Question

I do have following problem to solve: I have a simple but big gridmap (Size: mX-times-mY). I only distinguish between a tile being occupied by an obstacle or being free.

Now let’s say I use a relatively small sliding window of size n-times-n, so n << mX, n << mY. For ALL possible combinations of occupied and free tiles in that n-times-n window, I assign a identifier (let’s just say a number). Then I take a “snapshot” of a part of my big map in size of that window.

My Question: What is the easiest and/or fastest method to determine the identification number of the current pattern extracted from the map?

To demonstrate: I have a 2×2 Window (so a 2×2 Matrix). There are 2^(2×2)=16 possible combinations of patterns.

Pattern    #1    #2    #3    #4    #5    #6    #7    #8

           00    #0    0#    00    00    ##    #0    00
           00    00    00    #0    0#    00    #0    ##

Pattern    #9    #10   #11   #12   #13   #14   #15   #16

           0#    #0    0#    0#    #0    ##    ##    ##
           0#    0#    #0    ##    ##    0#    #0    ##

with # = obscured
     0 = free

Assuming I extract the pattern

#0
0#

from my map, how can I easily and quickly identify that this is pattern number 10 in aboves example?

My Ideas so far:

1: Simple Looping

Loop over all patterns until I find the correct one, maybe by checking if the difference of both patterns is a zero matrix.

2: Using Row-Sum and Column-Sum as identifier

I sum all n rows and all n columns and use those sums as subindices in an multidimensional array.
For above example: sumRow1 = 1 sumRow2 = 1 sumCol1 = 1 sumCol2 = 1
So my pattern is saved in patternNumbers[1][1][1][1] = 10. Another example would be pattern 16, which would be saved at patternNumbers[2][2][2][2] = 16.

Possible advantage for bigger windows: I only have to calculate 2*n sums and immediately can use those sums to address my pattern I was looking for.
Big, big disadvantage: I need an (2*n)-dimensional Array in which many empty entrys will be left. So relatively big overhead.

Has anyone of you done this before? Any ideas how to solve this? I also did not take into account any sort of symmetry (rotational or line).

Any help is greatly appreciated!

Answer 1

Well, looks like I found a relatively easy solution which was pointed out by someone else to me:

By taking all rows or all columns (doesn’t matter, your choice!) in a specific order – which has to be the same for all patterns – and align them as one long vector, then you get a regular binary expression. Just convert that binary number into a decimal number and you have your “unique identifier”.

Example 1:

00
#0

can also be written as
00
10

now reorder the rows to

0010 = 2

=============================

Example 2:

##
00

can also be written as
11
00

now reorder the rows to

1100 = 12