Add draft of first part of polynomials article
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---
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title: "Search as a Polynomial"
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date: 2022-10-22T14:51:15-07:00
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draft: true
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tags: ["Mathematics"]
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---
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Suppose that you're trying to get from city A to city B, and then from city B
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to city C. Also suppose that your trips are measured in one-hour intervals, and
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that trips of equal duration are considered equivalent.
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Given possible routes from A to B, and then given more routes from B to C, what
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are the possible routes from A to C you can build up?
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We can try with an example. Maybe there are two routes from A to B that take
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two hours each, and one "quick" trip that takes only an hour. On top of this,
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there's one three-hour trip from B to C, and one two-hour trip. Given these
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building blocks, the list of possible trips from A to C is as follows.
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{{< latex >}}
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\begin{aligned}
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\text{two}\ 2h\ \text{trips} \rightarrow \text{one}\ 3h\ \text{trip} = \text{two}\ 5h\ \text{trips}\\
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\text{two}\ 2h\ \text{trips} \rightarrow \text{one}\ 2h\ \text{trip} = \text{two}\ 4h\ \text{trips}\\
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\text{one}\ 1h\ \text{trip} \rightarrow \text{one}\ 3h\ \text{trip} = \text{one}\ 4h\ \text{trips}\\
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\text{one}\ 1h\ \text{trip} \rightarrow \text{one}\ 2h\ \text{trip} = \text{two}\ 3h\ \text{trips}\\
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\textbf{total:}\ \text{two}\ 5h\ \text{trips}, \text{three}\ 4h\ \text{trips}, \text{one}\ 3h\ \text{trip}
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\end{aligned}
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{{< /latex >}}
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Does this look a little bit familiar? We're combining every length of trips
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of A to B with every length of trips from B to C, and then totaling them up.
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In other words, we're multiplying two binomials!
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{{< latex >}}
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\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}
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{{< /latex >}}
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In fact, they don't have to be binomials. We can represent any combination
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of trips of various lengths as a polynomial. Each term \\(ax^n\\) represents
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\\(a\\) trips of length \\(n\\). As we just saw, multiplying two polynomials
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corresponds to "sequencing" the trips they represent -- matching each trip in
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one with each of the trips in the other, and totaling them up.
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What about adding polynomials, what does that correspond to? The answer there
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is actually quite simple: if two polynomials both represent (distinct) lists of
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trips from A to B, then adding them just combines the list. If I know one trip
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that takes two hours (\\(x^2\\)) and someone else knows a shortcut (\\(x\\\)),
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then we can combine that knowledge (\\(x^2+x\\)).
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Well, that's a neat little thing, and pretty quick to demonstrate, too. But
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we can push this observation a bit further. To generalize what we've already
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seen, however, we'll need to figure out "the bare minimum" of what we need to
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make polynomial multiplication work as we'd expect.
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### Polynomials over Semirings
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Let's watch what happens when we multiply two binomials, paying really close
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attention to the operations we're performing. The following (concrete)
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example should do.
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{{< latex >}}
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\begin{aligned}
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& (x+1)(1-x)\\
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=\ & (x+1)1+(x+1)(-x)\\
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=\ & x+1-x^2-x \\
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=\ & x-x+1-x^2 \\
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=\ & 1-x^2
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\end{aligned}
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{{< /latex >}}
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The first thing we do is _distribute_ the multiplication over the addition, on
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the left. We then do that again, on the right this time. After this, we finally
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get some terms, but they aren't properly grouped together; an \\(x\\) is at the
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front, and a \\(-x\\) is at the very back. We use the fact that addition is
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_commutative_ (\\(a+b=b+a\\)) and _associative_ (\\(a+(b+c)=(a+b)+c\\)) to
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rearrange the equation, grouping the \\(x\\) and its negation together. This
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gives us \\((1-1)x=0x=0\\). That last step is important: we've used the fact
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that multiplication by zero gives zero. We didn't use it in this example,
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but another important property we want is for multiplication to be associative,
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too.
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So, what if we didn't use numbers, but rather anything _thing_ with two
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operations, one kind of like \\((\\times)\\) and one kind of like \\((+)\\)?
