Elementary yet Precise Best-Case Analysis of MergeSort with an Application to the Sum of Digits Problem

Abstract

An exact formula $B\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-{\sum }_{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right),$ where $\mathit{Zigzag}\left(x\right)=min(x-\lfloor x\rfloor ,\lceil x\rceil -x),$ for the minimum number $B\left(n\right)$ of comparisons of keys performed by $\mathtt{MergeSort}$ on an n-element array is derived and analyzed. The said formula is less complex than any other known formula for the same and can be evaluated in $O\left({log}^c\right)$ time, where c is a constant. It is shown that there is no closed-form formula for the above. Other variants for $B\left(n\right)$ are described as well. Since the recurrence relation for the minimum number of comparisons of keys for $\mathtt{MergeSort}$ is identical with a recurrence relation for the number of 1s in binary expansions of all integers between 0 and n (exclusively), the above results extend to the sum of binary digits problem.

1. Introduction

“One Picture is Worth a Thousand Words”

[An advertisement for the San Antonio Light (1918)]

Teaching undergraduate Analysis of Algorithms has been a rewarding, although a bit taxing, experience. I was often surprised to learn that many basic problems that clearly belong to its core syllabus had been left unanswered or partially answered. Additionally, it seemed a bit odd to me that many otherwise decent texts offered unnecessarily imprecise computations (text [1] being a notable exception in this category) of several rather fundamental results.

In this article, I pursue a seemingly marginal—albeit fundamental—topic, a precise characterization of the best-case behavior of a well-known sorting algorithm, $\mathtt{MergeSort}$, whose pursuit, however, yields some interesting findings that could hardly be characterized as “marginal.” It turns out that—contrary to what a casual student of this subject might believe—computing the exact formula for the number of comparisons of keys that $\mathtt{MergeSort}$ performs on any n-element array in the best case is not a routine exercise (in fact, it is significantly more complicated than precise derivation of the well-known exact formula for the number of comparisons of keys performed by it in the worst case) and leads to an instance of a problem that gained some notoriety for being a hard nut to crack analytically: the sum of digits problem. Even more unexpectedly, the relatively straightforward (although not quite closed-form) formula for the said number of comparisons yields an improvement of a well-known answer to this instance of the sum of digits problem:

How many 1s appear in binary representations of all integers between (but not including) 0 and n?

2. Materials and Methods

Several derivations presented in this article may be characterized as experimental mathematics. Once they produced experimentally (using Wolfram Mathematica software) the desired (sought-after) formulas, finding analytic proofs of the said formulas became a relatively easy and straightforward exercise, and has been done in Section 9, Section 10 and Section 11. Those derivations provide an insightful exercise that demonstrates how certain, apparently, hard-to-solve recurrence relations pertaining to the complexity of divide-and-conquer algorithms can be practically and precisely solved by means of experiments and computations augmented with formal analytic proofs of the derived solutions.

No materials were used in the research described in this article.

3. Results

The main result of this article is the following formula of the Main Theorem that gives the exact number $B\left(n\right)$ of comparisons of keys performed by the classic $\mathtt{MergeSort}$ algorithm in the best case while sorting an n-element array.

$$B\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right),$$

where

$$\mathit{Zigzag}\left(x\right)=min(x-\lfloor x\rfloor ,\lceil x\rceil -x).$$

The above formula may be evaluated in sublinear $O\left(lo{g}^{c}n\right)$ time, where c is a constant, that is considerably shorter than the time needed for running $\mathtt{MergeSort}$ on its best-case input array (for instance, an array of consecutive integers) of n elements or for counting directly how many times digit 1 occurs in binary representations of all integers between 1 and n. This constitutes a significant improvement over other exact formula, attributed to J. R. Trollope and discussed in Section 8, for the said number of 1s that involves infinite summation. Even if only finitely many terms in that infinite summation are non-zero, the Trollope formula does not indicate how long the process of its computation will take for any value of n. For instance, the Trollope formula may possibly require more time to evaluate than counting directly how many times digit 1 occurs in binary representations of all integers between 1 and n. (See Note 1 at the end of this Section.)

The exactness of the formula derived and proved in this article allows for its definite experimental verification, which in general cannot be done for characterizations given by Big-O/Big-$\Omega$/Big-$\Theta$ notation. Such an experimental verification has been performed by the uthor using a Java program that incorporates the code for $\mathtt{Merge}$ visualized in Figure 1.

Moreover, the exact and practical way to compute the answer to the question of how many comparisons of key does $\mathtt{MergeSort}$ perform in the best case while sorting an n-element array fills a significant loophole in many computer science faculty and students’ knowledge of the time complexity of $\mathtt{MergeSort}$ and satisfies the cognitive curiosity of those who realize the advantages of knowing things exactly in exact sciences and mathematics. (See Note 2 at the end of this Section.)

It is also demonstrated (see Corollary 3) that there is no exact closed-form formula that gives the said number of comps and (in Section 7) that the said number is significantly larger (by a term of size $\Theta \left(n\right)$) than half of the number of comps performed by $\mathtt{MergeSort}$ in the worst case.

Note 1. 

Since the total number of significant bits in binary representations of all numbers between 1 and n is equal to ${\sum }_{i=1}^n(\lfloor lgi\rfloor$ + 1), the direct counting of 1s in all those numbers requires at least $lgn!$ or, by virtue of Stirling’s formula, $nlgn-nlge+O(lgn)$ steps that take $\Theta (nlogn)$ time to carry them on, which is significantly more than the $O\left(lo{g}^{c}n\right)$ time that is sufficient for the evaluation of Formula (15) derived and proved in this article.

Note 2. 

A student of mine was once trying to convince me that the number of comparisons of keys that $\mathtt{MergeSort}$ performs in the worst case was exactly equal to twice the number it performed in the best case. In order to show her why it was not exactly the case, I experimentally derived an exact formula for it. This is how I got into writing this article.

4. $\mathtt{MergeSort}$ and Its Best-Case Behavior

A call to $\mathtt{MergeSort}$ inherits an n-element array $\mathtt{A}$ of integers and sorts it non-decreasingly, following the steps described below.

A Java code of $\mathtt{Merge}$ is shown in Figure 1.

A typical measure of the running time of $\mathtt{MergeSort}$ is the number of comparisons of keys, which for brevity I call comps, that it performs while sorting array $\mathtt{A}$. Since no comps are performed outside $\mathtt{Merge}$, the running time of $\mathtt{MergeSort}$ can be computed as the sum of numbers of comps performed by all calls to $\mathtt{Merge}$ during the execution of $\mathtt{MergeSort}$.

Because all comps during the execution of $\mathtt{Merge}$ are performed within the $\mathtt{while}$-loop that begins in line 76 of the code shown in Figure 1, and the said loop does not terminate before the last element of one of the two arrays has been compared with an element of the other array, the number of comps performed by $\mathtt{Merge}$ on two arrays is never less than the size of any of the two arrays passed to it, which in the case of step 2c of Algorithm 1 is not less than $\lfloor {\displaystyle \frac{n}{2}}\rfloor$, thus making $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ a lower bound for the said number of comps. Since the number of comps performed by $\mathtt{Merge}$ in the case where all the elements of the first array of size $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ are less than all the elements of the second array of size $\lceil {\displaystyle \frac{n}{2}}\rceil$ is actually equal to the size of the first array, it follows that $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ is the minimum number of comps performed during the execution of $\mathtt{Merge}$ (and not just a lower bound of that number).

