MHB How can you solve the eight-digit challenge?

[soroban]

Place the digits 1 through 8 in the boxes
so that no two consecutive digits are adjacent
(not vertically, horizontally or diagonally).
. . \begin{array}{cccccccccc}&& * & - & * & - & * \\ && | && | && | \\ * &-& * &-& * &-& * &-& * \\ | && | && | && | && | \\ * &-& * &-& * &-& * &-& * \\ && | && | && | \\ && * & - & * & - & * \end{array}

 

[Nono713]

Spoiler

The two central boxes are adjacent to the most other boxes, so it makes sense to put 1 and 8 in them to maximize our options (since 0 and 9 are not valid digits). By horizontal symmetry, it doesn't really matter what order we put them in. So we start with:​

  ? ?
? 1 8 ?
  ? ?

Here we automatically deduce that 7 must go in the leftmost box and 2 in the rightmost box, otherwise there would be two adjacent consecutive numbers. Hence:

  ? ?
7 1 8 2
  ? ?

Hence we have 3, 4, 5, and 6 left to place. We see that 6 must be in one of two boxes on the right, and 3 must be in one of the boxes on the left. Furthermore, 4 and 5 cannot be put adjacent to one another. This forces the following configuration (up to symmetry):

  3 5
7 1 8 2
  4 6

And the puzzle is solved.