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Tough Sudoku for 27/October/2010

                 
                 
                 
                 
                 
                 
                 
                 
                 

Choose a number, and place it in the grid above.

  1 2 3 4 5 6 7 8 9
Check out the latest post in the Sudoku Forum

Sudoku Solving Techniques

Submitted by: sudoku lover

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Books! A cube of knowledge?
27/Oct/10 12:20 AM
too easy, use XW(4)df35 and SF(1)cfi356 and UP=81.
27/Oct/10 12:27 AM
"I don't care if you WANT to become an architect, Dear ... stop building things with your books and start doing your homework!"
27/Oct/10 12:33 AM
I just did yesterday's tough, no one found the one step solution (UP=26)

(5)a2=a9-(5=6)e9-(16=8)e12-e7=d7-*(23)d78=d2
=> a2<>28, d2<>6, UP=81.
27/Oct/10 12:49 AM

SE=3.8
27/Oct/10 1:17 AM
Alfred - note the manifest superiority of AIC chains in the above solution. It simply isn't possible for a forcing chain engineered from an UP guess to produce collinear deductions (such as a2<>8, d2<>6 above) because such chains have no symmetry.
27/Oct/10 1:30 AM
Thanks farpointer for posting a neat solution for yesterday's. I can see how a2<>28:
If a9=5,e9=6,e1=1,e2=8=d7,d2=2. But I can't quite see how you derive d2<>6. I would need another step:
If d2=6,e1=1,e2=8=d7;23 have no place in column d. It seems that this result is implicit in your chain, but my Eureka reading skills are not up to that.
27/Oct/10 9:08 AM
- sub chain (16)e12 == (2)d2 => d2<>6.
27/Oct/10 10:06 AM
to read and verify an AIC chain you needn't go to the trouble of setting values.
1.verify independantly each direct weak link (by inspection)
2.verify each direct strong link
3.for each deduction
--- verify that it is weak with two points, i, j on the chain i == x -- y == j
4.if the chain is looped, deductions can be made on any pair i, j
i -- x == y -- j or i -- j.
27/Oct/10 10:18 AM
you see what's happening - the chain is just a path, which can be read right to left or left to right - just using the current possibility grid as a reference. At each x--y, or x==y, simply refer to the position and follow the path, verifying by inspection.

each sub-chain is itself an More...
27/Oct/10 10:40 AM
Good morning all.
27/Oct/10 12:50 PM
Thanks farpointer. It takes a while to sink in.I can see: Whether e12=16;OR e2=8=d7,d2=2;d2<>6.
I've noticed that the second term is always the false value in an AIC chain [(5)a9 ] . If you start with (5)a9=a2-(6=5)e9, there is no proper
continuation. How do you know which way to start?
Regards, Alfred.
27/Oct/10 5:12 PM
12 minutes on the dot. Hi everyone.
27/Oct/10 8:24 PM
18:03
02/Nov/10 12:18 PM
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