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				|  Posted: Tue Jan 24, 2006 4:01 pm    Post subject: Sun 22-Jan-2006 Puzzle |   |  
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				| Can someone give me a tip.  Looks like I get 10 moves then I'm stuck.  Here is where I am. 
 5 _ 4 | _ 9 _ | _ _ _
 _ 9 3 | _ _ 2 | _ _ _
 2 8 6 | _ _ 3 | _ _ 9
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 _ 2 _ | 6 _ _ | _ 9 4
 9 6 1 | _ _ _ | 7 _ _
 4 3 _ | _ _ 9 | _ 6 _
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 _ 5 2 | 8 _ _ | 9 7 _
 _ 4 _ | 9 2 _ | 5 3 _
 _ _ 9 | _ 5 _ | 8 4 2
 
 I hope I drew the puzzle correctly.
 
 Thanks,
 Rick
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		| Steve R 
 
 
 Joined: 24 Oct 2005
 Posts: 289
 Location: Birmingham, England
 
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				|  Posted: Tue Jan 24, 2006 4:30 pm    Post subject: Sun 22-Jan-2006 Puzzle |   |  
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				| Rick 
 You did indeed get the puzzle right.
 
 I think the next step is to check where 7 fits in the top right-hand box. It must lie in row 1 or row 2. The same applies to the top left-hand box. So, in the top middle box, it must occupy row 3 and cannot fill r1c4.
 
 Excluding 7 from r1c4, leaves 1 as the only admissible entry.
 
 Over to you.
 
 Steve
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		| alanr555 
 
 
 Joined: 01 Aug 2005
 Posts: 198
 Location: Bideford Devon EX39
 
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				|  Posted: Tue Jan 24, 2006 11:18 pm    Post subject: Re: Sun 22-Jan-2006 Puzzle |   |  
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				| > I think the next step is to check where 7 fits in the top > right-hand box. It must lie in row 1 or row 2. The same
 > applies to the top left-hand box. So, in the top middle
 > box, it must occupy row 3 and cannot fill r1c4.
 
 > Excluding 7 from r1c4, leaves 1 as the only
 > admissible entry.
 
 Whilst this is all true it does assume that the human solver
 has already derived 17 as the candidate profile for r1c4.
 
 This IS a puzzle that needs more than just Mandatory Pairs
 to solve it - but there was more that the original poster on
 this topic could do before reverting to Candidate Profiles.
 
 The options for the '7' in boxes 1,2,3 are derived quite easily
 using Mandatory Pairs.
 
 Box 3 MP 7 in r1c9 and r2c9
 Box 3 r1c9 set to 8 by other logic
 Thus r2c9=7
 Box 1 has row 3 fully occupied
 Thus 7 goes in r1c2 (sole position in box)
 Then
 Line 1 has already reached 574 -9- 328 by other logic
 leaving 1 and 6 to fill the two places
 There is a 6 in r4c4 and so r1c4=1 and r1c6=6.
 
 This is an interpretation from my workings of the puzzle but as
 I did not record the sequence of resolution I may be mistaken.
 
 Anyway, I did resort to Candidate Profiles to get the full solution.
 
 Alan Rayner  BS23 2QT
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