In this instalment we’ll take a further look at **Monoid morphisms** to
try to cement our understanding of this concept. We’ve used monoids to
understand parallelism in the Map-Reduce style programming
model. We’ll continue by looking at applications in the related notion
of *Divide & Conquer* style problem solving.

To reiterate, monoid morphisms are structure-preserving maps between monoids. That is for two monoids \((M, \oplus, e)\) and \((N, \otimes, f)\) a monoid morphism \(h\) is a function satisfying

\[ \begin{aligned} h(e) & = f\\ h(a \oplus b) & = h(a) \otimes h(b) \end{aligned} \]

so a monoid morphism preserves identity and composition.

Below we will solve at a set of Divide & Conquer type problems and show that monoid morphisms often capture the divide and conquer steps.

Wishing to calculate the result of applying a function \(h\), we start by “dividing” it’s argument by writing it as a composition in terms of \(\oplus\).

\[ h(a \oplus b) \]

and by using the monoid morphism law, we rewrite in terms of \(\otimes\)

\[ h(a) \otimes h(b) \]

which captures the “conquering” — a re-composition step that recombines solved subproblems \(h(a)\) and \(h(b)\) into a full solution.

There is the ordinary monoid of lists, but there is a also different
monoid of *sorted* lists. The monoid of sorted lists does not have the
usual concatenation \(\doubleplus\) as it’s composition, as it does not
"maintain sortedness", i.e. \(xs \doubleplus ys\) need not be sorted just because
\(xs\) and \(ys\) are.

Instead we’ll use as composition the operation \(\texttt{merge}\) which we’ll denote with \(\odot\). It is clearly associative, and has the empty list as its identity.

The function \(sort\) is a monoid morphism from ordinary lists to the monoid of sorted lists

\begin{aligned} sort(\emptyList) & = \emptyList\\ sort(xs \doubleplus ys) & = sort(xs) \odot sort(ys) \end{aligned}

If \(sort\) is an \(n^2\) time sort, such as insertion-sort or bubble-sort, this monoid morphism encodes an optimization step. Since merging can be done in linear time, and the decomposition \(xs \doubleplus ys\) can be chosen to halve the length of the composite list, we can go from \(n^2\) to \((2(\frac{n}{2})^2 + n)\) time, which is a significant performance improvement.

By repeatedly applying this monoid morphism we derive the merge-sort algorithm.

The function \(exp\) is a monoid morphism from the monoid of numbers with addition to the monoid of numbers with multiplication. The meaning of this statement is that \(exp\) should satisfy

\[ exp(0) = 1\\ exp(b + c) = exp(b) * exp(c) \]

which we know to be true, since

\[ exp(a + b) = e^{a + b} = e^a e^b = exp(a) * exp(b) \]

This argument would work for any exponentiation function, not just with \(e\) as a base, so there is a monoid morphism of this form for any number \(a\), namely the function \(x \mapsto a^x\).

\[ a^0 = 1\\ a^{b + c} = a^b * a^c \]

We can take advantage of this identity if we’d like to calculate an exponential where the exponent is an even whole number.

\[ a^{2n} = a^{n} * a^{n} \]

Assuming we’re calculating this exponent in the naive way of performing \(2n\) multiplications, this rule describes an optimization step, resulting instead in \(n + 1\) multiplications, as we need only calculate \(a^n\) a single time.

A similar optimization can be performed for an odd whole number, leading to an algorithm using \(log_2(n)\) multiplications.

In interview question I came across ask one to find the number of trailing zeroes in a large factorial, let’s say \(100!\), without calculating this actual value, as it is very large.

Again, this problem can be solved through a divide & conquer style of problem solving using monoid morphisms.

There is a monoid morphism from numbers with multiplication, to the pointwise lifted monoid of numbers with addition.

The fact that such a morphism exists is known as the Fundamental Theorem of Arithmetic, that is, we can map a number to its prime factorization.

\[ primes(1) = \emptyset\\ primes(a * b) = primes(a) \cup_+ primes(b) \]

Here \(\cup_+\) denotes pointwise addition in the set of prime frequencies.

By applying this monoid morphism, we can find the prime factorization of \(100!\).

\begin{aligned} & primes(100!) \\ = \enspace & primes(100 * 99!) \\ = \enspace & primes(100) \cup_+ primes(99!) \\ = \enspace & primes(100) \cup_+ primes(99) \cup_+ \ldots \cup_+ primes(2) \end{aligned}

From this result it is easy to find the number of zeroes in \(100!\). We simply take the min prime frequency for numbers 2 and 5. Since finding the prime factorizations up to \(100\) is quickly done, this again describes an optimization.

One might note that we can in fact solve this problem even faster, by
simply counting the prime frequencies for 2 and 5 *only*. In order to
formulate this simplification in the language of monoid morphisms we
would need to introduce equivalence classes, and their interactions
with monoid structure, which we will not do or explain here.

