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Posted: Wed Jan 18, 2006 10:59 am Post subject: jan 17 06 


Jan 17 2006
x24 5xx x7x
156 x4x 2x9
x78 x1x xx4
xx9 x7x xxx
237 851 946
xx1 x9x 3x7
7x9 12x 465
4x5 x6x xx2
612 4x5 79x
trying to figure out how the bold x in col 8 row4 is equal to a 2
had no problem getting to this point and once i got past it it was no problem
thanks 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Jan 18, 2006 1:00 pm Post subject: There's a triplet in row 6 


I got stuck at the same spot!
Look at the middle right 3x3 box. Clearly the "2" must fit at r4c8 or at r6c8, because there's already a "2" in column 7 (r2c7) and in column 9 (r8c9).
Now look at the middle left 3x3 box. The pair {4, 6} must lie in r4c2 & r6c2, because of the pair {8, 9} that evidently lies in r7c2 & r8c2.
(Note that this paragraph has been edited  dotdot caught my mistake.)
Finally, concentrate on the middle center 3x3 box. Clearly the only possibilities at r6c4 & r6c6 are {2, 4, 6}, because there's already a "3" in row 6. But now we've identified a triplet {2, 4, 6} lying in r6c2, r6c4, & r6c6. So the "2" can't possibly fit at r6c8, and the only spot left for a "2" in the middle right 3x3 box is at r4c8. dcb
Last edited by David Bryant on Wed Jan 18, 2006 9:14 pm; edited 1 time in total 

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dotdot
Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Posted: Wed Jan 18, 2006 5:32 pm Post subject: Schroedinger pan 


Hallo David
I sailed through this one on the day and wondered why it wasn't classed as medium.
But it certainly didn't seem medium when I looked at it again today in the context of this thread.
This sort of inconsistency is a metapuzzle to me.
But I stuck at it, and here offer what I suspect I did originally:
Look at the Glassman pan whose lid is the shared triplet r4b6. There is an open position in its bottom at r6c8.
It has a pretty useless looking handle; BUT
the {5,8} at r4c1 is not accounted for (not even partially) in the resolved 9/4/6/3/7 cells. To make up for this r6c8 must be {5,8}.
Is this watertight?
P.S. Corrigenda: Guest's r7c3 should be 3 and your triplet starts with r6c2 (not r6c1). 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Jan 18, 2006 9:33 pm Post subject: Re: Schroedinger's pan 


Hi, dotdot! Thanks for catching my mistake. I shouldn't write these things so early in the morning. :shock:
I'm not quite sure what you mean by citing a "pan" here. But one can spot the "hidden pair" {5, 8} in r6c1 & r6c8 without thinking about the {2, 4, 6} triplet explicitly.
Both "5" and "8" are already present in the middle center 3x3 box, so the {5, 8} pair can't fit in either r6c4 or r6c6. And {5, 8} can't fit at r6c2 either  you can spot this as a consequence of the naked pair {5, 8} in r1c4 & r1c6, or as a consequence of the "5" at r2c2 and the naked pair {8, 9} at r7c2 & r8c2. So there are only two cells, r6c1 & r6c8, at which either "5" or "8" can be entered, and the hidden pair {5, 8} must lie in those two spots. dcb 

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dotdot
Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Posted: Wed Jan 18, 2006 10:56 pm Post subject: Re: Schroedinger's pan 


David Bryant wrote:  I'm not quite sure what you mean by citing a "pan" here.  Follow my link above (and read your comments).
The triplet r4c7..9 is shared by row4 and box6, so
the rest of the row (the handle) has the same values as
the rest of the box (the bowl of the pan).
The above situation prompts pan recognition because
 the lid is off (visual emphasis)
 the bowl is nearly complete.
Recognising this as a pan with a small hole led me so directly to r4c1=r6c8 that I had no feeling of having got past a sticky bit.
The Glassman pan specimens were built from resolved cells.
But in the above case the useful correspondence involves an unresolved cell with 2 possibilities,
which can be considered as involving two Glassman pans in superposition, i.e. a Schroedinger pan. 

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