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As long as these operations satisfy the properties we have used so far, we
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should be able to create polynomials using them, and do this same sort of
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"combining paths" we did earlier. Before we get to that, let me just say
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that "things with addition and multiplication that work in the way we
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described" have an established name in math - they're called semirings.
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A __semiring__ is a set equipped with two operations, one called
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"multiplicative" (and thus carrying the symbol \\(\\times)\\) and one
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called "additive" (and thus written as \\(+\\)). Both of these operations
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need to have an "identity element". The identity element for multiplication
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is usually
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{{< sidenote "right" "written-as-note" "written as \(1\)," >}}
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And I do mean "written as": a semiring need not be over numbers. We could
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define one over <a href="https://en.wikipedia.org/wiki/Graph">graphs</a>,
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sets, and many other things! Nevertheless, because most of us learn the
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properties of addition and multiplication much earlier than we learn about
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other more "esoteric" things, using numbers to stand for special elements
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seems to help use intuition.
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{{< /sidenote >}}
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and the identity element for addition is written
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as \\(0\\). Furthermore, a few equations hold. I'll present them in groups.
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First, multiplication is associative and multiplying by \\(1\\) does nothing;
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in mathematical terms, the set forms a [monoid](https://mathworld.wolfram.com/Monoid.html)
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with multiplication and \\(1\\).
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{{< latex >}}
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\begin{array}{cl}
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(a\times b)\times c = a\times(b\times c) & \text{(multiplication associative)}\\
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1\times a = a = a \times 1 & \text{(1 is multiplicative identity)}\\
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\end{array}
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{{< /latex >}}
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Similarly, addition is associative and adding \\(0\\) does nothing.
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Addition must also be commutative; in other words, the set forms a
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commutative monoid with addition and \\(0\\).
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{{< latex >}}
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\begin{array}{cl}
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(a+b)+c = a+(b+c) & \text{(addition associative)}\\
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0+a = a = a+0 & \text{(0 is additive identity)}\\
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a+b = b+a & \text{(addition is commutative)}\\
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\end{array}
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{{< /latex >}}
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Finally, a few equations determine how addition and multiplication interact.
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{{< latex >}}
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\begin{array}{cl}
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0\times a = 0 = a \times 0 & \text{(annihilation)}\\
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a\times(b+c) = a\times b + a\times c & \text{(left distribution)}\\
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(a+b)\times c = a\times c + b\times c & \text{(right distribution)}\\
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\end{array}
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{{< /latex >}}
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That's it, we've defined a semiring. First, notice that numbers do indeed
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form a semiring; all the equations above should be quite familiar from algebra
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class. When using polynomials with numbers to do our city path finding,
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we end up tracking how many different ways there are to get from one place to
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another in a particular number of hours. There are, however, other semirings
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we can use that yield interesting results, even though we continue to add
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and multiply polynomials.
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One last thing before we look at other semirings: given a semiring \\(R\\),
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the polynomials using that \\(R\\), and written in terms of the variable
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\\(x\\), are denoted as \\(R[x]\\).
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#### The Semiring of Booleans, \\(\\mathbb{B}\\)
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Alright, it's time for our first non-number example. It will be a simple one,
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though - booleans (that's right, `true` and `false` from your favorite
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programming language!) form a semiring. In this case, addition is the
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"or" operation (aka `||`), in which the result is true if either operand
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is true, and false otherwise.
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{{< latex >}}
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\begin{array}{c}
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\text{true} + b = \text{true}\\
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b + \text{true} = \text{true}\\
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\text{false} + \text{false} = \text{false}
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\end{array}
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{{< /latex >}}
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For addition, the identity element -- our \\(0\\) -- is \\(\\text{false}\\).
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Correspondingly, multiplication is the "and" operation (aka `&&`), in which the
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result is false if either operand is false, and true otherwise.
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{{< latex >}}
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\begin{array}{c}
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\text{false} \times b = \text{false}\\
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b \times \text{false} = \text{false}\\
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\text{true} \times \text{true} = \text{true}
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\end{array}
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{{< /latex >}}
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For multiplication, the identity element -- the \\(1\\) -- is \\(\\text{true}\\).
