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Earl · Joined: 30 May 2007 Posts: 677 Location: Victoria, KS

Truly VH. A three-stepper for me. Any single shots out there?

A Solution: X-wing <4>, xy-wing (168), xy-chain eliminates 7 in R8C8. The <38> UR seems ineffective

Early Earl

Marty R. · Posted: Fri May 01, 2009 4:39 am Post subject:

Do I hang my head in shame if I used five steps?

keith · Posted: Fri May 01, 2009 6:26 am Post subject:

Not using the UR, I had:

keith · Posted: Fri May 01, 2009 6:33 am Post subject:

At last, a "very hard" worthy of the name. I think the minimal path is an XYZ-wing followed by an XY-wing.

After basics:

nataraj · Posted: Fri May 01, 2009 7:22 am Post subject:

great puzzle. truly very hard

I could not believe I needed all of these steps (I did not use the 38 UR):

- x wing (4)
- xy-wing 18-68-16
- xyz 14-147-17
- xy 47-48-87

storm_norm · Joined: 18 Oct 2007 Posts: 1741

Earl wrote:

Ema Nymton · Joined: 17 Apr 2009 Posts: 89

.
Please I ask your indulgences. Like a school child, I follow my teacher's instruction without questioning because I believe that eventually understanding will come through use of the formula. (Sounding a bit 'Moody Blues'ish) ;-/

nataraj · Posted: Fri May 01, 2009 12:13 pm Post subject:

Ema,

you got to be careful with "rules".

The "XY-rule" does not say anything about "outside the # buddies".
The xy-wing you found (like any other xy- or w- or m-wing) only allows a deduction about those cells that see both ends ("pincers") of the wing: they cannot contain the candidate that is present in both pincers.

In this case, the pincers are r3c4 and r8c8, the common candidate is "8", and the only cell that sees both pincers is r8c4. The xy-wing asserts that r8c4 cannot be 8. Which is kind of trivial because we already know it is "1".

A wing like this is called "useless" or "flightless" in this forum. It might lead to other eliminations (by way of "transporting" or "coloring") but in itself does not help the solution.

The cell r2c7 (marked $ in your grid) has nothing to do with the xy-wing.

--------

On the other hand, columns 4 and 8 contain exactly two "4"s each, forming an "x-wing" pattern (and leading to the elimination of "4" from r3c6 and r3c7.

There is also an xyz wing present (17 in r2c9, 14 in r1c3 and 147 in r2c3) that removes "1" from r2c2, plus a "unique rectangle" of type 1 in r34c12 - but that is a different story altogether.

arkietech · Posted: Fri May 01, 2009 12:44 pm Post subject:

I used two steps x=wing 4 and the xyz-wing 147

Good puzzle Very Happy

Ema Nymton · Joined: 17 Apr 2009 Posts: 89

Clement · Posted: Fri May 01, 2009 1:43 pm Post subject: Daily Sudoku: Fri 1-May-2009 VH

I used the following:-
1) UR <38>r34c12
2) APE in cells r2c2&r2c3 no possibility of 1 being in r2c2.
3) X-Wing in 1.
4) Two XY- wings <168> & <478>.

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

I used the x wing (4), xyz (147), and xy (478). But unlike some of the former posters I had no qualms about using the UR on 38 because it was a Type 4 and you have to be pretty smart to use Type 4s !!

storm_norm · Joined: 18 Oct 2007 Posts: 1741

cgordon wrote:

keith · Posted: Fri May 01, 2009 10:45 pm Post subject:

storm_norm · Joined: 18 Oct 2007 Posts: 1741

ahh,
I see, the type 1 as it stands in the grid can also be looked at as a type 4 and a type 6. I see I see.

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

Marty R. · Posted: Sat May 02, 2009 3:48 am Post subject:

sudocraz · Joined: 28 Apr 2008 Posts: 53

This was a fun puzzle. My husband and I did and x-wing, then a UR Type 1, then an xywing, and xyz wing and another xy wing.

I believe that the UR was a type 1, because one of those four squares could not be a 38. There was only one square that qualified for this.