 9 . | 2 7 3 | 1 . 5 ------+-------+------ 3 . 6 | . . . | 7 . 9 . 5 . | . 3 9 | . 6 . 4 . 9 | . . 6 | 3 . . ------+-------+------ 9 3 1 | . . 8 | . 7 6 7 4 . | 3 6 . | 9 . . . 6 . | 9 . 7 | . 3 .

and here the 8's are even more interesting (frustrating...):

Code: Select all: 8 8 . | . 8 . | 8 . . . 8 8 | 8 . . | . . . . . 8 | . . . | . 8 . ------+-------+------ . 8 . | 8 8 . | .-8 . 8 .-8 |-8 . . | 8 . 8 . 8 . | 8 8 . | .-8 8 ------+-------+------ . . . | . . 8 | . . . . . 8 | . . . | . 8 8 8 . 8 | . . . | 8 . 8

Mr. Nishio says we can eliminate in R5C34 and R46C8... (marked with -)

Any suggestions how this might be done?

by **ravel** » Thu Mar 02, 2006 10:16 am

Thanks for the daily x-cycle exercise (Jeff, nice loop yesterday)

An 8 in r4c8 or r6c8 forces 8 in r1c7 (eliminates 8 in r1c25) and r2c4 (eliminates 8 in r2c2), which leads to an x-wing in r46c25, that erases both candidates in r46c8.
[r4c8|r6c8]-8-[r3cc8]=8=[r1c7]-8-[r1c25]=8=[r2c4]-8-[r2c2]=8=[almost x-wing:r46c25]-8-[r4c8|r6c8]

After this elimination:
Go through 7 boxes (box 6 prepares box 9):
[r5c3](-8-[r5c79]=8=[r6c9]-8-[r8c9])-8-[r89c3]=8=[r9c1]-[r9c79]=8=[r8c8]-8-[r3c8]=8=[r1c7]-8-[r1c25]
=8=[r2c4]-8-[r2c23]=8=[r3c3]-8-[r5c3]
Go through all 9 boxes:
[r5c4](-8-[r5c79]=8=[r6c9]-8-[r89c9])-8-[r2c4]=8=[r1c5](-8-[r1c12])-8-[r1c7]=8=[r3c8]-8-[r8c8]
=8=[r9c7]-8-[r9c13]=8=[r8c3]-8-[r23c3]=8=[r2c2]-8-[r46c2]=8=[r5c13]-8-[r5c4]

by **aeb** » Thu Mar 02, 2006 1:10 pm

Havard wrote:
Code: Select all
8 8 . | . 8 . | 8 . . . 8 8 | 8 . . | . . . . . 8 | . . . | . 8 . ------+-------+------ . 8 . | 8 8 . | .-8 . 8 .-8 |-8 . . | 8 . 8 . 8 . | 8 8 . | .-8 8 ------+-------+------ . . . | . . 8 | . . . . . 8 | . . . | . 8 8 8 . 8 | . . . | 8 . 8

Mr. Nishio says we can eliminate in R5C34 and R46C8... (marked with -)
Any suggestions how this might be done?

(5,3)8, (3,8)8, (9,1)8, (?,7)8 - kills (5,3)8.
([46],8)8, (1,7)8, (2,4)8, ([64],5)8, (?,2)8 - kills ([46],8)8.
(5,4)8, (1,5)8, (3,8)8, (6,9)8, (9,7)8, (?,1)8 - kills (5,4)8.

by **Ruud** » Thu Mar 02, 2006 1:27 pm

Would this be a valid grouped X-Cycle for the digit 4 elimination?

Ruud.

by **ravel** » Thu Mar 02, 2006 1:59 pm

As i see it, aeb's nice deductions can be expressed as (discontinuous) nice loops, eg this for the first:
[r5c3](-8-[r5c7])(-8-[r3c3]=8=[r3c8]-8-[r1c7])-8-[r89c3]=8=[r9c1]-8-[r9c7]
But of course they are no cycles (in this sense nice "loop" also is a misleading name)
[edit] Oops, i missed that Ruud said digit 4

Anyway, i think my chains above are also valid grouped X-Cycles (with multiple inferences)

by **Jeff** » Thu Mar 02, 2006 2:38 pm

Ruud wrote:Would this be a valid grouped X-Cycle for the digit 4 elimination?

