Edit and improve the polynomial search post

content/blog/search_polynomials.md

@@ -5,40 +5,53 @@ draft: true
 tags: ["Mathematics"]
 ---
 
-Suppose that you're trying to get from city A to city B, and then from city B
-to city C. Also suppose that your trips are measured in one-hour intervals, and
-that trips of equal duration are considered equivalent.
-Given possible routes from A to B, and then given more routes from B to C, what
-are the possible routes from A to C you can build up?
+I read a really neat paper some time ago, and I've been wanting to write about
+it ever since. The paper is called [Algebras for Weighted Search](https://dl.acm.org/doi/pdf/10.1145/3473577),
+and it is a tad too deep to dive into in a blog article -- readers of ICFP are
+rarely the target audience on this site. However, one particular insight I
+gleaned from the paper merits additional discussion and demonstration. I'm
+going to do that here.
 
-We can try with an example. Maybe there are two routes from A to B that take
-two hours each, and one "quick" trip that takes only an hour. On top of this,
-there's one three-hour trip from B to C, and one two-hour trip. Given these
-building blocks, the list of possible trips from A to C is as follows.
+We can start with something concrete. Suppose that you're trying to get from
+city A to city B, and then from city B to city C. Also suppose that your
+trips are measured in one-hour intervals, and that trips of equal duration are
+considered equivalent. Given possible routes from A to B, and then given more
+routes from B to C, what are the possible routes from A to C you can build up?
 
-{{< latex >}}
-\begin{aligned}
-    \text{two}\ 2h\ \text{trips} \rightarrow \text{one}\ 3h\ \text{trip} = \text{two}\ 5h\ \text{trips}\\
-    \text{two}\ 2h\ \text{trips} \rightarrow \text{one}\ 2h\ \text{trip} = \text{two}\ 4h\ \text{trips}\\
-    \text{one}\ 1h\ \text{trip} \rightarrow \text{one}\ 3h\ \text{trip} = \text{one}\ 4h\ \text{trips}\\
-    \text{one}\ 1h\ \text{trip} \rightarrow \text{one}\ 2h\ \text{trip} = \text{two}\ 3h\ \text{trips}\\
-    \textbf{total:}\ \text{two}\ 5h\ \text{trips}, \text{three}\ 4h\ \text{trips}, \text{one}\ 3h\ \text{trip}
-\end{aligned}
-{{< /latex >}}
+In many cases, starting with an example helps build intuition. Maybe there
+are two routes from A to B that take two hours each, and one "quick" trip
+that takes only an hour. On top of this, there's one three-hour trip from B
+to C, and one two-hour trip. Given these building blocks, the list of
+possible trips from A to C is as follows.
 
-Does this look a little bit familiar? We're combining every length of trips
-of A to B with every length of trips from B to C, and then totaling them up.
-In other words, we're multiplying two binomials!
+1. Two two-hour trips from A to B, followed up by the three-hour trip from B to
+C.
+2. Two two-hour trips from A to B, followed by the shorter two-hour trip from B
+to C.
+3. One one-hour trip from A to B, followed by the three-hour trip from B to C.
+4. One one-hour trip from A to B, followed by the shorter two-hour trip from B to C.
+
+In the above, to figure out the various ways of getting from A to C, we had to
+examine all pairings of A-to-B routes with B-to-C routes. But then, multiple
+pairings end up having the same total length: the second and third bullet
+points both describe trips that take four hours. Thus, to give
+our final report, we need to "combine like terms" - add up the trips from
+the two matching bullet points, ending up with total of three four-hour trips.
+
+Does this feel a little bit familiar? To me, this bears a rather striking
+resemblance to an operation we've seen in algebra class: we're multiplying
+two binomials! Here's the corresponding multiplication:
 
 {{< latex >}}
 \left(2x^2 + x\right)\left(x^3+x^2\right) = 2x^5 + 2x^4 + x^4 + x^3 = \underline{2x^5+3x^4+x^3}
 {{< /latex >}}
 
-In fact, they don't have to be binomials. We can represent any combination
-of trips of various lengths as a polynomial. Each term \\(ax^n\\) represents
-\\(a\\) trips of length \\(n\\). As we just saw, multiplying two polynomials
-corresponds to "sequencing" the trips they represent -- matching each trip in
-one with each of the trips in the other, and totaling them up.
+It's not just binomials that correspond to our combining paths between cities.
+We can represent any combination of trips of various lengths as a polynomial.
+Each term \\(ax^n\\) represents \\(a\\) trips of length \\(n\\). As we just
+saw, multiplying two polynomials corresponds to "sequencing" the trips they
+represent -- matching each trip in one with each of the trips in the other,
+and totaling them up.
 
 What about adding polynomials, what does that correspond to? The answer there
 is actually quite simple: if two polynomials both represent (distinct) lists of
@@ -46,10 +59,10 @@ trips from A to B, then adding them just combines the list. If I know one trip
 that takes two hours (\\(x^2\\)) and someone else knows a shortcut (\\(x\\\)),
 then we can combine that knowledge (\\(x^2+x\\)).
 