Algorithm 1 $\mathtt{MergeSort}$

To sort an n-element array $\mathtt{A}$, do the following: If $n\le 1$, then return $\mathtt{A}$ to the caller; If $n\ge 2$, then (a) Pass the first $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ elements of $\mathtt{A}$ to a recursive call to $\mathtt{MergeSort}$; (b) Pass the last $\lceil {\displaystyle \frac{n}{2}}\rceil$ elements of $\mathtt{A}$ to another recursive call to $\mathtt{MergeSort}$; (c) Linearly merge, by means of a call to $\mathtt{Merge}$, the non-decreasingly sorted arrays that were returned from those calls onto one non-decreasingly sorted array ${\mathtt{A}}^{\prime }$; (d) Return ${\mathtt{A}}^{\prime }$ to the caller.

Obviously, increasingly sorted array $\mathtt{A}$ of any size $n\ge 2$ produces only best-case scenarios for all subsequent calls to $\mathtt{Merge}$. Therefore, $\lfloor {\displaystyle \frac{k}{2}}\rfloor$—where $k\ge 2$ is the size of any sub-array $\mathtt{A}$ of size greater than or equal to 2 passed to any (recursive) call to $\mathtt{MergeSort}$—is the actual minimum of and not just a lower bound on the said number of comps performed by $\mathtt{Merge}$ invoked by the said call to $\mathtt{MergeSort}$. This fact allows for a rudimentary analysis of the recursion tree for $\mathtt{MergeSort}$ that easily yields—as we will see in the next two subsections—the exact Formula (7) for the minimum number of comps for the entire $\mathtt{MergeSort}$. Without said fact, one could only derive by means of analysis of the recursion tree a lower bound on the number of comps performed by $\mathtt{MergeSort}$, even though—as I have pointed out, above—the size $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ of the first sub-array passed to $\mathtt{Merge}$ is the actual minimum of comps performed at this level and not just a lower bound.

The problem arises when one tries to reduce the said formula, which naturally involves long summations, to one that can be evaluated in a logarithmic time.

4.1. Recursion Tree

The obvious recursion tree for $\mathtt{MergeSort}$ and sufficiently large n is shown in Figure 2 and explained in some detail in the caption below it.

A recursive application of the equality

$$\lceil {\displaystyle \frac{n}{2}}\rceil =\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$$

(1)

(which can be easily verified separately for odd and even values of n) allows for rewriting of that tree onto one whose first four levels are shown in Figure 3.

4.2. Recurrence Relation $B\left(n\right)$ for the Minimum Numbers of Comps

As I have noticed, before, the number of comps performed by a single call to $\mathtt{MergeSort}$ in the best case on an n-element array for $n\ge 2$ is equal to $\lfloor {\displaystyle \frac{n}{2}}\rfloor$. (Recall that all comps performed during a single call to $\mathtt{MergeSort}$ are done within the call to $\mathtt{Merge}$ that $\mathtt{MergeSort}$ invokes after both recursive calls to itself have been completed).

Therefore, the following recurrence relation for the minimum number $B\left(n\right)$ of comparisons of keys that $\mathtt{MergeSort}$ performs on any n-element array is straightforward to derive from its description given by Algorithm 1, as follows:

$$B\left(1\right)=0,$$

(2)

and for $n\ge 2$,

$$B\left(n\right)=\lfloor {\displaystyle \frac{n}{2}}\rfloor +B\left(\lfloor {\displaystyle \frac{n}{2}}\rfloor \right)+B\left(\lceil {\displaystyle \frac{n}{2}}\rceil \right).$$

(3)

Using the equality (1), the recurrence relation (3) is equivalent to

$$B\left(n\right)=\lfloor {\displaystyle \frac{n}{2}}\rfloor +B\left(\lfloor {\displaystyle \frac{n}{2}}\rfloor \right)+B\left(\lfloor {\displaystyle \frac{n+1}{2}}\rfloor \right).$$

(4)

A graph of $B\left(n\right)$ is shown in Figure 4.

Note 3. 

If n is a power of 2, that is, if $n=2^{\lceil lgn\rceil }$, then

$$B\left(n\right)={\displaystyle \frac{n}{2}}lgn.$$

Indeed, in such a case, recurrences (3) and (4) simplify to $B\left(n\right)={\displaystyle \frac{n}{2}}+2B\left({\displaystyle \frac{n}{2}}\right)$, so that $B\left(n\right)=2{\displaystyle \frac{n}{2}}+2B\left({\displaystyle \frac{n}{2^2}}\right)=\dots =\lceil lgn\rceil {\displaystyle \frac{n}{2}}+2B\left({\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}\right)=\lceil lgn\rceil {\displaystyle \frac{n}{2}}+2B\left(1\right)=$[since $\lceil lgn\rceil =lgn$ and, by the equality (2), $B\left(1\right)=0$]$={\displaystyle \frac{n}{2}}lgn$.

4.3. A Solution of Recurrence Relation $B\left(n\right)$

Unfolding the recurrence (4) allows for noticing that the minimum number $B\left(n\right)$ of comps performed by all calls to $\mathtt{Merge}$ is equal to the sum of all values shown at nodes highlighted yellow in the recursion tree T of Figure 3. They may be summed up level by level. One can notice from Figure 3 that the number of comps performed at any level k with the maximal number $2^k$ of nodes is given by this formula:

$$\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor .$$

(5)

What is not clear is whether all levels of the recursion tree T are maximal. Fortunately, the answer to this question does not depend on whether a given instance of $\mathtt{MergeSort}$ is running on a best-case array or on any other case of array. It has been known from a classic analysis of the worst-case running time of $\mathtt{MergeSort}$ that every level of its recursion tree T that contains at least one non-leaf, or—in other words—a node that shows the value $p\ge 2$, is maximal.Appendix A, contains a detailed derivation of that fact. Thus all levels 0 through $h-1$ of T are maximal. Therefore, Formula (5) gives the number of comps for every level $0\le k\le h-1$.

The last level h of T may be not maximal because the level $h-1$ may contain leaves, or —in other words—nodes that show the value $p=1$, where $p=\lfloor {\displaystyle \frac{n+i}{2^{h-1}}}\rfloor$ for some $0\le i\le 2^{h-1}-1$, and as such do not have any children in the level h. However, for each such node, the value of $\lfloor {\displaystyle \frac{p}{2}}\rfloor =\lfloor {\displaystyle \frac{n+i}{2^h}}\rfloor$ is 0, so it can be included in summation (5) without affecting its value even though the said value does not correspond to any node in the level h. Therefore, Formula (5) gives the number of comps for the level $k=h$.

Additionally, the depth h of T is known to be equal to $\lceil lgn\rceil$, as Theorem A2 in Appendix A states. Thus the minimum number of comps performed by $\mathtt{MergeSort}$ is given by this summation

$$\sum _{k=0}^{\lceil lgn\rceil }\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor .$$

(6)

Since $n-1<2^{\lceil lgn\rceil }$ so that $n+2^{\lceil lgn\rceil }-1<2^{\lceil lgn\rceil +1}$ and for every $0\le i\le 2^{\lceil lgn\rceil }-1$, $n+i<2^{\lceil lgn\rceil +1}$, thus making

$$\lfloor {\displaystyle \frac{n+i}{2^{\lceil lgn\rceil +1}}}\rfloor =0,$$

it follows that the value of the inner summation (5) in summation (6) for $k=h=\lceil lgn\rceil$ is 0. Therefore, Formula (6) is equal to this slightly shorter formula:

$$\sum _{k=0}^{\lfloor lgn\rfloor }\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor .$$

(7)

Unfortunately, summation (7) contains $n-1$ non-zero terms (each corresponding to one of $n-1$ internal nodes of 2-tree T), so it cannot be evaluated quickly in its present form. Fortunately, as I am going to demonstrate, its inner summation (5) can be reduced to a closed-form formula (13).

