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good analytical solution for a specific 4x4 difficult?
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Posted on: Tue Jul 01, 2014 5:21 pm

Posts: 2246
Joined: Thu May 12, 2011 11:58 pm
good analytical solution for a specific 4x4 difficult?
in particular, one of the harder ones of June: http://www.calcudoku.org/en/2014-06-09/4/3
(more than a week old, so that link won't give you the puzzle if you're not a subscriber)

Here it is:

I was thinking along these lines:
- d1d2 can either be 1,2 => not possible, because then (20 - 11 - 3) = 6 is impossible,
or 2,3 => then c1 = 4, or 3,4, then c1 = 2
- either way, there must be a 1 in d3 or d4
- if c1 = 4, then a1b1b2 = 1,2,1, which is impossible because of the 7+ cage
- so c1 = 2
- so d3d4 = 1/2, and d1d2 = 3/4
- a2 + a3 = 7, so a2a3 = 3/4
- so a1=1, a4=2
- if a3=3, then c2 = 1, if a3=4, then c2 = 2, which is impossible, so c2 = 1
- so b1b2 = 3,2
- etc.

Any cleaner paths possible?

Posted on: Tue Jul 01, 2014 8:00 pm

Posts: 225
Joined: Fri Jun 17, 2011 8:15 pm
Re: good analytical solution for a specific 4x4 difficult?
The first thing I did was note that 40 - 8 - 7 - 11 = 14. From that
- both a2a3 and d1d2 have to be 3,4.
- d3d4 is 1,2 ==> c2c3c4 = 8 thus 1,3,4 ==> c1 = 2
- a1 = 1, a4=2
- a2=3,4 and c2=1,3,4 and d2=3,4 ==> b2 = 2 and b1 = 3
- a4=2 ==> d4 = 1, d3 = 2
- b4 has to be 4
- etc.

Posted on: Tue Jul 01, 2014 8:05 pm

Posts: 2246
Joined: Thu May 12, 2011 11:58 pm
Re: good analytical solution for a specific 4x4 difficult?
jaek wrote:
The first thing I did was note that 40 - 8 - 7 - 11 = 14. From that
- both a2a3 and d1d2 have to be 3,4.

Ah... of course. I'm just using the "row/column adds up to 10" a few times,
but never considered "40 - '+ cages'"

Posted on: Tue Jul 01, 2014 9:00 pm

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: good analytical solution for a specific 4x4 difficult?
pnm wrote:
jaek wrote:
The first thing I did was note that 40 - 8 - 7 - 11 = 14. From that
- both a2a3 and d1d2 have to be 3,4.

Ah... of course. I'm just using the "row/column adds up to 10" a few times,
but never considered "40 - '+ cages'"

Same line as jaek, you can only have a total of 14 for both cages "1-" with the "maximums" [3,4] in both cages simultaneously, btw it's very frequent to add all cages in this type of 4x4's. Then, since 8 + 7 + 7 = 22, i.e., the excess over the sum of the two leftmost columns gives c1 = 2. Now, since a2 + c2 = 5 (two upmost rows: 20 - 8 - 7) and a2 = 2, c2 = 3 is not valid because a 2 is not inside cage "1-" in a2-a3 ... it must be a2 = 4, c2 = 1 and the rest of the puzzle is immediate.

Posted on: Wed Jul 02, 2014 12:39 am

Posts: 523
Location: Ladysmith, BC, Canada
Joined: Fri May 13, 2011 1:37 am
Re: good analytical solution for a specific 4x4 difficult?
Thank you jaek (as well as clm) as I used your method on today's puzzle and solved it quite quickly......in the past, these have been very bothersome for me and usually take in excess of 5 minutes to solve if I don't get a lucky guess with which to begin.

I was able to solve the yesterday's 6x6 using the same type of logic you presented for the 4x4 and once again a puzzle that would cause me to take a lot of time was solved quickly....thanx again.

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