ENIGMA 1520

The problem is to find how large a fishpond can be equipped with a surround 2 flags wide, so that no pattern of 4 flags repeats itself when there are just 3 different colours of slabs.
It was agreed that there are 24 different patterns of slabs, but the patterns overlap, so that a new pattern is involved in each slab-width of the pond’s perimeter. Thus the perimeter is 24 slab-widths, and 48 slabs are used . . .
. . . always assuming that it is actually possible to arrange the pattersn to fit.

Computer research shows that there are over a billion solutions, and here are 3 of them:

  3     1     2     3     3  
  1     3     3     1     3  
  2     1     2     2  
  3     2     2     2  
  1     1     2     3  
  3     1     2     3  
  2     2     3     2  
  1     2     3     3  
  1     1     3     3  
  1     1     1     1     2  
  3     3     2     2     1  
         
9 vertical3 horizontal
0: 2 1 3 3   12: 1 3 1 2
1: 3 3 3 3   13: 2 1 2 3
2: 3 3 3 2   14: 3 2 1 1
3: 2 3 2 3   15: 1 1 1 3
4: 3 2 2 3   16: 3 1 2 2
5: 3 2 2 2   17: 2 2 2 1
6: 2 2 2 2   18: 1 2 1 1
7: 2 2 1 3   19: 1 1 1 1
8: 3 1 3 3   20: 1 1 3 3
9: 3 1 3 2   21: 3 1 1 2
10: 2 3 3 1   22: 2 1 1 2
11: 1 3 1 3   23: 2 1 2 1
  1     3     2     3     3     2     2     2  
  3     2     1     3     1     2     2     3  
  1     1     2     3  
  2     2     3     2  
  1     3     3     3  
  3     1     3     3  
  1     2     3     3     1     1     1     2  
  2     1     1     1     1     1     2     2  
         
6 vertical6 horizontal
0: 2 1 3 3   12: 3 2 1 1
1: 3 3 3 3   13: 1 1 2 2
2: 3 3 3 2   14: 2 2 3 1
3: 2 3 2 3   15: 1 3 1 3
4: 3 2 2 3   16: 3 1 2 1
5: 3 2 2 2   17: 1 2 1 2
6: 2 2 2 2   18: 1 2 3 1
7: 2 2 1 3   19: 1 3 3 1
8: 3 1 3 3   20: 1 3 1 1
9: 3 3 1 2   21: 1 1 1 1
10: 2 1 2 3   22: 1 1 1 2
11: 3 2 3 1   23: 2 1 2 2
  3     2     1     3     3     2     2  
  1     3     2     3     1     2     2  
  3     1     2     3  
  3     2     2     3  
  1     1     3     2  
  2     2     3     3  
  1     3     3     3  
  2     1     2     3     1     1     1  
  2     2     1     1     1     1     2  
         
7 vertical5 horizontal
0: 1 1 3 3   12: 1 3 1 3
1: 3 3 3 3   13: 3 1 2 3
2: 3 3 3 2   14: 3 2 1 1
3: 2 3 2 3   15: 1 1 2 2
4: 3 2 2 3   16: 2 2 3 1
5: 3 2 2 2   17: 1 3 1 2
6: 2 2 2 2   18: 2 1 2 2
7: 2 2 1 3   19: 2 1 2 1
8: 3 1 3 3   20: 1 2 3 1
9: 3 3 2 1   21: 1 3 1 1
10: 1 2 3 2   22: 1 1 1 1
11: 2 3 1 3   23: 1 1 1 2

Here is the program that produces the solution.
Here is the program’s output. It was eventually abandonned after running for 3 or 4 days on an Intel Pentium III (664.56-MHz).