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LegionMammal978yesterday at 3:35 PM0 repliesview on HN

It looks like the author has a pretty simple procedure for computing the 'identity' sandpile (which they unfortunately don't describe at all):

1. Fill a grid with all 6s, then topple it.

2. Subtract the result from a fresh grid with all 6s, then topple it.

So effectively it's computing 'all 6s' - 'all 6s' to get an additive identity. But I'm not entirely sure how to show this always leads to a 'recurrent' sandpile.

EDIT: One possible route: The 'all 3s' sandpile is reachable from any sandpile via a sequence of 'add 1' operations, including from its own successors. Thus (a) it is a 'recurrent' sandpile, (b) adding any sandpile to the 'all 3s' sandpile will create another 'recurrent' sandpile, and (c) all 'recurrent' sandpiles must be reachable in this way. Since by construction, our 'identity' sandpile has a value ≥ 3 in each cell before toppling, it will be a 'recurrent' sandpile.