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May 23 very hard

 
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Bill Denholm



Joined: 24 Mar 2006
Posts: 17
Location: Mountain View, California

PostPosted: Wed May 24, 2006 3:17 am    Post subject: May 23 very hard Reply with quote

I was trying to play the May 23 puzzle, but I got stuck, so I asked for a hint. The program gave me: 2 at R6, C3. I don't see the logic (if I did, I wouldn't have needed a hint). Can somebody explain the logic to me? Thanks.

Bill
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Charles



Joined: 31 Mar 2006
Posts: 8
Location: Lawton, Oklahoma, USA

PostPosted: Wed May 24, 2006 4:35 am    Post subject: May 23 very hard Reply with quote

Bill: Two is restricted to R9C1 and R9C2 - thus other twos in block 7 can be excluded. Good Luck. Charles
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Marty R.



Joined: 12 Feb 2006
Posts: 5121
Location: Rochester, NY, USA

PostPosted: Wed May 24, 2006 3:33 pm    Post subject: Reply with quote

I agree with Charles. After eliminating those other 2s, the puzzle is broken wide open.
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Guest





PostPosted: Wed May 24, 2006 4:21 pm    Post subject: May 23 Reply with quote

There are 6 linked pairs

r1c9 5 8
r2c7 8 5

r8c7 5 2
r7c9 2 5

r4c7 2 8
r4c9 8 2

r1c5 8 1
r2c6 1 8

r5c5 1 2
r6c6 2 1

r8c5 2 8
r7c6 8 2

Either the first option (column) or second option has a 2 at r4, r7 and r8.
That leads to r6c3 2.
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Bill Denholm



Joined: 24 Mar 2006
Posts: 17
Location: Mountain View, California

PostPosted: Wed May 24, 2006 5:20 pm    Post subject: Reply with quote

I'll have to study it some more, since I still don't understand it. I agree with guest about the linked pairs, I found them, too.

Charles, the hint was for a 2 in block four, not block 7, but thanks, anyway.

Bill
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Charles



Joined: 31 Mar 2006
Posts: 8
Location: Lawton, Oklahoma, USA

PostPosted: Wed May 24, 2006 7:01 pm    Post subject: May 23 classic Very Hard Block 4/7 Reply with quote

Bill: The exclusion of twos in block 7 is what allowed/forced the choice for two in block 4.
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Guest





PostPosted: Thu May 25, 2006 12:33 am    Post subject: Reply with quote

Bill

I should explain the linked pair a little further. The linked pairs put
a 2 at row 4, 7 and 8. Now look at column 3: the only place the 2 can
go is at row 6
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Bill Denholm



Joined: 24 Mar 2006
Posts: 17
Location: Mountain View, California

PostPosted: Thu May 25, 2006 3:57 am    Post subject: Reply with quote

Charles,

I am still having problems seeing how the 2s in block 7 are restricted to R9C1 and R9C2. I think they probably are, but there seems to be a missing step somewhere. What I see when I look at my block 7 is this:

000 | 003 | 120
000 | 000 | 056
700 | 000 | 080

000 | 020 | 120
000 | 056 | 056
009 | 080 | 080

120 | 020 | 000
050 | 050 | 400
080 | 080 | 000

I assume that you see something else.

Bill
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David Bryant



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

PostPosted: Thu May 25, 2006 2:03 pm    Post subject: A picture is worth a thousand words Reply with quote

I'm puzzled by this entire discussion. I think that a picture is worth a thousand words.
Code:
 125   25     7     3    18     6     4     9    58
  4     9    15     2     7    18    58     3     6
 68    68     3     9     5     4     7     2     1
 56     1    56     4     3     9    28     7    28
 23     7     9     8    12     5    13     6     4
 238    4    28     7     6    12    13     5     9
  7     3   12568  156    4    28     9    18    25
  9   2568  12568  156   28     3    25     4     7
1258   258    4    15     9     7     6    18     3

I think this is the point where you must be stuck, Bill. If not, please clarify which of the resolved cells in this grid you haven't managed to solve yet.

From here there are at least two ways to proceed.

-- Clearly the "2" in row 9 must appear either at r9c1 or r9c2. Therefore there cannot be a "2" at r7c3, r8c2, or r8c3. This leaves the "2" at r6c3 as unique in column 6, so you can set r6c3 = 2.

-- Alternatively, one can spot the X-Wing formation in r19c12 and reason that there cannot be a "2" at r5c1, r6c1, or r8c2, because the "2"s in columns 1 & 2 must lie either in r1c1 & r9c2, or else at r1c2 & r9c1. This leaves the "2" at r6c3 as unique in the middle left 3x3 box.

I hope that clears things up for you, Bill. dcb
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