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today's 4x4 difficult http://www.calcudoku.org/forum/viewtopic.php?f=16&t=575 
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Author:  pnm [ Thu Mar 27, 2014 9:04 pm ] 
Post subject:  today's 4x4 difficult 
Of the regular puzzles I enjoy doing the 4x4 difficult, trying to find the shortest "reasoning only" path to a solution. So what is the shortest path for today's? 
Author:  clm [ Thu Mar 27, 2014 11:53 pm ] 
Post subject:  Re: today's 4x4 difficult 
pnm wrote: Of the regular puzzles I enjoy doing the 4x4 difficult, trying to find the shortest "reasoning only" path to a solution. So what is the shortest path for today's? .... Let's see this: Since a1 + a2 + a3 = 9 (addition rule to column 1) >>> b2 = 2 (= 11  9) >>> b4 = 3, c4 = 2. Since d2 + b3 + c3 + d3 = 10 (= 15  3  2) and, at the same time, a3 + b3 +c 3 + d3 = 10, we have d2 = a3 = 3, because the "3" is the only number that can be placed simultaneously in d2 and a3. The rest is very fast: row 2: c2 = 1, a2 =4 column 2: c1 = 3, c3 =4, ... , the rest is obvious. 
Author:  kozibrada [ Fri Mar 28, 2014 5:42 am ] 
Post subject:  Re: today's 4x4 difficult 
clm wrote: Since d2 + b3 + c3 + d3 = 10 (= 15  3  2) and, at the same time, a3 + b3 +c 3 + d3 = 10, we have d2 = a3 = 3, because the "3" is the only number that can be placed simultaneously in d2 and a3. I solved this step differently. b2 = 2, of course. Before b4 and c4 I focused on 2 upper cages. They have a sum 20 (= 2 rows), thus d2 = a3. However, clm’s practice is probably better of the analytical view (1 row vs. 2 "my" rows)… 
Author:  pnm [ Fri Mar 28, 2014 9:51 am ] 
Post subject:  Re: today's 4x4 difficult 
clm wrote: Let's see this: Since a1 + a2 + a3 = 9 (addition rule to column 1) >>> b2 = 2 (= 11  9) >>> b4 = 3, c4 = 2. Since d2 + b3 + c3 + d3 = 10 (= 15  3  2) and, at the same time, Those were my steps too, but after that I like: because those cells sum to 10, there can not be two "1"'s in there, so the 9+ cage must have two, so c2=1 
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