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It's not hard to see that _both_ operations are commutative - the first and
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second equations for addition, for instance, can be combined to get
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\\(\\text{true}+b=b+\\text{true}\\), and the third equation clearly shows
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commutativity when both operands are false. The other properties are
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easy enough to verify by simple case analysis (there are 8 cases to consider).
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The set of booleans is usually denoted as \\(\\mathbb{B}\\), which means
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polynomials using booleans are denoted by \\(\\mathbb{B}[x]\\).
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Let's try some examples. We can't count how many ways there are to get from
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A to B in a certain number of hours anymore: booleans aren't numbers!
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Instead, what we _can_ do is track _whether or not_ there is a way to get
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from A to B in a certain number of hours (call it \\(n\\)). If we can,
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we write that as \\(\text{true}\ x^n = 1x^n = x^n\\). If we can't, we write
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that as \\(\\text{false}\ x^n = 0x^n = 0\\). The polynomials corresponding
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to our introductory problem are \\(x^2+x^1\\) and \\(x^3+x^2\\). Multiplying
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them out gives:
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{{< latex >}}
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(x^2+x^1)(x^3+x^2) = x^5 + x^4 + x^4 + x^3 = x^5 + x^4 + x^2
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{{< /latex >}}
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And that's right; if it's possible to get from A to B in either two hours
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or one hour, and then from B to C in either three hours or two hours, then
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it's possible to get from A to C in either five, four, or three hours. In a
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way, polynomials like this give us _less_ information than our original ones
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(which were \\(\\mathbb{N}[x]\\), polynomials over natural numbers \\(\\mathbb{N} = \\{ 0, 1, 2, ... \\}\\)), so it's unclear why we'd prefer them. However,
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we're just warming up - there are more interesting semirings for us to
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consider!
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#### Polynomials over Sets of Paths, \\(\\mathcal{P}(\\Pi)\\)
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Until now, we explicitly said that "all paths of the same length are
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equivalent". If we're giving directions, though, we might benefit
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from knowing not just that there _is_ a way, but what roads that
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way is made up of!
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To this end, we define the set of paths, \\(\\Pi\\). This set will consist
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of the empty path (which we will denote \\(\\circ\\), why not?), street
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names (e.g. \\(\\text{Mullholland Dr.}\\) or \\(\\text{Sunset Blvd.}\\)), and
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concatenations of paths, written using \\(\\rightarrow\\). For instance,
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a path that first takes us on \\(\\text{Highway}\\) and then on
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\\(\\text{Exit 4b}\\) will be written as:
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{{< latex >}}
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\text{Highway}\rightarrow\text{Exit 4b}
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{{< /latex >}}
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Furthermore, it's not too much of a stretch to say that adding an empty path
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to the front or the back of another path doesn't change it. If we use
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the letter \\(\\pi\\) to denote a path, this means the following equation:
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{{< latex >}}
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\circ \rightarrow \pi = \pi = \pi \rightarrow \circ
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{{< /latex >}}
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{{< sidenote "right" "paths-monoid-note" "So those are paths." >}}
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In fact, if you clicked through the
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<a href="https://mathworld.wolfram.com/Monoid.html">monoid</a>
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link earlier, you might be interested to know that paths as defined here
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form a monoid with concatenation \(\rightarrow\) and the empty path \(\circ\)
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as a unit.
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{{< /sidenote >}}
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Paths alone, though, aren't enough for our polynomials; we're tracking
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different _ways_ to get from one place to another. This is an excellent
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use case for sets!
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Our next semiring will be that of _sets of paths_. Some elements
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of this semiring are \\(\\varnothing\\), also known as the empty set,
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\\(\\{\\circ\\}\\), the set containing only the empty path, and the set
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containing a path via the highway, and another path via the suburbs:
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{{< latex >}}
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\{\text{Highway}\rightarrow\text{Exit 4b}, \text{Suburb Rd.}\}
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{{< /latex >}}
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So what are the addition and multiplication on sets of paths? Addition
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is the easier one: it's just the union of sets:
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{{< latex >}}
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A + B \triangleq A \cup B
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{{< /latex >}}
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It's well known (and not hard to verify) that set union is commutative
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and associative. The additive identity \\(0\\) is simply the empty set
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\\(\\varnothing\\). Intuitively, adding "no paths" to another set of
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paths doesn't add anything, and thus leaves that other set unchanged.
