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 today's 4x4 difficult 
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Posted on: Thu Mar 27, 2014 9:04 pm




Posts: 2192
Joined: Thu May 12, 2011 11:58 pm
Post 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?

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Posted on: Thu Mar 27, 2014 11:53 pm




Posts: 690
Joined: Fri May 13, 2011 6:51 pm
Post 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.


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Posted on: Fri Mar 28, 2014 5:42 am




Posts: 37
Location: Karviná
Joined: Sun Feb 03, 2013 3:25 am
Post 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)…


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Posted on: Fri Mar 28, 2014 9:51 am




Posts: 2192
Joined: Thu May 12, 2011 11:58 pm
Post 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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