How to implement a union-find (disjoint set) data structure in Coq? - data-structures

Question

I am quite new to Coq, but for my project I have to use a union-find data structure in Coq. Are there any implementations of the union-find (disjoint set) data structure in Coq?
If not, can someone provide an implementation or some ideas? It doesn't have to be very efficient. (no need to do path compression or all the fancy optimizations) I just need a data structure that can hold an arbitrary data type (or nat if it's too hard) and perform: union and find.
Thanks in advance

Answer 1

If all you need is a mathematical model, with no concern for actual performance, I would go for the most straightforward one: a functional map (finite partial function) in which each element optionally links to another element with which it has been merged.
If an element links to nothing, then its canonical representative is itself.
If an element links to another element, then its canonical representative is the canonical representative of that other element.
Note: in the remaining of this answer, as is standard with union-find, I will assume that elements are simply natural numbers. If you want another type of elements, simply have another map that binds all elements to unique numbers.
Then you would define a function find : UnionFind → nat → nat that returns the canonical representative of a given element, by following links as long as you can. Notice that the function would use recursion, whose termination argument is not trivial. To make it happen, I think that the easiest way is to maintain the invariant that a number only links to a lesser number (i.e. if i links to j, then i > j). Then the recursion terminates because, when following links, the current element is a decreasing natural number.
Defining the function union : UnionFind → nat → nat → UnionFind is easier: union m i j simply returns an updated map with max i' j' linking to min i' j', where i' = find m i and j' = find m j.
[Side note on performance: maintaining the invariant means that you cannot adequately choose which of a pair of partitions to merge into the other, based on their ranks; however you can still implement path compression if you want!]
As for which data structure exactly to use for the map: there are several available.
The standard library (look under the title FSets) has several implementations (FMapList, FMapPositive and so on) satisfying the interface FMapInterface.
The stdpp libray has gmap.
Again if performance is not a concern, just pick the simplest encoding or, more importantly, the one that makes your proofs the simplest. I am thinking of just a list of natural numbers.
The positions of the list are the elements in reverse order.
The values of the list are offsets, i.e. the number of positions to skip forward in order to reach the target of the link.
For an element i linking to j (i > j), the offset is i − j.
For a canonical representative, the offset is zero.
With my best pseudo-ASCII-art skills, here is a map where the links are { 6↦2, 4↦2, 3↦0, 2↦1 } and the canonical representatives are { 5, 1, 0 }:
6 5 4 3 2 1 0 element
↓ ↓ ↓ ↓ ↓ ↓ ↓
/‾‾‾‾‾‾‾‾‾↘
[ 4 ; 0 ; 2 ; 3 ; 1 ; 0 ; 0 ] map
\ \____↗↗ \_↗
\___________/
The motivation is that the invariant discussed above is then enforced structurally. Hence, there is hope that find could actually be defined by structural induction (on the structure of the list), and have termination for free.
A related paper is: Sylvain Conchon and Jean-Christophe Filliâtre. A Persistent Union-Find Data Structure. In ACM SIGPLAN Workshop on ML.
It describes the implementation of an efficient union-find data structure in ML, that is persistent from the user perspective, but uses mutation internally. What may be more interesting for you, is that they prove it correct in Coq, which implies that they have a Coq model for union-find. However, this model reflects the memory store for the imperative program that they seek to prove correct. I’m not sure how applicable it is to your problem.

Answer 2

Maëlan has a good answer, but for an even simpler and more inefficient disjoint set data structure, you can just use functions to nat to represent them. This avoids any termination stickiness. In essence, the preimages of any total function form disjoint sets over the domain. Another way of looking at this is as representing any disjoint set G as the curried application find_root G : nat -> nat since find_root is the essential interface that disjoint sets provide.
This is also analogous to using functions to represent Maps in Coq like in Software Foundations. https://softwarefoundations.cis.upenn.edu/lf-current/Maps.html
Require Import Arith.
Search eq_nat_decide.
(* disjoint set *)
Definition ds := nat -> nat.
Definition init_ds : ds := fun x => x.
Definition find_root (g : ds) x := g x.
Definition in_same_set (g : ds) x y :=
eq_nat_decide (g x) (g y).
Definition union (g : ds) x y : ds :=
fun z =>
if in_same_set g x z
then find_root g y
else find_root g z.
You can also make it generic over the type held in the disjoint set like so
Definition ds (a : Type) := a -> nat.
Definition find_root {a} (g : ds a) x := g x.
Definition in_same_set {a} (g : ds a) x y :=
eq_nat_decide (g x) (g y).
Definition union {a} (g : ds a) x y : ds a :=
fun z =>
if in_same_set g x z
then find_root g y
else find_root g z.
To initialize the disjoint set for a particular a, you need an Enum instance for your type a basically.
Definition init_bool_ds : ds bool := fun x => if x then 0 else 1.
You may want to trade out eq_nat_decide for eqb or some other roughly equivalent thing depending on your proof style and needs.