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Multiplication is a little bit more interesting, and uses the path
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concatenation operation we defined earlier. We will use this
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operation to describe path sequencing; given two sets of paths,
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\\(A\\) and \\(B\\), we'll create a new set of paths
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consisting of each path from \\(A\\) concatenated with each
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path from \\(B\\):
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{{< latex >}}
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A \times B = \{ a \rightarrow b\ |\ a \in A, b \in B \}
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{{< /latex >}}
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The fact that this definition of multiplication on sets is associative
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relies on the associativity of path concatenation; if path concatenation
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weren't associative, the second equality below would not hold.
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{{< latex >}}
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\begin{array}{rcl}
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A \times (B \times C) & = & \{ a \rightarrow (b \rightarrow c)\ |\ a \in A, b \in B, c \in C \} \\
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& \stackrel{?}{=} & \{ (a \rightarrow b) \rightarrow c \ |\ a \in A, b \in B, c \in C \} \\
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& = & (A \times B) \times C
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\end{array}
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{{< /latex >}}
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What's the multiplicative identity? Well, since multiplication concatenates
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all the combination of paths from two sets, we could try making a set of
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elements that don't do anything when concatenating. Sound familiar? It should,
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that's \\(\\circ\\), the empty path element! We thus define our multiplicative
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identity as \\(\\{\\circ\\}\\), and verify that it is indeed the identity:
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{{< latex >}}
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\begin{gathered}
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\{\circ\} \times A = \{ \circ \rightarrow a\ |\ a \rightarrow A \} = \{ a \ |\ a \in A \} = A \\
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A \times \{\circ\}= \{ a\rightarrow \circ \ |\ a \rightarrow A \} = \{ a \ |\ a \in A \} = A
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\end{gathered}
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{{< /latex >}}
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It's not too difficult to verify the annihilation and distribution laws for
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sets of paths, either; I won't do that here, though. Finally, let's take
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a look at an example. Like before, we'll try make one that corresponds to
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our introductory description of paths from A to B and from B to C. Now we need
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to be a little bit creative, and come up with names for all these different
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roads between our hypothetical cities. Let's say that \\(\\text{Highway A}\\)
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and \\(\\text{Highway B}\\) are the two paths from A to B that take two hours
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each, and then \\(\\text{Shortcut}\\) is the path that takes one hour. As for
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paths from B to C, let's just call them \\(\\text{Long}\\) for the three-hour
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path, and \\(\\text{Short}\\) for the two-hour path. Our two polynomials
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are then:
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{{< latex >}}
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\begin{array}{rcl}
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P_1 & = & \{\text{Highway A}, \text{Highway B}\}x^2 + \{\text{Shortcut}\}x \\
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P_2 & = & \{\text{Long}\}x^3 + \{\text{Short}\}x^2
|
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|
\end{array}
|
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|
{{< /latex >}}
|
||||||
|
|
||||||
|
Multiplying them gives:
|
||||||
|
{{< latex >}}
|
||||||
|
\begin{array}{rl}
|
||||||
|
& \{\text{Highway A} \rightarrow \text{Long}, \text{Highway B} \rightarrow \text{Long}\}x^5\\
|
||||||
|
+ & \{\text{Highway A} \rightarrow \text{Short}, \text{Highway B} \rightarrow \text{Short}, \text{Shortcut} \rightarrow \text{Long}\}x^4\\
|
||||||
|
+ & \{\text{Shortcut} \rightarrow \text{Short}\}x^3
|
||||||
|
\end{array}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
This resulting polynomial gives us all the paths from city A to city C,
|
||||||
|
grouped by their length!
|
||||||
Loading…
Reference in New Issue
Block a user