11
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Here is an interesting game. You start with an empty 4x4 grid. At each turn you can choose an empty cell and place a value in it. The placed value is given by the following rules:

  • If the chosen cell has no neighboring (horizontal or vertical) values then the placed value is 1.
  • Otherwise the placed value is the sum of all neighboring (horizontal or vertical) values.

What is the largest value that you can achieve in this game?

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2
9
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The best solution my computer found is

41

I suspect this is optimal.

  1  1  2  1
  5  4  3  1
  5  9 26 27
  5 14 14 41

I won't list the order, as you can simply choose the squares according to the increasing order of the numbers.

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2
  • $\begingroup$ "as you can simply choose the squares according to the increasing order of the numbers." This doesn't seem to be true to me. In particular, I suspect which 5 you place first matters a lot. $\endgroup$
    – NPSF3000
    Dec 4 '20 at 6:22
  • $\begingroup$ Well done! I have verified that this is the optimal answer. There is another solution that achieves the same score, but uses a slightly different path that does not involve number 27. $\endgroup$ Dec 4 '20 at 6:53
10
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Here is my answer:

38

as here

enter image description here

the order is as below

enter image description here

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6
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My first 3 tries, in the order of increasing score:

Score 20. Simply going row-wise, which produces a part of Pascal's triangle:

1  1  1  1
1  2  3  4
1  3  6  10
1  4  10 20

Score 32. An approach similar to the above, but placing ones at (1,3) and (3,1) first to boost the middle:

1  2  1  1
2  4  5  6
1  5  10 16
1  6  16 32

Score 37. An approach that utilizes Fibonacci sequence. Place the numbers in this order

1  3  4  5
16 2  6  7
15 14 10 8
13 12 11 9

which gives this:

1  2  2  2
37 1  3  5
35 22 8  5
13 13 13 5

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  • 1
    $\begingroup$ Love your work! This puzzle turned out more interesting than I realized :) $\endgroup$ Dec 3 '20 at 4:52
  • 2
    $\begingroup$ Damn, I just got the same result, then looked up to see that you'd just posted this :-( $\endgroup$
    – lxop
    Dec 3 '20 at 4:52
1
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I found three solutions that all achieve the following score, with more or less the same chain but finishing in different positions:

41

corner:

 41 27  1  1
 14 26  3  2
 14  9  4  1
  5  5  5  1

edge:

 14 41  1  1
 14 26  3  2
 14  9  4  1
  5  5  5  1

interior:

 14 15  1  1
 14 41  3  2
 14  9  4  1
  5  5  5  1

A computer search determined that this score is optimal.

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1
  • $\begingroup$ Nice work! I haven't seen the interior solution before. $\endgroup$ Dec 6 '20 at 1:53

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