3
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enter image description here

  2 1 
 1 3 0
  2 1

Above honeycomb structure contains numbers.
A number at each cell represent a statement.

x : This cell is surrounded by x True statements.

So if the number is 3 means the cell is surrounded by 3 True statement
Some statements are true and some are not.

Create another honeycomb structure with boolean (T/F) input,
to show which statements are true, and which statements are false.

example

enter image description here

 0 2         F T
2 3 3   ->  F T T 
 0 2         F T

there is only 1 solution. Find the solution !

Note :

  • This is a variation from a puzzle I posted yesterday.
  • This puzzle is more simple, it designed to be solved without computer.
  • Credit only for answers with explaination how to solve it without computer.
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  • $\begingroup$ In the example, if I understand the rule, should the "left column" be 212 from top to bottom instead of 020? I mean, the cell in the extreme left is 2, but it touch a True on the right, and top right and bottom right are False, so it should be 1... I'm confused. $\endgroup$ – marcoresk Sep 21 '16 at 5:59
  • $\begingroup$ @marcoresk So they are false. (check the F sign in the solution) $\endgroup$ – Jamal Senjaya Sep 21 '16 at 6:01
  • $\begingroup$ Ok, I got it. Thank you! I wil try now $\endgroup$ – marcoresk Sep 21 '16 at 6:04
  • $\begingroup$ This looks a bit too easy for a puzzle. $\endgroup$ – Matsmath Sep 21 '16 at 6:05
  • $\begingroup$ Bite-size is fine, especially when introducing a new kind of puzzle. What's next, 3-D? $\endgroup$ – humn Sep 21 '16 at 18:52
4
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We can see that

They cant all be False, since 0 would then be True
It can't only be one that is True, since it would have to be the 0, but then the two neighbouring 1s would be True
It can't be exactly 2 that are True, since we'd need two neighbouring 1s
For exactly 3 to be True the middle would be impossible
For exactly 4 to be True the middle would have to be one of them
Now we can see that the three 1s and the 3 could all be True and satisfy the two 2s and the 0 all being False and we are done:

  F T 
 T T F
  F T

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  • $\begingroup$ Nice reasoning but why don't you discuss uniqueness? $\endgroup$ – Matsmath Sep 21 '16 at 6:14
  • $\begingroup$ @Matsmath The puzzle stated was to find the solution, not to show that it is indeed unique. $\endgroup$ – Jonathan Allan Sep 21 '16 at 6:16
1
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     2 1         F T
    1 3 0   ->  T T F 
     2 1         F T

Honestly, I guessed

based on central-symmetry considerations.

I don't know if there are other solutions.

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  • 1
    $\begingroup$ I'm late to post but yeah I got the same answer and I don't think it could work any other way. $\endgroup$ – Xenocacia Sep 21 '16 at 6:15
  • $\begingroup$ @Matsmath You are the first, but lack explaination, $\endgroup$ – Jamal Senjaya Sep 21 '16 at 6:18
1
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Answer:

F(2) T(1)
T(1) T(3) F(0)
F(2) T(1)

Proof of one solution.

Fine a cells's T/F

Assume the "2" in the upper left corner is "T". That can't be from both the 3 and the 1 (either 1), the 1 touches 2+3 so the 1 would then be false. Ergo both 1's would have to be true and the 3 false. The 1's are already satisfied by the original 2, ergo the 0 and the other 2 are false, and the other 1 is true (because of the 3).... however this conflicts with the 2 in the bottom left.

So the "2" in the upper left corner can NOT be "T", it must be "F".

Find another cell's T/F

Now assume the "2" in the bottom left corner is "T". Again, you can't do that with both a 1 and 3, so both 1s must be True and the 3 False. The 3 must be false but it's already touching three Trues, the 0 must be False (because the True 1 gets that from the 2), so the Upper Right 1 must be True... but it's not touching anything True so it must also be False.

So the "2" in the lower left corner can NOT be "T", it must be "F".

Find a third cell's T/F

Assume the leftmost "1" is False, both 2's are already False so the 3 must also be False. There are only three remaining cells, they can't all be False because of the 0, they can't all be True because that would make the 3 True, the 1's can't be True unless the Zero is True which is nonsense, so both 1's have to be False, which again means the Zero is touching nothing True but can't be True itself.

So the leftmost "1" is True.

And now find the rest.

Which makes the "3" True.

Which makes the "0" False.

There are only two remaining cells so both 1's must be True because of the 3.

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