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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 link3.for each deduction--- verify that it is weak with two points, i, j on the chain i == x -- y == j4.if the chain is looped, deductions can be made on any pair i, ji -- 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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