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Charles
Joined: 31 Mar 2006
Posts: 8
Location: Lawton, Oklahoma, USA
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 Posted: Mon May 22, 2006 4:23 am Post subject: May 20 Hard Classic |
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| Not much discussion in the last few days. I miss learning something in almost every topic discussed. I solved May 20 using "If Then" on the threes in R7C1 and R7C3 For either choice of three, R4C1 cannot be six. Does this concept have a name? How did others solve this one? Somehow I think I overlooked something. Thanks, Charles |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Mon May 22, 2006 10:12 am Post subject: Re: May 20 Hard Classic |
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| Charles wrote: | | Somehow I think I overlooked something. |
| Code: | ----------------------
4 3 . | 6 . . | 2 . 1
. 6 2 | . 1 4 | . . .
9 1 8 | . 3 . | 4 . .
----------------------
. 5 . | 1 4 . | . . .
8 7 4 | 5 6 3 | 9 1 2
. . 1 | . 7 8 | . 4 .
----------------------
. . . | . . . | 1 5 4
. 4 . | 3 . 1 | 8 . .
1 . . | 4 . 6 | . 2 .
----------------------
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Maybe you missed the hidden 36-pair in column 3 (rows 4 and 7).
There are 3 and 6 in row 1.
In row 8 there is a 3 and 6 is restricted to column 8 or 9 (in box 9).
In row 9 there is a 6 and 3 is restricted to column 7 or 9 (in box 9).
The pair gives you a single in row 4.
PS: Similarly there is a hidden 57-pair in column 1 (rows 2 and 8):
Both 5 and 7 are in box 4.
In row 7 there is a 5 and the 7 is restricted to r7c46 in box 8.
This gives you a single in row 8. |
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