-Well, that's a neat little thing, and pretty quick to demonstrate, too. But
-we can push this observation a bit further. To generalize what we've already
-seen, however, we'll need to figure out "the bare minimum" of what we need to
-make polynomial multiplication work as we'd expect.
+Well, that's a neat little thing. But we can push this observation a bit
+further. To generalize what we've already seen, however, we'll need to
+figure out "the bare minimum" of what we need to make polynomial
+multiplication work as we'd expect.
 
 ### Polynomials over Semirings
 Let's watch what happens when we multiply two binomials, paying really close
@@ -73,11 +86,10 @@ front, and a \\(-x\\) is at the very back. We use the fact that addition is
 _commutative_ (\\(a+b=b+a\\)) and _associative_ (\\(a+(b+c)=(a+b)+c\\)) to
 rearrange the equation, grouping the \\(x\\) and its negation together. This
 gives us \\((1-1)x=0x=0\\). That last step is important: we've used the fact
-that multiplication by zero gives zero. We didn't use it in this example,
-but another important property we want is for multiplication to be associative,
-too.
+that multiplication by zero gives zero. Another important property (though
+we didn't use it here) is that multiplication has to be associative, too.
 
-So, what if we didn't use numbers, but rather anything _thing_ with two
+So, what if we didn't use numbers, but rather any _thing_ with two
 operations, one kind of like \\((\\times)\\) and one kind of like \\((+)\\)?
 As long as these operations satisfy the properties we have used so far, we
 should be able to create polynomials using them, and do this same sort of
@@ -197,7 +209,34 @@ them out gives:
 And that's right; if it's possible to get from A to B in either two hours
 or one hour, and then from B to C in either three hours or two hours, then
 it's possible to get from A to C in either five, four, or three hours. In a
-way, polynomials like this give us _less_ information than our original ones
+way, polynomials like this give us
+{{< sidenote "right" "homomorphism-note" "less information than our original ones" >}}
+In fact, we can construct a semiring homomorphism (kind of like a
+<a href="https://en.wikipedia.org/wiki/Ring_homomorphism">ring homomorphism</a>,
+but for semirings) from \(\mathbb{N}[x]\) to \(\mathbb{B}[x]\) as follows:
+
+{{< latex >}}
+    \sum_{i=0}^n a_ix^i \mapsto \sum_{i=0}^n \text{clamp}(a_i)x^i
+{{< /latex >}}
+
+Where the \(\text{clamp}\) function checks if its argument is non-zero.
+In the case of city path search, \(\text{clamp}\) asks the questions
+"are there any routes at all?".
+
+{{< latex >}}
+\text{clamp}(n) = \begin{cases}
+    \text{false} & n = 0 \\
+    \text{true} & n > 0
+\end{cases}
+{{< /latex >}}
+
+We can't construct the inverse of the above homomorphism (a mapping
+that would undo our clamping, and take polynomials in \(\mathbb{B}[x]\) to
+\(\mathbb{N}[x]\)). This fact gives us a more "mathematical" confirmation
+that we lost information, rather than gained it, but switching to
+boolean polynomials: we can always recover a boolean polynomial from the
+natural number one, but not the other way around.
+{{< /sidenote >}}
 (which were \\(\\mathbb{N}[x]\\), polynomials over natural numbers \\(\\mathbb{N} = \\{ 0, 1, 2, ... \\}\\)), so it's unclear why we'd prefer them. However,
 we're just warming up - there are more interesting semirings for us to
 consider!
@@ -228,17 +267,17 @@ the letter \\(\\pi\\) to denote a path, this means the following equation:
 {{< /latex >}}
 
 {{< sidenote "right" "paths-monoid-note" "So those are paths." >}}
-In fact, if you clicked through the
+Actually, if you clicked through the
 <a href="https://mathworld.wolfram.com/Monoid.html">monoid</a>
 link earlier, you might be interested to know that paths as defined here
 form a monoid with concatenation \(\rightarrow\) and the empty path \(\circ\)
 as a unit.
 {{< /sidenote >}}
 Paths alone, though, aren't enough for our polynomials; we're tracking
-different _ways_ to get from one place to another. This is an excellent
+different ways to get from one place to another. This is an excellent
 use case for sets!
 
-Our next semiring will be that of _sets of paths_. Some elements
+Our next semiring will be that of _sets of paths_. Some example elements
 of this semiring are \\(\\varnothing\\), also known as the empty set,
 \\(\\{\\circ\\}\\), the set containing only the empty path, and the set
 containing a path via the highway, and another path via the suburbs:
@@ -248,7 +287,8 @@ containing a path via the highway, and another path via the suburbs:
 {{< /latex >}}
 
 So what are the addition and multiplication on sets of paths? Addition
-is the easier one: it's just the union of sets:
+is the easier one: it's just the union of sets (the "triangle equal sign"
+symbol means "defined as"):
 
 {{< latex >}}
 A + B \triangleq A \cup B
@@ -283,7 +323,7 @@ A \times (B \times C) & = & \{ a \rightarrow (b \rightarrow c)\ |\ a \in A, b \i
 {{< /latex >}}
 
 What's the multiplicative identity? Well, since multiplication concatenates
-all the combination of paths from two sets, we could try making a set of
+all the combinations of paths from two sets, we could try making a set of
 elements that don't do anything when concatenating. Sound familiar? It should,
 that's \\(\\circ\\), the empty path element! We thus define our multiplicative
 identity as \\(\\{\\circ\\}\\), and verify that it is indeed the identity: