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Write more on finite height lattices

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Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>
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Danila Fedorin 2024-07-05 12:09:29 -07:00
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content/blog/02_spa_agda_combining_lattices.md
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@ -425,7 +425,7 @@ we can create the infinite chain:
On the other hand, our sign lattice _is_ of finite height; the longest chains
we can make have three elements and two `<` signs. Here's one:
{#sign-length-three}
{#sign-three-elements}
{{< latex >}}
\bot < + < \top

230
content/blog/03_spa_agda_fixed_height.md
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@ -21,8 +21,8 @@ infinitely long:
There isn't a "biggest natural number"! On the other hand, we've seen that our
[sign lattice]({{< relref "01_spa_agda_lattices#sign-lattice" >}}) has a finite
height; the longest chain we can make is three elements long; I showed one
such chain (there are many chains of length three) in
[the previous post]({{< relref "02_spa_agda_combining_lattices#sign-length-three" >}}),
such chain (there are many chains of three elements) in
[the previous post]({{< relref "02_spa_agda_combining_lattices#sign-three-elements" >}}),
but here it is again:
{{< latex >}}
@ -38,11 +38,13 @@ chains have five elements. The following is one example:
{{< latex >}}
(\bot, \bot) < (\bot, +) < (\bot, \top) < (+, \top) < (\top, \top)
{{< /latex >}}
{#sign-prod-chain}
The fact that \(L_1\) and \(L_2\) are themselves finite-height lattices is
important; if either one of them is not, we can easily construct an infinite
chain of the products. If we allowed \(L_2\) to be natural numbers, we'd
end up with infinite chains like this one:
{#product-both-finite-height}
{{< latex >}}
(\bot, 0) < (\bot, 1) < (\bot, 2) < ...
@ -76,6 +78,230 @@ talk about them in a subsequent post.
In the meantime, let's dig deeper into the notion of finite height, and
the Agda proofs of the properties I've introduced thus far.
### Formalizing Finite Height
The formalization I settled on is quite similar to the informal description:
a lattice has a finite height of length \(h\) if the longest chain
of elements compared by \((<)\) is exactly \(h\). There's only a slight
complication: we allow for equivalent-but-not-equal elements in lattices.
For instance, for a map lattice, we don't care about the order of the keys:
so long as two maps relate the same set of keys to the same respective values,
we will consider them equal. This, however, is beyond the notion of Agda's
propositional equality (`_≡_`). Thus, we we need to generalize the definition
of a chain to support equivalences. I parameterize the `Chain` module
in my code by an equivalence relation, as well as the comparison relation `R`,
which we will set to `<` for our chains. The equivalence relation and `R`/`<`
are expected to play together nicely (if `a < b`, and `a` is equivalent to `c`,
then it should be the case that `c < b`).
{{< codelines "agda" "agda-spa/Chain.agda" 3 7 >}}
From there, the definition of the `Chain` data type is much like the definition
of a vector, but indexed by the endpoints, and containing witnesses of `R`/`<`
between its elements. The indexing allows for representing
the type of chains between particular lattice elements, and serves to ensure
concatenation and other operations don't merge disparate chains.
{{< codelines "agda" "agda-spa/Chain.agda" 19 21 >}}
In the `done` case, we create a single-element chain, which has no comparisons.
In this case, the chain starts and stops at the same element (where "the same"
is modulo our equivalence). The `step` case prepends a new comparison `a1 < a2`
to an existing chain; once again, we allow for the existing chain to start
with a different-but-equivalent element `a2'`.
With that definition in hand, I define what it means for a type of
chains between elements of the lattice `A` to have a maximum height; simply
put, all chains must have length less than or equal to the maximum.
{{< codelines "agda" "agda-spa/Chain.agda" 38 39 >}}
Though `Bounded` specifies _a_ bound on the length of chains, it doesn't
specify _the_ (lowest) bound. Specifically, if the chains can only have
length three, they are bounded by both 3, 30, and 300. To claim a lowest
bound, we need to show that a chain of that length actually exists (otherwise,
we could take the previous natural number, and it would be a bound as well).
Thus, I define the `Height` predicate to require that a chain of the desired
height exists, and that this height bounds the length of all other chains.
{{< codelines "agda" "agda-spa/Chain.agda" 47 48 >}}
Finally, for a lattice to have a finite height, the type of chains formed by using
its less-than operator needs to have that height (satisfy the `Height h` predicate).
To avoid having to thread through the equivalence relation, congruence proof,
and more, I define a specialized predicate for lattices specifically.
I do so as a "method" in my `IsLattice` record.
{{< codelines "agda" "agda-spa/Lattice.agda" 153 180 "hl_lines = 27 28">}}
Thus, bringing the operators and other definitions of `IsLattice` into scope
will also bring in the `FixedHeight` predicate.
### Fixed Height of the "Above-Below" Lattice
We've already seen intuitive evidence that the sign lattice --- which is an instance of
the ["above-below" lattice]({{< relref "01_spa_agda_lattices#the-above-below-lattice" >}}) ---
has a fixed height. The reason is simple: we extended a set of incomparable
elements with a single element that's greater, and a single element that's lower.
We can't make chains out of incomparable elements (since we can't compare them
using `<`); thus, we can only have one `<` from the new least element, and
one `<` from the new greatest element.
The proof is a bit tedious, but not all that complicated.
First, a few auxiliary helpers; feel free to read only the type signatures.
They specify, respectively:
1. That the bottom element \(\bot\) of the above-below lattice is less than any
concrete value from the underlying set. For instance, in the sign lattice case, \(\bot < +\).
2. That \(\bot\) is the only element satisfying the first property; that is,
any value strictly less than an element of the underlying set must be \(\bot\).
3. That the top element \(\top\) of the above-below lattice is greater than
any concrete value of the underlying set. This is the dual of the first property.
4. That, much like the bottom element is the only value strictly less than elements
of the underlying set, the top element is the only value strictly greater.
{{< codelines "agda" "agda-spa/Lattice/AboveBelow.agda" 315 335 >}}
From there, we can construct an instance of the longest chain. Actually,
there's a bit of a hang-up: what if the underlying set is empty? Concretely,
what if there were no signs? Then we could only construct a chain with
one comparison: \(\bot < \top\). Instead of adding logic to conditionally
specify the length, I simply require that the set is populated by requiring
a witness
{{< codelines "agda" "agda-spa/Lattice/AboveBelow.agda" 85 85 >}}
I use this witness to construct the two-`<` chain.
{{< codelines "agda" "agda-spa/Lattice/AboveBelow.agda" 339 340 >}}
The proof that the length of two -- in terms of comparisons -- is the
bound of all chains of `AboveBelow` elements requires systematically
rejecting all longer chains. Informally, suppose you have a chain of
three or more comparisons.
1. If it starts with \(\top\), you can't add any more elements since that's the
greatest element (contradiction).
2. If you start with an element of the underlying set, you could add another
element, but it has to be the top element; after that, you can't add any
more (contradiction).
3. If you start with \(\bot\), you could arrive at a chain of two comparisons,
but you can't go beyond that (in three cases, each leading to contradictions).
{{< codelines "agda" "agda-spa/Lattice/AboveBelow.agda" 342 355 "hl_lines=8-14">}}
Thus, the above-below lattice has a length of two comparisons (or alternatively,
three elements).
{{< codelines "agda" "agda-spa/Lattice/AboveBelow.agda" 357 358 >}}
And that's it.
### Fixed Height of the Product Lattice
Now, for something less tedious. We saw above that for a product lattice
to have a finite height,
[its constituent lattices must have a finite height](#product-both-finite-height).
The proof was by contradiction (by constructing an infinitely long product
chain given a single infinite lattice). As a result, we'll focus this
section on products of two finite lattices `A` and `B`. Additionally, for the
proofs in this section, I require element equivalence to be decidable.
{{< codelines "agda" "agda-spa/Lattice/Prod.agda" 115 117 >}}
Let's think about how we might go about constructing the longest chain in
a product lattice. Let's start with some arbitrary element \(p_1 = (a_1, b_1)\).