Yes, Ruud.

The nice loop notation is:

[r2c4]-4-[r2c3]=4=[r3c3]-4-[r3c8]=4=[r4c8]-4-[r4c456]=4=[r5c4]-4-[r2c4] => r2c4<>4

The loop can be easily seen if the weak inferences are drawn as broken lines.

by **Havard** » Thu Mar 02, 2006 3:06 pm

here is another one:

original puzzle: (menneske.no 6533286)

Code: Select all: . . . | . 9 . | . . . 7 6 . | 1 . 3 | 8 . . . 9 3 | 2 . 7 | . . . ------+-------+------ . 3 9 | . . . | 7 8 . 8 . . | . . . | . . 9 . 1 7 | . . . | 2 3 . ------+-------+------ . . . | 4 . 1 | 9 . . . . 6 | 9 . 5 | 3 4 . . . . | . 2 . | . 6 .

now after some normal deductions, let's look at the 5's!

(- shows eliminations predicted by Nishio)

Code: Select all: 5 5 5 | 5 . . |-5 . . . . 5 | . 5 . | . . . -5 . . | . . . | 5 5 5 ------+-------+------ 5 . . | 5 . . | . .-5 . 5 5 | . . . | 5 5 . 5 . . | . 5 . | . . 5 ------+-------+------ . 5 5 | . . . | . . 5 . . . | . . 5 | . . . . 5 5 | . . . | 5 . 5

Now first, one 5 falls to the Empty Rectangle!

Code: Select all: . . . | . . . | . . . . . . | . . . | . . . -5 . . | . . . | . 5 . ------+-------+---|-- 5 X X | . . . | . | . + 5 5 | . . . | . 5 . 5 X X | . . . | . . . ------+-------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . .

then this one goes because of the line-box just created!

Code: Select all: 5 5 5 | 5 . . |-5 . . . . 5 | . 5 . | . . . . . . | . . . | 5 5 5 ------+-------+------ 5 . . | 5 . . | . . 5 . 5 5 | . . . | 5 5 . 5 . . | . 5 . | . . 5 ------+-------+------ . 5 5 | . . . | . . 5 . . . | . . 5 | . . . . 5 5 | . . . | 5 . 5

Now for the last one, I would suggest something like this:

Code: Select all: A . . | C . . | . . . | . . | | . . | . . . | . . | | . . | . . . |-----+-|-----+------ B * * | D . . | * * * | . . | .\ . | . . . B'. . | . E . | . . . ------+-------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . .

which eliminates the last 5, and solves the puzzle after a XY-wing and a beautiful skyscraper in the 4's!

Any other suggestions how to kill off those 5's?

havard

by **ronk** » Thu Mar 02, 2006 4:03 pm

Ruud wrote:Would this be a valid grouped X-Cycle for the digit 4 elimination?

Yes, and you may consider box 5 to have two overlapping groups ... making Jeff's nice loop expression be:

[r2c4]-4-[r2c3]=4=[r3c3]-4-[r3c8]=4=[r4c8]-4-[r4c456]=4=[r45c4]-4-[r2c4] => r2c4<>4

IOW r4c4 may be considered part of either group ... almost as if it wasn't there. It is the Intersection (ERI) of Havard's Empty Rectangle (earlier on this thread) and the "hinge pin" of Rod Hagglund's hinge.

Ron

by **ravel** » Thu Mar 02, 2006 4:40 pm

Havard wrote:Any other suggestions how to kill off those 5's?

An X-Cycle for the 3rd one:
[r4c9]-5-[r4c78]=5=[r4c23]-5-[r46c1]=5=[r1c1]-5-[r1c4]=5=[r2c5]-5-[r6c5]=5=[r4c4]-5-[r4c9]
An aeb-like notation:
(4,9)5,(6,5)5,(1,4)5,(?1)5

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