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 Author: pnm  [ Tue Jul 01, 2014 5:21 pm ] Post subject: 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?

 Author: jaek  [ Tue Jul 01, 2014 8:00 pm ] Post subject: 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.

 Author: pnm  [ Tue Jul 01, 2014 8:05 pm ] Post subject: 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'"

 Author: clm  [ Tue Jul 01, 2014 9:00 pm ] Post subject: 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.

 Author: beaker  [ Wed Jul 02, 2014 12:39 am ] Post subject: 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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