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tlanglet · Posted: Fri Jan 30, 2009 5:27 am Post subject:

Yet another fun VH type puzzle. Minimal basics and several choices for later techniques. My four step solution was:

storm_norm · Joined: 18 Oct 2007 Posts: 1741

I call these the "Danny specials"
in the image below notice the UR {7,8}. we know from the rules of unique rectangles that you cannot be left with a 7 and a 8 in r1c7 and r3c7...
then notice the other 7 in column 7
and the other 8 in column 7
as in the image

so either the 7 in r8c7 is true or the 8 in r2c7 is true...A.K.A. Strong Inference.
because if both were false then the Deadly Pattern is forced to exist.
if the 7 in r8c7 is true, then the 7 in r8c1 is not true... or if the 7 in r8c7 is not true...
r2c7 = 8
r2c1 <> 8
r8c1 = 8
either way, the 7 in r8c1 is not true.
and solves the puzzle.
or you can view it by this chain.
UR78[(7)r8c7 = (8)r2c7] - (8)r2c1 = (8)r8c1; r8c1 <> 7

tlanglet · Posted: Fri Jan 30, 2009 1:22 pm Post subject:

Great find Norm. IMPRESSIVE Exclamation

Ted

storm_norm · Joined: 18 Oct 2007 Posts: 1741

thank you, Ted.
this particular move was devastating. I have noticed patterns like this in the other puzzles. Every once in a while one proves critically beneficial.

Kdelle · Joined: 20 Mar 2008 Posts: 59 Location: Hudson, NH

In Norm's diagram above, in order to avoid the deadly pattern 39 in rows 7 and 8, either r7c2 must be 5 or r8c9 must be 4, either of which forces r8c3 = 8 which also solves the puzzle. I think that's correct logic anyway....

Kathy

storm_norm · Joined: 18 Oct 2007 Posts: 1741

daj95376 · Joined: 23 Aug 2008 Posts: 3854

Norm, once you have the UR strong link, you can apply forcing chain logic -- my preference.

Marty R. · Posted: Fri Jan 30, 2009 11:33 pm Post subject:

How about an alternate solution? The 78 UR contains two Hidden URs. Because of the strong link on 7 in row 1, the 8 is removed from r3c7. Because of the strong link on 8 in row 3, the 7 is removed from r1c7.

The 39 UR is a standard Type 6 (X-Wing on 9), so the 9s must be present in both the bivalue cells. The puzzle is now reduced to a BUG+1, with r8c1 = 4 in order to squash the BUG.

Kdelle · Joined: 20 Mar 2008 Posts: 59 Location: Hudson, NH

Norm,

My thinking was that a 5 in r7c2 forced a 2 in r7c1 which forced a 3 in r7c3 and then a 4 i r5c3....eliminating a 4 in r8c3. I think that's probably what Danny was saying in his post....but I'm not sure I understand his language yet!!!!!!!!! I AM trying!

Kathy

storm_norm · Joined: 18 Oct 2007 Posts: 1741

Marty R. · Posted: Wed Feb 04, 2009 10:54 pm Post subject:

I just got around to doing this, even though I've already commented.

An XY-Wing on 283 took out a couple of 3s, then a pincer transport made more eliminations. A W-Wing on 34 with pincer transport finished it off.

Asellus · Posted: Wed Feb 04, 2009 11:42 pm Post subject:

Norm,

While the 78UR route to your elimination is fine, it isn't necessary:

(7)r8c1 - (7=3)r8c7 - (3=8)r2c7 - (8)r2c1=(8-7)r8c1; r8c1<>7

It is just a 2-cell XY-Chain extended with a conjugate link. Thus, a simple one-step solution without uniqueness assumptions.

daj95376 · Joined: 23 Aug 2008 Posts: 3854

This puzzle offers a host of solution paths ... and I don't mean all of the long XY-Chains.