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help for 6x6 difficult Oct04 http://www.calcudoku.org/forum/viewtopic.php?f=16&t=485 
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Author:  pluto [ Mon Oct 07, 2013 7:13 am ] 
Post subject:  help for 6x6 difficult Oct04 
d6=5, e6=2, f6=1. These are obvious, but how to get further without resorting to trial and error has been bothering me off and on for several days. If anyone would care to give me a hint I'd be most appreciative. Thank you. 
Author:  pnm [ Mon Oct 07, 2013 9:25 am ] 
Post subject:  Re: help for 6x6 difficult Oct04 
After plodding along for a bit I get to the fact that a4b4 has to be either (1, 2), (4, 5), or (5, 4). I don't have a deterministic next step (yet). 
Author:  pnm [ Mon Oct 07, 2013 1:05 pm ] 
Post subject:  Re: help for 6x6 difficult Oct04 
clm wrote: It is also obvious that f4f5 = [3,6] (addition in column f) and consequently d5f5 is also [3,6] (addition in row 5). This means that "11+" (a5b5c5) = [2,4,5]. yesyes :) clm wrote: b4 = 6 (being the 4's already in use in columns "a" and "b") (I am affraid this is not among your options, Patrick ). argh, yes, that was sloppy of me (not enough coffee ). I somehow "derived" that the 3,6 in f4f5, and the "3 or 6" in d5 meant that there could not be a 6 in a4b4 
Author:  pluto [ Tue Oct 08, 2013 7:35 am ] 
Post subject:  Re: help for 6x6 difficult Oct04 
Thank you clm, and Patrick too. At the hypothesis stage I wasn't able to decide between [1,2,5] or [1,3,6] for "2, but now I see why it must be the former, and why b6 cannot be =6. Thanks again. 
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