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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?

Image

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 [smile]

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