Instead we note that any partitioning of a set over an equvalence relation induces a monoid morphism to the set of partitions as long as the composition respects the equivalence and challenge the interested reader to find out more.

\[ \newcommand\andand{\kern0.8ex\texttt{&&}\kern0.8ex} \newcommand\oror{\kern0.8ex\texttt{||}\kern0.8ex} \]

The negation function \(\texttt{!}\) which negates a boolean proposition is a monoid morphism from \((Bool, \texttt{&&}, \texttt{true})\) to \((Bool, \texttt{||}, \texttt{false})\).

\begin{aligned} \texttt{!}\texttt{true} =& \enspace \texttt{false}\\ \texttt{!}(a \andand b) =& \enspace !a \oror !b \end{aligned}

It is also a monoid morphism going the other way, from \((Bool, \texttt{||}, \texttt{false})\) to \((Bool, \texttt{&&}, \texttt{true})\)

\begin{aligned} \texttt{!}\texttt{false} =& \enspace \texttt{true}\\ \texttt{!}(a \oror b) =& \enspace !a \andand !b \end{aligned}

The fact that \(\texttt{!}\) is a monoid morphism in these ways is known as De Morgan’s laws.

Stable sorting functions are monoid morphisms from the monoid of comparators to the monoid of functions with function-composition. Let \(sort_c\) be the sorting function that sorts by comparator \(c\).

Let \(\varepsilon\) be the identity-comparator, that compares all elements as equal, and \(\diamond\) be comparator-composition. Then a stable sort satisfies

\[ sort_\varepsilon = id \\ sort_{c \thinspace \diamond \thinspace d} = sort_c \circ sort_d \]

this encodes the observation that for a stable sort we can fuse two sorts by different comparators \(c\) and \(d\) into a single sorting pass that compares first by \(c\) and then by \(d\).

A weird but interesting construction on monoids is the opposite monoid. It is constructed from a given monoid by taking

\[ x \otimes_{op} y = y \otimes x \]

so this monoid is the same except the order of arguments is the reverse of the target monoid. The identity element stays the same.

The function \(reverse\) is an example of a monoid morphism. It goes from the monoid of lists to the opposite monoid of lists — this means it will satisfy the monoid morphism laws

\begin{equation} reverse(\emptyList) = \emptyList\\ \begin{aligned} & reverse(xs \doubleplus ys)\\ = \enspace & reverse(ys) \doubleplus reverse(xs)\\ = \enspace & reverse(xs) \doubleplusop reverse(ys) \end{aligned} \end{equation}

It it helps, here’s an example with concrete values

\begin{aligned} reverse(\texttt{[1, 2, 3, 4, 5]}) & =\\ reverse(\texttt{[1, 2, 3] ++ [4, 5]}) & = \\ reverse(\texttt{[4, 5]}) \doubleplus reverse(\texttt{[1, 2, 3]}) & =\\ reverse(\texttt{[1, 2, 3]}) \doubleplusop reverse(\texttt{[4, 5]}) &= \\ \texttt{[5, 4] ++ [3, 2, 1]} \end{aligned}

A common interview-question consists of reversing the order of words in a string without using any extra memory, a so-called in place algorithm.

\[ reverseWords(\texttt{"one two three"}) = \texttt{"three two one"} \]

Since the function \(reverse\) is easily written as an in-place algorithm, it is possible to solve this problem by expressing a solution as some number of substring reversals on the input string.

We can construct such an expression by starting with the solution and working backwards until the character appears in their original order. We’ll make use of the monoid morphism laws.

\begin{aligned} & reverseWords(\texttt{"one two three"})\\ = \enspace & \texttt{"three two one"}\\ = \enspace & reverse(reverse(\texttt{"three two one"}))\\ = \enspace & reverse(reverse(\texttt{"three" ++ " two " ++ " one"}))\\ = \enspace & reverse(reverse(\texttt{"three"}) \doubleplusop reverse(\texttt{" two "}) \doubleplusop reverse(\texttt{"one"}))\\ = \enspace & reverse(reverse(\texttt{"one"}) \doubleplus reverse(\texttt{" two "}) \doubleplus reverse(\texttt{"three"}))\\ \end{aligned}

So an in-place algorithm for word-reversal can be found by reversing each individual word in a string, then reversing the entire string.

A related problem to the above is the problem of cycling strings.

\[ cycle_2(\texttt{"abcdef"}) = \texttt{"cdefab"} \]

that is, \(cycle_n\) will “rotate” \(n\) characters from the front of the string to the end.

This problem too can be made into a tricky interview question by asking to perform it in-place, but we can conquer such difficulties in a similar way to the above.

\begin{aligned} & cycle_3(\texttt{"abcdefgh"})\\ = \enspace & \texttt{"defghabc"}\\ = \enspace & reverse(reverse(\texttt{"defghabc"}))\\ = \enspace & reverse(reverse(\texttt{"defgh" ++ "abc"}))\\ = \enspace & reverse(reverse(\texttt{"defgh"}) \doubleplusop reverse(\texttt{"abc"}))\\ = \enspace & reverse(reverse(\texttt{"abc"}) \doubleplus reverse(\texttt{"defgh"}))\\ \end{aligned}

and so cycling a list in place by \(n\) places can be achieved by reversing the first \(n\) elements, then the rest of the list, and finally reversing the entire list again.