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sideli123
Joined: 20 Dec 2005 Posts: 7

Posted: Tue Dec 20, 2005 6:05 am Post subject: someone please help 


cant seem to solve this puzzle. will someone plz help
The * is where hint sez a 4 goes. i dont get it. I have been playing months and i dont see this.
65  2
3 8 75
2 53 6
36 9 24
1  9
92 4 6
2 75 48
46 1 
5 * 12 

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dotdot
Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Posted: Tue Dec 20, 2005 11:21 am Post subject: a shared pair 


Hallo sideli123
Look at how row1 and row3 intersect box2:
they both exclude {6} and {2} so {2,6} must be in the remaining cells r2c5..6.
Now look at how col6 intersects box2:
it excludes {1} and {9} from r1c6, so {1,9} must be in the remaining cells r1c4..5. (Note that r2c5, being {2,6}, is not eligible for a {1} or a {9}.)
Look at box2 yet again:
we have (final or) provisional assignments for every cell except r1c6 and r3c6. These must accomodate the remaining values {4,7} to satisfy box2 being {1..9}.
In summary, there are three pairs in box2: {1,9}, {2,6} and {4,7}.
The pair {4,7} is shared with column 6 and thus excludes {4,7} from the rest of column 6. That means r9c6 can't be 4, leaving only r9c5 available for box8's 4.
That was the machine's hint.
But that hint doesn't have to be the next move:
if, on identifying the {1,9}, we immediately looked for its effects as a shared pair (within row1) we might see that it excludes {9} from the r1c7, leaving only r8c7 available for col7's 9. 

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keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Tue Dec 20, 2005 2:17 pm Post subject: My next move 


My next move would be:
C6 has a pair {47}, so R6C6 cannot be 7.
C8 has a triple {367} so R6C8 cannot be 7.
Now R6C1 is pinned, and must be a 7.
As already noted, the first pair clears 4's as possibilities in Block 2: R1C5 and R2C5 cannot be 4, so R9C5 must be 4 (in C5).
Keith 

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sideli Guest

Posted: Sat Dec 24, 2005 4:00 am Post subject: 


Thanks, I see it now said the blind man. 

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