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march 12 9x9 cages of 27+ and 31+
http://www.calcudoku.org/forum/viewtopic.php?f=3&t=170
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Author:  beaker  [ Tue Mar 13, 2012 1:39 am ]
Post subject:  march 12 9x9 cages of 27+ and 31+

Need a solving strategy for determination of cells in the 27+ and 31+ cages in the 9x9 of March 12.....any help would be appreciated........Ken

Author:  clm  [ Tue Mar 13, 2012 2:28 am ]
Post subject:  Re: march 12 9x9 cages of 27+ and 31+

beaker wrote:
Need a solving strategy for determination of cells in the 27+ and 31+ cages in the 9x9 of March 12.....any help would be appreciated........Ken


If you continue a little bit from the process I have sent (in the other thread) for that puzzle you will quickly determine those cages, i.e.: "14+" = [6,8] >>> d9 = 6, e9 = 8, e5 = 4 (d5 = 7) >>> "7+" = [1,6] >>> h5 = 6, i5 = 1, g1 = 6.
Also g2 = 3 (g3 = 8), c1 <> 4 (because it would not be possible c2 + d2 = 12, that is, the combinations 3-9, 4-8 or 5-7). Then the 4 of row 1 goes to h1.
Also c1 <> 9 (it is not possible to obtain 7 with c2 + d2 due to f2 = 2, g2 = 3 and h2 = 1) so i1 = 9. Now i2 + i3 = 8 >>> i2-i3 = [3,5] >>> i2 = 5, i3 = 3. You have totally finished the cage "27+". Additionally c1 = 3.
Now, i6-i7-i8 = [2,7,8] >>> g8 + h8 = 14 >>> g8-h8 = [5,9] (8's in g3 and h4) then h8 = 5, g8 = 9. Also i6 = 2, etc. ...(i.e., d3 = 1, d4 = 3, c4 = 7, ... ).

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