Obstruction inferral

[enadeau]

Address issue #52

I'm still working on all obstruction inferral and the tiling methods.

[enadeau]

I'm done with the work on this. However, the implementation for all obstruction inferral is implemented very naively at the moment. Any suggestion to improve it is welcome. I might have one that we can discuss but I'm not sure it is worth our time at the moment.

[christianbean]

I would suggest doing

yield from Tiling(obstructions).gridded_perms_of_length(length)

It should achieve the same thing with far less explosion.

You probably want to set the tiling initialiser flags to False.

[enadeau]

I think that this suggestion doesn't work. Take the following code

t = Tiling(obstructions=[
    Obstruction(Perm((0, 1, 2)), ((0, 0),)*3),
    Obstruction(Perm((0, 1, 2)), ((1, 0),)*3),
    Obstruction(Perm((0, 1)), ((0, 0), (1, 0))),
    Obstruction(Perm((1, 0)), ((0, 0), (1, 0))),
], requirements=[[
    Requirement(Perm((0,)), ((1, 0),)),
]])
print(t)
print('Contained gridded permutations of length 1')
for p in t.gridded_perms_of_length(1):
    print(p)

The output is

+-+-+
|1|2|
+-+-+
1: Av(012)
2: Av+(012)
Crossing obstructions:
01: (0, 0), (1, 0)
10: (0, 0), (1, 0)
Requirement 0:
0: (1, 0)
Contained gridded permutations of length 1
0: (1, 0)

With your suggestion we would never try to add the obstruction 0: (0, 0)

[christianbean]

We would, note, I am suggesting initialising without the requirements.

[christianbean]

A thought:

Let T = (O, R) be a tiling, and T' = (O, emptyset).

Is Grid(T') - Grid(T) precisely those we want to add?

[christianbean]

This is obviously wrong... (e.g every one by one positive tiling becomes empty...).

However, all obstructions we can add do satisfy o in Grid(T') and o not in Grid(T).

[enadeau]

I pushed a more clever filtering that includes what we've discussed

[ulfarsson]

Looks good.