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Sun 22-Jan-2006 Puzzle

 
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PostPosted: Tue Jan 24, 2006 4:01 pm    Post subject: Sun 22-Jan-2006 Puzzle Reply with quote

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
-----------------------
_ 2 _ | 6 _ _ | _ 9 4
9 6 1 | _ _ _ | 7 _ _
4 3 _ | _ _ 9 | _ 6 _
-----------------------
_ 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

PostPosted: Tue Jan 24, 2006 4:30 pm    Post subject: Sun 22-Jan-2006 Puzzle Reply with quote

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: 194
Location: Bideford Devon EX39

PostPosted: Tue Jan 24, 2006 11:18 pm    Post subject: Re: Sun 22-Jan-2006 Puzzle Reply with quote

> 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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