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Mike Guest
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Posted: Tue Oct 25, 2005 8:08 pm Post subject: Please help October 24 hard puzzle |
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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
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Posted: Tue Oct 25, 2005 10:17 pm Post subject: Oct 24th puzzle |
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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
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Posted: Tue Oct 25, 2005 10:30 pm Post subject: How the "2" gets in r8c9 |
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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 |
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Lulu
Joined: 21 Sep 2005 Posts: 11 Location: Manchester, England
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Posted: Tue Oct 25, 2005 10:35 pm Post subject: how to get the 2 in r8c9 |
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Silly me! I hadn't mapped out all the pairs at this time only marked the obvious placings. |
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Glassman
Joined: 21 Oct 2005 Posts: 50 Location: England
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Posted: Tue Oct 25, 2005 10:39 pm Post subject: |
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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 |
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Mike Guest
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Posted: Tue Oct 25, 2005 10:59 pm Post subject: 67 Logic? |
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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
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Posted: Tue Oct 25, 2005 11:09 pm Post subject: Thanks Lulu |
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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
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Posted: Tue Oct 25, 2005 11:12 pm Post subject: This is a "hidden pair" |
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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 |
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Glassman
Joined: 21 Oct 2005 Posts: 50 Location: England
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Posted: Tue Oct 25, 2005 11:15 pm Post subject: Re: 67 Logic? |
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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 |
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Mike
Joined: 25 Oct 2005 Posts: 6 Location: Connecticut
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Posted: Tue Oct 25, 2005 11:37 pm Post subject: 67 Logic |
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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
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Posted: Tue Oct 25, 2005 11:58 pm Post subject: Completed. Thank you all !! |
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This is my third day attempting SuDoku and I am totally enjoying these logic mazes !! Thanks for all your help. [/b] |
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