4.4. Zigzag Function

In order to reduce (5) to a closed form, I am going to use the function Zigzag defined by

$$\mathit{Zigzag}\left(x\right)=min(x-\lfloor x\rfloor ,\lceil x\rceil -x).$$

(8)

The following fact is instrumental for that purpose.

Theorem 1. 

For every natural number n and every positive natural number m,

$$\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =2m\times \mathit{Zigzag}\left({\displaystyle \frac{n}{2m}}\right),$$

(9)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

The equality (9) can be verified experimentally, for instance, with the help of software for symbolic computation; I used Wolfram Mathematica for that purpose. The analytic proof is deferred to Section 9. □

Corollary 1. 

For every natural number n and every positive natural number m,

$$\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor ={\displaystyle \frac{n}{2}}-m\times \mathit{Zigzag}\left({\displaystyle \frac{n}{2m}}\right),$$

(10)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

First, let us note (analytic proof of this fact is a straightforward exercise; see Appendix B) that

$$\sum _{i=0}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =n.$$

(11)

From (11), I conclude

$$\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor +\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =n.$$

(12)

Solving Equations (9) and (12) for ${\sum }_{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor$ yields (10). □

Here is the closed form of summation (5).

Corollary 2. 

For every natural number n and every natural number k,

$$\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor ={\displaystyle \frac{n}{2}}-2^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right),$$

(13)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

Substitute $m=2^k$ in (10). □

The following theorem yields Formula (14) for the minimum number $B\left(n\right)$ of comps performed by $\mathtt{MergeSort}$.

Theorem 2. 

For every natural number n,

$$\sum _{k=0}^{\lfloor lgn\rfloor }\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor ={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right),$$

(14)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

$$\sum _{k=0}^{\lfloor lgn\rfloor }\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor =\sum _{k=0}^{\lfloor lgn\rfloor }({\displaystyle \frac{n}{2}}-2^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right))=$$

$$={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right).$$

□

Formula (14), although not a quite closed form, comprises summation with only $\lfloor lgn\rfloor +1$ closed-form terms, so it may be evaluated in $O\left({log}^c\right)$ time, where c is a constant. I will show in Section 5 that (14) does not have a closed form. Graphs of both sides of equality (14) are shown in Figure 6. One can see that, for natural numbers n, they coincide with the solution $B\left(n\right)$ of recurrences (2) and (3) visualized in Figure 4.

Main Theorem. 

For every natural number n, the minimum number $B\left(n\right)$ of comps that $\mathtt{MergeSort}$ performs while sorting an n-element array is

$$B\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right),$$

(15)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

As it has been shown in Section 4.3, Formula (7) gives the minimum number $B\left(n\right)$ of comps performed by $\mathtt{MergeSort}$ on any n-element array. Thus substituting $B\left(n\right)$ for

$$\sum _{k=0}^{\lfloor lgn\rfloor }\sum _{i=0}^{2^k-1}\lfloor {\displaystyle \frac{n+i}{2^{k+1}}}\rfloor$$

in equality (14) of Theorem 2 yields (15). □

5. A Fractal in $\mathit{B}\left(\mathit{n}\right)$

A deceitfully simple expression

$$\sum _{k=0}^{\lfloor lgx\rfloor }{2}^{k+1}\mathit{Zigzag}\left({\displaystyle \frac{x}{2^{k+1}}}\right),$$

(16)

half of which occurs in Formula (15) of the Main Theorem, is a formidable adversary for those who may try to turn it into a closed form, although the time required for its evaluation for any given n is $O\left({log}^c\right)$ (so, for all practical purposes, some may consider (15) a closed-form formula). That does not come as a surprise, taking into account that its graph, shown in Figure 7, bears a resemblance of fractal. This can be easily seen as soon as a sawtooth function $2^{\lfloor lgx\rfloor +1}-x$ is subtracted from it, yielding the function $F\left(x\right)$ given by

$$F\left(x\right)=\sum _{k=0}^{\lfloor lgx\rfloor }{2}^{k+1}\mathit{Zigzag}\left({\displaystyle \frac{x}{2^{k+1}}}\right)-2^{\lfloor lgx\rfloor +1}+x.$$

(17)

Since ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{x}{2^{\lfloor lgx\rfloor +1}}}<1$, equality (8) implies

$$\mathit{Zigzag}\left({\displaystyle \frac{x}{2^{\lfloor lgx\rfloor +1}}}\right)=1-{\displaystyle \frac{x}{2^{\lfloor lgx\rfloor +1}}},$$

or

$$2^{\lfloor lgx\rfloor +1}\mathit{Zigzag}\left({\displaystyle \frac{x}{2^{\lfloor lgx\rfloor +1}}}\right)=2^{\lfloor lgx\rfloor +1}-x.$$

(18)

Equality (18) simplifies definition (17) of function F to

$$F\left(x\right)=\sum _{k=1}^{\lfloor lgx\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{x}{2^k}}\right),$$

(19)

visualized in Figure 8.

The function F is a fractal with quasi similarity that repeats at intervals of exponentially growing length. It is a union

$$F=\bigcup _{k=0}^{\infty }{f}_k$$

(20)

of functions $f_k$, each having an interval $[2^k,2^{k+1})$ as its domain. In other words, for every integer $k\ge 0$,

$$f_k=F\upharpoonright [2^k,2^{k+1}),$$

(21)

which, of course, yields (20).

Let ${\widehat{f}}_k$ be the normalized $f_k$ on interval $[0,1)$, defined by

$${\widehat{f}}_k\left(x\right)={\displaystyle \frac{1}{2^k}}{f}_k\left(2^k(x+1)\right),$$

(22)

and ${\tilde{f}}_k$ be the periodized ${\widehat{f}}_k$ by composing it with a sawtooth function $x-\lfloor x\rfloor$ (the fractional part of x) defined by

$${\tilde{f}}_k\left(x\right)={\widehat{f}}_k(x-\lfloor x\rfloor ).$$

(23)

Contracting definitions (21), (22), and (23) yields

$${\tilde{f}}_k\left(x\right)={\displaystyle \frac{1}{2^k}}F\left(2^k(x-\lfloor x\rfloor +1)\right).$$

(24)

One can compute (an elementary geometric argument based on the graph visualized in Figure 9 will do) from (24) the following alternative formula for ${\tilde{f}}_k\left(x\right)$:

$${\tilde{f}}_k\left(x\right)=\sum _{i=0}^{k-1}{\displaystyle \frac{1}{2^i}}\mathit{Zigzag}\left(2^{i}x\right).$$

(25)

Figure 9 shows functions ${\tilde{f}}_0,\dots ,{\tilde{f}}_6$ drawn on the same graph.

