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June 3 - Hard

 
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danowen



Joined: 04 Jun 2006
Posts: 1

PostPosted: Sun Jun 04, 2006 4:29 pm    Post subject: June 3 - Hard Reply with quote

Hi Everyone, just started Sudoku for the first time last night and would like to get some help.

2xx|8x6|xxx
x6x|xxx|x28
x81|2x4|x36

x2x|6xx|xxx
83x|x9x|x67
xx6|xx8|x5x

x4x|5x7|6xx
6xx|xxx|x1x
xxx|x63|xx2

The next logical step is a 9 in r2c6...can someone help me out on how to arrive at that? thanks!

daniel
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Sun Jun 04, 2006 6:24 pm    Post subject: June 3 - Hard Reply with quote

Hi Daniel.

The pair (37) in r26c4 excludes 9 from r2c4 and explains the hint. Back to you.

Steve
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sdtibbs



Joined: 08 May 2006
Posts: 6

PostPosted: Mon Jun 05, 2006 2:06 am    Post subject: Reply with quote


Hi Steve
Would you please tell me how you arrive at the pair of 37 in r6c4. My candidates are 1347. C4 is 8,1379,2,6,14,1347,5,49,149. I can see the combinations of 14,49,149 but do not know why I can eliminate the 9 in r2c4 or the 14 in r6c4. Probably the reasoning is staring me in the face but I need the explanation. Many thanks
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Angel



Joined: 26 Mar 2006
Posts: 31

PostPosted: Mon Jun 05, 2006 10:43 am    Post subject: Reply with quote

Daniel,

You can see the combinations 14, 49 and 149. So you have 3 cells with 3 possible values. This tells you that the numbers 1, 4 and 9 must in some order go in these three cells (r5c4, r8c4, r9c4). You don't know how they are distributed but they must go in there somehow. So you won't have 1, 4 or 9 left to go anywhere else in that column. Therefore you can eliminate these numbers from r2c4 and r6c4. This leaves just 3 and 7 as the candidates for these cells.

Back to you now.
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Mon Jun 05, 2006 1:04 pm    Post subject: Reply with quote

Hi SDT

There are two ways of looking at your list of candidates for the fourth column.

One is to spot that three cells take their entries from the set {1, 4, 9}, as Angel did. Since the three cells must have different entries, they use up all three possibilities between them and these candidates can be excluded from the rest of the column.

The other is two spot that two candidates, 3 and 7, can only be entered into two cells, here rows 2 and 6 or r26c4 for short. Only one of these candidates can go in each cell so the candidates 3 and 7 are used up by these cells.

It seems that people differ in that some spot one of these approaches more quickly than the other. Both are effective. One is called a naked subset and the other a hidden subset. I donít remember which is which because it never seemed to matter. All the same you might find SadManís site helpful on the subject.

Steve
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Tue Jun 06, 2006 9:23 am    Post subject: Reply with quote

Steve R wrote:


It seems that people differ in that some spot one of these approaches more quickly than the other.

2xx|8x6|xxx
x6x|xxx|x28
x81|2x4|x36

x2x|6xx|xxx
83x|x9x|x67
xx6|xx8|x5x

x4x|5x7|6xx
6xx|xxx|x1x
xxx|x63|xx2
Without candidate list the hidden pair 37 in r26c4 is easier to spot - you already have 3 and 7 in in row 5 and in box 8, so only 2 cells are left in column 4.

With candidate list
8,1379,2,6,14,1347,5,49,149
the naked triple is easier to spot.
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sdtibbs



Joined: 08 May 2006
Posts: 6

PostPosted: Wed Jun 07, 2006 4:56 pm    Post subject: Reply with quote


Dear Steve,
Thank you for your reply. I do see it and I appreciate your time in answering my question even though it probably seems so basic to you. :D
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