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help for 6x6 difficult Oct04
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Posted on: Mon Oct 07, 2013 7:13 am

Posts: 7
Joined: Sun Nov 25, 2012 6:41 am
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.

Posted on: Mon Oct 07, 2013 9:25 am

Posts: 2246
Joined: Thu May 12, 2011 11:58 pm
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).

Posted on: Mon Oct 07, 2013 12:10 pm

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: help for 6x6 difficult Oct04
pluto wrote:
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.

It is also obvious that f4-f5 = [3,6] (addition in column f) and consequently d5-f5 is also [3,6] (addition in row 5). This means that "11+" (a5-b5-c5) = [2,4,5].

Next, the only hypothesis that is really required is this: A 6 is not possible in b6 (the 6 for row 6) because >>> "2-" (b1-b2-b3) = [1,2,5] (unique) >>> b5 = 4 >>> c5 = 5 and a 6 would become impossible in column "c". Thus c6 = 6 and, applying the parity rule to column "c", c5 must be odd then c5 = 5 (among 2,4,5 in cage "11+").

The rest is very simple: a5-b5 = [2,4] and a6-b6 = [3,4]. If a 1 is not in "2-" (a1-a2) it would be a4 = 1, b4 =2, "2-" (b1-b2-b3) = [1,3,6] (unique) and no 5's would be possible in column "b" so a1 = 1, a2 = 3 and a4 = 5 (the 5 for column "a") >>> b4 = 6 (being the 4's already in use in columns "a" and "b") (I am affraid this is not among your options, Patrick ). Also "2-" (b1-b2-b3) = [1,2,5] (unique, being b4 = 6).

And: b4 = 6 >>> f4 =3 >>> f5 = 6 >>> d5 = 3. The rest is completed quickly.

The graphic shows the invalid hypothesis with a 6 in b6:

Posted on: Mon Oct 07, 2013 1:05 pm

Posts: 2246
Joined: Thu May 12, 2011 11:58 pm
Re: help for 6x6 difficult Oct04
clm wrote:
It is also obvious that f4-f5 = [3,6] (addition in column f) and consequently d5-f5 is also [3,6] (addition in row 5). This means that "11+" (a5-b5-c5) = [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

Posted on: Tue Oct 08, 2013 7:35 am

Posts: 7
Joined: Sun Nov 25, 2012 6:41 am
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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