Since each function $f_k$, and—therefore—each function ${\widehat{f}}_k$, and—therefore—each function ${\tilde{f}}_k$, are a result of smaller and smaller triangles piled, originating in the function Zigzag of definition (19) of function F, on one another as shown in Figure 9, for any integers $0\le i<j$, ${\tilde{f}}_i$ linearly interpolates ${\tilde{f}}_j$. Because of that, each ${\tilde{f}}_i$ linearly interpolates the limit $\tilde{F}$ of all ${\tilde{f}}_k$s defined by

$$\tilde{F}\left(x\right)=\underset{k\to \infty }{lim}{\tilde{f}}_k\left(x\right),$$

(26)

as Figure 10 illustrates. An application of (25) to (26) yields

$$\tilde{F}\left(x\right)=\sum _{i=0}^{\infty }{\displaystyle \frac{1}{2^i}}\mathit{Zigzag}\left(2^{i}x\right).$$

(27)

Since, for every integer n, k and $i\ge k$, $2^i{\displaystyle \frac{n}{2^k}}$ is integer, $\mathit{Zigzag}\left(2^i{\displaystyle \frac{n}{2^k}}\right)=0$. Therefore, by virtue of (25) and (27), for every non-negative integer k and n,

$$\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)={\tilde{f}}_k\left({\displaystyle \frac{n}{2^k}}\right).$$

(28)

This and (25) eliminate the need for infinite summation (as it appears in (27)) while computing $\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)$.

It can be shown that, although a continuous function, $\tilde{F}$—known under the name of the blancmange function or the Takagi fractal curve—is nowhere differentiable. As such, it does not have a closed-form formula as any closed-form formula on a real interval must define a function and have a derivative at every point of that interval, except for a non-dense set of its points. Since $\tilde{F}$ can be expressed in a function, described by a closed-form formula, of the right-hand side of Formula (14), the latter does not have a closed-form formula either.

Theorem 3. 

For every closed-form formula $\phi \left(n\right)$ there is a positive n such that

$$\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right)\ne \phi \left(n\right),$$

(29)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

Follows from the above discussion. A more detailed analytic proof is deferred to Section 10. □

This way, we arrived at the following conclusion.

Corollary 3. 

There is no closed-form formula for $B\left(n\right)$.

Proof. 

A closed-form formula for $B\left(n\right)$ would, by virtue of (15) yield a closed-form formula for ${\sum }_{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right)$, which by Theorem 3 does not exist. □

Note 4. 

One can apply the reverse transformations to those used in Section 5 on the function $\tilde{F}$ and construct a fractal function $\breve{F}$, shown in Figure 11, given by the equation

$$\breve{F}\left(x\right)=2^{\lfloor lgx\rfloor }\tilde{F}\left({\displaystyle \frac{x}{2^{\lfloor lgx\rfloor }}}\right),$$

(30)

which for every positive integer n satisfies

$$\breve{F}\left(n\right)=F\left(n\right),$$

(31)

where F is given by (19).

6. Computing $\tilde{\mathit{F}}\left(\mathit{x}\right)$ and $\mathit{B}\left(\mathit{n}\right)$ from One Another

Computing values of the function $\tilde{F}\left(x\right)$ does not have to be as complex as (or more complex than) definition (27) implies. Of course, for every integer n, $\tilde{F}\left(n\right)$= 0. One can apply some elementary arguments based on a structure visualized in Figure 10 to conclude that

$$\tilde{F}\left({\displaystyle \frac{2}{3}}\right)=\tilde{F}\left({\displaystyle \frac{1}{3}}\right)={\displaystyle \frac{2}{3}}$$

(32)

(the latter being the maximum of $\tilde{F}\left(x\right)$) or that, for every positive integer k,

$$\tilde{F}\left({\displaystyle \frac{1}{2^k}}\right)={\displaystyle \frac{k}{2^k}}.$$

(33)

It takes a bit more work to compute

$$\tilde{F}\left({\displaystyle \frac{3}{2^k}}\right)={\displaystyle \frac{3k-4}{2^k}}.$$

(34)

It turns out that computing values of the function $\tilde{F}\left(x\right)$ for every x that has a finite binary representation can be done easily if an oracle for computing the values of the function $B\left(n\right)$ defined by (2) and (4) is given, which is not that surprising after a glance in Figure 11. Once that is accomplished, since $\tilde{F}\left(x\right)$ is a continuous function and the set of numbers with finite binary representations is dense in the set $\mathfrak{R}$ of reals, it allows for fast approximations of $\tilde{F}\left(x\right)$ for every real x. (It helps to remember that $\tilde{F}$ is a periodic function with $\tilde{F}\left(x\right)=\tilde{F}(x-\lfloor x\rfloor )$.)

Theorem 4. 

For every positive integer n and integer k with $n\le 2^k$,

$$\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)={\displaystyle \frac{n\times k-2B\left(n\right)}{2^k}}.$$

(35)

Proof. 

Equality (35) can be verified experimentally, for instance, with the help of software for symbolic computation (Section 4.4). The analytic proof is deferred to Section 11. □

Theorem 4 allows for easy computing of $B\left(n\right)$ if $\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)$ is given for some $k\ge lgn$ using this form of (35):

Corollary 4. 

For every positive integer n and integer k with $n\le 2^k$,

$$B\left(n\right)={\displaystyle \frac{n\times k}{2}}-2^{k-1}\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right).$$

(36)

Proof. 

An obvious conclusion from (35). □

For instance, putting $k=\lfloor lgn\rfloor +1$ in (36) easily yields (15). For $k=\lceil lgn\rceil$, I obtain

$$B\left(n\right)={\displaystyle \frac{n\lceil lgn\rceil }{2}}-2^{\lceil lgn\rceil -1}\tilde{F}\left({\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}\right)=$$

[by (27)]

$$={\displaystyle \frac{n\lceil lgn\rceil }{2}}-2^{\lceil lgn\rceil -1}\sum _{i=0}^{\infty }{\displaystyle \frac{1}{2^i}}\mathit{Zigzag}\left(2^i{\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}\right)=$$

[since for $i\ge \lceil lgn\rceil$, $2^i{\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}$ is integer and $\mathit{Zigzag}\left(2^i{\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}\right)=0$]

$$={\displaystyle \frac{n\lceil lgn\rceil }{2}}-2^{\lceil lgn\rceil -1}\sum _{i=0}^{\lceil lgn\rceil -1}{\displaystyle \frac{1}{2^i}}\mathit{Zigzag}\left(2^i{\displaystyle \frac{n}{2^{\lceil lgn\rceil }}}\right)=$$

$$={\displaystyle \frac{n\lceil lgn\rceil }{2}}-{\displaystyle \frac{1}{2}}\sum _{i=0}^{\lceil lgn\rceil -1}{2}^{\lceil lgn\rceil -i}\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{\lceil lgn\rceil -i}}}\right).$$

Substituting k for $\lceil lgn\rceil -i$, I conclude

$$B\left(n\right)={\displaystyle \frac{n\lceil lgn\rceil }{2}}-{\displaystyle \frac{1}{2}}\sum _{k=1}^{\lceil lgn\rceil }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^k}}\right),$$

(37)

similar to the (15) characterization of $B\left(n\right)$.

7. Relationship Between the Best Case and the Worst Case

An attentive student of $\mathtt{MergeSort}$ tends to believe (taking into account that substituting $n-1$ for $\lfloor {\displaystyle \frac{n}{2}}\rfloor$ in the recurrence relation (3) for the best case yields the recurrence relation for the worst case) that its worst-case behavior is about twice as bad as its best-case behavior. This, of course, is only approximately true. In this section, I will derive the exact difference between $2B\left(n\right)$ and $W\left(n\right)$ using the function F defined by (17).

