dailysudoku.com

Clement · Posted: Thu Jun 23, 2011 10:39 pm Post subject: Jun 24 VH

XY-Wing 28 25 58 pivoted in r6c2 or r4c1 solve it.

kuskey · Joined: 10 Dec 2008 Posts: 141 Location: Pembroke, NH

Also a 258 xy-wing with pivot at r6c2 setting r4c6<>8.

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

My very clever solution was a Type 6 UR for 35.

There are only two 3's in C3 so remove the 3 from R4C7. There are only two 5's in R5 so remove the 5 from R4C7 - leaving a single 2.

Marty R. · Posted: Fri Jun 24, 2011 3:57 pm Post subject:

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

Marty R. · Posted: Fri Jun 24, 2011 7:13 pm Post subject:

Craig,

Type 6 does have diagonal pairs, but one of the numbers needs to be an X-Wing. Then the two bivalue cells can be solved for the number that forms the X-Wing.

But the eliminations you describe are valid. Ted would probably describe the pattern as an Almost Unique Rectangle or Almost Type 6.

I've never heard of a Type 7 or higher.

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

Marty: Sorry for being pedantic - but if it doesn’t need an xy wing and always works with just strong links – it deserves better than a “Type of Hidden” or an “Almost” Type 6 UR. Evil or Very Mad

.

Marty R. · Posted: Fri Jun 24, 2011 9:20 pm Post subject:

OK, give it a name. Laughing

tlanglet · Posted: Fri Jun 24, 2011 9:46 pm Post subject:

I agree with Marty that the (35)r45c37 pattern is not a Type 6 UR because it doesn't contain an overlaying x-wing for either digit 3 or 5. Thus the "rote rule" for a Type 6 UR doesn't apply. Also note that the Type 6 rule provides a deletion in the two polyvalued cells of the digit with the overlaying x-wing; different digits in the same cell are not valid.

However, some deletions are possible using this pattern. To prevent the "Deadly Pattern" the internal SIS (Strong Inference Set) consists of r4c7=2 and r5c3=7. I did not see a common result by applying these two constraints.

But looking at the external SIS r5c8=3 and r4c6=5 , I found two deletions.

Given r4c6=5, then r5c8=3 forces r5c7=5 which results in r4c7<>5
In eureka notation this could be written as: aur(35)r45c37[(5)r4c6=(3)r5c8]-(3=5)r5c7; r4c7<>5

Also, given r5c8=3, then r4c6=5 forces r4c3=3 which results in r5c3<>3
In eureka notation this could be written as aur(35)r45c37[(3)r5c8=(5)r4c6]-(5=3)r4c3; r5c3<>3

Together, these two deletions solve the puzzle.

Ted

Clement · Posted: Fri Jun 24, 2011 10:07 pm Post subject: Jun 24 VH

cgordon,
In principle it is a Hidden Unique Rectangle. The issue here I think is which type. As for all unique rectangles we are tying to avoid reducing this rectangle to 35 in all four cells.

keith · Posted: Sat Jun 25, 2011 12:56 am Post subject:

After basics:

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada