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George Woods
Joined: 28 Mar 2006 Posts: 291 Location: Dorset UK

Posted: Mon Aug 14, 2006 2:29 pm Post subject: Sunday Times 13 aug 06 


At this crunch point
600 304 750
300 020 690
007 060 380
009 002 100
023 401 570
000 600 200
976 010 423
518 243 967
030 976 815
A solution can be reached by considering the impact of a 1 at r6 c1.
immediately r6c6 becomes 7, r4 c4 becomes 8 and r5c5 9
now r5c1 is 8 and we have 89 in rows 4 and 5 as well as 89 in col7 all pressurising r6c9 to hold both 8 and 9 which it can't do
Hence 1 must lie ar r3c1 ...and so on.. to the solution.
So knowing that Sudoku pro's hate forcing functions, can anyone offer me the name and logic behind the solution you would use!!!!
Note : I have removed an 8 in r2c2 that was not apparent at the crunch point!
Last edited by George Woods on Mon Aug 14, 2006 11:56 pm; edited 1 time in total 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Mon Aug 14, 2006 2:41 pm Post subject: My observations 


In the position you posted, you can easily solve R5 beginning with <8> in C1. Then, you can solve the remaining <9>'s, and then the <8>'s. Then <7>, then <4>, then the rest.
It is all naked singles, except for one hidden single ("pin") on <9>.
Keith 

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George Woods
Joined: 28 Mar 2006 Posts: 291 Location: Dorset UK

Posted: Mon Aug 14, 2006 11:54 pm Post subject: a mistake in my original posting 


Sorry for wating your time maybe I made an error in my posting  the 8 in box1 should not be there (although curiously this 8 appears in the same place with either position of the 1 in col 1)  SO IS THIS A CLUE TO THE PROPER TECHNIQUE FOR SOLVING THIS ONE? 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Tue Aug 15, 2006 12:51 pm Post subject: 


The same thing applies: Solve row 5, then find the 8 in row 2, etc.
Keith 

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George Woods
Joined: 28 Mar 2006 Posts: 291 Location: Dorset UK

Posted: Tue Aug 15, 2006 2:36 pm Post subject: Yet another silly 


The solution as you point out is so easy that I thought "It can't be!" So to make sure I had recorded the crunch point correctly, I went back to the original puzzle and repeated the solution from scratch to the crunch point and guess what ANOTHER STUPID ERROR IS REVEALED ON MY PART  the 2 in box 4 should not be there !!!!!!!!!!! 

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

Posted: Tue Aug 15, 2006 4:26 pm Post subject: It's still pretty simple 


OK, George, I guess this is the "crunch point" (assuming I've followed all this correctly}.
Code:  600 304 750
300 020 690
007 060 380
009 002 100
003 401 570
000 600 200
976 010 423
518 243 967
030 976 815 
 The only apparent possibilities at r6c3 are {1, 4, 5}.
 The only possibilities at r6c8 are {3, 4}.
 The "8" in column 2 must lie in box 1. So the only possibilities at r6c2 are {4, 5}.
 There's a naked pair {8, 9} in column 5 (r1c5 & r5c5). This reveals a hidden pair {3, 5} in r4c5 & r6c5.
Counting this up we find the triplet {3, 4, 5} in r6c2, r6c5, and r6c8. So the sole candidate at r6c3 is a "1", and it's a piece of cake from there. dcb
PS Please post the original puzzle when you have a question like this one, George  it will be more fun for the rest of us that way. 

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George Woods
Joined: 28 Mar 2006 Posts: 291 Location: Dorset UK

Posted: Wed Aug 16, 2006 10:57 am Post subject: 


Thanks
Once again we find, unlikely though it seems, that the key lies in the least populated row,column, or box. I suppose seeing the very obvious 35 and 34 in the row should have prompted the though "Bet there is a 45 along here"  but it's so easy to see it after the event!!!! 

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