An exact formula for the number $W\left(n\right)$ of comparisons of keys performed by $\mathtt{MergeSort}$ in the worst case is known (see Formula (A7) in Appendix A) and is given for any positive integer n by the following equality:

$$W\left(n\right)=\sum _{i=1}^n\lceil lgi\rceil .$$

(38)

From (15) and (17), one can derive

$$2B\left(n\right)=n\lfloor lgn\rfloor -2^{\lfloor lgn\rfloor +1}+2n-F\left(n\right)=$$

[by ${\sum }_{i=1}^n\lceil lgi\rceil =n\lfloor lgn\rfloor -2^{\lfloor lgn\rfloor +1}+n+1$ from [3]

$$=\sum _{i=1}^n\lceil lgi\rceil -1+n-F\left(n\right)=$$

[by (38)]

$$=W\left(n\right)-1+n-F\left(n\right).$$

The above equality yields the following characterization.

Theorem 5. 

For every positive integer n, the difference between twice the number $B\left(n\right)$ of a comparison of keys performed in the best case and the number $W\left(n\right)$ of a comparison of keys performed in the worst case by $\mathtt{MergeSort}$ while sorting an n-element array is

$$2B\left(n\right)-W\left(n\right)=n-1-F\left(n\right),$$

(39)

where $F\left(n\right)$, visualized in Figure 8, is given by (19).

Proof. 

Follows from the above discussion. □

In particular, since, for every positive integer n,

$$0\le F\left(n\right)\le {\displaystyle \frac{n-1}{2}}$$

(40)

(see Figure 8 for an explanation), I conclude with the following tight linear bounds on $2B\left(n\right)-W\left(n\right)$.

Corollary 5. 

For every positive integer n, the difference between twice the minimum number $B\left(n\right)$ and the maximum number $W\left(n\right)$ of a comparison of keys performed in the worst case by $\mathtt{MergeSort}$ while sorting an n-element array satisfies this inequality:

$${\displaystyle \frac{n-1}{2}}\le 2B\left(n\right)-W\left(n\right)\le n-1.$$

(41)

Proof. 

Follows from (39) and (40). □

Obviously, $2B\left(n\right)-W\left(n\right)=n-1$ whenever $F\left(n\right)=0$, that is, whenever $n=2^{\lfloor lgn\rfloor }$. It can be shown that $2B\left(n\right)-W\left(n\right)={\displaystyle \frac{n-1}{2}}$ whenever n= ${\displaystyle \frac{1}{3}}(2^{k+1}+{(-1)}^k)$ for some integer $k\ge 0$. Therefore, the number of comps performed by $\mathtt{MergeSort}$ in the best case is larger than half the number of comps performed by it in the worst case by at least by ${\displaystyle \frac{n-1}{4}}$ and at most by ${\displaystyle \frac{n-1}{2}}$, which makes the former number significantly larger (by a term of size $\Theta \left(n\right)$) than half of the latter number.

A graph of $2B\left(n\right)-W\left(n\right)$ and its tight bounds are shown in Figure 12.

8. The Sum of Digits Problem

A known [but not to me during my derivation of Formula (15)] explicit formula, published in [4], for the total number $A(n,2)$ of 1s in binary representations of all integers between 0 and n (not including 0 and n) is expressed in terms of the function Zigzag (referred to as $2g$ in [4]). The following are verbatim quotations and screenshots from [4].

“If $\alpha (\kappa ,r)$ denotes the sum of the digits of $\kappa$ when $\kappa$ is represented in base r, then”

“Let $g\left(x\right)$ be periodic of period 1 and defined on $[0,1]$ by”

It has been shown in [5] that the recurrence relation for $A(n,2)$ is the same as the recurrence relation for $B\left(n\right)$ given by (2) and (3). Therefore, Formula (14) derived in this article is equivalent to $A(n,2)$ given above by the considerably more complicated definition. Interestingly, the above definition can be simplified to (14) along the lines of the elementary derivation of the alternative Formula (37) for $B\left(n\right)$.

Even more interestingly, if someone did succeed in simplifying Trollope’s formula in [4] then I am not aware of it. In particular, the most recent article [6] on the subject of sum of digits does not contain any hint that such a simplification has been found and published.

9. Proof of Theorem 1, Section 4.4

In this section, I provide an analytic proof of the experimentally derived Theorem 1, Section 4.4, which was instrumental for the derivation of a logarithmic-length Formula (15) for $B\left(n\right)$. The result and its proof have a flavor of Concrete Mathematics. Although they are interesting in their own right, they cannot be found in [7].

Theorem 6. 

(Same as Theorem 1.) For every natural number n and every positive natural number m,

$$\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =2m\times \mathit{Zigzag}\left({\displaystyle \frac{n}{2m}}\right),$$

(42)

where Zigzag is a function defined by (8) and visualized in Figure 5.

Proof. 

First, let us note that

$$\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i+m}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =$$

$$=\sum _{i=0}^{m-1}(\lfloor {\displaystyle \frac{n+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor ),$$

that is,

$$\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =\sum _{i=0}^{m-1}(\lfloor {\displaystyle \frac{n+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor ).$$

(43)

Let

$$n=k\times 2m+r,$$

(44)

where $0\le r<2m$, and let $0\le i<m$. We have

$$\lfloor {\displaystyle \frac{n+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor =\lfloor {\displaystyle \frac{k\times 2m+r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor =k+\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor$$

and

$$\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =\lfloor {\displaystyle \frac{k\times 2m+r+i}{2m}}\rfloor =k+\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor .$$

Thus, by virtue of (43),

$$\sum _{i=m}^{2m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor -\sum _{i=0}^{m-1}\lfloor {\displaystyle \frac{n+i}{2m}}\rfloor =\sum _{i=0}^{m-1}(\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor ).$$

(45)

We have

$$\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor =\left\{\begin{array}{c}1 \mathrm{if}{\displaystyle \frac{1}{2}}\le {\displaystyle \frac{r+i}{2m}}<1\hfill \\ \\ 0 \mathrm{otherwise},\hfill \end{array}\right.$$

(46)

because ${\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}<{\displaystyle \frac{3m}{2m}}+{\displaystyle \frac{1}{2}}=2$ so that $\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor \le 1$, and therefore, $\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor \le 1$.

Let I be defined as

$$I=\{i\in \mathbb{N}\mid {\displaystyle \frac{1}{2}}\le {\displaystyle \frac{r+i}{2m}}<1\}=\{i\in \mathbb{N}\mid m-r\le i<2m-r\}.$$

(47)

By virtue of (46), we have

$$\sum _{i=0}^{m-1}(\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor )=\sum _{i\in I}(\lfloor {\displaystyle \frac{r+i}{2m}}+{\displaystyle \frac{1}{2}}\rfloor -\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor )=\sum _{i\in I}1=\#I,$$

(48)

where $\#\left(I\right)$ denotes the cardinality of I.

If $r\le m$, then, by (47), $\#I=m-(m-r)=r$. If $r>m$, then, by (47), $\#I=2m-r$. In any case,

$$\#I=min(r,2m-r)=2mmin({\displaystyle \frac{r}{2m}},1-{\displaystyle \frac{r}{2m}})=$$

[since $0\le {\displaystyle \frac{r}{2m}}<1$ so that $\lfloor {\displaystyle \frac{r+i}{2m}}\rfloor =0$ and $\lceil {\displaystyle \frac{r+i}{2m}}\rceil =1$]

$$=2mmin({\displaystyle \frac{r}{2m}}-\lfloor {\displaystyle \frac{r}{2m}}\rfloor ,\lceil {\displaystyle \frac{r}{2m}}\rceil -{\displaystyle \frac{r}{2m}})=$$

[by definition (8) of the function Zigzag]

$$=2m\times Zigzag\left({\displaystyle \frac{r}{2m}}\right)=$$

[since Zigzag is a periodic function with period 1]

$$=2m\times Zigzag(k+{\displaystyle \frac{r}{2m}})=2m\times Zigzag\left({\displaystyle \frac{k\times 2m+r}{2m}}\right)=$$

[by (44)]

$$=2m\times Zigzag\left({\displaystyle \frac{n}{2m}}\right).$$

Thus

$$\#I=2m\times Zigzag\left({\displaystyle \frac{n}{2m}}\right).$$

(49)

From (43), (45), (48), and (49), I conclude (42). □

10. Proof of Theorem 3, Section 5

In this section, I provide an analytic proof of Theorem 3.

