CODEWITHSUNDEEP
javascriptjavac#arraysalgorithm
Chron Bag
Chron Bag

Reputation: 587

More elegant way of finding neighbors in 1D array grid with wrap-around?

Let's say I have a 1D game grid like the following:

1D game grid

As an example, the cell south of 27 will be cell 35, and the cell south of 59 will be 3 (because of wrap-around). This can be implemented like so:

var s = spot = 59
var r = row length = 8
var b = board size = 64
var south = (s+r) mod b

Okay, now let's try to find the cell east of another. The cell east of 27 is 28, and the cell east of 31 is 24 (also because of wrap-around). The best I could come up with is this:

var s = spot = 31
var r = row length = 8
var lc = left column = Math.floor(s / 8) * 8
var east = lc + ((s - lc + 1) % 8)

This is considerably more complex, which makes me think I'm missing something obvious. Is there not a better way to do this?

Also, I haven't yet implemented, but I'd imagine finding the diagonal cells like north-east and south-east even more complex still.

For the purposes of this question please assume this is limited to 1D arrays. Also, I'd imagine there is a bit-wise solution that could be more elegant for board sizes that are a power of two, but preferably the solution would work for any board size.

Upvotes: 3

Views: 1780

Answers (2)

Chron Bag
Chron Bag

Reputation: 587

Here's a list of different methods I found for calculating east (in Javascript):

east = s + 1 - (s + 1 & 7 ? 0 : 8)

east = s + 1 - ((s % 8) / 7 | 0) * 8

east = (s + 1) % 8 == 0 ? Math.floor(s / 8) * 8 : s + 1

east = (s & 0xFFF8) + ((s - (s & 0xFFF8) + 1) % 8)

lc = Math.floor(s / 8) * 8; east = lc + ((s - lc + 1) % 8)

Upvotes: 0

MBo
MBo

Reputation: 80287

east: 

i + 1 - ((i mod 8) div 7) * 8

west:

(i - 1) +  (((i - 1) mod 8) div 7) * 8
or
(i - 1) +  (((i + 7) mod 8) div 7) * 8
to avoid potential problems with negative dividend modulo in some languages

Upvotes: 1

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