We need to find another value that isn't equal to \(p_1\), because we'rebuilding
chains of the less-than operator \((<)\), and not the less-than-or-equal operator
\((\leq)\). As a result, we need to change either the first component, the second
component, or both. If we're building "to the right" (adding bigger elements),
the new components would need to be bigger. Suppose then that we came up
with \(a_2\) and \(b_2\), with \(a_1 < a_2\) and \(b_1 < b_2\). We could then
create a length-one chain:
{{< latex >}}
(a_1, b_1) < (a_2, b_2)
{{< /latex >}}
That works, but we can construct an even longer chain by increasing only one
element at a time:
{{< latex >}}
(a_1, b_1) < (a_1, b_2) < (a_2, b_2)
{{< /latex >}}
We can apply this logic every time; the conclusion is that when building
up a chain, we need to increase one element at a time. Then, how many times
can we increase an element? Well, if lattice `A` has a height of two (comparisons),
then we can take its lowest element, and increase it twice. Similarly, if
lattice `B` has a height of three, starting at its lowest element, we can
increase it three times. In all, when building a chain of `A × B`, we can
increase an element five times.
This gives us a recipe for constructing
the longest chain in the product lattice: take the longest chains of `A` and
`B`, and start with the product of their lowest elements. Then, increase
the elements one at a time according to the chains. The simplest way to do
that might be to increase by all elements of the `A` chain, and then
by all of the elements of the `B` chain (or the other way around). That's the
strategy I took when [constructing the \(\text{Sign} \times \text{Sign}\)
chain above](#sign-prod-chain).
To formalize this notion, a few lemmas. First, given two chains where
one starts with the same element another ends with, we can combine them into
one long chain.
{{< codelines "agda" "agda-spa/Chain.agda" 31 33 >}}
More interestingly, given a chain of comparisons in one lattice, we are
able to lift it into a chain in another lattice by applying a function
to each element. This function must be monotonic, because it must not
be able to reverse \(a < b\) such that \(f(b) < f(a)\). Moreover, this function
should be injective, because if \(f(a) = f(b)\), then a chain \(a < b\) might
be collapsed into \(f(a) \not< f(a)\), changing its length. Finally,
the function needs to produce equivalent outputs when giving equivalent inputs.
The result is the following lemma:
{{< codelines "agda" "agda-spa/Lattice.agda" 196 217 >}}
Given this, and two lattices of finite height, we construct the full product
chain by lifting the `A` chain into the product via \(a \mapsto (a, \bot)\),
lifting the `B` chain into the product via \(b \mapsto (\top, b)\), and
concatenating the results. This works because the first chain ends with
\((\top, \bot)\), and the second starts with it.
{{< codelines "agda" "agda-spa/Lattice/Prod.agda" 177 179 >}}
This gets us the longest chain; what remains is to prove that this chain's
length is the bound of all other changes. To do so, we need to work in
the opposite direction; given a chain in the product lattice, we need to
somehow reduce it to chains in lattices `A` and `B`, and leverage their
finite height to complete the proof.
The key idea is that for every two consecutive elements in the product lattice
chain, we know that at least one of their components must've increased.
This increase had to come either from elements in lattice `A` or in lattice `B`.
We can thus stick this increase into an `A`-chain or a `B`-chain, increasing
its length. Since one of the chains grows with every consecutive pair, the
number of consecutive pairs can't exceed the length of the `A` and `B` chains.
I implement this idea as an `unzip` function, which takes a product chain
and produces two chains made from its increases. By the logic we've described,
the length two chains has to bound the main one's. I give the signature below,
and will put the implementation in a collapsible detail block. One last
detail is that the need to decide which chain to grow --- and thus which element
has increased --- is what introduces the need for decidable equality.
{{< codelines "agda" "agda-spa/Lattice/Prod.agda" 158 158 >}}
{{< codelines "agda" "agda-spa/Lattice/Prod.agda" 158 172 "" "**(Click here for the implementation of `unzip`)**" >}}
Having decomposed the product chain into constituent chains, we simply combine
the facts that they have to be bounded by the height of the `A` and `B` lattices,
as well as the fact that they bound the combined chain.
{{< codelines "agda" "agda-spa/Lattice/Prod.agda" 174 183 "hl_lines = 8-9" >}}
This completes the proof!
{{< todo >}}
The rest of this.
{{< /todo >}}