I begin with a brief discussion/motivation of what can be generally considered a closed-form formula for a function from the set of real numbers, except, perhaps, a finite number of reals, into a set of real numbers.

The general concept of closed-form formula (cff) has not been precisely defined in the literature. In this article, it denotes a formula that is composed of constants, variables, and some “standard” arithmetic and logical operations that has a finite and fixed structure and length that do not depend on the values of any variables occurring in the said formula. More specifically, in this article, it is used in the following sense.

The arithmetic constants are any integers with fixed values. The logical constants are true and false.

The basic arithmetic functions are: binary addition, binary multiplication, binary maximum and minimum, subtraction, division, exponentiation, logarithm functions, and the floor function. The basic arithmetic relations are: the equality relation and the less-than relation. The basic logical operations are: negation and binary conjunction.

An atomic cff is an expression that is a constant or variable, or denotes a reference to a basic function or relation whose arguments are constants or variables. The set of cffs is the smallest set of expressions that contains all atomic formulas and is closed under any finite and syntactically correct applications of basic functions and relations and the conditional arithmetic operation by the statement if c then d else e, where c is a logical cff and d and e are arithmetic cffs.

Note 5. 

One could use a more inclusive definition of closed-form formula without invalidating the results presented in this article as long as the extra arithmetic expressions allowed by it are differentiable on their domains except, perhaps, on their non-dense subsets.

Since non-rational real numbers are defined as limits of infinite, Cauchy-convergent rationals, limits of certain infinite sequences of reals do implicitly occur in definitions of some functions the references to which are accepted as cffs. Below is a motivating example of the function $2^x:\mathbb{R}⟶\mathbb{R}$ as an insight of what is accepted as a cff for a continuous function—like, say, $\tilde{F}\left(x\right)$—on the set $\mathbb{R}$ of reals or on an interval thereof. One picks a dense (in the metric topology of $\mathbb{R}$) subset $\mathbb{Q}$ (in this example, the set of rationals) of $\mathbb{R}$, with a collection of mappings ${\rho }_x\left(i\right):\mathbb{N}⟶\mathbb{Q}$, where $x\in \mathbb{R}$, given by ${\rho }_x\left(i\right)={\displaystyle \frac{\lfloor i\times x\rfloor }{i}}$ so that ${lim}_{i\to \infty }{\rho }_x\left(i\right)=x$. Since, for any $x\in \mathbb{R}\setminus \mathbb{Q}$, $2^x$ has been defined as

$$2^x=\underset{i\to \infty }{lim}{2}^{{\rho }_x\left(i\right)}=\underset{i\to \infty }{lim}\sqrt[i]{2^{\lfloor i\times x\rfloor }},$$

(50)

${lim}_{i\to \infty }\sqrt[i]{2^{\lfloor i\times x\rfloor }}$ is considered a cff for the function $2^x:\mathbb{R}⟶\mathbb{R}$.

For the reader’s convenience, Theorem 3 is quoted below as Theorem 7.

Theorem 7. 

(Same as Theorem 3). There is no cff $\phi \left(n\right)$ the domain of which contains all positive integers and the values of which coincide with ${\sum }_{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right)$ for all positive integers n; that is, for every cff $\phi \left(n\right)$, there is a positive n such that

$$\sum _{k=0}^{\lfloor lgn\rfloor }{2}^k\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{k+1}}}\right)\ne \phi \left(n\right),$$

(51)

where Zigzag is a function defined by (8) and visualized in Figure 5.

The rest of this section constitutes the proof of Theorem 7.

Lemma 1. 

For every positive integer n,

$$\tilde{F}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\right)={\displaystyle \frac{n(\lfloor lgn\rfloor +2)-2B\left(n\right)}{2^{\lfloor lgn\rfloor }}}-2,$$

(52)

where the function F has been defined by equality (17); the function $\tilde{F}$, visualized in Figure 13 has been defined by equality (27); and the function $B\left(n\right)$ has been defined by Equations (2) and (4).

Proof. 

From (17) I compute

$$\sum _{i=0}^{\lfloor lgn\rfloor }{2}^{i+1}\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{i+1}}}\right)=F\left(n\right)+2^{\lfloor lgn\rfloor +1}-n,$$

that is,

$$\sum _{i=0}^{\lfloor lgn\rfloor }{2}^i\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{i+1}}}\right)={\displaystyle \frac{1}{2}}F\left(n\right)+2^{\lfloor lgn\rfloor }-{\displaystyle \frac{n}{2}}.$$

(53)

Applying equality (53) to equality (15) I conclude

$$B\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-({\displaystyle \frac{1}{2}}F\left(n\right)+2^{\lfloor lgn\rfloor }-{\displaystyle \frac{n}{2}}),$$

or

$$B\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-{\displaystyle \frac{1}{2}}F\left(n\right)-2^{\lfloor lgn\rfloor }+{\displaystyle \frac{n}{2}},$$

that is,

$${\displaystyle \frac{1}{2}}F\left(n\right)={\displaystyle \frac{n}{2}}(\lfloor lgn\rfloor +1)-B\left(n\right)-2^{\lfloor lgn\rfloor }+{\displaystyle \frac{n}{2}},$$

or

$$F\left(n\right)=n(\lfloor lgn\rfloor +2)-2B\left(n\right)-2^{\lfloor lgn\rfloor +1}.$$

(54)

On the other hand, by virtue of (19),

$$F\left(n\right)=\sum _{i=1}^{\lfloor lgn\rfloor }{2}^i\mathit{Zigzag}\left({\displaystyle \frac{n}{2^i}}\right)=$$

[putting $j=\lfloor lgn\rfloor -i$]

$$=\sum _{j=0}^{\lfloor lgn\rfloor -1}{2}^{\lfloor lgn\rfloor -j}\mathit{Zigzag}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor -j}}}\right)=2^{\lfloor lgn\rfloor }\sum _{j=0}^{\lfloor lgn\rfloor -1}{\displaystyle \frac{1}{2^j}}\mathit{Zigzag}\left({\displaystyle \frac{2^{j}n}{2^{\lfloor lgn\rfloor }}}\right)=$$

[since for $j\ge \lfloor lgn\rfloor$, ${\displaystyle \frac{2^{j}n}{2^{\lfloor lgn\rfloor }}}\in \mathbb{N}$ so that $\mathit{Zigzag}\left({\displaystyle \frac{2^{j}n}{2^{\lfloor lgn\rfloor }}}\right)=0$]

$$=2^{\lfloor lgn\rfloor }\sum _{j=0}^{\infty }{\displaystyle \frac{1}{2^j}}\mathit{Zigzag}\left({\displaystyle \frac{2^{j}n}{2^{\lfloor lgn\rfloor }}}\right)=$$

[by (27)]

$$=2^{\lfloor lgn\rfloor }\tilde{F}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\right).$$

Thus

$$F\left(n\right)=2^{\lfloor lgn\rfloor }\tilde{F}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\right)$$

(55)

or

$$\tilde{F}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\right)={\displaystyle \frac{1}{2^{\lfloor lgn\rfloor }}}F\left(n\right).$$

(56)

Combining equalities (54) and (56) yields

$$\tilde{F}\left({\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\right)={\displaystyle \frac{1}{2^{\lfloor lgn\rfloor }}}(n(\lfloor lgn\rfloor +2)-2B\left(n\right)-2^{\lfloor lgn\rfloor +1}),$$

or (52). □

Lemma 2. 

