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Please help October 24 hard puzzle

 
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Mike
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PostPosted: Tue Oct 25, 2005 8:08 pm    Post subject: Please help October 24 hard puzzle Reply with quote

Please help with logic for the October 24, 2005 hard puzzle. This is where I hit a wall . . .

470 203 900
060 810 020
021 079 500

500 001 000
200 000 005
000 700 006

702 900 450
090 547 032
054 102 079

Thanks for your help !
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Lulu



Joined: 21 Sep 2005
Posts: 11
Location: Manchester, England

PostPosted: Tue Oct 25, 2005 10:17 pm    Post subject: Oct 24th puzzle Reply with quote

I'm having problems working out how you got a 2 in r8c9 because at this position there are still 2 options for 2 in box 9 it could also be at r8c7.

Otherwise:-
Look at r8c3 this can only be an 8 because the 6 in column 3 must be at either r4 or r5 to satisfy box 4.

r1c3 5 is the only option

You should be able to complete box 2 from this stage.

See how you go

Lulu
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David Bryant



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

PostPosted: Tue Oct 25, 2005 10:30 pm    Post subject: How the "2" gets in r8c9 Reply with quote

Lulu wrote:
I'm having problems working out how you got a 2 in r8c9 because at this position there are still 2 options for 2 in box 9 it could also be at r8c7.

Ah, but there's already a {1, 8} pair in column 9 -- at r1c9 & r7c9 -- so the only possible value for r8c9 is a "2". dcb Smile
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Lulu



Joined: 21 Sep 2005
Posts: 11
Location: Manchester, England

PostPosted: Tue Oct 25, 2005 10:35 pm    Post subject: how to get the 2 in r8c9 Reply with quote

Silly me! I hadn't mapped out all the pairs at this time only marked the obvious placings. Embarassed
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Glassman



Joined: 21 Oct 2005
Posts: 50
Location: England

PostPosted: Tue Oct 25, 2005 10:39 pm    Post subject: Reply with quote

Mike I don't know what this is called, as I am new to this forum and the terminology, but there is one of this compiler's favourite helpful hints in this elegant puzzle, which I saw as soon as I looked at it. Two numbers, here 6 and 7, on intersecting rows and columns, helpfully isolated to highlight them. See what this tells you about the box where they intersect, and then work it through.

Glassman Cool
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Mike
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PostPosted: Tue Oct 25, 2005 10:59 pm    Post subject: 67 Logic? Reply with quote

Thanks all for trying to find the logic thread in this puzzle.

For Glassman - I'm not sure where to find the intersection strategy published by the author of the puzzle . . . could you point me in the right direction? Also, which 67's are you referring to?
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Mike



Joined: 25 Oct 2005
Posts: 6
Location: Connecticut

PostPosted: Tue Oct 25, 2005 11:09 pm    Post subject: Thanks Lulu Reply with quote

Lulu - your second comment is right on and now I can continue. Thanks so much!!
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David Bryant



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

PostPosted: Tue Oct 25, 2005 11:12 pm    Post subject: This is a "hidden pair" Reply with quote

Glassman wrote:
Mike I don't know what this is called, as I am new to this forum and the terminology, ...

People usually call this a "hidden pair", because the geometry of the intersection (in this case in the middle left 3x3 box) leaves only two cells in which the {6, 7} can fit. It's a "hidden pair" -- and not a "naked pair" -- because it appears on a cursory inspection that some other values (in this case {3, 8, 9}) might also be possible in those two cells. dcb Smile
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Glassman



Joined: 21 Oct 2005
Posts: 50
Location: England

PostPosted: Tue Oct 25, 2005 11:15 pm    Post subject: Re: 67 Logic? Reply with quote

Mike wrote:
I'm not sure where to find the intersection strategy published by the author of the puzzle . . . could you point me in the right direction?

Sorry, I don't follow you the helpful hint is in the puzzle, not published elsewhere.
Mike wrote:
... which 67's are you referring to?

Those in r6 and c2 (in the original puzzle), beautifully isolated except for a wayward 9 that just had to get in on the act and spoil the symmetry, typical of 9s.

Glassman Cool
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Mike



Joined: 25 Oct 2005
Posts: 6
Location: Connecticut

PostPosted: Tue Oct 25, 2005 11:37 pm    Post subject: 67 Logic Reply with quote

David and Glassman,

I understand this logic and had isolated 67 with other possibilities to r4c3 and r5c3. This is a nifty way to reduce the possibilities further. Lulu's logic provided the jump which produced a 7 in r4c9 and therefore set 6 in r4c3 and 7 in r5c3.

Thanks for your thoughts,

Mike
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Mike



Joined: 25 Oct 2005
Posts: 6
Location: Connecticut

PostPosted: Tue Oct 25, 2005 11:58 pm    Post subject: Completed. Thank you all !! Reply with quote

This is my third day attempting SuDoku and I am totally enjoying these logic mazes !! Thanks for all your help. Very Happy [/b]
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