If the function $B\left(n\right)$ defined by equations (2) and (4) has a cff $\beta :\mathbb{N}⟶\mathbb{N}$, then the function $\tilde{F}\left(x\right)$ defined by equation (27) has a cff $\phi :[1,2)⟶[0,{\displaystyle \frac{2}{3}}]$.

Proof. 

Let

$$D=\{{\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}\mid n\in \mathbb{N}\}$$

(57)

be the set of rationals in the interval $[1,2)$ with finite binary representations, enumerated by $\nu \left(n\right):\mathbb{N}⟶D$ given by $\nu \left(n\right)={\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}$ and visualized in Figure 14. (It is a trivial exercise to show that every real number with finite binary representation in the interval $[1,2)$ is of the form ${\displaystyle \frac{n}{2^{\lfloor lgn\rfloor }}}$ for some $n\in \mathbb{N}$, and it is obvious that every real number of that form has a finite binary representation and falls into that interval).

D is a dense subset of the interval $[1,2)$ of reals. Indeed, if $x\in [1,2)$, then, for every $n\in \mathbb{N}$, ${\displaystyle \frac{\lfloor 2^{n}x\rfloor }{2^n}}\in D$ and

$$\underset{n\to \infty }{lim}{\displaystyle \frac{\lfloor 2^{n}x\rfloor }{2^n}}=x.$$

(58)

Hence, for any $x\in [1,2)$, putting

$$n=\lfloor 2^{i}x\rfloor ,$$

(59)

so that

$$\lfloor lgn\rfloor =\lfloor lg\lfloor 2^{i}x\rfloor \rfloor =\lfloor lg{2}^{i}x\rfloor =\lfloor i+lgx\rfloor =i+\lfloor lgx\rfloor =i$$

[the last equality holds because $1\le x<2$ so that $0\le lgx<1$ and $\lfloor lgx\rfloor =0$], or

$$\lfloor lgn\rfloor =i,$$

(60)

I conclude, by virtue of (58),

$$\tilde{F}\left(x\right)=\tilde{F}(\underset{i\to \infty }{lim}{\displaystyle \frac{\lfloor 2^{i}x\rfloor }{2^i}})=$$

[by the continuity of $\tilde{F}\left(x\right)$]

$$=\underset{i\to \infty }{lim}\tilde{F}({\displaystyle \frac{\lfloor 2^{i}x\rfloor }{2^i}})=$$

[by equality (52) of Lemma 1]

$$=\underset{i\to \infty }{lim}{\displaystyle \frac{\lfloor 2^{i}x\rfloor (i+2)-2B\left(\lfloor 2^{i}x\rfloor \right)}{2^i}}-2=\underset{i\to \infty }{lim}{\displaystyle \frac{i\lfloor 2^{i}x\rfloor -2B\left(\lfloor 2^{i}x\rfloor \right)}{2^i}}+\underset{i\to \infty }{lim}2{\displaystyle \frac{\lfloor 2^{i}x\rfloor }{2^i}}-2=$$

[by equality (58)]

$$\underset{i\to \infty }{lim}{\displaystyle \frac{i\lfloor 2^{i}x\rfloor -2B\left(\lfloor 2^{i}x\rfloor \right)}{2^i}}+2x-2.$$

Thus for any $x\in [1,2)$,

$$\tilde{F}\left(x\right)=\underset{i\to \infty }{lim}{\displaystyle \frac{i\lfloor 2^{i}x\rfloor -2B\left(\lfloor 2^{i}x\rfloor \right)}{2^i}}+2x-2.$$

(61)

Equality (61) shows that if there is a cff $\beta :\mathbb{N}⟶\mathbb{N}$ for the function B defined by Equations (2) and (4) then there is a cff $\phi :[1,2)⟶[0,{\displaystyle \frac{2}{3}}]$ given by

$$\underset{i\to \infty }{lim}{\displaystyle \frac{i\lfloor 2^{i}x\rfloor -2\beta \left(\lfloor 2^{i}x\rfloor \right)}{2^i}}+2x-2$$

for $\tilde{F}\left(x\right)$. This completes the proof of Lemma 2. □

Should the nowhere differentiable function $\tilde{F}$ have a cff, it would be differentiable everywhere except, perhaps, on a non-dense subset of $\mathbb{R}$. The following inductive argument demonstrates that. All atomic cffs are differentiable except, perhaps, on a non-dense subset of $\mathbb{R}$. If a finite number of cffs are differentiable except, perhaps, on non-dense subsets of $\mathbb{R}$, then their composition is differentiable except, perhaps, on non-dense subsets of $\mathbb{R}$. (For instance, the function $\sqrt{x}Zigzag\left({\displaystyle \frac{1}{x}}\right)$ is differentiable on $(0,1)$, except for the non-dense set $\{{\displaystyle \frac{2}{n}}\mid n\in \mathbb{N}\}$.) Thus $\tilde{F}$ has no cff.

The above observation, together with Lemma 2, completes the proof of Theorem 7.

11. Proof of Theorem 4, Section 6

In this Section, I provide an analytic proof of experimentally derived Theorem 4, Section 6. This result, re-stated as Theorem 8 below, allows for practically efficient computations of values of the continuous blancmange function for reals with finite binary floating-point representations. I also provide some properties (Lemmas 3–5) of the Zigzag function, given by equality (8) and visualized in Figure 5, that are useful for a neat derivation of a formula for the blancmange function as the limit of a finite sum of some values of the Zigzag function.

Let the function $\tilde{F}$ (known as the blancmange function), visualized in Figure 13, be defined by (27) and $B\left(n\right)$, given by (2) and (4), be the least number of comparisons of keys that $\mathtt{MergeSort}$ performs while sorting an n-element array.

Theorem 8. 

(Same as Theorem 4.) For every positive integer n and integer k with $n\le 2^k$,

$$\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)={\displaystyle \frac{n\times k-2B\left(n\right)}{2^k}}.$$

(62)

The remainder of this section constitutes a proof of Theorem 8.

Note. The function Zigzag, visualized in Figure 5, has been defined by Equation (8).

Lemma 3. 

For every $k\ge \lfloor lgn\rfloor +2$,

$$2^{k}Zigzag\left({\displaystyle \frac{n}{2^k}}\right)=n.$$

(63)

Proof. 

Let $k\ge \lfloor lgn\rfloor +2$, or $2^k\ge 2\times 2^{\lfloor lgn\rfloor +1}>2n$, that is,

$$2^k>2n.$$

(64)

We have

$$0\le \lfloor {\displaystyle \frac{n}{2^k}}\rfloor \le$$

[by (64)]

$$\le \lfloor {\displaystyle \frac{n}{2n}}\rfloor =\lfloor {\displaystyle \frac{1}{2}}\rfloor =0$$

or

$$\lfloor {\displaystyle \frac{n}{2^k}}\rfloor =0.$$

(65)

Additionally,

$$1\le \lceil {\displaystyle \frac{n}{2^k}}\rceil \le$$

[by (64)]

$$\le \lceil {\displaystyle \frac{n}{2n}}\rceil =\lceil {\displaystyle \frac{1}{2}}\rceil =1$$

or

$$\lceil {\displaystyle \frac{n}{2^k}}\rceil =1.$$

(66)

Now,

$$2^{k}Zigzag\left({\displaystyle \frac{n}{2^k}}\right)=$$

[by (8)]

$$=2^{k}min\{{\displaystyle \frac{n}{2^k}}-\lfloor {\displaystyle \frac{n}{2^k}}\rfloor ,\lceil {\displaystyle \frac{n}{2^k}}\rceil -{\displaystyle \frac{n}{2^k}}\}=min\{n-2^k\lfloor {\displaystyle \frac{n}{2^k}}\rfloor ,2^k\lceil {\displaystyle \frac{n}{2^k}}\rceil -n\}=$$

[by (65) and (66)]

$$=min\{n,2^k-n\}=$$

[since by (64), $2^k-n\ge n$]

$$=n.$$

Hence, (63) holds. □

Lemma 4. 

For every $k\ge \lfloor lgn\rfloor +1$,

$$\sum _{i=\lfloor lgn\rfloor +2}^k{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)=n\times k-n(\lfloor lgn\rfloor +1).$$

(67)

Proof. 

By induction on k.

Basis step: $k=\lfloor lgn\rfloor +1$.

$$L=\sum _{i=\lfloor lgn\rfloor +2}^{\lfloor lgn\rfloor +1}{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)=0.$$

$$R=n(\lfloor lgn\rfloor +1)-n(\lfloor lgn\rfloor +1)=0.$$

Hence, $L=R$. This completes the basis step.

Inductive step: $k\ge \lfloor lgn\rfloor +2$.

Inductive hypothesis: (67).

$$\sum _{i=\lfloor lgn\rfloor +2}^{k+1}{2}^{i}Zigzag({\displaystyle \frac{n}{2^i}})=\sum _{i=\lfloor lgn\rfloor +2}^k{2}^{i}Zigzag({\displaystyle \frac{n}{2^i}})+2^{k}Zigzag({\displaystyle \frac{n}{2^k}})=$$

[by the inductive hypothesis and by equality (63) of Lemma 3]

$$=n\times k-n(\lfloor lgn\rfloor +1)+n=n\times (k+1)-n(\lfloor lgn\rfloor +1).$$

Thus

$$\sum _{i=\lfloor lgn\rfloor +2}^{k+1}{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)=n\times (k+1)-n(\lfloor lgn\rfloor +1).$$

This completes the inductive step. □

Lemma 5. 

For every $k\ge \lfloor lgn\rfloor +1$,

$$2^k\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)=\sum _{i=1}^k{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right).$$

(68)

Proof. 

By definition (27) of the function $\tilde{F}$, I get

$$2^k\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)=2^k\sum _{i=0}^{\infty }{\displaystyle \frac{1}{2^i}}Zigzag\left(2^i{\displaystyle \frac{n}{2^k}}\right)=$$

[since, for every integer x, $Zigzag\left(x\right)=0$, so that, for $i\ge k$, $Zigzag\left(2^i{\displaystyle \frac{n}{2^k}}\right)=0$]

$$=2^k\sum _{i=0}^{k-1}{\displaystyle \frac{1}{2^i}}Zigzag\left(2^i{\displaystyle \frac{n}{2^k}}\right)=\sum _{i=0}^{k-1}{2}^{k-i}Zigzag\left({\displaystyle \frac{n}{2^{k-i}}}\right)=$$

[putting $j=k-i$]

$$=\sum _{j=1}^k{2}^{j}Zigzag\left({\displaystyle \frac{n}{2^j}}\right),$$

which completes the proof of (68). □

At this point, I am ready to conclude the proof of Theorem 4.

By virtue of (15), e have

$$2B\left(n\right)=n(\lfloor lgn\rfloor +1)-\sum _{k=0}^{\lfloor lgn\rfloor }{2}^{k+1}Zigzag\left({\displaystyle \frac{n}{2^{k+1}}}\right)=$$

$$=n(\lfloor lgn\rfloor +1)-\sum _{i=1}^{\lfloor lgn\rfloor +1}{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)=$$

$$=n(\lfloor lgn\rfloor +1)-\sum _{i=1}^k{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)+\sum _{i=\lfloor lgn\rfloor +2}^k{2}^{i}Zigzag\left({\displaystyle \frac{n}{2^i}}\right)=$$

[by Lemmas 4 and 5]

$$=n(\lfloor lgn\rfloor +1)-2^k\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)+n\times k-n(\lfloor lgn\rfloor +1)=n\times k-2^k\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right),$$

that is,

$$2B\left(n\right)=n\times k-2^k\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right),$$

from which (63) follows.

This completes the proof of Theorem 4.

Note. A glance at the proof of Lemma 4 suffices to notice that it fails if $n>2^k$, and so does Theorem 4. In particular, for $k=\lfloor lgn\rfloor$, Lemma 1 yields

$$\tilde{F}\left({\displaystyle \frac{n}{2^k}}\right)={\displaystyle \frac{n\times k-2B\left(n\right)}{2^k}}+{\displaystyle \frac{2n}{2^k}}-2>{\displaystyle \frac{n\times k-2B\left(n\right)}{2^k}}$$

(69)

since, for $n>2^{\lfloor lgn\rfloor }$, ${\displaystyle \frac{2n}{2^{\lfloor lgn\rfloor }}}-2>0.$

12. Final Remarks

It appears that a lack of sufficient interaction between research related to the sum of digits problem and analysis of recursive algorithms had a detrimental impact on the progress of the said research. In particular, I found that some clever use of recursion trees that are often utilized in the analysis of algorithms but are relatively rarely used in pure mathematics allowed for a relatively easy derivation of a finitary Formula (7) that, due to its finiteness, is significantly simpler and more practical to compute (but still requiring the addition of $n-1$ terms) than an instance for $r=2$ of the infinite formula (1.1) derived in [4] and quoted in Section 8. I suppose that if the mathematicians interested in the sum of digits problem knew Formula (7), then some of them would simplify it to Formula (15) of the Main Theorem that can be evaluated in $O\left(lo{g}^{c}n\right)$ for some constant c.

A similar lack of communication between researchers has been hypothesized in [6]: “When reading the literature on the subject we have noted that the most recent papers do not always cite more ancient ones, confirming a remark of Stolarsky […]: Whatever its mathematical virtues, the literature on sums of digital sums reflects a lack of communication between researchers”.

On the other hand, those willing to find or quote from the literature a formula for the exact number of comparisons of keys performed by $\mathtt{MergeSort}$ in the best case on an n-element array might have been discouraged by finding Trollope’s infinite formula (1.1) that, for $r=2$, yields the count of 1s occurring in binary representations of all integers between 1 and n and has been proved, around 1974, to yield the said number of comps. To make things look even more discouraging, directly counting the said 1s requires $\Theta (nlgn)$ time that is about as long as running $\mathtt{MergeSort}$ on an increasingly sorted array of consecutive numbers from 1 to n and counting how many comps have been done. These could have been factors that stopped some of those analyzing $\mathtt{MergeSort}$ from trying to precisely characterize its best-case time complexity, thus leaving a significant loophole in many computer science faculty and students’ knowledge of the time complexity of $\mathtt{MergeSort}$.

Fortunately, this article fills this important loophole.